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Quickie Review of Chapter 6

Recall the following morphs of Maxwell's equations, this time with the sources and expressed in terms of potentials by means of the homogeneous equations. Gauss's Law for magnetism is: = 0 This is an identity if we define $ \vB = \curl \vA$ : () = 0

Similarly, Faraday's Law is + t & = & 0
+ t & = & 0
(+ t) & = & 0 and is satisfied as an identity by a scalar potential such that: + t & = & -&phis#phi;
& = & -&phis#phi;- t

Now we look at the inhomogeneous equations in terms of the potentials. Ampere's Law: & = & &mu#mu;(+ &epsi#epsilon;t)
() & = & &mu#mu;(+ &epsi#epsilon;t)
() - &nabla#nabla;^2& = & &mu#mu;+ &mu#mu;&epsi#epsilon;t
() - &nabla#nabla;^2& = & &mu#mu;- &mu#mu;&epsi#epsilon;&phis#phi;t - &mu#mu;&epsi#epsilon; t
&nabla#nabla;^2- &mu#mu;&epsi#epsilon;t & = & -&mu#mu; + (+ &mu#mu;&epsi#epsilon;&phis#phi;t)

Similarly Gauss's Law for the electric field becomes: & = & &rho#rho;&epsi#epsilon;
(-&phis#phi;- t) & = & &rho#rho;&epsi#epsilon;
&nabla#nabla;^2 &phis#phi;+ t & = & -&rho#rho;&epsi#epsilon;

In the the Lorenz gauge,

$\displaystyle \deldot {\bf A} + \mu\epsilon \partialdiv{\Phi}{t} = 0$ (13.3)

the potentials satisfy the following inhomogeneous wave equations:
$\displaystyle \nabla^2 \Phi - \mu\epsilon \partialdiv{^2\Phi}{t^2}$ $\displaystyle =$ $\displaystyle -
\frac{\rho}{\epsilon}$ (13.4)
$\displaystyle \nabla^2 {\bf A} - \mu\epsilon \partialdiv{^2{\bf A}}{t^2}$ $\displaystyle =$ $\displaystyle -\mu\vJ$ (13.5)

where $ \rho$ and $ \vJ$ are the charge density and current density distributions, respectively. For the time being we will stick with the Lorenz gauge, although the Coulomb gauge: = 0 is more convenient for certain problems. It is probably worth reminding y'all that the Lorenz gauge condition itself is really just one out of a whole family of choices.

Recall that (or more properly, observe that in its role in these wave equations) &mu#mu;&epsi#epsilon;= 1v^2 where $ v$ is the speed of light in the medium. For the time being, let's just simplify life a bit and agree to work in a vacuum: &mu#mu;_0&epsi#epsilon;_0 = 1c^2 so that: &nabla#nabla;^2 &Phi#Phi;- 1c^2^2&Phi#Phi;t^2 & = & - &rho#rho;&epsi#epsilon;_0
&nabla#nabla;^2 A - 1c^2 ^2At^2 & = & -&mu#mu;_0

If/when we look at wave sources embedded in a dielectric medium, we can always change back as the general formalism will not be any different.


next up previous contents
Next: Green's Functions for the Up: Maxwell's Equations, Yet Again Previous: Maxwell's Equations, Yet Again   Contents
Robert G. Brown 2017-07-11