electrodynamic vector potential wave equations in free space.


by Peeter
Tags: electrodynamic, equations, free, potential, space, vector, wave
Peeter
Peeter is offline
#1
Dec28-08, 10:32 PM
P: 294
In David Bohm's "Quantum Theory" (an intro topic building up to the Rayleigh-Jeans law), he states:

"We now show that in empty space the choice div a = 0 also leads to \phi = 0 ...

But since div a = 0, we obtain

[tex]
\nabla^2\phi = 0
[/tex]

This is, however, simply Laplace's equation. It is well known that the only solution of this equation that is regular over all space is \phi = 0. (All other solutions imply the existence of charge at some points in space and, therefore, a failure of Laplace's equation at these points.)"

Now, everything leading up to the Laplacian I understand fine, but I'm not clear on the argument that requires \phi = 0. In particular I'm not sure what regular means in this context.

Bohm is trying to arrive at the wave equation for the vector potential. I'd seen this done differently before by picking the gauge

[tex]
\partial_\mu A^\mu = 0
[/tex]

to arrive at the four wave equations

[tex]
\partial_\mu\partial^\mu A^\nu = J^\nu
[/tex]

and doing so one has no requirement for [itex]A^0 = 0[/itex] for the free space case. Are these two approaches equivalent in some not obvious to me?
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Phrak
Phrak is offline
#2
Dec30-08, 06:01 PM
P: 4,513
Quote Quote by Peeter View Post
Now, everything leading up to the Laplacian I understand fine, but I'm not clear on the argument that requires \phi = 0. In particular I'm not sure what regular means in this context.
It's taken me a few days to understand what your sticking point was, so you may already know the answer.

The only meaning of "regular", as applied to harmonic functions, that I have found, means infinitely differentiable. This doen't seem to fit--exactly.

I think Bohm is saying that no harmonic function exists over all R^3 space (including infinity) that is singularity free. Translated into physics, this means that phi is either constant over all space, or the space contains charge--even if that charge is located at infinity. (Bohm must have been mistaken, and should been speaking of grad(phi)=0, as this is the most general condition that allows space to be charge free.)

You might try Liouville's Boundedness Theorem if you can follow it.
http://mathworld.wolfram.com/Liouvil...ssTheorem.html
Peeter
Peeter is offline
#3
Dec31-08, 08:56 AM
P: 294
thanks Phrak. I think you are right. It must be grad(phi) = 0 here that he means. That makes two things make more sense. One is the wave equation result he's trying to get. If I do that calculation myself, using div(a) = 0, I get:

[tex]
\begin{align*}
0 &= \nabla^2 \phi \\
0 &= \nabla^2 \mathbf{a} - \partial_{00}{\mathbf{a}} - \partial_0 \nabla \phi \\
\end{align*}
[/tex]

So, with grad(phi) = 0 one gets the vector potential wave equation, and there isn't any requirement to make phi itself equal to zero.

With the correction of phi =0 -> grad(phi) =0, the use of the term regular makes some sense too. In the context of complex numbers , by writing i = e_1 e_2 as a Clifford product and factoring out one of the vectors, grad(phi) = 0 can be used as a compact way of expressing the Cauchy Reimann equations, the condition for a complex number to be differentiable at a point, ie: regular.

Phrak
Phrak is offline
#4
Jan4-09, 02:32 PM
P: 4,513

electrodynamic vector potential wave equations in free space.


You lost me with the clifford algebra, but good luck.
Peeter
Peeter is offline
#5
Jan4-09, 05:29 PM
P: 294
Quote Quote by Phrak View Post
You lost me with the clifford algebra, but good luck.
That part doesn't matter too much, but if you are curious a more complete description of what I was talking about is here:

http://sites.google.com/site/peeterj...y_gradient.pdf


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