# Understanding ultraviolet catastrophe

1. Apr 6, 2014

### Absentee

Hey guys. I'm struggling to picture ultraviolet catastrophe. The graph tells me that at certain temperature, intensity should exponentially increase with decrease of wavelength.
Of course, the theory is wrong, but how do i even PICTURE this with old model?

If i imagine atoms in the dense material as oscillators, and heat the material to certain temperature (lets say 500°C), at that certain temperature most of oscillators would have about the same kinetic energy, right? So even if i even manage to find some of the oscillators with very high frequency (which i guess is less likely) the overall intensity of that "light" should still be low (becouse i have only several of these high-energy electrons).

What i want to say is that, if i have a distribution of kinetic energies in the material, the highest energy electrons (whose energy is above average) produce highest frequencies but there is just few of them, so how can i have such great intensity at small wavelength?

I'm just getting into this so please don't be too mathy. I would like to get the "feel" for this.

Thanks.

2. Apr 6, 2014

### UltrafastPED

3. Apr 11, 2014

### Jano L.

The spectrum leading to UV catastrophe is not exponential. Intensity per unit wavelength interval depends on wavelength as $kT/\lambda^4$. The point is that if the energy of EM radiation behaves as assumed in the derivation of the Rayleigh-Jeans formula (that is, it is given by the Poynting formula and its distribution over EM modes obeys equipartition rule), the only way how a high - frequency mode could be in equilibrium with material oscillator of the same frequency (which is assumed to be present) is if they both have the same average energy $kT$. Since this is true for arbitrarily high frequency, we get diverging intensity as frequency increases.

4. Apr 11, 2014

### The_Duck

The point is that we can also regard *each mode of the electromagnetic field* as an oscillator. In a sufficiently large and opaque material these EM oscillators should come into thermal equilibrium with the oscillating atoms of the material. If there aren't many atoms vibrating at high frequencies, it may take a while for the high-frequency EM modes to come into thermal equilibrium but it should happen eventually. In thermal equilibrium each mode of the electromagnetic field should have about the same energy as the typical energy of a vibrating atom. But there are an infinite number of EM field modes, most of them with very very high frequencies. So this predicts that in thermal equilibrium the EM field should carry an infinite amount of energy, most of it in the form of very very high frequency light waves.

