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

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.

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 $$h\nu \ll kT$$.

• 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?
– user285848
May 27, 2023 at 14:07
• @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. May 27, 2023 at 15:03