Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++\((r, \phi, z)\). Relation to Cartesian coordinates: +
+ ++\[ +x = r \cos \phi, y = r \sin \phi, z = z +\label{Gr(1.74)} +\] +
+ ++The unit vectors are +
+ ++\[ +\hat{\bf r} = \cos \phi ~\hat{\bf x} + \sin \phi~\hat{\bf y}, +\hat{\boldsymbol \phi} = -\sin \theta ~\hat{\bf x} + \cos \phi~\hat{\bf y}, +\hat{\bf z} = \hat{\bf z}. +\label{Gr(1.75)} +\] +
+ ++Infinitesimal displacement: +
+ ++\[ +d{\bf l} = dr ~\hat{\bf r} + r d\phi~\hat{\boldsymbol \phi} + dz ~\hat{\bf z}. +\label{Gr(1.77)} +\] +
+ ++Volume element: +
+ ++\[ +d\tau = r dr d\phi dz +\label{Gr(1.78)} +\] +
+ ++Range of parameters: \(r \in [0, \infty[\), \(\phi \in [0, 2\pi[\) and \(z \in ]-\infty, \infty[\). +
++\[ +{\boldsymbol ∇} × {\bf v} = \left( \frac{1}{r} \frac{\partial v_z}{\partial \phi} - \frac{∂ vφ}{∂ z}\right) ~\hat{\bf r} +
++\tag{cyl_curl} +\label{cyl_curl} +\] +
++\[ +{\boldsymbol ∇}2 T = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial T}{\partial r} \right) +
++\label{Gr(1.82)} +\] +
+Created: 2022-02-07 Mon 08:02
+ +
+
++\((r, \theta, \phi)\). \(\theta\) is the polar angle, \(\phi\) the azimuthal angle. +
+ ++\[ +x = r \sin \theta \cos \phi, +y = r \sin \theta \sin \phi, +z = r \cos \theta. +\label{Gr(1.62)} +\] +
+ ++Unit vectors: \(\hat{\boldsymbol r}, \hat{\boldsymbol \theta}, \hat{\boldsymbol \phi}\). +
+ ++\[ +{\bf A} = A_r \hat{\bf r} + A_{\theta} \hat{\bf \theta} + A_{\phi} \hat{\boldsymbol \phi} +\label{Gr(1.63)} +\] +
+ ++In terms of Cartesian unit vectors: +
+ +\begin{align} +\hat{\boldsymbol r} &= \sin \theta \cos \phi \hat{\bf x} + \sin \theta \sin \phi \hat{\bf y} + \cos \theta \hat{\bf z}, \nonumber \\ +\hat{\boldsymbol \theta} &= \cos \theta \cos \phi \hat{\bf x} + \cos \theta \sin \phi \hat{\bf y} - \sin \theta \hat{\bf z}, \nonumber \\ +\hat{\boldsymbol \phi} &= -\sin \phi \hat{\bf x} + \cos \phi \hat{\bf y}. +\label{Gr(1.64)} +\end{align} + ++Careful: these unit vectors are direction dependent, i.e. we should really +write \(\hat{\boldsymbol r} (\theta, \phi), \hat{\boldsymbol \theta} (\theta, \phi), \hat{\boldsymbol \phi} (\theta, \phi)\). +
+ ++Infinitesimal displacement \(d{\bf l}\): +
+ ++\[ +d{\bf l} = dr \hat{\boldsymbol r} + r d\theta \hat{\boldsymbol \theta} + r\sin \theta d\phi \hat{\boldsymbol \phi}. +\label{Gr(1.68)} +\] +
+ ++Infinitesimal volume element: +
+ ++\[ +d\tau = dl_r dl_{\theta} dl_{\phi} = r^2 \sin \theta dr d\theta d\phi +\label{Gr(1.69)} +\] +
+ ++Infinitesimal surface element: depends on situation. +
++\[ +{\boldsymbol ∇} T = \frac{\partial T}{\partial r} \hat{\boldsymbol r} + \frac{1}{r} \frac{\partial T}{\partial \theta} \hat{\boldsymbol \theta} +
++\label{Gr(1.70)} +\] +
++\[ +{\boldsymbol ∇} ⋅ {\bf v} = \frac{1}{r^2} \frac{\partial}{\partial r} (r2 vr) + \frac{1}{r\sin \theta} \frac{\partial}{\partial \theta} (sinθ vθ) +
++\label{Gr(1.71)} +\] +
++\[ +{\boldsymbol ∇} × {\bf v} = \frac{1}{r\sin \theta} \left[ \frac{\partial}{\partial \theta} (sin θ vφ) - \frac{∂ vθ}{∂ φ} \right] \hat{\bf r} +
++\label{Gr(1.72)} +\] +
++\[ +{\boldsymbol ∇}2 T = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r2 \frac{\partial T}{\partial r}\right) +
++\label{Gr(1.73)} +\] +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++\({\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} T) \equiv {\boldsymbol \nabla}^2 T\) is called the Laplacian of the scalar field \(T\). +The Laplacian of a vector field \({\boldsymbol \nabla}^2 {\bf v}\) is also defined as the vector with components +given by the Laplacian of the corresponding vector elements. +
++This always vanishes. +
++\({\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf v})\) does not appear often in physics. No special name. +
++This always vanishes. +
++\[ +{\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf v}) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf v}) - {\boldsymbol \nabla}^2 {\bf v} +\label{Gr(1.47)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++The del/grad/nabla operator is defined as +
+ ++\[ +{\boldsymbol \nabla} = \hat{\bf x} \frac{\partial}{\partial x} + \hat{\bf y} \frac{\partial}{\partial y} + \hat{\bf z} \frac{\partial}{\partial z} +\label{Gr(1.39)} +\] +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Consider a function \(T\) of the 3 variables \(x, y, z\) noted as \(T({\bf r})\). This is commonly known +as a scalar field. +
+ ++Its change under a displacement by \(d{\bf l}\) is given by +
+ ++\[ +\Delta_{d{\bf l}} T = T({\bf r} + d{\bf l}) - T({\bf r}) = {\boldsymbol \nabla} T \cdot d{\bf l} + \mbox{O}(dl^2) +\label{Gr(1.35)} +\] +
+ ++in which +
+ ++\[ +{\boldsymbol \nabla} T \equiv \frac{\partial T}{\partial x} ~\hat{\bf x} + \frac{\partial T}{\partial y} ~\hat{\bf y} + \frac{\partial T}{\partial z} ~\hat{\bf z} +\label{Gr(1.36)} +\] +
+ ++is a vector called the gradient of \(T\). +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Consequences: for any smooth differentiable function \(f(x)\), +
+ +Created: 2022-02-07 Mon 08:02
+ ++\[ +\delta^{(3)} ({\bf r} - {\bf r}') = \delta (x - x') \delta (y - y') \delta (z - z') +\label{Gr(1.96)} +\] +
+ ++\[ +\int d\tau f({\bf r}) \delta^{(3)} ({\bf r} - {\bf a}) = f({\bf a}) +\label{Gr(1.97)} +\] +
+ ++Resolution of divergence of \(\hat{\bf r}/r^2\) paradox: +
+ ++\[ +{\boldsymbol \nabla} \cdot \left(\frac{\hat{\bf r}}{r^2} \right)= 4\pi \delta^{(3)} ({\bf r}). +\label{Gr(1.99)} +\] +
+ ++More generally, +
+ ++\[ +\boxed{ +{\boldsymbol \nabla} \cdot \left(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} \right)= 4\pi \delta^{(3)} ({\bf r}). +}\label{Gr(1.100)} +\] +
+ ++Since +
+ ++\[ +{\boldsymbol \nabla}_1 \left(\frac{1}{r_{12}}\right) = -\frac{\hat{\bf r}_{12}}{r_{12}^2} +\label{Gr(1.101)} +\] +
+ ++we have that +
+ ++\[ +{\boldsymbol \nabla}^2 \left( \frac{1}{|{\bf r} - {\bf r}'|} \right) = -4\pi \delta^{(3)} ({\bf r} - {\bf r}') +\label{Gr(1.102)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Try to calculate this directly: +
+ ++\[ +{\boldsymbol \nabla} \cdot \frac{\hat{\bf r}}{r^2} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{1}{r^2}\right) =^{?} 0 +\label{Gr(1.84)} +\] +
+ ++Now apply the divergence theorem. +Integrate over a sphere of radius \(R\) centered at the origin (Prob. 1.38b): +
+ + ++\[ +\oint \frac{\hat{\bf r}}{r^2} \cdot d{\bf a} = \int \left(\frac{1}{R^2} \hat{\bf r} \right) \cdot \left( R^2 \sin \theta d\theta d\phi ~\hat{\bf r} \right) += \int_0^{\pi} d\theta \sin \theta \int_0^{2\pi} d\phi = 4\pi +\label{Gr(1.85)} +\] +
+ ++Problem: in (1.84), we've divided by zero when \(r = 0\). +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++\[ +\int_a^b \frac{df}{dx} dx = f(b) - f(a) +\label{Gr(1.54)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++\[ +{\int_{\bf a}^{\bf b}}_{\cal P} ({\boldsymbol \nabla} T) \cdot d{\bf l} = T({\bf b}) - T({\bf a}). +\label{Gr(1.55)} +\] +
+ ++Corollary 1: \(\int_{\bf a}^{\bf b} ({\boldsymbol \nabla} T) \cdot d{\bf l}\) is independent of the path taken from \({\bf a}\) to \({\bf b}\). +
+ ++Corollary 2: \(\oint ({\boldsymbol \nabla} T) \cdot d{\bf l} = 0\) since the beginning and end points coincide. +
+Created: 2022-02-07 Mon 08:02
+ ++\[ +\int_{\cal V} ({\boldsymbol \nabla} \cdot {\bf v}) d\tau = \oint_{\cal S} {\bf v} \cdot d{\bf a}. +\label{Gr(1.56)} +\] +
+ ++This is know either as Gauss' theorem, Green's theorem or the divergence theorem. +
+Created: 2022-02-07 Mon 08:02
+ ++Example: \({\boldsymbol \nabla} \cdot (f{\bf A}) = f({\boldsymbol \nabla} \cdot {\bf A}) + {\bf A} \cdot ({\boldsymbol \nabla} f)\) implies (via Gauss' theorem) +
+ ++\[ +∫\cal V {\boldsymbol ∇} ⋅ (f{\bf A}) dτ = ∫\cal V f({\boldsymbol ∇} ⋅ {\bf A}) dτ +
++\] +
+ ++or in other words +
+ ++\[ +∫\cal V f({\boldsymbol ∇} ⋅ {\bf A}) dτ = -∫\cal V {\bf A} ⋅ ({\boldsymbol ∇} f) dτ +
++\label{Gr(1.59)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++\[ +{\int_{\bf a}^{\bf b}}_{\cal P} {\bf v} \cdot d{\bf l} +\label{Gr(1.48)} +\] +
+ ++where \({\cal P}\) is a path from point \({\bf a}\) to point \({\bf b}\). +
+ ++Example: work done by a force, \(W = \int {\bf F} \cdot d{\bf l}\). +
+ ++Integral over a closed loop: +
+ ++\[ +\oint {\bf v} \cdot d{\bf l} +\label{Gr(1.49)} +\] +
++\[ +\int_{\cal S} {\bf v} \cdot d{\bf a} +\label{Gr(1.50)} +\] +
+ ++Over a closed surface: +
+ ++\[ +\oint {\bf v} \cdot d{\bf a} +\] +
++\[ +\int_{\cal V} T d\tau +\label{Gr(1.51)} +\] +
+ ++In Cartesian coordinates: +
+ ++\[ +d\tau = dx dy dz +\label{Gr(1.52)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++\[ +\int_{\cal S} ({\boldsymbol \nabla} \times {\bf v}) \cdot d{\bf a} = \oint_{\cal P} {\bf v} \cdot d{\bf l}. +\label{Gr(1.57)} +\] +
+ ++Corollary 1: \(\int ({\boldsymbol \nabla} \times {\bf v}) \cdot d{\bf a}\) depends only on the boundary line +and not on the particular surface used. +
+ ++Corollary 2: \(\oint ({\boldsymbol \nabla} \times {\bf v}) \cdot d{\bf a}= 0\) for any closed surface, since +the boundary shrinks to a point. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +
+\[
+ {\bf A} × {\bf B} = \left| \begin{array}{ccc} \hat{x} & \hat{y} & \hat{z}
+ Ax & Ay & Az \\ Bx & By & Bz \end{array} \right|
+ = \left( Ay Bz - Az By \right) \hat{x} + \left( Bz Ax - Bx Az \right) \hat{y}
+
+\] +
+ ++The cross product is distributive: +
+ ++\[ + {\bf A} \times ({\bf B} + {\bf C}) = {\bf A} \times {\bf B} + {\bf A} \times {\bf C} +\] +
+ ++The cross-product is anti-commutative: +
+ ++\[ + {\bf A} \times {\bf B} = - {\bf B} \times {\bf A} +\] +
+ ++this relation making plain that +
+ ++\[ + {\bf A} \times {\bf A} = 0. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Vectors will always be denoted by bold symbols, e.g. \({\bf A}\). +Component notation (in \({\mathbb R}^3\) with cartesian coordinates): +
+ ++\[ + {\bf A} = \left( \begin{array}{c} + A_x \\ A_y \\ A_z \end{array}\right) +\] +
+ ++Commutativity and associativity of vector addition +
+ ++\[ + {\bf A} + {\bf B} = {\bf B} + {\bf A}, \hspace{10mm} + ({\bf A} + {\bf B}) + {\bf C} = {\bf A} + ({\bf B} + {\bf C}) +\] +
+ ++Distributivity of multiplication by a scalar +
+ ++\[ + c ({\bf A} + {\bf B}) = c{\bf A} + c{\bf B} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Cartesian coordinates. Position vector: +
+ ++\[ +{\bf r} \equiv x ~\hat{\bf x} + y ~\hat{\bf y} + z ~\hat{\bf z} +\label{Gr(1.19)} +\] +
+ ++Magnitude +
+ ++\[ +r = \left[ x^2 + y^2 + z^2 \right]^{1/2} +\label{Gr(1.20)} +\] +
+ ++Unit radial vector: +
+ ++\[ +\hat{\bf r} = \frac{\bf r}{r} +\label{Gr(1.21)} +\] +
+ ++Infinitesimal displacement vector: +
+ ++\[ +d{\bf l} = dx ~\hat{\bf x} + dy ~\hat{\bf y} + dz ~\hat{\bf z} +\label{Gr(1.22)} +\] +
+ ++Separation vector: for two position vectors \({\bf r}_a, {\bf r}_b\), +
+ ++\[ +{\bf r}_{ab} \equiv {\bf r}_a - {\bf r}_b +\label{Gr(1.23)NOT} +\] +
+ + ++NOTE: I WILL NOT USE THE …ING GRIFFITHS CURLY \(r\) notation! +
+ +Created: 2022-02-07 Mon 08:02
+ ++Cartesian coordinates in \({\mathbb R}^3\): +
+ ++\[ + {\bf A} \cdot {\bf B} = + \left( \begin{array}{ccc} A_x & A_y & A_z \end{array} \right) + \left(\begin{array}{c} B_x \\ B_y \\ B_z \end{array} \right) = A_x B_x + A_y B_y + A_z B_z +\] +
+ ++The scalar product is commutative and distributive: +
+ ++\[ + {\bf A} \cdot {\bf B} = {\bf B} \cdot {\bf A}, \hspace{10mm} + {\bf A} \cdot ({\bf B} + {\bf C}) = {\bf A} \cdot {\bf B} + {\bf A} \cdot {\bf C} +\] +
+ ++In a general coordinate system with metric \(g\), +
+ ++\[ + {\bf A} \cdot {\bf B} = \sum_{\mu, \nu} g^{\mu \nu} A_\mu B_\nu +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Scalar triple product +
+ ++\(| {\bf A} \cdot ({\bf B} \times {\bf C}) |\) is the volume of the parallelepiped subtended by \({\bf A}, {\bf B}\) and \({\bf C}\). +
+ ++The scalar triple product is preserved under cyclic reordering: +
+ ++\[ + {\bf A} \cdot ({\bf B} \times {\bf C}) = {\bf B} \cdot ({\bf C} \times {\bf A}) = {\bf C} \cdot ({\bf A} \times {\bf B}) + \label{VectAn:TripleProduct} + %\label{Gr(1.15)} +\] +
+ ++Useful form: +
+ ++\[ + {\bf A} \cdot ({\bf B} \times {\bf C}) = \left| \begin{array}{ccc} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{array} \right| + \label{VectAn:TripleProductAsDet} + %\label{Gr(1.16)} +\] +
+ ++The order can be interchanged: +
+ ++\[ +{\bf A} \cdot ({\bf B} \times {\bf C}) = ({\bf A} \times {\bf B}) \cdot {\bf C} +\label{Gr(1.16)} +\] +
+ ++Vector triple product +
+ ++\[ +{\bf A} \times ({\bf B} \times {\bf C}) = {\bf B} ({\bf A} \cdot {\bf C}) - {\bf C} ({\bf A} \cdot {\bf B}) +\label{Gr(1.17)} +\] +
+ ++Nota bene: +
+ ++\[ +({\bf A} \times {\bf B}) \times {\bf C} = -{\bf A} ({\bf B} \cdot {\bf C}) + {\bf B} ({\bf A} \cdot {\bf C}) +\] +
+ ++is a different vector. The cross-product is {\bf not} associative in general +but obeys the more general relation +
+ ++\[ + {\bf A} \times ({\bf B} \times {\bf C}) + {\bf B} \times ({\bf C} \times {\bf A}) + {\bf C} \times ({\bf A} \times {\bf B}) = 0 +\] +
+ ++All higher vector products can be reduced to combinations of single vector product terms. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Consider a fixed vector field \({\bf F} = {\bf F}(x, y, z) = {\bf F}({\bf r})\). Assume that +the functions \({\boldsymbol \nabla} \cdot {\bf F} = D({\bf r})\) and \({\boldsymbol \nabla} \times {\bf F} = {\bf C} ({\bf r})\) +are given to us everywhere within a finite volume \({\cal V}\) (for consistency, \({\bf C}\) must be divergenceless, +\({\boldsymbol \nabla} \cdot {\bf C} = 0\)). Then, +
+ ++\[ +{\bf F} ({\bf r}) = -{\boldsymbol \nabla} \phi ({\bf r}) + {\boldsymbol \nabla} \times {\bf A} ({\bf r}) +\] +
+ ++where +
+ ++\[ +\phi({\bf r}) = \frac{1}{4\pi} \int_{\cal V} \frac{D({\bf r}') d\tau'}{|{\bf r} - {\bf r}'|}, +\hspace{1cm} +{\bf A} ({\bf r}) = \frac{1}{4\pi} \int_{\cal V} \frac{{\bf C}({\bf r}') d\tau'}{|{\bf r} - {\bf r}'|}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++If the curl of a vector field \({\bf F}\) vanishes, then \({\bf F}\) can be written as the gradient of a +scalar potential \(V\): +
+ ++\[ +{\boldsymbol \nabla} \times {\bf F} = 0 \Longleftrightarrow {\bf F} = -{\boldsymbol \nabla} V +\label{Gr(1.103)} +\] +
++The following conditions are equivalent: +
+ ++If the divergence of a vector field \({\bf F}\) vanishes, then \({\bf F}\) can be expressed as the curl +of a {\bf vector potential} \({\bf A}\). +
++The following conditions are equivalent: +
+ +Created: 2022-02-07 Mon 08:02
+ ++This chapter is a module-by-module list of the core concepts which +you should be comfortable with. Use it as a diagnostics guide to assess how +well you understand the material. +
+ ++So my suggestion is: take a number of blank sheets of paper, +and try to do the things that are suggested from the top of your head, +i.e. without looking +at any of your notes, books, internet or whatever. +This should help you identify holes in your comprehension and thus +clarify what your study priorities are. +
+ ++Note/warning: this is in some sense the minimal +stuff you should have assimilated. +The things listed here are by no means exhaustive of the +subject matter, or of the full examinable material. +If you do understand everything here and are able to apply it, +I have little doubt you'll pass the course. +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
+ ++Applications: +As a strict minimum, you should be able to: +
+ +Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
+ ++Applications: +As a strict minimum, you should be able to: +
+ +Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ ++Things you should be completely at ease with: +
+ ++Fundamentals: +Things you should be able to do (ideally: from scratch, on a blank sheet of paper): +
+ +Created: 2022-02-07 Mon 08:02
+ ++Fundamentals: +After properly studying this module, you should be able to: +
++Applications: +As a strict minimum, you should be able to: +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++1831: 3 experiments by Faraday (according to Griffiths! but it's historically incorrect) +\paragraph{1)} Pull a loop of wire through a magnetic field. +\paragraph{2)} Move magnet around a still loop. +\paragraph{3)} Change strength of field, holding magnet and loop still. +
+ + ++Actually, historically, things didn't happen like that. +The first experiment that Faraday performed (1831) involved two metal coils wound +on opposite sides of a metal ring. When a current was turned on through the first +coil, it generated a transient current in the second coil (as measured by a +galvanometer). Within the next few months he had come up with lots of variations +on this idea. Faraday observed transient current in a circuit when: +
+ ++Faraday's big insight was to summarize these effects by noticing that +
+ ++\[ + \boxed{ + \mbox{\bf A changing magnetic field induces an electric field} + } +\] +
+ ++Empirically: the changing magnetic field induces an electric current around +the circuit. This current is really driven by an electric field having a component +along the wire. The line integral of this field is called the +
++Electromotive force (or electromotance), + \[ + {\cal E} \equiv \oint_{\cal P} {\bf E} \cdot d{\bf l}. + \] +
+ ++You can think of the emf in different ways. It's the energy accumulated as a unit charge is moved around the circuit; altenatively, if you cut the wire, it would be the voltage you would measure between the two ends. +
+ ++The precise statement is that the electromotive force is proportional +to the rate of change of the magnetic flux, +\[ +{\cal E} = \oint_{\cal P} {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt} +\label{Gr(7.14)} +\] +so we obtain +
++Faraday's law (integral form N.B.: for a stationary loop) + \[ + \oint_{\cal P} {\bf E} \cdot d{\bf l} = -\int_{\cal S} \frac{\partial {\bf B}}{\partial t} \cdot d{\bf a} + \label{Gr(7.15)} + \] +
+ ++Note that Faraday's law is valid +for any loop (on a wire or not). Using Stokes' theorem, +\[ +\oint_{\cal P} {\bf E} \cdot d{\bf l} = \int_{\cal S} ({\boldsymbol \nabla} \times {\bf E}) \cdot d{\bf a}, +\] +we obtain +
++Faraday's law (differential form) + \[ + {\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} + \label{Gr(7.16)} + \] +
+ ++Right-hand rule always sorts signs out. Easier rule: {\bf Lenz's law}, which +states that {\bf nature resists a change in flux}. This is in fact just +{\bf Le Ch\atelier's principle} of any action at an equilibrium point leading +to an opposing counter-reaction. +
+Created: 2022-02-07 Mon 08:02
+ ++Work per unit time in current loop: +\[ +\frac{dW}{dt} = -{\cal E} I = L I \frac{dI}{dt} +\] +Start from zero current, integrate in time: +\[ +W = \frac{1}{2} L I^2 +\label{Gr(7.29)} +\] +Nicer way (generalizable to surface and volume currents): from (\ref{Gr(7.25)}), flux through loop is \(\Phi = L I\). But +\[ +\Phi = \int_{\cal S} {\bf B} \cdot d{\bf a} = \int_{\cal S} ({\boldsymbol \nabla} \times {\bf A}) \cdot d{\bf a} += \oint_{\cal P} {\bf A} \cdot d{\bf l}, +\] +so +\[ +LI = \oint {\bf A} \cdot d{\bf l} +\] +and +\[ +W = \frac{1}{2} I \oint {\bf A} \cdot d{\bf l} = \frac{1}{2} \oint ({\bf A} \cdot {\bf I}) dl +\label{Gr(7.30)} +\] +Generalization to volume currents: +\[ +W = \frac{1}{2} \int_{\cal V} ({\bf A} \cdot {\bf J}) d\tau +\label{Gr(7.31)} +\] +Even better: use Ampère, \({\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}\): +\[ +W = \frac{1}{2\mu_0} \int_{\cal V} {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) d\tau +\label{Gr(7.32)} +\] +Integrate by parts using product rule 6: +\[ +{\boldsymbol ∇} ⋅ ({\bf A} × {\bf B}) = {\bf B} ⋅ ({\boldsymbol ∇} × {\bf A}) +
++\] +so +\[ +{\bf A} ⋅ ({\boldsymbol ∇} × {\bf B}) = {\bf B} ⋅ {\bf B} +
++\] +Then, +\[ +W = \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \int_{\cal V} d\tau {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) \right] += \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \oint_{\cal S} d{\bf a} \cdot ({\bf A} \times {\bf B}) \right] +\label{Gr(7.33)} +\] +We can integrate over all space: after neglecting boundary terms (assuming fields fall to zero at infinity), we are left with +
++\[ + W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2 + \label{Gr(7.34)} + \] +
+ ++Summary: energy in electric and magnetic fields: +
+\begin{align} +W_{elec} = \frac{1}{2} \int d\tau V\rho = \frac{\varepsilon_0}{2} \int d\tau E^2, \hspace{2cm} +\mbox{(2.43 and 2.45)}, \\ +W_{mag} = \frac{1}{2} \int d\tau ({\bf A} \cdot {\bf J}) = \frac{1}{2\mu_0} \int d\tau B^2, +\hspace{2cm} \mbox{(7.31 and 7.34)} +\end{align} + ++\paragraph{Example 7.13:} coaxial cable (inner cylinder radius \(a\), outer \(b\)) carries current \(I\). +Find energy stored in section of length \(l\). +\paragraph{Solution:} from Ampère, +\[ + {\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \phi}, \hspace{1cm} a < s < b, \hspace{1cm} + {\bf B} = 0, \hspace{1cm} s < a ~\mbox{or}~ s > b. + \] +Energy is thus +\[ + W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\phi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2 + = \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}. + \] +
+ ++Note: gives easy way to find inductance, since \(W = \frac{1}{2} L I^2\). +
+Created: 2022-02-07 Mon 08:02
+ ++Thinking of Faraday's experiments. +Two loops of wire. From Biot-Savart, flux \(\Phi_2\) of \({\bf B}_1\) through loop 2 +is (using fact that \({\bf B}_1\) is proportional to \(I_1\)) +\[ +\Phi_2 = \int {\bf B}_1 \cdot d{\bf a}_2 \Longrightarrow +\Phi_2 = M_{21} I_1 +\] +where \(M_{21}\) is the {\bf mutual inductance} of the two loops. +
+ ++Useful formula: +\[ +\Phi_2 = \int {\bf B}_1 \cdot d{\bf a}_2 = \int ({\boldsymbol \nabla} \times {\bf A}_1) \cdot d{\bf a}_2 += \oint {\bf A}_1 \cdot d{\bf l}_2 +\] +But from (\ref{Gr(5.63)}), +\[ +{\bf A}_1 ({\bf r}) = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_1} \frac{d{\bf l}_1}{|{\bf r} - {\bf r}_1|} +\] +so +\[ +\Phi_2 = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_2} d{\bf l}_2 \cdot +\left(\oint_{{\cal P}_1} \frac{d{\bf l}_1 }{|{\bf r}_2 - {\bf r}_1|}\right) +\] +and we can write the mutual inductance as the {\bf Neumann formula}, +\[ +M_{21} = \frac{\mu_0}{4\pi} \oint_{{\cal P}_1} \oint_{{\cal P}_2} \frac{d{\bf l}_1 \cdot d{\bf l}_2} +{|{\bf r}_1 - {\bf r}_2|} +\label{Gr(7.22)} +\] +Two things: +first, \(M_{21}\) is purely geometrical. Second, +\[ +M_{12} = M_{21} +\label{Gr(7.23)} +\] +
+ ++\paragraph{Example 7.10:} +short solenoid (length \(l\), radius \(a\), \(n_1\) turns per unit length) lies concentrically inside +a very long solenoid (radius \(b\), \(n_2\) turns per unit length). Current \(I\) in short solenoid. +What is flux through long solenoid ? +\paragraph{Solution:} complicated to calculate \({\bf B}_1\). Use mutual inductance, starting from +the reverse situation: current \(I\) on outer solenoid, calculate flux through inner one. +Field of outer solenoid: from (\ref{Gr(5.57)}), +\[ + B = \mu_0 n_2 I + \] +so flux through a single loop of inner solenoid is +\[ + B \pi a^2 = \mu_0 n_2 I \pi a^2. + \] +For \(n_1 l\) turns in total, total flux through inner solenoid is +\[ + \Phi = \mu_0 \pi a^2 n_1 n_2 l I. + \] +Same as flux through outer solenoid if inner one has current \(I\). Mutual inductance is here +\[ + M = \mu_0 \pi a^2 n_1 n_2 l. + \] +
+ ++What if we vary current in loop 1? Flux in 2 will vary. Induces EMF in loop 2: +\[ +{\cal E} = -\frac{d\Phi_2}{dt} = -M \frac{dI_1}{dt}. +\label{Gr(7.24)} +\] +Changing current also induces EMF in the source loop itself: +\[ +\Phi = L I +\label{Gr(7.25)} +\] +where \(L\) is the {\bf self-inductance} (or inductance) of the loop. Depends only on +geometry. Changing current induces EMF of +\[ +{\cal E} = -L \frac{dI}{dt} +\label{Gr(7.26)} +\] +Inductance: measured in {\bf henries} (\(H\)). \(H = V s/A\). +
+ + ++\paragraph{Example 7.11:} find self-inductance of toroidal coil with +rectangular cross-section (inner radius \(a\), outer radius \(b\), height \(h\)) +which carries total of \(N\) turns. +\paragraph{Solution:} magnetic field inside toroid is (\ref{Gr(5.58)}) +\[ + B = \frac{\mu_0 NI}{2\pi s} + \] +Flux through single turn: +\[ + \int {\bf B} \cdot d{\bf a} = \frac{\mu_0 N I}{2\pi} h \int_a^b \frac{ds}{s} + = \frac{\mu_0 N I h}{2\pi} \ln \frac{b}{a}. + \] +Total flux: \(N\) times this, so self-inductance is +\[ + L = \frac{\mu_0 N^2 h}{2\pi} \ln \frac{b}{a} + \label{Gr(7.27)} + \] +
+ ++Inductance (like capacitance) is intrinsically positive. Use Lenz law. Think of {\bf back EMF}. +
+ ++\paragraph{Example 7.12:} circuit with inductance \(L\), resistor \(R\) and battery \({\cal E}_0\). +What is the current ? +\paragraph{Solution:} +Ohm's law: +\[ + {\cal E}_0 - L \frac{dI}{dt} = IR \Longrightarrow I(t) = \frac{{\cal E}_0}{R} + k e^{-(R/L)t}. + \] +If initial condition: \(I(0) = 0\), then +\[ + I(t) = \frac{{\cal E}_0}{R} \left[ 1 - e^{-(R/L)t} \right] + \label{Gr(7.28)} + \] +where \(\tau \equiv L/R\) is the {\bf time constant} of the circuit. +
+ +Created: 2022-02-07 Mon 08:02
+ ++Two sources of electric fields: electric charges, and changing magnetic fields. +
+ ++Electric fields induced by a changing magnetic field are determined in an exactly +parallel way as magnetostatic fields from the current: exploit parallel +between Ampère and Faraday! +\[ +{\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} +\hspace{3cm} +{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}. +\] +For electric field induced by changing magnetic field: use tricks of Ampère's +law in integral form: +\[ +\oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} +\hspace{3cm} +\oint_{\cal P} {\bf E} \cdot d{\bf l} = -\frac{d}{dt} \Phi +\] +
+ + + ++{\bf Example 7.7:} +\({\bf B}(t)\) points up in circular region of radius \(R\). What is the induced \({\bf E}(t)\) ? +\paragraph{Solution:} +amperian loop of radius \(s\), apply Faraday: +\[ + \oint {\bf E} \cdot d{\bf l} = E (2\pi s) = -\frac{d\Phi}{dt} = -\pi s^2 \frac{dB}{dt} + \Rightarrow {\bf E} = -\frac{s}{2} \frac{dB}{dt} \hat{\boldsymbol \phi}. + \] +Increasing \({\bf B}\): clockwise (viewed from above) \({\bf E}\) from Lenz. +
+ ++{\bf Example 7.8:} wheel or radius \(b\) with line charge \(\lambda\) on the rim. +Uniform magnetic field \({\bf B}_0\) in central region up to \(a < b\), +pointing up. Field turned off. What happens ? +\paragraph{Solution:} the wheel starts spinning to compensate the reduction of field. +Faraday: +\[ + \oint {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt} = - \pi a^2 \frac{dB}{dt} + \Rightarrow {\bf E} = -\frac{a^2}{2b} \frac{dB}{dt} \hat{\boldsymbol \phi}. + \] +Torque on segment \(d{\bf l}\): \(|{\bf r} \times {\bf F}| = b \lambda E dl\). +Total torque: +\[ + N = b\lambda \oint E dl = -b \lambda \pi a^2 \frac{dB}{dt} + \] +so total angular momentum imparted is +\[ + \int N dt = -\lambda \pi a^2 b \int_{B_0}^0 dB = \lambda \pi a^2 b B_0. + \] +
+ ++The precise way the field is turned off doesn't matter. Only electric field does work. +
+ ++{\bf N.B.:} we use magnetostatic formulas for changing fields. This is +called the {\bf quasistatic} approximation, and works provided we deal with +'slow enough' phenomena. +
+ ++{\bf Example 7.9:} infinitely long straight wire carries \(I(t)\). Find +induced \({\bf E}\) field as a function of distance \(s\) from wire. +\paragraph{Solution:} quasistatic: magnetic field is \(B = \frac{\mu_0 I}{2\pi s}\) +and circles the wire. Like \({\bf B}\) field of solenoid, \({\bf E}\) runs parallel +to wire. Amperian loop with sides at distances \(s_0\) and \(s\): +\[ + \oint {\bf E} \cdot d{\bf l} = E(s_0)l - E(s)l = -\frac{d}{dt} \int {\bf B} \cdot d{\bf a} + = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \int_{s_0}^s \frac{ds'}{s'} + = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \ln(s/s_0). + \] +So: +\[ + {\bf E} (s) = \left[ \frac{\mu_0}{2\pi} \frac{dI}{dt} \ln s + K \right] \hat{\bf x} + \label{Gr(7.19)} + \] +where \(K\) is a constant (depends on the history of \(I(t)\)). +
+ ++{\bf N.B.:} this can't be true always, since it blows up as \(s \rightarrow \infty\). +Reason: in this case, we've overstepped the quasistatic limit. We need +\(s \ll c\tau\) where \(\tau\) is a typical time scale for change of \(I(t)\). +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Full set of equations for the electromagnetic field: +
++{\bf Maxwell's equations} {\it (in vacuum)} +
+\begin{align} + (i) {\boldsymbol \nabla} \cdot {\bf E} &= \frac{\rho}{\varepsilon_0}, \hspace{1cm} &\mbox{Gauss}, \nonumber \\ + (ii) {\boldsymbol \nabla} \cdot {\bf B} &= 0, \hspace{1cm} &\mbox{anonymous} \nonumber \\ + (iii) {\boldsymbol \nabla} \times {\bf E} &= -\frac{\partial {\bf B}}{\partial t}, \hspace{1cm} &\mbox{Faraday}, \nonumber \\ + (iv) {\boldsymbol \nabla} \times {\bf B} &= \mu_0 {\bf J} + \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t}, \hspace{1cm} &\mbox{Ampère + Maxwell}. + \label{Gr(7.39)} +\end{align} + ++Complement: +
++{\bf Force law} +\[ + {\bf F} = q ({\bf E} + {\bf v} \times {\bf B}). + \label{Gr(7.40)} + \] +
+ ++These equations summarize the {\bf entire content of classical electrodynamics}. +
+ ++\paragraph{Note:} even the continuity equation can be derived from Maxwell's equations: +take divergence of \((iv)\). +
+ + + ++Better way of writing: all fields on left, all sources on right, +
+Created: 2022-02-07 Mon 08:02
+ ++The term which should be zero (but isn't) in (\ref{Gr(7.35)}) can be rewritten using +the continuity equation as +\[ +{\boldsymbol \nabla} \cdot {\bf J} = -\frac{\partial \rho}{\partial t} = - \frac{\partial}{\partial t} +(\varepsilon_0 {\boldsymbol \nabla} \cdot {\bf E}) = -{\boldsymbol \nabla} \cdot \left( +\varepsilon_0 \frac{\partial {\bf E}}{\partial t} \right). +\] +The extra term would thus be eliminated if we were to put +
++\[ + {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} + \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} + \label{Gr(7.36)} + \] +
+ ++\paragraph{Note:} this changes nothing in magnetostatics. Aesthetic appeal: +\[ + \boxed{ + \mbox{A changing electric field induces a magnetic field.} + } +\] +Real confirmation of Maxwell's theory: 1888, Hertz's experiments on propagation of electromagnetic waves. +
+ + + ++Maxwell baptized this term the +
++{\bf Displacement current} +\[ + {\bf J}_d \equiv \varepsilon_0 \frac{\partial {\bf E}}{\partial t}. + \label{Gr(7.37)} + \] +
