Why Can We Freely Choose the Divergence of A in Electromagnetic Theory?

[adphysics]

I understand the concept of a gauge transform, and I understand why it is that the magnetic field would be unchanged with the addition of the gradient of an arbitrary scalar potential onto the magnetic vector potential A, and I understand why the electric field E would be invariant under the following pair of gauge transforms:
phi=>phi+ (d(psi)/dt) and A=>A-grad(psi) where psi and phi are scalar potentials, A is the magnetic vector potential, and E=-grad(phi)-dA/dt.

What I don't understand is why we are completely free to choose the divergence of A in the time dependant case. It won't affect the magnetic field, but surely it will affect the electric field?

 

[CompuChip]

No, the electric field will not be affected.
You will see this when you write out the equation. Basically, the extra term you get in the electric field is
$$-\nabla(\frac{\partial\psi}{\partial t} + \frac{\partial}{\partial t} (\nabla\psi)$$

 

[adphysics]

Sorry, I should have been more clear. That part I understand. I understand why, if we begin with E=-grad(phi)-dA/dt and make the simultaneous gauge transforms, we have the same electric field. What I don't understand is the following:

Take the divergence of the electric field:

div E=-Laplacian(phi)-d/dt (div A)

By Gauss' Law:

div E=rho/epsilon

So my question is how I can freely choose div A when it affects an actual physical quantity (charge density). If I choose the Coulomb gauge, I get Poisson's equation. If I choose the Lorenz gauge, I get the d'Alembert wave equation. I understand why the Lorenz gauge is more convenient and useful in this case, since the Coulomb gauge specifies phi completely but makes A hard to calculate. I just don't understand why we are free to choose divA.

 

[CompuChip]

I don't quite see what you mean, sorry.
If the electric field does not change under a gauge transformation on (phi, A), then its divergence doesn't change either, does it?
I.e. if the change
$$\Delta E = -\nabla(\frac{\partial\psi}{\partial t}) + \frac{\partial}{\partial t} (\nabla\psi)$$
is zero, then
$$\nabla \cdot (\Delta E) = \nabla \cdot 0 = 0$$
vanishes as well, doesn't it.

 

[adphysics]

I know. I'm just having trouble understanding how, say, the Lorenz gauge is actually related to the simultaneous gauge transforms shown above. In all the derivations I've encountered of the wave equations for E and B using the Lorenz gauge, like this one:

http://farside.ph.utexas.edu/teaching/em/lectures/node47.html

The author begins by saying that they are free to choose divergence, and then substituting the Lorenz gauge into div E=-Laplacian (phi)-d divA/dt. But he doesn't change the scalar potential. That's what I'm having trouble understanding. The gauge transforms require that both A and phi be changed, don't they?

 

[CompuChip]

Yes. Let's take a simple example, where phi = 0 and A = (x^2, y^2, 0).
The divergence is div(A) = 2x + 2y.

Let f(x, y) = x^2 y + y^2 x
Then grad f = (2 x y + y^2, x^2 + 2 x y, 0)
A - grad(f) = div(x^2 - 2 x y - y^2, y^2 - 2 x y - x^2, 0)
of which the divergence is
div(A - grad(f)) = (2x - 2y) + (2y - 2x) = 0.

In this case there is no explicit time dependence, so the scalar potential will not change, but in general it will.
The whole construction is set up in such a way, however, that E and B do not change.

Did I understand your question now?

 

[siddharth]

Sorry, I should have been more clear. That part I understand. I understand why, if we begin with E=-grad(phi)-dA/dt and make the simultaneous gauge transforms, we have the same electric field. What I don't understand is the following:

Take the divergence of the electric field:

div E=-Laplacian(phi)-d/dt (div A)

By Gauss' Law:

div E=rho/epsilon

So my question is how I can freely choose div A when it affects an actual physical quantity (charge density). If I choose the Coulomb gauge, I get Poisson's equation. If I choose the Lorenz gauge, I get the d'Alembert wave equation. I understand why the Lorenz gauge is more convenient and useful in this case, since the Coulomb gauge specifies phi completely but makes A hard to calculate. I just don't understand why we are free to choose divA.

If you make the simultaneous gauge transformations,
$$\vec{A}^\prime=\vec{A} + \nabla \psi$$

$$\phi^\prime=\phi - \frac{\partial \psi}{\partial t}$$

then,

$$\vec{E}^\prime=-\nabla \phi^\prime - \frac{\partial \vec{A}^\prime}{\partial t}$$

$$= -\nabla (\phi - \frac{\partial \psi}{\partial t}) - \frac{\partial (\vec{A} + \nabla \psi)}{\partial t}$$

$$= -\nabla \phi - \frac{\partial \vec{A}}{\partial t}$$

$$= \vec{E}$$

so the electric field doesn't change

EDIT: Ah, nevermind. I read "I understand why" as "I don't understand why"