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Manchot
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I'm reteaching electrodynamics to myself on a more rigorous footing, and I'm trying to prove to myself that setting the divergence of the vector potential is justified using a gauge shift. I could use the Helmholtz theorem to do this, but the problem with this from my perspective is that I haven't actually justified the full version of the theorem, only the weaker version which requires that a vector function decay to zero faster than 1/r at infinity. This isn't a problem for the field quantities (since all physical fields decay like 1/r^2); nevertheless, they do pose a problem for the potential quantities (which in general will not even decay). Basically, given a vector potential [tex]\vec{A}[/tex], I want to show that fixing the divergence of the gauge-shifted potential [tex]\vec{A}\prime[/tex] to some scalar function [tex]D[/tex] is equivalent to adding the gradient of some some scalar function [tex]\phi[/tex].
[tex]\nabla \cdot \vec{A}' = D[/tex]
[tex]\nabla \cdot (\vec{A} + \nabla \phi) = D[/tex]
[tex]{\nabla}^2 \phi = D - \nabla \cdot \vec{A}[/tex]
Since the right-hand side is just some function of position, proving that the divergence can be adjusted by adding the gradient of a scalar amounts to proving that Poisson's equation has a solution for an arbitrary source term. No boundary conditions are specified, so I would expect that there are actually an infinite number of solutions; however, I cannot prove this. Does anyone have any insights? Thanks.
[tex]\nabla \cdot \vec{A}' = D[/tex]
[tex]\nabla \cdot (\vec{A} + \nabla \phi) = D[/tex]
[tex]{\nabla}^2 \phi = D - \nabla \cdot \vec{A}[/tex]
Since the right-hand side is just some function of position, proving that the divergence can be adjusted by adding the gradient of a scalar amounts to proving that Poisson's equation has a solution for an arbitrary source term. No boundary conditions are specified, so I would expect that there are actually an infinite number of solutions; however, I cannot prove this. Does anyone have any insights? Thanks.
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