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## The ultraviolet catastrophe

In this blog I mentioned the Rayleigh-Jeans law for blackbody radiation, which predicted that the energy density (energy per unit volume) of the radiation emitted by a blackbody as a function of frequency varies as the square of the frequency. This was derived by Lord Rayleigh in 1900, and then more rigorously by Rayleigh and Sir James Jeans working together in 1905.

They found that the energy density (energy per unit volume) of the radiation coming from a blackbody varies as the square of the frequency of the radiation. This is the Rayleigh-Jeans law, and mathematically we can write this as

$\epsilon(\omega) \propto \omega^{2} \text{ or, using an alternative nomenclature, } u(\nu) \propto \nu^{2}$.

The black curve is the observed variation of energy density as a function of frequency, the purple curve is the prediction of the Rayleigh-Jeans law, leading to the so-called “ultraviolet” catastrophe”.

Lord Rayleigh

Sir James Jeans

This law led to what became known as the ultraviolet catastrophe, as it predicted that blackbodies would get brighter and brighter at higher frequencies of radiation, and that the total power radiated per unit area of the blackbody would be infinite. It was in trying to resolve this absurdity in 1900 that Max Planck came up with the idea of the quantisation of energy, which was the first step in what would later become quantum mechanics, an entirely new description of the sub-atomic world. But how was the Rayleigh-Jeans law derived? In order to properly understand what Planck did in 1900, we first of all need to properly understand what Rayleigh and Jeans derived using so-called classical physics.

## Radiation and gas in a cavity

We are going to consider, in particular, oscillating electrons which produce Electromagnetic (EM) radiation (light) (I will in a future blog go over the classical theory derived by J.J. Thomson which explains why oscillating electrons radiate EM waves).

Suppose these oscillating electrons are mixed with a very thin gas. As the electrons radiate away they will lose energy, and so as they come back into equilibrium with the gas molecules they will do so at a lower average energy, and hence a lower temperature. This is why a glowing furnace cools over time, the radiation takes energy away from the gas or solid producing the radiation.

But, if we enclose the radiation in a box with perfectly reflective walls, the radiation has nowhere to go. Hence the radiation and the gas molecules will come into thermal equilibrium with each other. This is what is meant by Blackbody radiaton, it is when the radiation is in thermal equilibrium with the object producing the radiation.

Rayleigh and Jeans used such an idealised cavity to derive their eponymous law. Here we will go through the steps of their derivation, which will lead to the law which was clearly wrong. In a future blog I will explain what modifications Planck made to the steps laid out here to produce the correct blackbody radiation law, and hence avoid the ultraviolet catastrophe.

## Writing the Electromagnetic Field in space and time

We need to satisfy the three dimensional wave equation for the Electric field in a cubic cavity, the length of each side being $L$. The electric field can be written as

$\vec{E} = E_{x} + E_{y} + E_{z}$

and each component can be written as e.g. $E_{x} = E \sin(kx - \omega t), E_{y} = E \sin(ky - \omega t)$ etc. as the EM wave varies in both space (x,y,z) and time (t). The $\omega$ in the above equation is a quantity we have come across before in a previous blog, it is the angular frequency in time of the wave, and is related to the time frequency via the equation $\omega = 2\pi \nu$ where $\nu$ is the time frequency measured in Hertz. The $k$ in the above equation is the spacial equivalent of $\omega \text{, } k$ is called the wave number and is defined as the number of wavelengths per unit wavelength. We can write that $k = \frac{2 \pi}{\lambda} \text{ where } \lambda \text{ is the wavelength of the wave }$.

The magnetic field can similarly be written as $\vec{B} = B_{x} + B_{y} + B_{z}$, but according to Maxwell’s equations $\frac{ E_{x} }{ B_{x} } = \frac{ E_{y} }{ B_{y} } = \frac{ E_{z} }{ B_{z} } = c \text{, the speed of light}$ and so the magnetic component is much less than the electrical component, and we can ignore it in this derivation.

## The 3-dimensional wave equation

You may remember from AS-level physics the relationship between wavelength, frequency and the speed of a wave, which is written at AS-level as $f \lambda = v \text{ where } f \text{ is the frequency of the wave, } \lambda$ is its wavelength and $v$ is the speed of the wave [for some reason that I have never fully understood, university level physics uses $\nu \text{ rather than } f$ to represent the frequency of a wave]. If we are talking about EM waves, as we are here, then the speed of the waves is of course $c \text{, the speed of light }$.

This equation can be derived from the so-called “wave equation”, which is a second order differential equation relating the spatial variation of the wave to its temporal (time) variation. For the electric field, if we just consider the x-dimension for now, we can write

$\frac{ d^{2}E_{x} }{ dx^{2} } = \frac{ 1 }{c^{2} } \frac{ d^{2}E_{x} }{ dt^{2} }$

we can show that this comes back to our expression $\nu \lambda =v \text{ (or }c \text{ if we are talking about EM waves)}$. Writing $E_{x} = E \sin(kx - \omega t)$ we can write for the spatial component

$\frac{ dE_{x} }{ dx } = E k \cos(kx - \omega t) \text{ and } \frac{ d^{2}E_{x} }{ dx^{2} } = -E k^{2} \sin(kx - \omega t) = -k^{2}E_{x}$.

Now looking at the temporal component, $\frac{ dE_{x} }{ dt } = -\omega E \cos(kx - \omega t) \text{ and } \frac{ d^{2} E_{x} }{ dt^{2} } = -\omega^{2} E \sin(kx - \omega t) = -\omega^{2} E_{x}$.
So, if $\frac{ d^{2}E_{x} }{ dx^{2} } = \frac{ 1 }{c^{2} } \frac{ d^{2}E_{x} }{ dt^{2} }$ we can write $-k^{2}E_{x} = \frac{ 1 }{ c^{2} } (-\omega^{2} E_{x}) \rightarrow k^{2} = \frac{ \omega^{2} }{ c^{2} }$. But, $k = \frac {2\pi }{ \lambda } \text{ and } \omega = 2 \pi \nu$ so $k^{2} = \frac{ \omega^{2} }{ c^{2} } \text{ can be written as } \frac{ 4\pi^{2} }{ \lambda^{2} } = \frac{ 4\pi^{2} \nu^{2} }{ c^{2} } \rightarrow c^{2} = \nu^{2} \lambda^{2}$ which gives us $\nu \lambda = c$, just as we wanted.

In 3-dimensions the 1-dimensional wave equation becomes

$\frac{ \partial^{2} E }{ \partial x^2 } + \frac{ \partial^{2} E }{ \partial y^{2} } + \frac{ \partial^{2} E }{ \partial z^{2} } = \frac{1}{c^2} \frac{ \partial^{2} E }{ \partial t^2}$

where we need to use the partial derivatives as we need to differentiate the x-component, the y-component and the z-component of $\vec{E} = E_{x} + E_{y} + E_{z}$.

In part 2 of this blog I will show what conditions the 3-D Wave Equation needs to satisfy because it is enclosed in our previously-mentioned cubical cavity (with each side of length $L$), and how this determines the number of modes the Electric Field can have.

Part 2 of this blog is here.