Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question

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!

share|cite|improve this answer
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
For the first equation, you have done the summation incorrectly in the first term. Try writing it out more explicitly. You can simplify the second term with the sum further. Pressure is a scalar quantity! I'm sorry if my hint was confusing. I've rephrased it for you. – jeffdk Mar 5 '13 at 18:11
To answer your second question, yes, that is what the Kronecker delta means. – jeffdk Mar 5 '13 at 18:20
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.