# Thermodynamics and electromagnetic fields

The energy density in an electromagnetic field is given by: $u = (1/2) <\vec{E}^2 + \vec{B}^2>$ where $<,>$ denotes the average over time. In a cavity it holds that $u= u(T) = \frac{1}{V}U(T,V)$ The force acting on a surface element is given by $d\vec{F_i} = \sum_{j=1}^3 \sigma_{ij} d\vec{o_j}$ where $\vec{o_j}$ denotes the infinitesimal surface element. It also holds that $\sigma_{ij} = \frac{1}{2}(\vec{E}^2 + \vec{B}^2)\delta_{ij} - E_i E_j - B_i B_j$. Since we are in a cavity the radiation is isotropic and the tensor $\sigma$ only has values on the diagonal and it holds that $\sigma_{ij} = p \delta_{ij}$

I'm now supposed to show that the following equation holds:

$p(T) = \frac{1}{3} u(T)$

This should probably be really easy but I didn't get any hints for this exercise and I just can't seem to find a way to start. If someone could just give me a hint about how I should start this problem, I'd greatly appreciate it.

Cheers!

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See if you can use the equation for the differential force, $dF_i$, in terms of the energy density, $u$. You should be able to do this using the expression for the electromagnetic stress, $\sigma_{ij}$.
Also note that if you have a pressure $p$, (and this is your only stress; a pressure is basically an isotropic stress) then the force on any surface element of any orientation is $dF = p do$. In this case, I'm just using $F$ for the magnitude of the force on the element of surface area $o$. The $d$ in this context is just used to signify that the surface area element is an infinitesimally small one so that the pressure $p$ is constant for that whole area. Use this to come up with an expression for the pressure in terms of the force density equation.
Ok so starting with $dF_i = \sum_{j=1}^3 \sigma_{ij} d\vec{\sigma_j} = \frac{1}{2} (\vec{E}^2 + \vec{B}^2)d\vec{\sigma_i} - \sum_{j=1}^3 (E_i E_j + B_i B_j)d\vec{\sigma_j}$ Since the kronecker delta kills all the terms where j isn't equal to i. Can you help me interpret the equation $<\sigma_{ij}> = p \delta_{ij}$? shouldn't $p$ be a vector? But this equation implies it's a scalar... I'm really confused:( –  user17574 Mar 5 '13 at 17:47
Wait does the second equation mean that the ij-th entry of the matrix $\sigma_{ij}$ simply is $p$ if $i=j$ and $0$ else? –  user17574 Mar 5 '13 at 18:07
Why is the first term incorrect? i is either 1,2 or 3, so all the $\frac{1}{2}(\vec{E}^2 + \vec{B}^2) \delta_{ij}$ drop out if $j$ isn't equal to $i$, and only $\frac{1}{2}(\vec{E}^2 + \vec{B}^2)d\vec{\sigma_i}$ is left over. I feel really dumb right now... –  user17574 Mar 5 '13 at 18:48