What was the ultraviolet catastrophe

1. Feb 11, 2012

CrimpJiggler

I'm trying to get my head around exactly what this ultraviolet catastrophe is. So the early 20th century scientists were studying black body emission spectra (at constant temperatures) and plotted the intensity of the emitted light, against the frequency/wavelength. From what I've read, classical physics theory predicts that the intensity should continue to increase as the frequency of the light increases. Thats the concept I'm stuck on. So heres a black body emission spectrum:

The black line is what classical theory predicts, the other lines are what really happens. Sorry if this is a difficult question to answer but why does classical theory predict that the emitted light should rise in intensity like that?

2. Feb 11, 2012

Staff: Mentor

What you're looking for is a derivation of the Rayleigh-Jeans law (the black line on your graph). Google may be of some help here.

3. Feb 11, 2012

Ken G

The classical theory just counts the number of possible "modes" in each frequency interval in your graph, and treats each mode like it was a simple harmonic oscillator at that resonant frequency. Classical thermodynamics says you'll get kT of energy in each mode, so the energy in the spectrum amounts to simply counting how many modes there are. Hence, the "UV catastrophe" comes from the fact that there are an infinite number of modes as you go to higher and higher frequency (this is most easily seen if you use frequency rather than wavelength as your independent variable, because there are infinitely many modes as you go to infinitely high frequencies).

The resolution, of course, is that those infinitely many modes don't have kT of energy in them, because you can't partially excite a mode-- you need a full "quantum" of energy, proportional to the frequency, and when the quantum is much more than KT, it's so unlikely to get the full quantum that you don't end up with kT expectation value in the high frequency modes.

4. Feb 11, 2012

gulfcoastfella

jtbell,

I've seen you make a lot of really helpful posts on Physics Forums. Thanks for taking the time to guide us in the right direction.

5. Feb 11, 2012

CrimpJiggler

Whats really confusing me is why they postulated that the number of modes increases as the frequency of the radiation increases. Was the theory that the higher frequency radiation is emitted by a larger combination of the substances degrees of freedom and thus, should also be of greater intensity since there are more modes involved? For example, lets say a single rotational degree of freedom causes the emission of infrared while emitting ultraviolet requires a combination of the substances rotational, vibrational and translational degrees of freedom. Is that the theory?

6. Feb 11, 2012

Jano L.

The number of modes increases with frequency because the frequency is due to Maxwell's equations determined by the type of the mode of the electromagnetic field. Any type of mode of oscillation is defined by a triple of integers $(n_1, n_2, n_3)$ and the frequency of such mode is $\omega = \frac{2\pi c}{L} \sqrt{n_1^2+n_2^2+n_3^2}$. You can picture these triples as points in one octant of the xyz coordinate system in space, $n_1, n_2, n_3$ used as xyz coordinates. Now imagine a thin spherical shell that contains some of the points. The radius of the shell is proportional to $\omega$, so the larger frequency of a mode, the larger is the radius of the shell. Now calculate the number of points inside this spherical shell. If the shell is small (the frequency is low), there are just a few points inside. But if the shell is large (high frequency), the number of points inside it is great (proportional to its surface, or $\omega$.

In the Rayleigh - Jeans calculation, they attempted to find a thermal equilibrium of the field by applying equipartition formula to modes of the electromagnetic field only. The absurd results can be interpreted in various ways.

Rayleigh thought the law of equipartition cannot hold for ether. Jeans concluded that common thermal radiation is not in equilibrium (he thought it would take immensely long for the molecules to get equilibrated with the ether).

7. Feb 11, 2012

CrimpJiggler

Thats an interesting visual analogy you used. I get it now but theres one thing really boggling my mind now. I came across this explanation:
and am trying to make sense of it. How can a travelling, transverse electromagnetic wave be thought of as a standing wave? I was also getting the concept of modes mixed up. I was thinking of translational, rotational and vibrational degrees of freedom but I've since discovered that the word modes, when talking about standing waves, actually refers to this:

which makes a lot more sense. With that in mind, I can see how your octant point encompassing sphere analogy applies because I can see how the higher modes oscillate at higher frequencies. Do they mean you can consider the source of the electromagnetic waves to be a standing wave? In that case I can see how the modes of the oscillator would influence the intensity of the light produced, otherwise I'm completely lost.

8. Feb 11, 2012

Jano L.

Actually the idea of modes goes like this: the electromagnetic field is described by electric and magnetic field. These are functions of spatial position and time $\mathbf E(\mathbf x, t), \mathbf B(\mathbf x, t)$. These functions are resolved into their Fourier components, for example the electric field is written as sum

$$E_x = \sum_{n_1,n_2,n_3} E(n_1, n_2, n_3) \cos \frac{\pi}{L} n_1 x \sin \frac{\pi}{L} n_2 y \sin \frac{\pi}{L} n_3 z \cos \omega t.$$

The number $E(n_1, n_2, n_3)$ is the amplitude (intensity) of oscillation of the mode $n_1, n_2, n_3$. The thermal electromagnetic radiation is supposed to be irregular, chaotic, which leads to many non-zero amplitudes. Thus many modes are vibrating at the same time.

Based on Poynting's theorem, each mode is ascribed energy proportional to the square of its amplitude. Sum the energies of all the modes with similar frequency and the sum is what is usually meant as the intensity of radiation at this frequency.

9. Feb 11, 2012

Ken G

It all comes down to the "superposition principle." This means that all the different amplitudes in the electromagnetic fields can simply be added together to get the total, they don't interact with each other in any way. Hence you are free to break them up (above was mentioned "Fourier analyze") any way you like, sort of like if you were counting beans, you could first count the larger beans and then the smaller, or the red beans and then the white. When we refer to modes as "standing waves", it means we are choosing to count together the pairs of oppositely propagating waves at the same frequency. When you pair them up like that, they become standing waves, at least in a statistical average-- it doesn't matter if the actual fields have nodes like that (they won't-- the presence of all those frequencies, and randomly varying strengths in opposite directions, means that no points will have zero field for any length of time). It's just one way to do the counting of the modes.

10. Aug 6, 2012

vijayan_t

the intensity is decreasing for higher frequency. see this explanation

Last edited by a moderator: Sep 25, 2014
11. Aug 7, 2012

USeptim

Let the energy-momentum vector be [E, px, py, pz] where E is the total energy and px, py and pz are the components of the momentum.

For photons this tensor takes the form E [1, k$_{x}$, k$_{y}$, k$_{z}$ ] where kx, ky and kz are de direction of propagation of the phonton and their square must be one: (kx^2 + ky^2 + kz^2)^(1/2) = 1.

Here the phase space would be formed by the variables kx, ky and kz that depend on E. Given an energy E the values that can take the phase are represented by a spherical shell with radius E/c.

The density of states is the "volume" of the phase modes that can be chosen with an energy between E and E+dE, when dE tends to zero, this is simply 4πE^2*dE/c^3, so the density of states grows with E^2 and therefore with w^2, the photon's frequency.