Mandl and Shaw 5.1

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Main Question or Discussion Point

[SOLVED] Mandl and Shaw 5.1

To show

[itex]-\frac{1}{2}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2 \quad \mathrm{and} \quad -\frac{1}{2}\partial_\nu A_\mu \partial^\nu A^\mu[/itex]

represent the same Lagrangian it suffices to show that

[itex]\partial_\nu A_\mu\partial^\mu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu[/itex] is at most a 4-divergence.

The trouble is, I have no idea why this would be the case. Is this a matter of utilizing the product rule in some clever way?

Edit: Yes it is: factor out [itex]\partial_\nu[/itex].
 
Last edited:

Answers and Replies

  • #2
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hi
does anybody have any suggestion to solve 2.4 too?!
 
  • #3
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To show

[itex]-\frac{1}{2}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2 \quad \mathrm{and} \quad -\frac{1}{2}\partial_\nu A_\mu \partial^\nu A^\mu[/itex]

represent the same Lagrangian it suffices to show that

[itex]\partial_\nu A_\mu\partial^\mu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu[/itex] is at most a 4-divergence.

The trouble is, I have no idea why this would be the case. Is this a matter of utilizing the product rule in some clever way?

Edit: Yes it is: factor out [itex]\partial_\nu[/itex].
Can this be done without using the Lorentz gauge ([itex]\partial_\mu A^\mu = 0[/itex]) or is it necessary imposed ?
 

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