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## Blackbody radiation and the “ultraviolet catastrophe”

In Physics, we refer to an object which radiates perfectly as a blackbody. The name is a little misleading, as there is nothing about a blackbody which is black in colour. An example of a blackbody is a heated cannon ball, or the surface of stars like the Sun. It has been known since time immemorial that if you heat a metal object, such as a cannon ball, it will start to glow; it actually starts giving off its own visible light rather than just reflecting light. We now know that, even at room temperature, all objects are giving off their own light, but it is infrared light, a type of light to which are eyes are not sensitive and is thus invisible unless we use a thermal imaging camera.

## Wien’s Displacement Law

When an object glows, we can analyse which parts of the spectrum are giving off how much energy. To put it another way, how bright the object is at different wavelengths. When we do this for any object which is radiating like a blackbody we find that the spectra all have the same shape, independent of the temperature of the object or its composition.

The figure below shows the spectra for three different blackbodies. The only difference between the three is their temperature. One is at a temperature of 3,000K, the second one is at 4,000K and the third one is at 5,000K. These three curves explain why an object does not give off visible light when it is at low temperature. 3,000K is pretty hot (nearly 2,700 C), but an object at this temperature is barely giving off any visible light, the peak of the curve is in the infrared. If we came down to room temperature, about 300K, there would be no radiation short-wards of 700nm, so no visible light is to be seen from a room temperature object. These three curves also explain why an object changes colour as we heat it, it goes from red-hot to white-hot to blue-hot. It is why stars have different colours, the cool ones appear red, the hotter ones white and the hottest of all appear blue.

Blackbody spectra for three different temperatures, 3000K (red), 4000K (green) and 5000K (blue). Notice the wavelength of the peak of each curves moves as the temperature changes. This is Wien’s displacement law. The x-axis is in nanometres, on this scale the visible part of the spectrum is from about 400 to 700 nm.

As you can see from the figure, the size of the curve changes as we change the temperature, and also the wavelength of the peak changes, moving to shorter wavelengths (to the left) as we increase the temperature. The scale on the x-axis is nanometres, so on this scale the kind of wavelengths we can see with our eyes (the visible part of the spectrum) is from about 400 to 700 nm.

In 1893 the German physicist Wilhelm Wien came up with a mathematical formula which allowed one to calculate the wavelength of this peak if one knew the temperature. This equation is known as Wien’s displacement law. It is an empirical law, in the sense that Wien had no physical explanation for why the equation worked, but it did. The physics behind why Wien’s law works did not come until seven years later in 1900 with the work of Max Planck.

We can state Wien’s displacement law as

$\lambda \text{ (in metres) } = \frac{2.898 \times 10^{-3} }{T}$

where $T$ is the temperature in Kelvin. Another law, called the Steffan-Boltzmann law, gives the total amount of power per unit area (Watts per metre squared) that is being emitted at all wavelengths by the blackbody (mathematically this is the same as the area under these curves).

$P = \sigma T^{4}$

where $\sigma$ is known as the Steffan-Boltzmann constant, and once again $T$ is the temperature of the object in Kelvin.

## The “ultraviolet catastrophe”

To understand the physics of blackbody spectra, physicists imagined the electromagnetic waves bouncing around inside a closed box, what is often referred to as a cavity. The cavity has a small hole in it, which allows the radiation to escape, this is the radiation we observe. We can see from the curves above that the spectrum (either expressed against wavelength $\lambda$ or against frequency $\nu$ (one can go from wavelength to frequency and back by using the wave equation. For electromagnetic radiation, this can be written

$c = \lambda \ \nu$
where $c$ is the speed of light ($3 \times 10^{8} \text{ m/s}$).

The electromagnetic (EM) radiation is produced by an EM field in the cavity. But, there are EM waves with different frequencies. The ensuing EM field is due to the superposition (addition) of all the possible EM waves in the cavity. We can think of the EM waves like pieces of string between two fixed points (the sides of the cavity). So we can draw an analogy to the musical notes that e.g. a guitar string can produce. With a piece of string, only certain standing waves can exist between two fixed points, you cannot get vibrations of any wavelength. Let us suppose the fixed points are a distance $L$ apart. As the vibrations need to be zero at the fixed ends, the only wavelengths which can be produced by such a string have to have wavelengths $\lambda$ so that $L = \frac{n\lambda}{2}$ where $n=1,2,3,4....... \text{ etc }$.