5. May 28, 2015

### EM_Guy

To understand the classical derivation of the Rayleigh-Jeans law, it seems to me that the key is grasping the whole equipartition of energy concept. According to classical physics, energy must be conserved. So, classical physics should not (in my mind) predict obtaining an infinite amount of energy from a finite amount of energy. That would be a perpetual motion machine, no? So, we have our “Rayleigh-Jeans” cube. In it, we have some dust particles of an ideal blackbody. The blackbody dust has a certain temperature, which is to say that the blackbody has a certain finite energy. The blackbody is radiating. Ordinarily, by the conservation of energy, this radiation would cause the temperature (and internal energy) of the blackbody to decrease. However, we are in a “Rayleigh-Jeans” cube, so the radiation is bouncing all over the place and is not absorbed by the walls of the cube. After a transient period of time, the blackbody and the radiation are in thermal equilibrium. Again, no energy has left the cube, and there is a finite amount of energy in the cube. So, it seems to me that the equipartition of energy would suggest that the energy radiated from the blackbody is partitioned equally over all possible modes (field configurations) supported by the cavity. There are an infinite number of discrete modes supported by the cavity, because there is always that next harmonic. Thus, the energy in each mode would be Etot/infinity, no? This looks like zero, but we would be dealing with “levels of zero” and “levels of infinity.” So that the infinite number of modes times the energy in each mode equals the total finite energy in the cube. In other words: (infinity # of modes)*(Etot/infinity) = Etot. i.e. Infinities cancel, and the conservation of energy (predicted by classical physics) holds.
But in the Rayleigh-Jeans derivation, it predicts that the energy in each mode is proportional to temperature (since we are in thermal equilibrium). Emode = kT. What I don’t quite understand is why this k is specifically Boltzmann’s constant. Also, since k is finite and positive and T is finite and positive, then there is a nonzero (non-epsilon) energy in each mode. But I just said that the total energy must be partitioned by an infinite number of modes. So Emode = Etot/(infinite # of modes).
If we say that the number of modes (which is understood to be infinite) is N, then we would say that Emode = Etot/N. Or equivalently, Etot = N*Emode = NkT. But again, I’m having a conservation of energy crisis here. N is infinite. And kT is finite. So, NkT is infinite. Classical physics should not predict a violation of the conservation of energy. How is this issue resolved?
Let’s say that we don’t know Boltzmann’s constant. Then, maybe we can say that Etot = NkT such that N is very large, and kT is very small. Or equivalently, Emode = kT.
Okay, so N is an infinite quantity, but there are levels of infinity.
As has been derived (which I think I follow)
N = 8pi*L^3*f^3/(3c^3).
As f increases to infinity, N increases to infinity. That is, if we consider the total number of frequencies up to a certain maximum frequency, the number of modes supported by the cavity (over which the finite energy in the cavity would have to be partitioned) is given by the expression above. As this maximum frequency increases to infinity, the number of modes increases to infinity.
From here, by doing some straightforward calculus, we arrive at the Rayleigh-Jeans law, which is giving us the volumetric energy density per unit frequency, or we can say the specific energy per unit frequency. This energy density per frequency increases as the square of frequency. But this is an energy density per differential frequency. du/df. The energy density per frequency explodes to infinity, but when we multiply infinity by a differential (df), we get a finite value, no? What I mean to say (I think) is that the catastrophe of the ultraviolet catastrophe is not a violation of the conservation of energy. Classical physics should not and would not predict a violation of the conservation of energy, right?
However, I think I’m still running into a problem with a finite positive value for Boltzmann’s constant. If Boltzmann’s constant could be arbitrarily small, then I think that I could affirm with more assurance that the ultraviolet catastrophe is not a violation of the conservation of energy law. However, since Boltzmann’s constant is a definite positive value, it does seem like the Rayleigh-Jeans law predicts that if you have a blackbody in a Rayleigh-Jeans cube, you end up with infinite energy – the most absurd of all perpetual motion machines (which are all absurd anyway).

Perhaps the problem is that I don't quite understand Boltzmann's constant: http://en.wikipedia.org/wiki/Boltzmann_constant

6. May 28, 2015

### Jano L.

The original problem is with the idea that each mode contributes with the same value of expected average energy. This idea is based on a formal analogy of an EM field in a cavity modes with a set of harmonic oscillators. Since the expected average energy per harmonic oscillator is $k_B T$, one would get infinite total energy for EM field, because there is infinity of modes.

Total energy is of course best thought of as finite, but then the equipartition makes no sense - whatever non-zero value chosen for energy per mode, sum of energies of all modes cannot be finite.

It is a long time since it became clear that equipartition is not a good assumption for advancing our understanding of spectrum of thermal radiation. Already Jeans, Lorentz and Planck knew this.

Search equipartition theorem and statistical physics.

For example, by discarding the idea that energy can be expressed as sum over modes. Or at least of the idea that all modes have the same energy, which is apparently absurd.

It is not the spectral density $\rho(\omega)$ that explodes to infinity, but the integral

$$u = \int_0^\infty \rho(\omega)\,d\omega$$

that should give the spatial d ensity of energy. If the Rayleigh-Jeans formula was valid for how great so ever $\omega$, energy in how small so ever cube of space would come out of this formula as infinite.

Is is too crude to say classical physics somehow necessarily leads to ultraviolet catastrophe. The Rayleigh-Jeans formula depends on severe simplifying and unrealistic assumptions. If we remove those, we are left with classical physics still, yet free of the Rayleigh-Jeans formula and the problems it manufactured.

