Questions on derivation of Rayleigh-Jeans equation

So it starts by assuming a cube cavity with length $$L$$ that acts approximately as a black body. [note: the cube length is $$a$$ but I'm using $$L$$ instead] Standing waves are formed inside the cavity, the equation for a standing wave in 3D is $$E(x,t) = E_0 \sin(k_x x) \sin(k_y y) \sin(k_z z) \sin ( 2 \pi \nu t),$$ where $$k_x = \frac{n_x \pi }{L}$$ and so on.

The wave vector $$\vec k = k_x \hat i + k_y \hat j + k_z \hat k$$

Where the magnitude of $$k = \frac { 2 \pi}{ \lambda}$$

Then $$k^2 = k^{2}_x + k^{2}_y + k^{2}_z = (\frac{n_x \pi }{L})^2 + (\frac{n_y \pi }{L})^2 + (\frac{n_z \pi }{L})^2 = \frac { 4 \pi^2}{ \lambda^2}$$

Therefore,

$$n^{2}_x + n^{2}_y + n^{2}_z = \frac { 4 L^2}{ \lambda^2} = \frac { 4 L^2 \nu^2}{c^2}$$

$$\implies \nu^2 = \frac { c^2 n^2}{ 4 L^2}$$

Now here comes the first part that I don't understand:

'Volume of the n space is a sphere of volume $$\frac {4}{3} \pi n^3$$ which gives the number of nodes.

What does that volume represent, how does it give the number of nodes?

Continuing the proof:

Since $$n$$ is a positive integer, we divide the volume by 8, then the new volume $$V = \frac { \pi}{6} n^3 = \frac { \pi}{6} ( \frac {2L \nu}{c})^{3}$$

Here comes the second part that doesn't make sense to me:

We got the volume $$V = \frac { \pi}{6} ( \frac {2L \nu}{c})^{3}$$

He now differentiates both sides

$$dV = 3 \nu^2 \frac {4}{3} \pi \frac {L^3}{c^3} d \nu = 4 \pi \frac {L^3}{c^3} \nu^2 d \nu$$

We know that $$n$$ is a positive integer that can't take on continuous values, then what does differentiating the volume with respect to the frequency mean? I hope my questions were clear, thank you.

I can see you're using HyperPhysics' derivation of the Reyleigh-Jeans Equation. I've got some issues with how correct some parts of the derivation are but it is good enough for our purposes. I'll try my best to explain.

First Question:

The equation $$n^{2}_x + n^{2}_y + n^{2}_z = \frac { 4 L^2}{ \lambda^2} = \frac { 4 L^2 \nu^2}{c^2}$$ relates the number of nodes (points where the wave doesn't oscillate) of the standing wave inside the box to the frequency of that wave. It essentially allows you to say "This standing wave has $$n_x$$ nodes in the $$x$$-direction, $$n_y$$ nodes in the $$y$$-direction, and $$n_z$$ nodes in the $$z$$-direction, so then it's frequency must be such-and-such. That I assume is pretty clear to you

Different combinations of $$n_x$$, $$n_y$$, and $$n_z$$ are called normal modes or just modes(not nodes, as you wrote). As you can see from the equation, there can be multiple modes with the same frequency. And any mode in which $$n^{2}_x + n^{2}_y + n^{2}_z = \frac { 4 L^2 \nu^2}{c^2} \tag{1}\label{eq1}$$ is a mode of frequency $$\nu$$. Now imagine we represent the nodes in Cartesian space. The $$x$$, $$y$$, and $$z$$-axes represent $$n_x$$, $$n_y$$, and $$n_z$$ respectively. Every point with coordinates $$(n_x, n_y,n_z)$$ in whics the $$n$$'s are integers will represent a normal mode of the cavity. Since equation $$\eqref{eq1}$$ is equivalent to that of a sphere of radius $$R=\frac{2L\nu}{c}$$ you can see that a sphere of such radius will touch all points corresponding to normal modes of frequency $$\nu$$, inside it will lie all points corresponding to modes of frequency below $$\nu$$, and outside of it all points corresponding to modes of frequency above $$\nu$$. You can also see that since all $$n$$'s are positive integers, only one eighth of this '$$n$$-space' will be occupied by points corresponding to nodes, and they will have a density of one node per unit volume.

