diff --git a/build/c_m_cs_sph.html b/build/c_m_cs_sph.html
index 4b25f18..2a57f3a 100644
--- a/build/c_m_cs_sph.html
+++ b/build/c_m_cs_sph.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1631,14 +1631,14 @@ which \(r\) is the distance from the chosen origin,
The usual Cartesian coordinates relate to spherical coordinates
according to
-
+
-
+
@@ -1663,14 +1663,14 @@ A generic vector can be expressed as
where the explicit relation between spherical and
Cartesian unit vectors is
-
+
-
+
@@ -1693,14 +1693,14 @@ and \(\hat{\boldsymbol \varphi} (\theta, \varphi)\).
An infinitesimal displacement \(d{\bf l}\) can be written as
-
+
-
+
@@ -1716,14 +1716,14 @@ d{\bf l} = dr ~\hat{\boldsymbol r} + r d\theta ~\hat{\boldsymbol \theta} + r\sin
Infinitesimal volume element:
-
+
-
+
@@ -1744,14 +1744,14 @@ Infinitesimal surface element: depends on situation.
\({\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} T) \equiv {\boldsymbol \nabla}^2 T\) 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
@@ -1634,44 +1634,44 @@ given by the Laplacian of the corresponding vector elements.
-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.
+Around 1831, Faraday performed a number of experiments pertaining to
+the effects of time-dependent fields.
-
-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
@@ -1647,27 +1643,39 @@ on this idea. Faraday observed transient current in a circuit when:
Faraday's big insight was to summarize these effects by noticing that
-
-
-
\[
- \boxed{
- \mbox{\bf A changing magnetic field induces an electric field}
- }
+\boxed{
+\mbox{A changing magnetic field induces an electric field.}
+}
\]
-
-
-
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
@@ -1676,21 +1684,56 @@ You can think of the emf in different ways. It's the energy accumulated as a uni
-The precise statement is that the electromotive force is proportional
+The precise statement associated to Faraday's observations
+is that the electromotive force is proportional
to the rate of change of the magnetic flux,
+
+
+
+
+
+
+
+
Gr (7.14)
+
+
+
+
+
+
\[
{\cal E} = \oint_{\cal P} {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt}
-\label{Gr(7.14)}
+\tag{Fl_flux}\label{Fl_flux}
\]
so we obtain
-
+
Faraday's law (integral form N.B.: for a stationary loop)
- \[
- \oint_{\cal P} {\bf E} \cdot d{\bf l} = -\int_{\cal S} \frac{\partial {\bf B}}{\partial t} \cdot d{\bf a}
- \label{Gr(7.15)}
- \]
+
-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
+Right-hand rule always sorts signs out. Easier rule: Lenz's law, which
+states that physical systems naturally resist a change in flux.
+This is in fact just
+Le Châtelier's principle of any action at an equilibrium point leading
to an opposing counter-reaction.
@@ -1739,7 +1798,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_Fl_e.html b/build/emd_Fl_e.html
index 750df8a..11abc16 100644
--- a/build/emd_Fl_e.html
+++ b/build/emd_Fl_e.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1631,7 +1631,7 @@ 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
+Nicer way (generalizable to surface and volume currents): from PLI, flux through loop is \(\Phi = L I\). But
\[
\Phi = \int_{\cal S} {\bf B} \cdot d{\bf a} = \int_{\cal S} ({\boldsymbol \nabla} \times {\bf A}) \cdot d{\bf a}
= \oint_{\cal P} {\bf A} \cdot d{\bf l},
@@ -1647,16 +1647,16 @@ W = \frac{1}{2} I \oint {\bf A} \cdot d{\bf l} = \frac{1}{2} \oint ({\bf A} \cdo
\]
Generalization to volume currents:
-
+
-
+
-
gr (7.31)
+
Gr (7.31)
@@ -1664,31 +1664,22 @@ Generalization to volume currents:
+{\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) =
+{\bf B} \cdot {\bf B} - {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}).
