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

In summary, the conversation discusses using the Euler-Lagrange equation to obtain Maxwell's equations from the EM Lagrangian density, and addresses the question of how to show that the derivative with respect to the scalar potential A for the first term in the density is zero. The conclusion is that it is trivially zero as A_\mu and \partial_\nu A_\mu are treated as independent quantities.
  • #1
neu
230
3

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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  • #2
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.
 
  • #3
Thank you, that's very helpful.
 

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

What is the Euler-Lagrange 2nd derivative?

The Euler-Lagrange 2nd derivative, also known as the second variation or the second functional derivative, is a mathematical concept used in the calculus of variations. It is a measure of the sensitivity of a functional to changes in the input variables or functions.

How is the Euler-Lagrange 2nd derivative calculated?

The Euler-Lagrange 2nd derivative is calculated by taking the second derivative of the functional with respect to the input variables or functions. It is often represented by the symbol δ2F/δu2, where F is the functional and u is the input variable or function.

What is the significance of the Euler-Lagrange 2nd derivative?

The Euler-Lagrange 2nd derivative is important in the calculus of variations because it helps determine the extrema or critical points of a functional. This allows for the optimization of a system or process, which is useful in many scientific and engineering applications.

Can the Euler-Lagrange 2nd derivative be negative?

Yes, the Euler-Lagrange 2nd derivative can be negative. This indicates that the functional has a concave shape at that point, and it is a local maximum. It can also be positive, indicating a convex shape and a local minimum. The sign of the second derivative is important in determining the nature of the critical point.

How is the Euler-Lagrange 2nd derivative used in physics?

In physics, the Euler-Lagrange 2nd derivative is often used in the Lagrangian formulation of classical mechanics. It helps derive the equations of motion for a system, taking into account constraints and conservation laws. It is also used in quantum field theory to calculate the variation of the action with respect to fields.

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