# Variation in action for modified EM Field Action

• maverick280857
In summary: This result is significant because it shows that the equations of motion, which are derived from the Lagrangian, are also invariant under this gauge transformation. This means that our theory is consistent and we can use this freedom to simplify calculations. Also, it is interesting to note that this transformation does not change the physical content of the theory, only the mathematical form. This is a fundamental concept in Lagrangian Field Theory and is crucial in understanding the behavior of particles and fields. In summary, the conversation discusses a question and solution related to Lagrangian Field Theory. The question involves determining the variation in the action for a given potential and the solution shows that the variation is equal to zero. This result is significant because it demonstrates gauge invar

#### maverick280857

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

Solution:

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

$$\delta F_{\mu\nu}F^{\mu\nu} &=& \delta(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})F^{\mu\nu}$$
$$\implies = (\partial_{\mu}\partial_{\nu}\Phi - \partial_{\nu}\partial_{\mu}\Phi)F^{\mu\nu} = 0$$

Similarly $F_{\mu\nu}\delta F^{\mu\nu} = 0$. Therefore $\delta S = 0$.

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

#### Attachments

• qftquest2.pdf
34.6 KB · Views: 205
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.

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