Energy of electromagnetic field

[VVS]

Hey everyone!

I am supposed to calculate the energy contribution of the magnetic field term of an electromagnetic field.

Basically the term is the following:

[itex]\int_\Omega dx^3 (curl(\vec{A}))^2[/itex]

And we can use the following two equations for simplifying:

[itex]div(A)=0[/itex]

and

[itex]\Box A=0[/itex]

So basically what I did was express the curl in terms of the levi civita tensor.
Then you can simplify the volume integral to:

[itex]\int_\Omega dx^3 \sum_{m,l}\frac{\partial A_m}{\partial x_l} (\frac{\partial A_m}{\partial x_l}-\frac{\partial A_l}{\partial x_m})[/itex]

Then I can use a trick and take out the first partial derivative with respect to [itex]x_l[/itex]:

[itex]\int_\Omega dx^3 \sum_{m,l}\frac{\partial}{\partial x_l} \left(A_m(\frac{\partial A_m}{\partial x_l}-\frac{\partial A_l}{\partial x_m})\right)-\left(A_m(\frac{\partial^2 A_m}{\partial x_l^2}-\frac{\partial^2 A_l}{\partial^2 x_mx_l})\right)[/itex]

The term [itex]\frac{\partial^2 A_l}{\partial^2 x_mx_l}[/itex] vanishes since it contains the divergence of [itex]\vec{A}[/itex] and the term [itex]A_m\frac{\partial^2 A_m}{\partial x_l^2}[/itex]is exactly the result we want: [itex]\frac{1}{c^2}\vec{A}\cdot \ddot{\vec{A}}[/itex] because of the d'Alembertian relation.
So what is left to be shown is that the first term [itex]\int_\Omega dx^3 \sum_{m,l}\frac{\partial}{\partial x_l} \left(A_m(\frac{\partial A_m}{\partial x_l}-\frac{\partial A_l}{\partial x_m})\right)[/itex] vanishes, which I have no clue how to show. Hope someone can help me.

 

[mark726]

Hello there,

Thank you for sharing your approach for calculating the energy contribution of the magnetic field term in an electromagnetic field. It seems like you have made some progress in simplifying the integral, but you are stuck on how to show that the first term in your expression vanishes.

One possible approach to show this is to use the fact that the divergence of the curl of a vector field is always zero, i.e. div(curl(\vec{A}))=0. This can be proven using vector calculus identities and the fact that the divergence operator is a scalar operator. Since we are assuming that div(\vec{A})=0, this means that the term inside the integral, \sum_{m,l}\frac{\partial}{\partial x_l} \left(A_m(\frac{\partial A_m}{\partial x_l}-\frac{\partial A_l}{\partial x_m})\right), is equivalent to div(curl(\vec{A})). Therefore, it also evaluates to zero and the first term in your expression vanishes.

I hope this helps. Let me know if you have any further questions. Good luck with your calculations!