diff --git a/build/ems_ms_ce.html b/build/ems_ms_ce.html new file mode 100644 index 0000000..22831b4 --- /dev/null +++ b/build/ems_ms_ce.html @@ -0,0 +1,1702 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+

Charge Conservation and the Continuity Equation + + +ems.ms.ce

+
+

+As an aside, now that we have defined our volume current density vcurd, we can obtain +one of the most important and useful equations in electrodynamics. +

+ +

+Consider a generic surface \({\cal S}\). The total current \(I\) going through it is +\[ +I = \int_{\cal S} {\bf J} \cdot d{\bf a} +\] +Specializing to a closed surface, we can use the divergence theorem to rewrite this as +\[ +\oint_{\cal S} d{\bf a} \cdot {\bf J} = \int_{\cal V} d\tau ~{\boldsymbol \nabla} \cdot {\bf J} +\] +Let us now invoke the principle of charge conservation. By its definition, the total current going +through the surface is (minus) the change of charge contained inside the surface per unit time, +\[ +\int_{\cal V} d\tau ~{\boldsymbol \nabla} \cdot {\bf J} = -\frac{d}{dt} \int_{\cal V} d\tau ~\rho += - \int_{\cal V} d\tau ~\frac{\partial \rho}{\partial t}. +\] +Since this is valid for any volume, we get the +

+
+

+Continuity Equation +

+
+

+ + + + +

+
+
    +
  • FLS II (13.8)
  • +
  • Gr (5.29)
  • +
  • PM (4.10)
  • +
+ +
+ +
+

+\[ +{\boldsymbol \nabla} \cdot {\bf J} + \frac{\partial \rho}{\partial t} = 0 +\tag{conteq}\label{conteq} +\] +

+ +
+ +

+Although you might think that we've made assumptions (e.g. about the current being steady), +it turns out (as we will see later after studying the Maxwell equations) +that the continuity equation is a fundamental consequence of the basic laws of +electrodynamics, and is always obeyed. It is a simple consequence of charge conservation +and locality. +

+
+
+ + + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/ems_ms_dcB_c.html b/build/ems_ms_dcB_c.html new file mode 100644 index 0000000..1471d71 --- /dev/null +++ b/build/ems_ms_dcB_c.html @@ -0,0 +1,1916 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Curl of \({\bf B}\) from Biot-Savart; Ampère's Law + + +ems.ms.dcB.c
+
+

+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 curl_xprod +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}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) ++ ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} +\end{align} +

+The first term can be reworked into +\[ +-{\bf J} ({\bf r}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) += -{\bf J} ({\bf r}') \left({\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|}\right) = -{\bf J} ({\bf r}') \left(-4\pi \delta^{(3)} ({\bf r} - {\bf r}') \right) +\label{Gr(5.51)} +\] +where we have used Lap1or. Back in Biot-Savart, 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 that for a symmetric function of \({\bf r} - {\bf r}'\), +changing from \(\nabla\) to \(\nabla'\) simply changes the sign, i.e. +\[ +({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} += -({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}') {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} +\] +This allows us 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 div_prod +(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 +

+
+

+Ampère's law +

+
+

+ + + + +

+
+
    +
  • FLS II (13.13)
  • +
  • Gr (5.56)
  • +
  • W (15-12)
  • +
+ +
+ +
+

+\[ +{\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} +\tag{Amp}\label{Amp} +\] +

+ +
+

+(in differential form). Using Stokes' theorem, +\[ +\int_{\cal S} d{\bf a} \cdot {\boldsymbol \nabla} \times {\bf B} = \oint_{\cal P} d{\bf l} \cdot {\bf B} = \mu_0 \int_{\cal S} d{\bf a} \cdot {\bf J} +\] +yielding +

+
+

+Ampère's law (integral form) +

+
+

+ + + + +

+
+
    +
  • FLS II (13.16)
  • +
  • Gr (5.57)
  • +
  • PM (6.25)
  • +
+ +
+ +
+

+\[ +\oint_{\cal P} d{\bf l} \cdot {\bf B} = \mu_0 I_{enc} \hspace{2cm} +I_{enc} = \int_{\cal S} d{\bf a} \cdot {\bf J}. +\label{Gr(5.55)} +\] +

+ +
+

+where \(I_{enc}\) is the current enclosed in the 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. +

+ +
+

+Example: infinite wire +

+ +

+Same as direct integration we did before, but let's now use Ampère. +

+ +

+Solution: by symmetry, \({\bf B}\) is circumferential and can only depend on \(r\). Then, +choosing an amperian loop at a fixed radius \(r\), we get +\[ +\oint d{\bf l} \cdot {\bf B} = B 2\pi s = \mu_0 I ~~\Rightarrow~~ B = \frac{\mu_0 I}{2\pi r} +\] +

+ +
+ +
+

+Example: uniform sheet current +

+ +

+Consider an infinite sheet with uniform surface current \({\bf K} = K \hat{\bf x}\) flowing in \(xy\) plane. +

+ +

+Task: calculate \({\bf B}\). +

+ +

+Solution: from Biot-Savart BiotSavart_s, \({\bf B}\) 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: +\[ +\oint {\bf B} \cdot d{\bf l} = 2B l = \mu_0 I_{enc} = \mu_0 K l +~~\Rightarrow~~ +{\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0}{2} K \hat{\bf y}, & z < 0, \\ +-\frac{\mu_0}{2} K \hat{\bf y}, & z > 0. \end{array} \right. +\label{Gr(5.56)} +\] +

+ +
+ +
+

+Example: solenoid +

+ +

+Consider an infinite solenoid along \(\hat{\bf z}\), made of a wire carrying current \(I\) +doing \(n\) turns per unit length on a cylinder of radius \(R\). +

