I am currently reading a derivation of Rayleigh-Jeans law for cavity radiation from Eisberg and Resnick 1 . The authors derive the law by considering a cavity with metallic walls. In the book, the authors state

Now, since electromagnetic radiation is a transverse vibration with the electric field vector $\mathbf{E}$ perpendicular to the propagation direction, and since the propagation direction for this component is perpendicular to the wall in question, its electric field vector $\mathbf{E}$ is parallel to the wall. A metallic wall cannot, however, support an electric field parallel to the surface, since charges can always flow in such a way as to neutralize the electric field. Therefore, $\mathbf{E}$ for this component must always be zero at the wall. That is, the standing wave associated with the $x$-component of the radiation must have a node (zero amplitude) at $x = 0$.

But blackbodies need not be made of metallic walls. The walls can be insulating as well which can support a non-zero electric field. Then how does Rayleigh-Jeans law hold for non-metallic blackbodies (eg. stars)?

Also, the authors consider a sinusoidal wave-function for the electric field standing wave in the cavity.

The electric field for one-dimensional electromagnetic standing waves can be described mathematically by the function $$E\,(x, t) = E_0 \sin {(2 \pi x / \lambda)} \sin {(2 \pi v t)} \tag{1-6}$$ where $\lambda$ is the wavelength of the wave, $v$ is its frequency, and $E_0$ is its maximum amplitude.

Why do we need that the time and space dependence of the electric field to be sinusoidal? Shouldn't any standing wave which has nodes at the same points suffice?

Since any wave can be described as a sum of sinusoidal components using Fourier analysis, why don't we write the electric field as such and the proceed further?


  1. Eisberg, R.; Resnick, R. Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2nd ed.; Wiley: Hoboken, NJ, 1985, pp. 6-12.

2 Answers 2


The logic is as follows.

First, one can show by a very general thermodynamic argument that, in the limit where the linear dimensions of the cavity are large compared to the wavelengths in the radiation, then in thermal equilibrium conditions the spectrum of energy density per unit frequency interval ($\rho(\omega))$ in the radiation is independent of the shape and nature of the cavity. We only require that the walls are opaque; they can be of any material and surface quality. (The reasoning is associated with Kirchhoff though I guess others contributed too).

With this prediction in our back pocket, as it were, we can now go ahead and try to calculate $\rho(\omega)$ for one specific type of cavity, such as a metallic conducting one with a simple treatment of the walls. We can be sure that the answer we get will in fact be more general!

The use of sine waves in the calculation is a form of Fourier analysis. It is saying that any distribution at all can be mathematically expressed as a sum of sine waves.

By the way, this is a nice example of the way thermodynamic reasoning plays a significant role in physics. It became fashionable during recent decades to regard thermodynamics as somehow a 'second-class-citizen,' as if all we need is quantum theory and Boltzmann's definition of entropy. This is a mistake because in fact thermodynamics offers constraints and symmetry principles which can play a powerful role in understanding many situations in thermal physics. In the present example the roles are:

  1. quantum theory of radiation in a solvable case tells what happens in that case
  2. thermodynamic reasoning says that the prediction thus obtained can also be applied more generally (in thermal equilibrium)
  • $\begingroup$ Sorry for the late reply. I also understand that in the further derivation of the Rayleigh-Jeans law, we do not really use the sinusoidal nature of the waves. We just use the fact there are nodes are particular points. But if we were to really write an equation of the electric field, shouldn't we actually represent the field as the sum of sinusoidal waves and then carry on with the further derivation? $\endgroup$ Jun 2, 2019 at 6:14
  • $\begingroup$ Also, can you suggest some books which explain the thermodynamic arguments you mentioned in your answer (and preferably are understandable at a freshman level)? $\endgroup$ Jun 2, 2019 at 6:16
  • 1
    $\begingroup$ @ApoorvPotnis I learned these points from a thermodynamics textbook by Adkins (undergraduate physics level); I have since written an undergraduate textbook at the same level (Oxford University Press) and obviously this is the book which I would recommend, since it presents the material as I think best. $\endgroup$ Jun 2, 2019 at 16:35

The problem solved is : with a metallic bound what is the spectrum of electromagnetic radiation within a cavity. The classical electromagnetic theory gives the ultraviolet divergence for this experimental setup, but the quantized black body formula fits the data. It can be derived for a metal cavity.

Then the hypothesis is extended to any body of matter, that it will have the same functional behavior, and this hypothesis was , with emissivity and absoptivity constants more or less fulfilled. An experimental fact, look at the yellow in this curve for the sun.

Here is a publication that says the geometry plays a role in the real spectra, not only absorption and emission.


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