\]
Then,
\[
@@ -1698,12 +1689,27 @@ W = \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \int_{\cal V} d\tau {\bold
\]
We can integrate over all space: after neglecting boundary terms (assuming fields fall to zero at infinity), we are left with
@@ -1713,26 +1719,38 @@ We can integrate over all space: after neglecting boundary terms (assuming fiel
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)}
+W_{elec} &= \frac{1}{2} \int d\tau ~V\rho &= \frac{\varepsilon_0}{2} \int d\tau ~E^2, \\
+W_{mag} &= \frac{1}{2} \int d\tau ~({\bf A} \cdot {\bf J}) &= \frac{1}{2\mu_0} \int d\tau ~B^2,
\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,
+which are equations W_vcd, W_intEsq, W_intAJ and W_intBsq.
+
+
+
+
+Example: energy in coaxial cable
+
+
+
+Consider a coaxial cable with inner cylinder radius \(a\), outer \(b\),
+carrying current \(I\).
+
+
+
+Task: find the energy stored in a section of length \(l\).
+
+
+
+Solution: from Ampère,
\[
- {\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a < s < b, \hspace{1cm}
- {\bf B} = 0, \hspace{1cm} s < a ~\mbox{or}~ s > b.
- \]
+{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a < s < b, \hspace{1cm}
+{\bf B} = 0, \hspace{1cm} s < a ~\mbox{or}~ s > b.
+\]
Energy is thus
\[
- W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
- = \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}.
- \]
+W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \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}.
+\]
@@ -1758,7 +1776,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_Fl_i.html b/build/emd_Fl_i.html
index 261d578..16cf28d 100644
--- a/build/emd_Fl_i.html
+++ b/build/emd_Fl_i.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1629,7 +1629,7 @@ 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.
+where \(M_{21}\) is the mutual inductance of the two loops.
@@ -1638,7 +1638,7 @@ 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)}),
+But from A_CoulG,
\[
{\bf A}_1 ({\bf r}) = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_1} \frac{d{\bf l}_1}{|{\bf r} - {\bf r}_1|}
\]
@@ -1647,44 +1647,89 @@ 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},
+and we can write the mutual inductance as the Neumann formula,
+
-\paragraph{Example 7.10:}
-short solenoid (length \(l\), radius \(a\), \(n_1\) turns per unit length) lies concentrically inside
+Example: solenoid in solenoid
+
+
+
+Consider a short solenoid (length \(l\), radius \(a\), \(n_1\) turns per unit length)
+which 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
+
+
+
+Task: compute the flux through the long solenoid.
+
+
+
+Solution: it's 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)}),
+Field of outer solenoid: from Amp_int,
\[
- B = \mu_0 n_2 I
- \]
+B = \mu_0 n_2 I
+\]
so flux through a single loop of inner solenoid is
\[
- B \pi a^2 = \mu_0 n_2 I \pi a^2.
- \]
+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
\[
- \Phi = \mu_0 \pi a^2 n_1 n_2 l I.
- \]
+\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
\[
- M = \mu_0 \pi a^2 n_1 n_2 l.
- \]
+M = \mu_0 \pi a^2 n_1 n_2 l.
+\]
@@ -1697,62 +1742,101 @@ What if we vary current in loop 1? Flux in 2 will vary. Induces EMF in loop 2:
\label{Gr(7.24)}
\]
Changing current also induces EMF in the source loop itself:
+
+
+
+
+
+
+
+
Gr (7.25)
+
+
+
+
+
+
\[
\Phi = L I
-\label{Gr(7.25)}
+\tag{PLI}\label{PLI}
\]
-where \(L\) is the {\bf self-inductance} (or inductance) of the loop. Depends only on
+where \(L\) is the 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} (\(H\)). \(H = V s/A\).
+Inductance: measured in henries (\(H\)). \(H = V s/A\).
-
+
-\paragraph{Example 7.11:} find self-inductance of toroidal coil with
+Example: self-inductance of toroidal coil
+
+
+
+Consider a 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)})
+
+
+
+Task: find its self-inductance
+
+
+
+Solution: magnetic field inside toroid is Btor
\[
- B = \frac{\mu_0 NI}{2\pi s}
- \]
+B = \frac{\mu_0 NI}{2\pi s}
+\]
Flux through single turn:
\[
- \int {\bf B} \cdot d{\bf a} = \frac{\mu_0 N I}{2\pi} h \int_a^b \frac{ds}{s}
- = \frac{\mu_0 N I h}{2\pi} \ln \frac{b}{a}.
- \]
+\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
\[
- L = \frac{\mu_0 N^2 h}{2\pi} \ln \frac{b}{a}
- \label{Gr(7.27)}
- \]
+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}.
