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

  • I
  • Thread starter Thread starter michael879
  • Start date Start date
  • Tags Tags
    Eom Lagrangian
AI Thread Summary
Plugging the equations of motion (EOM) into the Lagrangian can yield unexpected results, as demonstrated with the electromagnetic Lagrangian. The original Lagrangian leads to EOM that, when substituted back, results in a discrepancy of a factor of two. The expectation that the terms would cancel to produce a trivial action is challenged by the realization that the numerical value alone does not capture the Lagrangian's dependence on its variables. This highlights the complexity of manipulating Lagrangians and emphasizes the need for careful consideration of dependencies. Understanding these nuances is crucial for accurate applications in theoretical physics.
michael879
Messages
696
Reaction score
7
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?
 
Physics news on Phys.org
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.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top