+ ++Resolves the charging capacitor plate problem: if plates close together, +field is +\[ +E = \frac{\sigma}{\varepsilon_0} = \frac{1}{\varepsilon_0} \frac{Q}{A} +\] +where \(A\) is the area. Between the plates, +\[ +\frac{\partial E}{\partial t} = \frac{1}{\varepsilon_0 A} \frac{dQ}{dt} = \frac{1}{\varepsilon_0 A} I. +\] +Checking (\ref{Gr(7.36)}), +\[ +\oint {\bf B} \cdot d{\bf l} = \mu_0 I_{\mbox{enc}} + \mu_0 \varepsilon_0 \int d{\bf a} \cdot \frac{\partial {\bf E}}{\partial t} +\label{Gr(7.38)} +\] +Flat surface: OK, \(E = 0\) and \(I_{\mbox{enc}} = I\). Balloon surface: \(I = 0\) but \(\int d{\bf a} \cdot (\partial {\bf E}/\partial t) = I/\varepsilon_0\). Answers are consistent. +
+Created: 2022-02-07 Mon 08:02
+ ++We've encountered: +
+\begin{align} +(i) &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0}, \hspace{1cm} &\mbox{Gauss}, \nonumber \\ +(ii) &{\boldsymbol \nabla} \cdot {\bf B} = 0, \hspace{1cm} &\mbox{anonymous} \nonumber \\ +(iii) &{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \hspace{1cm} &\mbox{Faraday}, \nonumber \\ +(iv) &{\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}, \hspace{1cm} &\mbox{Ampère}. +\end{align} ++Fatal inconsistency: div of curl must always vanish. Check on \((iii)\): +\[ +{\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} \times {\bf E}) += {\boldsymbol \nabla} \cdot \left( -\frac{\partial {\bf B}}{\partial t} \right) = -\frac{\partial}{\partial t} ({\boldsymbol \nabla} \cdot {\bf B}) = 0. +\] +But: try same with \((iv)\): +\[ +{\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} \times {\bf B}) = \mu_0 {\boldsymbol \nabla} \cdot {\bf J} +\label{Gr(7.35)} +\] +LHS must be zero, but RHS is not zero for non-steady currents. Cannot be right ! +
+ ++Other way of seeing that Ampère's law must fail for non-steady currents: suppose we're charging a capacitor. +In integral form, +\[ +\oint {\bf B} \cdot d{\bf l} = \mu_0 I_{\mbox{enc}}. +\] +But the surface can either cut the charging wire, or not (by going 'around' the capacitor plate). +So for non-steady currents, the 'current enclosed by a loop' is ill-defined. +
+Created: 2022-02-07 Mon 08:02
+ ++In free space, where \(\rho\) and \({\bf J}\) vanish: +
+\begin{align} +(i) &{\boldsymbol \nabla} \cdot {\bf E} = 0, +&(iii) {\boldsymbol \nabla} \times {\bf E} + \frac{\partial {\bf B}}{\partial t} = 0, \\ +(ii) &{\boldsymbol \nabla} \cdot {\bf B} = 0, +&(iv) {\boldsymbol \nabla} \times {\bf B} - \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} = 0, +\end{align} ++Symmetry: replace \({\bf E}\) by \({\bf B}\) and \({\bf B}\) by \(-\mu_0 \varepsilon_0{\bf E}\) in the first pair. +They turn into the second pair. This symmetry is spoiled by \(\rho\) and \({\bf J}\). What if we had +a truly symmetric situation, {\it i.e.} +
+\begin{align} +(i) &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho_e}{\varepsilon_0}, +&(iii) {\boldsymbol \nabla} \times {\bf E} = -\mu_0 {\bf J}_m -\frac{\partial {\bf B}}{\partial t}, \\ +(ii) &{\boldsymbol \nabla} \cdot {\bf B} = \mu_0 \rho_m, +&(iv) {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}_e + \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t}, +\label{Gr(7.43)} +\end{align} ++where \(\rho_m\) would represent the density of magnetic charge and \({\bf J}_m\) would be the current +of magnetic charge. Both charges would be conserved: +\[ +{\boldsymbol \nabla} \cdot {\bf J}_m = -\frac{\partial \rho_m}{\partial t}, \hspace{1cm} +{\boldsymbol \nabla} \cdot {\bf J}_e = -\frac{\partial \rho_e}{\partial t}. +\label{Gr(7.44)} +\] +Maxwell's equations {\bf beg} for magnetic charges. But we've never found any! +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++The angular momentum of EM fields is directly given by +
++{\bf Angular momentum of EM fields} +\[ + {\boldsymbol l} = {\boldsymbol r} \times {\boldsymbol g} + = \varepsilon_0 {\boldsymbol r} \times + \left({\boldsymbol E} \times {\boldsymbol B}\right) + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Very important distinction: {\bf global} versus {\bf local} conservation of charge. +
+ ++Charge in a volume {\cal V}: +\[ +Q_{\cal V} (t) = \int_{\cal V} d\tau \rho ({\bf r}, t) +\label{Gr(8.1)} +\] +Current \({\bf J}\) flowing out through boundary \({\cal S}\) of \({\cal V}\): conservation of charge means +\[ +\frac{dQ_{\cal V}}{dt} = -\oint_{\cal S} d{\bf a} \cdot {\bf J} +\label{Gr(8.2)} +\] +This means that +\[ +\int_{\cal V} d\tau \frac{\partial \rho}{\partial t} = -\int_{\cal V} d\tau {\boldsymbol \nabla} \cdot {\bf J} +\label{Gr(8.3)} +\] +Since this is true for any volume, we have (re)derived the +
++{\bf Continuity equation} +\[ + \frac{\partial \rho}{\partial t} + {\boldsymbol \nabla} \cdot {\bf J} = 0 + \label{Gr(8.4)} + \] +
+ ++Therefore, conservation of charge is a direct consequence of Maxwell's equations. +
+ ++One thing to note: we have viewed \(\rho\) and \({\boldsymbol J}\) as sources +(the ''right-hand side'') of Maxwell's equations. The continuity equation thus +imposes a functional constraint on these sources: not {\it any} \(\rho\) and +\({\boldsymbol J}\) will do the trick, only the ones what obey it. +
+Created: 2022-02-07 Mon 08:02
+ ++From Newton's second law, +\[ + {\boldsymbol F} = \frac{d {\boldsymbol p}_{\tiny \mbox{mech}}}{dt} +\] +we have +\[ +\frac{d {\boldsymbol p}_{\tiny \mbox{mech}}}{dt} = -\varepsilon_0 \mu_0 \frac{d}{dt} \int_{\cal V} {\boldsymbol S} d\tau + \oint_S {\boldsymbol T} \cdot d{\boldsymbol a} +\] +in which the first integral can be interpreted as the momentum stored in the EM fields, and the second is the momentum per unit time flowing in through the surface. +
+ ++This is thus simply a conservation law for momentum, with +
++{\bf Momentum density in the EM fields} +\[ + {\boldsymbol g} = \varepsilon_0 \mu_0 {\boldsymbol S} = \varepsilon_0 {\boldsymbol E} \times {\boldsymbol B} + \] +
+ ++In a region in which the mechanical momentum is not changing due to external influences, we then have the +
++{\bf Continuity equation for EM momentum} +\[ + \frac{\partial}{\partial t} {\boldsymbol g} - {\boldsymbol \nabla} \cdot {\boldsymbol T} = 0 + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Total EM force on charges in volume \({\cal V}\): +\[ + {\boldsymbol F} = \int_{\cal V} \left( {\boldsymbol E} + {\boldsymbol v} \times {\boldsymbol B} \right) \rho ~d\tau = \int_{\cal V} \left( \rho {\boldsymbol E} + {\boldsymbol J} \times {\boldsymbol B} \right) d\tau +\] +Force per unit volume: +\[ + {\boldsymbol f} \equiv \rho {\boldsymbol E} + {\boldsymbol J} \times {\boldsymbol B}. +\] +Substitute for \(\rho\) and \({\boldsymbol J}\) using Maxwell (Gauss and Ampère-Maxwell): +\[ + {\boldsymbol f} = \varepsilon_0 ({\boldsymbol \nabla} \cdot {\boldsymbol E}) {\boldsymbol E} + \left( \frac{1}{\mu_0} {\boldsymbol \nabla} \times {\boldsymbol B} - \varepsilon_0 \frac{ \partial {\boldsymbol E}}{\partial t} \right) \times {\boldsymbol B}. +\] +
+ ++On the other hand we have +\[ + \frac{\partial }{\partial t} \left( {\boldsymbol E} × {\boldsymbol B} \right) + = \frac{∂ {\boldsymbol E}}{∂ t} × {\boldsymbol B} +
++\] +Using Faraday to substitute for \(\frac{\partial {\boldsymbol B}}{\partial t}\), +\[ + \frac{ ∂ {\boldsymbol E}}{∂ t} × {\boldsymbol B} + = \frac{\partial }{\partial t} \left( {\boldsymbol E} × {\boldsymbol B}\right) +
++\] +so +\[ + {\boldsymbol f} = \varepsilon_0 \left( \left( {\boldsymbol \nabla} \cdot {\boldsymbol E} \right) {\boldsymbol E} - {\boldsymbol E} \times \left( {\boldsymbol \nabla} \times {\boldsymbol E} \right) \right) - \frac{1}{\mu_0} \left( {\boldsymbol B} \times \left( {\boldsymbol \nabla} \times {\boldsymbol B} \right) \right) - \varepsilon_0 \frac{\partial}{\partial t} \left( {\boldsymbol E} \times {\boldsymbol B} \right). +\] +Since \({\boldsymbol \nabla} \cdot {\boldsymbol B} = 0\), we can symmetrize the expression in \({\boldsymbol E}\) and \({\boldsymbol B}\). Moreover, by product rule 4, +\[ + \frac{1}{2}{\boldsymbol \nabla} \left( E^2 \right) = \left( {\boldsymbol E} \cdot {\boldsymbol \nabla} \right) {\boldsymbol E} + {\boldsymbol E} \times \left( {\boldsymbol \nabla} \times {\boldsymbol E} \right) +\] +so +\[ + {\boldsymbol E} \times \left( {\boldsymbol \nabla} \times {\boldsymbol E} \right) = \frac{1}{2} {\boldsymbol \nabla} E^2 - \left({\boldsymbol E} \cdot {\boldsymbol \nabla} \right) {\boldsymbol E} +\] +and similarly for \({\boldsymbol B}\). We thus get +
+\begin{align} + {\boldsymbol f} =& \varepsilon_0 \left( \left( {\boldsymbol \nabla} \cdot {\boldsymbol E} \right) {\boldsymbol E} + \left( {\boldsymbol E} \cdot {\boldsymbol \nabla} \right) {\boldsymbol E} \right) + \frac{1}{\mu_0} \left( \left( {\boldsymbol \nabla} \cdot {\boldsymbol B} \right) {\boldsymbol B} + \left( {\boldsymbol B} \cdot {\boldsymbol \nabla} \right) {\boldsymbol B} \right) \\ + &- \frac{1}{2} {\boldsymbol \nabla} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \varepsilon_0 \frac{\partial}{\partial t} \left( {\boldsymbol E} \times {\boldsymbol B} \right). +\end{align} ++This expression can be greatly simplified by introducing the +
++{\bf Maxwell stress tensor} +\[ + Tij ≡ ε0 \left( Ei Ej - \frac{1}{2} δij E2\right) +
++\] +
+ ++The element \(T_{ij}\) represents the force per unit area in the $i$th direction acting on a surface element oriented in the $j$th direction. Diagonal elements are pressures, off-diagonal elements are shears. +
+ + + ++We then obtain +
++{\bf EM force per unit volume} +\[ + {\boldsymbol f} = {\boldsymbol \nabla} \cdot {\boldsymbol T} - \varepsilon_0 \mu_0 \frac{\partial {\boldsymbol S}}{\partial t} + \] +
+ ++where \({\boldsymbol S}\) is the Poynting vector. Integrating, we obtain the +
++{\bf Total force on charges in volume} +\[ + {\boldsymbol F} = \oint_S {\boldsymbol T} \cdot d{\boldsymbol a} - \varepsilon_0 \mu_0 \frac{d}{dt} \int_{\cal V} {\boldsymbol S} d\tau. + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Earlier: work done to assemble a static charge distribution: +\[ +W_e = \frac{\varepsilon_0}{2} \int d\tau ~E^2 +\tag{\ref{eq:Energy_as_int_E2}} +\] +Work necessary to get currents going: +\[ +W_m = \frac{1}{2\mu_0} \int d\tau ~B^2 +\label{Gr(7.34)} +\] +Total energy should be sum of these two. Derivation from scratch. +
+ ++Suppose that at time \(t\), we have fields \({\bf E}\) and \({\bf B}\) produced by some charge +and current distributions \(\rho\) and \({\bf J}\). In an interval \(dt\), how much work is +done by EM forces ? From Lorentz force law: +\[ +{\bf F} \cdot d{\bf l} = q({\bf E} + {\bf v} \times {\bf B}) \cdot {\bf v} dt = q ~{\bf E} \cdot {\bf v} dt +\] +Really, we're looking at a small volume element \(d\tau\) carrying charge \(\rho d\tau\), moving +at velocity \({\bf v}\) such that \({\bf J} = \rho {\bf v}\). Thus, +\[ +\frac{dW}{dt} = \int_{\cal V} d\tau ~ {\bf E} \cdot {\bf J} +\label{Gr(8.6)} +\] +The integrand is the work done per unit time, per unit volume, {\it i.e.} the power delivered per unit volume. +In terms of fields alone: use Ampère-Maxwell: +\[ +{\bf E} \cdot {\bf J} = \frac{1}{\mu_0} {\bf E} \cdot ({\boldsymbol \nabla} \times {\bf B}) - \varepsilon_0 {\bf E} \cdot \frac{\partial {\bf E}}{\partial t} +\] +Using product rule 6, +\[ +{\boldsymbol ∇} ⋅ ({\bf E} × {\bf B}) = {\bf B} ⋅ ({\boldsymbol ∇} × {\bf E}) +
++\] +Invoking Faraday \({\boldsymbol \nabla} \times {\bf E} = - \partial {\bf B}/\partial t\), +\[ +{\bf E} ⋅ ({\boldsymbol ∇} × {\bf B}) = - {\bf B} ⋅ \frac{∂ {\bf B}}{∂ t} +
++\] +But obviously, +\[ +{\bf B} \cdot \frac{\partial {\bf B}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} B^2, \hspace{1cm} +{\bf E} \cdot \frac{\partial {\bf E}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} E^2 +\label{Gr(8.7)} +\] +so we get +\[ +{\bf E} ⋅ {\bf J} = -\frac{1}{2} \frac{\partial}{\partial t} \left( ε0 E2 + \frac{1}{\mu_0} B2 \right) +
++\label{Gr(8.8)} +\] +Substituting this in \ref{Gr(8.6)} and using the divergence theorem, +we obtain +
++{\bf Poynting's theorem} +\[ + \frac{dW}{dt} = -\frac{d}{d t} ∫\cal V dτ \frac{1}{2} \left( ε0 E2 + \frac{1}{\mu_0} B2 \right) +
++ \label{Gr(8.9)} +\] +
+ ++First term on RHS: total energy in EM fields. Second term: rate at which +energy is carried by EM fields out of \({\cal V}\) across its boundary surface. +
+ + + ++Energy per unit time, per unit area carried by EM fields: +
++{\bf Poynting vector} +\[ + {\bf S} \equiv \frac{1}{\mu_0} ({\bf E} \times {\bf B}) + \label{Gr(8.10)} + \] +
+ ++We can thus express Poynting's theorem more compactly: +
++{\bf Poynting's theorem} +\[ + \frac{dW}{dt} = - \frac{dU_{em}}{dt} - \oint_{\cal S} d{\bf a} \cdot {\bf S}. + \label{Gr(8.11)} + \] +
+ ++where we have defined the total +
++{\bf Energy in electromagnetic fields} +\[ + U_{em} \equiv \frac{1}{2} \int d\tau \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) + \label{Gr(8.5)} + \] +
+ ++We can rewrite the energies in terms of densities, +\[ +U_{em} = \int d\tau ~u_{em}, \hspace{1cm} +u_{em} \equiv \frac{1}{2} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) +\label{Gr(8.13)} +\] +Then, +\[ +\frac{d}{dt} \int_{\cal V} d\tau ~u_{em} = -\oint_{\cal S} d{\bf a} \cdot {\bf S} = -\int_{\cal V} d\tau ~({\boldsymbol \nabla} \cdot {\bf S}) +\] +so we get the +
++{\bf Poynting theorem (differential form)} +\[ + \frac{\partial}{\partial t} u_{em} + {\boldsymbol \nabla} \cdot {\bf S} = 0 + \label{Gr(8.14)} + \] +
+ ++and has a similar for to the continuity equation +(charge density goes into energy density, charge current goes into Poynting vector). +
+ + + ++\paragraph{Example 8.1} Current in a wire: Joule heating. Energy per unit time delivered to wire: from Poynting. +Assuming that the field is uniform, the electric field parallel to the wire is +\[ + {\boldsymbol E} = \frac{V}{L} \hat{\boldsymbol x}, + \] +where \(V\) is the potential difference between the ends ald \(L\) is the length. Magnetic field is circumferential: +wire of radius \(a\), +\[ + {\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \phi} + \] +Poynting: +\[ + {\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \phi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s} + \] +and points radially inwards. Energy per unit time passing surface of wire: +\[ + \int d{\bf a} \cdot {\bf S} = S (2\pi a L) = -V I + \] +where the minus sign means energy is flowing {\it in} (the wire heats up), +and the value is as expected. +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Energy density in EM field: +\[ +u = \frac{1}{2} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) +\label{Gr(9.53)} +\] +For a monochromatic EM plane wave, +\[ +B^2 = \frac{1}{c^2} E^2 = \mu_0 \varepsilon_0 E^2 +\label{Gr(9.54)} +\] +so the electric and magnetic contributions to energy density are equal +and +\[ + u = \varepsilon_0 E^2 = \varepsilon_0 E_0^2 \cos({\boldsymbol k} \cdot {\boldsymbol r} - \omega t + \delta) +\] +
+ ++Poynting vector: +\[ +{\bf S} = \frac{1}{\mu_0} {\bf E} \times {\bf B}, +\label{Gr(9.56)} +\] +so for a monochromatic EM plan wave, +\[ + {\bf S} = c \varepsilon_0 E_0^2 \cos^2 ({\boldsymbol k} \cdot {\boldsymbol r} - \omega t + \delta) ~\hat{\boldsymbol n} \times (\hat{\boldsymbol k} \times \hat{\boldsymbol n}) = +\label{Gr(9.57)} +\] +or more succinctly: +
++{\bf Poynting vector of a monochromatic EM wave} +\[ + {\boldsymbol S} = c u ~\hat{\boldsymbol k} + \] +
+ ++This has a transparent physical interpretation: the energy density \(u\) flows with velocity \(c\) in direction of the wave. +
+ ++Similary, we get the +
++{\bf Momentum density of a monochromatic EM wave} +\[ + {\boldsymbol g} = \frac{1}{c^2} {\boldsymbol S} = \frac{u}{c} ~\hat{\boldsymbol k} + \] +
+ ++Time averages: integrating over a (integer number of) cycle(s), we have +\[ + \langle u \rangle = \frac{\varepsilon_0}{2} E_0^2, \hspace{5mm} + \langle {\boldsymbol S} \rangle = \frac{c \varepsilon_0}{2} E_0^2 ~\hat{\boldsymbol k}, \hspace{5mm} + \langle {\boldsymbol g} \rangle = \frac{\varepsilon_0}{2c} E_0^2 ~\hat{\boldsymbol k}. +\] +
+ ++The average power per unit time per unit area transported by an EM wave is called the {\bf Intensity} +\[ + I \equiv \langle S \rangle = \frac{c\varepsilon_0}{2} E_0^2 +\] +
+ ++The {\it radiation pressure} is the momentum transfer per unit area per unit of time +\[ + P = \frac{1}{A}\frac{\Delta p}{\Delta t} = \frac{\langle g \rangle A c \Delta t}{A \Delta t} = \frac{\varepsilon_0}{2} E_0^2 = \frac{I}{c}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++A {\it monochromatic} wave is one having a single frequency in its temporal dependence. Say that the propagation direction is \(\hat{\boldsymbol z}\): we'd then have +\[ + {\bf E} (z, t) = {\bf E}_0 e^{i(k z - \omega t)}, \hspace{1cm} + {\bf B} (z,t) = {\bf B}_0 e^{i(k z - \omega t)} +\] +Maxwell's equations impose constraints. Since \({\boldsymbol \nabla} \cdot {\bf E} = 0\) and \({\boldsymbol \nabla} \cdot {\bf B} = 0\), +\[ +(E_0)_z = 0 = (B_0)_z +\label{Gr(9.44)} +\] +so {\bf electromagnetic waves are transverse}. +
+ ++From Faraday: \({\boldsymbol \nabla} \times {\bf E} = -\partial {\bf B}/\partial t\), we then get +\[ +-k(E_0)_y = \omega (B_0)_x, \hspace{1cm} k (E_0)_x = \omega (B_0)_y \Longrightarrow +{\bf B}_0 = \frac{k}{\omega} ~\hat{\bf z} \times {\bf E}_0 +\label{Gr(9.46)} +\] +so \({\bf E}\) and \({\bf B}\) are mutually perpendicular, and +\[ +B_0 = \frac{k}{\omega} E_0 = \frac{1}{c} E_0. +\label{Gr(9.47)} +\] +
+ ++Generalizing to propagation in the direction of an arbitrary wavevector +\({\boldsymbol k}\) and (transverse) polarization vector \(\hat{\boldsymbol n}\), we have the +
++{\bf E and B fields for a monochromatic EM plane wave} +\[ + {\boldsymbol E} ({\boldsymbol r},t ) = E_0 e^{i ({\boldsymbol k} \cdot {\boldsymbol r} - \omega t)} ~\hat{\boldsymbol n}, + \hspace{10mm} + {\boldsymbol B} ({\boldsymbol r}, t) = \frac{E_0}{c} e^{i({\boldsymbol k} \cdot {\boldsymbol r} - \omega t)} ~\hat{\boldsymbol k} \times \hat{\boldsymbol n} + = \frac{1}{c} ~\hat{\boldsymbol k} \times {\boldsymbol E} ({\boldsymbol r}, t) + \] +with the transversality condition +\[ + \hat{\boldsymbol k} \cdot \hat{\boldsymbol n} = 0 + \] +
+ ++or if you prefer explicit real parts (adding a possible phase shift \(\delta\)): +\[ + {\boldsymbol E} ({\boldsymbol r},t ) = E_0 \cos({\boldsymbol k} \cdot {\boldsymbol r} - \omega t + \delta) \hat{\boldsymbol n}, + \hspace{10mm} + {\boldsymbol B} ({\boldsymbol r}, t) = \frac{E_0}{c} \cos ({\boldsymbol k} \cdot {\boldsymbol r} - \omega t + \delta) ~\hat{\boldsymbol k} \times \hat{\boldsymbol n} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++\subsubsection*{The wave equation for \({\bf E}\) and \({\bf B}\)} +Take Maxwell's equations in vacuum: +
+\begin{align} +(i) &{\boldsymbol \nabla} \cdot {\bf E} = 0, +&(iii) {\boldsymbol \nabla} \times {\bf E} + \frac{\partial {\bf B}}{\partial t} = 0, \\ +(ii) &{\boldsymbol \nabla} \cdot {\bf B} = 0, +&(iv) {\boldsymbol \nabla} \times {\bf B} - \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} = 0, +\end{align} ++These take the form of coupled first-order partial differential equations for \({\bf E}\) and \({\bf B}\). They can be decoupled: simply take the curl of \((iii)\) and \((iv)\): +
+\begin{align} +{\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf E}) += {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf E}) - {\boldsymbol \nabla}^2 {\bf E} += {\boldsymbol \nabla} \times \left( -\frac{\partial {\bf B}}{\partial t} \right) += -\frac{\partial}{\partial t} ({\boldsymbol \nabla} \times {\bf B}) += -\mu_0 \varepsilon_0 \frac{\partial^2 {\bf E}}{\partial t^2}, \\ +{\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf B}) += {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf B}) - {\boldsymbol \nabla}^2 {\bf B} += {\boldsymbol \nabla} \times \left(\mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} \right) += \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} += -\mu_0 \varepsilon_0 \frac{\partial^2 {\bf B}}{\partial t^2}. +\end{align} ++Since \({\boldsymbol \nabla} \cdot {\bf E} = 0\) and \({\boldsymbol \nabla} \cdot {\bf B} = 0\), +we get the +
++{\bf Wave equations for electric and magnetic fields in vacuum} +\[ + {\boldsymbol \nabla}^2 {\bf E} = \mu_0 \varepsilon_0 \frac{\partial^2 {\bf E}}{\partial t^2}, + \hspace{1cm} + {\boldsymbol \nabla}^2 {\bf B} = \mu_0 \varepsilon_0 \frac{\partial^2 {\bf B}}{\partial t^2}. + \label{Gr(9.41)} + \] +
+ ++The equations for the electric and magnetic fields are now decoupled, at the price of becoming second-order equations. +
+ ++In vacuum, the cartesian components of the fields thus obey the three-dimensional +wave equation +\[ +{\boldsymbol \nabla}^2 f = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}. +\] +Maxwell's equations therefore support solutions in terms of waves travelling at a speed +\[ +c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 299 792 458 ~m/s. +\] +That is, a form +\[ +{\bf E} ({\bf r},t) = {\bf E}_0 e^{i ({\bf k} \cdot {\bf r} - \omega t)}, \hspace{1cm} +{\bf B} ({\bf r},t) = {\bf B}_0 e^{i ({\bf k} \cdot {\bf r} - \omega t)}, +\] +solves (\ref{Gr(9.41)}) for \(\omega = c |{\bf k}|\). +Here and under, we use complex exponentials for convenience, remembering that +the actual electric and magnetic fields are given by the real part. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Maxwell's equations in vacuum: +
+\begin{align} +(i)~~ &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0}, \hspace{1cm} &\mbox{Gauss}, \nonumber \\ +(ii)~~ &{\boldsymbol \nabla} \cdot {\bf B} = 0, \hspace{1cm} &\mbox{anonymous} \nonumber \\ +(iii)~~ &{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \hspace{1cm} &\mbox{Faraday}, \nonumber \\ +(iv)~~ &{\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} + \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t}, \hspace{1cm} &\mbox{Ampère + Maxwell}. +\label{Gr(7.39)} +\end{align} ++are complete as they stand. In presence of matter: more convenient to write sources in terms of +free charges and currents. +
+ ++From static case: electric polarization \({\bf P}\) produces bound charge density (\ref{Gr(4.12)}) +\[ +\rho_b = -{\boldsymbol \nabla} \cdot {\bf P} +\label{Gr(7.46)} +\] +and magnetization \({\bf M}\) produces bound current density (\ref{Gr(6.13)}) +\[ +{\bf J}_b = {\boldsymbol \nabla} \times {\bf M} +\label{Gr(7.47)} +\] +One new feature in nonstatic case: change in electric polarization involves flow of bound charge +(call it \({\bf J}_p\)) which must be included in total current. +Consider a small chunk of polarized material. +Polarization induces charge density \(\sigma_b = P\) at one end and \(-\sigma_b\) at other. If \(P\) +increases a bit, charge increases, giving net current +\[ +dI = \frac{\partial \sigma_b}{\partial t} da_{\perp} = \frac{\partial P}{\partial t} da_{\perp}. +\] +We therefore have the +
++{\bf Polarization current density} +\[ + {\bf J}_p = \frac{\partial {\bf P}}{\partial t} + \label{Gr(7.48)} + \] +
+ ++otherwise simply called the {\bf polarization current}. +This has nothing to do with the bound current \({\bf J}_b\) +(the latter is associated to magnetization; +the polarization current is the result of linear motion of charge when +polarization changes). We can check consistency with the continuity equation +associated to the conservation of bound charges: +
+ ++\[ +{\boldsymbol \nabla} \cdot {\bf J}_p = {\boldsymbol \nabla} \cdot \frac{\partial {\bf P}}{\partial t} += \frac{\partial}{\partial t} ({\boldsymbol \nabla} \cdot {\bf P}) = -\frac{\partial \rho_b}{\partial t} +~~\longrightarrow~~\frac{\partial \rho_b}{\partial t} + {\boldsymbol \nabla} \cdot {\bf J}_p = 0 +\] +so OK, continuity equation is satisfied. +\({\bf J}_p\) is essential to account for conservation of bound charge. +Changing magnetization does not lead to analogous accumulation of charge and current. +
+ + + ++In view of this: total charge density can be separated into 2 parts, +{\it free} and {\it bound}: +
++\[ + \rho = \rho_f + \rho_b = \rho_f - {\boldsymbol \nabla} \cdot {\bf P} + \label{Gr(7.49)} + \] +
+ ++and current can be separated into three parts, {\it free}, {\it bound} and +{\it polarization}: +
++\[ + {\bf J} = {\bf J}f + {\bf J}b + {\bf J}p = {\bf J}f + {\boldsymbol ∇} × {\bf M} +
++ \label{Gr(7.50)} +\] +
+ ++Gauss's law: can be rewritten +\[ +{\boldsymbol \nabla} \cdot {\bf E} = \frac{1}{\varepsilon_0} (\rho_f - {\boldsymbol \nabla} \cdot {\bf P}) +\hspace{5mm}\longrightarrow \hspace{5mm} +{\boldsymbol \nabla} \cdot {\bf D} = \rho_f +\label{Gr(7.51)} +\] +where (as in static case) +
++\[ + {\bf D} \equiv \varepsilon_0 {\bf E} + {\bf P} + \label{Gr(7.52)} + \] +
+ ++Ampère's law including Maxwell's term: +\[ +{\boldsymbol ∇} × {\bf B} = μ0 \left( {\bf J}f + {\boldsymbol ∇} × {\bf M} +
++\] +or +\[ +{\boldsymbol \nabla} \times {\bf H} = {\bf J}_f + \frac{\partial {\bf D}}{\partial t} +\label{Gr(7.53)} +\] +where as before +
++\[ + {\bf H} \equiv \frac{1}{\mu_0} {\bf B} - {\bf M} + \label{Gr(7.54)} + \] +
+ ++Faraday's law and \({\boldsymbol \nabla} \cdot {\bf B} = 0\) remain +unaffected by our separation into free and +bound parts, since they don't involve \(\rho\) or \({\bf J}\). +
+ ++In terms of free charges and currents, we thus get +
++{\bf Maxwell's equations {\it (in matter)}} +
+\begin{align} + (i)~~ &{\boldsymbol \nabla} \cdot {\bf D} = \rho_f, \nonumber \\ + (ii)~~ &{\boldsymbol \nabla} \cdot {\bf B} = 0, \nonumber \\ + (iii)~~ &{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \nonumber \\ + (iv)~~ &{\boldsymbol \nabla} \times {\bf H} = {\bf J}_f + \frac{\partial {\bf D}}{\partial t}. + \label{Gr(7.55)} +\end{align} + ++Last term: {\bf displacement current}, +\[ +{\bf J}_d = \frac{\partial {\bf D}}{\partial t} +\label{Gr(7.58)} +\] +
+ ++Must be complemented by the {\bf constitutive relations} giving \({\bf D}\) and \({\bf H}\) +in terms of \({\bf E}\) and \({\bf B}\). +For the restricted case of linear media: +
++\[ + {\bf P} = \varepsilon_0 \chi_e {\bf E}, \hspace{1cm} + {\bf M} = \chi_m {\bf H} + \label{Gr(7.56)} + \] +so +\[ + {\bf D} = \varepsilon {\bf E}, \hspace{1cm} + {\bf H} = \frac{1}{\mu} {\bf B}, + \label{Gr(7.57)} + \] +where \(\varepsilon \equiv \varepsilon_0(1 + \chi_e)\) and \(\mu \equiv \mu_0 (1 + \chi_m)\). +
+ +Created: 2022-02-07 Mon 08:02
+ ++Discontinuities between different media, deduced from +
++{\bf Maxwell's equations {\it (in matter)}, integral form} +
+\begin{align} + (i)~~ &\oint_{\cal S} {\bf D} \cdot d{\bf a} = Q_{f_{enc}}, \nonumber \\ + (ii)~~ &\oint_{\cal S} {\bf B} \cdot d{\bf a} = 0 \nonumber \\ + (iii)~~ &\oint_{\cal P} {\bf E} \cdot d{\bf l} = -\frac{d}{dt} \int_{\cal S} {\bf B} \cdot d{\bf a}, \nonumber \\ + (iv)~~ &\oint_{\cal P} {\bf H} \cdot d{\bf l} = I_{f_{enc}} + \frac{d}{dt} \int_{\cal S} {\bf D} \cdot d{\bf a}. +\end{align} + ++Applying \((i)\) to wafer-thin Gaussian pillbox straddling boundary between 2 materials: +\({\bf D}_1 \cdot {\bf a} - {\bf D}_2 \cdot {\bf a} = \sigma_f a\) so +\[ +\boxed{ +D^{\perp}_1 - D^{\perp}_2 = \sigma_f +} +\label{Gr(7.59)} +\] +Same reasoning applied to \((ii)\) gives +\[ +\boxed{ +B^{\perp}_1 - B^{\perp}_2 = 0 +} +\label{Gr(7.60)} +\] +For \((iii)\): Amperian loop straddling surface: \({\bf E}_1 \cdot {\bf l} - {\bf E}_2 \cdot {\bf l} = +-\frac{d}{dt} \int_{\cal S} {\bf B} \cdot d{\bf a}\). Limit of small loop: flux vanishes, therefore +\[ +\boxed{ +{\bf E}_1^{\parallel} - {\bf E}_2^{\parallel} = 0 +} +\label{Gr(7.61)} +\] +Similarly, \((iv)\) implies \({\bf H}_1 \cdot {\bf l} - {\bf H}_2 \cdot {\bf l} = I_{f_{enc}}\). +No volume current can contribute, but a surface current can. Can write +\(I_{f_{enc}} = {\bf K}_f \cdot (\hat{\bf n} \times {\bf l}) = ({\bf K}_f \times \hat{\bf n}) \cdot {\bf l}\) +and thus +\[ +\boxed{ +{\bf H}_1^{\parallel} - {\bf H}_2^{\parallel} = {\bf K}_f \times \hat{\bf n} +} +\label{Gr(7.62)} +\] +These are the general boundary conditions for electrodynamics. +
+ ++In case of linear media: can be expressed in terms of \({\bf E}\) and \({\bf B}\) alone: +
+\begin{align} +(i)~~ &\varepsilon_1 E_1^{\perp} - \varepsilon_2 E_2^{\perp} = \sigma_f, \nonumber \\ +(ii)~~ &B_1^{\perp} - B_2^{\perp} = 0, \nonumber \\ +(iii)~~ &{\bf E}_1^{\parallel} - {\bf E}_2^{\parallel} = 0, \nonumber \\ +(iv)~~ &\frac{1}{\mu_1} {\bf B}_1^{\parallel} - \frac{1}{\mu_2} {\bf B}_2^{\parallel} = {\bf K}_f \times \hat{\bf n}. +\label{Gr(7.63)} +\end{align} ++If there is no free charge and no free current at boundary: +
+\begin{align} +(i)~~ &\varepsilon_1 E_1^{\perp} - \varepsilon_2 E_2^{\perp} = 0, \nonumber \\ +(ii)~~ &B_1^{\perp} - B_2^{\perp} = 0, \nonumber \\ +(iii)~~ &{\bf E}_1^{\parallel} - {\bf E}_2^{\parallel} = 0, \nonumber \\ +(iv)~~ &\frac{1}{\mu_1} {\bf B}_1^{\parallel} - \frac{1}{\mu_2} {\bf B}_2^{\parallel} = 0. +\label{Gr(7.64)} +\end{align} ++These are basis of theory of reflection and refraction. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++We now consider EM waves inside a bulk conductor and will consider nonvanishing free charge \(\rho_f\) and current \({\boldsymbol J}_f\) densities. By Ohm's law, we have +\[ + {\boldsymbol J}_f = \sigma {\boldsymbol E} +\] +which means that the Maxwell equations reduce to +
+\begin{align} + {\boldsymbol \nabla} \cdot {\boldsymbol E} &= \frac{\rho_f}{\varepsilon}, + \hspace{10mm} & + {\boldsymbol \nabla} \cdot {\boldsymbol B} &= 0, \nonumber\\ + {\boldsymbol \nabla} \times {\boldsymbol E} &= -\frac{\partial {\boldsymbol B}}{\partial t} & + {\boldsymbol \nabla} \times {\boldsymbol B} &= \mu \sigma {\boldsymbol E} + \mu \varepsilon \frac{\partial {\boldsymbol E}}{\partial t}. +\end{align} ++Putting together the continuity equation for free charge +\[ + \frac{\partial \rho_f}{\partial t} + {\boldsymbol \nabla} \cdot {\boldsymbol J}_f = 0 +\] +with Ohm's law and Gauss's law yields +\[ + \frac{\partial \rho_f}{\partial t} = -\sigma ({\boldsymbol \nabla} \cdot {\boldsymbol E}) = -\frac{\sigma}{\varepsilon} \rho_f + \longrightarrow + \rho_f (t) = \rho_f (0) e^{-\frac{\sigma}{\varepsilon} t} +\] +so any initial free charge dissipates with characteristic time \(\tau \equiv \frac{\varepsilon}{\sigma}\). +
+ ++After the free charge has dissipated, we have +
