Maxwell's equations in Lagrangian classical field theory

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Homework Help Overview

The discussion revolves around the derivation and manipulation of the Maxwell Lagrangian in the context of classical field theory. Participants are exploring the implications of the Lagrangian, particularly in relation to the equations of motion and alternative formulations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to differentiate the Lagrangian with respect to the field derivatives and are questioning the correctness of their steps. There is also a focus on rewriting the Lagrangian in a more compact form and discussing the implications of total derivatives in the context of the action.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning and calculations. Some have confirmed the correctness of earlier steps, while others are refining their approaches to ensure clarity and accuracy in their derivations.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of assistance they can provide to one another. There is an emphasis on ensuring that the derivations respect the underlying principles of field theory.

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Homework Statement



Given the Maxwell Lagrangian ##\mathcal{L} = -\frac{1}{2} (\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}) + \frac{1}{2} (\partial_{\mu}A^{\mu})^{2}##,

show that

(a) ##\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}A_{\nu})} = - \partial^{\mu}A^{\nu}+(\partial_{\rho}A^{\rho})\eta^{\mu\nu}## and hence obtain the equations of motion ##\partial_{\mu}F^{\mu\nu}=0##.

(b) we may rewrite the Maxwell Lagrangian (up to an integration by parts) in the compact form ##\mathcal{L} = - \frac{1}{4} F_{\mu\nu}F^{\mu\nu}##.

Homework Equations



The Attempt at a Solution



(a) ##\mathcal{L} = -\frac{1}{2} (\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}) + \frac{1}{2} (\partial_{\mu}A^{\mu})(\partial_{\mu}A^{\mu})##

##= -\frac{1}{2} (\partial_{\mu}A_{\nu})(\partial_{\rho}A_{\sigma})(\eta^{\rho\mu}\eta^{\sigma\nu}) + \frac{1}{2} (\partial_{\mu}A_{\rho})(\partial_{\mu}A_{\sigma})(\eta^{\rho\mu}\eta^{\sigma\mu})##

Am I on the right track? Do I now differentiate each of the terms using the product rule?
 
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(a) Let me redo this part of the question.

The Lagrangian ##\mathcal{L}## is given by ##\mathcal{L} = -\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+\frac{1}{2}(\partial_{\mu}A^{\mu})^{2}##.

Now,

##\frac{\partial}{\partial(\partial_{\rho}A_{\sigma})}\Big(-\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})\Big)##

##=\frac{\partial}{\partial(\partial_{\rho}A_{\sigma})}\Big(-\frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta}(\partial_{\mu}A_{\nu})(\partial_{\alpha}A_{\beta})\Big)##

##=-\frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta}\frac{\partial}{\partial(\partial_{\rho}A_{\sigma})}\Big((\partial_{\mu}A_{\nu})(\partial_{\alpha}A_{\beta})\Big)##

##=-\frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta}({\eta^{\rho}}_{\mu}{\eta^{\sigma}}_{\nu}\partial_{\alpha}A_{\beta}+{\eta^{\rho}}_{\alpha}{\eta^{\sigma}}_{\beta}\partial_{\mu}A_{\nu})##

##=-\frac{1}{2}({\eta^{\rho}}_{\mu}\eta^{\mu\alpha}{\eta^{\sigma}}_{\nu}\eta^{\nu\beta}\partial_{\alpha}A_{\beta}+{\eta^{\rho}}_{\alpha}\eta^{\alpha\mu}{\eta^{\sigma}}_{\beta}\eta^{\beta\nu}\partial_{\mu}A_{\nu})##

##=-{\eta^{\rho}}_{\mu}\eta^{\mu\alpha}{\eta^{\sigma}}_{\nu}\eta^{\nu\beta}\partial_{\alpha}A_{\beta}##

##=-\eta^{\rho\alpha}\eta^{\sigma\beta}\partial_{\alpha}A_{\beta}##

##=-\partial^{\rho}A^{\sigma}##

Am I correct so far?
 
That looks good to me.
 
