# Boltzmann distribution for particle energy

1. Mar 9, 2013

### Alfred Cann

I'm puzzled by the appearance in the literature of 2 conflicting forms:

P(E)=√(E)*exp(-E), which I understand as derived from the Maxwell distribution for speed.
It is a chi -square distribution with 3 degrees of freedom.

P(E)=exp(-E), which seems wrong to me.

But it is not simply a mistake because numerous reputable sources give each form. What is the relationship?

2. Mar 9, 2013

### Pagan Harpoon

The first expression takes account of the density of states. As you correctly pointed out, it is for three dimensions. In three dimensions, the number of possible directions for a particle's momentum increases in proportion with the square root of its energy, so its density of states has the same behaviour.

The second expression would be valid where that density of states doesn't apply. For example if you're considering excitations of a quantum oscillator, the energy levels are ħω(1/2 + n), the density of states doesn't vary with E (or maybe I should just say there's no degeneracy, since it's a discrete system), so the occupation probability is just the exponential factor on its own.

More generally, the occupation probability is:

P(E) = g(E)*e^(-E/kT)

where g is the density of states.

3. Mar 9, 2013

### Jorriss

I don't know quite what you mean but I assume the difference is the difference between summing over individual states and summing over levels.

4. Mar 9, 2013

### Pagan Harpoon

e^(-E/kT) gives you the probability to occupy a particular state that has energy of E. But in order to get the occupation probability, you also need to consider the number (or density) of states with energy E. Consider that if I tell you a particle has E = 0, then there is only one possible state... it's stationary; since there is only one state, it's pretty unlikely the particle will have E=0. However, if I tell you it has energy of 1 eV, there are lots of possible states - it could be moving in any direction, so it is more likely to be here than with E=0. As E gets larger, the number of possible states that have that energy increases (in proportion with √E) so the probability to find a particle with that energy also increases.

Edit - Oh, I mistook Joriss's post for the OP answering me. Oh well, maybe what I wrote will be useful anyway.

Last edited: Mar 9, 2013
5. Mar 9, 2013

### Jorriss

A quote on my behalf probably would of been good.

6. Mar 10, 2013

### Alfred Cann

Clarification

I'm concerned only with classical particles of a gas in a box. Its velocity space has 3 dimensions.
There is no degeneracy because energy is not quantized. Whatever energy level a particle has, it can own that; another particle whose energy is infinitesimally close to it can have its own energy level. Therefore, there are no multiple states with the same energy.
I still feel the first equation is right, as borne out by numerous websites and books.
But the second equation has equally eminent supporters.
They can't both be right.

Last edited: Mar 10, 2013
7. Mar 10, 2013

### Jorriss

Degeneracy does not require energy quantization.

8. Mar 10, 2013

### Pagan Harpoon

The number of dimensions is always 3, what I said was that the number of possible directions increases. If I tell you a particle has energy E, then you know that the magnitude of the momentum is |p| = √(2mE). You then know that the vector momentum, (px, py, pz) lies somewhere on the surface of a sphere with radius |p|. As |p| gets larger, the surface area of the sphere gets larger, so there are more possible states with that energy.

Ultimately, you want to know what is the probability for a particle to have energy E? There are two things that go into that...

1. How many possible states are there that have energy E?

As I explained in the start of this post, the number of possibilities gets larger as E gets larger, so you can expect some function that increases with E to be involed. To get the actual √E dependence, you can look at a derivation of the maxwell-boltzmann distribution for energies in a gas.

2. What is the probability to occupy each of those possible states? The probability to occupy one particular state with energy E is e^(-E/kT).

The probability to find a particle with energy E is (density of states with energy E)*(probability to occupy each individual state with energy E). So you just multiply the two factors discussed above.

9. Mar 13, 2013

### Alfred Cann

I guess I misread pagan harpoon; you were just adding more explanation to the case I already understood, with the √E included. This is the case for particles of a classical gas in a box.
What I don't understand are the references in the literature to a simple exponential. I now realize they must apply to a different case. But what case?

10. Mar 13, 2013

### Pagan Harpoon

Lots of cases. I mentioned the 1d quantum oscillator, that's one. The most general expression is:

P(E) = g(E)*e^(-E/kT)

where g is the density of states. So anywhere where the density of states doesn't vary with energy, the distribution would just be the exponential part.

Edit - I suspect that the distribution for a 2d classical gas would also just be the exponential factor. That's just from working in my head just now, though. So don't take my word for it.

Last edited: Mar 13, 2013
11. Mar 13, 2013

### Alfred Cann

I think I understand what you said; the simple exponential is for cases quite different from the molecules of a classical gas in a box.
What has me confused is instances in the literature that seem to refer to that same case, yet use the simple exponential. Two examples:

1. gallileo.phys.virginia.edu/classes/252/home.html (or just google physics 252)
Select: kinetic theory of gases>what about potential energy>last 2 par.

2. hyperphysics.phy-astr.gsu.edu/hbase/quantum/disfcn.html (or google boltzmann distr>hyperphysics)
Scroll to maxwell-boltzmann distribution>classical applications.

