Lagrangian Density Problem

[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.

 

[sgd37]

they key concept here is the implementation of the coulomb gauge i.e. $$\partial_{\mu} A^{\mu} = 0$$ so that

$$\frac{1}{2}\partial_\mu A_\nu \partial^\nu A^\mu = \frac{1}{2}\partial_\mu (A_\nu \partial^\nu A^\mu)$$

since the second term after applying the differential operator is 0 due to the gauge condition

The equations of motion are thus unchanged by this and in general Lagrangian theory Lagrangians that differ by a total derivative lead to the same equations of motion. Since if

$$L' = L + \partial_{\mu} f(q(x^{\mu}), x^{\mu})$$ then the action is given by

$$S' = S + f(q_{f}(x^{\mu}), x_{f}^{\mu}) - f(q_{i}(x^{\mu}), x_{i}^{\mu})$$

the extra term gives zero under variation such that

$$\delta S' = \delta S$$

Consequently the action is left invariant if the potential can be assumed to be 0 at the initial and final instances otherwise it changes by a constant

 

[jmlaniel]

From what you say, the function f here is :

$f^\mu = A_\nu \partial^\nu A^\mu$

I notice that f here is a function of the field ($A_\nu$) but also function of the derivative of the field ($\partial^\nu A^\mu$).

I have seen that the Lagrangian is not altered when f is a function of the field, but not if it is function of the derivative of the field. This is also reflected in the way you define the function f in the Lagrangian.

I am myself trying to prove that :

$L' = L - \frac{1}{2} \partial_\nu ( -A_\mu \partial^\mu A^\nu + A^\nu \partial_\mu A^\mu)$

where the last term does not contribute to the Lagrangian so that $L' = L$. I see that the second term can at least be removed using Lorentz gauge $\partial_\mu A^\mu = 0$, but the first one is identical to the one discussed here.

So my question is the following one : can your argument work even if you have derivatives of the field inside f?