# If the Rayleigh-Jeans Law is derived considering waves contained in a cavity, how does it give us the emitted wavelengths of a blackbody

I believe that I understand the mathematics in the derivation of the Rayleigh-Jeans law which results in the equation:

$${{du_v} \over dv } = {8πkTν^2\over c^3}$$

where u is the average energy at the frequency v, and the other symbols have the normal definitions for the Rayleigh-Jeans law.

However, this was derived considering the standing waves in a cubic cavity of sides of length L, so I am struggling to relate this to the energy density of the wavelengths emitted by a black body.

Firstly, how does this cavity represent the conditions of a black body? Is it due to the fact that a blackbody absorbs all wavelengths, hence the waves are not able to escape and for this to occur and obey the wave equation the waves must exist as standing waves?

Secondly, if we have modelled these wavelengths as being unable to escape the cavity, then how do they relate to the wavelengths emitted by a blackbody? How do these contained electromagnetic waves relate to the continuous spectra observed from black bodies.

I do know that the model fails at small wavelengths and that quantisation is required to avoid the ultraviolet catastrophe. I am just struggling to see how the equation derived relates to the physical situation at hand.

• I remember a lecturer explaining how if you open a tiny aperture in the cube, that aperture will (seen from the outside) be a blackbody, because light entering will be totally absorbed (it will not find its way out the tiny hole by reflection) and the steady-state radiation inside will be what you see if you look at the aperture. No doubt somebody else will explain it more correctly. Mar 2 at 12:49

I always find Eisberg's explanation more complete than that of other books, perhaps you can find it useful too...

In essence, the small hole can serve just a peeking mechanism into de cavity and does not disturb the conditions: thermal equilibrium and walls that reflects light perfectly for all wavelengths. Also, a light ray that enters through the hole will bounce millions of times at increasingly arbitrary angles against the walls before being able to reach the hole again, thus keeping the equilibrium condition unchanged and the blackness of the body since it absorbs anything and emits very little.

Now, light that escapes the cavity carries energy, the higher the energy density in the cavity the more energy per unit area of the hole flows out of the cavity, thus the proportionality given the constancy of $$c$$. The energy carried out by emission through the hole is not a big proportion of all energy stored and does not break the conditions of the problem much.

As for the continuity of the spectrum, you can have standing waves which are of all wavelengths; for instance, the electric field of the light wave can be studied as a vector $$(E_x, E_y, E_z)$$ with each component being a standing wave in the corresponding direction, although only some modes are allowed by the geometry, combining $$(n_x,n_y,n_z)$$ modes will give you almost any wavelength: $$\lambda^2_{tot} = \lambda^2_{x} + \lambda^2_{y} +\lambda^2_{z}, \quad \quad n_i \lambda_{i} = L.$$ Adding the imperfections in the cavity will complete the continuity.

Finally, since there is thermal equilibrium, light rays reach the hole at random thus being a random sample of what is inside.

The model is not perfect but is good approximation.

As I recall, black-body radiation is defined as what you find inside a closed cavity at equilibrium. The radiation emitted by an object under other conditions may (or may not) be a close approximation to black-body radiation.