7. May 29, 2015

### EM_Guy

First, my apologies... I don't know how to write equations with proper formatting here.

But that spectral density is proportional to the square of frequency, so, as frequency approaches infinity, the spectral density would approach infinity too. I get it that the integral explodes, but it looks to me that the spectral density explodes too.

Why is kT the expected average energy per harmonic oscillator? I see how kT would be the expected average energy of each molecule of gas in the cube, and I see can see that the cube might contain some finite number of molecules. COE seems to dictate then that for this energy to be partitioned equally over all the modes of the cavity, that finite energy would be divided equally among the modes of the cavity - which is absurd (catastrophic), because then each mode would get zero energy.

I'm still puzzling over thermal equilibrium. According to the zeroth law of thermodynamics, two bodies that have the same temperature are in thermal equilibrium with each other. In this case, if the two bodies are in contact, heat does not flow. In our case, we are talking about the need for gas molecules to be in thermal equilibrium with standing electromagnetic waves. Can we even speak of a wave or a field having a temperature? I get it that a field has energy. I suppose then that the "temperature" of the field is proportional to its energy. For the standing waves to be in thermal equilibrium with the gas though, do the temperatures need to be equal, or their energies? According to the zeroth law of thermodynamics, we say that two bodies that have the same temperature are in thermal equilibrium, but we have in mind (I think) that the internal energies of each body are proportional to their temperatures, and that the constant of proportionality if the Boltzmann constant (or some number of molecules times the Boltzmann constant - or some fraction times that - a little handwaving here). So, I think that the zeroth law really is saying that if we have two adjacent bodies that have the same internal energy (aka - the same temperature), then heat will not flow between them. The point then, I think, is that a standing wave mode must have the same "temperature" (aka energy) as its adjacent gas molecules to be in thermal equilibrium with it. And this would require (according to classical physics) the mode to have an average thermal energy of kT. Here then is the catastrophe - right? There are many more modes supported by the cavity than there are molecules of gas in the cavity, so the modes supported by the cavity can never get to a state of thermal equilibrium with the gas in the cavity.

So, by abandoning classical assumptions and embracing quantum assumptions, do we get thermal equilibrium between the gas molecules and the excited modes? I still don't think this happens. Some modes have greater energy than other modes. However, I think we can say that the total energy provided by the gas molecules equals the sum total of energy of all excited modes, yes? Thus the COE law is "saved" by quantum considerations? And yet, thermal equilibrium is still never achieved?

8. May 30, 2015

### Jano L.

That's true, the spectral function diverges to +infinity for increasing $\omega$. That already implies the integral diverges, which is wrong in physics.

This follows from the Boltzmann probability distribution $\rho(q,p)=e^{-H(q,p)/(k_BT)}/Z$ applied to an harmonic oscillator, which has

$$H(q,p) = aq^2 + bp^2.$$

The expected average energy is calculated as

$$\langle E \rangle = \int_{-\infty}^{\infty}dq \int_{-\infty}^{\infty}dp H(q,p)\rho(q,p)$$

the calculation leads to expected average $k_B T$, i.e. independent of $a,b$.

It is not necessary to think of thermodynamic equilibrium as being between material bodies and equilibrium radiation. It is possible to view it as equilibrium between material bodies (the original meaning in thermodynamics and its 0th law). The equilibrium radiation is thermal radiation with characteristic spectrum that allows the bodies to be in mutual thermodynamic equilibrium (thermal radiation with different spectrum would upset the equilibrium).

The equilibrium radiation is not standing wave. It can be described as sum of incoherent (with random phases) standing waves. But the sum itself is chaotic oscillations, it is not a standing wave.

