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

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SUMMARY

The discussion centers on the complexities of substituting equations of motion (EOM) into the Lagrangian for electromagnetism, specifically using the Lagrangian L = -\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu} - J^\mu A_\mu. The participant encounters a discrepancy, finding a factor of 2 difference when comparing the expected result after substitution with the derived Lagrangian L \equiv \dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}. The key takeaway is that while substituting EOM into the Lagrangian is valid, the resulting expression retains the same numerical value but may differ in its functional form.

PREREQUISITES
  • Understanding of Lagrangian mechanics and its application in electromagnetism.
  • Familiarity with the equations of motion derived from the Lagrangian formalism.
  • Knowledge of tensor notation and the electromagnetic field tensor F_{\mu\nu}.
  • Basic grasp of variational principles in classical field theory.
NEXT STEPS
  • Study the derivation and implications of the Euler-Lagrange equations in field theory.
  • Explore the concept of total derivatives in the context of Lagrangian mechanics.
  • Investigate the role of gauge invariance in electromagnetism and its effects on the Lagrangian.
  • Learn about the implications of substituting EOM into the Lagrangian in other field theories.
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This discussion is beneficial for theoretical physicists, graduate students in physics, and anyone studying classical field theory, particularly in the context of electromagnetism and Lagrangian mechanics.

michael879
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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?
 
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michael879 said:
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.
 

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