Gauges in Classical EM

1. Nov 12, 2009

JustinLevy

Everytime I try to work out the Lagrangian for EM in different gauges, it gets messy really quick. Maybe there is some trick to simplify the process that I do not know, but either way I'd appreciate some suggestions.

Starting point:

For a point particle a (non-relativistic) Lagrangian that gives classical electrodynamics is:
$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\phi + q\dot{\vec{x}} \cdot \vec{A} - \frac{1}{4\mu_0} \int F_{\mu\nu} F^{\mu\nu} \ d^3r$$
The coordinates are $x$ and $A^\mu$, with the fields being a function of position.

I can obtain the two Maxwell source equations and the Lorentz force law with this (the remaining two Maxwell equations follow directly from the definition of the electric and magnetic fields in terms of the potential $A^\mu$).

If instead of going all the way to Maxwell's equations, I just solve for the evolution of A^\mu, I get the source equations:
$$\partial_\nu F^{\mu \nu} = \mu_0 J^\mu$$
Which is true in any gauge.

Question 1]
Due to the j.A term, the Lagrangian itself is not gauge invariant (right?) even though the evolution equations resulting from it may be.
So what "gauge" is this Lagrangian in? How do I determine it?

----

Now if I want to use as coordinates $A^\mu$ as a function of momentum space instead, I can write the following (deriving got a bit messy at times, so I won't show that):

$$\mathcal{L} = \frac{1}{2}m\dot{\vec{x}}^2 - q\int \phi(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} + q\dot{\vec{x}} \cdot \int \vec{A}(\vec{k})e^{+i\vec{k}\cdot\vec{x}} \frac{d^3k}{(2\pi)^3} - \frac{1}{2} \int [ \frac{k^2}{\mu_0} \vec{A}(\vec{k})^2 - \epsilon_0 \dot{\vec{A}}(\vec{k})^2 - \epsilon_0 k^2 \phi(\vec{k})^2 + \frac{1}{\mu_0} \dot{\phi}(\vec{k})^2 ] \frac{d^3k}{(2\pi)^3}$$
Now the coordinates are $x$ and $A^\mu$, with the fields being a function of momentum space.
You can derive:
$$(-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\phi(k) = -\rho(k)/\epsilon_0$$
$$(-k^2 -\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\vec{A}(k) = -\mu_0 \vec{j}(k)$$
Which are maxwell's equations in terms of the potentials in the Lorenz gauge, $\partial_\mu A^\mu = 0$. (Correct?)

Question 2]
How did the evolution equations for $A^\mu$ somehow become gauge dependent now? Is there something I'm missing in the Lagrangian (probably the free field part)?

Question 3]
What does this look like in the Coulomb gauge? (Coulomb gauge, $$\partial_\mu A^\mu = \frac{1}{c^2}\frac{\partial}{\partial t}\phi$$)
All the most helpful sources I've found only do this for the completely free field, and so also add the constraint that $\phi=0$ which can't be enforced in the interacting case so screws things up for me here. When ever I try to work it out without that constraint the extra little piece of $\phi$ floating around gets annoying quick.

Thanks everyone.

Last edited: Nov 12, 2009
2. Nov 12, 2009

hamster143

1. It is gauge invariant as long as charge is conserved.

2. You may have gotten the momentum space Lagrangian wrong. The $$\dot{\phi}$$ term shouldn't be there, because it corresponds to $$d_0 A^0$$, which is not present in $$F^{\mu\nu}$$.

3. Nov 12, 2009

JustinLevy

Electric charge? or is there some kind of other "Noether charge" relating to gauge invariance?

I got the correct Maxwell's equations and Lorentz force law, right?
So the $$\dot{\phi}$$ term needs to be there.

As for the last comment:
$$A^{\mu} = (\phi/c,\vec{A})$$
$$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$$

So I don't understand your objection. There is definitely a $$\partial^0 A^0 = - \frac{1}{c^2} \dot{\phi}$$ piece somewhere in all that.

Remember, in terms of the potentials (as a function of position), Maxwell's equations are:
$$\partial_\nu F^{\mu\nu} = \mu_0 j^\mu$$
So
$$\partial_\nu F^{\mu\nu} = \partial_\nu\partial^\mu A^\nu - \partial_\nu\partial^\nu A^\mu = \mu_0 j^\mu$$
In the Lorenz gauge we have $$\partial_\nu A^\nu=0$$, so the equations simplify to:
$$\partial_\nu\partial^\nu A^\mu = -\mu_0 j^\mu$$

Since in that equation, all the components of A^\mu are treated equally, then of course there should be a time derivative of phi if there is a time derivative of A^1 (as there is for A^2 and A^3).

4. Nov 12, 2009

hamster143

1. Yes, electric charge. If you take

$$\mathcal{L} = \int [- \phi + q\dot{\vec{x}} \cdot \vec{A}] q \delta^3(x-y) d^3 y$$

and perform a global gauge transformation $$\vec{A} \rightarrow \vec{A}+\nabla f$$, $$\phi \rightarrow \phi - \partial f / \partial t$$

you should be able to show that the lagrangian changes by a total derivative (I think ...)

2. $$\partial^0 A^0$$ can only come from the term $$F^{00}$$ and you can clearly see that $$F^{00} = 0$$ .

Last edited: Nov 12, 2009
5. Nov 12, 2009

JustinLevy

Ahh, okay thanks. I'll play with that.

No, look at Maxwell's equations again:
$$\partial_\nu F^{\mu\nu} = \mu_0 j^\mu$$
The derivative can come from terms like $$\partial_0 F^{10}$$.

So I really think the $$\dot{\phi}$$ terms need to be in there.

6. Nov 12, 2009

hamster143

They can come back into Maxwell's equations through gauge fixing, but they aren't present in the basic version of the EM lagrangian as you wrote it down in the first post.

Momentum space version should contain terms $$(\dot{\vec{A}})^2$$, $$k^2\phi^2$$, $$\vec{k} \dot{\vec{A}} \phi$$, and $$(\vec{k} \times \vec{A})^2$$.

7. Nov 12, 2009

JustinLevy

Hmm... Okay, looking at it as
$$F^{\mu\nu}F_{\mu\nu} = 2(B^2 - \frac{1}{c^2}E^2)$$
that makes sense.

So you are saying that the "Lorenz gauge fixing" occurred when $$\vec{k} \dot{\vec{A}} \phi$$, and $$(\vec{k} \times \vec{A})^2$$ got set to terms with just $$\vec{k}^2 \vec{A}^2$$ and $$\dot{\phi}^2$$?

I guess that seem reasonable.
I think I need to play with it a bit to convince myself of what is going on.

Thanks again.

8. Nov 12, 2009

hamster143

Exactly.

$$B^2 - \frac{1}{c^2}E^2 = (\vec{k} \times \vec{A})^2 - \frac{1}{c^2}(\vec{k}^2 \phi^2 + 2\vec{k} \dot{\vec{A}}\phi + \dot{\vec{A}}^2)$$

All I'm saying is that I don't know how you managed to get the $$\dot{\phi}^2$$ term in there.

9. Nov 12, 2009

JustinLevy

Well, the Lagrangian I wrote appears to give the correct equations in the Lorenz gauge. I guess I just accidentally "gauge fixed" it or something. The coordinates do look much more like oscillator equations in this form. Is this the form they normally use on the way to quantizing for QED? The Lorenz gauge has the nice feature in that it still holds after a Lorentz transformation.

Without a $$\dot{\phi}$$ term, what is the conjugate momentum for that coordinate?

10. Nov 12, 2009

hamster143

There isn't any. You get a constraint equation instead. That's why you need gauge fixing!