What is the expected result of plugging equations of motion into the Lagrangian?

[michael879]

I know that in general plugging the EOM into the Lagrangian is tricky, but it should be perfectly valid if done correctly. Can someone help me see what I'm doing wrong here? I know I'm doing something dumb but I've been staring at it for too long

Start with the E&M Lagrangian:
$L = -\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu} - J^\mu A_\mu$
which gives the EOM
$\partial_\nu F^{\mu\nu} = -J^\mu$
Plugging this back into the Lagrangian
$L = -\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_\mu\partial_\nu F^{\mu\nu}$
$A_\mu\partial_\nu F^{\mu\nu} = -\partial_\nu A_\mu F^{\mu\nu}$ + total derivative
$= \dfrac{1}{2}F_{\mu\nu} F^{\mu\nu}$ + total derivative
from which we find
$L \equiv \dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}$
which is just the negative free-field Lagrangian...

I would have expected those two terms to exactly cancel, leaving just the total derivative, but no matter how I look at it I'm off by a factor of 2.. If you plug the sole EOM back into the Lagrangian, you would expect the result to be a trivial action right?

 

[stevendaryl]

I know that in general plugging the EOM into the Lagrangian is tricky, but it should be perfectly valid if done correctly. Can someone help me see what I'm doing wrong here? I know I'm doing something dumb but I've been staring at it for too long

I'm not sure what you expect to be the result of plugging the equations of motion back into the Lagrangian. You're guaranteed to get something that has the same numerical value as the original Lagrangian, but it's not just the value of the Lagrangian that is important, but also its dependence on its arguments.