+ +

+Task: calculate \({\bf B}\). +

+ +

+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: +

+
+

+ + + + +

+
+
    +
  • Gr (5.59)
  • +
+ +
+ +
+

+\[ +\oint d{\bf l} \cdot {\bf B} = Bl = \mu_0 I_{enc} = \mu_0 I n l ~~\Rightarrow~~ +{\bf B} = \left\{ \begin{array}{cc} \mu_0 I n ~\hat{\bf z}, & r < R, \\ +0, & r > R \end{array} \right. +\tag{Bsol}\label{Bsol} +\] +

+ +
+ +

+N.B.: the situations in which Ampère's law can be useful are those involving some +form of symmetry, for example +i) infinite straight lines, ii) infinite planes, iii) infinite solenoids, iv) toroids. +

+ +
+

+Example: toroidal coil +

+ +

+Consider a toroidal coil +(it doesn't matter what the cross-sectional shape, as long as it is rotationally symmetric). +

+ +

+Task: calculate \({\bf B}\). +

+ +

+Solution: by rotational symmetry, the magnetic field is circumferential everywhere. +Outside the coil, the field must again be zero. +

+ +

+Consider now an Amperian loop which is just a circle at radius \(r\): +

+
+

+ + + + +

+
+
    +
  • Gr (5.60)
  • +
+ +
+ +
+

+\[ +B 2\pi r = \mu_0 I_{enc} ~~\Rightarrow~~ +{\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0 N I}{2\pi r} \hat{\boldsymbol \varphi}, & \mbox{inside coil}, \\ +0, & \mbox{outside} \end{array} \right. +\tag{Btor}\label{Btor} +\] +

+ +
+
+
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/ems_ms_dcB_d.html b/build/ems_ms_dcB_d.html new file mode 100644 index 0000000..61508d4 --- /dev/null +++ b/build/ems_ms_dcB_d.html @@ -0,0 +1,1702 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Divergence of \({\bf B}\) from Biot-Savart + + +ems.ms.dcB.d
+
+

+For a completely generic situation, we can more formally compute the +divergence and curl of \({\bf B}\) directly from Biot-Savart (here taking for example the volume current +density version BiotSavart_v) +

+ +

+\[ +{\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} +\] +Applying the divergence gives +\[ +{\boldsymbol \nabla} \cdot {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~{\boldsymbol \nabla} \cdot +\left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) +\label{Gr(5.46)} +\] +Note that we can write (using div1or) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). Substituting this in and using product rule div_xprod, 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 +

+
+
+

+ + + + +

+
+
    +
  • FLS II (13.12)
  • +
  • Gr (5.50)
  • +
  • PM (6.28)
  • +
+ +
+ +
+

+\[ +{\boldsymbol \nabla} \cdot {\bf B} = 0 +\tag{divB0}\label{divB0} +\] +

+ +
+

+The divergence of a magnetic field is always zero. +Said equivalently (and thinking of the divergence theorem and the magnetic equivalent of +Gauss's law): there are no magnetic charges. +

+
+
+ + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/ems_ms_dcB_iw.html b/build/ems_ms_dcB_iw.html new file mode 100644 index 0000000..5a93f15 --- /dev/null +++ b/build/ems_ms_dcB_iw.html @@ -0,0 +1,1714 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Simplistic case: infinite wire + + +ems.ms.dcB.iw
+
+

+Let us start with the simple case of the field \({\bf B}\) coming from an infinite wire, as +given by Bwire1, rewritten here in cylindrical coordinates along the axis of the wire: +

+
+

+ + + + +

+
+
    +
  • Gr (5.39)
  • +
+ +
+ +
+

+\[ +{\bf B}_\mbox{wire} = \frac{\mu_0 I}{2\pi r} \hat{\boldsymbol \varphi} +\tag{Bwire}\label{Bwire} +\] +

+ +

+The first observation is that since the field is purely circumferential, its divergence +cyl_div manifestly vanishes, +\[ +\nabla \cdot {\bf B}_\mbox{wire} = 0 +\] +

+ +

+Looking for the curl, we might first want to apply cyl_curl on Bwire. The only term we should worry about is +\[ +\frac{1}{r} \frac{\partial}{\partial r} \left(r \times \frac{\mu_0 I}{2\pi r}\right) +\] +which is manifestly zero for \(r \neq 0\), but has an as-yet-unresolved \(0/0\) singular form at the origin. +To treat this term correctly, (and keeping Stokes in mind), we can take a step back and +calculate the line integral of \({\bf B}\) along a circular path of radius \(r\) centered on the wire. +For a generic path, +\(d{\bf l} = dr ~\hat{\bf r} + s d\varphi ~\hat{\boldsymbol \varphi} + dz ~\hat{\bf z}\) +so for a loop encircling the wire once, +\[ +\oint d{\bf l} \cdot {\bf B}_\mbox{wire} = \frac{\mu_0 I}{2\pi} \int_0^{2\pi} d\varphi \frac{1}{r} r = \mu_0 I. +\] +

+ +

+This immediately generalizes to a collection of straight wires: +\[ +\oint_{\cal P} d{\bf l} \cdot {\bf B} = \mu_0 I_{enc} +\label{Gr(5.42)} +\] +where \(I_{enc}\) is the current flow through a surface \({\cal S}\) defined by the closed path \({\cal P}\): +

+ +

+\[ +I_{enc} = \int_{\cal S} d{\bf a} \cdot {\bf J} +\] +Applying Stokes' theorem then immediately yields +

+ +

+\[ +{\boldsymbol \nabla} \times {\bf B}_\mbox{wire} = \mu_0 {\bf J} +\] +

+
+
+ + + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/ems_ms_lf_sc.html b/build/ems_ms_lf_sc.html new file mode 100644 index 0000000..53583f4 --- /dev/null +++ b/build/ems_ms_lf_sc.html @@ -0,0 +1,1804 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Steady Currents + + +ems.ms.lf.sc
+
+