Alright, then. Now,

##\frac{\partial}{\partial(\partial_{\rho}A_{\sigma})} \Big( \frac{1}{2}(\partial_{\mu}A^{\mu})(\partial_{\nu}A^{\nu})\Big)~##

##~=~ \frac{\partial}{\partial(\partial_{\rho}A_{\sigma})} \Big( \frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta}(\partial_{\mu}A_{\alpha})(\partial_{\nu}A_{\beta})\Big)~##

##~=~ \frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta} \frac{\partial}{\partial(\partial_{\rho}A_{\sigma})} \Big( (\partial_{\mu}A_{\alpha})(\partial_{\nu}A_{\beta})\Big)~##

##~=~ \frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta} \Big( {\eta^{\rho}}_{\mu}{\eta^{\sigma}}_{\alpha}(\partial_{\nu}A_{\beta})+{\eta^{\rho}}_{\nu}{\eta^{\sigma}}_{\beta}(\partial_{\mu}A_{\alpha}))\Big)~##

##~=~ \frac{1}{2} \Big( {\eta^{\rho}}_{\mu}\eta^{\mu\alpha}{\eta^{\sigma}}_{\alpha}\eta^{\nu\beta}(\partial_{\nu}A_{\beta})+{\eta^{\rho}}_{\nu}\eta^{\nu\beta}{\eta^{\sigma}}_{\beta}\eta^{\mu\alpha}(\partial_{\mu}A_{\alpha}))\Big)~##

##~=~ {\eta^{\rho}}^{\alpha}{\eta^{\sigma}}_{\alpha}(\partial_{\nu}A^{\nu})##

##~=~ {\eta^{\sigma}}_{\alpha}{\eta^{\alpha}}^{\rho}(\partial_{\nu}A^{\nu})##

##~=~ {\eta^{\sigma}\rho}(\partial_{\nu}A^{\nu})##

##~=~ (\partial_{\nu}A^{\nu}){\eta^{\sigma}\rho}##.

Therefore,

##\partial_{\rho}(\partial^{\rho}A^{\sigma}+(\partial_{\nu}A^{\nu})\eta^{\rho\sigma})~=~0~##

##-\partial_{\rho}\partial^{\rho}A^{\sigma}+\partial^{\sigma}(\partial_{\nu}A^{\nu})~=~0~##

##\partial_{\rho}\partial^{\rho}A^{\sigma}-\partial^{\sigma}(\partial_{\nu}A^{\nu})~=~0~##

##\partial_{\mu}\partial^{\mu}A^{\nu}-\partial^{\nu}(\partial_{\mu}A^{\mu})~=~0~##

##\partial_{\mu}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})~=~0~##

##\partial_{\mu}F^{\mu\nu}~=~0~##.

Am I correct?
 
Good. (A minus sign was left out in the first equation after the "therefore", but you have it back in the rest of the derivation.)
 
(b) ##\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}##

##=-\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})##

##=-\frac{1}{4}[(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})-(\partial_{\mu}A_{\nu})(\partial^{\nu}A^{\mu})-(\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu})+(\partial_{\nu}A_{\mu})(\partial^{\nu}A^{\mu})]##

##=-\frac{1}{4}[(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})-(\partial_{\mu}A_{\nu})(\partial^{\nu}A^{\mu})-(\partial_{\mu}A_{\nu})(\partial^{\nu}A^{\mu}))##

##=-\frac{1}{2}[(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})-(\partial_{\mu}A_{\nu})(\partial^{\nu}A^{\mu})]##

##=-\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+\frac{1}{2}[\partial^{\nu}\{A^{\mu}(\partial_{\mu}A_{\nu}\})-A^{\mu}(\partial^{\nu}\partial_{\mu}A_{\nu})]##

##=-\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+\frac{1}{2}[\partial^{\nu}\{A^{\mu}(\partial_{\mu}A_{\nu})\}-A^{\mu}(\partial_{\mu}\partial^{\nu}A_{\nu})]##

##=-\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+\frac{1}{2}[\partial^{\nu}\{A^{\mu}(\partial_{\mu}A_{\nu})\}+(\partial_{\mu}A^{\mu})(\partial^{\nu}A_{\nu})-\partial_{\mu}\{A^{\mu}(\partial^{\nu}A_{\nu})\}]##

##=-\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+\frac{1}{2}(\partial_{\mu}A^{\mu})^{2}+\partial^{\nu}[\frac{A^{\mu}}{2}(\partial_{\mu}A_{\nu})]-\partial_{\mu}[\frac{A^{\mu}}{2}(\partial^{\nu}A_{\nu})]##

The last two terms are total derivatives. Therefore, when integrating the last two terms over the entire region of Minkowski spacetime, the condition that the field ##A^{\mu}(x)## vanishes at spatial infinity and at the initial and final times ensures that the integrals of the last two terms are zero.

Therefore, the action is unchanged under the addition of the total derivatives to the Lagrangian. Therefore, the terms with total derivatives can be omitted from the Lagrangian to obtain

##\mathcal{L}=-\frac{1}{2}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu})+\frac{1}{2}(\partial_{\mu}A^{\mu})^{2}##

Am I correct?
 
Looks very good!
 

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