+Inductance (like capacitance) is intrinsically positive. Use Lenz law.
+Think of back EMF.
-
+
-\paragraph{Example 7.12:} circuit with inductance \(L\), resistor \(R\) and battery \({\cal E}_0\).
-What is the current ?
-\paragraph{Solution:}
+Example: circuit
+
+
+
+Consider a circuit with inductance \(L\), resistor \(R\) and battery \({\cal E}_0\).
+
+
+
+Task: find the current
+
+
+
+Solution:
Ohm's law:
\[
- {\cal E}_0 - L \frac{dI}{dt} = IR \Longrightarrow I(t) = \frac{{\cal E}_0}{R} + k e^{-(R/L)t}.
- \]
+{\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.
+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 time constant of the circuit.
@@ -1777,7 +1861,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_Fl_ief.html b/build/emd_Fl_ief.html
index dd2af9b..5c6e2a3 100644
--- a/build/emd_Fl_ief.html
+++ b/build/emd_Fl_ief.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1628,7 +1628,7 @@ 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!
+between Ampère and Faraday
\[
{\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}
\hspace{3cm}
@@ -1645,42 +1645,60 @@ law in integral form:
-
+
-{\bf Example 7.7:}
-\({\bf B}(t)\) points up in circular region of radius \(R\). What is the induced \({\bf E}(t)\) ?
-\paragraph{Solution:}
+Example: loop with time-dependent flux
+
+
+
+Consider a time-dependent magnetic field \({\bf B}(t)\) directed vertically
+through a horizontal circular region of radius \(R\).
+
-{\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.
+Example: wheel with charged rim traversed by flux
+
+
+
+Consider a wheel of radius \(b\) with line charge \(\lambda\) on the rim.
+A uniform magnetic field \({\bf B}_0\) pointing up is traversing the central region
+up to radius \(a < b\). The field is then turned off. What happens?
+
+
+
+Solution: the wheel starts spinning to compensate the reduction of field.
Faraday:
\[
- \oint {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt} = - \pi a^2 \frac{dB}{dt}
- \Rightarrow {\bf E} = -\frac{a^2}{2b} \frac{dB}{dt} \hat{\boldsymbol \varphi}.
- \]
+\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 \varphi}.
+\]
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
+N = b\lambda \oint E dl = -b \lambda \pi a^2 \frac{dB}{dt}
+\]
+so total angular momentum imparted to the wheel is
\[
- \int N dt = -\lambda \pi a^2 b \int_{B_0}^0 dB = \lambda \pi a^2 b B_0.
- \]
+\int N dt = -\lambda \pi a^2 b \int_{B_0}^0 dB = \lambda \pi a^2 b B_0.
+\]
@@ -1690,33 +1708,44 @@ The precise way the field is turned off doesn't matter. Only electric field doe
-{\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.
+N.B.: we use magnetostatic formulas for changing fields. This is
+called the 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 \(B = \frac{\mu_0 I}{2\pi s}\)
+Example: field from wire with time-dependent current
+
+
+
+Consider an infinitely long straight wire which carries current \(I(t)\).
+
+
+
+Task: find the induced \({\bf E}\) field as a function of distance \(s\) from wire.
+
+
+
+Solution: assuming we can use the quasistatic approximation, the
+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 \(s\):
\[
- \oint {\bf E} \cdot d{\bf l} = E(s_0)l - E(s)l = -\frac{d}{dt} \int {\bf B} \cdot d{\bf a}
- = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \int_{s_0}^s \frac{ds'}{s'}
- = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \ln(s/s_0).
- \]
+\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:
\[
- {\bf E} (s) = \left[ \frac{\mu_0}{2\pi} \frac{dI}{dt} \ln s + K \right] \hat{\bf x}
- \label{Gr(7.19)}
- \]
+{\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 \(s \rightarrow \infty\).
+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 \(I(t)\).
@@ -1743,7 +1772,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
-These equations summarize the {\bf entire content of classical electrodynamics}.
+These equations contain the entirety of pre-quantum electrodynamics.
-\paragraph{Note:} even the continuity equation can be derived from Maxwell's equations:
-take divergence of \((iv)\).
+Note: even the continuity equation can be derived from Maxwell's equations:
+take divergence of \((iv)\) and use \((i)\).