+Consider an infinitesimal surface \(\Delta {\cal S}\) with normal unit vector \(\hat{\bf n}\). +

+ +

+In the region containing this surface, let there be a flow of charge described by a charge +density \(\rho\) moving at velocity \({\bf v}\). +

+ +

+The charge flowing through \(\Delta {\cal S}\) +in one unit of time is thus \(\Delta q = {\bf J} \cdot {\bf n} ~\Delta {\cal S} ~\Delta t\). +

+ +

+This leads us to define a current density \({\bf J}\) as the vector representing the +charge flowing through a unit area per unit time, with direction along the motion of the charges. +Then, \(\Delta q = \rho {\bf v} \cdot {\bf n} ~\Delta {\cal S} ~\Delta t\) so we can identify +

+
+

+ + + + +

+
+
    +
  • Gr (5.26)
  • +
  • PM (4.4)
  • +
+ +
+ +
+

+\[ +{\bf J} = \rho {\bf v} +\tag{vcurd}\label{vcurd} +\] +

+ +

+Similarly, a current running down a wire can be described by a linear charge density \(\lambda\) +moving at velocity \({\bf v}\), giving the current +

+
+

+ + + + +

+
+
    +
  • Gr (5.14)
  • +
+ +
+ +
+

+\[ +{\bf I} = \lambda {\bf v} +\tag{lcurd}\label{lcur} +\] +

+ +

+For charge flowing on a surface, we can describe the surface current density as a surface charge +density \(\sigma\) moving at velocity \({\bf v}\), +

+
+

+ + + + +

+
+
    +
  • Gr (5.23)
  • +
+ +
+ +
+

+\[ +{\bf K} = \sigma {\bf v} +\tag{scurd}\label{scurd} +\] +

+ +
+

+Magnetic forces on wire, surface and volume carrying current densities: +

+
+

+ + + + +

+
+
    +
  • Gr (5.16)
  • +
+ +
+ +
+

+\[ +{\bf F}_{mag} = \int {\bf v} \times {\bf B} ~\lambda dl = \int |{\bf I}| ~d{\bf l} \times {\bf B} +\tag{Fmagl}\label{Fmagl} +\] +

+
+

+ + + + +

+
+
    +
  • Gr (5.24)
  • +
+ +
+ +
+

+\[ +{\bf F}_{mag} = \int {\bf v} \times {\bf B} ~\sigma da = \int {\bf K} \times {\bf B} ~da +\tag{Fmags}\label{Fmags} +\] +

+
+

+ + + + +

+
+
    +
  • Gr (5.27)
  • +
+ +
+ +
+

+\[ +{\bf F}_{mag} = \int {\bf v} \times {\bf B} ~\rho d\tau = \int {\bf J} \times {\bf B} ~d\tau +\tag{Fmagv}\label{Fmagv} +\] +

+ +
+
+
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/ems_ms_vp_A.html b/build/ems_ms_vp_A.html new file mode 100644 index 0000000..11e700d --- /dev/null +++ b/build/ems_ms_vp_A.html @@ -0,0 +1,1897 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Definition; Gauge Choices + + +ems.ms.vp.A
+
+

+Since \({\boldsymbol \nabla} \cdot {\bf B} = 0\) in magnetostatics, following Helmholtz's theorem +(see c_m_vf_helm) we can write +

+
+
+

+ + + + +

+
+
    +
  • Gr (5.61)
  • +
+ +
+ +
+

+\[ +{\bf B} = {\boldsymbol \nabla} \times {\bf A} +\tag{BcurlA}\label{BcurlA} +\] +

+ +
+ +

+Substituting this in Ampère's law and using curlcurl gives +

+ +

+\[ +{\boldsymbol \nabla} \times {\bf B} = {\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf A} +) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf A}) - {\boldsymbol \nabla}^2 {\bf A} += \mu_0 {\bf J} +\] +

+ +

+In the case of electrostatics, we could add any constant to electrostatic potential +without affecting the physics. Here, in magnetostatics, we can +add any curlless function (said otherwise: the gradient of any scalar field) to the vector potential, +without changing the magnetic field. Making a specific choice is called making a gauge choice. +For example, we can always choose to eliminate the divergence of \({\bf A}\), giving us the example of the +

+
+

+Coulomb gauge +

+
+

+ + + + +

+
+
    +
  • Gr (5.63)
  • +
+ +
+ +
+

+\[ +{\boldsymbol \nabla} \cdot {\bf A} = 0. +\tag{CoulG}\label{CoulG} +\] +

+ +
+

+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, because it's mathematically exactly the same as +solving the Poisson equation 🐟 with p_vcd. Therefore, +provided \({\boldsymbol \nabla} \cdot {\bf A}_0\) goes to +zero at infinity, +\[ +\lambda ({\bf r}) = \frac{1}{4\pi} \int d\tau' \frac{{\boldsymbol \nabla}' \cdot{\bf A}_0 ({\bf r}')}{|{\bf r} - {\bf r}'|}. +\] +(exercise: prove this in one line using the Laplacian and Lap1or). +

+ +

+Under this gauge choice, Ampère's law becomes +

+
+
+

+ + + + +

+
+
    +
  • Gr (5.64)
  • +
+ +
+ +
+

+\[ +{\boldsymbol \nabla}^2 {\bf A} = -\mu_0 {\bf J} +\tag{Amp_CoulG}\label{Amp_CoulG} +\] +

+ +
+

+Again this is precisely like the Poisson equation, with one separate equation for each component. +For currents falling off sufficiently rapidly at infinity, we thus have +

+
+
+

+ + + + +

+
+
    +
  • Gr (5.65)
  • +
+ +
+ +
+

+\[ +{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{J({\bf r}')}{|{\bf r} - {\bf r}'|} +\tag{A_CoulG}\label{A_CoulG} +\] +