+\begin{align} + {\boldsymbol \nabla} \cdot {\boldsymbol E} &= 0, + \hspace{10mm} & + {\boldsymbol \nabla} \cdot {\boldsymbol B} &= 0, \nonumber\\ + {\boldsymbol \nabla} \times {\boldsymbol E} &= -\frac{\partial {\boldsymbol B}}{\partial t} & + {\boldsymbol \nabla} \times {\boldsymbol B} &= \mu \sigma {\boldsymbol E} + \mu \varepsilon \frac{\partial {\boldsymbol E}}{\partial t} +\end{align} ++Applying ${\boldsymbol ∇} × $ to the curl equations gives the modified wave equations +\[ + {\boldsymbol \nabla}^2 {\boldsymbol E} = \mu \varepsilon \frac{\partial^2 {\boldsymbol E}}{\partial t^2} + \mu \sigma \frac{\partial {\boldsymbol E}}{\partial t}, \hspace{10mm} + {\boldsymbol \nabla}^2 {\boldsymbol B} = \mu \varepsilon \frac{\partial^2 {\boldsymbol B}}{\partial t^2} + \mu \sigma \frac{\partial {\boldsymbol B}}{\partial t} +\] +with plane-wave solutions +\[ + {\boldsymbol E} ({\boldsymbol r}, t) = {\boldsymbol E}_0 e^{i \tilde{\boldsymbol k} \cdot {\boldsymbol r} - \omega t} +\] +with +\[ + \tilde{k}^2 \equiv \tilde{\boldsymbol k} \cdot \tilde{\boldsymbol k} = \mu \varepsilon \omega^2 + i\mu \sigma \omega. +\] +We can thus write +\[ + \tilde{k} \equiv k + i\kappa, \hspace{5mm} + k = k_+, \hspace{3mm}\kappa = k_-, \hspace{5mm} + k_{\pm} \equiv \omega \sqrt{\frac{\mu\varepsilon}{2}} \left[\sqrt{1 + \left(\frac{\sigma}{\varepsilon \omega}\right)^2} \pm 1 \right]^{1/2}. +\] +The wave can thus be written (letting \(\hat{\boldsymbol k}\) represent the direction of propagation and using the notations \({\boldsymbol k} \equiv k \hat{\boldsymbol k}\) and \({\boldsymbol \kappa} \equiv \kappa \hat{\boldsymbol k}\)) +\[ + {\boldsymbol E} ({\boldsymbol r}, t) = {\boldsymbol E}_0 e^{-{\boldsymbol \kappa} \cdot {\boldsymbol r}} e^{i ({\boldsymbol k} \cdot {\boldsymbol r} - \omega t)} +\] +with a similar solution for \({\boldsymbol B}\). +The quantity \(d = \frac{1}{\kappa}\) is known as the {\bf skin depth}. +
+ ++For simplicity, we will put the propagation direction along \(z\) +and the polarization along \(x\): +\[ + {\boldsymbol E} (z,t) = E_0 e^{-\kappa z} e^{i(kz - \omega t)} \hat{\boldsymbol x}. +\] +From Maxwell (iii) we get +\[ + {\boldsymbol B} (z,t) = \frac{\tilde{k}}{\omega} E_0 e^{-\kappa z} e^{i(kz - \omega t)} \hat{\boldsymbol y}. +\] +
+ ++To simplify things further, we write the complex wavevector in polar form: +\[ + \tilde{k} = K e^{i\phi}, \hspace{5mm} + K = \sqrt{k^2 + \kappa^2} = \omega \sqrt{\mu\varepsilon \sqrt{1 + \left( \frac{\sigma}{\varepsilon \omega}\right)^2}}, + \hspace{5mm} + \phi = \mbox{atan}~\frac{\kappa}{k}. +\] +The electric and magnetic fields thus develop a relative phase: +if we write \(E_0 = E e^{i \delta_E}\) and \(B_0 = B e^{i\delta_B}\) then +\[ + \delta_B - \delta_E = \phi +\] +so the magnetic field lags behind the electric field. The real amplitudes are related by +\[ + \frac{B}{E} = \frac{K}{\omega} = \sqrt{\mu\varepsilon \sqrt{1 + \left( \frac{\sigma}{\varepsilon \omega}\right)^2}} +\] +so the final form of the fields is +\[ + {\boldsymbol E} (z,t) = E e^{-\kappa z} \cos (kz - \omega t + \delta_E) \hat{\boldsymbol x}, \hspace{5mm} + {\boldsymbol B} (z,t) = B e^{-\kappa z} \cos (kz - \omega t + \delta_E + \phi) \hat{\boldsymbol y}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++In matter regions without free charge and free current: Maxwell's equations are +
+\begin{align} +(i)~~ &{\boldsymbol \nabla} \cdot {\bf D} = 0, \nonumber \\ +(ii)~~ &{\boldsymbol \nabla} \cdot {\bf B} = 0, \nonumber \\ +(iii)~~ &{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \nonumber \\ +(iv)~~ &{\boldsymbol \nabla} \times {\bf H} = \frac{\partial {\bf D}}{\partial t}. +\label{Gr(9.65)} +\end{align} ++For linear medium: +\[ +{\boldsymbol D} = \varepsilon {\boldsymbol E}, \hspace{1cm} +{\boldsymbol H} = \frac{1}{\mu} {\boldsymbol B}. +\label{Gr(9.66)} +\] +If the medium is homogeneous (no spatial dependence of \(\varepsilon\) or \(\mu\)), +
+\begin{align} +(i)~~ &{\boldsymbol \nabla} \cdot {\bf E} = 0, \nonumber \\ +(ii)~~ &{\boldsymbol \nabla} \cdot {\bf B} = 0, \nonumber \\ +(iii)~~ &{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \nonumber \\ +(iv)~~ &{\boldsymbol \nabla} \times {\bf B} = \mu \varepsilon \frac{\partial {\bf E}}{\partial t}. +\label{Gr(9.65)} +\end{align} ++These are the same equations as in vacuum, except for the substitution of \(\mu_0 \varepsilon_0\) by \(\mu \varepsilon\). +
+ ++Speed of propagation: +\[ +v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{n} +\label{Gr(9.68)} +\] +where the index of refraction of the material is defined as +
++{\bf Index of refraction} +\[ + n \equiv \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}} + \label{Gr(9.69)} + \] +
+ ++Fact: for most materials, \(\mu\) is very close to \(\mu_0\), so +\[ +n \simeq \sqrt{\varepsilon_r} +\label{Gr(9.70)} +\] +with \(\varepsilon_r\) being the dielectric constant \ref{Gr(4.34)}. +
+ ++Energy density: +\[ +u = \frac{1}{2} \left( \varepsilon E^2 + \frac{1}{\mu} B^2 \right) +\label{Gr(9.71)} +\] +Poynting vector: +\[ +{\boldsymbol S} = \frac{1}{\mu} {\boldsymbol E} \times {\boldsymbol B} +\label{Gr(9.72)} +\] +Wave intensity +\[ +I = \frac{1}{2} \varepsilon v E_0^2 +\label{Gr(9.73)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Interesting question: what happens to an EM wave as it passes from one medium to another? +Incident wave: produces reflected and transmitted waves. +Detailed study: starts from boundary conditions \ref{Gr(7.64)}, +
+\begin{align} +(i)~~ &\varepsilon_1 E_1^{\perp} - \varepsilon_2 E_2^{\perp} = 0, \nonumber \\ +(ii)~~ &B_1^{\perp} - B_2^{\perp} = 0, \nonumber \\ +(iii)~~ &{\bf E}_1^{\parallel} - {\bf E}_2^{\parallel} = 0, \nonumber \\ +(iv)~~ &\frac{1}{\mu_1} {\bf B}_1^{\parallel} - \frac{1}{\mu_2} {\bf B}_2^{\parallel} = 0. +\label{eq:EMBdryCondAtMediumInterface} +\end{align} +Created: 2022-02-07 Mon 08:02
+ ++Boundary between two media: \(x-y\) plane. Plane wave of frequency \(\omega\) travelling in \(z\) direction and polarized +in \(x\) direction approaches surface: +\[ +{\boldsymbol E}_I (z,t) = E_{0_I} e^{i (k_1 z - \omega t)} \hat{\boldsymbol x}, \hspace{1cm} +{\boldsymbol B}_I (z,t) = \frac{1}{v_1} E_{0_I} e^{i (k_1 z - \omega t)} \hat{\boldsymbol y} +\label{Gr(9.75)} +\] +(in which we have used \(B_0 = \frac{1}{v} E_0\), \ref{Gr(9.47)}). +
+ ++Reflected wave: +\[ +{\boldsymbol E}_R (z,t) = E_{0_R} e^{i (-k_1 z - \omega t)} \hat{\boldsymbol x}, \hspace{1cm} +{\boldsymbol B}_R (z,t) = \frac{-1}{v_1} E_{0_R} e^{i (-k_1 z - \omega t)} \hat{\boldsymbol y} +\label{Gr(9.76)} +\] +(minus sign is conventional here, but handy for inverting Poynting vector direction). +
+ ++For the transmitted wave, we put +\[ +{\boldsymbol E}_T (z,t) = E_{0_T} e^{i (k_2 z - \omega t)} \hat{\boldsymbol x}, \hspace{1cm} +{\boldsymbol B}_T (z,t) = \frac{1}{v_2} E_{0_T} e^{i (k_2 z - \omega t)} \hat{\boldsymbol y}. +\label{Gr(9.75)} +\] +
+ ++Our boundary is by choice of coordinate system at \(z = 0\). Our setup calls for solving the boundary conditions \ref{eq:EMBdryCondAtMediumInterface} with \({\boldsymbol E}_I + {\boldsymbol E}_R\) and \({\boldsymbol B}_I + {\boldsymbol B}_R\) on one side, and \({\boldsymbol E}_T\) and \({\boldsymbol B}_T\) on the other. +
+ ++At normal incidence, there are no perpendicular components of the fields relative to the surface, so \ref{eq:EMBdryCondAtMediumInterface} (i) and (ii) are obeyed. (iii) means that +\[ +E_{0_I} + E_{0_R} = E_{0_T} +\label{Gr(9.78)} +\] +while (iv) means that +\[ +\frac{1}{\mu_1} \frac{1}{v_1} (E_{0_I} - E_{0_R}) = \frac{1}{\mu_2} \frac{1}{v_2} E_{0_T}, +\label{Gr(9.79)} +\] +which can be rewritten as +\[ +E_{0_I} - E_{0_R} = \beta E_{0_T}, \hspace{1cm} +\beta \equiv \frac{\mu_1 v_1}{\mu_2 v_2} = \frac{\mu_1 n_2}{\mu_2 n_1}. +\label{Gr(9.80)} +\] +
+ ++Solving these coupled equations, we can write +outgoing amplitudes in terms of incident ones: +\[ +E_{0_R} = \frac{1 - \beta}{1 + \beta} E_{0_I}, \hspace{10mm} +E_{0_T} = \frac{2}{1 + \beta} E_{0_I}. +\label{Gr(9.82)} +\] +
+ ++The intensity of reflected and transmitted waves is the power per unit area transported by the wave. In vacuum, this was +\[ +I \equiv \langle S \rangle = \frac{1}{2} c \varepsilon_0 E_0^2. +\label{Gr(9.63)} +\] +Here, this becomes +\[ +I = \frac{1}{2} v \varepsilon E_0^2. +\label{Gr(9.73)} +\] +The fraction of incident intensity in the reflected intensity is +\[ +R \equiv \frac{I_R}{I_I} = \frac{E^2_{0_R}}{E^2_{0_I}} = \left( \frac{1 - \beta}{1 + \beta} \right)^2 +\label{Gr(9.86)} +\] +while the transmitted intensity is +\[ +T \equiv \frac{I_T}{I_I} = \frac{v_2 \varepsilon_2}{v_1 \varepsilon_1} \frac{E^2_{0_T}}{E^2_{0_I}} += \sqrt{\frac{\mu_1 \varepsilon_2}{\mu_2 \varepsilon_1}} \frac{E^2_{0_T}}{E^2_{0_I}} = \frac{4 \beta}{(1 + \beta)^2}. +\label{Gr(9.86)} +\] +We thus have that the {\bf reflection} and {\bf transmission coefficients} satisfy +\[ +R + T = 1. +\label{Gr(9.88)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++We now consider an interface between two media. Media \(1\) (for \(z < 0\)) has light velocity \(v_1\) while \(2\) (for \(z > 0\)) has \(v_2\). +
+ ++Incident wave: +\[ +{\boldsymbol E}_I ({\boldsymbol r},t) = {\boldsymbol E}_{0_I} e^{i ({\boldsymbol k}_I \cdot {\boldsymbol r} - \omega t)}, \hspace{1cm} +{\boldsymbol B}_I ({\boldsymbol r},t) = \frac{1}{v_1} \hat{\boldsymbol k}_I \times {\boldsymbol E}_{I} ({\boldsymbol r}, t). +\] +Reflected wave: +\[ +{\boldsymbol E}_R ({\boldsymbol r},t) = {\boldsymbol E}_{0_R} e^{i ({\boldsymbol k}_R \cdot {\boldsymbol r} - \omega t)}, \hspace{1cm} +{\boldsymbol B}_R ({\boldsymbol r},t) = \frac{1}{v_1} \hat{\boldsymbol k}_R \times {\boldsymbol E}_{R} ({\boldsymbol r}, t). +\] +Transmitted wave: +\[ +{\boldsymbol E}_T ({\boldsymbol r},t) = {\boldsymbol E}_{0_T} e^{i ({\boldsymbol k}_T \cdot {\boldsymbol r} - \omega t)}, \hspace{1cm} +{\boldsymbol B}_T ({\boldsymbol r},t) = \frac{1}{v_2} \hat{\boldsymbol k}_T \times {\boldsymbol E}_{T} ({\boldsymbol r}, t). +\] +All waves have the same frequency \(\omega\). Since \(\omega = k v\), the three wavevectors are related by +\[ + k_I v_1 = k_R v_1 = k_T v_2 ~~\longrightarrow~~ k_I = k_R = \frac{v_2}{v_1} k_T = \frac{n_1}{n_2} k_T + \label{eq:RTObliquek} +\] +
+ ++These forms for incident, reflected and transmitted wave can be substituted in the boundary conditions (\ref{eq:EMBdryCondAtMediumInterface}). Since these must be valid for any \(x\) and \(y\) (on the interface at \(z=0\)), we must have that the \(x\) and \(y\) components of the wavevectors coincide for all the waves: +\[ + k_{I_x} = k_{R_x} = k_{T_x}, \hspace{10mm} + k_{I_y} = k_{R_y} = k_{T_y} +\] +
+ ++From now on we will orient the axes so that \({\boldsymbol k}_I\) lies in the \(xz\) plane. This means that \({\boldsymbol k}_R\) and \({\boldsymbol k}_T\) also lie in that plane. This is the +
++{\bf First law of reflection:} +the incident, reflected and transmitted wave vectors form a plane (called the plane of incidence) which also includes the normal to the surface. +
+ ++Specializing (\ref{eq:RTObliquek}) to our notations, we have +\[ + k_I \sin \theta_I = k_R \sin \theta_R = k_T \sin \theta_T +\] +with the incidence (\(\theta_I\)) and reflection (\(\theta_R\)) angles +and the angle of refraction (\(\theta_T\)) obey the following laws: +
++{\bf Law of reflection} +\[ + \theta_I = \theta_R + \] +{\bf Law of refraction (Snell's law)} +\[ + n_1 \sin \theta_I = n_2 \sin \theta_T + \] +
+ ++This takes care of the spatially-dependent exponential factors in the boundary conditions. The coefficients must further obey +
+\begin{align} + \varepsilon_1 \left({\boldsymbol E}_{0_I} + {\boldsymbol E}_{0_R} \right)_z &= \varepsilon_2 \left({\boldsymbol E}_{0_T} \right)_z, + & \hspace{10mm} + \left({\boldsymbol B}_{0_I} +{\boldsymbol B}_{0_R} \right)_z &= \left({\boldsymbol B}_{0_T}\right)_z, \nonumber \\ + \left( {\boldsymbol E}_{0_I} + {\boldsymbol E}_{0_R} \right)_{x,y} &= \left({\boldsymbol E}_{0_T}\right)_{x,y}, + & \frac{1}{\mu_1} \left({\boldsymbol B}_{0_I} + {\boldsymbol B}_{0_R} \right)_{x,y} &= \frac{1}{\mu_2} \left({\boldsymbol B}_{0_T}\right)_{x,y}. + \label{eq:EMBdryCondAtMediumInterface:amp} +\end{align} + ++Further treatment depends on the polarization of the incoming wave. +The two cases of polarization parallel and perpendicular to the plane of incidence must be treated separately. We thus divide our incident electric field as follows: +\[ + {\boldsymbol E}_{0_I} = {\boldsymbol E}_{0_I}^{\parallel} + {\boldsymbol E}_{0_I}^{\perp}. +\] +
+ ++\paragraph{Polarization in plane of incidence:} +in this case the first equation of (\ref{eq:EMBdryCondAtMediumInterface:amp}) gives +\[ + \varepsilon_1 \left(-E_{0_I} \sin \theta_I + E_{0_R} \sin \theta_R \right) = -\varepsilon_2 E_{0_T} \sin \theta_T. +\] +The second equation is a trivial \(0=0\). The third is +\[ + E_{0_I} \cos \theta_I + E_{0_R} \cos \theta_R = E_{0_T} \cos \theta_T +\] +while the fourth gives +\[ + \frac{1}{\mu_1 v_1} \left( E_{0_I} - E_{0_R} \right) = \frac{1}{\mu_2 v_2} E_{0_T} +\] +Given the laws of reflection and refraction, the first and fourth equations are the same and reduce to +\[ + E_{0_I} - E_{0_R} = \beta E_{0_T}, +\] +while the third equation becomes +\[ + E_{0_I} + E_{0_R} = \alpha E_{0_T}, \hspace{10mm} \alpha \equiv \frac{\cos \theta_T}{\cos \theta_I} +\] +Writing everything in terms of the incident amplitude, we get +
++{\bf Fresnel's equations for reflection and transmission amplitudes (parallel case)} +\[ + E_{0_R} = \frac{\alpha - \beta}{\alpha + \beta} E_{0_I}, + \hspace{10mm} + E_{0_T} = \frac{2}{\alpha + \beta} E_{0_I} + \] +
+ ++Amplitudes for transmitted and reflected wave: depend on angle of incidence: +\[ + \alpha = \frac{\sqrt{1 - \sin^2 \theta_T}}{\cos \theta_I} = \frac{\left[1 - \left(\frac{n_1}{n_2}\right)^2 \sin^2 \theta_I\right]^{1/2}}{\cos \theta_I} +\] +Behaviour: for \(\theta_I = 0\) we recover (\ref{Gr(9.82)}). +For grazing waves \(\theta_I \rightarrow \pi/2\) we have that \(\alpha \rightarrow \infty\) and the wave is totally reflected. The most interesting angle is the one at which \(\alpha = \beta\) and the reflected wave has zero amplitude. This is known as +
++{\bf Brewster's angle {\it (at which the reflected wave amplitude vanishes)}} + \[ + \theta_B = \arcsin \left[ \frac{1 - \beta^2}{(n_1/n_2)^2 - \beta^2} \right]^{1/2} + \] +
+ ++Power per unit area striking the interface: \({\boldsymbol S} \cdot \hat {\boldsymbol n}\) and thus +\[ + I_I = \frac{\varepsilon_1 v_1}{2} E_{0_I}^2 \cos \theta_I, + \hspace{5mm} + I_R = \frac{\varepsilon_1 v_1}{2} E_{0_R}^2 \cos \theta_R, + \hspace{5mm} + I_T = \frac{\varepsilon_2 v_2}{2} E_{0_T}^2 \cos \theta_T. +\] +
+ ++Reflection and transmission coefficients: +\[ + R \equiv \frac{I_R}{I_I} = \frac{E_{0_R}^2}{E_{0_I}^2} = \left( \frac{\alpha - \beta}{\alpha + \beta} \right)^2, \hspace{5mm} + T \equiv \frac{I_R}{I_I} = \frac{\varepsilon_2 v_2}{\varepsilon_1 v_1} \frac{E_{0_T}^2 \cos \theta_T}{E_{0_I}^2 \cos \theta_I} = \alpha \beta \left( \frac{2}{\alpha + \beta} \right)^2. +\] +Of course, we get \(R + T = 1\) as expected. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++We now consider a cylindrical geometry, formed by two concentric conducting cylinders. We let \(a\) (resp. \(b\)) represent the radius of the inner (resp. outer) cylinder. +
+ ++In this case, we can find nontrivial solutions to Maxwell's equations for \(E_z = 0 = B_z\). Simple inspection gives \(k = \frac{\omega}{c}\) and +
+\begin{align} + c B_y &= E_x, &\hspace{10mm} + \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= 0, &\hspace{10mm} + \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= 0, \nonumber \\ + c B_x &= -E_y, & + \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} &= 0, & + \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} &= 0 +\end{align} ++The derivative equations are precisely the electrostatic and magnetic equations for empty space with cylindrical symmetry. The solution is thus that of an infinite line charge and an infinite straight current: +\[ + {\boldsymbol E}_0 (s, \phi) = \frac{A}{s} \hat{\boldsymbol s}, \hspace{10mm} + {\boldsymbol B}_0 (s, \phi) = \frac{A}{cs} \hat{\boldsymbol \phi} +\] +where \(A\) is a constant amplitude. Substituting and taking the real part, +\[ + {\boldsymbol E} (s, \phi, z, t) = \frac{A}{s} \cos (kz - \omega t) \hat{\boldsymbol s}, \hspace{10mm} + {\boldsymbol B} (s, \phi, z, t) = \frac{A}{cs} \cos (kz - \omega t) \hat{\boldsymbol \phi}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++A waveguide is a hollow structure in which EM waves can propagate. For simplicity here, we will assume that the boundaries of the waveguide are perfect conductors. Since the electric and magnetic fields vanish inside the walls of the waveguide, the boundary conditions at the inner walls are +\[ + {\boldsymbol E}^{\parallel} = 0, \hspace{10mm} B^\perp = 0. +\] +Free charges and currents will circulate on the surfaces in order to satisfy these constraints. +
+ ++We are interested in monochromatic waves propagating down the waveguide and will thus take the generic ansatz +\[ + {\boldsymbol E} (x, y, z, t) = {\boldsymbol E}_0 (x,y) e^{i(kz - \omega t)}, \hspace{10mm} + {\boldsymbol B} (x, y, z, t) = {\boldsymbol B}_0 (x,y) e^{i(kz - \omega t)}. +\] +These fields must obey Maxwell's equations. Unlike waves in vacuum, it will turn out that EM waves in waveguides are not purely transverse. We will thus start with a general ansatz for the fields: +\[ + {\boldsymbol E}_0 = \sum_i E_i \hat{\boldsymbol i}, \hspace{10mm} + {\boldsymbol B}_0 = \sum_i B_i \hat{\boldsymbol i}, \hspace{10mm} + i = x, y, z. +\] +Maxwell's equations (iii) and (iv) then give +
+\begin{align} + \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= i\omega B_z, + &\hspace{10mm} + \frac{\partial B_y}{\partial x} - \frac{\partial B_x}{\partial y} &= -i \frac{\omega}{c^2} E_z, \nonumber \\ + \frac{\partial E_z}{\partial y} - ik E_y &= i\omega B_x, + &\hspace{10mm} + \frac{\partial B_z}{\partial y} - ik B_y &= -i \frac{\omega}{c^2} E_x, \nonumber \\ + ik E_x - \frac{\partial E_z}{\partial x} &= i\omega B_y, + &\hspace{10mm} + ik B_x - \frac{\partial B_z}{\partial x} &= -i \frac{\omega}{c^2} E_y. +\end{align} ++The transverse components can be isolated by simple algebra: +
+\begin{align} + E_x &= \frac{i}{(\omega/c)^2 - k^2} \left( k \frac{\partial E_z}{\partial x} + \omega \frac{\partial B_z}{\partial y} \right), \nonumber \\ + E_y &= \frac{i}{(\omega/c)^2 - k^2} \left( k \frac{\partial E_z}{\partial y} - \omega \frac{\partial B_z}{\partial x} \right), \nonumber \\ + B_x &= \frac{i}{(\omega/c)^2 - k^2} \left( k \frac{\partial B_z}{\partial x} - \frac{\omega}{c^2} \frac{\partial E_z}{\partial y} \right), \nonumber \\ + B_y &= \frac{i}{(\omega/c)^2 - k^2} \left( k \frac{\partial B_z}{\partial y} + \frac{\omega}{c^2} \frac{\partial E_z}{\partial x} \right). +\end{align} ++Putting these back into Maxwell (i) and (ii) gives the decoupled equations +\[ + \left[ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \left(\frac{\omega}{c}\right)^2 - k^2 \right] E_z = 0, +\] +with an identical equation for \(B_z\). If \(E_z = 0\) the waves are called {\bf TE} waves (for {\it transverse electric}), and if \(B_z = 0\) they are called {\bf TM} (for {\it transverse magnetic}) waves. If \(E_z = 0 = B_z\) they are called {\bf TEM waves}. The latter cannot occur in a hollow waveguide (simple proof: Gauss + Faraday). +
+Created: 2022-02-07 Mon 08:02
+ ++Assume a waveguide with rectangular cross-section of width \(a\) along \(\hat{\boldsymbol x}\) and \(b\) along \(\hat{\boldsymbol y}\). Without loss of generality (and to follow common historical convention) we will assume \(a \geq b\). +
+ ++In view of this geometry, we will use separation of variables in Cartesian coordinates. We thus put +\[ + B_z (x,y) = X(x) Y(y) +\] +so +\[ + Y \frac{d^2 X}{dx^2} + X \frac{d^2 Y}{dy^2} + \left[(\omega/c)^2 - k^2 \right] XY = 0. +\] +Invoking separation of variables, we obtain +\[ + \frac{d^2 X}{dx^2} = -k_x^2 X, \hspace{5mm} + \frac{d^2 Y}{dy^2} = -k_y^2 Y, \hspace{5mm} + k_x^2 + k_y^2 + k^2 = \frac{\omega^2}{c^2}. +\] +The general solution for \(X\) is +\[ + X(x) = A \sin (k_x x) + B \cos (k_x x) +\] +and to fit the boundary conditions we must have that \(B_x = 0\) so +\(\partial B_x/\partial x\) must vanish at \(x=0, a\), +so \(dX/dx\) must also vanish at \(x = 0, a\) and thus \(A = 0\) and +\[ + k_x = m \pi/a, \hspace{5mm} m = 0, 1, \cdots +\] +The same reasoning applies to \(Y\) so we get +\[ + B_z (x, y) = B_0 \cos \frac{m\pi x}{a} \cos \frac{n \pi y}{b} +\] +which is called the TE\(_{mn}\) mode. Its wavenumber is +\[ + k = \sqrt{\left(\frac{\omega}{c}\right)^2 - \pi^2 \left[\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 \right]}. +\] +In view of this, if the frequency is low enough, namely +\[ + \omega < c\pi \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2} \equiv \omega_{mn} +\] +then the wavenumber is imaginary and the travelling wave is exponentially attenuated. This frequency is called the {\bf cutoff frequency} for this mode. +
+ ++In terms of the cutoff frequency, we have wavevector +\[ + k = \frac{1}{c} \sqrt{\omega^2 - \omega_{mn}^2} +\] +and wave velocity +\[ + v = \frac{\omega}{k} = \frac{c}{\sqrt{1 - \left(\frac{\omega_{mn}}{\omega}\right)^2}} +\] +which is {\it greater} than \(c\). The energy of the wave however propagates at the {\bf group velocity} +\[ + v_g = \frac{d\omega}{dk} = c \sqrt{1 - \left(\frac{\omega_{mn}}{\omega}\right)^2}. +\] +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++When rewriting our fields into potentials, we passed from six unknowns (the three components of the electric and magnetic fields) to four (the scalar potential, and the three components of the vector potential). +
+ ++There is however a certain ambiguity left when representing fields in terms of potentials. Namely, it is easy to verify that new fields +\[ + V^\prime = V - \frac{\partial \lambda}{\partial t}, \hspace{10mm} + {\boldsymbol A}^\prime = {\boldsymbol A} + {\boldsymbol \nabla} \lambda +\] +give the same \({\boldsymbol E}\) and \({\boldsymbol B}\) fields as \(V, {\boldsymbol A}\) for any scalar function \(\lambda ({\boldsymbol r}, t)\). +
+ ++This very interesting fact that electrodynamics is invariant under +the gauge choice is called gauge freedom. Going from one choice to another is done by +implementing a gauge transformation. +
+Created: 2022-02-07 Mon 08:02
+ ++The {\bf Coulomb Gauge} is specified by taking +\[ + {\boldsymbol \nabla} \cdot {\boldsymbol A} = 0 +\] +in which case (\ref{eq:LaplacianV}) becomes simply +\[ + {\boldsymbol \nabla}^2 V = -\frac{\rho}{\varepsilon_0} +\] +{\it i.e.} Poisson's equation, whose solution we already know: +\[ + V({\boldsymbol r}, t) = \frac{1}{4\pi \varepsilon_0} \int d\tau' \frac{\rho({\boldsymbol r}^\prime, t)}{|{\boldsymbol r} - {\boldsymbol r}^\prime|} +\] +Note that this is an equal-time relationship (it does not mean instantaneous action at a distance, since \(V\) by itself is not physically measurable). +
+ ++Although Gauss's law looks nice in the Coulomb gauge, Amp{\`e}re-Maxwell does not: +\[ + {\boldsymbol \nabla}^2 {\boldsymbol A} - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2} = -\mu_0 {\boldsymbol J} + \mu_0 \varepsilon_0 {\boldsymbol \nabla} \left( \frac{\partial V}{\partial t} \right). +\] +
+Created: 2022-02-07 Mon 08:02
+ ++A more aesthetic choice is the {\bf Lorenz gauge}: +\[ + {\boldsymbol \nabla} \cdot {\boldsymbol A} + \mu_0 \varepsilon_0 \frac{\partial V}{\partial t} = 0 + \label{eq:LorenzGauge} +\] +which is chosen to put the second term in the left-hand side of (\ref{eq:LaplacianA}) to zero. What remains is then +\[ + {\boldsymbol \nabla}^2 {\boldsymbol A} - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2} = -\mu_0 {\boldsymbol J} +\] +while the equation for \(V\) becomes +\[ + {\boldsymbol \nabla}^2 V - \mu_0 \varepsilon_0 \frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\varepsilon_0}. +\] +These can be written compactly upon introducing a new operator: the +
++{\bf d'Alembertian operator} +\[ + \square^2 \equiv {\boldsymbol \nabla}^2 - \mu_0 \varepsilon_0 \frac{\partial^2}{\partial t^2} + \label{eq:dAlembertian} + \] +
+ ++so we get the +
++{\bf Inhomogeneous Maxwell equations (Lorenz gauge)} +\[ + \square^2 V = -\frac{\rho}{\varepsilon_0}, \hspace{10mm} + \square^2 {\boldsymbol A} = -\mu_0 {\boldsymbol J} + \label{eq:InhomogeneousMaxwellLorenzGauge} + \] +
+ ++This gauge is especially nice in the context of special relativity. +The whole of electrodynamics has thus reduced to solving the inhomogeneous +wave equations (\ref{eq:InhomogeneousMaxwellLorenzGauge}) +in terms of specified sources. +
+ + ++Without choosing the Lorenz gauge, we can still write the inhomogeneous +Maxwell equations in a simpler form. Defining +\[ + L \equiv {\boldsymbol \nabla} \cdot {\boldsymbol A} + \mu_0 \varepsilon_0 \frac{\partial V}{\partial t}, +\] +we have by direct inspection +\[ + \square^2 V + \frac{\partial L}{\partial t} = -\frac{\rho}{\varepsilon_0}, \hspace{10mm} + \square^2 {\boldsymbol A} - {\boldsymbol \nabla} L = -\mu_0 {\boldsymbol J}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Solving these for generale time-dependent sources \(\rho({\boldsymbol r}, t)\) and \({\boldsymbol J} ({\boldsymbol r}, t)\) is not an easy task. +
+ ++Useful strategy: represent fields in terms of potentials. +
+ ++Easiest: +
++\[ + {\boldsymbol B} = {\boldsymbol \nabla} \times {\boldsymbol A} + \] +
+ ++Putting this into Faraday's law gives +\[ + {\boldsymbol \nabla} \times \left({\boldsymbol E} + \frac{\partial {\boldsymbol A}}{\partial t} \right) = 0 +\] +so this can be written as the gradient of a scalar (by choice: \(-{\boldsymbol \nabla} V\)) so we get +
++\[ + {\boldsymbol E} = -{\boldsymbol \nabla} V - \frac{\partial {\boldsymbol A}}{\partial t} + \label{eq:E_from_Potentials} + \] +
+ ++Using this potential representation for \({\boldsymbol E}\) and \({\boldsymbol B}\) automatically fulfills the two homogeneous Maxwell equations. For the inhomogeneous equations, substituting (\ref{eq:E_from_Potentials}) into Gauss's law gives +
++\[ + {\boldsymbol \nabla}^2 V + \frac{\partial}{\partial t} {\boldsymbol \nabla} \cdot {\boldsymbol A} = -\frac{\rho}{\varepsilon_0} + \label{eq:LaplacianV} + \] +
+ ++whereas Amp{\`ere}-Maxwell becomes +\[ + {\boldsymbol \nabla} \times \left({\boldsymbol \nabla} {\boldsymbol A}\right) = \mu_0 {\boldsymbol J} - \mu_0 \varepsilon_0 {\boldsymbol \nabla} \left(\frac{\partial V}{\partial t}\right) - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2} +\] +which becomes after simple rearrangement and use of the identity \({\boldsymbol \nabla} \times \left({\boldsymbol \nabla} \times {\boldsymbol A}\right) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\boldsymbol A}) - {\boldsymbol \nabla}^2 {\boldsymbol A}\), +
++\[ + \left( {\boldsymbol ∇}2 {\boldsymbol A} - μ0 ε0 \frac{∂2 {\boldsymbol A}}{∂ t2} \right) +
++ \label{eq:LaplacianA} +\] +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++A generic configuration of static charges coupled via the Coulomb interaction +defines an electrostatic problem, whose solution is in principle obtained +from calculating either the field according to (\ref{eq:E_from_rho}), +
++\[ + {\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\mathbb{R}^3} d\tau' \rho({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} + \tag{\ref{eq:E_from_rho}} + \] +
+ ++or (often simpler) by calculating the electrostatic potential, using either the +explicit construction (\ref{eq:V_from_rho}) +
++\[ + V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' \frac{\rho({\bf r}')}{|{\bf r} - {\bf r}'|}. + \tag{\ref{eq:V_from_rho}} + \] +
+ ++Alternately, we have also seen that the two fundamental equations for the +electrostatic field, Gauss's law (\ref{Gr(2.14)}) and the no-perpetual-machine (vanishing curl) +condition (\ref{Gr(2.20)}) can be expressed as the single +'local' (differential) condition (Poisson's equation) (\ref{eq:Poisson}) +
+ ++\[ + {\boldsymbol \nabla}^2 V = -\frac{\rho}{\varepsilon_0}. + \tag{\ref{eq:Poisson}} + \] +
+ ++In the specific case where the charge density vanishes, we fall back onto the simpler +Laplace equation +
++\[ + {\boldsymbol \nabla}^2 V = 0 + \tag{\ref{eq:Laplace}} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Of course, the simplest situation is to start by looking at the region of space +where there is no charge density. The potential then solves Laplace's equation. How can it possibly look ? +
++\[ +\frac{d^2 V(x)}{dx^2} = 0 \Longrightarrow V(x) = mx + b +\label{Gr(3.6)} +\] +Properties: +\paragraph{1.} \(V(x)\) is the average of \(V(x + a)\) and \(V(x - a)\) for any \(a\). +\paragraph{2.} Solutions to Laplace's equation have no local maxima or minima. +
+ ++Boundary conditions: always work: two end values, one end value + same end derivative value. +Not always: one end value + derivative value at other end, two end derivative values. +
++\[ +\frac{d^2 V}{dx^2} + \frac{d^2 V}{dy^2} = 0. +\] +Properties: +\paragraph{1.} The value of \(V(x,y)\) equals the average value around the point: +\[ +V(x,y) = \frac{1}{2\pi R} \oint V dl +\] +\paragraph{2.} \(V\) has no local maxima or minima. All extrema occur at the boundaries. +
++\[ +{\boldsymbol \nabla}^2 V = 0 +\] +Properties: +\paragraph{1.} \(V({\bf r})\) is the average value of \(V\) over any spherical surface +centered at \({\bf r}\): +\[ +V({\bf r}) = \frac{1}{4\pi R^2} \oint V da +\] +\paragraph{2.} \(V\) can have no local maxima or minima. All extrema occur at the boundaries. +
+ ++\paragraph{Another way of seeing this} is to write the second derivatives as +\[ +\frac{\partial^2 V({\bf r})}{\partial x^2} = f_x ({\bf r}), \hspace{5mm} +\frac{\partial^2 V({\bf r})}{\partial y^2} = f_y ({\bf r}), \hspace{5mm} +\frac{\partial^2 V({\bf r})}{\partial z^2} = f_z ({\bf r}), \hspace{5mm} +f_x + f_y + f_z = 0. +\] +The \(f_a ({\bf r})\) represent the three components of the curvature of \(V({\bf r})\). +An extremum of \(V\) at \({\bf r}_e\) would be characterized by \({\boldsymbol \nabla} V |_{{\bf r}_e} \cdot \delta{\bf r} = 0\) +for any infinitesimal displacement \(\delta{\bf r}\) around the extremum point. For a local +minimum, the second derivative form should be greater than zero, \(\sum_{i,j} \frac{\partial^2 V}{\partial r_i \partial r_j} \delta r_i \delta r_j > 0\) +for any displacement vector. Choosing alternately displacements along the three axes, +the form becomes \(f_x (\delta x)^2\), \(f_y (\delta y)^2\) or \(f_z (\delta z)^2\). Since the squared displacements +are necessarily positive, we thus require \(f_x > 0\), \(f_y > 0\) and \(f_z > 0\). This is impossible in view +of the \(f_x + f_y + f_z = 0\) condition above. +
+ +
+Earnshaw's theorem
+Since solutions to Laplace's equation have no local minimum,
+it is impossible to find a static distribution of charges which generates an electrostatic field
+with a stable equilibrium position for a test charge.
