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Question on 1st order Lagrangian Derivation in Faddeev-Jackiw Formalism

+ 2 like - 0 dislike
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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.

This post imported from StackExchange Physics at 2014-03-22 17:26 (UCT), posted by SE-user twistor59
asked Feb 11, 2012 in Theoretical Physics by twistor59 (2,490 points) [ no revision ]
Looks like the term you are worried about just comes from a partial integration of the $(\partial_k A_0) F^{0k}$ term.

This post imported from StackExchange Physics at 2014-03-22 17:26 (UCT), posted by SE-user Olaf
D'oh ! I knew I'd kick myself !!

This post imported from StackExchange Physics at 2014-03-22 17:26 (UCT), posted by SE-user twistor59

1 Answer

+ 1 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:26 (UCT), posted by SE-user Qmechanic
answered Feb 11, 2012 by Qmechanic (2,580 points) [ no revision ]
Thanks ! In my defence I haven't done any physics since 1984 !!

This post imported from StackExchange Physics at 2014-03-22 17:26 (UCT), posted by SE-user twistor59
@twistor59. You are welcome. Thank you for the Faddeev link!

This post imported from StackExchange Physics at 2014-03-22 17:26 (UCT), posted by SE-user Qmechanic

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