+ +
+

+For line and surface currents, +

+
+
+

+ + + + +

+
+ +
+ +
+

+\[ +{\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I ({\bf r}')}}{|{\bf r} - {\bf r}'|}, +\tag{A_CoulG_l}\label{A_CoulG_l} +\] +

+
+

+ + + + +

+
+ +
+ +
+

+\[ +{\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K ({\bf r}')}}{|{\bf r} - {\bf r}'|}. +\tag{A_CoulG_s}\label{A_CoulG_s} +\] +

+ +
+ + +
+

+Example: infinite solenoid +

+ +

+Task: find the vector potential of an infinite solenoid with \(n\) turns +pet unit length, radius \(R\) and current \(I\). +

+ +

+Solution: one cannot use Amp_CoulG since the current extends to infinity +(comment: you are encouraged to check that the integral converges anyway, by combining the integrals +for \(z > 0\) and \(z < 0\) into one). +

+ +

+Nice trick: notice that +

+ +

+\[ +\oint d{\bf l} \cdot {\bf A} = \int d{\bf a} \cdot ({\boldsymbol \nabla} \times {\bf A}) += \int d{\bf a} \cdot {\bf B} = \Phi. +\] +This is reminiscent of Ampère's law in integral form Amp_int, +\[ +\oint d{\bf l} \cdot {\bf B} = \mu_0 I_{enc}. +\] +Mathematically, it's the same equation! +Replacement: \({\bf B} \rightarrow {\bf A}\) and \(\mu_0 I_{enc} \rightarrow \Phi\). +And to paraphrase Feynman's lectures: the same equations have the same solutions. +

+ +

+Use symmetry: vector potential can only be cicumferential. Using an amperian loop at a radius +\(r\) inside the solenoid, and the fact that the field inside a solenoid is \(\mu_0 n I\) +(as per Btor), we get +\[ +\oint d{\bf l} \cdot {\bf A} = A (2\pi r) = \int d{\bf a} \cdot {\bf B} = \mu_0 n I (\pi r^2), +\] +so +

+ +

+\[ +{\bf A} = \frac{\mu_0 n I}{2} r ~\hat{\boldsymbol \varphi}, \hspace{1cm} r < R. +\] +For an amperian loop outside, the flux is always \(\mu_0 n I (\pi R^2)\), so +

+ +

+\[ +{\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{r} \hat{\boldsymbol \varphi}, \hspace{1cm} r > R. +\] +Exercise: check that \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) and that +\({\boldsymbol \nabla} \cdot {\bf A} = 0\). +

+ +
+
+
+ + + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_ld.html b/build/emsm_esm_ld.html new file mode 100644 index 0000000..d95cd15 --- /dev/null +++ b/build/emsm_esm_ld.html @@ -0,0 +1,1649 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ + +

Linear Dielectrics + + +emsm.esm.ld

+ +
+
+ + +
In this section:
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_ld_bvp.html b/build/emsm_esm_ld_bvp.html new file mode 100644 index 0000000..a7321a5 --- /dev/null +++ b/build/emsm_esm_ld_bvp.html @@ -0,0 +1,1885 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Boundary Value Problems with Linear Dielectrics + + +emsm.esm.ld.bvp
+
+

+In homogeneous linear dielectric, the bound charge density is directly +proportional to the free charge: +

+
+

+ + + + +

+
+
    +
  • Gr (4.39)
  • +
+ +
+ +
+

+\[ +\rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = - {\boldsymbol \nabla} \cdot \left(\varepsilon_0 \frac{\chi_e}{\varepsilon} {\bf D}\right) += -\left( \frac{\chi_e}{1 + \chi_e} \right) \rho_f +\tag{rhob_ld}\label{rhob_ld} +\] +If \(\rho = 0\), then any net charge is located on the surface, +and the potential then obeys the Laplace equation. +

+ +

+It is convenient to rewrite our boundary conditions in terms of free charge densities: +from Dperp_disc, +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.40)
  • +
+ +
+ +
+

+\[ +\varepsilon_{above} E^{\perp}_{above} - \varepsilon_{below} E^{\perp}_{below} = \sigma_f +\tag{Edisc_f}\label{Edisc_f} +\] +

+ +
+

+or in terms of the potential, +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.41)
  • +
+ +
+ +
+

+\[ +\varepsilon_{above} \frac{\partial \phi_{above}}{\partial n} - +\varepsilon_{below} \frac{\partial \phi_{below}}{\partial n} = -\sigma_f +\tag{dpdisc_f}\label{dpdisc_f} +\] +

+ +
+

+The potential itself is continuous, +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.42)
  • +
+ +
+ +
+

+\[ +\phi_{above} = \phi_{below} +\tag{pcont}\label{pcont} +\] +

+ +
+ + +
+

+Example: sphere of homogeneous dielectric material in uniform electric field +

+ +

+Consider a sphere made of a linear isotropic dielectric material and subjected to +an external electrostatic field \({\bf E}_0\). +

+ +

+Task: find the electric field inside the sphere. +

+ +

+Solution: resembles the case of the conducting sphere, here cancellation is not total. +

+ +

+Need to solve Laplace's equation for \(\phi(r, \theta)\) with boundary conditions +

+\begin{align*} +&(i)~~\phi_{in} (R,\theta) = \phi_{out} (R, \theta), \nonumber \\ +&(ii)~~\varepsilon \frac{\partial \phi_{in} (R,\theta)}{\partial n} += \varepsilon_0 \frac{\partial \phi_{out} (R,\theta)}{\partial n}, \nonumber \\ +&(iii)~~\phi_{out} (r) \rightarrow -E_0 r \cos \theta, ~~r \gg R. +\end{align*} +