+
+Going back to Poisson's equation, we can make a few comments: +
+ ++We therefore want to ask the question: under what conditions can an electrostatic problem be fully +defined by solving Poisson's equation ? We start by mentioning some cases, and interpreting them thereafter. +
+ ++First case: the electrostatic potential is uniquely determined +if \(\rho({\bf r})\) is given throughout all space. In this case, Poisson's equation +is explicitly solved by (\ref{eq:V_from_rho}). Explicit check: +\[ +{\boldsymbol \nabla}^2 V ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' \rho({\bf r}') {\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|} += \frac{1}{4\pi \varepsilon_0} \int_{\mathbb{R}^3} d\tau' (-4\pi) \delta ({\bf r} - {\bf r}') = -\frac{\rho ({\bf r})}{\varepsilon_0}. +\] +where we have used (\ref{Gr(1.102)}), and the fact that the delta function is always resolved since we +integrate over all space. Note: it is implicitly assumed that the integral in (\ref{eq:V_from_rho}) +converges, i.e. that the charge density \(\rho({\bf r})\) is sufficiently well-behaved (does not +become singular). +
+ ++First case (corollary): the electrostatic potential is uniquely determined +in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density +\(\rho ({\bf x})\) is given everywhere within \({\cal V}\), vanishes outside of \({\cal V}\), +and the value of the surface charge density \(\sigma\) is given everywhere on the boundary \({\cal S}\). +Of course, \({\cal S}\) need not be a connected surface. +
+ ++This is obvious: we know where all the charges are, so this is really the same as the first case. +
+ ++Second case: the electrostatic potential is uniquely determined +in a certain volume \({\cal V}\) bounded by boundary \({\cal S}\), provided the charge density +\(\rho ({\bf x})\) is given everywhere within \({\cal V}\), and the value of \(V\) is given everywhere on the +boundary \({\cal S}\). Of course, \({\cal S}\) need not be a connected surface. +
+ ++Here, the logic is quite simple: since the electrostatic potential is known on all the surface enclosing +the space \({\cal V}\), and since Poisson's equation is local, we need not consider anything outside of \({\cal V}\) +to obtain \(V\) within \({\cal V}\). +
+ ++Given a solution \(V_1 ({\bf r})\), we can easily show that it is unique. Suppose there was another solution +\(V_2 ({\bf r})\). Look at the difference, \(U \equiv V_1 - V_2\). In the bulk, \(U\) obeys the Laplace +equation +\[ +{\boldsymbol \nabla}^2 U = {\boldsymbol \nabla}^2 V_1 - {\boldsymbol \nabla}^2 V_2 = -\frac{\rho}{\varepsilon_0} + \frac{\rho}{\varepsilon_0} = 0. +\] +Moreover, \(U ({\bf r}) = 0\) for \({\bf r} \in {\cal S}\). Since solutions to the Laplace equation take +their maximal and minimal value on the boundary, we must have \(U = 0\) \(\forall {\bf r} \in {\cal V}\) +(Griffiths' proof). +
+ ++This all feels a bit amateurish and not very systematic. Can we be more precise and general? What kinds of boundary information do we really need to specify the solution uniquely ? +
+Created: 2022-02-07 Mon 08:02
+ ++George Green +
+ ++[1793-1841] was a self-taught English mathematician whose academic life story +did not follow the normal path: he was a miller! Despite having no formal education, in 1828, +he published at his own expense what was to become a truly lasting +contribution to science: An Essay on the Application of Mathematical Analysis +to the Theories of Electricity and Magnetism, in which the results discussed here can +be found. He was later admitted, at the young age of +40, as an undergraduate student in Cambridge where he remained until his early demise at age 47. +
+ ++We can provide a very precise statement about uniqueness of solutions to Poisson's (or Laplace's) +equation with some very basic considerations starting from the divergence theorem +\[ +\int_{\cal V} d\tau {\boldsymbol \nabla} \cdot {\bf F} = \oint_{\cal S} da ~{\bf F} \cdot {\bf n} +\] +Let \({\bf F} = \phi {\boldsymbol \nabla} \psi\), where \(\phi\) and \(\psi\) are scalar fields. We +can then write +\[ +{\boldsymbol \nabla} \cdot (\phi {\boldsymbol \nabla} \psi) = \phi {\boldsymbol \nabla}^2 \psi + {\boldsymbol \nabla} \phi \cdot {\boldsymbol \nabla} \psi +\] +and +\[ +\phi {\boldsymbol \nabla} \psi \cdot {\bf n} = \phi \frac{\partial \psi}{\partial n}. +\] +Substituting this in the divergence theorem gives {\bf Green's first identity} +\[ +\int_{\cal V} d\tau ~(\phi {\boldsymbol \nabla}^2 \psi + {\boldsymbol \nabla} \phi \cdot {\boldsymbol \nabla} \psi) = \oint_{\cal S} da ~\phi \frac{\partial \psi}{\partial n}. +\label{eq:GreensFirstIdentity} +\] +This first identity will prove crucial in the argument that follows. +As an aside for now, for completeness, if we do the same thing again but with \(\phi\) and \(\psi\) +interchanged, and subtract the result, we obtain another useful result known as +{\bf Green's second identity} or {\bf Green's theorem} +\[ +\int_{\cal V} d\tau (\phi {\boldsymbol \nabla}^2 \psi - \psi {\boldsymbol \nabla}^2 \phi) += \oint_{\cal S} da \left(\phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi}{\partial n} \right). +\label{eq:GreensTheorem} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Suppose now that we have two solutions to the Poisson equation, \(V_1 ({\bf r})\) and \(V_2 ({\bf r})\). +Defining \(U = V_1 - V_2\), we see that \(U\) manifestly obeys Laplace +within \({\cal V}\), \({\boldsymbol \nabla}^2 U = 0\). +We can now use Green's first identity (\ref{eq:GreensFirstIdentity}) to shed some light on the boundary +problem for the electrostatic potential. Namely, put \(\phi = \psi = U\). This yields +\[ +\int_{\cal V} d\tau \left( U {\boldsymbol \nabla}^2 U + {\boldsymbol \nabla} U \cdot {\boldsymbol \nabla} U \right) += \oint_{\cal S} da ~U \frac{\partial U}{\partial n}. +\] +The first term on the left-hand side vanishes since \(U\) satisfies Laplace. +The right-hand side can be made to vanish if \(U\) obeys either +
+\begin{align} + &U|_{\cal S} = 0 &\mbox{({\bf Dirichlet})} \label{eq:Dirichlet}\\ + \mbox{or}& & \nonumber \\ + &\frac{\partial U}{\partial n}|_{\cal S} = 0 &\mbox{({\bf Neumann})} \label{eq:Newmann} +\end{align} ++boundary conditions on each individual boundary surface. In those cases, we are left with +\[ +\int_{\cal V} d\tau \left|{\boldsymbol \nabla} U \right|^2 = 0, \longrightarrow {\boldsymbol \nabla} U = 0. +\] +\(U\) is thus constant. For Dirichlet, \(U = 0\) throughout \({\cal V}\), and thus \(V_2 = V_1\) and the solution +is unique. For Neumann, the solution is unique apart from an unimportant constant. +
+ ++We can thus finally state the +
++Uniqueness Theorem +
+ ++The solution to Poisson's equation \({\boldsymbol \nabla}^2 V = -\frac{\rho}{\varepsilon_0}\) inside a volume \({\cal V}\) +bounded by a (in general disconnected) surface \({\cal S}\) is uniquely defined provided either Dirichlet \(V |_{{\cal S}_i}\) or Neumann +\(\frac{\partial V}{\partial n} |_{{\cal S}_i}\) boundary conditions are used on each individual surface. +
+ ++Note that these types of boundary conditions can be mixed, i.e. Dirichlet on some surfaces, +Neumann on others). +
+ ++Existence of solutions: this is another matter. +Intuitively, from our first case: +the solution always exists for Dirichlet boundary conditions. +
+ ++Link to earlier cases: the 'second case' above, in which the potential is specified on the +boundaries, is the case of Dirichlet boundary conditions. +The 'first case corollary', where the normal derivative of the potential is given, is a subcase involving +Neumann boundary conditions (subcase, because we could imagine other charges living outside volume \({\cal V}\), +whereas our first case corollary involved only surface charges). +
+ ++\paragraph{Note on Griffiths' presentation of uniqueness theorem(s):} +we have used Green's identity to provide a general statement on uniqueness. Griffiths +might mislead you into thinking that there are numerous cases and corollaries. +Here (in italics) is his (confused) way of thinking about it (see Comment/warning below): +
+ ++\subsubsection*{\it Boundary Conditions and Uniqueness Theorem} +\paragraph{\it First uniqueness theorem:} {\it The solution to Laplace's equation in some volume +\({\cal V}\) is uniquely determined if \(V\) is specified on the boundary surface \({\cal S}\). +{\bf Corollary:} the potential in a volume \({\cal V}\) is uniquely determined if +a) the charge density throughout the region and b) the value of \(V\) on all boundaries +are specified.} +
+ ++\subsubsection*{\it Conductors and the Second Uniqueness Theorem} +\paragraph{\it Second uniqueness theorem:} {\it In a volume \({\cal V}\) surrounded by conductors +and containing a specified charge density \(\rho\), the electric field is uniquely determined +if the total charge on each conductor is given.} +
+ ++{\it Comments: this is the same uniqueness as before, in view of the fact that conductors are +equipotentials, and capacitance relates the charge to the potential. +} +
+ +
+{\bf Comment/warning: {\color{blue} uniqueness theorem on uniqueness theorems}}
+Do not be misled by Griffiths: there is a {\it unique} uniqueness theorem for the
+solution of Poisson's equation, namely the one we have stated starting from Green's first identity.
+
Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Put \({\bf d}\) along \(\hat{\bf z}\). Then, (\ref{eq:electric_dipole}) becomes +\[ +V_{\mbox{\tiny di}}({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{p \cos \theta}{r^2} +\label{Gr(3.102)} +\] +Taking the gradient, +\[ +E_r = -\frac{\partial V}{\partial r} = \frac{1}{4\pi \varepsilon_0} \frac{2p\cos \theta}{r^3}, \hspace{1cm} +E_\theta = -\frac{1}{r} \frac{\partial V}{\partial \theta} = \frac{1}{4\pi \varepsilon_0} \frac{p \sin \theta}{r^3}, \hspace{1cm} +E_\phi = -\frac{1}{r \sin \theta} \frac{\partial V}{\partial \phi} = 0, +\] +we get +\[ +{\bf E}_{\mbox{\tiny di}} (r, \theta) = \frac{1}{4\pi \varepsilon_0} \frac{p}{r^3} (2\cos \theta \hat{\bf r} + \sin \theta \hat{\bf \theta}) +\label{Gr(3.103)} +\] +or in a better coordinate-free form +\[ + {\bf E}_{\mbox{\tiny di}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{1}{r^3} \left[3 ({\bf p} \cdot \hat{\bf r}) \hat{\bf r} - {\bf p}\right] + \label{eq:dipole_field} +\] +
+ ++Dipole energy +
+Created: 2022-02-07 Mon 08:02
+ ++Axially symmetric quadrupole +
+ ++Quadrupole energy +
+Created: 2022-02-07 Mon 08:02
+ ++Let's consider the spatial function in the potential for a single point source charge: +\[ +\frac{1}{|{\bf r} - {\bf r}_s|} +\] +How does this look when we're at large distances \(|{\bf r}| \gg |{\bf r}_s|\) ? +We can formally expand this in powers of \(|{\bf r}_s|/|{\bf r}|\). For simplicity, let's start by +putting \({\bf r} = r \hat{\bf z}, r > 0\) and \({\bf r}_s = r_s \hat{\bf z}\), with \(|r_s| < r\). +Then, by Taylor expanding, we get +\[ +\frac{1}{|{\bf r} - {\bf r}_s|} = \frac{1}{r - r_s} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r_s}{r}\right)^l. +\] +Formally, we could do this for any vector \({\bf r}_s\) such that \(|{\bf r}_s| < |{\bf r}|\) by +Taylor expanding with the \({\boldsymbol \nabla}\) operator, +\[ +\frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{1}{l!} \left( - {\bf r}_s \cdot {\boldsymbol \nabla} \right)^l \frac{1}{r} += \frac{1}{r} - {\bf r}_s \cdot {\boldsymbol \nabla} \frac{1}{r} + \frac{1}{2} \left({\bf r}_s \cdot {\boldsymbol \nabla} \right)^2 \frac{1}{r} + ... +\] +However, it is more practical to exploit the fact that in the configuration above, +the problem has azimuthal symmetry (everything on the \(\hat{\bf z}\) axis), and therefore +the potential takes the form of the general solution of Laplace's equation (\ref{Gr(3.65)}) with \(\theta = 0\). +Reading the parameters, we get \(A_l = 0\), \(B_l = r_s^l\). Putting back a generic angle, +we thus get +\[ +\frac{1}{|{\bf r} - {\bf r}_s|} = \sum_{l=0}^{\infty} \frac{r_s^l}{r^{l+1}} P_l (\cos \theta), +\hspace{1cm} \cos \theta \equiv \hat{\bf r} \cdot \hat{\bf r}_s, \hspace{1cm} r_s < r. +\label{Gr(3.94)} +\] +We thus see that our beloved Legendre polynomials are quite handy beasts indeed. +Considering an arbitrary charge distribution over a volume \({\cal V}\), +we can expand the potential at a point \({\bf r}\) outside \({\cal V}\) according to +(here, we put the origin of our coordinate system closer to all points in \({\cal V}\) than to \({\bf r}\) +to ensure convergence) +\[ +V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} +\int_{\cal V} d\tau_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s), +\hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal V}. +\label{Gr(3.95)} +\] +This is an exact expansion of our potential if we keep all terms. However, the power of +this comes from the fact that truncating the series gives a very good approximation to the +original potential, as long as we are sufficiently far away from the source charges. +
+ ++\paragraph{%DO NOT DO IT LIKE THIS ! +Ex. 3.10:} an {\bf electric dipole} consists of two equal and opposite charges \(\pm q\) separated +by a distance \(d\). For definiteness: we put \(q\) at position \({\bf d}/2\) and \(-q\) at \(-{\bf d}/2\). +The potential is +\[ +V({\bf r}) = \frac{q}{4\pi \varepsilon_0} \left( \frac{1}{|{\bf r} - {\bf d}/2|} - \frac{1}{|{\bf r} + {\bf d}/2|} \right). +\] +From the law of cosines, \(|{\bf r} \pm {\bf d}/2|^2 = r^2 + (d/2)^2 \mp rd \cos \theta += r^2 \left( 1 \mp \frac{d}{r} \cos \theta + \frac{d^2}{4r^2}\right)\) where \(\theta\) is the angle +between \(\hat{\bf r}\) and \(\hat{\bf d}\). +
+ ++For \(r \gg d\), we can expand, +\[ +\frac{1}{|{\bf r} \mp {\bf d}/2|} \simeq \frac{1}{r} \left( 1 \pm \frac{d}{r} \cos \theta \right)^{1/2} +\simeq \frac{1}{r} \left( 1 \pm \frac{d}{2r} \cos \theta \right). +\] +Putting things together, +\[ +V({\bf r}) \simeq \frac{1}{4\pi \varepsilon_0} \frac{q d \cos \theta}{r^2} +\label{Gr(3.90)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++The next terms in the expansion are obtained similarly: the {\bf quadrupole term} is +\[ +V_{\mbox{\tiny quad}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{1}{r^3} \int_{\cal V} d\tau_s r_s^2 P_2 (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s) += \sum_{a,b = x,y,z} \frac{r_a r_b}{r^5} \int_{\cal V} d\tau_s \frac{1}{2} (3 r_{s,a} r_{s,b} - r_s^2 \delta_{a,b}) \rho ({\bf r}_s) +\] +and can be rewritten as +
++\[ + V_{\mbox{\tiny quad}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{1}{2} \sum_{a,b} \frac{r_a r_b}{r^5} Q_{ab} + \] +
+ ++in terms of the {\bf quadrupole moment} +
++\[ + Q_{ab} = \int_{\cal V} d\tau_s (3 r_{s,a} r_{s,b} - r_s^2 \delta_{a,b}) \rho ({\bf r}_s). + \] +
+ ++This is a symmetric rank \(2\) tensor, \(Q_{ab} = Q_{ba}\). +Moreover, it is traceless, \(\sum_a Q_{aa} = 0\). It therefore has \(5\) independent components. +
+ + + ++Our expansion for the potential thus looks like +
++\[ + V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q}{r} + \sum_a \frac{r_a}{r^3} p_a + \frac{1}{2} \sum_{a,b} \frac{r_a r_b}{r^5} Q_{ab} + ... \right) + \] +
+ ++This can be carried further if we feel like it, with the {\bf octopole}, {\bf hexadecapole}, {\bf triacontadipole}, {\bf hexecontatetrapole}, … terms (see info box). +
+ ++\paragraph{Important property:} +the leading nonvanishing multipole moment is independent of the chosen location for +the origin of the coordinate system (see Jackson Prob. 4.4). +
+ ++{\bf A consistent nomenclature for the multipole expansion?}\\\\ +You all know the terms {\bf monopole}, {\bf dipole} and {\bf quadrupole}, +and perhaps also the less frequently used {\bf octupole}, {\bf hexadecapole} [16], +{\bf triacontadipole} (or {\bf dotriacontapole}) [32] and {\bf tetrahexacontapole} +(or {\bf hexacontatetrapole}) [64]. +Physicists are clearly insufficiently educated in the humanities: +these terms sound very fancy and their choice seems to make sense, but it doesn't. +{\bf Mono-} is derived from the Greek +{\it monos} ('alone'); {\bf di-} is derived from the Greek {\it dis} ('twice'); +{\bf quadru-} is a fake Latin prefix ({\it quadri-} would be genuine) +meaning 'something to do with the number 4', and +{\bf octu-} is another (fake) Latin prefix (both Greek and Latin have {\it octo/okto} for 8, +but Greek makes compounds with {\it octo-} or {\it octa-}, never {\it octu-}). +
+ +
+A more consistent nomenclature would be to go either fully Greek {\it or} Latin,
+yielding:
+
+\begin{tabular}{r|ll}
+ & Greek-inspired & Latin-inspired
+ \hline
+ 1 & monopole & unipole
+ 2 & dipole & duopole
+ 4 & tetrapole & quadrupole
+ 8 & octopole & octopole
+\end{tabular}
+
+Irrespective of whether you have a predilection for the Greek or Latin version,
+you can go wild and ask how this could generalize. The way to do this is not
+uniquely defined; here is a set of possibilities for
+more terms than you might ever (hopefully) need:
+
+\begin{tabular}{r|ll}
+ 16 & hexadecapole & sexdecapole
+ 32 & triacontadipole & trigentiduopole
+ 64 & hexecontatetrapole & sexagintiquadrupole
+ 128 & hecatonikosioctopole & viginticentioctopole
+ 256 & diacosipentecontahexapole & ducentiquinquagintisexapole
+\end{tabular}
+
Created: 2022-02-07 Mon 08:02
+ ++The series (\ref{Gr(3.95)}) is organized in increasing powers of inverse distance. +The leading term is called the {\bf monopole} term, and is +
++\[ + V_{\mbox{\tiny mono}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r}, \hspace{1cm} + Q = \int_{\cal V} d\tau_s \rho({\bf r}_s). + \label{Gr(3.97)} + \] +
+ ++For a point charge, the monopole term gives the exact potential. +
+ ++The next term is the {\bf dipole} term: by using \(P_1 (x) = x\), we have +\[ +V_{\mbox{\tiny di}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{1}{r^2} \int_{\cal V} d\tau_s \hat{\bf r} \cdot {\bf r}_s \rho({\bf r}_s) +\] +This can be written +
++\[ + V_{\mbox{\tiny di}} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{\hat{\bf r} \cdot {\bf p}}{r^2}, + \hspace{10mm}{\bf p} \equiv \int_{\cal V} d\tau_s ~{\bf r}_s ~\rho({\bf r}_s). + \label{eq:electric_dipole} + \] +
+ ++in terms of the {\bf dipole moment} \({\bf p}\). +Note that the dipole moment is an internal property of the source charges, and that it +in general depends on the chosen point of origin (more on this later). +
+ ++Since dipole moments are vectors, they are summed following vector addition rules. +
+ ++{\bf Pure dipole:} two charges closer and closer together, but charges higher and higher such that \({\bf p}\) remains finite. +
+Created: 2022-02-07 Mon 08:02
+ ++The fact that conductors are equipotentials means that we can fomally solve +loads of electrostatic problems involving conductors of various shapes, by +looking at combinations of point charges. +
+ ++Let's consider the simplest electrostatic problem above a single point charge: +a system of two point charges. For definiteness, we put a charge \(q\) at coordinate +\(d \hat{z}\), and a charge \(-q\) at \(-d \hat{z}\). By superposition, we have that +\[ +V(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x2 + y2 + (z-d)2]1/2} +
++\label{Gr(3.9)} +\] +Now it's obvious that \(V = 0\) when \(z = 0\). Therefore, in the region \(z > 0\), this problem +is completely equivalent to a second problem: a point charge \(q\) at \(d \hat{z}\), +{\it and a grounded conductor on the whole plane \(z = 0\)}. Yet another equivalent +problem in the region \(z < 0\) is that of a charge \(-q\) at \(-d \hat{z}\) with a grounded +conductor on the plane \(z = 0\). +
+ ++We can go one step further. In the two point charges problem (dipole), we could put +a conductor on {\it any} of the equipotential lines, and still have an explicit solution +for the potential in terms of the potential from the original two point charges. +
+ ++In the Image problem, we might want to know what the induced surface charge is. +
+Created: 2022-02-07 Mon 08:02
+ ++Force is the same as case of two point charges: +\[ +{\bf F} = -\frac{1}{4\pi \varepsilon_0} \frac{q^2}{(2d)^2} \hat{\bf z} +\label{Gr(3.12)} +\] +Energy is however not the same: for two point charges, +(\ref{Gr(2.42)}) gives +\[ +W = -\frac{1}{4\pi \varepsilon_0} \frac{q^2}{2d} +\label{Gr(3.13)} +\] +whereas for the single charge and conducting plane, +\[ +W = -\frac{1}{4\pi \varepsilon_0} \frac{q^2}{4d} +\label{Gr(3.14)} +\] +Half: \(E^2\) integrated over half the space in (\ref{eq:Energy_as_int_E2}). +
+Created: 2022-02-07 Mon 08:02
+ ++Use (\ref{Gr(2.49)}), where normal direction is now \(\hat{\bf z}\). +\[ +\sigma(x, y) = \frac{-qd}{2\pi (x^2 + y^2 + d^2)^{3/2}} +\label{Gr(3.10)} +\] +Total induced charge: \(Q = \int \sigma da\). +Use planar coordinates: \(\sigma(r) = \frac{-qd}{2\pi (r^2 + d^2)^{3/2}}\), so +\[ +Q = \int_0^{2\pi} d\phi \int_0^{\infty} dr r \frac{-qd}{2\pi (r^2 + d^2)^{3/2}} += \frac{qd}{\sqrt{r^2 + d^2}}|_0^{\infty} = -q +\label{Gr(3.11)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++We can also play around differently. Elaborating on the grounded conducting plane above: +if we just change \(d\), the problem stays of the same nature. Now if, however, we change +the value of the charge at \(z = -d\) from \(q\) to \(q'\): what do we get ? The potential is +\[ +V(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x2 + y2 + (z-d)2]1/2} +
++\] +This vanishes when +
+\begin{align} +\left(\frac{q}{q'}\right)^2 \frac{ x^2 + y^2 + z^2 + d^2 + 2dz}{x^2 + y^2 + z^2 + d^2 - 2dz} = 1, +\rightarrow +x^2 + y^2 + z^2 + d^2 + 2d \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1} z = 0 +\nonumber \\ +\rightarrow x^2 + y^2 + (z + d\alpha)^2 = d^2 (\alpha^2 - 1), \hspace{1cm} \alpha \equiv \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1}. +\end{align} ++This only makes sense if \(\alpha > 1\), {\it i.e.} \(q' < q\), otherwise the radius is not positive definite. +Therefore, the equipotential \(V = 0\) is a sphere of radius \(R = d\sqrt{\alpha^2 - 1}\) centered at \(z = -d\alpha\). +The problem is therefore equivalent to a grounded metal sphere of that radius centered at this position, +with a single point charge \(q\) at \(d\hat{z}\) ! +(Griffiths pulls that out of a hat, but you now see where it comes from). Correspondence with +Griffiths Ex 3.2: his \(a-b\) is my \(2d\), but his \(b\) is (3.16) \(R^2/a\), so we get \(a - R^2/a = 2d\), +whose solution is \(a = d[1 + \alpha]\). +
+Created: 2022-02-07 Mon 08:02
+ ++In many cases, we can't just guess the form of the potential. However, we can often use +some symmetry of the problem to greatly simplify the form of solution to Laplace's equation. +
+Created: 2022-02-07 Mon 08:02
+ +
+{\bf Example: separation of variables (Cartesian coordinates)}
+Two infinite grounded metal plates parallel to the \(xz\) plane, one at \(y = 0\) and the
+other at \(y = a\). End at \(x = 0\) closed off with infinite insulated strip maintained
+at potential \(V_0 (y)\). Find potential inside the slot.
+\paragraph{Solution:} indep of \(z\), so 2d problem. Solve
+\[
+\frac{\partial^2 V}{\partial^2 x} + \frac{\partial^2 V}{\partial^2 y} = 0
+\label{Gr(3.20)}
+\]
+\[
+(i) V(x, y = 0) = 0, \hspace{5mm} (ii)V(x, y = a) = 0, \hspace{5mm} (iii) V(0, y) = V_0 (y), \hspace{5mm} (iv) V (x \rightarrow \infty) \rightarrow 0.
+\label{Gr(3.21)}
+\]
+Look for solutions of form
+\[
+V(x,y) = X(x) Y(y), \hspace{1cm}
+\frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 0
+\label{Gr(3.23)}
+\]
+Choose
+\[
+\frac{d^2 X_n}{dx^2} = k_n^2 X_n, \hspace{1cm} \frac{d^2Y_n}{dy^2} = -k_n^2 Y_n
+\label{Gr(3.26)}
+\]
+where \(k_n\) is some real number. We can linearly combine solutions of (\ref{Gr(3.26)})
+for different \(k_n\) and still get a solution to (\ref{Gr(3.23)}).
+Let's look first of all at the solutions of (\ref{Gr(3.26)}) for a given \(k_n\).
+Since this is a second-order linear differential equation, there are two linearly
+independent solutions. Most general solution:
+\[
+X_n(x) = Ae^{k_nx} + Be^{-k_nx}, \hspace{1cm} Y_n(y) = C \sin k_ny + D \cos k_ny
+\label{Gr(3.27)}
+\]
+Fix constants: from \((iv)\), \(A = 0\). From \((i)\), D = 0. Left with
+\[
+V(x,y) = C_n e^{-k_nx} \sin k_n y
+\label{Gr(3.28)}
+\]
+Then, \((ii)\) requires
+\[
+k_n = \frac{n\pi}{a}, \hspace{1cm} n = 1, 2, 3, ...
+\label{Gr(3.29)}
+\]
+But we can use as solution any linear combination of the functions defined
+by these momenta. Fix coefficients with Fourier series:
+\[
+V(x,y) = \sum_{n=1}^{\infty} C_n e^{-n\pi x/a} \sin (n\pi x/a)
+\label{Gr(3.30)}
+\]
+Needed:
+\[
+\int_0^a dy \sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2}
+\label{Gr(3.33)}
+\]
+so
+\[
+C_n = \frac{2}{a} \int_0^a dy V_0(y) \sin(n\pi y/a)
+\label{Gr(3.34)}
+\]
+
+\paragraph{Specific example:} say that \(V_0(y) = V_0\), some constant. Then, +\[ +C_n = \frac{2V_0}{a} \int_0^a dy \sin(n\pi y/a) = \frac{2V_0}{n\pi} (1 - \cos n\pi) = \frac{4V_0}{n\pi} \delta_{n, odd} +\label{Gr(3.35)} +\] +
+ ++Applicable provided {\bf completeness} and {\bf orthogonality}: +
+ ++Completeness: +\[ +f(x) = \sum_{n=1}^{\infty} C_n f_n (y), \hspace{1cm} \forall f \in C^{\infty} +\label{Gr(3.38)} +\] +
+ ++Orthogonality: +\[ +\int_0^a dx f_n (x) f_{n'} (x) = 0, \hspace{1cm} n \neq n' +\label{Gr(3.39)} +\] +
+ ++The solution for the specific case \(V_0 (y) = V_0\) therefore is +\[ +V(x,y) = \frac{4V_0}{\pi}\sum_{n=1, 3, 5, ...}^{\infty} \frac{1}{n} e^{-n\pi x/a} \sin (n\pi x/a) +\] +
+ + +
+ {\bf Example: rectangular pipe}
+ Two infinitely long grounded plates at \(y = 0,a\) are connected at \(x = \pm b\)
+to metal strips maintained at constant \(V = V_0\). Find the potential in the resulting rectangular pipe.
+\paragraph{Solution:} indep of \(z\). Laplace: \(\partial^2 V/\partial x^2 + \partial^2 V/\partial y^2 = 0\),
+boundary conditions
+\[
+(i) V (y = 0) = 0, \hspace{5mm} (ii) V (y = a) = 0, \hspace{5mm} (iii) V(x = b) = 0, \hspace{5mm} (iv) V(x = -b) = 0.
+\label{Gr(3.40)}
+\]
+Generic solution: as (\ref{Gr(3.27)}),
+\[
+V(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky).
+\]
+By symmetry, \(V(x,y) = V(-x,y)\) so \(A = B\). Now \(e^{kx} + e^{-kx} = 2\cosh kx\). Generic solution becomes
+(redefining \(C\) and \(D\))
+\[
+V(x,y) = \cosh kx (C\sin ky + D\cos ky).
+\]
+Boundary conditions \((i)\) and \((ii)\) require \(D = 0\), \(k = n\pi/a\) so
+\[
+V(x,y) = C \cosh(n\pi x/a) \sin(n\pi y/a)
+\label{Gr(3.41)}
+\]
+with \((iv)\) already satisfied if \((iii)\) is. Full solution is linear combination of complete set of functions,
+\[
+V(x,y) = \sum_{n=1}^{\infty} C_n \cosh(n\pi x/a) \sin(n\pi y/a).
+\]
+Coefficients: chosen such that \((iii)\) is fulfilled, \(V(b,y) = V_0\). From (\ref{Gr(3.35)}):
+\[
+V(x,y) = \frac{4V_0}{\pi} \sum_{n = 1, 3, 5, ...} \frac{1}{n} \frac{\cosh (n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a).