-
Better way of writing: all fields on left, all sources on right,
-The term which should be zero (but isn't) in (\ref{Gr(7.35)}) can be rewritten using
-the continuity equation as
+The term which should be zero (but isn't) in divcurlB can be rewritten using the continuity equation as
\[
{\boldsymbol \nabla} \cdot {\bf J} = -\frac{\partial \rho}{\partial t} = - \frac{\partial}{\partial t}
(\varepsilon_0 {\boldsymbol \nabla} \cdot {\bf E}) = -{\boldsymbol \nabla} \cdot \left(
@@ -1631,37 +1630,67 @@ the continuity equation as
\]
The extra term would thus be eliminated if we were to put
-\paragraph{Note:} this changes nothing in magnetostatics. Aesthetic appeal:
+Note: this changes nothing in magnetostatics. Aesthetic appeal:
\[
- \boxed{
- \mbox{A changing electric field induces a magnetic field.}
- }
+\boxed{
+\mbox{A changing electric field induces a magnetic field.}
+}
\]
Real confirmation of Maxwell's theory: 1888, Hertz's experiments on propagation of electromagnetic waves.
Fatal inconsistency: div of curl must always vanish. Check on \((iii)\):
@@ -1637,11 +1637,28 @@ Fatal inconsistency: div of curl must always vanish. Check on \((iii)\):
= {\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 \((iv)\):
+
+
+
+
+
+
+
+
Gr (7.35)
+
+
+
+
+
+
\[
{\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} \times {\bf B}) = \mu_0 {\boldsymbol \nabla} \cdot {\bf J}
-\label{Gr(7.35)}
+\tag{divcurlB}\label{divcurlB}
\]
-LHS must be zero, but RHS is not zero for non-steady currents. Cannot be right !
+LHS must be zero, but RHS is not zero for non-steady currents. Cannot be right!
@@ -1673,7 +1690,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_Me_mc.html b/build/emd_Me_mc.html
index 499a8a7..c62a9e2 100644
--- a/build/emd_Me_mc.html
+++ b/build/emd_Me_mc.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1633,7 +1633,7 @@ In free space, where \(\rho\) and \({\bf J}\) vanish:
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.}
+a truly symmetric situation, i.e.
\begin{align}
(i) &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho_e}{\varepsilon_0},
@@ -1650,7 +1650,7 @@ of magnetic charge. Both charges would be conserved:
{\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!
+Maxwell's equations beg for magnetic charges. But we've never found any!
@@ -1671,7 +1671,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
-Very important distinction: {\bf global} versus {\bf local} conservation of charge.
+Very important distinction: global versus local conservation of charge.
-Charge in a volume {\cal V}:
+Charge in a volume \({\cal V}\):
\[
-Q_{\cal V} (t) = \int_{\cal V} d\tau \rho ({\bf r}, t)
+Q_{\cal V} (t) = \int_{\cal V} d\tau~ \rho ({\bf r}, t)
\label{Gr(8.1)}
\]
Current \({\bf J}\) flowing out through boundary \({\cal S}\) of \({\cal V}\): conservation of charge means
@@ -1643,13 +1643,12 @@ This means that
\]
Since this is true for any volume, we have (re)derived the
@@ -1660,7 +1659,7 @@ 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
+imposes a functional constraint on these sources: not any \(\rho\) and
\({\boldsymbol J}\) will do the trick, only the ones what obey it.
