Maxwell's equations in Lagrangian classical field theory

The only problem is that you started with the wrong Lagrangian. It should have been ##\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}##. But your calculations are correct, they just show that there are two different Lagrangians for the same physical theory; one of them is not really a Lagrangian at all, just a total derivative as you showed.
  • #1
spaghetti3451
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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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  • #2
(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
That looks good to me.
 
  • #4
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
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
(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
Looks very good!
 

1. What are Maxwell's equations?

Maxwell's equations are a set of four partial differential equations that describe the behavior and interactions of electric and magnetic fields. These equations were developed by physicist James Clerk Maxwell in the 19th century and are the foundation of classical electromagnetism.

2. What is Lagrangian classical field theory?

Lagrangian classical field theory is a mathematical framework used to describe the dynamics of a physical system in terms of continuous fields, rather than individual particles. It is based on the principle of least action, which states that the actual path taken by a system is the one that minimizes the action, a quantity defined by the Lagrangian function.

3. How are Maxwell's equations related to Lagrangian classical field theory?

Maxwell's equations can be derived from the Lagrangian density of the electromagnetic field using the principle of least action. In this formulation, the equations represent the Euler-Lagrange equations of motion for the electromagnetic field. This approach provides a more elegant and unified description of electromagnetism.

4. What is the significance of Maxwell's equations in physics?

Maxwell's equations are fundamental to our understanding of electromagnetism and have numerous applications in many areas of physics, including optics, electronics, and telecommunications. They also played a crucial role in the development of Einstein's theory of general relativity and the concept of electromagnetic waves.

5. Are Maxwell's equations still relevant in modern physics?

Yes, Maxwell's equations are still relevant and widely used in modern physics. They have been extensively tested and proven to accurately describe the behavior of electric and magnetic fields in a wide range of physical phenomena. In addition, they have been extended and refined to incorporate new discoveries, such as quantum mechanics and special relativity.

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