+\label{Gr(3.42)}
+\]
+
Created: 2022-02-07 Mon 08:02
+ ++In spherical coordinates, the Laplace equation takes the following form: +
++ +
+ +\begin{equation} +\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial V}{\partial r}\right) + + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial V}{\partial \theta}\right) + + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 V}{\partial \phi^2} = 0 +\label{Gr(3.53)} +\end{equation} + ++If you are dealing with a problem having azimuthal symmetry, +\(V\) is independent of \(\phi\) and the equation simplifies to: +
+ +\begin{equation} +\frac{\partial}{\partial r} \left(r^2 \frac{\partial V}{\partial r}\right) ++ \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial V}{\partial \theta}\right) += 0. +\label{Gr(3.54)} +\end{equation} + ++Look for solution in form +
+ ++\[ +V(r, \theta) = R(r) \Theta (\theta). +\label{Gr(3.55)} +\] +
+ ++Put in (\ref{Gr(3.54)}), divide by \(V\): +
+ ++\[ +\frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) + \frac{1}{\Theta \sin \theta} \frac{d}{d\theta} +\left(\sin \theta \frac{d\Theta}{d\theta} \right) = 0. +\label{Gr(3.56)} +\] +
+ ++Separation of variables logic: each term must be a constant, +
+ ++\[ +\frac{1}{R} \frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1), \hspace{1cm} +\frac{1}{\Theta \sin \theta} \frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) +\label{Gr(3.57)} +\] +
+ ++so the problem, originally involving {\it partial} differentials, now is given by +{\it ordinary} differential equations. Radial equation: +
+ ++\[ +\frac{d}{dr} \left( r^2 \frac{dR}{dr} \right) = l(l+1) R +\label{Gr(3.58)} +\] +
+ ++Search for solution of form \(r^\alpha\): \(\frac{d}{dr} (r^2 \alpha r^{\alpha - 1}) = \alpha (\alpha + 1) r^{\alpha} = l(l+1) r^{\alpha}\) +so we get \(\alpha = l\) or \(-(l+1)\). (\ref{Gr(3.58)}) thus has the general solution +
+ ++\[ +R(r) = A r^l + \frac{B}{r^{l+1}} +\label{Gr(3.59)} +\] +
+ ++Angular equation: +
+ ++\[ +\frac{d}{d\theta} \left(\sin \theta \frac{d\Theta}{d\theta} \right) = -l(l+1) \sin \theta \Theta +\label{Gr(3.60)} +\] +
+ ++has solutions in terms of {\bf Legendre polynomials} of the variable \(\cos \theta\): +\[ +\Theta(\theta) = P_l (\cos \theta) +\label{Gr(3.61)} +\] +\(P_l(x)\): convenient formula is the {\bf Rodrigues formula}: +
+ ++\[ +P_l(x) = \frac{1}{2^l l!} \left( \frac{d}{dx} \right)^l (x^2 - 1)^l +\label{Gr(3.62)} +\] +
+ ++Actually, a more practical formula is Bonnet's recursion relation +
+ ++\[ +(l + 1) P_{l+1} (x) = (2l + 1) x P_l (x) - l P_{l-1} (x) +\] +
+ ++First few examples: +
+ +\begin{align} +P_0 (x) &= 1 \nonumber \\ +P_1 (x) &= x \nonumber \\ +P_2 (x) &= \frac{1}{2} (3x^2 - 1) \nonumber \\ +P_3 (x) &= \frac{1}{2} (5x^3 - 3x) \nonumber \\ +P_4 (x) &= \frac{1}{8} (35x^4 - 30x^2 + 3) \nonumber \\ +P_5 (x) &= \frac{1}{8} (63x^5 - 70x^3 + 15x). +\end{align} + ++Prefactor: chosen such that +
+ ++\[ +P_l(1) = 1 +\label{Gr(3.63)} +\] +
+ ++Angular equation (\ref{Gr(3.60)}) is second order: should be 2 solutions ! +These other solutions blow up at \(\theta = 0\) and/or \(\theta = \pi\) (unacceptable on physical grounds). +Ex.: second solution for \(l = 0\) is +
+ ++\[ +\Theta (\theta) = \ln \left( \tan \frac{\theta}{2} \right) +\label{Gr(3.64)} +\] +
+ ++General solution to problem with azimuthal symmetry: +
++\[ + V(r,\theta) = \sum_{l=0}^{\infty} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta) + \label{Gr(3.65)} + \] +
+ ++Example: separation of variables (spherical coordinates) +
+ ++The potential \(V_0 (\theta)\) is specified on the surface of a hollow +sphere of radius \(R\). Find potential inside sphere. +\paragraph{Solution:} (this is a case of Dirichlet boundary conditions) +here, \(B_l = 0\) \(\forall l\) since potential cannot diverge at origin. Formal solution: +
+ ++\[ +V(r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta) +\label{Gr(3.66)} +\] +
+ ++Boundary condition: +
+ ++\[ +V(R,\theta) = \sum_{l=0}^\infty A_l R^l P_l (\cos \theta) = V_0 (\theta) +\label{Gr(3.67)} +\] +
+ ++Use fact that Legendre polynomials are orthogonal functions: +
+ ++\[ +\int_{-1}^1 dx P_l (x) P_{l'} (x) = \frac{2}{2l + 1} \delta_{l l'}, +\label{LegendreOrthogonality} +\] +
+ ++or in other words +
+ ++\[ +\int_0^\pi d\theta \sin \theta P_l (\cos \theta) P_{l'} (\cos \theta) = \frac{2}{2l + 1} \delta_{l l'} +\label{Gr(3.68)} +\] +
+ ++We thus get +
+ ++\[ +A_l = \frac{2l + 1}{2R^l} \int_0^\pi d\theta \sin \theta P_l (\cos \theta) V_0 (\theta). +\label{Gr(3.69)} +\] +
+ ++Specific example: choose +
+ ++\[ +V_0 (\theta) = k \sin^2 (\theta/2) +\label{Gr(3.70)} +\] +
+ ++This is \(V_0 (\theta) = \frac{k}{2} (1 - \cos \theta) = \frac{k}{2} (P_0 (\cos \theta) - P_1 (\cos \theta))\). +
+ ++Thus, \(A_0 = k/2\), \(A_1 = -k/2\), and all others are zero, so +
+ ++\[ +V(r, \theta) = \frac{k}{2} (1 - \frac{r}{R} \cos \theta). +\label{Gr(3.71)} +\] +
+ + ++Example: surface charge density on sphere +
+ ++A surface charge density \(\sigma_0 (\theta)\) is glued over +the surface of a spherical shell of radius \(R\). Find \(V\) inside and outside sphere. +\paragraph{Solution:} (this is a case of Neumann boundary conditions) Could use direct integration. Try separation of variables. In interior: +
+ ++\[ +V^ + +
++(other terms blow up as \(r \rightarrow 0\), so need \(B_l^< = 0\) here). Exterior: +
+ ++\[ +V^>(r, \theta) = \sum_{l=0}^{\infty} \frac{B_l^>}{r^{l+1}} P_l (\cos \theta), \hspace{1cm} r \geq R +\label{Gr(3.79)} +\] +
+ ++(other terms blow up as \(r \rightarrow \infty\), so need \(A_l^> = 0\) here). Since the potential +is continuous at \(r = R\): +
+ ++\[ +\sum_{l=0}^{\infty} A_l^< R^l P_l (\cos \theta) = \sum_{l=0}^{\infty} \frac{B_l^>}{R^{l+1}} P_l (\cos \theta) +\label{Gr(3.80)} +\] +
+ ++Since Legendre polynomials are orthogonal, +\[ +B_l^> = A_l^< R^{2l + 1}. +\label{Gr(3.81)} +\] +Surface charge induces discontinuity in derivative of \(V\) from (\ref{Gr(2.36)}): +
+ ++\[ +\left( \frac{\partial V^>}{\partial r} - \frac{\partial V^ + +
++so +
+ ++\[ +-\sum_{l=0}^\infty (l+1) \left(\frac{B_l^>}{R^{l+2}} + l A_l^< R^{l-1} \right) P_l (\cos \theta) = -\frac{\sigma_0 (\theta)}{\varepsilon_0}, +\rightarrow +\sum_{l=0}^\infty (2l+1) A_l^< R^{l-1} P_l (\cos \theta) = \frac{\sigma_0 (\theta)}{\varepsilon_0} +\label{Gr(3.83)} +\] +
+ ++Coefficients: from orthogonality relation (\ref{Gr(3.68)}), +
+ ++\[ +A_l^< = \frac{1}{2\varepsilon_0 R^{l-1}} \int_0^\pi d\theta \sin \theta \sigma_0 (\theta) P_l (\cos \theta). +\label{Gr(3.84)} +\] +
+ ++Specific case: choose +
+ ++\[ +\sigma_0 (\theta) = k \cos \theta = k P_1 (\cos \theta) +\label{Gr(3.85)} +\] +
+ ++All \(A_l^< = 0\) except for \(l = 1\), in which case +
+ ++\[ +A_1^< = \frac{k}{2\varepsilon_0} \int_0^\pi d\theta \sin \theta [P_l(\cos \theta)]^2 = \frac{k}{3\varepsilon_0}. +\] +
+ ++Potential inside/outside the sphere are then: +
+ ++\[ +V^< (r,\theta) = \frac{k}{3\varepsilon_0} r\cos \theta \hspace{3mm}\mbox{for}~ r \leq R, +\hspace{10mm} +V^> (r, \theta) = \frac{k R^3}{3\varepsilon_0} \frac{\cos \theta}{r^2} \hspace{3mm}\mbox{for}~ r \geq R. +\label{eq:PotentialUniformlyPolarizedSphere} +\] +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Two conductors, one at charge \(Q\) and the other at charge \(-Q\). Potential difference is +unambiguous: +
+ ++\[ +V = V_+ - V_- = -\int_{(-)}^{(+)} {\bf E} \cdot d{\bf l} +\] +
+ ++Capacitance +
+ ++\[ +C \equiv \frac{Q}{V} +\label{Gr(2.53)} +\] +
+ ++Example 2.11: +find the capacitance of two concentric spherical metallic shells with radii \(a\) and \(b\). +
+ ++\[ +V = -\int_b^a {\bf E} \cdot d{\bf l} = \frac{Q}{4\pi \varepsilon_0} (\frac{1}{a} - \frac{1}{b}), \rightarrow +C = \frac{Q}{V} = 4\pi \varepsilon_0 \frac{ab}{b - a}. +\] +
+ ++Work to charge up capacitor: \(dW = \frac{q}{C} dq\). Integrating, +
+ ++\[ +W = \frac{1}{2} C V^2 +\label{Gr(2.55)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Charge in hollow conductor. +Induced surface charges on inside and outside surfaces. +
+ ++Example 2.9: field outside a spherical conductor centered at origin, having +a cavity of a weird shape with a point charge \(q\) somewhere inside. What is the field +outside the sphere ? +Answer: ${\bf E} = \frac{1}{4\pi\varepsilon_0} q \frac{\hat{\bf r}}{r2}.$ +
+ ++Namely, the internal surface completely screens the point charge, leaving the field +to be exactly zero (as it should) within the conductor +(that's the idea behind a Faraday cage). By charge conservation, the conductor +remains neutral, so a charge \(q\) distributes itself uniformly over its external surface. +
+Created: 2022-02-07 Mon 08:02
+ ++Basic Properties of a Conductor +
++Note that (iv) links directly with our second case of solving Poisson's equation above: +for a system of conductors held at fixed potentials, the solution is unique. +
+Created: 2022-02-07 Mon 08:02
+ ++For a conductor, we can exploit the fact that electrical fields vanish on the inside to +get a proper boundary condition for the potential. Namely, here, boundary condition +\ref{Gr(2.33)} yields +\[ +{\bf E} = \frac{\sigma}{\varepsilon_0} \hat{\bf n} +\label{Gr(2.48)} +\] +which in terms of potential reads +\[ +\sigma = -\varepsilon_0 \frac{\partial V}{\partial n}. +\label{Gr(2.49)} +\] +The force per unit area on the surface of an object is +\[ +{\bf f} = \sigma {\bf E}_{\mbox{average}} = \frac{\sigma}{2} ({\bf E}_{\mbox{above}} + {\bf E}_{\mbox{below}}) +\label{Gr(2.50)} +\] +For a conductor, \({\bf E}_{\mbox{below}} = 0\), \({\bf E}_{\mbox{above}} = \frac{\sigma}{\varepsilon_0} \hat{\bf n}\), so +\[ +{\bf f} = \frac{\sigma^2}{2\varepsilon_0} \hat{\bf n} +\label{Gr(2.51)} +\] +amounting to an outward electrostatic pressure. In terms of the field, +\[ +P = \frac{\varepsilon_0}{2} E^2 +\label{Gr(2.52)} +\] +which can also be obtained from the principle of virtual work. +
+Created: 2022-02-07 Mon 08:02
+ ++We have already seen that the electrostatic potential is the potential energy of a unit charge brought from infinity. +
+ ++Since the force is proportional to the charge we're moving, +we can calculate the work per unit charge/ \(W_u\) (so \(W = q W_u\)), +
+ ++\[ +W_u = -\int_{{\bf a}}^{{\bf b}} {\bf E} \cdot d{\bf l} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++(i) An 'inconsistency' ? No. Self-energy is simply excluded in \ref{Gr(2.42)}, but included +in \ref{eq:Energy_as_int_E2}. +(ii) Is the energy stored in the fields, or in the charges ? Meaningless question at this stage. +Radiation theory points to the preferable interpretation that the energy is stored in the field. +(iii) Superposition. Double the charge: quadruple the energy. +
+Created: 2022-02-07 Mon 08:02
+ ++For this case, \ref{Gr(2.42)} becomes +
+ ++\[ +W = \frac{1}{2} \int \rho({\bf r}) V({\bf r}) ~d\tau +\label{Gr(2.43)} +\] +
+ ++Eliminate \(\rho\) and \(V\) in favor of \({\bf E}\): +
+ ++\[ +\rho = \varepsilon_0 {\boldsymbol \nabla} \cdot {\bf E} +\longrightarrow +W = \frac{\varepsilon_0}{2} \int ({\boldsymbol \nabla} \cdot {\bf E}) ~V d\tau. +\] +
+ ++Doing integration by parts and using \({\boldsymbol \nabla} V = -{\bf E}\), +
+ ++\[ +W = \frac{\varepsilon_0}{2} \left( \int_{\cal V} E^2 d\tau + \oint_{\cal S} V {\bf E} \cdot d{\bf a} \right) +\label{Gr(2.44)} +\] +
+ ++Integrating over all space, +
++ +
+ ++\[ + W = \frac{\varepsilon_0}{2} \int E^2 d\tau + \label{eq:Energy_as_int_E2} + \] +
+ ++Example 2.8: Find the energy of a uniformly charged spherical shell of total charge \(q\) and radius \(R\). +
+ ++Solution 1: use \ref{Gr(2.43)} in version for surface charges: +
+ ++\[ +W = \frac{1}{2} \int \sigma V da. +\] +
+ ++The potential at the surface of the sphere is constant, \(V = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\), so +
+ ++\[ +W = \frac{1}{8\pi \varepsilon_0} \frac{q}{R} \int \sigma da = \frac{1}{8\pi \varepsilon_0} \frac{q^2}{R}. +\] +
+ ++Solution 2: use \ref{eq:Energy_as_int_E2}. Inside sphere: \({\bf E} = 0\); outside, +
+ ++\[ +{\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2} \rightarrow E^2 = \frac{q^2}{(4\pi \varepsilon_0)^2 r^4}. +\] +
+ ++so +\[ + W = \frac{\varepsilon_0}{2} \int_{outside~sphere} \frac{q^2}{(4\pi \varepsilon_0)^2 r^4} r^2 \sin \theta dr d\theta d\phi += \frac{1}{32 \pi^2 \varepsilon_0} q^2 4\pi \int_R^{\infty} \frac{dr}{r^2} = \frac{1}{8\pi \varepsilon_0} \frac{q^2}{R}. +\] +
+ +Created: 2022-02-07 Mon 08:02
+ ++The last version can also be rewritten in terms of the potential, +
+ ++\[ +W = \frac{1}{2} \sum_{i=1}^m q_i V({\bf r}_i) +\label{Gr(2.42)} +\] +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++We have seen that the curl of the electrostatic field vanishes. +In view of the Helmhotz Theorem, this is but one (actually: 3, +since it's a vector equation) of the differential equations we need +to fix \({\bf E}\). We also need the divergence of \({\bf E}\). +
+ ++Feynman's 'bullets flying out'. +'Conservation' related to \(1/r^2\) (think of volume section +in space with walls radially oriented. Flux. +
+ ++If there is no charge within a volume \({\cal V}\) encompassed within the closed +surface \({\cal S}\), +
+ ++\[ +\oint {\bf E} \cdot d{\bf a} = 0. +\] +
+ ++However, if there is a charge inside, we don't get zero. Considering a little +sphere of radius \(r\) around the charge, +
+ ++\[ +\oint {\bf E} \cdot d{\bf a} = \frac{1}{4\pi\varepsilon_0} \int_{\cal S} \frac{q}{r^2} \hat{\bf r} \cdot \hat{\bf r} +r^2 \sin \theta d\theta d\phi = \frac{q}{\varepsilon_0} +\label{Gr(2.12)} +\] +
+ ++so by superposition, we obtain +
++Gauss' law (in integral form) +
+ ++\[ + \oint_{\cal S} {\bf E} \cdot d{\bf a} = \frac{Q_{\mbox{enc}}}{\varepsilon_0} + \label{Gr(2.13)} + \] +
+ ++where \(Q_{\mbox{enc}}\) is the total charge enclosed by \({\cal S}\). +
+ ++By applying the divergence theorem, +
+ ++\[ +\oint_{\cal S} {\bf E} \cdot d{\bf a} = \int_{\cal V} {\boldsymbol \nabla} \cdot {\bf E} ~d\tau +\] +
+ ++and using \(Q_{\mbox{enc}} = \int_{\cal V} \rho d\tau\), and using the fact the the choice of volume +is arbitrary, we get +
++Gauss' law in differential form +
+ ++\[ + {\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0}. + \label{Gr(2.14)} + \] +
+ ++We can also compute the divergence of \({\bf E}\) directly from \ref{eq:E_from_rho}: +
+ ++\[ +{\boldsymbol \nabla} \cdot {\bf E} = \frac{1}{4\pi \varepsilon_0} \int d\tau' \rho({\bf r}') {\boldsymbol \nabla} \cdot \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}. +\] +
+ ++Using \ref{Gr(1.100)}, +
+ ++\[ +{\boldsymbol \nabla} \cdot {\bf E} = \frac{1}{4\pi \varepsilon_0} \int d\tau' \rho({\bf r}') 4\pi \delta ({\bf r} - {\bf r}') = \frac{\rho({\bf r})}{\varepsilon_0} +\label{Gr(2.16)} +\] +
+ ++which is Gauss's law in differential form. +
+ ++As a simple self-consistency exercise, check that you can recover the integral +form by integrating and applying Gauss's divergence theorem. +
++Gauss's law in integral form is really useful when there is spherical, +cylindrical or plane symmetry. +
+ ++Gaussian surfaces: respectively, concentric sphere, coaxial cylinder, pillbox. +
+ ++Example 2.2: Field outside a uniformly charged sphere of radius \(R\) and total charge \(q\). +
+ ++Solution: consider a Gaussian surface which is a sphere of radius \(r > R\). Use spherical symmetry: +
+ ++\[ +\oint_{\cal S} {\bf E} \cdot d{\bf a} = \oint_{\cal S} |{\bf E}| da = |{\bf E}| 4\pi r^2 +\] +
+ ++since \({\bf E} = |{\bf E}| \hat{\bf r}\) and \(d{\bf a} = da \hat{\bf r}\). Therefore, +
+ ++\[ +{\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2}. +\] +
+ ++Same as point charge at origin! +
+ ++Example 2.3: infinitely long cylinder carrying charge density \(\rho = k s\) for some constant \(k\). Find \({\bf E}\) within the cylinder. +
+ ++Solution: Gaussian cylinder of length \(l\) and radius \(s\). Enclosed charge: +
+ ++\[ +Q_{\mbox{enc}} = \int d\tau \rho = \int_0^l dz \int_0^{2\pi} d\phi \int_0^s ds' (k s') s' = 2\pi k l \int_0^s ds' s'^2 = \frac{2\pi}{3} kls^3. +\] +
+ ++Symmetry: \({\bf E}\) must be radially outward, so \({\bf E} = |{\bf E}| \hat{\bf s}\). The surface integral +has no contribution from the ends of the cylinder, and +
+ ++\[ +\oint {\bf E} \cdot d{\bf a} = \int |{\bf E}| da = |{\bf E}| \int da = |{\bf E}| \int_0^l dz \int_0^{2\pi} d\phi s = |{\bf E}| ~2\pi l s. +\] +
+ ++Therefore, +
+ ++\[ +{\bf E} = \frac{1}{3\varepsilon_0} ks^2 \hat{\bf s}. +\] +
+ ++Example 2.4: infinite plane (defined by \(z = 0\)) with uniform surface charge density \(\sigma\). Find \({\bf E}\). +
+ ++Solution: Gaussian pillbox of area \(A = l_x l_y\) extending equal distances above and below plane. +
+ ++Enclosed charge: \(Q_{\mbox{enc}} = \sigma A\). Top and bottom surfaces yield \(\int {\bf E} d{\bf a} = 2A |{\bf E}|\). Sides contribute nothing. Thus, +
+ ++\[ +{\bf E} = \frac{\sigma}{2\varepsilon_0} \hat{\bf n} = \frac{\sigma}{2\varepsilon_0} \frac{z}{|z|} \hat{\bf z} +\label{Gr(2.17)} +\] +
+ ++where \(\hat{\bf n}\) is a unit vector extending away from the plane. Independent of distance from plane! +
+ ++Example 2.5: two infinite planes (put them vertical) carrying equal but opposite uniform surface charge densities \(\pm \sigma\). +
+ ++Solution: fields cancel to left and right of both planes. Between planes: field is \(\frac{\sigma}{\varepsilon_0}\) and points from \(-\) surface to \(+\) one. +
+ +Created: 2022-02-07 Mon 08:02
+ ++As a consequence of \ref{Gr(2.22)}, we immediately see that the work done when travelling +in any closed path vanishes, that is: +
+ + ++ +
+ ++\[ + \oint {\bf E} \cdot d{\bf l} = 0 + \tag{ointE0}\label{ointE0} + \] +
+ ++which by Stokes' theorem implies that the electrostatic field is curlless, +
+ + ++\[ + {\boldsymbol \nabla} \times {\bf E} = 0. + \tag{curlE0}\label{curlE0} + \] +
+ ++By the superposition principle, this therefore applies to any combination +of electrostatic fields. +
+ ++One important point to make is that this fact is only related to the fact that +electrostatic forces are central forces, i.e. they act purely radially +along the line connecting the point charges. It has nothing to do with the fact +that the force falls off with \(1/r^2\): any central force, irrespective of how it +falls off, has the property that the work done is independent of the path. +
+Created: 2022-02-07 Mon 08:02
+ ++The electric field generated by a continuous charge density \(\rho({\bf r})\) in volume \({\cal V}\) can be easily +calculated from Coulomb's law using the superposition principle. Since each infinitesimal +volume element \(d\tau' = dx' dy' dz'\) contains a charge \(dq' = \rho({\bf r}') d\tau'\), we have +
+ + ++ +
+ ++\[ +{\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \rho({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} +\tag{E_vcd}\label{E_vcd} +\] +
+ ++Similarly, if the charge is spread out over a two-dimensional surface \({\cal S}\) with (surface) charge density +\(\sigma({\bf r})\), we have over an infinitesimal area \(da'\) a charge \(dq' = \sigma({\bf r}') da'\), so +
+ + ++ +
+ ++\[ + {\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal S} da' \sigma({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} + \tag{E_scd}\label{E_scd} + \] +
+ ++Finally, for a line path \({\cal P}\) with linear charge density \(\lambda({\bf r}')\), +
+ + ++ +
+ ++\[ + {\bf E} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal P} dl' \lambda({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} + \tag{E_lcd}\label{E_lcd} + \] +
+ ++Example +
+ ++Find the electric field a distance \(z\) above the midpoint of a straight line segment +of length \(2L\) carrying a uniform line charge \(\lambda\). +
+ ++Solution: placing the line on the \(x\) axis, we have +
+ +\begin{align*} + {\bf r} &= z \hat{\bf z}, {\bf r}' = x \hat{\bf x}, dl' = dx, + |{\bf r} - {\bf r}'| = \sqrt{z^2 + {x}^2}, +\end{align*} + ++so we have +
+ +\begin{align*} + {\bf E} = \frac{1}{4\pi \varepsilon_0} \int_{-L}^L dx' \lambda \frac{z \hat{\bf z} - x \hat{\bf x}}{(z^2 + x^2)^{3/2}} + = \frac{\lambda}{4\pi \varepsilon_0} \left[ z \hat{\bf z} \int_{-L}^L dx \frac{1}{(z^2 + x^2)^{3/2}} - \hat{\bf x} \int_{-L}^L dx \frac{x}{(z^2 + x^2)^{3/2}} \right] +\end{align*} + ++The second integral vanishes by symmetry, whereas the first can be done in many ways, +most easily by observing that \(\frac{d}{dx} \left( \frac{x}{\sqrt{z^2 + x^2}} \right) + = \frac{1}{\sqrt{z^2 + x^2}} - \frac{x^2}{(z^2 + x^2)^{3/2}} = \frac{z^2}{(z^2 + x^2)^{3/2}}\), +leading to +
+ + +\begin{align} + {\bf E} = \frac{\lambda}{4\pi \varepsilon_0} z \hat{\bf z} \left. \left( \frac{x}{z^2\sqrt{z^2 + x^2}} \right) \right|_{-L}^L + = \frac{1}{4\pi \varepsilon_0} \frac{2\lambda L}{z \sqrt{z^2 + L^2}} \hat{\bf z}. +\end{align} + ++For large distances \(z \gg L\), this looks like the field of a point charge \(q\lambda L\): +
+ ++\({\bf E} = \frac{1}{4\pi \varepsilon_0} \frac{2\lambda L}{z^2} \hat{\bf z}\), +
+ ++whereas for short distances \(z \ll L\) the field looks like that of an infinite wire: +
+ ++\({\bf E} = \frac{1}{4\pi \varepsilon_0} \frac{2\lambda}{z}\). +
+ +Created: 2022-02-07 Mon 08:02
+ ++Our tendency is to think that force acts locally, not remotely. +It is thus natural to imagine a "vehicle" through which one charge exerts +a force on another. This leads us to an important first abstraction, that of +the electrostatic field. +
+ ++Going back to our source and test charge setup, we shall thus think of the +force on the test charge as being given by its coupling to +an electric field originating from the source charge. Invoking the +superposition principle, we can thus write +
+ ++ +
+ ++\[ + {\bf F}_t = q_t {\bf E}({\bf r}_t) + \label{Gr(2.3)} + \] +
+ ++where +
++ +
+ ++\[ + {\bf E} ({\bf r}) \equiv \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i \frac{{\bf r} - {\bf r}_i}{|{\bf r} - {\bf r}_i|^3} + \label{Gr(2.4)} + \] +
+ ++The electric field \({\bf E} ({\bf r})\) is thus the force per unit charge that would +be exerted if you put a test charge at position \({\bf r}\). +
+ ++Example +
+ ++Find the field at distance \(z\) above the midpoint between two equal +charges \(q\) placed a distance \(d\) apart from each other. +
+ ++Solution: by symmetry, only the \(\hat{\bf z}\) part of the field is nonvanishing. +We have \(|{\bf r} - {\bf r}'| = \sqrt{z^2 + (d/2)^2}\) for both charges and \(\cos \theta = z/|{\bf r} - {\bf r}'|\) so +
+ ++\[ + {\bf E} = \frac{1}{4\pi \varepsilon_0} \frac{2q z}{|{\bf r} - {\bf r}'|^{3/2}} \hat{\bf z} +\] +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++The force \(F_{t\leftarrow s}\) exerted by a point source charge \(q_s\) sitting at \({\bf r}_s\) +on a point test charge \(q_t\) sitting at \({\bf r}_t\) is given by Coulomb's law, +
+ ++\[ +{\bf F}_{t\leftarrow s} = \frac{q_t q_s}{4\pi \varepsilon_0} +\frac{{\bf r}_t - {\bf r}_s}{|{\bf r}_t - {\bf r}_s|^3} += \frac{q_t q_s}{4\pi \varepsilon_0} \frac{\hat{\bf r}_{ts}}{r_{ts}^2} +\tag{Cl}\label{Cl} +\] +
+ ++in which \(\varepsilon_0\), called the permittivity of free space +(or alternately vacuum permittivity or sometimes electric constant), takes the value +
+ ++\[ + \varepsilon_0 \equiv \frac{1}{\mu_0 c^2} = \frac{1}{35 950 207 149.472 7056 \pi} F/m \\ + \simeq 8.854 187 817 ... \times 10^{-12} F/m +\] +
+ ++in which \(\mu_0\) is the vacuum permeability (or alternately permeability of free space, +permeability of vacuum or magnetic constant), +
+ ++\[ + \mu_0 = 1.25663706212(19)x10^{-6} H/m +\] +
+ ++with the henry \(H = kg~m^2 / s^2 A^2\) being the unit for inductance. +
+ ++Info: the International System of Units +
+ ++Throughout these notes we will use the International System of Units (SI). +Units are perhaps less set in stone than you might think, because of their interdependence. +For example, the speed of light is a measurable constant of nature, and therefore one must +be careful when defining the metre and the second. +
+ ++Since 1960, the speed of light is defined to be exactly \(299 792 458 ~m/s\). +Since 1983, the metre is defined as the distance travelled by light in +\(1/299 792 458 s\). +
+ ++In the good old days (namely up to 2019), the vacuum permeability was defined +to have the handy value \(\mu_0 = 4\pi \times 10^{-7} H/m\). This is no longer true. +
+ ++The 2019 SI redefinition uses seven base units: second (s), metre (m), +kilogram (kg), Ampere (A), kelvin (K), mole (mol) +and candela (cd), all defined in terms of invariant constants of nature. +
+ ++The vacuum permeability is thus no longer a defined constant and must be measured +experimentally (it depends on the fine structure constant +as \(\mu_0 = 2\alpha \frac{h}{e^2 c}\)). Its currently measured value is given in the main text. +
+ +Created: 2022-02-07 Mon 08:02
+ ++Consider some distribution of charge which produces an electric field. +How much work is needed to carry a small test charge \(q_t\) from one place to another, i.e. from point \({\bf a}\) to point \({\bf b}\) ? +
+ ++The work done in carrying this charge along some path is the negative of +the electrical force in the direction of motion, so +
+ ++\[ +W = -\int_{{\bf a}}^{{\bf b}} {\bf F} \cdot d{\bf l} +\] +
+ ++Let's consider for simplicity a fixed source particle of charge \(q_s\) at position \({\bf r}_s \equiv {\bf 0}\). The work done against electrical forces when moving a unit charge test particle from \({\bf a}\) to \({\bf b}\) is then: +
+ ++\[ +-\int_{\bf a}^{\bf b} {\bf F} \cdot d{\bf l} += -\frac{q_t q_s}{4\pi \varepsilon_0} \int_{\bf a}^{\bf b} \frac{\bf r}{r^3} \cdot d{\bf r} +\] +
+ ++For the path, the 'angular' part is not contributing (see drawings in +FLS II 4-3). The integral is thus purely radial, +
+ ++As should be clear by now, this result does not depend on the path +(if it did, we'd have a perpetuum mobile when going from \({\bf r}_a\) +to \({\bf r}_a\) one way, and coming back another). +
+ ++When thinking about the energy of this pair of charges, we think +of starting from an initial configuration where the charges are infinitely +distance (which we associate to zero energy), and thus set \({\bf r}_a = \infty\). +We can thus write +
+ ++\[ +W = \frac{1}{4\pi \varepsilon_0} \frac{q_t q_s}{|{\bf r}_t - {\bf r}_s|} +\] +
+ + ++Example: fission +
+ ++By the superposition principle, the energy of a generic assembly of +charges \(\{ q_i \}\), \(i = 1, .. n\) (sitting at positions \({\bf r}_i\)) +is obtained by the pairwise sum (counting each pair only once) +of pairwise energies: +
+ ++\[ +W = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n \sum_{j > i}^n \frac{q_i q_j}{|{\bf r}_i - {\bf r}_j|} += \frac{1}{8\pi \varepsilon_0} \sum_{i=1}^n \sum_{j \neq i}^n \frac{q_i q_j}{|{\bf r}_i - {\bf r}_j|} +\label{Gr(2.41)} +\] +
+ ++where in the second equality we have symmetrized the sums for convenience. +
++PM3:1.6 +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Electrical forces obey the principle of superposition: +the electrostatic interaction between two charged particles is left entirely +unaffected by the presence of other charges. The total force on a +test charge \(q_t\) generated by a number of static point charges \(q_i\), \(i = 1, ..., n\) is thus +
+ ++\[ +{\bf F}_t = \sum_{i=1}^n {\bf F}_{t \leftarrow i} +\] +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Our two fundamental equations for the electrostatic field are +\[ +{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0} +\hspace{2cm} +{\boldsymbol \nabla} \times {\bf E} = 0. +\] +For the electrostatic potential, Gauss' law becomes +
++The Poisson equation +\[ + {\boldsymbol \nabla}^2 V = -\frac{\rho}{\varepsilon_0} + \label{eq:Poisson} + \] +
+ ++When the charge density vanishes, it becomes more simply +
++The Laplace Equation +\[ + {\boldsymbol \nabla}^2 V = 0 + \label{eq:Laplace} + \] +
+ ++Since the curl of a gradient is always zero, we by construction have +\({\boldsymbol \nabla} \times {\bf E} = - {\boldsymbol \nabla} \times ({\boldsymbol \nabla} V) = 0\). +
+Created: 2022-02-07 Mon 08:02
+ ++For a surface, Gauss's law states +\[ +\oint_{\cal S} {\bf E} \cdot d{\bf a} = \frac{Q_{\mbox{enc}}}{\varepsilon_0} = \frac{1}{\varepsilon_0} \sigma A +\] +where \(A\) is the area of the Gaussian pillbox and \(\sigma\) the surface charge density. +The sides contribute nothing if the pillbox is thin. Taking its area very small, we get +\[ +{\bf E}^{\perp}_{\mbox{above}} - {\bf E}^{\perp}_{\mbox{below}} = \frac{\sigma}{\varepsilon_0}, +\label{Gr(2.31)} +\] +so the normal component of \({\bf E}\) is discontinuous at the boundary by an amount \(\sigma/\varepsilon_0\). +
+ ++The tangential component is continuous: from the curlless condition \ref{Gr(2.19)} applied to +a small loop straddling the surface, +\[ +{\bf E}^{\parallel}_{\mbox{above}} = {\bf E}^{\parallel}_{\mbox{below}} +\label{Gr(2.32)} +\] +
+ ++Put together: +\[ +{\bf E}_{\mbox{above}} - {\bf E}_{\mbox{below}} = \frac{\sigma}{\varepsilon_0} \hat{\bf n} +\label{Gr(2.33)} +\] +with \(\hat{\bf n}\) a unit vector normal to the surface, pointing 'out'. +
+ ++The potential is continuous across any boundary: since +\[ +V_{\mbox{above}} - V_{\mbox{below}} = -\int_{\bf a}^{\bf b} {\bf E} \cdot d{\bf l} +\] +where the path shrinks to zero, +\[ +V_{\mbox{above}} = V_{\mbox{below}} +\label{Gr(2.34)} +\] +The gradient however inherits the discontinuity of the electrostatic field, since +\({\bf E} = -{\boldsymbol \nabla} V\): +\[ +{\boldsymbol \nabla} V_{\mbox{above}} - {\boldsymbol \nabla} V_{\mbox{below}} = -\frac{\sigma}{\varepsilon_0} \hat{\bf n} +\label{Gr(2.35)} +\] +or +\[ +\frac{\partial V_{\mbox{above}}}{\partial n} - \frac{\partial V_{\mbox{below}}}{\partial n} = -\frac{\sigma}{\varepsilon_0} +\label{Gr(2.36)} +\] +where +\[ +\frac{\partial V}{\partial n} = {\boldsymbol \nabla} V \cdot \hat{\bf n} +\label{Gr(2.37)} +\] +is the normal derivative of the potential. +
+ ++This is the kind of boundary condition that we need to fix a unique solution to Poisson's equation: +our only problem is that \ref{Gr(2.36)} gives the change of the normal derivative of \(V\), not +its value. However, if we assume (as in our first case corollary) that there are no charges living outside +of our volume \({\cal V}\), we find that \ref{Gr(2.36)} fully specifies the potential's normal derivative +if the surface charge is known. +
+Created: 2022-02-07 Mon 08:02
+ ++(i) The name. It's fine by me. I don't see what Griffiths' problem is +(more precisely: Griffiths sees a problem where there isn't one). +(ii) Advantages. In view of \ref{Gr(2.23)}, it's much simpler to first calculate +the potential (which is a scalar field), and to then calculate the electrostatic field +(which is a vector field). +
+ ++But how can a vector field (3 components, so 3 functions) be generated by a scalar field +(one component, so one function) ? Griffiths mentions that \ref{Gr(2.20)} explains this. +For curious students, resolve this paradox: \ref{Gr(2.20)}, when written out in +components, gives 3 functional constraints on the electric field. We might thus think that +we have no freedom left. How come we thus end up with a 'one function' degree of freedom ? +(iii) The reference point. Not a problem to put it at infinity if we only use +charge distributions which don't go to infinity. Otherwise, choose a different point! +(iv) Superposition principle. Already mentioned. +(v) Units. Newton-meters per coulomb or joules per coulomb, which is called a volt. +In the SI system where the ampere is a base unit, a volt is a \(kg m^2/s^3 A\). +
+Created: 2022-02-07 Mon 08:02
+ ++Since this work is independent of the path chosen, we can define a function called the electrostatic potential +
+ ++\[ +V({\bf r}) \equiv -\int_{\cal O}^{\bf r} {\bf E} \cdot d{\bf l} +\tag{es_pot}\label{es_pot} +\] +where \({\cal O}\) is some chosen reference point. The potential difference between two points is +well-defined without the need to specify the reference point, +
++\[ + V({\bf b}) - V({\bf a}) = - \int_{\bf a}^{\bf b} {\bf E} \cdot d{\bf l} + \label{Gr(2.22)} + \] +
+ ++Thus, the electrostatic potential is interpreted as the potential energy which a unit charge would +have obtained if brought to the specified point from the reference point, in other words the work you need to +do on the unit charge to bring it there. +Often, we put the reference point at infinity. The electrostatic potential coming from a single +point charge \(q\) at the origin then becomes +\[ +V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{q}{r} +\label{Gr(2.26)} +\] +The electrostatic potential moreover inherits the superposition principle from the electric field, +so for a distribution of point charges \(q_i\) at positions \({\bf r}_i\), we have +
++\[ + V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i} \frac{q_i}{|{\bf r} - {\bf r}_i|} + \label{Gr(2.27)} + \] +
+ ++For a continuous charge density in a volume \({\cal V}\), we have +
++\[ + V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\cal V} d\tau' \frac{\rho({\bf r}')}{|{\bf r} - {\bf r}'|}, + \label{eq:V_from_rho} + \] +
+ ++whereas for a surface or line charge distribution, respectively, +
++\[ + V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\cal S} da' \frac{\sigma({\bf r}')}{|{\bf r} - {\bf r}'|}, + \hspace{2cm} + V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int_{\cal P} dl' \frac{\lambda({\bf r}')}{|{\bf r} - {\bf r}'|}. + \label{Gr(2.30)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Example 2.6: find the potential inside and outside a spherical shell or radius \(R\) +centered at origin, which carries uniform surface charge. Ref point at infinity. +Solution: from Gauss's law, +\[ +{\bf E} = \frac{1}{4\pi \varepsilon_0} q \frac{\hat{\bf r}}{r^2}, \hspace{5mm} r > R, \hspace{2cm} +{\bf E} = 0, \hspace{5mm} r < R. +\] +Thus: using spherical symmetry, for \(r > R\): +\[ +V(r > R) = -\int_{{\cal O}: \infty}^{\bf r} {\bf E} \cdot d{\bf l} = -\frac{1}{4\pi \varepsilon_0} \int_{\infty}^r q \frac{dr'}{r'^2} += \frac{1}{4\pi \varepsilon_0} \frac{q}{r}, \hspace{5mm} r > R. +\] +For \(r < R\), the integrand vanishes for the part \(r < R\), and +\[ +V(r < R) = V(R) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}. +\] +
+ ++Example 2.7: find the potential of a uniformly charged spherical shell of radius \(R\) (same problem as 2.6, but now done with \ref{Gr(2.30)}). +Solution: Use +\[ +V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \int da' \frac{\sigma}{|{\bf r} - {\bf r}'|}. +\] +Put \({\bf r}\) on \(\hat{\bf z}\) axis so \({\bf r} = z \hat{\bf z}\), use law of cosines: +\[ +
