Saturday, 26 April 2008

Classical Electrodynamics

(Without proof) Maxwell's equations in Integral form are:

\oint \mathbf{E}\cdot d\mathbf{S} = \frac{\sum{Q}}{\epsilon_0} ~~~~\mathrm{(Gauss)}

\oint \mathbf{E}\cdot d\mathbf{L} = -\frac{\partial\Phi}{\partial t} = \int -\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{S} ~~~~\mathrm{(Faraday)}

\oint\mathbf{B}\cdot d\mathbf{S} = 0 ~~~~ \mathrm{(No~magnetic~monopoles)}

\oint \mathbf{H}\cdot d\mathbf{L} = \mathbf{I}_c + \mathbf{I}_d ~~~~\mathrm{(Ampere)}

Here,
\mathbf{H}=\frac{\mathbf{B}}{\mu_0}-\mathbf{M}
and Ampere's law has been written with the Displacement Current.

In differential form (again without proof), these can be written:

\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0} ~~~~ \mathrm{(Gauss)}

\nabla\cdot\mathbf{B} = 0 ~~~~ \mathrm{(No~magnetic~monopoles)}

\nabla\times\mathbf{E} + \frac{\partial\mathbf{B}}{\partial t} = 0 ~~~~\mathrm{(Faraday)}

\nabla\times\mathbf{B} = \mu_0\mathbf{j} + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}

The continuity equation:

\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j} = 0

follows from the Inhomogeneous Maxwell equations (
\nabla\cdot\mathbf{E}~,~\nabla\times\mathbf{B}
).

Electrostatics
In electrostatics, one can ignore the time-derivates, leaving:

\begin{align*}
\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}\\
\nabla\times\mathbf{E}=0\\
\nabla\cdot\mathbf{B}=0\\
\nabla\times\mathbf{B}=\mu_0\mathbf{j}
\end{align*}

\nabla\times\mathbf{E} = 0 \Rightarrow \mathbf{E} = -\nabla V
and V satisfies
\nabla^2 V = -\frac{\rho}{\epsilon_0}~~~(\star)

\nabla\cdot\mathbf{B}=0\Rightarrow \mathbf{B}=\nabla\times\mathbf{A}
and A satisfies
\nabla^2\mathbf{A} = -\mu_0\mathbf{j}


(\star)
is the Poisson equation, and the steady field solutions for V and A are:

\begin{align*}
V=\int\frac{\rho\,d\tau}{4\pi\epsilon_0 r}\\
\mathbf{A}=\int\frac{\mu_0\mathbf{j}\,d\tau}{4\pi r}
\end{align*}


Electromagnetic Waves
In vacuum, without sources, the Maxwell equations are:

\begin{array}{ll}
\displaystyle \nabla\cdot\mathbf{E} = 0 & \nabla\cdot\mathbf{E} = 0 \nonumber\\
\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} & \nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t} \nonumber
\end{array}


Now,

\begin{align*}
\nabla\times\nabla\times\mathbf{E}=-\nabla\times\frac{\partial\mathbf{B}}{\partial t} \\
= -\frac{\partial}{\partial t}(\nabla\times\mathbf{B}) \\
= -\epsilon_0\mu_0\frac{\partial^2\mathbf{E}}{\partial t^2} \\
\Rightarrow \nabla^2\mathbf{E} - \mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2} = 0\\
\nabla^2\mathbf{B} - \mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2} = 0
\end{align*}

The last two lines are three-dimensional wave equations with velocity
c=\frac{1}{\sqrt{\epsilon_0\mu_0}}
.

With Sources: Retarded Potentials
Including source terms, the Maxwell equations are:

\begin{array}{ll}
\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0} & \nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} \nonumber \\
\nabla\cdot\mathbf{B} = 0 & \nabla\times\mathbf{B} = \frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}+\mu_0\mathbf{j} \nonumber
\end{array}

One can still define
\mathbf{B}=\nabla\times\mathbf{A}
since
\nabla\cdot\mathbf{B}=0
, but now
\mathbf{E}=-\frac{\partial\mathbf{A}}{\partial t} - \nabla V
.

\begin{align*}
\frac{\partial}{\partial t}\nabla\cdot\mathbf{A} + \nabla^2V = \frac{\rho}{\epsilon_0} \\
\Rightarrow \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} \\
= -\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}-\frac{1}{c^2}\nabla\left(\frac{\partial V}{\partial t}\right) + \mu_0\mathbf{j}
\end{align*}


If we choose
\nabla\cdot\mathbf{A} = -\frac{1}{c^2}\frac{\partial V}{\partial t}
(i.e. the Lorenz Gauge), then the equations uncouple to:

\begin{align*}
\nabla^2 V = -\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon_0}\\
\nabla^2 \mathbf{A} = -\frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{j}
\end{align*}

These can be thought of as either Poisson equations with the addition of
\frac{\partial^2}{\partial t^2}
, or as wave equations with sources. The general solution of the Poisson equation for steady
\rho
(as has already been given) is:

V = \int\frac{\rho\,d\tau}{4\pi\epsilon_0 r}

Replace
\rho
by
\rho\left(t-\frac{r}{c}\right)
, rather than by
\rho(t)~\Rightarrow
Retarded Potentials (i.e. the potential lags the source by r/c).

\begin{align*}
V=\int\frac{\rho\left(t-\frac{r}{c}\right)\,d\tau}{4\pi\epsilon_0 r} \\
\mathbf{A}=\int\frac{\mu_0\mathbf{j}\left(t-\frac{r}{c}\right)\,d\tau}{4\pi\epsilon_0 r}
\end{align*}


Electromagnetic Energy: The Poynting Vector
For an unbounded plane wave,

\begin{align*}
\frac{\partial E_y}{\partial x}=-\frac{\partial B_z}{\partial t} \\
\frac{\partial B_z}{\partial x}=-\frac{1}{c^2}\frac{\partial E_y}{\partial t}
\end{align*}

The energy density transported by an electromagnetic wave is:

u_{EM}=\frac{1}{2}\epsilon_0E^2 + \frac{B^2}{2\mu_0}

The transported energy is thus

U_{EM} = \int u_{EM} \,dx\,dy\,dz


The rate of energy loss (e.g. from an aerial) is
-\frac{\partial U_{EM}}{\partial t}
.

\begin{align*}
-\frac{\partial U_{EM}}{\partial t} = \int -\frac{\partial}{\partial t}\left( \frac{B_z^2}{2\mu_0}+\frac{\epsilon_0 E_y^2}{2} \right)\,dx\,dy\,dz \\
= \int\left[ -\left(\frac{B_z}{\mu_0}\right)\frac{\partial B_z}{\partial t} - (\epsilon_0 E_y)\frac{\partial E_y}{\partial t}\right] \,dx\,dy\,dz \\
= \int\left[ \frac{B_z}{\mu_0}\frac{\partial E_y}{\partial x}+\frac{E_y}{\mu_0}\frac{\partial B_z}{\partial x}\right]\,dx\,dy\,dz
\end{align*}

Since
H_z=\frac{B_z}{\mu_0}
, then:

\begin{align*}
\int_{\mathrm{volume}}\frac{\partial}{\partial x}(E_yH_z)\,dx\,dy\,dz \\
= \int_{\mathrm{surface}}\left[(E_yH_z)_{x_1}-(E_yH_z)_{x_2}\right]\,dy\,dz
\end{align*}

This suggests that a vector N of the form below correctly describes both the magnitude and direction of energy flow. This is the Poynting Vector:

\mathbf{N}=\mathbf{E}\times\mathbf{H}

No comments:

Post a Comment