Why does Rayleigh-Jeans law agree with Planck's at long wavelengths?

Question

Why does Rayleigh-Jeans law agree with the Plancks distribution at high wavelengths? I know mathematically, that at higher wavelengths, we can approximate the exponential in the denominator by $1+ \frac{hc}{\lambda k_bT}$ which gives back the Rayleigh's law, but I do not understand what is the Physics behind this. Why does the classical result agree at high wavelength, i.e. low energy.

Answer 1

Planck's postulate is, an electromagnetic oscilllator mode of frequency $\nu$ can only have energies of $E=nh\nu$ (with $n=0,1,2,3,\dots$). That means you have ladder-like energy levels, with small steps at low frequencies, and big steps at high frequencies. You can visualize these energie levels for some example frequencies $\nu$ and by how many oscillators they are occupied:

_{(image from my answer to another related question)}

Doing the math, Planck derived the average energy of an oscillator mode at temperature $T$ to be $$E_\text{avg}(\nu,T)=\frac{h\nu}{e^{h\nu/kT}+1} \tag{1}$$

And now ansering your question: At the left part of the image above (i.e. for low frequencies $\nu$, or equivalently for long wavelengths $\lambda$), the fine-grained energy levels can be well approximated by an energy continuum.

This leaves the distribution of the oscillators among the energy levels, and hence the average energy nearly unchanged. $$E_\text{avg}(T) \approx kT. \tag{2}$$ You can derive this average either directly from Boltzmann's distribution (like Rayleigh-Jeans did) or as an approximation of (1) for $h\nu \ll kT$.