How Can Maxwell's Equations Be Derived Using the Euler-Lagrange Equation?

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SUMMARY

This discussion focuses on deriving Maxwell's equations using the Euler-Lagrange equation with the electromagnetic Lagrangian density defined as L = -1/4 FμνFμν - jμAμ. The electromagnetic field tensor is expressed as Fμν = ∂μAν - ∂νAμ. The main challenge discussed is demonstrating that the derivative with respect to the scalar potential A for the first term in the density is zero, as it only contains derivatives. The resolution emphasizes treating Aμ and ∂νAμ as independent quantities.

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  • Understanding of the Euler-Lagrange equation
  • Familiarity with electromagnetic theory and Maxwell's equations
  • Knowledge of Lagrangian mechanics
  • Basic concepts of tensor calculus
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neu
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Homework Statement


I'm asked to get Maxwell's equations using the Euler-lagrange equation:

[tex]\partial\left(\frac{\partial L}{\partial\left\partial_{\mu}A_{\nu}\right)}\right)-\frac{\partial L}{\partial A_{\nu}}=0[/tex]

with the EM Langrangian density:

[tex]L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-j_{\mu}A^{\mu}[/tex]

where the electromagnetic field tensor is:

[tex]F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}[/tex]

The Attempt at a Solution


I'm able to multiply out the density with the full form of the tensor F to get:

[tex]\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}\partial_{\mu}A_{\nu}\left(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right)=\frac{1}{2}\left(\partial_{\mu}A_{\nu}F^{\mu\nu}\right)[/tex]

My problem is that I know that the derivative w.r.t the scalar potential A for the 1st term in the density is zero as it only contains derivatives. i.e

[tex]\frac{\partial }{\partial A_{\mu}}\left(\partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu}-\partial_{\mu}A_{\nu}\partial^{\nu}A^{\mu}\right)=0[/tex]

But I'm unable to show it explicity
 
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neu said:

Homework Statement


I'm asked to get Maxwell's equations using the Euler-lagrange equation:

[tex]\partial\left(\frac{\partial L}{\partial\left\partial_{\mu}A_{\nu}\right)}\right)-\frac{\partial L}{\partial A_{\nu}}=0[/tex]

with the EM Langrangian density:

[tex]L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-j_{\mu}A^{\mu}[/tex]

where the electromagnetic field tensor is:

[tex]F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}[/tex]

The Attempt at a Solution


I'm able to multiply out the density with the full form of the tensor F to get:

[tex]\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}\partial_{\mu}A_{\nu}\left(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right)=\frac{1}{2}\left(\partial_{\mu}A_{\nu}F^{\mu\nu}\right)[/tex]

My problem is that I know that the derivative w.r.t the scalar potential A for the 1st term in the density is zero as it only contains derivatives. i.e

[tex]\frac{\partial }{\partial A_{\mu}}\left(\partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu}-\partial_{\mu}A_{\nu}\partial^{\nu}A^{\mu}\right)=0[/tex]

But I'm unable to show it explicity

I am not sure what you mean by "explicitly". As in other applications of the Lagrange formulation, you must treat [tex]A_\mu[/tex] and [tex]\partial_\nu A_\mu[/tex] independent quantities. So the derivative you wrote is trivially zero.
 
Thank you, that's very helpful.
 

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