Maxwell's equations in Lagrangian classical field theory

  • #1
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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?
 

Answers and Replies

  • #2
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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?
 
  • #3
TSny
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That looks good to me.
 
  • #4
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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?
 
  • #5
TSny
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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.)
 
  • #6
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(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?
 
  • #7
TSny
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Looks very good!
 

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