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?
