The Rayleigh-Jeans formula, which gives the energy per unit volume (energy density) of blackbody radiation emitted inside a cavity which has been heated to a temperature T in the frequency interval \(v\) to \(v + dv\) is proportional to v squared

$$\rho_T (v)dv=\frac{8\pi v^2kT}{c^3}dv$$Hence this equation predicts that the higher the frequency of the radiation emitted, the higher its energy density should be. Whereas the experimentally observed energy density signature of blackbody radiation as a function of frequency for a given temperature does not continually grow with increasing frequency, rather it peaks and then falls back. The graph below illustrates this discrepancy, with the results from the Rayleigh-Jeans formula given by the dashed line and the solid line is the experimental results for a Temperature of 1500K

So how is the above catastrophe resolved...

Classical physics tells us that for a system with many entities which are in thermal equilibrium at temperature T, their kinetic energies will have a specific distribution, the Boltzmann distribution. For which the average kinetic energy per degree of freedom is given by 1/2kT, where k is Boltzmann's constant. Thus according to classical physics, the average kinetic energy of the waves of thermal radiation being emitted from the internal surface of a cavity, the cavity having been heated to a temperature T, is given by this equation - each wave is in thermal equilibrium with every other one and they only have one degree of freedom, the amplitude of their electric field. A common property of physical systems is that their total energy is twice their kinetic energy, thus each of the waves of the thermal radiation has the average total energy:

$$\bar\epsilon = kT$$

This equation tells us that each wave has the same average total energy, and the value is independent of frequency. If fact, this equation is an integral part of the Rayleigh-Jeans formula that gives the energy density of the thermal radiation. It is simply the product of the average total energy of each wave, as given above, multiplied by the number of waves of radiation that can fit within the cavity within the frequency interval \(v\) and \(v + dv\).

From the graph above it can be seen that for waves that have small frequencies, the prediction of the Rayleigh-Jeans formula and the experimental results are in close agreement, that is, a wave's average total energy tends to \(kT\) as its frequency tends to 0

\(\bar\varepsilon \xrightarrow [\ v \rightarrow 0]{}kT\) (1)

However, the experimental results show that the average total energy of the waves tend to zero as their frequency tends to very high values, that is

\(\bar\varepsilon \xrightarrow[\ v \rightarrow \infty]{}0\) (2)

From these two tendencies it is clear to see that the average total energy of each wave is, in fact, dependent upon its frequency, that is each wave does not have the same total energy.

To determine the actual energy of each wave it is necessary to look at how the \(\bar\varepsilon=kT\) result is determined.

$$P(\varepsilon)=\frac{e^{-\varepsilon/{kT}}}{kT}$$

The above equation characterises a special form of the Boltzmann Distribution, where \(P(\varepsilon)d\varepsilon\) is the probability of finding a body of a system with an energy within the interval \(\varepsilon\) to \(\varepsilon + d\varepsilon\), where the number of states for that body within the interval is independent of \(\varepsilon\). Using this function, the average total energy \(\varepsilon\) can be found as follows:

$$\bar\varepsilon=\int_0^\infty \varepsilon P(\varepsilon) d\varepsilon$$

As carrying out an integral is simply determining the area under a graph, it is worth considering the function \(\varepsilon P(\varepsilon)\) graphically:

It was the great scientist, Planck, who first realised that the relationship between \(\varepsilon\) and \(v\) as set out by (1) and (2) could be achieved by treating \(\varepsilon\) as a discrete variable rather than a continuous one, that is by assuming that the energies within the energy distribution of the wave can only have discrete values, that is

$$\varepsilon=0,\Delta\varepsilon,2\Delta\varepsilon,3\Delta\varepsilon,4\Delta\varepsilon...$$

The effect of treating \(\varepsilon\) as a discrete variable rather than a continuous variable can be seen for small and large values of \(\Delta\varepsilon\) in the following graphs

Where \(\Delta\varepsilon\) is small, the area under the graph (shaded) is very similar to that where \(\varepsilon\) is treated as a continuous variable, that is the area tends to \(kT\). Where \(\Delta\varepsilon\) is large the area under the graph falls below that where \(\varepsilon\) is treated as a continuous variable. It is easy to see that as \(\Delta\varepsilon\) is made very large the area under the graph tends to 0, that is

$$\bar\varepsilon \xrightarrow[\ \Delta\varepsilon \rightarrow 0]{}kT$$

$$\bar\varepsilon \xrightarrow[\ \Delta\varepsilon \rightarrow \infty]{}0$$

These two tendencies are very similar to the tendencies set out by (1) and (2), in fact they can be related by simply stating that:

$$\Delta\varepsilon \propto v$$

To convert this from a proportionality to a relationship a constant of proportionality, \(h\), is required. This constant is the well known Planck's constant.

Thus the Ultraviolet catastrophe is resolved by assuming that waves of blackbody radiation can only have discrete values of energy within their energy distribution, and that the discrete values of energy are a dependent on the wave's frequency. These concepts were used by Planck to derive his blackbody spectrum formula

$$\rho_T (v)dv=\frac{8\pi v^2kT}{c^3} \frac{hv}{e^{hv/kT}-1}$$

The derivation of this formula will be the subject of a separate post.