Say we take the volume of this one-eighth of a sphere. Since the points corresponding to modes have a density of one, this volume will be equivalent to the total number of modes at or below frequency $$\nu$$ (approximately, but since for light we usually work with very high frequencies, the error is negligibly small). Thus the number of modes below frequency $$\nu$$ is $$N(\nu)=\frac{1}{8}\cdot\frac{4}{3}\pi R^3=\frac{\pi}{6}\cdot\left(\frac{2L\nu}{c}\right)^3=\frac{\pi}{6}\cdot\frac{8L^3\nu^3}{c^3}=\frac{4}{3}\frac{\pi L^3\nu^3}{c^3}$$

Second Question

The equation we previously derived isn't very useful as it is. It shows the number of modes of frequecy equal or below $$\nu$$. For blackbody analysis, it's be more convenient to work with the number of modes in a small interval of frequencies. This will be $$dN=\frac{dN}{d\nu}d\nu$$ And $$\frac{dN}{d\nu}$$ is the mode density per unit frequency. So we differentiate and obtain $$dN=4\pi\frac{L^3\nu^2}{c^3}d\nu$$ But something we've neglected is that any mode of radiation is actually two separate modes, because it can be polarized in one of two directions. So we multiply by two and change the notation a little bit, the number of modes in a cubical cavity of length $$L$$ of frequencies in between $$\nu$$ and $$\nu+d\nu$$ is $$N(\nu\to \nu+d\nu)=\frac{8\pi\nu^3L^3}{c^3}d\nu$$

• Why is $N(\nu)=\frac{4}{3}\frac{\pi L^3\nu^3}{c^3}$ useless, I can just plug any frequency that I want and come up with the number of modes corresponding to that frequency, that gives me information of how intense the radiation is for every frequency that I plug in? – khaled014z Mar 1 at 15:07
• As I said, it will not give you the number of modes at a certain frequency. Since inside the sphere are contained all the points corresponding to frequencies below $\nu$, you'll get the amount of modes corresponding to that frequency and all frequencies below. – user140323 Mar 1 at 21:30
• No problem, glad I could help! – user140323 Mar 2 at 2:51

"What does that volume represent"

Imagine 3 axes at right angles, labelled $$n_x, n_y, n_z.$$ points are plotted for all possible integral trios of $$n_x, n_y, n_z.$$, for example, (1, 1, 1), (1, 2, 1). The space will be filled with a regular lattice of points. A spherical surface of radius $$n$$ will enclose all those points for which $$n_x^2+n_y^2+ n_z^2\leq n^2.$$

"We know that 𝑛 is a positive integer that can't take on continuous values, then what does differentiating the volume with respect to the frequency mean?"

You have to accept that $$n$$ will be a huge number. Therefore a $$\Delta n$$ of millions will represent the addition of only a very thin spherical shell. Although $$n$$ can only changes in a jerky fashion, the individual jerks make virtually no difference to the thickness of the shell, whose volume is, to an excellent approximation, $$\frac{d}{dn}(\frac43 \pi n^3) \Delta n=4\pi n^2 \Delta n$$. We then multiply by $$\frac {dn}{d \nu}$$ (using the chain rule) and divide by 8, as you've explained. This gives the number of points with all positive $$n_x, n_y, n_z$$ per unit frequency range centred on frequency $$\nu.$$

• I understood the first part, but why are we taking the derivative of n, does it have something to do with spectral radiance? – khaled014z Mar 1 at 16:43
• You've understood that $\frac 18$ of the shell volume, $4\pi n^2 \Delta n,$ gives the number of modes with $n$ values between $n$ and $n + \Delta n?$ But we're trying to find the number of nodes with frequencies between $\nu$ and $\nu +\Delta \nu.$ We know that $\Delta n \approx \frac{dn}{d\nu} \Delta \nu.$ So the number of nodes between $\nu$ and $\nu +\Delta \nu.$ is $4\pi n^2 \frac{dn}{d\nu} \Delta \nu.$ – Philip Wood Mar 1 at 18:44