# Energy of electromagnetic field

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1. May 14, 2015

### 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:

$\int_\Omega dx^3 (curl(\vec{A}))^2$

And we can use the following two equations for simplifying:

$div(A)=0$

and

$\Box A=0$

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

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

Then I can use a trick and take out the first partial derivative with respect to $x_l$:

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

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

2. May 19, 2015