+Inside and outside sphere: +\[ +\phi_{in} (r,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos \theta), \hspace{1cm} +\phi_{out} (r,\theta) = -E_0 r \cos \theta + \sum_{l=0}^\infty \frac{B_l}{r^{l+1}} P_l (\cos \theta). +\] +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, +\[ +\phi_{in} (r, \theta) = -\frac{3E_0}{\varepsilon_r + 2} r \cos \theta = -\frac{3E_0}{\varepsilon_r + 2} z, +\hspace{1cm} +{\bf E} = \frac{3}{\varepsilon_r + 2} {\bf E}_0. +\] +

+ +
+ + +
+

+Example: point charge above dielectric plane +

+ +

+Suppose that the whole region below \(z = 0\) is filled with a uniform linear dielectric with susceptibility \(\chi_e\). +

+ +

+Task: calculate the force on a point charge \(q\) situated a distance \(d\) above the origin. +

+ +

+Solution: bound surface charge is of opposite sign, force is attractive. No bound volume charge because of rhob_ld. Using sigmab and PchiE, +\[ +\sigma_b = {\bf P} \cdot \hat{\bf n} = P_z = \varepsilon_0 \chi_e E_z +\] +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 \(\pi\) -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 \((0,0,-d)\): +\[ +\phi (x,y,z>0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{\sqrt{x^2 + y^2 + (z-d)^2}} + \frac{q_b}{\sqrt{x^2 + y^2 + (z+d)^2}}\right] +\] +A charge \(q + q_b\) at \((0,0,d)\) gives +\[ +\phi (x,y,z<0) = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q + q_b}{\sqrt{x^2 + y^2 + (z-d)^2}}\right] +\] +Putting these two together yields a solution to Poisson going to zero at infinity, and is therefore the unique solution. +Correct discontinuity at \(z = 0\): +\[ +-\varepsilon_0 \left( \frac{\partial \phi}{\partial z}|_{z = 0^+} - \frac{\partial \phi}{\partial z}|_{z = 0^-} \right) += -\frac{1}{2\pi} \left(\frac{\chi_e}{\chi_e + 2}\right) \frac{qd}{(r^2 + d^2)^{3/2}}. +\] +Force on \(q\): +\[ +{\bf F} = \frac{1}{4\pi\varepsilon_0} \frac{q q_b}{4d^2} \hat{\bf z} = -\frac{1}{4\pi\varepsilon_0} \left( +\frac{\chi_e}{\chi_e + 2} \right) \frac{q^2}{4d^2} \hat{\bf z} +\label{Gr(4.54)} +\] +

+ +
+
+
+ + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_ld_e.html b/build/emsm_esm_ld_e.html new file mode 100644 index 0000000..66403a7 --- /dev/null +++ b/build/emsm_esm_ld_e.html @@ -0,0 +1,1713 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Energy in Dielectric Systems + + +emsm.esm.ld.e
+
+

+To charge a capacitor: +\[ +W = \frac{1}{2} C V^2 +\] +If capacitor filled with linear isotropic dielectric: +\[ +C = \varepsilon_r C_{vac} +\] +From electrostatics: +\[ +W = \frac{\varepsilon_0}{2} \int d\tau E^2 +\] +How is this changed? Counting only energy in fields: should decrease by +factor \(1/\varepsilon_r^2\). However, this would neglect the 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): +\[ +\Delta W = \int d\tau (\Delta \rho_f ({\bf r})) \phi ({\bf r}). +\label{Gr(4.56)} +\] +But \(\rho_f = {\boldsymbol \nabla} \cdot {\bf D}\) so \(\Delta \rho_f = {\boldsymbol \nabla} \cdot (\Delta {\bf D})\), so +\[ +\Delta W = \int d\tau ({\boldsymbol \nabla} \cdot (\Delta {\bf D})) \phi += \int d\tau {\boldsymbol \nabla} \cdot ((\Delta {\bf D}) \phi) + \int d\tau (\Delta {\bf D}) \cdot {\bf E} +\] +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: +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.58)
  • +
+ +
+ +
+

+\[ +W = \frac{1}{2} \int d\tau {\bf D} \cdot {\bf E} +\tag{WintDE}\label{WintDE} +\] +

+ +
+
+
+ + + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_ld_f.html b/build/emsm_esm_ld_f.html new file mode 100644 index 0000000..d453856 --- /dev/null +++ b/build/emsm_esm_ld_f.html @@ -0,0 +1,1757 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Forces on Dielectrics + + +emsm.esm.ld.f
+
+

+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 fringing field around edges. +

+ +

+Better: reason from energy. Pull dielectric out by \(dx\). Energy change equal to work done: +

+ +

+\[ +dW = F_{me} dx +\] +where \(F_{me}\) is mechanical force exerted by external agent. \(F_{me} = -F\), where \(F\) is +electrical force on dielectric. Electrical force on slab: +

+ +

+\[ +F = -\frac{dW}{dx} +\] +Energy stored in capacitor: +

+
+

+ + + + +

+
+
    +
  • Gr (4.61)
  • +
+ +
+ +
+

+\[ +W = \frac{1}{2} C V^2 +\tag{WcapV}\label{WcapV} +\] +Capacitance in configuration considered: +

+ +

+\[ +C = \frac{\varepsilon_0 w}{d} (\varepsilon_r l - \chi_e x) +\] +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 \(Q\), +

+
+

+ + + + +

+
+
    +
  • Gr (4.63)
  • +
+ +
+ +
+

+\[ +W = \frac{1}{2} \frac{Q^2}{C} +\tag{WcapQ}\label{WcapQ} +\] +so +

+ +

+\[ +F = -\frac{dW}{dx} = \frac{1}{2} \frac{Q^2}{C^2} \frac{dC}{dx} = \frac{1}{2} V^2 \frac{dC}{dx}. +\] +But +\[ +\frac{dC}{dx} = -\frac{\varepsilon_0 \chi_e w}{d} +\] +so +

