How do we obtain the Lagrangian of the EM field in Ryder QFT, 2nd Edition?

[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.

 

[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)<br /> -\left(\partial_{\mu}A_{\nu}\right)\left(\partial^{\mu}A^{\nu}\right)\right]

Now move the derivatives and discard the 4-divergences.

 

[PhysiSmo]

Thank you for your answer! Ok, here we go:

(\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?

 

[dextercioby]

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

L=L&#039;+ 4div

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

 

[PhysiSmo]

Very well then! Thanks again for your help!