Thermodynamics and electromagnetic fields

Question

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!

Answer 1

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.

Now you should be able to equate your two expressions and obtain the desired answer. Good luck!