+ +

+\[ +F = -\frac{\varepsilon_0 \chi_e w}{2d} V^2. +\] +

+ +

+Common mistake: to use WcapV (for \(V\) constant) instead of WcapQ +(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 +

+ +

+\[ +dW = F_{me} dx + V dQ +\] +and +

+ +

+\[ +F = -\frac{dW}{dx} + V \frac{dQ}{dx} = -\frac{1}{2} V^2 \frac{dC}{dx} + V^2 \frac{dC}{dx} = \frac{1}{2} V^2 \frac{dC}{dx} +\] +so like before but with the correct sign. +

+
+
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_ld_sp.html b/build/emsm_esm_ld_sp.html new file mode 100644 index 0000000..a4aecdc --- /dev/null +++ b/build/emsm_esm_ld_sp.html @@ -0,0 +1,1848 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Susceptibility, Permittivity, Dielectric Constant + + +emsm.esm.ld.sp
+
+

+For many substances: polarization is proportional to field, at least if the latter isn't too strong: +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.30)
  • +
+ +
+ +
+

+\[ +{\bf P} = \varepsilon_0 \chi_e {\bf E} +\tag{PchiE}\label{PchiE} +\] +

+ +
+

+Constant \(\chi_e\): called the electric susceptibility of the medium. Since \(\varepsilon_0\) is +there, \(\chi_e\) is dimensionless. Materials that obey PchiE are called +linear dielectrics. +

+ +

+Note: \({\bf E}\) on the RHS of PchiE is the total electric field, +due to free charges and to the polarization itself. Putting a dielectric in field \({\bf E}_0\), +we can't compute \({\bf P}\) directly from PchiE. +

+ +

+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 +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.32)
  • +
+ +
+ +
+

+\[ +{\bf D} = \varepsilon {\bf E} +\tag{DepsE}\label{DepsE} +\] +

+ +
+

+where \(\varepsilon\) is called the permittivity of the material. +In vacuum, susceptibility is zero, +permittivity is \(\varepsilon_0\). Also, +

+
+

+ + + + +

+
+
    +
  • Gr (4.34)
  • +
+ +
+ +
+

+\[ +\varepsilon_r \equiv 1 + \chi_e = \frac{\varepsilon}{\varepsilon_0} +\tag{epsr}\label{epsr} +\] +is called the relative permittivity or dielectric constant of the material. +This is all just nomenclature, everything is already in PchiE. +

+ + + +
+

+Example: metal sphere with insulator shell +

+ +

+Consider a metal sphere of radius \(a\) carrying charge \(Q\), surrounded out to radius \(b\) by +a linear dielectric material of permittivity \(\varepsilon\). +

+ +

+Task: find the potential at the center (relative to infinity). +

+ +

+Solution: we 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 GlD_int: +\[ +{\bf D} = \frac{Q}{4\pi r^2} \hat{\bf r}, \hspace{1cm} r > a. +\] +Inside sphere, \({\bf E} = {\bf P} = {\bf D} = 0\). Find \({\bf E}\) using DepsE: +\[ +{\bf E} = \left\{ \begin{array}{cc} +\frac{Q}{4\pi \varepsilon r^2} \hat{\bf r}, & a < r < b, \\ +\frac{Q}{4\pi \varepsilon_0 r^2} \hat{\bf r}, & r > b. \end{array} \right. +\] +The potential is thus +\[ +\phi = -\int_\infty^0 d{\bf l} \cdot {\bf E} = -\int_\infty^b dr \frac{Q}{4\pi\varepsilon_0 r^2} - \int_b^a dr \frac{Q}{4\pi\varepsilon r^2} += \frac{Q}{4\pi} \left( \frac{1}{\varepsilon_0 b} + \frac{1}{\varepsilon a} - \frac{1}{\varepsilon b} \right). +\] +

+ +

+It was thus not necessary to compute the polarization or the bound charge explicitly. +We can however extract those from the polarization, +\[ +{\bf P} = \varepsilon_0 \chi_e {\bf E} = \frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon r^2} \hat{\bf r}, +\] +giving +\[ +\rho_b = -{\boldsymbol \nabla} \cdot {\bf P} = 0, \hspace{1cm} +\sigma_b = {\bf P} \cdot \hat{\bf n} = \left\{ \begin{array}{cc} +\frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon b^2}, & \mbox{outer surface} \\ +-\frac{\varepsilon_0 \chi_e Q}{4\pi \varepsilon a^2}, & \mbox{inner surface} \end{array} \right. +\] +

+ +

+A dielectric is thus like an imperfect conductor: the charge \(Q\) is 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 \({\bf E}\)? No: if there is a boundary +between two materials with different dielectric constants, then a closed loop integral +of e.g. \({\bf P}\) would not vanish. +

+ +

+The only situation in which this parallel works is when space is +entirely filled with a homogeneous linear dielectric, but then this is just like +a vacuum with a different permittivity. +

+ +
+

+Example: parallel-plate capacitor filled with dielectric material +

+ +

+Consider a parallel-plate capacitor filled with insulating material of +dielectric constant \(\varepsilon_r\). +

+ +

+Task: determine the effect on the capacitance. +

+ +

+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 +

+ +

+\[ +C = \varepsilon_r C_{vac} +\] +

+ +
+
+
+ + + + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_mE.html b/build/emsm_esm_mE.html new file mode 100644 index 0000000..4572fc2 --- /dev/null +++ b/build/emsm_esm_mE.html @@ -0,0 +1,1647 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ + +

