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

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.

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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

## 1 Answer

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.

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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