Different formulations of the covariant EM Lagrangian

[Gan_HOPE326]

Homework Statement

I'm reading through A. Zee's "Quantum Field Theory in a nutshell" for personal learning and am a bit confused about a passage he goes through when discussing field theory for the electromagnetic field. I am well versed in non relativistic quantum mechanics but have no professional experience with relativity, so the notation is a bit obscure to me. Basically the text moves through one formulation to a different one of the Lagrangian and I can't properly understand what is going on. If you have the book, this is on pp. 32-33 (second edition).

Homework Equations

The EM Lagrangian is defined first as

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

with

$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$

Fair enough. I can follow up to this point. What confuses me however is that in the next formula he writes down the action, and in this one the Lagrangian is expressed as

$$\frac{1}{2}A_{\mu}[(\partial^2+m^2)g^{\mu\nu}-\partial^{\mu}\partial^{\nu}]A_{\nu}$$

plus an interaction term. Don't bother too much with the mass term, which was introduced artificially in order to remove it later so that gauge invariance needn't be treated. I'm confused however by the differential operators. What's going on? These all look like second derivatives, but I was convinced that for how they were presented in the original Lagrangian all that had to appear were squares of first derivatives, at most.

The Attempt at a Solution

I tried developing the original expression for the Lagrangian by replacing the expression for the tensor F, but with little luck. I don't know how to multiply the derivatives, I guess, and am not sure what the author means by his notation. Perhaps ∂² is actually supposed to be the square of the derivative rather than the second derivative? Also, shouldn't all the diagonal terms go to zero anyway?

Thanks a lot!

 

[Orodruin]

These all look like second derivatives, but I was convinced that for how they were presented in the original Lagrangian all that had to appear were squares of first derivatives, at most.

Did you try partial integration?

 

[Gan_HOPE326]

Did you try partial integration?

What should I integrate? The formulas both represent the Lagrangian in theory. I don't think there's any integration going on between them.

I'll point out that while the second expression is from the action, it is still the Lagrangian. It is the kernel of the integral, I left the integral sign out of the formula because it didn't matter.

 

[Orodruin]

I don't think there's any integration going on between them.

Of course there is. Lagrangians appear in action integrals. If they only differ up to a total derivative they are equivalent and give the same equations of motion.

 

[Gan_HOPE326]

Of course there is. Lagrangians appear in action integrals. If they only differ up to a total derivative they are equivalent and give the same equations of motion.

Not sure I'm following. As I explained above, first the book defines the Lagrangian. Then it writes the action integral as

$$S = \int\mathcal{L}dx^4$$

and then in the next passage replaces L with the formulation I posted above, while still keeping the integral sign. So the action integral hasn't been carried out yet at that point. Why would there be some variables that have been integrated?

 

[Orodruin]

There are no variables that have been integrated. The basic fundamental quantity is the action and you can have several equivalent Lagrangians that give the same action. You can relate them by partial integration (neglecting the boundary term).

 

[Gan_HOPE326]

There are no variables that have been integrated. The basic fundamental quantity is the action and you can have several equivalent Lagrangians that give the same action. You can relate them by partial integration (neglecting the boundary term).

Oh, ok, got it. I found a similar example on a scalar field earlier. I'm still not sure how to work it out with tensor notation but at least that's a direction to look in. Thanks!