9. Jun 1, 2015

### EM_Guy

I'm not quite following you here. We are dealing with a cavity with a blackbody dust cloud within it. For the sake of argument, can we simply assume that this dust cloud is comprised of atoms of a single element. So, we have one material body (one cloud). We are also discussing the various modes (field configurations) supported by the cavity. And we are talking about thermal equilibrium.

I'm fine with the idea that the equilibrium radiation is the sum of standing waves - and not itself a standing wave.

If the radiation and cloud are in thermal equilibrium with each other (and it seems like you are resisting me even framing this thought with your talk of it being unnecessary to view the thermal equilibrium as being between materials bodies and equilibrium radiation), then the cloud has an average internal energy of kT, no? And each mode of the radiation would have an average internal energy of kT as well - according to classical physics - right? But there are an infinite number of discrete modes that the cavity, in principle, could support. And there simply is not an infinite amount of energy that would be needed to spread this energy over each mode - with each mode taking kT.

So, Planck simply postulated that energy is discretized. And that alone resolved the problem, right? Or was there something else also wrong with the Rayleigh-Jeans formula?

10. Jun 1, 2015

### Jano L.

I prefer the term thermodynamic equilibrium - it is really about the basic assumption for validity of thermodynamics, not about heat balance. But I suppose thermal equilibrium is often used to mean the same thing.

One important thing to realize is that it is not necessary to think of the equilibrium radiation as sum of standing waves. Standing waves only occur in cavity with walls made of perfect conductor (low-resistance metal gets close for low-frequency part of the spectrum). In reality, there is field even on the metallic surface and the standing wave is not really appropriate for high frequency components of the field.

That is not a big deal, because one can use travelling waves instead. These can be chosen to form a complete basis for any regular EM field (not too close to point particles).

Also, sinusoidal waves are not the only way to describe any given radiation; there are other systems of functions that could serve as basis. In other words, the decomposition of the radiation into Fourier standing waves is by no means unique.

No, there is no reason for that. The material cloud can have any energy depending on the number of its particles, whether they interact etc. The idea of using equipartition is not that the whole energy is $k_B T$. It is that energy per one abstract harmonic oscillator defined via Fourier decomposition is to be given energy $k_B T$.

It is what I prefer in this context where thermodynamic point of view plays great role, but I understand you when you say matter interacts with radiation.

According to the Rayleigh idea of equipartition applied to abstract oscillator defined by Fourier decomposition of the EM radiation. Not according to classical physics. Classical physics is a great body of knowledge and works well without the former idea.

Planck invented several different ways to arrive at a spectral function for equilibrium radiation. He was not attempting to resolve any problem - he did not consider Rayleigh-Jeans to be really an important problem (the term "ultraviolet catastrophe" did not exist back then). He was interested in the blackbody radiation independently and sought a way to combine thermodynamics with EM theory to obtain the spectral function.

The most characteristic part of his writings on the problem of spectrum of equilibrium radiation is the idea that energy transferred between a material oscillator and an EM radiation occurs by parcels of magnitude $h\nu$, where $\nu$ is the natural frequency of the oscillator and $h$ is some constant. He derived spectral function and fit to the experimental curve lead to the value of $h \approx 6E-34$Js.

11. Jun 1, 2015

### EM_Guy

Argh... I just wrote a response and then lost it.

I'm sorry. I meant to say that the energy in the cloud is on average (within a certain probability distribution) the (number of atoms in the cloud) times kT - not just kT. kT I think is the average energy of any particular atom.

I get it that a Fourier decomposition of the radiated energy can be described as the linear combination of a number of different basis functions. Is the idea that the cloud has within it a harmonic oscillator for each one of these basis functions? Just for clarity, physically, what I have in mind is a cloud in a closed cavity; that's it. So, whatever radiation is in the cavity is originating with the cloud.

So, you are separating the Rayleigh-Jeans law from the rest of classical physics. And of course, there is nothing in classical physics (neither the Rayleigh-Jeans law nor anything else) that gives a theoretical explanation for the blackbody spectral radiation observed.