# Use of the Lorenz Gauge

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?

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CompuChip
Homework Helper
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
$$-\nabla(\frac{\partial\psi}{\partial t} + \frac{\partial}{\partial t} (\nabla\psi)$$

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
Homework Helper
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.

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
Homework Helper
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
Homework Helper
Gold Member
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"