Ryder QFT, 2nd Edition

1. Sep 15, 2007

PhysiSmo

We have the Lagrangian of EM field: $$L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

Variation of Lagrangian give Maxwell's equations:

$$\partial_{\mu} F^{\mu\nu}=0$$.

or

$$(g_{\mu\nu}\partial_{\mu}\partial^{\mu}-\partial_{\mu}\partial_{\nu})A^{\mu}=0.$$ (equation 7.3, p.241)

Ryder, then, claims that after partial integration, and discarding of surface terms, Lagrangian may be written

$$L=\frac{1}{2}A^{\mu}[g_{\mu\nu}\partial_{\mu}\partial^{\mu}-\partial_{\mu}\partial_{\nu}]A^{\nu}$$.

I simply can't figure out this last derivation. Which quantity has to be integrated to give such result? How do we obtain this particular form? I played with various quantities and integrals, but can't prove it..thanx in advance.

2. Sep 15, 2007

dextercioby

$$L=-\frac{1}{4}\left(\partial_{[\mu}A_{\nu]}\right)\left(\partial^{[\mu}A^{\nu]\right) =\frac{1}{2}\left[\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right) -\left(\partial_{\mu}A_{\nu}\right)\left(\partial^{\mu}A^{\nu}\right)\right]$$

Now move the derivatives and discard the 4-divergences.

3. Sep 15, 2007

PhysiSmo

$$(\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu})=\partial_{\nu}(A_{\mu}\partial^{\mu}A^{\nu})-A_{\mu}\partial_{\nu}\partial^{\mu}A^{\nu}=$$

$$=\partial_{\nu}(A_{\mu}\partial^{\mu}A^{\nu})-A^{\mu}\partial_{\nu}\partial_{\mu}A^{\nu}$$

Similarly,

$$(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})=\partial_{\mu}(A_{\nu}\partial^{\mu}A^{\nu})-A^{\mu}(g_{\mu\nu}\partial_{\mu}\partial^{\mu})A^{\nu}$$.

I can see that discarding the first terms of each equation yields the correct result. But why one should discard these terms? I understand that when integrated, it can be converted to a surface integral and thus equal to zero. But why do we act so in this particular form, without integration?

4. Sep 15, 2007

dextercioby

Nope, discarding only takes place when both 4-divergences are under the integral sign . We act without integration and simply write

$$L=L'+ 4div$$

And by integration we see that L and L' are physically equivalent, since the lagrangian action is the same.

5. Sep 16, 2007

PhysiSmo

Very well then! Thanks again for your help!