Matter Bathed in E Fields; Polarization + + +emsm.esm.mE

+ +
+
+ + +
In this section:
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_mE_P.html b/build/emsm_esm_mE_P.html new file mode 100644 index 0000000..b3bfa93 --- /dev/null +++ b/build/emsm_esm_mE_P.html @@ -0,0 +1,1654 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Polarization + + +emsm.esm.mE.P
+
+

+In a macroscopic sample subjected to an external electrostatic field, a complicated microscopic dance +is induced among the constituent charges. As a result, local induced/permanent dipole moments tend to somewhat cancel each other, leaving a residual polarization. The leading remnant can be captured in a vector +called the polarization and defined as +\[ +{\bf P} \equiv ~\mbox{dipole moment per unit volume} +\] +

+ +

+Though it's important to remember that one could also talk about higher moments than dipole moments, +for virtually all practical purposes the polarization is sufficient. +

+
+
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_mE_o.html b/build/emsm_esm_mE_o.html new file mode 100644 index 0000000..a158c41 --- /dev/null +++ b/build/emsm_esm_mE_o.html @@ -0,0 +1,1825 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Overview + + +emsm.esm.mE.o
+
+

+Let us briefly discuss how electric fields affect (charged) matter. +

+ +

+Effect on fundamental particles: admittedly a complicated business, for which quantum mechanics +is required. The assumption we make is that the charges of the particles we're dealing with are never +changed by anything that happens to them. This is however not true in the real world, because the +charges we measure are usually the remnants of some complicated intermingling of constituent particles. +For example, 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 much beyond the current course, and needs the whole machinery +of 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. +

+ +

+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 Schrödinger 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 Châtelier's principle tells us that the equilibrium position moves linearly with the strength of the +perburtation. +

+ +

+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 +(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 atomic polarizability. +

+ +

+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 polarizability tensor of the molecule. Always possible to use 'principal' axes +such that all but 3 of the terms cancel. +

+ +

+Polar molecules: unlike individual atoms, molecules can have a permanent dipole moment. These are called +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: +

+
+

+Torque on a dipole +

+
+

+ + + + +

+
+
    +
  • Gr (4.4)
  • +
+ +
+ +
+

+\[ +{\bf N} = {\bf p} \times {\bf E} +\tag{di_N}\label{di_N} +\] +

+ +
+

+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 +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.5)
  • +
+ +
+ +
+

+\[ +{\bf F} = ({\bf p} \cdot {\boldsymbol \nabla}) {\bf E} +\tag{di_F}\label{di_F} +\] +

+ +
+

+The energy of a dipole in an electric field can be shown to be +

+
+

+ + + + +

+
+
    +
  • Gr (4.6)
  • +
+ +
+ +
+

+\[ +U = -{\bf p} \cdot {\bf E} +\tag{di_E}\label{di_E} +\] +

+ +

+Many atoms: gases and liquids: 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. +

+ +

+Many atoms: solids: 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 ferroelectricity (after ferromagnetism, which is the correspdonding magnetic phenomenon happening with iron). +

+
+
+ + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_po.html b/build/emsm_esm_po.html new file mode 100644 index 0000000..30c2300 --- /dev/null +++ b/build/emsm_esm_po.html @@ -0,0 +1,1836 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ + +

Polarized Objects; Bound Charges + + +emsm.esm.po

+ +
+

+What is the electric field produced by an object with polarization \({\bf P}\)? +Let us try to address this question by Working with the potential. +For a single dipole, we have p_di +\[ +\phi({\bf r}) = \frac{1}{4\pi \varepsilon_0} \frac{({\bf r} - {\bf r}') \cdot {\bf p}}{|{\bf r} - {\bf r}'|^3} +\label{Gr(4.8)} +\] +Considering a dipole moment per unit volume \({\bf P}\) (our definition of the polarization), we get +

+
+

+ + + + +

+
+
    +
  • Gr (4.9)
  • +
+ +
+ +
+

+\[ +\phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P}({\bf r}')}{|{\bf r} - {\bf r}'|^3} +\tag{p_P}\label{p_P} +\] +This equation is perfectly acceptable and usable as it is. +There exists however another convenient representation. Starting from +\[ +{\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} +\] +we can write (using product rule div_prod) +

+\begin{align*} +\phi({\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 +

+\begin{equation*} +\phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint_{\cal S} d{\bf a}' \cdot \frac{{\bf P} ({\bf r}')}{|{\bf r} - {\bf r}'|} - \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \cdot {\bf P}({\bf r}')}{|{\bf r} - {\bf r}'|}. +\end{equation*} +

+Interpretation: first terms is like contribution of a surface charge, +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.11)
  • +
+ +
+ +
+

+\[ +\sigma_b({\bf r}) = {\bf P} ({\bf r}) \cdot \hat{\bf n} +\tag{sigmab}\label{sigmab} +\] +

+ +
+

+and second term looks like contribution of a volume charge, +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.12)
  • +
+ +
+ +
+

+\[ +\rho_b ({\bf r}) = -{\boldsymbol \nabla} \cdot {\bf P} ({\bf r}) +\tag{rhob}\label{rhob} +\] +

+ +
+

+Using these definitions, +

+
+
+

+ + + + +

+
+
    +
  • Gr (4.13)
  • +
+ +
+ +
+

+\[ +\phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \oint_{\cal S} da' \frac{\sigma_b ({\bf r}')}{|{\bf r} - {\bf r}'|} + \frac{1}{4\pi \varepsilon_0} \int_{\cal V} d\tau' \frac{\rho_b ({\bf r}')}{|{\bf r} - {\bf r}'|}. +\tag{p_bound}\label{p_bound} +\] +

+ +
+

+These bound charges faithfully represent the object's sources for electrical fields. +

+ + +
+

+Example: uniformly polarized sphere +

+ +

+Task: compute the electric field produced by a uniformly polarized sphere of radius \(R\). +

