Gauge Theory - Having Trouble with V' = 0

[Michael King]

Homework Statement

Show that it is always possible to pick a gauge so V' = 0.

Homework Equations

We weren't given any, but I've been working with:
(a)$$\vec{E} + \frac{\partial\vec{A}}{\partial{t}} = -\nabla{V}$$
(b)$$\vec{A'} = \vec{A} + \nabla{\psi}$$ (where $$\psi$$ is a scalar function)
(c)$$V' = V - \frac{\partial{\psi}}{\partial{t}}$$

The Attempt at a Solution

The way I've attempted it is not to choose the Lorentz or Coulomb gauge, but to make
$$\nabla\cdot{\vec{A}} + \mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}} = x$$ (where $$x$$ is a scalar function)
Furthermore, from Griffths solutions I found:
$$\nabla\cdot{\vec{A'}} = - \mu_{0}\epsilon_{0}\frac{\partial{V'}}{\partial{t}}$$
In our case, the above equals 0 as V' = 0.

I took the divergence of (b):
$$\nabla\cdot\{vec{A'}} = \nabla\cdot\{\vec{A}} + \nabla^{2}{\psi}$$
$$0 = \nabla\cdot\{\vec{A}} + \nabla^{2}{\psi}$$
$$- \nabla\cdot\{\vec{A}} = \nabla^{2}{\psi}$$
$$- (x - \mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}}) = \nabla^{2}{\psi}$$
$$-x + \mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}} = \nabla^{2}{\psi}$$
$$\mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}} = \nabla^{2}{\psi} + x$$
$$\mu_{0}\epsilon_{0}\frac{\partial^{2}{\psi}}{\partial{t^{2}}} = \nabla^{2}{\psi} + x$$
... which gives me an inequality (i.e. an extra x, and also an extra factor of mu_0 and epsilon_0).

I've also been trying to use the solution to Griffths problem 10.7 as reference but I am really stuck on the steps to take.

 

[Ben Niehoff]

The easiest way to prove this would be to take your equations for a gauge transformation (b and c), set V' to zero, and show that given any A and V, there exists some psi which effects the desired gauge transformation.