+| {\bf r} - {\bf r}' | += R2 + z2 - 2Rz cos θ' | +
+\] +Element of surface area: \(R^2 \sin \theta' d\theta' d\phi'\), so +
+\begin{align} +4\pi \varepsilon_0 V(z) &= \sigma \int_0^{\pi} d\theta' \int_0^{2\pi} d\phi' \frac{R^2 \sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\ +&= 2\pi R^2 \sigma \int_0^{\pi} d\theta' \frac{\sin \theta'}{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}} \nonumber \\ +&= 2\pi R^2 \sigma \left. \left( \frac{\sqrt{R^2 + z^2 - 2Rz \cos \theta'}}{Rz} \right)\right|_0^{\pi} \nonumber \\ +&= \frac{2\pi R\sigma}{z} \left(\sqrt{R^2 + z^2 + 2Rz} - \sqrt{R^2 + z^2 - 2Rz} \right) \nonumber \\ +&= \frac{2\pi R \sigma}{z} \left( \sqrt{(R + z)^2} - \sqrt{(R - z)}^2 \right). +\end{align} ++Take positive root: +
+\begin{align} +V(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (z-R)) = \frac{R^2 \sigma}{\varepsilon_0 z}, \hspace{1cm} \mbox{outside}, \nonumber \\ +V(z) = \frac{R\sigma}{2\varepsilon_0 z} ((R+z) - (R-z)) = \frac{R \sigma}{\varepsilon_0}, \hspace{1cm} \mbox{inside}. +\end{align} ++In terms of total charge on shell, \(q = 4\pi R^2 \sigma\), \(V(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{z}\) outside, +and \(V(z) = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}\) inside. +
+ +Created: 2022-02-07 Mon 08:02
+ ++Relation espot gives us the potential in terms of the field. Let us now try to find +a relationship giving the field in terms of the potential. For this, we look again at +\ref{Gr(2.22)}. By the fundamental theorm for gradients, this also equals +\[ +V({\bf b}) - V({\bf a}) = \int_{\bf a}^{\bf b} ({\boldsymbol \nabla} V) \cdot d{\bf l} +\] +so +\[ +\int_{\bf a}^{\bf b} ({\boldsymbol \nabla} V) \cdot d{\bf l} = - \int_{\bf a}^{\bf b} {\bf E} \cdot d{\bf l} +\] +Since this is true for any choice of \({\bf a}\) and \({\bf b}\), we have +
++\[ + {\bf E} = -{\boldsymbol \nabla} V + \label{Gr(2.23)} + \] +
+ ++We can explicitly check \ref{Gr(2.23)} by 'extracting' the \({\boldsymbol \nabla}\) operator from the +integral in the definition of the field in e.g. \ref{Gr(2.4)}: +\[ +{\bf E} ({\bf r}) = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i \frac{{\bf r} - {\bf r}_i}{|{\bf r} - {\bf r}_i|^3} += \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i (-1) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}_i|} += - {\boldsymbol \nabla} \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{|{\bf r} - {\bf r}_i|} += - {\boldsymbol \nabla} V +\] +by using \ref{Gr(1.101)} and \ref{Gr(2.27)}. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Steady currents lead to constant magnetic fields: {\bf magnetostatics}. +
+ ++Then, since \(\frac{\partial \rho}{\partial t} = 0\), the continuity equation leads to +\[ +{\boldsymbol \nabla} \cdot {\bf J} = 0 +\label{Gr(5.31)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++This is given experimentally (around 1820) by the +
++{\bf Biot-Savart law} +\[ + {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} + \label{eq:BiotSavart} + \] +
+ ++in which \(\mu_0\) is the vacuum permeability (or alternately permeability of free space, +permeability of vacuum or magnetic constant), +\[ + \mu_0 = 1.25663706212(19)x10^{-6} H/m +\] +with the {\it henry} \(H = kg m^2 / s^2 A^2\) being the unit for inductance. +
+ ++For surface and volume density currents: +
++\[ + {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3}, + \hspace{2cm} + {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{{\bf J} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} + \label{Gr(5.39)} + \] +
+ ++{\bf N.B.:} there is no Biot-Savart law for a single point charge. Not steady current ! +
+ + ++The {\bf superposition principle} applies here as well: collection of currents generates a \({\bf B}\) field which is +the vector sum of the fields generated by the individual currents. +
+ ++\paragraph{Example 5.5:} find \({\bf B}\) a distance \(s\) from a long straight wire carrying steady current \(I\). +\paragraph{Solution:} {\bf Gr Fig 5.18}: +\(dl {\bf I} \times ({\bf r} - {\bf r}')\) points out of the page (blackboard !), and has +magnitude \(dl' \sin \alpha = dl' \cos \theta\). But \(l' = s \tan \theta\) so \(dl' = \frac{s}{\cos^2 \theta} d\theta\), +and \(s = |{\bf r} - {\bf r}'| \cos \theta\). Then, +\[ + B = \frac{\mu_0}{4\pi} I \int_{\theta_1}^{\theta_2} d\theta \cos \theta \frac{\cos^2 \theta}{s^2} \frac{s}{\cos^2 \theta} + = \frac{\mu_0 I}{4\pi s} (\sin \theta_2 - \sin \theta_1) + \label{Gr(5.35)} + \] +For infinite wire: \(\theta_1 = -\pi/2\), \(\theta_2 = \pi/2\), so +\[ + B = \frac{\mu_0 I}{2\pi s} + \label{Gr(5.36)} + \] +
+ ++Immediate consequence: force per unit length between two wires with currents \(I_1\) and \(I_2\) separated by distance \(d\): +\[ +f = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d} +\label{Gr(5.37)} +\] +(like currents attract). +
+ ++\paragraph{Example 5.6:} find {\bf B} a distance \(z\) above the center of a circular loop of radius \(R\), +carrying a steady counterclockwise current \(I\). +\paragraph{Solution:} Gr Fig 5.21. By symmetry, only the vertical component doesn't cancel. +\[ + B(z) = \frac{\mu_0 I}{4\pi} \int dl' \frac{\cos \theta}{|{\bf r} - {\bf r}'|} + = \frac{\mu_0 I}{4\pi} \frac{\cos \theta}{R^2 + z^2} \int dl' = \frac{\mu_0 I}{2} \frac{R^2}{(R^2 + z^2)^{3/2}} + \label{Gr(5.38)} + \] +(since \(\cos \theta = R/\sqrt{R^2 + z^2}\)). +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++\[ +{\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' \frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} +\label{Gr(5.45)} +\] +Apply the divergence: +\[ +{\boldsymbol \nabla} \cdot {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\boldsymbol \nabla} \cdot +\left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) +\label{Gr(5.46)} +\] +Note that we can write (using \ref{Gr(1.101)}) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). Substituting this in and using product rule number 6, +\[ +{\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf A}) - {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}), +\] +we get +
+\begin{align} +{\boldsymbol \nabla} \cdot \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) +&= -{\boldsymbol \nabla} \cdot \left( {\bf J}({\bf r}') \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \right) \nonumber \\ +&= -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \cdot \bigl({\boldsymbol \nabla} \times {\bf J} ({\bf r}') \bigr) ++ {\bf J} ({\bf r}') \cdot \left({\boldsymbol \nabla} \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) +\label{Gr(5.47)} +\end{align} ++But \({\bf J}\) depends only on \({\bf r}'\) so \({\boldsymbol \nabla} \times {\bf J} ({\bf r}') = 0\), and since +the curl of a gradient always vanishes, we obtain +
++\[ + {\boldsymbol \nabla} \cdot {\bf B} = 0 + \label{eq:DivBisZero} + \] +
+ ++The divergence of a magnetic field is always zero. +Said equivalently: there are no magnetic charges. +
+ ++We can play the same trick with the curl: +\[ +{\boldsymbol \nabla} \times {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~{\boldsymbol \nabla} \times +\left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) +\label{Gr(5.49)} +\] +Do as above but now use product rule 8 +\[ +{\boldsymbol \nabla} \times ({\bf A} \times {\bf B}) = ({\bf B} \cdot {\boldsymbol \nabla}) {\bf A} - ({\bf A} \cdot {\boldsymbol \nabla}) {\bf B} + {\bf A} ({\boldsymbol \nabla} \cdot {\bf B}) - {\bf B} ({\boldsymbol \nabla} \cdot {\bf A}) +\] +and (dropping terms involving derivatives of \({\bf J}({\bf r}')\) with respect to \({\bf r}\), +which vanish): +
+\begin{align} +{\boldsymbol \nabla} \times \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) +&= -{\boldsymbol \nabla} \times \left( {\bf J}({\bf r}') \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \right) \nonumber \\ +&= ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} +-{\bf J} ({\bf r}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) +\label{Gr(5.50)} +\end{align} ++Last term: +\[ +-{\bf J} ({\bf r}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) += -{\bf J} ({\bf r}') \left({\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|}\right) = -{\bf J} ({\bf r}') \left(-4\pi \delta^{(3)} ({\bf r} - {\bf r}') \right) +\label{Gr(5.51)} +\] +where we have used \ref{Gr(1.102)}. This term thus integrates to +\[ +\frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf J} ({\bf r}') 4\pi \delta^{(3)} ({\bf r} - {\bf r}') = \mu_0 {\bf J}({\bf r}). +\] +To treat the other term, we use +\[ +({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} += -({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}') {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} +\] +to rewrite the second part of the integral as +\[ +-\frac{\mu_0}{4\pi} {\boldsymbol \nabla} \int_{\cal V} d\tau' ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}') \frac{1}{|{\bf r} - {\bf r}'|} += \frac{\mu_0}{4\pi} {\boldsymbol \nabla} \int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \cdot {\bf J} ({\bf r}')}{|{\bf r} - {\bf r}'|} += 0 +\] +where in the second step we have used integration by parts using product rule 5 +\[ +{\boldsymbol \nabla} \cdot (f {\bf A}) = f ({\boldsymbol \nabla} \cdot {\bf A}) + {\bf A} \cdot ({\boldsymbol \nabla} f) +\] +(the surface term vanishes because we take \({\bf J} \rightarrow 0\) +at infinity), and in the third step we have used the assumption of steady-state so \({\boldsymbol \nabla}' \cdot {\bf J} = 0\). +
+ ++We thus obtain in total +
++Ampère's law + \[ + {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} + \label{eq:Ampere} + \] +
+ ++(in differential form). Using Stokes' theorem, +\[ +\int_{\cal S} {\boldsymbol \nabla} \times {\bf B} \cdot d{\bf a} = \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 \int_{\cal S} {\bf J} \cdot d{\bf a} +\] +so +
++\[ + \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \hspace{2cm} + I_{enc} = \int_{\cal S} {\bf J} \cdot d{\bf a}. + \label{Gr(5.55)} + \] +
+ ++where \(I_{enc}\) is the current enclosed in the {\bf amperian loop} \({\cal P}\) which defines the boundary of surface \({\cal S}\). +
+ ++Sign ambiguity: resolved by right-hand rule as usual. +
+ ++Ampère's law in magnetostatics takes a parallel role to Gauss's law in electrostatics. +
+ ++\paragraph{Example 5.7:} same as Example 5.5, but now with Ampère. +\paragraph{Solution:} by symmetry, \({\bf B}\) is circumferential and can only depend on \(s\). Then, +choosing an amperian loop at a fixed radius \(s\), we get +\[ + \oint {\bf B} \cdot d{\bf l} = B 2\pi s = \mu_0 I ~~\Rightarrow~~ B = \frac{\mu_0 I}{2\pi s} + \] +
+ ++\paragraph{Example 5.8:} uniform surface current \({\bf K} = K \hat{\bf x}\) flowing in \(xy\) plane. +\paragraph{Solution:} Biot-Savart: \({\bf B}\) must be perpendicular to \({\bf K}\). Intuition: +\({\bf B}\) cannot have a component along \(\hat{\bf z}\). Right-hand rule: \({\bf B}\) along \(-\hat{\bf y}\) for \(z > 0\), +and along \(\hat{\bf y}\) for \(z < 0\). Amperian loop of width \(l\) punching through surface: +\[ + \oint {\bf B} \cdot d{\bf l} = 2B l = \mu_0 I_{enc} = \mu_0 K l + ~~\Rightarrow~~ + {\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0}{2} K \hat{\bf y}, & z < 0, \\ + -\frac{\mu_0}{2} K \hat{\bf y}, & z > 0. \end{array} \right. + \label{Gr(5.56)} + \] +
+ ++\paragraph{Example 5.9:} solenoid along \(\hat{\bf z}\), wire carrying current \(I\) doing \(n\) turns per unit length on cylinder of radius \(R\). +\paragraph{Solution:} by symmetry, \({\bf B}\) must be along axis of solenoid. Outside: infinitely far away, \({\bf B}\) must vanish. +But an amperian loop outside gives zero always, so \({\bf B}\) vanishes everywhere outside the solenoid. +Amperian loop of length \(l\), half-inside and half-outside: +\[ + \oint {\bf B} \cdot d{\bf l} = Bl = \mu_0 I_{enc} = \mu_0 I n l ~~\Rightarrow~~ + {\bf B} = \left\{ \begin{array}{cc} \mu_0 I n \hat{\bf z}, & s < R, \\ + 0, & s > R \end{array} \right. + \label{Gr(5.57)} + \] +
+ ++\paragraph{Situations where Ampère's law can be useful:} +i) infinite straight lines, ii) infinite planes, iii) infinite solenoids, iv) toroids. +
+ ++\paragraph{Example 5.10:} toroidal coil (no matter the shape, as long as it is rotationally symmetric). +\paragraph{Solution:} magnetic field is circumferential everywhere. Outside coil, field again zero. +Amperian loop half inside, half outside: +\[ + B 2\pi s = \mu_0 I_{enc} ~~\Rightarrow~~ + {\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0 N I}{2\pi s} \hat{\boldsymbol \phi}, & \mbox{inside coil}, \\ + 0, & \mbox{outside} \end{array} \right. + \label{Gr(5.58)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Infinite straight wire: calculate line integral of \({\bf B}\) along circular path of radius \(s\) centered on wire: +from \ref{Gr(5.36)}, +\[ +\oint {\bf B} \cdot d{\bf l} = \oint \frac{\mu_0 I}{2\pi s} dl = \mu_0 I. +\] +Any loop will do: +\[ +{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \phi} +\label{Gr(5.41)} +\] +and \(d{\bf l} = ds \hat{\bf s} + s d\phi \hat{\boldsymbol \phi} + dz \hat{\bf z}\) so for a loop encircling the wire once, +\[ +\oint {\bf B} \cdot d{\bf l} = \frac{\mu_0 I}{2\pi} \oint \frac{1}{s} s d\phi = \mu_0 I. +\] +
+ ++For a general collection of straight wires: +\[ +\oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} +\label{Gr(5.42)} +\] +where \(I_{enc}\) is the current flow through a surface \({\cal S}\) defined by the closed path \({\cal P}\): +\[ +I_{enc} = \int_{\cal S} {\bf J} \cdot d{\bf a} +\label{Gr(5.43)} +\] +Applying Stokes' theorem: +
++\[ + {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} + \label{Gr(5.44)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Consider a little surface \(\Delta {\cal S}\) having a normal unit vector \(\hat{\bf n}\). +
+ ++We start by defining a {\bf current density} \({\bf J}\) as the vector representing the amount of +charge flowing through a unit area per unit time. Its direction is along the motion of the charges. +
+ ++The charge flowing through \(\Delta {\cal S}\) in one unit of time is +thus \(\Delta q = {\bf J} \cdot {\bf n} ~\Delta {\cal S} ~\Delta t\). +
+ ++Let the flow of charge be given by a charge density \(\rho\) moving at velocity \({\bf v}\). +Then, \(\Delta q = \rho {\bf v} \cdot {\bf n} ~\Delta {\cal S} ~\Delta t\) so we can identify +\[ +{\bf J} = \rho {\bf v} +\label{Gr(5.26)} +\] +The total current \(I\) going through a surface \({\cal S}\) is +\[ +I = \int_{\cal S} {\bf J} \cdot d{\bf a} +\label{Gr(5.28)} +\] +Over a closed surface, we can use the divergence theorem, +\[ +\oint_{\cal S} {\bf J} \cdot d{\bf a} = \int_{\cal V} d\tau {\boldsymbol \nabla} \cdot {\bf J} +\] +Since charge is conserved, +\[ +\int_{\cal V} d\tau {\boldsymbol \nabla} \cdot {\bf J} = -\frac{d}{dt} \int_{\cal V} d\tau \rho += - \int_{\cal V} d\tau \frac{\partial \rho}{\partial t}. +\] +Since this is valid for any volume, we get the +
++continuity equation +\[ + {\boldsymbol \nabla} \cdot {\bf J} + \frac{\partial \rho}{\partial t} = 0 + \label{eq:continuity} + \] +
+ ++Current down a wire: linear charge density \(\lambda\) moving at velocity \({\bf v}\) means current +\[ +{\bf I} = \lambda {\bf v} +\label{Gr(5.14)} +\] +Current on a surface: {\bf surface current density} made up of surface charge density \(\sigma\) +moving at velocity \({\bf v}\) +\[ +{\bf K} = \sigma {\bf v} +\label{Gr(5.23)} +\] +
+ ++Forces on wire, surface and volume carrying current densities: +\[ + {\bf F}_{mag} = \int {\bf v} \times {\bf B} ~\lambda dl = \int |{\bf I}| ~d{\bf l} \times {\bf B} + \label{Gr(5.16)} + \] +\[ + {\bf F}_{mag} = \int {\bf v} \times {\bf B} ~\sigma da = \int {\bf K} \times {\bf B} ~da + \label{Gr(5.24)} + \] +\[ + {\bf F}_{mag} = \int {\bf v} \times {\bf B} ~\rho d\tau = \int {\bf J} \times {\bf B} ~d\tau + \label{Gr(5.27)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Force on a point charge \(q\) moving at velocity \({\bf v}\) in magnetic field \({\bf B}\): +
++\[ + {\bf F}_{mag} = q {\bf v} \times {\bf B} + \label{eq:LorentzForce} + \] +
+ ++Units of \({\bf B}\): \(N/(A~m)\) is called a {\bf tesla} (symbol: \(T\)). Total electromagnetic force: +
++\[ + {\bf F}_{mag} = q ({\bf E} + {\bf v} \times {\bf B}) + \label{eq:EMForce} + \] +
+ ++\paragraph{Example 5.1:} cyclotron motion. Field \({\bf B}\) pointing into page. Charge \(q > 0\) moves +counterclockwise with speed \(v\) on a circle of radius \(R\). Magnetic force points inwards. +Equating, obtain the {\bf cyclotron formula} +\[ + q v B = m \frac{v^2}{R} ~~\rightarrow~~ p = mv = q B R. + \label{Gr(5.3)} + \] +The {\bf cyclotron frequency} is +\[ + \omega = 2\pi \frac{v}{2\pi R} = \frac{q B}{m} + \label{Gr(5.4)} + \] +
+ ++\paragraph{Example 5.2:} cycloid motion. Recommendation: {\it look at it!!} +
+ ++Important point: {\bf magnetic forces do no work}. Work: +\[ +dW_{mag} = {\bf F}_{mag} \cdot d{\bf l} = q ({\bf v} \times {\bf B}) \times {\bf v} dt = 0 +\label{Gr(5.11)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Since \({\boldsymbol \nabla} \cdot {\bf B} = 0\) in magnetostatics, following Helmholtz's theorem we can write +
++\[ + {\bf B} = {\boldsymbol \nabla} \times {\bf A} + \label{Gr(5.59)} + \] +
+ ++Connection with Ampère's law: +\[ +{\boldsymbol \nabla} \times {\bf B} = {\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf A} +) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf A}) - {\boldsymbol \nabla}^2 {\bf A} += \mu_0 {\bf J} +\label{Gr(5.60)} +\] +Electrostatics: you could add any constant to electrostatic potential. Here: you can +add any curlless function (so gradient of a scalar field) to the vector potential, +without changing the magnetic field. This is called a {\bf gauge choice} in electrodynamics. +For example, we can {\bf always} eliminate the divergence of \({\bf A}\), +
++{\bf Example gauge choice:} +\[ + {\boldsymbol \nabla} \cdot {\bf A} = 0. + \label{Gr(5.61)} + \] +
+ ++Proof: suppose our starting \({\bf A}_0\) is not divergenceless. We add \({\boldsymbol \nabla} \lambda\) +to the vector potential, so \({\bf A} = {\bf A}_0 + {\boldsymbol \nabla} \lambda\). Then, +\[ +{\boldsymbol \nabla} \cdot {\bf A} = {\boldsymbol \nabla} \cdot {\bf A}_0 + {\boldsymbol \nabla}^2 \lambda. +\] +The scalar field then obeys a Poisson-like equation, +\[ +{\boldsymbol \nabla}^2 \lambda = -{\boldsymbol \nabla} \cdot {\bf A}_0, +\] +whose solution we know how to find. Provided \({\boldsymbol \nabla} \cdot {\bf A}_0\) goes to +zero at infinity, +\[ +\lambda ({\bf r}) = \frac{1}{4\pi} \int d\tau' \frac{{\boldsymbol \nabla}' \cdot{\bf A}_0 ({\bf r}')}{|{\bf r} - {\bf r}'|}. +\] +{\it Provide proof in one line using Laplacian leading to delta function.} +
+ ++Under this gauge choice, Ampère's law becomes +
++\[ + {\boldsymbol \nabla}^2 {\bf A} = -\mu_0 {\bf J} + \label{Gr(5.62)} + \] +
+ ++Note: this is a Poisson equation for each component. +For currents falling off sufficiently rapidly at infinity, +
++\[ + {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{J({\bf r}')}{|{\bf r} - {\bf r}'|} + \label{Gr(5.63)} + \] +
+ ++For line and surface currents, (beware Griffiths' horrendous notation) +
++\[ + {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I ({\bf r}')}}{|{\bf r} - {\bf r}'|}, + \hspace{2cm} + {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K ({\bf r}')}}{|{\bf r} - {\bf r}'|}. + \label{Gr(5.64)} + \] +
+ ++\paragraph{Example 5.11:} a spherical shell of radius \(R\), carrying a uniform surface charge +\(\sigma\), is set spinning at angular velocity \(\omega\). Find the vector potential at \({\bf r}\). +\paragraph{Solution:} do it by yourselves. Fun conclusion: the field inside +the sphere is uniform ! +\[ + {\bf B} = \frac{2}{3} \mu_0 \sigma R {\boldsymbol \omega}. + \label{Gr(5.68)} + \] +
+ ++\paragraph{Example 5.12:} find the vector potential of an infinite solenoid with \(n\) turns +pet unit length, radius \(R\) and current \(I\). +\paragraph{Solution:} cannot use \ref{Gr(5.64)} since the current extends to infinity +({\bf Comment:} check that the integral converges anyway, by combining the integrals +for \(z > 0\) and \(z < 0\) into one). +
+ ++{\bf Nice trick:} notice that +\[ + \oint {\bf A} \cdot d{\bf l} = \int ({\boldsymbol \nabla} \times {\bf A}) \cdot d{\bf a} + = \int {\bf B} \cdot d{\bf a} = \Phi. + \label{Gr(5.69)} + \] +This is reminiscent of Ampère's law in integral form, \ref{Gr(5.55)}, +\[ + \oint {\bf B} \cdot d{\bf l} = \mu_0 I_{enc}. + \] +It's the same equation ! Replacement: \({\bf B} \rightarrow {\bf A}\) and \(\mu_0 I_{enc} \rightarrow \Phi\). +And to paraphrase Feynman's lectures: {\it the same equations have the same solutions.} +
+ ++Use symmetry: vector potential can only be cicumferential. Using an 'amperian' loop at a radius +\(s\) {\it inside} the solenoid, and the fact that the field inside a solenoid is \(\mu_0 n I\) +(\ref{Gr(5.57)}), we get +\[ + \oint {\bf A} \cdot d{\bf l} = A (2\pi s) = \int {\bf B} \cdot d{\bf a} = \mu_0 n I (\pi s^2), + \] +so +\[ + {\bf A} = \frac{\mu_0 n I}{2} s \hat{\boldsymbol \phi}, \hspace{1cm} s < R. + \label{Gr(5.70)} + \] +For an 'amperian' loop outside, the flux is always \(\mu_0 n I (\pi R^2)\), so +\[ + {\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{s} \hat{\boldsymbol \phi}, \hspace{1cm} s > R. + \label{Gr(5.71)} + \] +\paragraph{Exercise:} check that \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) and that +\({\boldsymbol \nabla} \cdot {\bf A} = 0\). +
+ +Created: 2022-02-07 Mon 08:02
+ +
+\paragraph{Problem 5.33:} similarly to the electrostatic dipole,
+the magnetic field of a dipole can be written in coordinate-free form.
+For this: we define a useful object, the
+
+{\bf Levi-Civita symbol} +\[ + \epsilon_{ijk} = \left\{ \begin{array} {cc} + 1, & i,j,k ~\mbox{even permutation of }~ x, y, z \\ + -1, & i,j,k ~\mbox{odd permutation of }~ x, y, z \\ + 0, & ~\mbox{otherwise}. \end{array} \right. + \label{LeviCivita} + \] +
+ ++Using this: we can for example rewrite the cross product as +\[ +({\bf A} \times {\bf B})_i = \epsilon_{ijk} A_j B_k. +\] +Important identity: +\[ +\boxed{ +\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl} +} +\label{sumLeviCivita} +\] +The curl of a vector is thus +\[ +({\boldsymbol \nabla} \times {\bf A})_i = \epsilon_{ijk} \partial_j A_k. +\] +
+ ++We can now easily do problem 5.33 (we write the dipole moment \(m\) as \(M\) to make notation clear) +\[ +(B_{di})_i = \epsilon_{ijk} \partial_j \left( \frac{\mu_0}{4\pi} \frac{{\bf M} \times {\bf r}}{r^3} \right)_k += \frac{\mu_0}{4\pi} \epsilon_{ijk} \partial_j \left( \epsilon_{klm} M_l \frac{r_m}{r^3} \right) += \frac{\mu_0}{4\pi} \epsilon_{ijk} \epsilon_{klm} M_l \partial_j \frac{r_m}{r^3} +\] +But +\[ +\partial_j \frac{r_m}{r^3} = \partial_j \frac{r_m}{[\sum_l r_l^2]^{3/2}} = \frac{\delta_{jm}}{r^3} - 3 \frac{r_j r_m}{r^5} +\] +and \(\epsilon_{ijk} \epsilon_{klm} = \epsilon_{ijk} \epsilon_{lmk} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}\), so we get +
+\begin{align} +({\bf B}_{di})_i &= \frac{\mu_0}{4\pi} (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) M_l \left(\frac{\delta_{jm}}{r^3} - 3 \frac{r_j r_m}{r^5}\right) +\nonumber \\ +&= \frac{\mu_0}{4\pi} \left( \frac{M_i}{r^3} (3 - 3) - M_j (\frac{\delta_{ij}}{r^3} - 3 \frac{r_i r_j}{r^5}) \right) \nonumber \\ +&= \frac{\mu_0}{4\pi} \left( 3 \frac{ r_i (M_j r_j)}{r^5} - \frac{M_i}{r^3} \right) +\end{align} ++Putting back the vector notation, and the \(m\) notation for the magnetic dipole (instead of \(M\)), we obtain +\[ +\boxed{ +{\bf B}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{1}{r^3} [3({\bf m} \cdot \hat{\bf r}) \hat{\bf r} - {\bf m}] +} +\label{Gr(5.87)} +\] +
+ + ++\paragraph{Exercise:} rederive all derivative and product rules using Levi-Civita. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +
+Electrostatic fields: discontinuous at location of suface charge.
+Magnetostatic fields: discontinuous at location of surface current.
+
+Equation \ref{eq:DivBisZero}: \(\oint {\bf B} \cdot d{\bf a} = 0\) applied to wafer-thin +pillbox straddling surface: normal component +\[ +B^{\perp}_{above} = B^{\perp}_{below}. +\label{Gr(5.72)} +\] +Tangential component: amperian loop of side length \(l\) perpendicular to surface current: +\[ +\oint {\bf B} \cdot d{\bf l} = (B^{\parallel}_{above} - B^{\parallel}_{below}) l = \mu_0 I_{enc} = \mu_0 K l, +\] +and therefore +\[ +B^{\parallel}_{above} - B^{\parallel}_{below} = \mu_0 K +\label{Gr(5.73)} +\] +So: component of \({\bf B}\) that is parallel to surface but perpendicular to current flow +is discontinuous, whereas the one parallel to the flow is continuous. In vector notation: +
++\[ + {\bf B}_{above} - {\bf B}_{below} = \mu_0 {\bf K} \times \hat{\bf n}, + \label{Gr(5.74)} + \] +
+ ++where \(\hat{\bf n}\) points upwards. For the vector potential, the relations are +
++\[ + {\bf A}_{above} = {\bf A}_{below} + \label{Gr(5.75)} + \] +
+ ++This can be seen first from the condition \({\boldsymbol \nabla} \cdot {\bf A} = 0\), which +guarantees that the normal component is continuous. Second, \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) +leads to +\[ +\oint {\bf A} \cdot d{\bf l} = \int {\bf B} \cdot d{\bf a} = \Phi, +\] +where the loop is vanishingly small and straddles the surface. Since the flux then goes to zero, +the tangential components of \({\bf A}\) are also continuous. +
+ ++However, the derivative of \({\bf A}\) inherits the discontinuity of \({\bf B}\): explicitly, +
+\begin{align} +{\bf B}_{above} - {\bf B}_{below} &= {\boldsymbol \nabla} \times {\bf A}_{above} - {\boldsymbol \nabla} \times {\bf A}_{below} +\nonumber \\ +&= \left| \begin{array}{ccc} \hat{\bf x} & \hat{\bf y} & \hat{\bf z} \\ \partial_x & \partial_y & \partial_z \\ +A_{x, above} & A_{y, above} & A_{z, above} \end{array} \right| - +\left| \begin{array}{ccc} \hat{\bf x} & \hat{\bf y} & \hat{\bf z} \\ \partial_x & \partial_y & \partial_z \\ +A_{x, below} & A_{y, below} & A_{z, below} \end{array} \right| +\end{align} ++We put the normal direction along \(\hat{\bf z}\) and the current along \(\hat{\bf x}\). +Since the normal component is continuous, \(\partial_{x,y} A_z\) is the same above and below, and we can +drop these terms. +The \(\hat{\bf z}\) term vanishes since it just involves the difference of \(B^{\perp}\), which is the same above and below. +Similary, the \(\hat{\bf x}\) term is the magnetic field parallel to the surface current, which also isn't discontinuous. +What is left is +\[ +%\hat{\bf x} (-\partial_z A_{y,above} + \partial_z A_{y, below}) + +\hat{\bf y} (\partial_z A_{x, above} - \partial_z A_{x, below}) +%+ \hat{\bf z} (\partial_x A_{y, above} - \partial_x A_{y below} - \partial_y A_{x, above} + \partial_y A_{x, below}) += \mu_0 K \hat{\bf x} +\] +Therefore, reidentifying the normal component, we get +
++\[ + \frac{\partial {\bf A}_{above}}{\partial n} - \frac{\partial {\bf A}_{below}}{\partial n} + = -\mu_0 {\bf K} + \label{Gr(5.76)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Remember our expansion for the electrostatic field (\ref{Gr(3.94)}) +\[ +\frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{[r^2 + (r')^2 - 2r r' \cos \theta']^{1/2}} += \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r'}{r}\right)^l P_l (\cos \theta'). +\label{Gr(5.77)} +\] +The expansion for the vector potential for a current loop carrying current \(I\) over path \({\cal P}\) +can thus be written +\[ +{\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{r^{l+1}} \oint_{\cal P} d{\bf l}' +(r')^l P_l (\cos \theta') +\label{Gr(5.78)} +\] +or (in my notations) +\[ +{\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} +\oint_{\cal P} d{\bf l}_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s), +\hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal P}. +\] +Explicitly, the first few terms are +\[ +{\bf A}({\bf r}) = \frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint\cal P d{\bf l}' +
++\label{Gr(5.79)} +\] +Again, these are known as the {\bf monopole}, {\bf dipole}, {\bf quadrupole} terms. +
+ ++{\bf Note:} {\bf the magnetic monopole term always vanishes.} This is simply because the total vector +displacement on a closed loop is zero, \(\oint d{\bf l}' = 0\), or in other words: there are no magnetic +monopoles (also from Maxwell's equation \({\boldsymbol \nabla} \cdot {\bf B} = 0\)). +
+ ++The dominant term is thus the dipole, +\[ +{\bf A}_{di} ({\bf r}) = \frac{\mu_0 I}{4\pi} \frac{1}{r^2} \oint d{\bf l}' (\hat{\bf r} \cdot {\bf r}') +\label{Gr(5.81)} +\] +Using equation \ref{Gr(1.108)} from Problem 1.61, +\[ +\oint_{\cal P} ({\bf c} \cdot {\bf r}) d{\bf l} = {\bf a} \times {\bf c}, \hspace{1cm} +{\bf a} = \int_{\cal S} d{\bf a} +\label{Gr(1.108)} +\] +with \({\bf c} = \hat {\bf r}\), +
+ ++\paragraph{Parenthesis: Problem 1.61 (e)} Start from Stokes' theorem, +\[ + \int_{\cal S} {\boldsymbol \nabla} \times {\bf V} \cdot d{\bf a} = \oint_{\cal P} {\bf V} \cdot d{\bf l} + \] +Let \({\bf V} = {\bf c} T\), where \({\bf c}\) is constant. +On the left-hand side: +\[ + LHS = \int_{\cal S} T ({\boldsymbol \nabla} \times c) \cdot d{\bf a} - \int_{\cal S} {\bf c} \times ({\boldsymbol \nabla} T) \cdot d{\bf a} + \] +The first term is zero since \({\bf c}\) is constant. For the second term: use vector identity nr 1, +\(({\bf c} \times ({\boldsymbol \nabla} T)) \cdot d{\bf a} = {\bf c} \cdot ({\boldsymbol \nabla}T \times d{\bf a})\). The second +term thus becomes +\[ +
++\] +Treating the right-hand side of the original equation now, +\[ + RHS = {\bf c} \cdot \oint_{\cal P} T d{\bf l} + \] +so we get (since this is valid for any \({\bf c}\)) +\[ + \int_{\cal S} {\boldsymbol \nabla} T \times d{\bf a} = -\oint_{\cal P} T d{\bf l}. + \] +Now put \(T = {\bf c} \cdot {\bf r}\) in this: conclusion is \ref{Gr(1.108)}. +
+ ++Back to our problem: +\[ +\oint d{\bf l}' (\hat{\bf r} \cdot {\bf r}') = -\hat{\bf r} \times \int_{\cal S} d{\bf a}' +\label{Gr(5.82)} +\] +and defining the +
++magnetic dipole moment + \[ + {\bf m} \equiv I \int_{\cal S} d{\bf a} = I {\bf a} + \label{Gr(5.84)} + \] +
+ ++we obtain the convenient expression for the +
++{\bf dipole term of the vector potential} +\[ + {\bf A}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times \hat{\bf r}}{r^2} + \label{Gr(5.83)} + \] +
+ ++\paragraph{Example 5.13:} find magnetic dipole moment of bookend shape of Gr. Fig. 5.52. +All sides have length \(w\) and carry current \(I\). +\paragraph{Solution:} combine two loops, use \ref{Gr(5.83)} +\[ + {\bf m} = I w^2 \hat{\bf y} + I w^2 \hat{\bf z} + \] +
+ ++{\bf Note:} {\it the magnetic dipole moment is independent of the choice of origin.} +
+ ++{\bf Note:} does there exist a {\it pure magnetic dipole} ? Well, yes, but it's an +infinitely small loop carrying an infinitely large current, so that the dipole term +is finite. +
+ ++In practice: dipole approximation often good enough when far away from source on a +scale of the source's current loops. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Field due to polarization: effectively comes from bound surface and volume charges, +
++\[ + \rho_b = -{\boldsymbol \nabla} \cdot {\bf P}, \hspace{1cm} \sigma_b = {\bf P} \cdot \hat{\bf n}. + \] +
+ ++Rest of charges: {\bf free charge} (electrons in conductors, ions embedded in dielectric, +{\it i.e.} any charge which doesn't come from polarization). +
+ ++Within dielectric: +\[ +\rho = \rho_b + \rho_f +\label{Gr(4.20)} +\] +Gauss's law: +\[ +\varepsilon_0 {\boldsymbol \nabla} \cdot {\bf E} = \rho = \rho_b + \rho_f = -{\boldsymbol \nabla} \cdot {\bf P} + \rho_f. +\] +Convenient way of writing: +\[ +{\boldsymbol \nabla} \cdot (\varepsilon_0 {\bf E} + {\bf P}) = \rho_f. +\] +Defining the +
++Electric displacement \({\bf D}\) + \[ + {\bf D} \equiv \varepsilon_0 {\bf E} + {\bf P} + \label{Gr(4.21)} + \] +
+ ++Gauss's law becomes +
++\[ + {\boldsymbol \nabla} \cdot {\bf D} = \rho_f + \label{Gr(4.22)} + \] +
+ ++or in integral form +
++\[ + \oint d{\bf a} \cdot {\bf D} = Q_{f_{enc}} + \label{Gr(4.23)} + \] +
+ ++\paragraph{Example 4.4:} long straight wire, uniform line charge \(\lambda\), surrounded by rubber insulation +to radius \(a\). Find \({\bf D}\). +\paragraph{Solution:} cylindrical Gaussian surface, radius \(s\) and length \(L\). Applying \ref{Gr(4.23)}, +\[ + D 2\pi s L = \lambda L, + \] +so +\[ + {\bf D} = \frac{\lambda}{2\pi s} \hat{\bf s} + \label{Gr(4.24)} + \] +This holds within insulation and outside of it. Outside, \({\bf P} = 0\) so +\[ + {\bf E} = \frac{1}{\varepsilon_0} {\bf D} = \frac{\lambda}{2\pi \varepsilon_0 s} \hat{\bf s}, + \hspace{1cm} s > a. + \] +Inside: can't know the electric field, since \({\bf P}\) isn't known. +
+ ++\paragraph{Note of caution:} while \ref{Gr(4.23)} might lead you to think that +\[ +{\bf D} ({\bf r}) = \frac{1}{4\pi} \int_{\cal V} d\tau' \rho_f ({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} +\] +but this is {\bf not} correct. In particular, the curl of the displacement isn't always zero, +\[ +{\boldsymbol \nabla} \times {\bf D} = \varepsilon_0 ({\boldsymbol \nabla} \times {\bf E} + {\boldsymbol \nabla} \times {\bf P}) + = {\boldsymbol \nabla} \times {\bf P} +\label{Gr(4.25)} +\] +There is not 'potential' for \({\bf D}\). +
+Created: 2022-02-07 Mon 08:02
+ ++From \ref{Gr(4.23)}: +\[ +D^{\perp}_{above} - D^{\perp}_{below} = \sigma_f +\label{Gr(4.26)} +\] +Then, \ref{Gr(4.25)} give +
++\[ + {\bf D}^{\parallel}_{above} - {\bf D}^{\parallel}_{below} = {\bf P}^{\parallel}_{above} - {\bf P}^{\parallel}_{below}. + \label{Gr(4.27)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Let's quickly pause to give a classification of the types of responses that might be expected from +materials. +\subsubsection*{Permanent polarization:} here, \({\bf P} \neq 0\) even if \({\bf E} = 0\). +These are the {\bf electrets}. +