12. Mar 13, 2013

### Pagan Harpoon

1: Note that at the start of the section entitled "What about potential energy?", in the first sentence where that exponential factor is introduced there is an arrow over the v, indicating that it refers to one specific velocity state. The probability to find a particle in that one state of (vx, vy, vz) is indeed just the exponential factor. But this excludes other states with the same velocity magnitude in a different direction. That argument I mentioned before with the sphere doesn't apply because you have nailed down not just the magnitude of the velocity, but all three of its components.

So P(v) ~ e^(-E/kT), (note that I'm using underline to indicate a vector).

But P(|v|) ~ |v|2e^(-E/kT), because the area of a sphere of radius |v| is proportional to |v|2.

As a sort of aside, you may notice that the velocity distribution is proportional to |v|2 (and therefore E) whereas the energy distribution was proportional to √E. This is because to get from the v distribution to the E distribution, you also need to divide by dE/dv, the details of this will be discussed in a derivation of the maxwell boltzmann distribution for energy.

Later in that same paragraph, the author is again using just the e^(-E/kT) formula to talk about variation of the density of a gas with height. This works because the potential is just U = mgh, there are the same number of position states at every h, so the density of states is constant with respect to h. If, for example, you were considering a situation where a planet has an extremely thick atmosphere, then you would need to consider it because the number of available position states will scale with (h+R)2 where R is the planet's radius. In the case of the Earth, R is about 6,000 km and most of the atmosphere is below 10 km, so it's okay to ignore the effect of varying density of states (the fact that temperature goes down with altitude will be more significant anyway).

I think that the final few sentences of that section are a little misleading. The author writes:
I don't agree with this line, I think it would be more correct to say:
The way it is written suggests that it applies when considering all states with energy E, which I don't think is correct.

---

2: In the second link you posted, the author mostly discusses Boltzmann statistics in very general terms (if we are looking at the same part of it). The case of velocities of gas particles is only mentioned quite briefly, and it is explained that the origin of that extra v2 term is indeed the density of states:

Last edited: Mar 13, 2013
13. Mar 14, 2013

### Alfred Cann

I still don't have a good physical feel for what kind of situation would have the exponential distribution. Can you give me a real-world example, as close as possible to a gas in a box?

From my communications background, I know that a narrowband noise has a rayleigh distributed amplitude and exponential distributed power. The underlying voltage distribution is a bivariate gaussian because the signal can be described in the complex plane (rotating at the center frequency) by an amplitude and phase or by 2 quadrature components. This makes me think that the underlying distribution of whatever has an exponential energy distribution must also be 2-dimensional. Am I misleading myself?

14. Mar 14, 2013

### Alfred Cann

Furthermore, I think Phys 252 is not just misleading, but wrong, because now he considers only the potential (gravitational) energy and neglects the kinetic (thermal) energy, though he still includes that term in his equation.
Expanding what I wrote about the complex plane, the I and Q components are normally distributed, the amplitude is rayleigh distributed, and the phase angle is uniformly distributed.

15. Mar 14, 2013

### Jorriss

16. Mar 15, 2013

### Pagan Harpoon

I'm afraid I can't really respond to what you've said about electrical noise, because I don't understand it well enough.

In the physics 252 link, the author first considers what the distribution is with respect to height, ignoring velocity. He obtains the exponential dependence. Then, at the end he combines that result with the one for the velocity distribution. It's just the product of the two exponentials. He does not include a density of states with the exponential velocity distribution because, as I previously discussed, he uses a vector velocity rather than its magnitude. The only problem I have with it is that I think he should have made it more clear that the expression he obtained is not a distribution for energies, it is a distribution for the six-component space (x, y, z, vx, vy, vz). To go from that to an energy distribution would require the inclusion of the density of states.

We have so far encountered several cases where the exponential function on its own is the correct distribution for energies. Excitations of a quantum oscillator is one (for example, vibrations of a diatomic molecule). The probability for it to have En is proportional to e^(-En/kT), same goes for any discrete system without degeneracy. The distribution for potential energy in the potential U = mgh is another one, the probability for a particle to have potential energy of mgh is proportional to e^(-mgh/kT), note that this expression says nothing about the particle's kinetic energy.

If you want a case close to the 3d classical gas where the distribution for total energy is just the exponential factor, I believe that a 2d classical gas is one.

17. Mar 18, 2013

### Philip Wood

I've enjoyed reading this high quality discussion. The only claim which I'm not happy about is that there are lots of systems in which the probability of the system having energy E is given by the pure exponential; in other words in which the density of states is independent of energy. The only (somewhat artificial?) system that springs to mind immediately is a single harmonic oscillator. My advice to the OP would be to stop worrying about trying to find more such cases. It won't help your understanding of statistical mechanics.

The key thing, as already explained very competently by others on this thread, is that the pure exponential form gives the probability of the system being in SOME PARTICULAR STATE with energy E, whereas the exponential multiplied by a density of states function, g(E) gives you the probability of the system having energy E (i.e. being in any one of many states with energy E).