What is the difference between two Lagrangian densities in electrodynamics?

[jameson2]

Homework Statement

Given the Lagrangian density:
L= -\frac{1}{2} \partial_{\mu}A_\nu \partial^{\mu}A^\nu -\frac{1}{c}J_\mu A^\mu

(a) find the Euler Lagrange equations of motion. Under what assumptions are they the Maxwell equations of electrodynamics?

(b) Show that this Lagrangian density differs from
L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{c}J_\mu A^\mu
by a 4-divergence.
Does the added 4-divergence affect the action? Does it affect the equations of motion?

Homework Equations

F^{\mu\nu}= \partial^\mu A^\nu -\partial^\nu A^\mu

The Attempt at a Solution

(a) I worked out the equation of motion to be
\partial_\mu \partial^\mu A^\nu = \frac{1}{c} J^\nu
For the second part, I'm not sure. Since the Maxwell equations come from this equation of motion:
\partial _\mu F^{\mu\nu}=\frac{1}{c}J_\nu
I think I just compare the two expressions (and expanding F as above), so they are the same if
\partial_\mu \partial^\nu A^\mu = 0

I'm not sure if this is right though.

(b) I'm less sure of this part. First I found the difference between the two Lagrangian densities to be
\frac{1}{2}\partial_\mu A_\nu \partial^\nu A^\mu
and I'm not sure how to show this is a 4-divergence.

I'd assume that it does affect the action, but I don't know how to show it.

I'm fairly sure it does affect the equations of motion, as the first Lagrangian results in
\partial_\mu \partial^\mu A^\nu = \frac{1}{c} J^\nu
while the second results in
\partial _\mu F^{\mu\nu}=\frac{1}{c}J_\nu
so they're obviously different. But this seems a little easy, as it seems as if this had already been shown?

Thanks for any help.

 

[hunt_mat]

The first part looks good, for the second part, you have shown what they're asking as that is the definition of the four divergence, as to whether the four diveregence effects the equations of motion, I don't think they will, this should be a standard result though.

 

[chwie]

Adding a four divergence to the Lagrangian can't affect the equation of motion. Then both solution should be equal. The idea is that as the volume integral of the four divergence can be transform to a surface integral and by assumption the field doesn't varied at the surface then this term doesn't contribute. The proof is in any book of field theory.

I know probably you already handed out the homework, but for future reference.

The \partial_\mu \partial^\nu A^\mu = 0 is true if you used lorenz gauge, but is not true in general. That is also the answer to the part a) ( you need to work in lorenz gauge.)

In that case u have \partial_\mu A_\nu \partial^\nu A^\mu=\partial_\mu A_\nu \partial^\nu A^\mu+A_\nu\partial_\mu \partial^\nu A^\mu=\partial_\mu(A_\nu\partial^\nu A^\mu)

This is your four divergence. I used lorenz gauge to add the extra term which is zero ( remember lorenz gauge is \partial_\mu A^\mu=0)

Now the rest you have it done, because if you assume lorenz gauge both equation of motion are equal. The idea is that if we assume lorenz gauge the difference between both lagrangian is a four divergence, then the equation of motion should be the same if we keep assuming lorenz gauge.