+ ++\subsubsection*{Nonlinear dielectrics:} +\[ +P_i = \sum_j \alpha_{ij} E_j + \sum_{jk} \beta_{ijkl} E_j E_k E_l + ... +\] +(even-order terms usually absent, but could be there in principle). +\subsubsection*{Linear dielectrics:} only linear term is there. Most general: +\[ +P_i = \varepsilon_0 \chi_{ij} E_j +\label{Gr(4.38)} +\] +where \({\bf \chi}\) is the {\bf electric susceptibility tensor} (it's a tensor of rank 2), +which cannot depend on \({\bf E}\) (otherwise we go back to the nonlinear case). +Common in crystals, leading to for example double refraction. +\subsubsection*{Linear isotropic dielectrics:} at a given point, the electrical properties of +the dielectric are independent of the direction of \({\bf E}\) (isotropy). Liquids fall into +this category. We here have +
++\[ + {\bf P} = \varepsilon_0 \chi_e {\bf E} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++For many substances: polarization is proportional to field, if the latter isn't too strong: +
++\[ + {\bf P} = \varepsilon_0 \chi_e {\bf E} + \label{Gr(4.30)} + \] +
+ ++Constant \(\chi_e\): called the {\bf electric susceptibility} of the medium. Since \(\varepsilon_0\) is +there, \(\chi_e\) is dimensionless. Materials that obey \ref{Gr(4.30)} are called +{\bf linear dielectrics}. +
+ ++\paragraph{Note:} \({\bf E}\) on the RHS of \ref{Gr(4.30)} is the {\bf total} electric field, +due to free charges and to the polarization itself. Putting dielectric in field \({\bf E}_0\), +we can't compute \({\bf P}\) directly from \ref{Gr(4.30)}. Better way: compute \({\bf D}\). +
+ ++In linear dielectrics: +\[ +{\bf D} = \varepsilon_0 {\bf E} + {\bf P} = \varepsilon_0 {\bf E} + \varepsilon_0 \chi_e {\bf E} += \varepsilon_0 (1 + \chi_e) {\bf E} +\label{Gr(4.31)} +\] +so +
++\[ + {\bf D} = \varepsilon {\bf E} + \label{Gr(4.32)} + \] +
+ ++where \(\varepsilon\) is called the {\bf permittivity} of the material. In vacuum, susceptibility is zero, +permittivity is \(\varepsilon_0\). Also, +\[ +\varepsilon_r \equiv 1 + \chi_e = \frac{\varepsilon}{\varepsilon_0} +\label{Gr(4.34)} +\] +is called the {\bf relative permittivity} or {\bf dielectric constant} of the material. +This is all just nomenclature, everything is already in \ref{Gr(4.30)}. +
+ + + ++\paragraph{Example 4.5:} metal sphere of radius \(a\) carrying charge \(Q\), surrounded out to radius \(b\) by +a linear dielectric material of permittivity \(\varepsilon\). Find potential at center (relative to infinity). +\paragraph{Solution:} need to know \({\bf E}\). Could try to locate bound charge: but we don't know \({\bf P}\) ! +What we do know: free charge, situation is spherically symmetric, so can calculate \({\bf D}\) +using \ref{Gr(4.23)}: +\[ + {\bf D} = \frac{Q}{4\pi r^2} \hat{\bf r}, \hspace{1cm} r > a. + \] +Inside sphere, \({\bf E} = {\bf P} = {\bf D} = 0\). Find \({\bf E}\) using \ref{Gr(4.32)}: +\[ + {\bf E} = \left\{ \begin{array}{cc} + \frac{Q}{4\pi \varepsilon r^2} \hat{\bf r}, & a < r < b, \\ + \frac{Q}{4\pi \varepsilon_0 r^2} \hat{\bf r}, & r > b. \end{array} \right. + \] +The potential is thus +\[ + V = -\int_\infty^0 d{\bf l} \cdot {\bf E} = -\int_\infty^b dr \frac{Q}{4\pi\varepsilon_0 r^2} - \int_b^a dr \frac{Q}{4\pi\varepsilon r^2} + = \frac{Q}{4\pi} \left( \frac{1}{\varepsilon_0 b} + \frac{1}{\varepsilon a} - \frac{1}{\varepsilon b} \right). + \] +
+ ++It was thus not necessary to compute the polarization or the bound charge explicitly. This can be done: +\[ + {\bf P} = \varepsilon_0 \chi_e {\bf E} = \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon r^2} \hat{\bf r}, + \] +so +\[ + \rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = 0, \hspace{1cm} + \sigma_b = {\bf P} \cdot \hat{\bf n} = \left\{ \begin{array}{cc} + \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon b^2}, & \mbox{outer surface} \\ + -\frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon a^2}, & \mbox{inner surface} \end{array} \right. + \] +
+ ++Dielectric thus like an imperfect conductor: charge \(Q\) not fully screened. +
+ ++In linear dielectrics, the parallel between \({\bf E}\) and \({\bf D}\) is also not perfect. +Remark: since \({\bf P}\) and \({\bf D}\) are both proportional to \({\bf E}\) inside the dielectric, +does it mean that their curl vanishes like for \({\bf E}\) ? {\bf No}: if there is a boundary +between two materials with different dielectric constants, then a closed loop integral +of {\it e.g.} \({\bf P}\) would not vanish. +
+ ++Only case where parallel works: space entirely filled with homogeneous linear dielectric. +
+ ++\paragraph{Example 4.6:} parallel-plate capacitor filled with insulating material of +dielectric constant \(\varepsilon_r\). What is the effect on the capacitance ? +\paragraph{Solution:} field confined between plates, and reduced by factor \(1/\varepsilon_r\). +Potential difference \(V\) also reduced by same factor. Since \(Q = C/V\), capacitance +is increased by factor of \(\varepsilon_r\), so +\[ + C = \varepsilon_r C_{vac} + \label{Gr(4.37)} + \] +
+ ++In homogeneous linear dielectric: +\[ +\rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = - {\boldsymbol \nabla} \cdot \left(\varepsilon_0 \frac{\chi_e}{\varepsilon} {\bf D}\right) += -\left( \frac{\chi_e}{1 + \chi_e} \right) \rho_f +\label{Gr(4.39)} +\] +If \(\rho = 0\), any net charge is on surface, potential then obeys Laplace. +
+ ++Convenient to rewrite boundary conditions in terms of free charge: from \ref{Gr(4.26)}, +
++\[ + \varepsilon_{above} E^{\perp}_{above} - \varepsilon_{below} E^{\perp}_{below} = \sigma_f + \label{Gr(4.40)} + \] +
+ ++or in terms of the potential, +
++\[ + \varepsilon_{above} \frac{\partial V_{above}}{\partial n} - + \varepsilon_{below} \frac{\partial V_{below}}{\partial n} = -\sigma_f + \label{Gr(4.41)} + \] +
+ ++Potential itself is continuous, +
++\[ + V_{above} = V_{below} + \label{Gr(4.42)} + \] +
+ ++\paragraph{Example 4.7:} sphere of homogeneous dielectric material in uniform electric field \({\bf E}_0\). +Find electric field inside sphere. +\paragraph{Solution:} resembles Example 3.8 (conducting sphere), here cancellation is not total. +
+ ++Need to solve Laplace's equation for \(V(r, \theta)\) with boundary conditions +
+\begin{align} + &(i)~~V_{in} (R,\theta) = V_{out} (R, \theta), \nonumber \\ + &(ii)~~\varepsilon \frac{\partial V_{in} (R,\theta)}{\partial n} + = \varepsilon_0 \frac{\partial V_{out} (R,\theta)}{\partial n}, \nonumber \\ + &(iii)~~V_{out} (r) \rightarrow -E_0 r \cos \theta, ~~r \gg R. +\end{align} ++Inside and outside sphere: +\[ + V_{in} (r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta), \hspace{1cm} + V_{out} (r,\theta) = -E_0 r \cos \theta + \sum_{l=0}^\infty \frac{B_l}{r^{l+1}} P_l (\cos \theta). + \] +Boundary condition \((i)\) imposes +\[ + \sum_{l=0}^\infty A_l R^l P_l(\cos \theta) = -E_0 R \cos \theta + \sum_{l=0}^\infty \frac{B_l}{R^{l+1}} P_l (\cos \theta) + \] +so +\[ + A_l R^l = \frac{B_l}{R^{l+1}}, ~~ l \neq 1, \hspace{1cm} + A_1 R = -E_0 R + \frac{B_1}{R^2}. + \] +Boundary condition \((ii)\): +\[ + \varepsilon_r \sum_{l=0}^\infty l A_l R^{l-1} P_l (\cos \theta) = - E_0 \cos \theta - \sum_{l=0}^\infty \frac{(l+1)B_l}{R^{l+2}} P_l (\cos \theta) + \] +so +\[ + \varepsilon_r l A_l R^{l-1} = -\frac{(l+1)B_l}{R^{l+2}}, ~~ l \neq 1, \hspace{1cm} + \varepsilon_r A_1 = -E_0 - \frac{2B_1}{R^3}. + \] +We then have +\[ + A_l = 0 = B_l, ~~ l \neq 1, \hspace{1cm} + A_l = -\frac{3}{\varepsilon_r + 2} E_0, ~~ B_1 = \frac{\varepsilon_r - 1}{\varepsilon_r + 2} R^3 E_0. + \] +Thus, +\[ + V_{in} (r, \theta) = -\frac{3E_0}{\varepsilon_r + 2} r \cos \theta = -\frac{3E_0}{\varepsilon_r + 2} z, + \hspace{1cm} + {\bf E} = \frac{3}{\varepsilon_r + 2} {\bf E}_0. + \] +
+ ++\paragraph{Example 4.8:} suppose region below \(z = 0\) is filled with uniform linear dielectric with susceptibility \(\chi_e\). +Calculate force on point charge \(q\) situated a distance \(d\) above origin. +\paragraph{Solution:} bound surface charge is of opposite sign, force is attractive. No bound volume charge +because of \ref{Gr(4.39)}. Using \ref{Gr(4.11)} and \ref{Gr(4.30)}, +\[ + \sigma_b = {\bf P} \cdot \hat{\bf n} = P_z = \varepsilon_0 \chi_e E_z + \] +where \(E_z\) is the z-component of total field just below surface of dielectric (due to \(q\) and to bound charge). +Contribution from charge \(q\) from Coulomb (careful: \(\theta\) is $π$-rotated as compared to spherical coord so \(\theta = 0\) represents \(-\hat{z}\)) +\[ + -\frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + d^2} \cos \theta = -\frac{1}{4\pi\varepsilon_0} \frac{qd}{(r^2 + d^2)^{3/2}}, \hspace{1cm} + r = \sqrt{x^2 + y^2}. + \] +$z$-component of field from bound charge: \(-\sigma_b/2\varepsilon_0\), so +\[ + \sigma_b = \varepsilon_0 \chi_e \left[ -\frac{1}{4\pi\varepsilon_0} \frac{qd}{(r^2 + d^2)^{3/2}} - \frac{\sigma_b}{2\varepsilon_0} \right], + \] +so +\[ + \sigma_b = -\frac{1}{2\pi} \left(\frac{\chi_e}{\chi_e + 2}\right) \frac{qd}{(r^2 + d^2)^{3/2}}. + \label{Gr(4.50)} + \] +As per conducting plane, except for factor \(\chi_e/(\chi_e + 2)\). Total bound charge: +\[ + q_b = -\left(\frac{\chi_e}{\chi_e + 2}\right) q. + \] +Field: by direct integration, or more nicely by method of images: replace dielectric by single point charge +\(q_b\) at \((0,0,-d)\): +\[ + V (x,y,z>0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{\sqrt{x^2 + y^2 + (z-d)^2}} + \frac{q_b}{\sqrt{x^2 + y^2 + (z+d)^2}}\right] + \] +A charge \(q + q_b\) at \((0,0,d)\) gives +\[ + V (x,y,z<0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q + q_b}{\sqrt{x^2 + y^2 + (z-d)^2}}\right] + \] +Putting these two together yields a solution to Poisson going to zero at infinity, and is therefore the unique solution. +Correct discontinuity at \(z = 0\): +\[ + -\varepsilon_0 \left( \frac{\partial V}{\partial z}|_{z = 0^+} - \frac{\partial V}{\partial z}|_{z = 0^-} \right) + = -\frac{1}{2\pi} \left(\frac{\chi_e}{\chi_e + 2}\right) \frac{qd}{(r^2 + d^2)^{3/2}}. + \] +Force on \(q\): +\[ + {\bf F} = \frac{1}{4\pi\varepsilon_0} \frac{q q_b}{4d^2} \hat{\bf z} = -\frac{1}{4\pi\varepsilon_0} \left( + \frac{\chi_e}{\chi_e + 2} \right) \frac{q^2}{4d^2} \hat{\bf z} + \label{Gr(4.54)} + \] +
+ ++To charge a capacitor: +\[ +W = \frac{1}{2} C V^2 +\] +If capacitor filled with linear isotropic dielectric (Ex. 4.6): +\[ +C = \varepsilon_r C_{vac} +\] +From Chap. 2: +\[ +W = \frac{\varepsilon_0}{2} \int d\tau E^2 +\] +How is this changed ? Counting only energy in fields: should decrease by +factor \(1/\varepsilon_r^2\). However, this would neglect the {\bf strain energy} +associated to the distortion of the polarized constituents of the dielectric medium. +
+ ++Derivation from scratch: bring in free charge: \(\rho_f\) increased by \(\Delta \rho_f\), polarization changes +(also bound charge distribution). Work done on free charges (only that matters): +\[ +\Delta W = \int d\tau (\Delta \rho_f ({\bf r})) V ({\bf r}). +\label{Gr(4.56)} +\] +But \(\rho_f = {\boldsymbol \nabla} \cdot {\bf D}\) so \(\Delta \rho_f = {\boldsymbol \nabla} \cdot (\Delta {\bf D})\), so +\[ +\Delta W = \int d\tau ({\boldsymbol \nabla} \cdot (\Delta {\bf D})) V += \int d\tau {\boldsymbol \nabla} \cdot ((\Delta {\bf D}) V) + \int d\tau (\Delta {\bf D}) \cdot {\bf E} +\] +First integral: divergence theorem changes it to a surface integral which vanishes when integrating over all space. +Therefore, +\[ +\Delta W = \int d\tau (\Delta {\bf D}) \cdot {\bf E} +\label{Gr(4.57)} +\] +This applies to any material. +
+ ++Special case of linear isotropic dielectric: \({\bf D} = \varepsilon {\bf E}\), so +\[ +\Delta W = \Delta \left( \frac{1}{2} \int d\tau {\bf D} \cdot {\bf E} \right) +\] +Total work done: +
++\[ + W = \frac{1}{2} \int d\tau {\bf D} \cdot {\bf E} + \label{Gr(4.58)} + \] +
+ ++As for a conductor: a dielectric is attracted into an electric field. Calculations can be +complicated: parallel plate capacitor with partially inserted dielectric: force comes +from {\bf fringing field} around edges. +
+ ++Better: reason from energy. Pull dielectric out by \(dx\). Energy change equal to work done: +\[ +dW = F_{me} dx +\label{Gr(4.59)} +\] +where \(F_{me}\) is mechanical force exerted by external agent. \(F_{me} = -F\), where \(F\) is +electrical force on dielectric. Electrical force on slab: +\[ +F = -\frac{dW}{dx} +\label{Gr(4.60)} +\] +Energy stored in capacitor: +\[ +W = \frac{1}{2} C V^2 +\label{Gr(4.61)} +\] +Capacitance in configuration considered: +\[ +C = \frac{\varepsilon_0 w}{d} (\varepsilon_r l - \chi_e x) +\label{Gr(4.62)} +\] +where \(l\) is the length of the plates, and \(w\) is their width. Assume total charge \(Q\) on each +plate is held constant as \(x\) changes. In terms of \(Q\), +\[ +W = \frac{1}{2} \frac{Q^2}{C} +\label{Gr(4.63)} +\] +so +\[ +F = -\frac{dW}{dx} = \frac{1}{2} \frac{Q^2}{C^2} \frac{dC}{dx} = \frac{1}{2} V^2 \frac{dC}{dx}. +\label{Gr(4.64)} +\] +But +\[ +\frac{dC}{dx} = -\frac{\varepsilon_0 \chi_e w}{d} +\] +so +\[ +F = -\frac{\varepsilon_0 \chi_e w}{2d} V^2. +\label{Gr(4.65)} +\] +
+ ++Common mistake: to use \ref{Gr(4.61)} (for \(V\) constant) instead of \ref{Gr(4.63)} +(for \(Q\) constant) in computing the force. In this case, sign is reversed, +\[ +F = -\frac{1}{2} V^2 \frac{dC}{dx}. +\] +Here, the battery also does work, so +\[ +dW = F_{me} dx + V dQ +\label{Gr(4.66)} +\] +and +\[ +F = -\frac{dW}{dx} + V \frac{dQ}{dx} = -\frac{1}{2} V^2 \frac{dC}{dx} + V^2 \frac{dC}{dx} = \frac{1}{2} V^2 \frac{dC}{dx} +\label{Gr(4.67)} +\] +so like before but with the correct sign. +
+Created: 2022-02-07 Mon 08:02
+ ++For a macroscopic moment: local induced/permanent dipole moments tend to somewhat cancel each other, leaving a residual +polarization. Convenient measure: +\[ +{\bf P} \equiv ~\mbox{dipole moment per unit volume} +\] +called the {\bf polarization}. +What is the electric field produced by an object with polarization \({\bf P}\) ? Work with the potential. +For a single dipole, \ref{eq:electric_dipole}: +\[ +V({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{({\bf r} - {\bf r}') \cdot {\bf p}}{|{\bf r} - {\bf r}'|^3} +\label{Gr(4.8)} +\] +With dipole moment per unit volume \({\bf P}\), we get +\[ +V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P}({\bf r}')}{|{\bf r} - {\bf r}'|^3} +\label{Gr(4.9)} +\] +This is correct as it is. There exists however another convenient representation. We know that +\[ +{\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} +\] +so we can write (using product rule number 5, ${\boldsymbol ∇} ⋅ (f {\bf A}) = f ({\boldsymbol ∇} ⋅ {\bf A}) + {\bf A} ⋅ {\boldsymbol ∇} f$~) +
+\begin{align} +V({\bf r}) &= \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' {\bf P} ({\bf r}') \cdot +{\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} \nonumber \\ +& = \frac{1}{4\pi \varepsilon_0} \left[ \int_{\cal V} d\tau' {\boldsymbol \nabla}' \cdot \left( \frac{{\bf P} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right) +- \int_{\cal V} d\tau' \frac{1}{|{\bf r} - {\bf r}'|} {\boldsymbol \nabla}' \cdot {\bf P} ({\bf r}') \right] +\end{align} ++Using the divergence theorem, this becomes +\[ +V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint\cal S d{\bf a}' ⋅ \frac{{\bf P} ({\bf r}')}{|{\bf r} - {\bf r}'|} +
++\label{Gr(4.10)} +\] +Interpretation: first terms is like contribution of a surface charge, +
++\[ + \sigma_b({\bf r}) = {\bf P} ({\bf r}) \cdot \hat{\bf n} + \label{Gr(4.11)} + \] +
+ ++and second term looks like contribution of a volume charge, +
++\[ + \rho_b ({\bf r}) = -{\boldsymbol \nabla} \cdot {\bf P} ({\bf r}) + \label{Gr(4.12)} + \] +
+ ++Using these definitions, +
++\[ + V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint\cal S d{\bf a}' ⋅ \frac{σb ({\bf r}')}{|{\bf r} - {\bf r}'|} +
++ \label{Gr(4.13)} +\] +
+ ++These {\bf bound charges} faithfully represent the object's sources for electrical fields. +
+ + ++\paragraph{Example 4.2:} electric field produced by uniformly polarized sphere of radius \(R\). +\paragraph{Solution:} put \(z\) axis along \({\bf P}\). Since \({\bf P}\) is uniform, \(\rho_b = 0\). +Surface charge: +\[ + \sigma_b ({\bf r}) = {\bf P} \cdot \hat{\bf n} = P \cos \theta. + \] +This was computed in Example: surface charge density on a sphere (eq. \ref{eq:PotentialUniformlyPolarizedSphere}): +\[ + V(r, \theta) = \left\{ \begin{array}{cc} + \frac{P}{3\varepsilon_0} r\cos \theta, & r \leq R \\ + \frac{P}{3\varepsilon_0} \frac{R^3}{r^2} \cos \theta, & r \geq R. + \end{array} \right. + \] +But \(r\cos \theta = z\), so the field inside the sphere is uniform, +\[ + {\bf E} = -{\boldsymbol \nabla} V = -\frac{P}{3\varepsilon_0} \hat{\bf z} = -\frac{1}{3\varepsilon_0} {\bf P}, + \hspace{1cm} r < R. + \label{Gr(4.14)} + \] +Outside sphere, potential is identical to that of pure point dipole at origin, +\[ + V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \frac{{\bf p} \cdot {\hat {\bf r}}}{r^2}, + \hspace{1cm} r > R + \label{Gr(4.15)} + \] +where the total dipole moment is simply the integral over the polarization, +\[ + {\bf p} = \frac{4}{3} \pi R^3 {\bf P} + \label{Gr(4.16)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Outside dielectric: OK, don't need microscopic details of field. +
+ ++Inside dielectric: microscopic field is very complicated. Macroscopic field is what we should +concentrate on. Defined as average field over region containing many constituents. +
+ ++Suppose we want to calculate field at point \({\bf r}\) inside dielectric. Consider a sphere around +\({\bf r}\), such that many atoms are within the sphere. Total macroscopic electric field: +\[ +{\bf E} = {\bf E}_{out} + {\bf E}_{in}. +\] +From Problem 3.47, average field (over a sphere) produced by charges outside sphere is +equal to field they produce at center of sphere. Therefore, \({\bf E}_{out}\) is field at \({\bf r}\) +from dipoles exterior to sphere, for which we can use \ref{Gr(4.9)}: +\[ +V_{out} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{out} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} +{|{\bf r} - {\bf r}'|^3} +\label{Gr(4.17)} +\] +For dipoles inside: cannot be treated in this fashion. But their average field is all we need, and +this is given by \ref{Gr(3.105)} (see Example below) +\[ +{\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{\bf p}{R^3} +\label{Gr(3.105)} +\] +irrespective of charge distribution within sphere. +
+ ++\paragraph{Problem 3.47:} show that the average field inside sphere of radius \(R\) due to all charges within +the sphere is +\[ + {\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{{\bf p}}{R^3} + \label{Gr(3.105)} + \] +where \({\bf p}\) is the dipole moment. Many ways to do this. +\paragraph{a)} show that average field due to single charge \(q\) at \({\bf r}\) inside sphere is same as +field at \({\bf r}\) due to uniformly charged sphere with \(\rho = -q/(\frac{4}{3} \pi R^3)\), +\[ + \frac{1}{4\pi \varepsilon_0} \frac{1}{\frac{4}{3} \pi R^3} \int d\tau' q \frac{{\bf r}' - {\bf r}}{|{\bf r} - {\bf r}'|^3}. + \] +\paragraph{Solution}: average field due to point charge \(q\) at \({\bf r}\): +\[ + {\bf E}_{av} = \frac{1}{\frac{4}{3} \pi R^3} \int d\tau' {\bf E} ({\bf r}'), \hspace{1cm} + {\bf E}({\bf r}') = \frac{q}{4\pi \varepsilon_0} \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} + \] +so +\[ + {\bf E}_{av} = \frac{1}{\frac{4}{3} \pi R^3} \frac{q}{4\pi\varepsilon_0} \int d\tau' \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} + \] +Field at \({\bf r}\) due to uniformly charged sphere: +\[ + {\bf E}_s ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int d\tau' \rho({\bf r}') \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} + \] +If \(\rho = -q/(\frac{4}{3} \pi R^3)\), the two expressions coincide. +
+ ++\paragraph{b)} +Express the latter in terms of the dipole moment. +\paragraph{Solution:} from Problem 2.12, field inside uniformly charged sphere is +\[ + {\bf E} ({\bf r}) = \frac{1}{3\varepsilon_0} \rho \hat{\bf r} + \] +so +\[ + {\bf E}_{s} ({\bf r}) = \left. \frac{1}{3\varepsilon_0} \rho \hat{\bf r} \right|_{\rho = \frac{-q}{\frac{4}{3}\pi R^3}} = -\frac{q}{4\pi\varepsilon_0} \frac{\hat{\bf r}}{R^3} = -\frac{{\bf p}}{4\pi\varepsilon_0 R^3}. + \] +\paragraph{c)} Arbitrary charge distribution +\paragraph{Solution:} just use superposition ! +\paragraph{d)} Show that the average field over the volume of a sphere due to all the charges outside is the same as the field they produce at the center: +\paragraph{Solution:} +Field at \({\bf r}'\) due to charge at \({\bf r}\) outside the sphere: +\[ + {\bf E} ({\bf r}') = \frac{q}{4\pi\varepsilon_0} \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} + \] +Averaged over sphere: +\[ + {\bf E}_{av} = \frac{1}{\frac{4}{3} \pi R^3} \frac{q}{4\pi\varepsilon_0} \int d\tau' \frac{{\bf r'} - {\bf r}}{|{\bf r} - {\bf r}'|^3} + \] +This is same as field produced at \({\bf r}\) outside sphere, by uniformly charged sphere with \(\rho = -q/(\frac{4}{3} \pi R^3)\). +We know that this is simply +\[ + {\bf E}_s = \frac{1}{4\pi\varepsilon_0} (-q) \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} + \] +where the sphere is centered on \({\bf r}'\). But this is precisely the field produced at the center of the sphere \({\bf r}'\), by +a charge \(q\) sitting at \({\bf r}\). +
+ ++We thus have +\[ +{\bf E}_{in} = -\frac{1}{4\pi\varepsilon_0} \frac{\bf p}{R^3} +\] +Since \({\bf p} = \frac{4}{3} \pi R^3 {\bf P}\), we get +\[ +{\bf E}_{in} = -\frac{1}{3\varepsilon_0} {\bf P} +\label{Gr(4.18)} +\] +
+ ++Now: by assumption, \({\bf P}\) does not vary significantly over volume of sphere. +Term left out of \ref{Gr(4.17)} is thus field at center of uniformly polarized sphere. +But this is precisely equal to the \({\bf E}_{in}\) contribution, so macroscopic field is +given by the potential +\[ +V({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} +{|{\bf r} - {\bf r}'|^3} +\label{Gr(4.19)} +\] +where \({\cal V}\) is the entire volume of the dielectric. +
+Created: 2022-02-07 Mon 08:02
+ ++Surface charge and volume charge from polarization. +
+ ++Consider an initially neutral small volume \({\cal V}\) with closed surface \({\cal S}\). When the external electric +field is applied, the material in \({\cal V}\) will distort, and some charges will move across +\({\cal S}\). The leftover charge in \({\cal V}\) is simply the bound charge, and is +the negative of the charge that has moved across the boundary \({\cal S}\), +\[ +\int_{\cal V} d\tau \rho_b = -\oint_{\cal S} d{\bf a} \cdot {\bf P} +\] +Using the divergence theorem, and since this must hold for any volume, we recover +\(\rho_b = - {\boldsymbol \nabla} \cdot {\bf P}\). +
+ ++Surface charge: draw pictures of material near its boundary, with cancellations everywhere except on boundary. +
+ ++Volume charge: drawing of diverging polarization. +
+Created: 2022-02-07 Mon 08:02
+ ++How Electric Fields Influence Matter +
+ + + ++\paragraph{E fields on fundamental particles:} complicated business. +Quantum mechanics is obviously important here. +In fact, the measured charge of an electron is a 'renormalized' charge +(it depends on how closely you probe the electron, in other words at what collision/scattering +energies you probe it). This goes beyond the current course, and needs the whole machinery +of {\bf quantum electrodynamics}. +
+ ++For us here, we will assume that fundamental particles remain 'unchanged' in the presence of an E field, +irrespective of how strong the latter is. +
+ ++\paragraph{Simple atoms:} for example Hydrogen. +We simply have a nucleus (proton) with an orbiting electron. +The mass ratio between these is about 1800 to 1. To treat this, we'd need to start from the +nonrelativistic Schr\"odinger equation for the electron (assuming we're in the center-of-mass frame) +in the presence of a constant (for simplicity) external electric field \({\bf E}\): +\[ +-\frac{\hbar^2}{2m} {\boldsymbol \nabla}^2 \psi - \frac{e^2}{4\pi \varepsilon_0 r} \psi - e {\bf E} \cdot {\bf r} ~\psi = E \psi. +\] +As compared to the zero-field case, the energy levels are modified changed +(by the Stark effect, linear and nonlinear (latter for case of hydrogen in fundamental level); +see Landau Lifschitz, vol 3 nr 77). +
+ ++If the field is small, one can use perturbation theory. This gives an electric dipole moment of: +\[ +{\bf p} = \langle \psi | (-e {\bf r}) | \psi \rangle = ... = \frac{9}{2} (4\pi \varepsilon_0 a_B^3) {\bf E} +\] +where \(a_B\) is the Bohr radius, \(a_B = \hbar^2/m e^2\). +
+ ++Although the numerical factor is not guessable, the overall form is: +Le Ch\atelier's principle tells us that the equilibrium position moves linearly with the strength of the +perburtation. +
+ ++\paragraph{More complex atoms:} we face a similar scenario. +The nucleus is now relatively even heavier than each electron. +At small fields, we can neglect nonlinear effects +({\it e.g.} a given electron orbital change leading to changes in other orbitals). +We still expect to have some induced dipole moment which increases linearly with the external field, +\[ +{\bf p} = \alpha {\bf E} +\] +except that now we have to solve a much more complicated QM problem. +The factor \(\alpha\) is an atom-specific number called the {\bf atomic polarizability}. +
+ ++\paragraph{Molecules:} atoms can now 'share' electrons, so the charge distribution can become nontrivial. +Example: carbon dioxide, \(O - C - O\). Higher polarizability along axis than perpendicular to axis. +In totally non-symmetric case: expect +\[ +p_i = \sum_{j = x,y,z} \alpha_{ij} E_j +\] +where \(\alpha_{ij}\) is the {\bf polarizability tensor} of the molecule. Always possible to use 'principal' axes +such that all but 3 of the terms cancel. +
+ ++\paragraph{Polar molecules:} unlike individual atoms, molecules can have a permanent dipole moment. These are called +{\bf polar molecules}. Example: \(H Cl\) has elecronic density more closely bound on \(Cl\) than \(H\), so has a dipole +moment pointing from \(Cl\) to \(H\). Other example: water, with \(105^\circ\) angle between the \(H^+\) and \(O^-\), +dipole moment pointing from \(O^-\) along bisector. +
+ ++Torque on dipole: if field is uniform, overall force on dipole cancels, but torque remains: +
++{\bf Torque on a dipole:} +\[ + {\bf N} = {\bf p} \times {\bf E} + \label{Gr(4.4)} + \] +
+ ++If field is non-uniform, +\[ + {\bf F} = {\bf F}_+ + {\bf F}_- = q({\bf E}_+ - {\bf E}_-) = q ({\bf d} \cdot {\boldsymbol \nabla}) {\bf E} + \] +so +
++\[ + {\bf F} = ({\bf p} \cdot {\boldsymbol \nabla}) {\bf E} + \label{Gr(4.5)} + \] +
+ ++Energy of dipole in electric field (Problem 4.7): +\[ +\boxed{ +U = -{\bf p} \cdot {\bf E} +} +\label{Gr(4.6)} +\] +
+ ++\paragraph{Many atoms: gas and liquid phases:} here, atoms or molecules are still more or less free from each other's influence +as far as polarization is concerned. Random thermal motion, +external field gives preferential direction to polarization. +The relationship is still linear. +
+ ++\paragraph{Many atoms: solid phase:} here, things can be more complicated. Material can be insulating or conducting. +If conducting: external field makes charges move such that interior becomes equipotential. +If insulating: each constituent atom/molecule can pick up an induced polarization, polar molecules can tend to +line up, crystal structure can be deformed, … +
+ ++For zero field, the solid can have either zero or nonzero polarization. If nonzero: we call this spontaneous polarization, +or rather {\bf ferroelectricity} (after {\bf ferromagnetism}, which is the correspdonding magnetic phenomenon happening with iron). +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Nomenclature: as in electric case, we have bound currents, and everything else, which we call the {\bf free current}. +Total current: +\[ +{\bf J} = {\bf J}_b + {\bf J}_f +\label{Gr(6.17)} +\] +Ampère's law: +\[ +\frac{1}{\mu_0} ({\boldsymbol \nabla} \times {\bf B}) = {\bf J} = {\bf J}_f + {\bf J}_b += {\bf J}_f + ({\boldsymbol \nabla} \times {\bf M}), +\] +so we can define +
++\[ + {\bf H} \equiv \frac{1}{\mu_0} {\bf B} - {\bf M} + \label{Gr(6.18)} + \] +
+ ++and rewrite Ampère's law as +
++\[ + {\boldsymbol \nabla} \times {\bf H} = {\bf J}_f + \label{Gr(6.19)} + \] +
+ ++or in integral form, +
++\[ + \oint {\bf H} \cdot d{\bf l} = I_{f_{enc}} + \label{Gr(6.20)} + \] +
+ ++\({\bf H}\) in magnetostatics: parallel role to \({\bf D}\) in electrostatics. Allows us to +rewrite Ampère's law in terms of free currents alone. Bound currents come along for the ride. +
+ ++\paragraph{Example 6.2:} long copper rod radius \({\bf R}\) carries uniformly distributed free current \(I\). +Find \({\bf H}\) inside and outside rod. +\paragraph{Solution:} copper weakly diamagnetic: dipoles line up opposite the field. +Bound currents antiparallel to \(I\) in bulk and parallel at surface. All currents longitudinal +so \({\bf B}, {\bf M}, {\bf H}\) are circumferential. Apply integral form of Ampère's law with +radius \(s < R\): \(H (2\pi s) = I_{f_{enc}} = I \frac{\pi s^2}{\pi R^2}\) so +\[ + {\bf H} = \frac{I s}{2\pi R^2} \hat{\boldsymbol \phi}, \hspace{5mm} s \leq R + \label{Gr(6.21)} + \] +Outside, +\[ + {\bf H} = \frac{I}{2\pi s} \hat{\boldsymbol \phi}, \hspace{5mm} s \geq R. + \label{Gr(6.22)} + \] +There, \({\bf M} = 0\) so \({\bf B} = \mu_0 {\bf H}\). +
+ ++Similarly to electric case: cannot assume that \({\bf H}\) is like \({\bf B}\). +\({\bf H}\) might have a divergence, +\[ +{\boldsymbol \nabla} \cdot {\bf H} = -{\boldsymbol \nabla} \cdot {\bf M} +\label{Gr(6.23)} +\] +
++Recall \ref{Gr(7.31)}, magnetic energy of system of free currents: +\[ +W_{mag} = \frac{1}{2} \int_{\cal V} d\tau {\bf A} \cdot {\bf J}_f +\] +Similarly to the electric case, we can consider the presence of linear media. +Then, work necessary to increase flux is (from EMF) \(\Delta W_{mag} = V_{ext} \Delta q = V I \Delta t = I \Delta \phi\) so +\[ +\Delta W_{mag} = I \Delta \phi +\] +In terms of current density: use \(\phi = \int_{\cal S} (\boldsymbol \nabla \times {\bf A}) \cdot d{\bf a} = \int_{\cal C} {\bf A} \cdot d{\bf s}\), +move to volume currents: +\[ +\Delta W_{mag} = \int_{\cal V} d\tau {\bf J}_f \cdot \Delta {\bf A} = \int_{\cal V} d\tau ({\boldsymbol \nabla} \times {\bf H}) \cdot \Delta {\bf A} +\] +But \({\boldsymbol \nabla} \times (\Delta {\bf A}) = \Delta {\bf B}\) and +\[ +({\boldsymbol ∇} × {\bf H}) ⋅ Δ {\bf A} = {\bf H} ⋅ ({\boldsymbol ∇} × Δ {\bf A}) +
++\] +Integrating, we get +\[ +\Delta W_{mag} = \int_{all~space} d\tau {\bf H} \cdot \Delta {\bf B} +\] +Case of linear isotopic homogeneous medium: +\[ +W_{mag} = \int_{all~space} d\tau \frac{1}{2} {\bf H} \cdot {\bf B} +\] +
+ + + ++\subsubsection*{Boundary conditions} +Can rewrite BCs in terms of \({\bf H}\): from \ref{Gr(6.23)}, +\[ +H^{\perp}_{above} - H^{\perp}_{below} = -(M^{\perp}_{above} - M^{\perp}_{below}) +\label{Gr(6.24)} +\] +while \ref{Gr(6.19)} gives +\[ +{\bf H}^{\parallel}_{above} - {\bf H}^{\parallel}_{below} = {\bf K}_f \times \hat{\bf n} +\label{Gr(6.25)} +\] +These are more useful than BCs on \({\bf B}\), \ref{Gr(5.72)} and \ref{Gr(5.73)}: +\[ +B^{\perp}_{above} = B^{\perp}_{below}. +\label{Gr(6.26)} +\] +and +\[ +{\bf B}^{\parallel}_{above} - {\bf B}^{\parallel}_{below} = \mu_0 K \times \hat{\bf n} +\label{Gr(6.27)} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Orbital motion of electrons: so fast that 'current' is for most purposes steady: +\(I = e/T = \frac{ev}{2\pi R}\). Orbital dipole moment (\(I \pi R^2\)) is thus +\[ +{\bf m} = -\frac{evR}{2} \hat{\bf z} +\label{Gr(6.4)} +\] +In practice: harder to tilt orbit than spin, so this represents a small paramagnetic contribution. +
+ ++More significant: speeding up or slowing down of electron on its orbit. Here: handwaving +calculation. Centripetal +acceleration: \(v^2/R\) usually sustained by electrical forces alone, +\[ +\frac{1}{4\pi \varepsilon_0} \frac{e^2}{R^2} = m_e \frac{v^2}{R} +\label{Gr(6.5)} +\] +In presence of magnetic field (say perpendicular to plane of orbit): +\[ +\frac{1}{4\pi \varepsilon_0} \frac{e^2}{R^2} + e \bar{v} B = m_e \frac{\bar{v}^2}{R} +\label{Gr(6.6)} +\] +New speed: \(e \bar{v} B = \frac{m_e}{R} (\bar{v}^2 - v^2) = \frac{m_e}{R} (\bar{v} + v) (\bar{v} - v)\), +or (assuming small change) +\[ +\Delta v = \frac{eRB}{2m_e} +\label{Gr(6.7)} +\] +So electron speeds up when \({\bf B}\) is turned on. Change in magnetic dipole moment: +\[ +\Delta {\bf m} = -\frac{1}{2} e(\Delta v) R \hat{\bf z} = -\frac{e^2 R^2}{4m_e} {\bf B} +\label{Gr(6.8)} +\] +so change in \({\bf m}\) is opposite to change in \({\bf B}\). Universal phenomenon, leading +to diamagnetism. Usually much weaker than paramagnetism. Observed in substances having even number of +electrons (where paramagnetism is usually absent). +
+ + ++Fun fact: a paramagnet attracted into the field, whereas a diamagnet is repelled away. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Suppose we have a piece of material with known magnetization \({\bf M}\). +What is the field produced by this object? +For a single dipole: refer to \ref{Gr(5.83)} (vector potential of single dipole): +\[ +{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} +\label{Gr(6.10)} +\] +For a chunk of material with local magnetization \({\bf M} ({\bf r})\), +by the principle of superposition we thus have: +
++\[ + {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~\frac{{\bf M} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} + \label{Gr(6.11)} + \] +
+ ++In principle, this is all that is needed. +As in electric case however, a more illuminating version of this equation can be given +by using some simple identities: +using \({\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}\), +\[ +{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf M} ({\bf r}') \times \left( {\boldsymbol \nabla}' +\frac{1}{|{\bf r} - {\bf r}'|} \right), +\] +we can further integrate by parts and use product rule: \({\boldsymbol \nabla} \times (f {\bf A}) = f ( {\boldsymbol \nabla} \times {\bf A}) - {\bf A} \times ({\boldsymbol \nabla} f)\), we get +\[ +{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \left\{ +∫\cal V dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} +
++\right\} +\] +Problem 1.61 b) (p.56): leads to +\[ + {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} + ∫\cal V dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} +
++ \label{Gr(6.12)} +\] +Reinterpretation: first term: potential from volume current, +
++\[ + {\bf J}_b = {\boldsymbol \nabla} \times {\bf M} + \label{Gr(6.13)} + \] +
+ ++second term: potential from surface current, +
++\[ + {\bf K}_b = {\bf M} \times \hat{\bf n} + \label{Gr(6.14)} + \] +
+ ++With these definitions, +
