In David Bohm's "Quantum Theory" (an intro topic building up to the Rayleigh-Jeans law), he states:(adsbygoogle = window.adsbygoogle || []).push({});

"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 whatregularmeans 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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# Electrodynamic vector potential wave equations in free space.

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