The same was believed to be true for EM waves inside a cavity. The EM waves had to have a complete number of half-cycles within the box. Each of these allowed vibrations of the EM field is called a mode.

The number of modes of a particular frequency $\omega$ in the frequency interval $\omega$ to $\omega + d\omega$ (where $d\omega$ is a small increment) is found to be

$N(\omega)d\omega \propto \omega^{2}d\omega$

(this comes about because we can think of the vibrations of the EM field like a harmonic oscillator).
In classical physics, in the 19th century, a theory called the equipartition theorem had been developed, and this stated that each mode would have an energy of $E = \frac{1}{2}kT$ where $k$ is Boltzmann’s constant. This equipartition theorem had allowed physicists to work out the kinetic energy of molecules of a gas at any given temperature. This was an attempt to apply this equipartition law to EM radiation, which were thought of as waves.

If the energy density (which is the same as the power per unit area) of each mode is $\frac{1}{2}kT$, the energy $\epsilon$ as a function of frequency in a small frequency interval from $\omega \text{ to } \omega+d\omega$ would be given by
$\epsilon(\omega) \propto \omega^{2} kT d\omega$

This is the so-called Rayleigh-Jeans law, and notice that is says that the energy density in a frequency interval is proportional to the square of the frequency. If we plot the energy density at a particular frequency $\epsilon (\omega)$ as a function of frequency $\omega$ we get a graph which looks like this

The so-called “ultraviolet catastrophe”. The black curve shows the observed blackbody curve, the red curve shows the prediction of the Rayleigh-Jeans law.

The black curve in this plot shows the observed spectrum of a blackbody (note, this plot is against frequency not wavelength, so it is backwards from the figure above with the three different spectra. In this plot against frequency, frequency increases to the right, which means we are moving to shorter wavelength).

Now look at the red curve, it is going off to infinity!!! That is to say, the classical theory (the Rayleigh-Jeans law) predicts that there will be an infinite number of modes allowed in the cavity as we go to higher and higher frequency. This means the curve will go off up to infinity as shown, because the power per unit area at each frequency $\epsilon(\omega)$ is proportional to $\omega^{2}$, so more and more energy exists in each frequency interval as we go to higher and higher frequencies (shorter and shorter wavelengths). The total area under the curve, the power per unit area, of the blackbody would be infinite and it would be infinitely bright at ultraviolet wavelengths (high frequencies)!!!!! This clearly doesn’t happen, the theory had completely broken down.

This disagreement of the observed behaviour with the prediction of the Rayleigh-Jeans law came to be known as the ultraviolet catastrophe. It showed that there was something deeply flawed in our understanding of physics. Max Planck would solve the problem in 1900, and usher in a complete overthrowing of our understanding of physics, bringing in the age of quantum physics and leading to the demise of classical physics.

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.

### 12 Responses

1. Curiously, I have just been writing about this in a chapter of my latest book. I will now shamelessly steal anything I missed from you!

• Only the good bits I hope 😛

2. […] 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. […]

3. […] These differences in colour are due to stars having different surface temperatures. As the figure below shows, the blackbody curve peaks at different wavelengths depending on a star’s temperature, so a red star is cooler than a white star, which in turn is cooler than a blue star. This is because of Wien’s displacement law, which I discussed in this blog. […]

4. […] very important fact in its interpretation as being due to radiation from the hot, early Universe), here I blogged about blackbody radiation and the ultraviolet catastrophe, and here I showed how we can use the fact that stars radiate as blackbodies to determine their […]

5. […] a blackbody will also radiate at other wavelengths(see my blog here to remind yourself of the shape of a blackbody curve), so such accretion disks will also radiate visible light, infrared light, and even radio emission. […]

6. […] RhEvans (21 de Octubre de 2013) Re: Blackbody radiation and the “ultraviolet catastrophe” [Mensaje de un blog]. Recuperado de https://thecuriousastronomer.wordpress.com/2013/10/21/blackbody-radiation-and-the-ultraviolet-catast&#8230; […]

7. on 08/12/2017 at 20:04 | Reply tibebu solomon

thank u. gained great knowledge , so I rejoice so much and I hope that my joy would be ur satisfication.

8. on 13/12/2017 at 19:01 | Reply Dr sudhakarreddy

Wonderful explanation
Even a non physics person like me could understand

9. […] المصدر الثاني […]