# Reasoning in Deriving the Ultraviolet Catastrophe

1. Sep 12, 2014

### center o bass

When one derives the ultraviolate catasrophe one looks at a cavity of perfectly conducting walls and with standing electromagnetic waves. Since these waves must vanish at the walls of the cavity, the wavelengths of the radiation is restricted according to the dimensions of the cavity: for example for a cubic cavity of length $L$ we must have
$$n_i \frac{\lambda_i}{2} = L; i = 1,2,3;$$
where $n_i$ must be positive integers.
In other words, the walls of the container makes each mode of the electromagnetic waves be defined by a vector of positive integer components: $\vec n = n_1 \vec e_1 + n_2 \vec e_2 + n_3 \vec e_3$.

The way one now proceeds is to find how many modes there are in the volume of an octant shell of witdth $dn$ in $\vec n$-space, which expressed in terms of frequency is given by
$$dN(\nu) = \frac{4 \pi V \nu^2 }{ c^3 },$$
where $V=L^3$ which by associating $1/2 k T$ of energy to each degree of freedom, gives an energy density per unit frequency of
$$u(\nu) = \frac{8 \pi \nu^2}{c^3}.$$

The Ultraviolet catastrophe now arises by going to the total energy density in the cavity given by
$$\int_0^\infty \frac{8 \pi \nu^2}{c^3} d\nu = \infty.$$

The quantum mechanical calculation saves us from this result by the fact that it is not equally probable for each mode to exist: in fact the probability of a mode of frequency $\nu$ is given by
$$\frac{1}{e^{h \nu/kT} -1 }.$$

Now here comes the question: in the QM calculation the modes are not assumed to be equally populated but are cut off by the factor $\frac{1}{e^{h \nu/kT} -1 }$ at high frequencies. What is the reason for assuming equally populated modes in the classical calculation?

2. Sep 12, 2014

### Staff: Mentor

3. Sep 12, 2014

### The_Duck

As jtbell says, the equipartition theorem shows that (classically) thermal equilibrium can only be reached when every mode has an equal average energy. Therefore another way of stating the ultraviolet catastrophe is that according to classical mechanics a cavity with finite energy can never reach thermal equilbrium; rather, thermal interactions should distribute the electromagnetic energy in the cavity to higher and higher frequencies over time.

4. Sep 12, 2014

### quo

Probably none, esspecialy the infinite frequency is rather quite nonsense, thus unavoidably leads to the catastrophe.

5. Sep 13, 2014

### center o bass

Are you then counting every sinusoidal frequency as an individual degree of freedom?

6. Sep 13, 2014

### center o bass

But even though every mode has an equal average energy, it does not follow that every mode exist within the cavity. Why might it not be more likely, as in the QM calculation, that a mode of higher frequency is not as common as the lower frequency modes?