++\[ + {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} ∫\cal V dτ' \frac{{\bf J}b ({\bf r}')}{|{\bf r} - {\bf r}'|} +
++ \label{Gr(6.15)} +\] +
+ ++so the field produced by the material is the same as that produced by {\bf bound currents} +in the volume and surface of the material. +
+ + + ++\paragraph{Example 6.1:} find field of uniformly magnetized sphere. +\paragraph{Solution:} put z axis along \({\bf M}\). +\[ + {\bf J}_b = {\boldsymbol \nabla} \times {\bf M} = 0, + \hspace{1cm} + {\bf K}_b = {\bf M} \times \hat{\bf n} = M \sin \theta \hat{\boldsymbol \phi}. + \] +Rotating spherical shell of uniform surface charge \(\sigma\): surface current density +\[ + {\bf K} = \sigma {\bf v} = \sigma \omega R \sin \theta \hat{\boldsymbol \phi}. + \] +Same if \(\sigma R {\boldsymbol \omega} = {\bf M}\). Refer to Example 5.11,% ({\it not done in class !}), +\[ + {\bf B} = \frac{2}{3} \mu_0 {\bf M} + \label{Gr(6.16)} + \] +inside sphere, whereas outside: pure dipole field, with +\[ + {\bf m} = \frac{4}{3} \pi R^3 {\bf M}. + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++As per electric case. Reasoning done in Problem 6.11. +
+Created: 2022-02-07 Mon 08:02
+ ++As electric case: bound currents are real things. +Current loops inside matter: cancelling more or less, +unless non-uniform, or at the edge. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Ferromagnets. Magnetic domains, {\bf hysteresis}. {\bf Curie point}: iron goes from ferromagnetic +to paramagnetic at about \(770^\circ C\). {\bf Phase transitions}. +
+Created: 2022-02-07 Mon 08:02
+ ++In para/diamagnets: when \({\bf B}\) is removed, \({\bf M}\) disappears. For not too strong +fields, proportionality. Custom (slightly different than for dielectrics): +{\bf magnetic susceptibility} \(\chi_m\) defined as +
++\[ + {\bf M} = \chi_m {\bf H} + \label{Gr(6.29)} + \] +
+ ++(and not \({\bf M} = \frac{1}{\mu_0} \chi_m {\bf B}\) had the electrostatics parallel +been followed historically). Materials that obey this are called linear media. Then, +\[ +{\bf B} = \mu_0 ({\bf H} + {\bf M}) = \mu_0 (1 + \chi_m) {\bf H} +\label{Gr(6.30)} +\] +and thus +\[ +{\bf B} = \mu {\bf H}, \hspace{1cm} \mu \equiv \mu_0 (1 + \chi_m) +\label{Gr(6.31)} +\] +where \(\mu\) is called the {\bf permeability} of the material. +
+ ++Again, although \({\bf M}\) and \({\bf H}\) are proportional to \({\bf B}\), it doesn't follow +that their divergence vanishes. At the boundary between two different media, they can +have nonzero divergence. +
+ ++In linear medium: bound volume current proportional to free current, +\[ +{\bf J}_b = {\boldsymbol \nabla} \times {\bf M} = {\boldsymbol \nabla} \times (\chi_m {\bf H}) += \chi_m {\bf J}_f. +\label{Gr(6.33)} +\] +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++{\bf Magnetization}: magnetic dipole moment per unit volume, denoted by \({\bf M}\). +
+ +
+\({\bf M}\) parallel to \({\bf B}\): {\bf paramagnets}
+\({\bf M}\) antiparallel to \({\bf B}\): {\bf diamagnets}
+\({\bf M} \neq 0\) even if \({\bf B} = 0\): {\bf ferromagnets}
+
Created: 2022-02-07 Mon 08:02
+ ++Torque on magnetic dipole: +
++\[ + {\bf N} = {\bf m} \times {\bf B} + \label{Gr(6.1)} + \] +
+ ++Tends to align magnetic dipoles with field. Accounts for paramagnetism. +Exact for extended dipole in uniform field, or for point dipole in non-uniform field. +Similar to electric analogue, \ref{Gr(4.4)}, \({\bf N} = {\bf p} \times {\bf E}\). +
+ ++In a non-uniform field: (to be done in Problem 6.4): +
++\[ + {\bf F} = {\boldsymbol \nabla} ({\bf m} \cdot {\bf B}) + \label{Gr(6.3)} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++These lecture notes provide a full treatment of classical (pre-quantum) +electromagnetic phenomena (static and dynamic). +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Throughout these lecture notes, boxes with contextual colors are used, +serving specific purposes. These are: +
+ ++Specific prerequisites for understanding the upcoming material +
++Objectives for this part: what you should learn by reading this +
++Core material: you have to know this by heart, and how to use it. +
++Main matter which you should know how to use. +
++Example of the concepts just covered. +
++Additional (contextual) information. +
++Additional historical context. +
+Created: 2022-02-07 Mon 08:02
+ +Online Lecture Notes
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++We have seen that a four-vector transforms according to +\[ + \bar{a}^\mu = \Lambda_\nu^\mu a^\nu +\] +in which \(\Lambda\) is a matrix representing the Lorentz transformation. +The form this matrix takes depends on the actual transformation: for the specific +case of motion in the \(x\) direction with velocity \(v\), +\[ + \Lambda = \left( \begin{array}{cccc} + \gamma & -\gamma \beta & 0 & 0 \\ + -\gamma \beta & \gamma & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 \end{array} \right). +\] +A four-vector is synonymous to a {\bf rank-one tensor}. +Higher-rank tensors are simply objects carrying more indices. +For example, a {\bf rank-two tensor} transforms as +\[ + \bar{t}^{\mu \nu} = \Lambda^\mu_\lambda \Lambda^\nu_\sigma t^{\lambda \sigma}. +\] +Such a rank-two tensor can be represented similarly to a matrix: +\[ + t^{\mu \nu} = \left( \begin{array}{cccc} + t^{00} & t^{01} & t^{02} & t^{03} \\ + t^{10} & t^{11} & t^{12} & t^{13} \\ + t^{20} & t^{21} & t^{22} & t^{23} \\ + t^{30} & t^{31} & t^{32} & t^{33} \end{array} \right). +\] +Special cases include symmetric \(t_s^{\mu \nu} = t_s^{\nu \mu}\) and +antisymmetric \(t_a^{\mu \nu} = -t_a^{\nu \mu}\) tensors. The latter contains +six independent elements: +\[ + t_a^{\mu \nu} = \left( \begin{array}{cccc} + 0 & t_a^{01} & t_a^{02} & t_a^{03} \\ + -t_a^{01} & 0 & t_a^{12} & t_a^{13} \\ + -t_a^{02} & -t_a^{12} & 0 & t_a^{23} \\ + -t_a^{03} & -t_a^{13} & -t_a^{23} & 0 \end{array} \right) +\] +Under the Lorentz transformation along \(x\) defined above, we can work out +how the nonvanishing elements of an antisymmetric tensor transform: +
+\begin{align} + \bar{t}_a^{01} &= \Lambda^0_\mu \Lambda^1_\nu t_a^{\mu \nu} + = \Lambda^0_0 \Lambda^1_0 t_a^{00} + \Lambda^0_1 \Lambda^1_0 t_a^{10} + + \Lambda^0_0 \Lambda^1_1 t_a^{01} + \Lambda^0_1 \Lambda^1_1 t_a^{11} \\ + & = (\Lambda^0_0 \Lambda^1_1 - \Lambda^0_1 \Lambda^1_0) t_a^{01} + = \gamma^2 (1 - \beta^2) t_a^{01} = t_a^{01}, \\ + \bar{t}_a^{02} &= \Lambda^0_\mu \Lambda^2_\nu t_a^{\mu \nu} + = \Lambda^0_0 \Lambda^2_2 t_a^{02} + \Lambda^0_1 \Lambda^2_2 t_a^{12} + = \gamma (t_a^{02} - \beta t_a^{12}), \\ + \bar{t}_a^{03} &= \Lambda^0_\mu \Lambda^3_\nu t_a^{\mu \nu} + = \Lambda^0_0 \Lambda^3_3 t_a^{03} + \Lambda^0_1 \Lambda^3_3 t_a^{13} + = \gamma (t_a^{03} - \beta t_a^{13}), \\ + \bar{t}_a^{12} &= \Lambda^1_\mu \Lambda^2_\nu t_a^{\mu \nu} + = \Lambda^1_0 \Lambda^2_2 t_a^{02} + \Lambda^1_1 \Lambda^2_2 t_a^{12} + = \gamma (t_a^{12} - \beta t_a^{02}), \\ + \bar{t}_a^{13} &= \Lambda^1_\mu \Lambda^3_\nu t_a^{\mu \nu} + = \Lambda^1_0 \Lambda^3_3 t_a^{03} + \Lambda^1_1 \Lambda^3_3 t_a^{13} + = \gamma (t_a^{13} - \beta t_a^{03}), \\ + \bar{t}_a^{23} &= \Lambda^2_\mu \Lambda^3_\nu t_a^{\mu \nu} + = \Lambda^2_2 \Lambda^3_3 t_a^{23} + = t_a^{23}. +\end{align} + + + ++Comparing with the transformation rules for the electromagnetic field which we obtained in (\ref{eq:EMFieldsLorentzTransfo}), we can define the +
++{\bf Electromagnetic Field Tensor} +\[ + F^{\mu \nu} = \left( \begin{array}{cccc} + 0 & E_x/c & E_y/c & E_z/c \\ + -E_x/c & 0 & B_z & -B_y \\ + -E_y/c & -B_z & 0 & B_x \\ + -E_z/c & B_y & -B_x & 0 \end{array} \right) + \] +
+ ++together with the handy +
++{\bf Dual Field Tensor} +\[ + G^{\mu \nu} = \left( \begin{array}{cccc} + 0 & B_x & B_y & B_z \\ + -B_x & 0 & -E_z/c & E_y/c \\ + -B_y & E_z/c & 0 & -E_x/c \\ + -B_z & -E_y/c & E_x/c & 0 \end{array} \right) + \] +
+ ++obtained from the field tensor by the substitution +\({\boldsymbol E}/c \rightarrow {\boldsymbol B}\), +\({\boldsymbol B} \rightarrow -{\boldsymbol E}/c\). +
+ + + ++Our electromagnetic field transformation laws then become the simple +
++{\bf Lorentz Transformation Rules for EM Fields} +\[ + \bar{F}^{\mu \nu} = \Lambda^\mu_\lambda \Lambda^\nu_\sigma F^{\lambda \sigma} + \] +
+ +Created: 2022-02-07 Mon 08:02
+ ++Now that we know that what one observer sees as an electric field +can be view by another as a magnetic field, we can ask the general +question of how fields transform upon Lorentz transformations. +
+ ++Let's start with what is perhaps the simplest case: the electric field +between the plates of an infinite parallel-plate capacitor. +For a surface charge density \(\sigma_0\) on bottom +and \(-\sigma_0\) on top plates (putting the plates perpendicular to \(\hat{\boldsymbol y}\)), this is +\[ + {\boldsymbol E}_0 = \frac{\sigma_0}{\varepsilon_0} \hat{\boldsymbol y}. +\] +
+ ++Let us now assume that we move to a reference frame moving +at velocity \(v_0\) in direction \(\hat{\boldsymbol x}\). +In this frame, the field between the plates +will still be along \({\boldsymbol y}\). In terms of the surface +charge density per plate \(\sigma\) as measured in this frame, +\[ + {\boldsymbol E} = \frac{\sigma}{\varepsilon_0} \hat{\boldsymbol y}. +\] +Lorentz contraction affects the +length scale longitudinal to the motion (it does not +affect the perpendicular length scales) so the surface charge density in the +moving frame becomes +\[ + \sigma = \gamma_0 \sigma_0, \hspace{10mm} + \gamma_0 = \frac{1}{\sqrt{1 - v_0^2/c^2}}. +\] +In the two frames, the electric fields perpendicular to the direction of motion are thus related by +\[ + {\boldsymbol E}^\perp = \gamma_0 {\boldsymbol E}_0^\perp. +\] +For the field parallel to the motion, we can simply repeat the +argument but now with motion along \({\boldsymbol y}\). +Since Lorentz contraction does not affect the surface charge +density, we get +\[ + E^\parallel = E_0^\parallel. +\] +
+ ++Going back to our setup with plates in the \(xz\) plane which we started from, +in the moving frame, there is now a magnetic field due to surface currents: +\[ + {\boldsymbol K}_{\mbox{\tiny top}} = \sigma v_0 \hat{\boldsymbol x} + = -{\boldsymbol K}_{\mbox{\tiny bot}}. +\] +This magnetic field between the plates is thus +\[ + {\boldsymbol B} = -\mu_0 \sigma v_0 ~\hat{\boldsymbol z}. +\] +
+ ++If we now have a further referential frame \(\bar{S}\) moving at velocity +\(\bar{v}\) with respect to the original one, we'd have +\[ + \bar{E}_y = \frac{\bar{\sigma}}{\varepsilon_0}, \hspace{10mm} + \bar{B}_z = - \mu_0 \bar{\sigma} \bar{v} +\] +where +\[ + \bar{v} = \frac{v + v_0}{1 + v v_0/c^2}, \hspace{10mm} + \bar{\sigma} = \bar{\gamma} \sigma, \hspace{10mm} + \bar{\gamma} = \frac{1}{1 - \bar{v}^2/c^2}. +\] +
+ + + ++We now want to express \(\bar{\boldsymbol E}, \bar{\boldsymbol B}\) in terms of +\({\boldsymbol E}, {\boldsymbol B}\) and other data in frame \({\cal S}\) +(here: \(v\)). To start, we have +\[ + \bar{E}_y = \frac{\bar{\gamma}}{\gamma_0} \frac{\sigma}{\varepsilon_0}, + \hspace{10mm} + \bar{B}_z = - \frac{\bar{\gamma}}{\gamma_0} \mu_0 \sigma \bar{v}. +\] +The ratio of contraction factors is +\[ + \frac{\bar{\gamma}}{\gamma_0} = \frac{\sqrt{1 - v_0^2/c^2}}{\sqrt{1 - \bar{v}^2/c^2}} + = \frac{\sqrt{c^2 - v_0^2} ~(1 + v v_0/c^2)}{\sqrt{c^2 (1 + vv_0/c^2)^2 - (v + v_0)^2}} + = \frac{1 + v v_0/c^2}{\sqrt{1 - v^2/c^2}} = \gamma (1 + vv_0/c^2). +\] +We can thus write +\[ + \bar{E}_y = \gamma (1 + v v_0/c^2) \frac{\sigma}{\varepsilon_0} + = \gamma \left( E_y - \frac{v}{c^2 \varepsilon_0 \mu_0} B_z \right) + = \gamma \left( E_y - v B_z \right) +\] +and +\[ + \bar{B}_z = -\gamma (1 + vv_0/c^2) \mu_0 \sigma \frac{v + v_0}{1 + vv_0/c^2} + = \gamma (B_z - \varepsilon_0 \mu_0 v E_y) + = \gamma (B_z - \frac{v}{c^2} E_y). +\] +
+ ++To do \(E_z\) and \(B_y\), simply put the capacitor in the \(xy\) plane. +Following the same argument, this gives +\[ + \bar{E}_z = \gamma (E_z + v B_y), \hspace{10mm} + \bar{B}_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right). +\] +We already know that \(\bar{E}_x = E_x\). For \(B_x\), we consider a +solenoid with axis along \(x\). The windings get tighter, +\(\bar{n} = \gamma n\) but the clock goes slower so \(\bar{I} = \frac{1}{\gamma} I\). +These factors cancel so \(\bar{B}_x = B_x\). +
+ + + ++We thus obtain the +
++{\bf EM field transformation laws (motion along \(x\) with velocity \(v\))} +
+\begin{align} + \bar{E}_x &= E_x, \hspace{10mm} & + \bar{E}_y &= \gamma (E_y - v B_z), \hspace{10mm} & + \bar{E}_z &= \gamma (E_z + v B_y), \\ + \bar{B}_x &= B_x, & + \bar{B}_y &= \gamma \left( B_y + \frac{v}{c^2} E_z \right), & + \bar{B}_z &= \gamma \left( B_z - \frac{v}{c^2} E_y \right) + \label{eq:EMFieldsLorentzTransfo} +\end{align} + ++Two special cases can be mentioned: +
+ ++\paragraph{If \({\boldsymbol B} = 0\) in \({\cal S}\):} +Then, \(\bar{\boldsymbol B} = \gamma \frac{v}{c^2} (E_z \hat{\boldsymbol y} - E_y \hat{\boldsymbol z}) = \frac{v}{c^2} (\bar{E}_z \hat{\boldsymbol y} - \bar{E}_y \hat{\boldsymbol z})\) so +\[ + \bar{\boldsymbol B} = -\frac{1}{c^2} {\boldsymbol v} \times \bar{\boldsymbol E}. +\] +
+ ++\paragraph{If \({\boldsymbol E} = 0\) in \({\cal S}\):} +Then, \(\hat{\boldsymbol E} = -\gamma v (B_z \hat{\boldsymbol y} - B_y \hat{\boldsymbol z}) = -v (\bar{B}_z \hat{\boldsymbol y} - \bar{B}_y \hat{\boldsymbol z})\) +so +\[ + \bar{\boldsymbol E} = {\boldsymbol v} \times \bar{\boldsymbol B}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Let us consider a charge density of moving sources. +For an infinitesimal cloud of volume \(V\) containing charge \(Q\) +moving at velocity \({\boldsymbol u}\), we have +\[ + \rho = \frac{Q}{V}, \hspace{10mm} + {\boldsymbol J} = \rho {\boldsymbol u}. +\] +In the rest frame of the charges, we have the proper density +\[ + \rho_0 = \frac{Q}{V_0} +\] +in terms of the rest volume \(V_0\), which is related to the perceived +volume \(V\) in the original frame by +\[ + V = \sqrt{1 - u^2/c^2} V_0. +\] +We thus obtain +\[ + \rho = \rho_0 \frac{1}{\sqrt{1 - u^2/c^2}}, \hspace{10mm} + {\boldsymbol J} = \rho_0 \frac{{\boldsymbol u}}{\sqrt{1 - u^2/c^2}}. +\] +Recognizing the proper velocity, we thus get that charge density and current density can together form the +
++{\bf Current density 4-vector} +\[ + J^\mu = \rho_0 \eta^\mu, \hspace{10mm} + J^\mu = \left( c\rho, J_x, J_y, J_z \right) + \] +
+ ++The continuity equation (\ref{eq:continuity}) takes the simple form +
++{\bf Continuity equation} +\[ + \frac{\partial J^\mu}{\partial x^\mu} = 0 + \] +
+ ++while similarly the notation simplifies for +
++{\bf Maxwell's equations} +\[ + \frac{\partial F^{\mu \nu}}{\partial x^\nu} = \mu_0 J^\mu, + \hspace{10mm} + \frac{\partial G^{\mu \nu}}{\partial x^\nu} = 0 + \] +
+ ++In terms of \(F^{\mu \nu}\) and the proper velocity \(\eta^\mu\), we also have the +
++{\bf Minkowski force on a charge \(q\)} +\[ + K^\mu = q F^{\mu \nu} \eta_\nu + \] +
+ ++whose vector components are +\[ + {\boldsymbol K} = \frac{q}{\sqrt{1 - u^2/c^2}} \left( + {\boldsymbol E} + {\boldsymbol u} \times {\boldsymbol B} \right) +\] +which becomes the Lorentz force law when remembering +(\ref{eq:MinkowskiForce}). +
+ + ++We had +\[ + {\boldsymbol E} = -{\boldsymbol \nabla} V - \frac{\partial {\boldsymbol A}}{\partial t}, \hspace{10mm} + {\boldsymbol B} = {\boldsymbol \nabla} \times {\boldsymbol A}. +\] +We can group the potentials together into a 4-vector: +\[ + A^\mu = \left( V/c, A_x, A_y, A_z \right). +\] +The field tensor is then expressed as +\[ + F^{\mu \nu} = \frac{\partial A^\nu}{\partial x_\mu} - \frac{\partial A^\mu}{\partial x_\nu}. +\] +The inhomogeneous Maxwell equation becomes +\[ + \frac{\partial F^{\mu \nu}}{\partial x^\nu} = \mu_0 J^\mu ~~\longrightarrow~~ + \frac{\partial}{\partial x_\mu} \left(\frac{\partial A^\nu}{\partial x^\nu} \right) - \frac{\partial}{\partial x_\nu} \left( \frac{\partial A^\mu}{\partial x^\nu} \right) = \mu_0 J^\mu +\] +We can now exploit gauge invariance +\[ + A^\mu ~~\longrightarrow~~ {A^\mu}^\prime = A^\mu + \frac{\partial \lambda}{\partial x_\mu} +\] +which leaves \(F^{\mu \nu}\) invariant. In particular, we can choose +the Lorenz gauge (\ref{eq:InhomogeneousMaxwellLorenzGauge}) here expressed as +\[ + \frac{\partial A^\mu}{\partial x^\mu} = 0. +\] +Defining the +
++{\bf d'Alembertian operator} +\[ + \square^2 \equiv \frac{\partial}{\partial x_\nu} \frac{\partial}{\partial x^\nu} = {\boldsymbol \nabla}^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} + \] +
+ ++we obtain the final form of +
+ ++{\bf Maxwell's equations (Lorenz gauge, 4-vector notation)} +\[ + \square^2 A^\mu = -\mu_0 J^\mu + \] +
+ ++which is the most aesthetically pleasing and practical form +of these equations. +
+Created: 2022-02-07 Mon 08:02
+ ++Electromagnetism is by construction fully compatible with special relativity. +Going further, special relativity means that the existence of electricity +implies the existence of magnetism, and vice-versa. +
+ ++To illustrate this, we take a simple example +
+ ++Consider a wire at rest in the lab frame. This wire carries a current made of a (positive) line charge density \(\lambda\) moving towards the right at velocity \(v\), and a (negative) line charge density \(-\lambda\) moving towards the left at velocity \(-v\). In the lab frame, the current is thus +\[ + I_{\mbox{\tiny lab}} = 2 \lambda v +\] +Additionally, in the lab frame, there is a charge \(q\) situated +at a distance \(s\) from the cable and moving with a velocity \(u < v\) parallel to the wire. In the lab frame, there is no electrical force between the wire and the charge, since the wire carries no net charge. +
+ ++Let us now examine this situation in the rest frame of the moving particle. +In this frame, the velocities of the line charges in the wire are given by +Einstein's velocity addition rule: +\[ + v_\pm = \frac{v \mp u}{1 \mp uv/c^2} +\] +Since \(v_- > v_+\), Lorentz contraction is stronger for the negative line charges +than for the positive ones. In this frame, the wire thus carries a nonzero net charge. The densities are +\[ + \lambda_\pm = \pm \gamma_\pm ~\lambda_0 +\] +where \(\lambda_0\) is the line charge density in the rest frame of the line charges, and +\[ + \gamma_\pm = \frac{1}{\sqrt{1 - v_\pm^2/c^2}}. +\] +The relationship between \(\lambda\) and \(\lambda_0\) is +\[ + \lambda = \gamma \lambda_0, \hspace{10mm} + \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. +\] +The dilation factors in the test particle's frame are thus +
+\begin{align} + \gamma_\pm &= \frac{1}{\sqrt{1 - (v \mp u)^2/(c^2 \mp uv)^2}} + = \frac{c^2 \mp uv}{\sqrt{(c^2 \mp uv)^2 - c^2 (v \mp u)^2}} \nonumber \\ + &= \frac{c^2 \mp uv}{\sqrt{(c^2 - v^2)(c^2 - u^2)}} + = \gamma \frac{1 \mp uv/c^2}{\sqrt{q - u^2/c^2}} +\end{align} ++so the resultant line charge in the test frame is +\[ + \lambda_{\mbox{\tiny test}} = \lambda_+ + \lambda_- + = \lambda_0 ( \gamma_+ - \gamma_-) + = -\frac{2\lambda u v}{c^2 \sqrt{1 - u^2/c^2}}. +\] +Therefore, as a result of Lorentz contraction, a current-carrying wire which is +neutral in one reference frame, can appear to be charged in another. +
+ ++In the frame of the test particle, there is an electric field +equal to that of a uniformly charged wire: +\[ + E = \frac{\lambda_{\mbox{\tiny test}}}{2\pi \epsilon_0 s} +\] +so the force (in the test frame) is +\[ + F_{\mbox{\tiny test}} = q E = -\frac{\lambda v}{\pi \epsilon_0 c^2 s} \frac{q u}{\sqrt{1 - u^2/c^2}}. +\] +If there is a force on our test charge in this test frame, there must +also be one in the lab frame. Using equation (\ref{eq:ForceTransfoSimple}), +\[ + F = \sqrt{1 - u^2/c^2} ~F_{\mbox{\tiny test}} + = -\frac{\lambda v}{\pi \varepsilon_0 c^2} \frac{qu}{s} +\] +which upon recognizing \(c^2 = \frac{1}{\varepsilon_0 \mu_0}\) +and the current \(I = 2 \lambda v\) becomes +\[ + F = - q u ~\frac{\mu_0 I}{2\pi s} +\] +which you will recognize as the magnetic part of the Lorentz force +for a charge \(q\) moving at velocity \(v\) in the presence of the +magnetic field of a long straight wire carrying current \(I\). +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++Newton's second law remains valid provided we use the relativistic momentum: +
++{\bf Newton's law (relativistic case)} +\[ + {\boldsymbol F} = \frac{d{\boldsymbol p}}{dt} + \] +
+ ++\paragraph{Example 12.10: Motion under a constant force.} A particle +of mass \(m\) is subjected to a constant force \(F\). If it starts at the +origin at \(t=0\), what is it's position as a function of time? +\paragraph{Solution:} +\[ + \frac{dp}{dt} = F, ~~p(0) = 0 ~~\longrightarrow~~ + p = Ft + \] +and thus +\[ + p(t) = \frac{m u(t)}{1 - u(t)^2/c^2} = Ft + ~~\longrightarrow~~ u(t) = \frac{(F/m)t}{\sqrt{1 + (Ft/mc)^2}}. + \] +Integrating again to get the displacement, +\[ + x(t) = \frac{F}{m}\int_0^t dt' \frac{t'}{\sqrt{1 + (Ft'/mc)^2}} + = \frac{mc^2}{F} \left[ \sqrt{1 + (Ft/mc)^2} - 1 \right]. + \] +The particle's world line thus shows {\bf hyperbolic motion}. +
+ ++\paragraph{Work and energy} +In the context of relativity, work is still the line integral of the force: +\[ + W \equiv \int {\boldsymbol F} \cdot d{\boldsymbol l} + \label{eq:RelativisticWork} +\] +The relationship between work done and increased energy also still holds: +\[ + W = \int d{\boldsymbol l} \cdot \frac{d{\boldsymbol p}}{dt} + = \int dt \frac{d {\boldsymbol l}}{dt} \cdot \frac{d{\boldsymbol p}}{dt} + = \int dt {\boldsymbol u} \cdot \frac{d{\boldsymbol p}}{dt} +\] +but +\[ + \frac{d{\boldsymbol p}}{dt} \cdot {\boldsymbol u} + = \frac{d}{dt} \left( \frac{m {\boldsymbol u}}{\sqrt{1 - u^2/c^2}} \right) \cdot {\boldsymbol u} + = \frac{d}{dt} \left( \frac{m c^2}{\sqrt{1 - u^2/c^2}} \right) + = \frac{dE}{dt} +\] +and we thus get +\[ + W = \int dt \frac{dE}{dt} = \Delta E +\] +
+ + + ++\paragraph{Force} +Since \({\boldsymbol F}\) involves a derivative with respect to ordinary time, +it does not transform well under Lorentz transformations. +For the example of motion along \(\hat{\boldsymbol x}\), +the transverse components transform similarly: +\[ + \bar{F}_y = \frac{d\bar{p}_y}{d\bar{t}} + = \frac{dp_y}{\gamma dt - \frac{\gamma \beta}{c} dx} + = \frac{dp_y/dt}{\gamma \left( 1 - \frac{\beta}{c} \frac{dx}{dt}\right)} + = \frac{F_y}{\gamma (1 - \beta u_x/c)}, + \hspace{6mm} + \bar{F}_z = \frac{F_z}{\gamma (1 - \beta u_x/c^2)} +\] +whereas the longitudinal component transforms in a complicated way: +\[ + \bar{F}_x = \frac{d\bar{p}_x}{d\bar{t}} + = \frac{\gamma dp_x - \gamma \beta dp^0}{\gamma dt - \frac{\gamma \beta}{c}dx} + = \frac{F_x - \frac{\beta}{c} \frac{dE}{dt}}{1 - \beta u_x/c} + = \frac{F_x - \beta({\boldsymbol u} \cdot {\boldsymbol F})/c}{1 - \beta u_x/c}. +\] +For the specific case where the particle is instantaneously at rest in +the original frame, then +\[ + \bar{\boldsymbol F}_\perp = \frac{1}{\gamma} {\boldsymbol F}_\perp, + \hspace{10mm} + \bar{F}_\parallel = F_\parallel. + \label{eq:ForceTransfoSimple} +\] +
+ ++The way to avoid complicated transformation rules is to define a four-vector +as the derivative of momentum with respect to proper time, which leads to +the definition of the +
++{\bf Minkowski force} +\[ + {\boldsymbol K} = \frac{d{\boldsymbol p}}{d\tau} + = \left( \frac{dt}{d\tau} \right) \frac{d{\boldsymbol p}}{dt} + = \frac{1}{\sqrt{1 - u^2/c^2}} {\boldsymbol F} + \label{eq:MinkowskiForce} + \] +
+ ++whose zeroth component is defined as \(1/c\) times the rate of increase of energy, +\[ + K^0 = \frac{dp^0}{d\tau} = \frac{1}{c} \frac{dE}{d\tau}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++If you move at velocity \(u\), then, as compared to a ground clock, +your time will go slower. For a ground clock time interval \(dt\), +your {\bf proper time} interval is +\[ + d\tau = \sqrt{1 - u^2/c^2}~dt. +\] +For the observer on the ground, your velocity is the rate of change +of your position \({\boldsymbol l}\) with respect to ground time: +\[ + {\boldsymbol u} = \frac{d {\boldsymbol l}}{dt} +\] +and this is called the {\bf ordinary velocity}. +The {\bf proper velocity} is defined as the hybrid-frame quantity +(distance as measured on the ground) divided by (proper time interval): +
++{\bf Proper velocity} +\[ + {\boldsymbol \eta} \equiv \frac{d {\boldsymbol l}}{d\tau} + \] +
+ ++Proper and ordinary velocity are thus related by +\[ + {\boldsymbol \eta} = \frac{1}{\sqrt{1 - u^2/c^2}} {\boldsymbol u}. +\] +The nice thing about proper velocity (as compared to ordinary velocity) +is that it transforms simply from one inertial system to another. +By adding the zeroth component +\[ + \eta^0 = \frac{dx^0}{d\tau} = c \frac{dt}{d\tau} = \frac{c}{\sqrt{1-u^2/c^2}} +\] +we can define the +
++{\bf four-velocity} or {\bf proper velocity four-vector} +\[ + \eta^\mu \equiv \frac{dx^\mu}{d\tau} + \] +which transforms as +\[ + \bar{\eta}^\mu = \Lambda^\mu_\nu \eta^\nu + \] +
+ ++(the transformation is simple because \(d\tau\) in the denominator is an +invariant). +
+ ++By contrast, the ordinary velocities obey cumbersome transformation rules: +for our usual relative frame velocity of \(v\) in the \(x\) direction, +\[ + \bar{u}_x = \frac{d\bar{x}}{dt} = \frac{u_x - v}{1-v u_x/c^2}, \hspace{10mm} + \bar{u}_y = \frac{d\bar{y}}{dt} = \frac{u_y}{1-v u_x/c^2}, \hspace{10mm} + \bar{u}_z = \frac{d\bar{z}}{dt} = \frac{u_z}{1-v u_x/c^2}. +\] +
+Created: 2022-02-07 Mon 08:02
+ ++The {\bf relativistic momentum} \({\boldsymbol p}\) is defined as +\[ + {\boldsymbol p} \equiv m {\boldsymbol \eta} = + \frac{m {\boldsymbol u}}{1 - u^2/c^2}. + \] +The {\bf relativistic energy} is defined as +\[ + E \equiv \frac{m c^2}{\sqrt{1 - u^2/c^2}}. + \] +These can be combined into the {\bf energy-momentum four-vector} +\[ + p^\mu \equiv m \eta^\mu. + \] +
+ ++When the object is stationary, its energy is the +
++{\bf Rest energy} +\[ + E_{\mbox{\tiny rest}} \equiv m c^2. + \] +
+ ++When moving, the difference between relativistic and rest energies +is the +
++{\bf Kinetic energy} +\[ + E_{\mbox{\tiny kin}} \equiv E - mc^2 = mc^2 \left( \frac{1}{\sqrt{1-u^2/c^2}} - 1 \right). + \] +
+ ++For velocities much smaller than the speed of light, we can expand +this to +\[ + E_{\mbox{\tiny kin}} = \frac{1}{2} mu^2 + \frac{3}{8} \frac{mu^2}{c^2} + ... +\] +
+ ++In a closed system, +
++{\bf Total relativistic energy and momentum is conserved} +\[ + E^2 - c^2 p^2 = m^2 c^4 + \] +
+ ++N.B.: don't confuse an {\bf invariant} quantity with a {\bf conserved} quantity. +
+Created: 2022-02-07 Mon 08:02
+ +Created: 2022-02-07 Mon 08:02
+ ++\paragraph{Four-vectors.} Let's introduce the standard notations +\[ + x^0 \equiv ct, \hspace{10mm} \beta \equiv \frac{v}{c}, + \hspace{10mm} x^1 = x, ~~x^2 = y, ~~x^3 = z. +\] +The Lorentz transformation then reads +
++{\bf Lorentz transformation (motion along \(x\) at velocity \(v\))} +\[ + \bar{x}^0 = \gamma \left( x^0 - \beta x^1 \right), + ~~~~\bar{x}^1 = \gamma \left( x^1 - \beta x^0 \right), + ~~~~\bar{x}^2 = x^2, + ~~~~\bar{x}^3 = x^3 + \] +or in matrix form +\[ + \left( \begin{array}{c} \bar{x}^0 \\ \bar{x}^1 \\ \bar{x}^2 \\ \bar{x}^3 + \end{array} \right) + = \left( \begin{array}{cccc} + \gamma & -\gamma \beta & 0 & 0 \\ + -\gamma \beta & \gamma & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 \end{array} \right) + \left( \begin{array}{c} x^0 \\ x^1 \\ x^2 \\ x^3 \end{array} \right) + \] +
+ ++This can be compactly written as +\[ + \bar{x}^\mu = \sum_{\nu = 0}^3 \Lambda^\mu_\nu x^\nu. +\] +
+ ++\paragraph{Covariant and contravariant vectors.} Four-vectors with +upper index are called {\it contravariant}. Their lower-index +counterparts are called {\it covariant} vectors and are obtained by +using the Minkowski metric \(g_{\mu \nu}\) according to +
++\[ + a_\mu = \sum_{\nu = 0}^3 g_{\mu \nu} a^\nu, \hspace{10mm} + g_{\mu \nu} = \left( \begin{array}{cccc} + -1 & 0 & 0 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 \end{array} \right) + \] +
+ ++\paragraph{Scalar products} are defined as the in-product of covariant/contravariant four-vectors, +
++\[ + \sum_{\mu = 0}^3 a^\mu b_\mu \equiv a^\mu b_\mu + \] +
+ ++where in the right-hand side we have introduced the +{\bf Einstein summation convention}, namely that any repeated index +is implicitly summed over. As you can trivially check, it doesn't matter +which vector is co/contravariant: \(a^\mu b_\mu = a_\mu b^\mu\). +Scalar products are Lorentz-invariant and thus take the same value in +all inertial systems. +
+ ++\paragraph{Invariant intervals.} Generalizing the notion of the norm of +a vector, the scalar product of a four-vector with itself is known as +the invariant interval. Because of the geometry of spacetime, the invariant +can take positive or negative values. The nomenclature goes as follows: +
+\begin{center} + \begin{tabular}{cc} + $a^\mu a_\mu > 0$ & $a^\mu$ is {\it spacelike} \\ + $a^\mu a_\mu < 0$ & $a^\mu$ is {\it timelike} \\ + $a^\mu a_\mu = 0$ & $a^\mu$ is {\it lightlike} + \end{tabular} +\end{center} ++For two events \(A\) and \(B\), the difference +\[ + \Delta x^\mu \equiv x_A^\mu - x_B^\mu +\] +is called the {\bf displacement four-vector} and its self-scalar product +is the {\bf invariant interval} between the two events: +\[ + I \equiv \Delta x^\mu \Delta x_\mu = -c^2 \Delta t^2 + |{\boldsymbol x}|^2 +\] +where \(t\) is the time difference between the events and \({\boldsymbol x}\) +is their spatial separation vector. +
+ ++\paragraph{Spacetime diagrams.} These are also know as +{\it Minkowski diagrams}. Time is on the vertical axis, space on the +horizontal one. The trajectory of a particle is known as its +{\bf world line}. Light is represented as propagating at lines at +45 degrees, defining the {\bf forward} and +{\bf backward light cones}. Lorentz transformations, which preserve +all invariant intervals, move spacetime points around but leave them +on the same hyperboloid. +
+Created: 2022-02-07 Mon 08:02
+ ++To talk about coordinate transformations, it is necessary to talk about +{\it events}, namely occurrences at a specific point in space and time. +
+ ++We will use two inertial frames of reference: \({\cal S}\) and \(\bar{\cal S}\). +We will assume that \(\bar{\cal S}\) is moving relative to \({\cal S}\) +at velocity \(v\) in the positive \(x\) direction. +
+ ++Imagine we have an event \(E\) occurring at \((t, x, y, z)\). +We would like to know the coordinates \((\bar{t}, \bar{x}, \bar{y}, \bar{z})\) +in system \(\bar{\cal S}\). +
+ ++If the clock is started at the moment at which the origins \({\cal O}\) +and \(\bar{\cal O}\) pass each other, then after a time \(t\), +\(\bar{\cal O}\) will be at a distance \(vt\) from \({\cal O}\) +and we would have +\[ + x = d + vt, +\] +where \(d\) is the distance from \(\bar{\cal O}\) to \(\bar{\cal A}\) +(\(\bar{\cal A}\) being the point on the \(\bar{x}\) axis that is even +with \(E\) when the event occurs). +
+ ++If Galilean physics is used, then \(d = \bar{x}\) and the transformation +rules are +\[ + \bar{t} = t, ~~\bar{x} = x - vt, ~~\bar{y} = y, ~~\bar{z} = z. +\] +If however we properly take into account Lorentz contraction, we +must set +\[ + d = \frac{1}{\gamma} \bar{x} ~~\longrightarrow~~ \bar{x} = \gamma (x - vt). +\] +
+ ++Doing the same argument from the point of view of an observer at +rest in frame \(\bar{S}\), we would write +\[ + \bar{x} = \bar{d} - v \bar{t} +\] +where \(\bar{d}\) is the distance from \({\cal O}\) to \({\cal A}\) at time +\(\bar{t}\) (\({\cal A}\) being the point on the \(x\) axis that is even with +\(E\) when the event occurs). Again invoking Lorentz, +\[ + \bar{d} = \frac{1}{\gamma} x ~~\longrightarrow~~ x = \gamma(\bar{x} + v \bar{t}). +\] +Solving these relations yields the dictionary for +
++{\bf Lorentz transformations (motion along \(x\) at velocity \(v\))} +
+\begin{align} + \bar{t} &= \gamma \left( t - \frac{v}{c^2} x \right), & \bar{y} &= y, \nonumber\\ + \bar{x} &= \gamma \left( x - vt \right), & \bar{z} &= z +\end{align} + ++\paragraph{Einstein's velocity addition rule.} Using these rules, +one can show that velocities add as +\[ + v_{13} = \frac{v_{12} + v_{23}}{1 + v_{12}v_{23}/c^2} +\] +
+Created: 2022-02-07 Mon 08:02
+ ++Einstein's postulates: +
+ ++This has a number of important consequences. +
+ ++\paragraph{Relativity of simultaneity:} two events which are simultaneous in one reference frame, are not necessarily simultaneous in another one. +
+ ++\paragraph{Time dilation:} example of a light ray in a travelling train car. +For the observer inside the car: \(\Delta t_{\mbox{\tiny car}} = h/c\). +For an observer on the ground, if the train is moving at velocity \(v\), +then \(\Delta t_{\mbox{\tiny gr}} = \sqrt{h^2 + v^2 \Delta t_{\mbox{\tiny gr}}^2}/c\) so +\[ + \Delta t_{\mbox{\tiny gr}} = \frac{h}{c} \frac{1}{\sqrt{1 - v^2/c^2}} +\] +and we get +\[ + \Delta t_{\mbox{\tiny tr}} = \sqrt{1 - v^2/c^2}~ \Delta t_{\mbox{\tiny gr}} +\] +so the time interval in the train is shorter, namely there is a +
++{\bf Time dilation factor} +\[ + \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} + \label{eq:Gamma} + \] +
+ ++\paragraph{Lorentz contraction:} lengths are also modified. +Back to our train, with a mirror on one end. A light signal +is sent from the opposite end, and the time for the round-trip +of the light is measured. For the observer on the +train, the time is \(\Delta t_{\mbox{\tiny tr}} = 2 \Delta x_{\mbox{\tiny tr}}/c\) +with \(\Delta x_{\mbox{\tiny tr}}\) being the length of the train car. +For the observer on the ground, the total time is made up of the +back and forth journey of the light, with times +\[ + \Delta t_{\mbox{\tiny gr,1}} = \frac{\Delta x_{\mbox{\tiny gr}} + v \Delta t_{\mbox{\tiny gr,1}}}{c}, \hspace{10mm} + \Delta t_{\mbox{\tiny gr,2}} = \frac{\Delta x_{\mbox{\tiny gr}} - v \Delta t_{\mbox{\tiny gr,2}}}{c} +\] +so +\[ + \Delta t_{\mbox{\tiny gr,1}} = \frac{\Delta x_{\mbox{\tiny gr}}}{c-v}, \hspace{10mm} + \Delta t_{\mbox{\tiny gr,2}} = \frac{\Delta x_{\mbox{\tiny gr}}}{c+v} +\] +and thus +\[ + \Delta t_{\mbox{\tiny gr}} = + \Delta t_{\mbox{\tiny gr,1}} + \Delta t_{\mbox{\tiny gr,2}} + = \frac{2 \Delta x_{\mbox{\tiny gr}}}{c} \frac{1}{1 - v^2/c^2}. +\] +Using the time dilation relation then gives +
++{\bf Lorentz contraction} +\[ + \Delta x_{\mbox{\tiny tr}} = \frac{1}{\sqrt{1 - v^2/c^2}} \Delta x_{\mbox{\tiny gr}} + \] +
+ ++Note that a moving object is only contracted in its direction of motion. +
+Created: 2022-02-07 Mon 08:02
+ +