@@ -1684,7 +1683,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
From Newton's second law,
\[
- {\boldsymbol F} = \frac{d {\boldsymbol p}_{\tiny \mbox{mech}}}{dt}
+{\boldsymbol F} = \frac{d {\boldsymbol p}_{\tiny \mbox{mech}}}{dt}
\]
we have
\[
@@ -1636,24 +1636,54 @@ in which the first integral can be interpreted as the momentum stored in the EM
This is thus simply a conservation law for momentum, with
@@ -1677,7 +1707,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_ce_mst.html b/build/emd_ce_mst.html
index fcb15eb..c8cd786 100644
--- a/build/emd_ce_mst.html
+++ b/build/emd_ce_mst.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1638,30 +1638,27 @@ Substitute for \(\rho\) and \({\boldsymbol J}\) using Maxwell (Gauss and Ampère
On the other hand we have
-\[
- \frac{\partial }{\partial t} \left( {\boldsymbol E} × {\boldsymbol B} \right)
- = \frac{∂ {\boldsymbol E}}{∂ t} × {\boldsymbol B}
@@ -1742,7 +1778,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_ce_poy.html b/build/emd_ce_poy.html
index fbfea9b..d62a2a8 100644
--- a/build/emd_ce_poy.html
+++ b/build/emd_ce_poy.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1638,38 +1638,45 @@ 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:
+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,
+
+
+
+
+
+
+
+
Gr (8.6)
+
+
+
+
+
+
\[
\frac{dW}{dt} = \int_{\cal V} d\tau ~ {\bf E} \cdot {\bf J}
-\label{Gr(8.6)}
+\tag{dWdt_intEJ}\label{dWdt_intEJ}
\]
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,
+Using div_xprod,
\[
-{\boldsymbol ∇} ⋅ ({\bf E} × {\bf B}) = {\bf B} ⋅ ({\boldsymbol ∇} × {\bf E})
-
+{\bf E} \cdot {\bf J} = -\frac{1}{2} \frac{\partial}{\partial t} \left( \varepsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \frac{1}{\mu_0} {\boldsymbol \nabla} \cdot ({\bf E} \times {\bf B}).
\label{Gr(8.8)}
\]
-Substituting this in \ref{Gr(8.6)} and using the divergence theorem,
+Substituting this in dWdt_intEJ and using the divergence theorem,
we obtain
@@ -1781,26 +1863,38 @@ and has a similar for to the continuity equation
-
+
+
+Example: Joule heating
+
+
+
+Task: characterize the energy flow for a current-carrying wire.
+
+
+
+Solution: the energy per unit time delivered to wire the wire can
+be obtained from Poynting's theorem.
+
+
-\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
\[
- {\boldsymbol E} = \frac{V}{L} \hat{\boldsymbol x},
- \]
+{\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 \(a\),
\[
- {\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \varphi}
- \]
+{\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \varphi}
+\]
Poynting:
\[
- {\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \varphi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s}
- \]
+{\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \varphi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s}
+\]
and points radially inwards. Energy per unit time passing surface of wire:
\[
- \int d{\bf a} \cdot {\bf S} = S (2\pi a L) = -V I
- \]
+\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.
@@ -1826,7 +1920,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
@@ -1674,7 +1674,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_emw_ep.html b/build/emd_emw_ep.html
index 3fc6cb3..d5e1548 100644
--- a/build/emd_emw_ep.html
+++ b/build/emd_emw_ep.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1652,7 +1652,7 @@ so for a monochromatic EM plan wave,
\]
or more succinctly:
-
+
{\bf Poynting vector of a monochromatic EM wave}
\[
@@ -1668,7 +1668,7 @@ This has a transparent physical interpretation: the energy density \(u\) flows w
Similary, we get the
-
+
{\bf Momentum density of a monochromatic EM wave}
\[
@@ -1719,7 +1719,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_emw_mpw.html b/build/emd_emw_mpw.html
index 474d7a1..64a7e25 100644
--- a/build/emd_emw_mpw.html
+++ b/build/emd_emw_mpw.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1653,7 +1653,7 @@ B_0 = \frac{k}{\omega} E_0 = \frac{1}{c} E_0.
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}
\[
@@ -1697,7 +1697,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
Author: Jean-Sébastien Caux
-
Created: 2022-03-01 Tue 08:14
+
Created: 2022-03-02 Wed 15:45
diff --git a/build/emd_emw_we.html b/build/emd_emw_we.html
index f21e8ac..47983d8 100644
--- a/build/emd_emw_we.html
+++ b/build/emd_emw_we.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1650,7 +1650,7 @@ These take the form of coupled first-order partial differential equations for \(
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}
\[
@@ -1706,7 +1706,7 @@ target="_blank">Creative Commons Attribution 4.0 International License.
diff --git a/build/emdm_Me_Mem.html b/build/emdm_Me_Mem.html
index 171e7b0..c1e1f09 100644
--- a/build/emdm_Me_Mem.html
+++ b/build/emdm_Me_Mem.html
@@ -1,7 +1,7 @@
-
+
Pre-Quantum Electrodynamics
@@ -1657,7 +1657,7 @@ dI = \frac{\partial \sigma_b}{\partial t} da_{\perp} = \frac{\partial P}{\partia
\]
We therefore have the
-
+
{\bf Polarization current density}
\[
@@ -1675,7 +1675,7 @@ 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: