Variation in action for modified EM Field Action

  • #1
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Main Question or Discussion Point

Hi everyone

I am teaching myself QFT, and am currently learning Lagrangian Field Theory. Here is a question I am trying to solve, and I am not absolutely sure if my solution is correct because I am new to this notation and material. I would be grateful if someone could go over it and let me know if its correct. This isn't homework.

Also, I'd like to know what the significance of this result is, if it is indeed true.

Question: Consider the action [itex]S = \frac{1}{4}\int d^{4}x F_{\mu\nu}F^{\mu\nu}[/itex]. Vary the potential according to [itex]A_{\mu} \rightarrow A_{\mu} + \partial_{\mu}\Phi[/itex] where [itex]\Phi[/itex] is a scalar field. Determine the variation in the action.

Solution:

[tex]\delta S = \frac{1}{4}\int d^{4}x (\delta F_{\mu\nu}F^{\mu\nu} + F_{\mu\nu}\delta F^{\mu\nu})[/tex]

[tex]\delta F_{\mu\nu}F^{\mu\nu} &=& \delta(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})F^{\mu\nu}[/tex]
[tex]\implies = (\partial_{\mu}\partial_{\nu}\Phi - \partial_{\nu}\partial_{\mu}\Phi)F^{\mu\nu} = 0[/tex]

Similarly [itex]F_{\mu\nu}\delta F^{\mu\nu} = 0[/itex]. Therefore [itex]\delta S = 0[/itex].

I have attached a pdf file, which contains this question and my working.
 

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Answers and Replies

  • #2
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I did not read your pdf-file, but your conclusion is correct. delta_S is equal to zero. Moreover, F_mu_nu does not vary. It is know as a "gauge" liberty, invariance of transformations. A_mu may be replaced by another A_mu in a certain way, but this does not change fields and particle trajectories. It is similar to the potential energy shifts U -> U+const in CM. The only thing that counts is the exchange, not the absolute value.

Bob.
 
  • #3
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Thanks bob. I see, so this is a gauge transformation under which the Lagrangian is invariant.
 

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