+ +

+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. p_uni_ch_sph): +\[ +\phi(r, \theta) = \left\{ \begin{array}{cc} +\frac{P}{3\varepsilon_0} r\cos \theta, & r \leq R \\ +\frac{P}{3\varepsilon_0} \frac{R^3}{r^2} \cos \theta, & r \geq R. +\end{array} \right. +\] +

+ +

+But \(r\cos \theta = z\), so the field inside the sphere is uniform, +

+ +

+\[ +{\bf E} = -{\boldsymbol \nabla} \phi = -\frac{P}{3\varepsilon_0} \hat{\bf z} = -\frac{1}{3\varepsilon_0} {\bf P}, +\hspace{1cm} r < R. +\] +Outside the sphere, the potential is identical to that of pure point dipole at origin, +

+ +

+\[ +\phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \frac{{\bf p} \cdot {\hat {\bf r}}}{r^2}, +\hspace{1cm} r > R +\] +where the total dipole moment is simply the integral over the polarization, +

+ +

+\[ +{\bf p} = \frac{4}{3} \pi R^3 {\bf P} +\] +

+ +
+
+ + +
In this section:
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_po_fid.html b/build/emsm_esm_po_fid.html new file mode 100644 index 0000000..c63d964 --- /dev/null +++ b/build/emsm_esm_po_fid.html @@ -0,0 +1,1832 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
The Field Inside a Dielectric + + +emsm.esm.po.fid
+
+

+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: +\[ +{\bf E} = {\bf E}_{out} + {\bf E}_{in}. +\] +

+ +

+In p_ball_avg, we established that +the average field (over a sphere) produced by charges outside the sphere is +equal to the field they produce at center of sphere. +Therefore, \({\bf E}_{out}\) is the field at \({\bf r}\) +from dipoles exterior to the sphere, for which we can use p_P: +

+ +
+

+ + + + +

+
+
    +
  • Gr (4.17)
  • +
+ +
+ +
+

+\[ +\phi_{out} ({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{out} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} +{|{\bf r} - {\bf r}'|^3} +\tag{p_out}\label{p_out} +\] +The dipoles inside the sphere however cannot be treated in this fashion. +However, their average field is all we need, and +this is given by (see Example below for the derivation) +

+
+

+ + + + +

+
+
    +
  • Gr (3.105)
  • +
+ +
+ +
+

+\[ +{\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{\bf p}{R^3} +\tag{Eavg_p}\label{Eavg_p} +\] +irrespective of charge distribution within sphere. +

+ +
+

+Average field inside a sphere +

+ +

+Task: show that the average field inside a sphere of radius \(R\) due to all charges within +the sphere is +\[ +{\bf E}_{av} = -\frac{1}{4\pi\varepsilon_0} \frac{{\bf p}}{R^3} +\] +where \({\bf p}\) is the dipole moment, irrespective of what the charge distribution is. +

+ +

+Solution: there are many ways to do this. +

+ +

+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 \(\rho = -q/(\frac{4}{3} \pi R^3)\), +\[ +\frac{1}{4\pi \varepsilon_0} \frac{1}{\frac{4}{3} \pi R^3} \int d\tau' q \frac{{\bf r}' - {\bf r}}{|{\bf r} - {\bf r}'|^3}. +\] +

+ +

+Solution a): 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. +

+ +

+b) Express the latter in terms of the dipole moment. +

+ +

+Solution b): from E_uni_sph, the field inside a uniformly charged sphere is +\[ +{\bf E} ({\bf r}) = \frac{1}{3\varepsilon_0} \rho \hat{\bf r} +\] +so +\[ +{\bf E}_{s} ({\bf r}) = \left. \frac{1}{3\varepsilon_0} \rho \hat{\bf r} \right|_{\rho = \frac{-q}{\frac{4}{3}\pi R^3}} = -\frac{q}{4\pi\varepsilon_0} \frac{\hat{\bf r}}{R^3} = -\frac{{\bf p}}{4\pi\varepsilon_0 R^3}. +\] +

+ +

+c) Generalize to an arbitrary charge distribution +

+ +

+Solution c): just use superposition! +

+ +

+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: +

+ +

+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 the same as the 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 +\[ +{\bf E}_{in} = -\frac{1}{3\varepsilon_0} {\bf P} +\label{Gr(4.18)} +\] +

+ +

+By assumption, \({\bf P}\) does not vary significantly over the volume of sphere. +The term left out of p_out is thus the field at the center of a uniformly polarized sphere. +But this is precisely equal to the \({\bf E}_{in}\) contribution, so macroscopic field is +given by the potential +

+ +

+\[ +\phi({\bf r}) = \frac{1}{4\pi\varepsilon_0} \int_{\cal V} d\tau' \frac{({\bf r} - {\bf r}') \cdot {\bf P} ({\bf r}')} +{|{\bf r} - {\bf r}'|^3} +\] +where \({\cal V}\) is the entire volume of the dielectric. +

+
+
+ + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file diff --git a/build/emsm_esm_po_pibc.html b/build/emsm_esm_po_pibc.html new file mode 100644 index 0000000..40dde5c --- /dev/null +++ b/build/emsm_esm_po_pibc.html @@ -0,0 +1,1665 @@ + + + + + + +Pre-Quantum Electrodynamics + + + + + + + + + +
+
+

+Pre-Quantum Electrodynamics +

+
+ +
+
Physical Interpretation of Bound Charges + + +emsm.esm.po.pibc
+
+

+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. +

+
+
+ + + +
+
+
+
+ +Creative Commons License + +Except where otherwise noted, all content is licensed under a +Creative Commons Attribution 4.0 International License. +
+
+

Author: Jean-Sébastien Caux

+

Created: 2022-03-15 Tue 08:10

+

+
+ + +
+ \ No newline at end of file