Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm looking at this reference (sorry it's a postscript file, but I can't find a pdf version on the web. This paper describes a similar procedure).

The topic is the Faddeev-Jackiw treatment of Lagrangians which are singular (Hessian vanishes) - similar to what Dirac does, but without the need to differentiate between first and second class constraints. Just looking at classical stuff here, no quantization.

Starting with the Maxwell Lagrangian

$$\mathcal{L}=F_{\mu\nu}F^{\mu\nu}$$

where

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

we see that it's second order in time derivatives acting on A.

We choose to write it in first order form like this

$$\mathcal{L}=(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})F^{\mu\nu}-{1\over{2}}F_{\mu\nu}F^{\mu\nu}$$

where we're treating $F^{\mu\nu}$ now as an auxilliary, independent variable. Having defined this, Faddeev says

"we rewrite (the last equation) as:

$$\mathcal{L}=(\partial_{0}A_{k})F^{0k}+A_{0}(\partial_{k}F^{0k})-F^{ik}(\partial_{i}A_{k}-\partial_{k}A_{i})-{1\over{2}}(F^{0k})^2-{1\over{2}}(F^{ik})^2$$"

My question is how does he arrive at this from the previous equation ? I don't see how just expanding the indices into time and space values ever gets me to $A_{0}(\partial_{k}F^{0k})$

I can see how there's something special about $A_{0}$, since when I write out the EOM for the first order Lagrangian, $A_{0}$ drops out, which indeed it should do because we'll end up with it being a Lagrange multiplier. I just can't see how you end up with that term, with $A_{0}$ multiplying $\partial_{k}F^{0k}$.

It's clearly correct since $A_{0}divE$ is just the Gauss law constraint.

share|improve this question
    
Looks like the term you are worried about just comes from a partial integration of the $(\partial_k A_0) F^{0k}$ term. –  Olaf Feb 11 '12 at 14:37
    
D'oh ! I knew I'd kick myself !! –  twistor59 Feb 11 '12 at 14:44
add comment

1 Answer

up vote 1 down vote accepted

Faddeev has implicitly dropped a total 4-divergence term $d_{\mu}(A_0 F^{0\mu})$ in the Lagrangian density ${\cal L}$. This does not affect the equations of motion, i.e., Maxwell's equations.

share|improve this answer
    
Thanks ! In my defence I haven't done any physics since 1984 !! –  twistor59 Feb 11 '12 at 14:47
    
@twistor59. You are welcome. Thank you for the Faddeev link! –  Qmechanic Feb 11 '12 at 14:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.