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I'm following this derivation of the Rayleigh-Jeans equation.

So, I will state the facts relevant to my question.

  • Cube of side length L.
  • Radiation forms standing waves with nodes at the faces of the cube.

It follows that, for axis-parallel radiation, the length of the cube will be a multiple of half the wavelength. $ m\frac{\lambda}2 = L $

For radiation in arbitrary direction, decomposing in components:

  • $ \lambda^2 = \lambda_x^2 + \lambda_y^2 + \lambda_z^2 $
  • $ m_x\frac{\lambda_x}{2} = m_y\frac{\lambda_y}{2} = m_z\frac{\lambda_z}{2} = L $
  • $ \lambda_x=\frac{2L}{m_x} $ (...)

The wavenumber (q) is the number of radians per unit distance. $ q = \frac{2\pi}{\lambda} $

So I get $ q^2 = (2\pi)^2(\lambda^2)^{-1} = (2\pi)^2(\lambda_x^2 + \lambda_y^2 + \lambda_z^2)^{-1} = (2\pi)^2[(\frac{2L}{m_x})^2 + (\frac{2L}{m_y})^2 + (\frac{2L}{m_z})^2]^{-1} $

Cancelling the factor $ 2^2 $ we have: $ q^2 = \pi^2\left[(\dfrac{L}{m_x})^2 + (\dfrac{L}{m_y})^2 + (\dfrac{L}{m_z})^2\right]^{-1} $

But this last equation is different than equation 9 in the derivation and I don't see how one can transform into the other.

Equation 9 in the derivation: $ q^2 = π^2\left[ \left(\dfrac{m_X}{L}\right)^2 + \left(\dfrac{m_Y}{L}\right)^2 + \left(\dfrac{m_Z}{L}\right)^2\right] $

Where did I make a mistake here?

DeltA
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    Your equation $\lambda^2 = \lambda_x^2 + \lambda_y^2 + \lambda_z^2$ is not correct. It should rather be $\frac{1}{\lambda^2} = \frac{1}{\lambda_x^2} + \frac{1}{\lambda_y^2} + \frac{1}{\lambda_z^2}$. May be my answer to Significance of wave number? is helpful. – Thomas Fritsch Sep 21 '21 at 21:55
  • @ThomasFritsch Your answer is helpful, thanks a lot. However, I still don't quite get it. Say we have a 2D square as the cavity (for simplicity). The numbers $m_x$ and $m_y$ indicate the number of nodes (half waves) in each direction. Isn't the half wave length = $ ((L/m_x)^2 + (L/m_y)^2)^{1/2} $? – DeltA Sep 21 '21 at 23:06
  • No. This would make $\lambda/2$ longer than $L/m_x$ and $L/m_y$. But actually $\lambda/2$ is shorter than these. – Thomas Fritsch Sep 21 '21 at 23:23
  • I fail to see why. Perhaps I'm looking at it from the wrong perspective. I'll re-think all from the start. I would apreciate if you could explain me this with a bit more detail. – DeltA Sep 21 '21 at 23:42
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    Here's some intuition that might wrongly lead you to $\lambda^2=\lambda_x^2+\lambda_y^2+\lambda_z^2$. Imagine computing $\lambda$ for a plane wave traveling in some random direction. You might say is the distance between peaks can be computed via the Pythagorean theorem as the distance between peaks along coordinate directions. But this is wrong, and "the distance between peaks along coordinate directions" doesn't really make sense here. Instead you should write $e^{i \vec k \cdot \vec x}$ as a product of three terms $e^{i k_x x} e^{i k_y y} e^{i k_z z}$. This leads you to use the wavenumber. – Andrew Sep 22 '21 at 03:29
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    The intuitive wrong argument boils down to "wavelength is a length; combining a length and the wave's direction gives me a vector, and I can resolve a vector into components". But really you want to resolve the wave into mode functions. The wavenumber appears in the argument of the mode function in a simple way $e^{i k x}$, while the wavelength doesn't. This is a sign that the mode function wants you to work with $k$ over $\lambda$... more rigorously, you can work through the math of relating the magnitude of $\vec{k}$ to $k_x, k_y, k_z$ and you'll see it's exactly what @ThomasFritsch said. – Andrew Sep 22 '21 at 03:38
  • Thanks a lot, I get it now. My mental model was wrong. Thinking about the exponential forms made me see it. I think this kind of things should be explained better in the books. I didn't found a single source about this derivation that touched this topic. – DeltA Sep 22 '21 at 06:01

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