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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.

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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:

energy levels with percentages
(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.

enter image description here

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 $h\nu \ll kT$.

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  • $\begingroup$ Sir, exactly what I was looking for. Thankyou. Just one thing I have to ask, in your answer to the related question you mentioned; you said that the distribution of the oscillators in the energy levels would be governed by MB statistics. But if this is quantum mechanics, shouldn't we use BE statistics? $\endgroup$
    – user285848
    May 27, 2023 at 14:07
  • $\begingroup$ @Shikhar To get the probability that the oscillator of frequency $\nu$ has energy $E_n=nh\nu$) you use the Boltzmann distribution: $p_n\propto \exp(-E_n/kT)$. See The Derivation of the Planck Formula (page 9). This is conceptually different from asking how many photons have an energy between $E$ and $E+dE$. There you use the Bose-Einstein distribution. $\endgroup$ May 27, 2023 at 15:03

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