What is the action for E-M in terms of E & B?

  • Thread starter Ankerbrau
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Main Question or Discussion Point

Typically the action for E-M is

[tex]F_{\mu\nu}F^{\mu \nu}[/tex]
where
[tex]F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu[/tex]
since the equations of motion for
[tex]A_{\mu}[/tex]
are the inhomogenous Maxwell equations.

However, here comes my problem:
If one expresses this action in terms of the electric and magnetic
field E and B
[tex]F_{\mu\nu}F^{\mu \nu}=B^2-E^2[/tex]
the equations of motion for those fields
would be
E=0
and
B=0.

So, where is the trick and what is the correct action
for the fields E and B?

Thanks in advance for your ideas and comments!
 

Answers and Replies

  • #2
fzero
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The correct canonical variables for electrodynamics are the components of 4-vector potential and their derivatives. Perhaps the best way to see this is to include the Lagrangian for a test particle, which requires coupling the momentum of the particle to the 4-vector potential. It's therefore obvious that the E and B fields cannot be canonical variables. Landau and Lifschitz, The Classical Theory of Fields, is a good reference for this.
 
  • #3
dextercioby
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[itex] \mathcal{L} = E^2 - B^2 [/itex] is not a valid Lagrangian density, because it contains no derivatives.
 
  • #4
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Thanks, obviously

[tex]E^2-B^2[/tex] is not the right Lagrangian.
My question was more:
Is there any Lagrangian at all in terms of E and B,
no matter how awkward it looks?

Today a colleague explained me that
experimentally the Aharonov-Bohm effect showed
that the real physical can only be [tex]A_\mu[/tex],
theory wise it is connected to gauge invariant quantities
that do not depend on E and B like
[tex]\int A_\mu dx^\mu[/tex]
 
  • #5
fzero
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I would say that [tex]E^2-B^2[/tex] is the right Lagrangian (density), but the variational principle must involve the potentials [tex]A_\mu[/tex] as canonical variables. As I said, if you couple electrodynamics to matter, this is more obviously forced on you.
 

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