Is it possible to prove the Blackbody radiation law using the fluctuation-dissipation theorem?

Has it been done, or is there some reason why it wouldn't work? I would appreciate if you could point me towards some resources.

I know that Boltzmann first derived it on purely thermodynamic grounds (although that didn't give him the constant of proportionality in $u = a T^4$). That can be obtained with the following derivation, based on the quantisation of energy:

  1. Start from the Boltzmann probability:

$$p(n) = \frac{ \displaystyle e^{-E_n / k_B T}}{ \displaystyle \sum_{n=0}^{\infty}{e^{-E_n / k_B T}}} \, , $$

where $E_n = n \hbar \omega $

  1. Mean energy per mode:

$$\overline{E} = \sum_{n=0}^{\infty}{E_n \, p(n)} = \frac{\displaystyle \sum_{n=0}^{\infty} E_n \, \displaystyle e^{-E_n / k_B T}}{ \displaystyle \sum_{n=0}^{\infty}{e^{-E_n / k_B T}}} \, .$$

Using summation rules, we find:

$$\overline{E} = \frac{\hbar \omega}{e^{\hbar \omega / k_B T} -1} \, .$$

  1. Number of modes per frequency interval is:

$$n(\omega) d \omega = \frac{\omega^2 k_B T}{\pi^2 c^3} d \omega \, .$$

  1. Hence, the energy density of radiation is:

$$u(\omega) d\omega = \frac{\omega^2 k_B T}{\pi^2 c^3} \frac{\hbar \omega}{e^{\hbar \omega / k_B T} -1} d \omega \, , $$

which is the Planck distribution function.

  1. Integrate to obtain the Stefan-Boltzmann Law:

$$u = a\, T^4$$ .

  • $\begingroup$ FD theorem relates spectral intensity to response function of some system. So maybe we could derive the black body radiation spectrum if we knew some response function of the black body or black body radiation to some perturbation. But this seems like deriving a simple known thing from a complicated unknown thing. $\endgroup$ Mar 28, 2022 at 18:52
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    $\begingroup$ There is indeed a derivation of the law from fluctuation-dissipation theorem search for Near-field radiative heat transfer and Rytov theory $\endgroup$
    – Mauricio
    Mar 29, 2022 at 10:11
  • $\begingroup$ @Mauricio Thanks for your reply! I still can't seem to find the proof however, can you link a paper? (if there is any). $\endgroup$ Mar 29, 2022 at 12:38
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    $\begingroup$ I suggest checking this derivation - it may be a bit high-ended, but in reality the two are rather straightforward statements about the product of two operators averaged over the Boltzmann distribution. It is worth adding the Lehmann representation in the mix. $\endgroup$
    – Roger V.
    Mar 29, 2022 at 15:15

1 Answer 1


The calculation is rather technical and it is usually only carried out when you want to involve the response function (dielectric function) of the material or work at short distances (Wikipedia: near field radiative heat transfer).

For a paper you can check D.Polder and M. Van Hove, Phys.Rev.B 1971 (doi:10.1103/PhysRevB.4.3303), based on Rytov calculations. You can also find a version of it in L.Novotny and B.Hetch "Principles of Nano-Optics" Cambridge.

The idea is the following:

  • Currents (or dipoles) in the material have a quantum fluctuation–dissipation relation, something of the form

$$\langle j_\alpha(x,t) [j'_\beta(x',t')]^*\rangle\propto\operatorname{Im}(\epsilon)\frac{\hbar\omega^2}{\exp(\hbar\omega/k_{\rm B}T)-1}\delta(x-x')\delta(\omega-\omega')\delta_{\alpha\beta}$$ where $\alpha,\beta=x,y,z$, $\vec{j}$ are the currents, $\epsilon$ is the dielectric function, $\omega$ is frequency and the rest is the usual physical constants.

  • Relate the currents to the electric and magnetic fields by using dyadic Green functions. Example for the electric field:

$$ E_\alpha=\int \mathrm{d}^3r\int \mathrm{d}\omega' \; G_{\alpha\beta}j_\beta(\mathbf r,\mathbf r ',\omega')j_\beta(\mathbf{r}',\omega') $$

where $G_{\alpha,\beta}$ is the dyadic Green tensor for the electric field.

  • Calculate the Poyting vector, $\langle S\rangle =\langle \vec E \times \vec H\rangle $ where $\vec{H}$ is the magnetic field strenght. Relate each to the currents and use the fluctuation-dissipation relation. Take into account the symmetries of the problem

  • For the case of two semi-infinite spaces separated by a distance $d$ (for the final formulas, see Wikipedia article on NFRHT), you find that if $\epsilon\to 1$ and $d\to\infty$, you recover that the flux is $\sigma (T^4_1-T^4_2)$, where $T_1,T_2$ are the the temperatures of each semi-space.

Interestingly for some materials this formalism can also recover the emissivity of gray bodies and more complicated responses.


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