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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”.

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 3-D electric field is a function of space $(x,y,z)$ and of time $t$, so it can be written as $\vec{E}(x,y,z,t)$

To make things simpler we will start off by considering a 1-dimensional travelling wave in e.g. the x-direction. For this we can write $\vec{E}(x,t) = E \sin (kx - \omega t)$

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 $\vec{B}(x,t) = B \sin (kx - \omega t)$, but according to Maxwell’s equations $\frac{ E }{ B } = c \text{, the speed of light}$ 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, the z-component and the time-component of $\vec{E}(x,y,z,t)$.

In part 2 of this blog I will show what form the 3-D equation of a travelling wave has, and 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.

### 11 Responses

1. […] Here is the first part of a series of blogs where I will derive the Rayleigh-Jeans law from first principles, after which I will derive the Planck law. […]

2. […] In part 1 of this blog I showed that the 1-dimensional (let’s say the x-direction) wave equation for the electric field can be generalised to the 3-dimensional wave equation […]

3. […] part 1 of this blog, I showed that the 3-dimensional wave equation for an electromagnetic (EM) wave can be written […]

4. on 18/12/2013 at 20:28 | Reply Jim Rogers

Have you written, or will you write, or is it already written.what prompted Plank’s course of investigation that led to the assumption that the energy of each ocsillator does not variy continuously, but in steps. Either he was just fooling around or he had a monumental insight.

• on 20/07/2017 at 16:47 | Reply RhEvans

It’s written. He was fooling around, and found that his assumption of harmonic oscillators fitted the observed BB curve.

5. […] the series I did on the derivation of the Rayleigh-Jeans law, which I posted in three parts (part 1 here, part 2 here and part 3 here). I have had many thousands of hits on this series, but several people […]

6. on 23/03/2017 at 13:19 | Reply ADOLPH MACHOGU

EXCELLENT WORK

7. on 05/06/2017 at 11:43 | Reply noumenon0

Is this the derivation of Rayleigh/Jeans’s paper of 1905? If so, this is after Planck’s mathematical solution of 1900. So, did Planck make use of a forced harmonic oscillator derivation with radiation damping term, or did he make use of the above ‘mode-counting’ derivation (?), ….. (which may have had the advantage of retrospective considerations). Thank you for you website.

• on 05/06/2017 at 14:45 | Reply RhEvans

I have done a separate derivation of Planck’s formula, following the method he used (there are other ways of doing it too). Yes, he imagined massless forced harmonic oscillators in his derivation.

• on 05/06/2017 at 15:46 noumenon0

Okay, thank you, I will read it.

8. on 02/01/2018 at 08:12 | Reply Jason

Thank you very much for this careful derivation, it’s the most careful I have found online.

I have a question about the very first step though. What is the motivation for the assumption of a cubical cavity? The factor of 8π that appears in the final expression seems to depend on this choice of shape. If the choice of a spherical cavity were made, wouldn’t you get something different? I think so, because in that case satisfying the boundary conditions leads to transcendental equations, and not a simple, integrable relation like what you obtain at the end of “part 2.” It seems likely that the generic solution du/dν ~ ν^2 would still apply, but that factor of 8π seems important, as it is related to the value of the Stefan-Boltzmann constant.

I got on this track because I’m interested in the connection between the ideal cavity and a real thermal body. Take the sun or a hot coal for instance, which to a very good approximation emit black body radiation. Where do I put a cavity, cubical or any other shape, to compute how the black body radiation arises? I considered drawing some imaginary volume within some small region near a point on the surface (the one whose emission I’m trying to evaluate), but then why would the boundary condition E=0 have to apply? And it doesn’t seem to make sense to consider the entire volume, since a lump of coal or the sun wouldn’t make a very good “cavity.”

I tried to come up with a different argument that avoids a cavity altogether, I’m interested in your feedback. Consider a generic, small volume near the surface of the body. Being a black body, that patch contains a very large number of oscillators that can emit in any frequency range (or wavelength). Nearby oscillators are independent from each other if they emit at different frequencies (ν), in a different direction, in a different polarization, or if they are separated from each other by more than a wavelength. Let σ (a vector) be the linear wavenumber, i.e. c*|σ| = ν. Then you have 2 * 4πσ^2 d|σ| independent oscillators per unit volume (the first factor of 2 is for the polarization). Thermal equilibrium between that patch and the rest of the black body means that the energy in each mode is kT. Switching from |σ| to ν, this gives the energy per unit volume directly as du = (8πkT/c^3) ν^2 dν. Derivation of Planck’s Law immediately follows as before by replacing the average energy per mode of kT with hν / [e^(hν/kT) – 1], as derivable from the assumption that the probability of emitting n photons of frequency ν is proportional to e^(-nhν/kT).