# Electromagnetic fields in a cubical cavity

I'm trying to solve the standing electromagnetic modes in a cubical cavity problem without using separation of variables. The cube is a perfect conductor, and hence the boundary conditions are $E_{\parallel} = 0$ and $B_{\bot} = 0$.

Now, I try to solve the wave equation (there will be a total of six of them, but that's not my concern):

I start by taking the equation to Fourier domain, giving me, for instance for $E_{x}$:

$$E_x(\textbf{r},t) = \int_{\bf{R^3}} A(\textbf{k}) e^{i(\textbf{k} \textbf{r} - \omega t)} d \textbf{k}$$

where I've dropped the wave traveling back in time. This is a pretty standard result stating that any well behaved electromagnetic field can be expanded in a plane wave basis.

Now, I want to solve for standing waves, hence I want the exponential time factor to come out of the integral. This reduces my integral over $\bf{R^3}$ to a set of $\bf{k}$ vectors with a constant length, i.e., over a sphere of a given radius. This is because then $\omega = c*|k|$ will be constant, thus giving us:

$$E_x(\textbf{r})e^{-i \omega t} = e^{-i \omega t} \int_{\textbf{|k|}=R} A(\textbf{k}) e^{i\textbf{k} \textbf{r}} d \textbf{k}$$

But I do not know how to apply the boundary condition on $E_x(\textbf{r})$ and get the following condition on $\bf{k}$:

$$k_x = \frac {\pi n_x} {L}, k_y = \frac {\pi n_y} {L}, k_z = \frac {\pi n_z} {L}$$

Or is my attempt to not use separation of variables totally useless?

• Considering how easy separation of variables really is, I'd say any other attempt is probably wasting your time. – Gert Nov 9 '16 at 0:00

By using plane wave solutions in the three coordinate directions you have already used the separation of variables method for the solution of the three dimensional wave equation $$∆\vec E=\frac{∂^2 \vec E}{c^2∂t^2}$$ Also, for finding the solutions satisfying the boundary conditions, you have to take discrete Fourier series not the Fourier integrals.

• But I am not taking the plane waves in only the three coordinate directions. The $\bf{k}$ vectors span a spherical shell for a given magnitude $|k|$. Also, why the Fourier series? My fields are localized in space, hence I will need to use the Fourier integral. – Sidd Nov 9 '16 at 2:50
• @Sidd - If you take plane wave in any direction its wave vector can always be decomposed into wave vectors in the coordinate directions. – freecharly Nov 9 '16 at 2:59
• @Sidd - You need particular solutions satisfying the boundary conditions to get the general solution by superposition. This is done by Fourier series, even though this means that this also solves the periodic cube problem. Fourier integrals are not helpful here unless you use them to get discrete Fourier series in the end. – freecharly Nov 9 '16 at 3:07
• I understand your point about decomposition into three coordinate vectors, but since that's in the exponential, it gives us multiplicative terms. Also, about the series, can you show or hint towards how that would solve the problem. I'm just unable to see it. – Sidd Nov 9 '16 at 3:09
• @Sidd - You assume cosine and sine functions wave functions with the wave vectors $k_x, k_y$ and $k_z$ so that they have nodes at the cube walls. Then you make a 3D Fourier series out of the products of the particular solutions. – freecharly Nov 9 '16 at 3:16

"Or is my attempt to not use separation of variables totally useless?"

Yes and no. See, you're doing separation of variables, you're just doing it out of order. As @freecharly noted, you're starting with the solutions that comes from performing separation of variables in 3-d without boundary conditions.

That said, the trick to translating your boundary conditions from real space to Fourier space is to lean on the linear independence of the Fourier modes (especially as a function of time).

For instance, we want to specify one of the boundary conditions completely. So, for $$y=0$$ $$E_{||}=0$$ is the same as saying \begin{align} E_x(x,0,z,t) & = 0\ \forall x,z,t \\ E_z(x,0,z,t) & = 0\ \forall x,z,t. \end{align} Because each $$e^{i\omega t}$$ that has a different $$\omega$$ is linearly independent of every other group of $$\omega$$, and you need the equation to be true for all $$t$$, you can separate out each mode/frequency. Similarly, the $$\forall x,z$$ will allow you to focus in on the $$e^{ik_y y}$$ factors. Then, when you combine the boundary conditions on the opposite face of the box, you'll be forced to only consider quantized $$k_y$$ and the combinations of sine and cosine that have the right zeros.

Ending you up in the same place as separation of variables, but the logic is a little more complicated.