# Boltzmann distribution

1. Mar 15, 2012

### aaaa202

Consider a resevoir of N atoms in contact with a single atom. Obviously, if the atom is in a high energy state then the multiplicity left for the resevoir is significantly lower. So this is in agreement with the fact that looking at the single atom, the probability for the ground state is very high. But now suppose that there is alot of energy among the atoms. If the ground state is overwhelmingly probable what happens to all this energy in the boltzmann distribution, where does it go?

My question is probably a bit confusing, but it is just weird for me, that you get that the ground state is overwhelmingly probable for each atom in the resevoir, even if the energy density in the resevoir is enormous - because you do get that for low temperatures right?

2. Mar 15, 2012

### Khashishi

For many systems, the density of states increases with increasing energy. In this case, although the ground state is more likely than any individual excited state, there's so many excited states that most of the particles will be in some excited state. For example, consider the speed distribution of gas particles in 3 dimensions (Maxwell-Boltzmann distribution). The most likely speed is not zero, but rather sqrt(2kT/m). That's because the number of states of a given speed scales as v^2.

3. Mar 15, 2012

### Rap

It's not overwhelmingly probable, just the most probable.

4. Mar 16, 2012

### aaaa202

hmm.. my confusion is still there..
Like suppose you have A LOT of energy, more than enough energy to excite every state more than one time. If the temperature is low then there will be one billion ground state atoms for every 1 excited state atom. But that's just weird, because where would all the energy go in that case.

It's like this: Suppose you have a box of 100 atoms and want to distribute 200 energy units among them. Then I am pretty sure you will not get that the most probable microstate is the one, where most of the atoms are in the ground state. But if you look at it from one atoms point of view, it is clear that the multiplicity is highest in the case, where it is in the ground state since there is more energy left to be distributed among all the others atoms. But clearly only one of the distributions may be right. Can someone explain why they should be boltzmann distributed anyways. I hope you understand my question this time :)

5. Mar 16, 2012

### Rap

Temperature and energy are related. Temperature is proportional to the energy per atom. So you can't really say "suppose you have a lot of energy but a low temperature". If you have a lot of energy, then you can calculate the temperature, and it will be high. If you have a box of 100 atoms and want to distribute 200 energy units (say an energy unit is kT), Boltzmann says if $N_\epsilon$ is the number of particles with energy $\epsilon=0,1,2,3,...$, then

$$\frac{N_\epsilon}{N}=\frac{e^{-\epsilon/kT}}{\sum_{\epsilon=0}^\infty e^{-\epsilon/kT}}$$

The sum in the denominator is called the partition function:

$$Z=\sum_{\epsilon=0}^\infty e^{-\epsilon/kT}$$

The total energy is

$$U=\sum_{\epsilon=0}^\infty \epsilon N_\epsilon$$

If you solve these equations, using N=100 and U=200, you get $kT=1/\ln(3/2)$ or about 2.466. Now you can calculate Z:

$$Z=\sum_{i=0}^\infty e^{-\epsilon/kT}=3$$
So now you can calculate the number of atoms in the ground state ($\epsilon=0$)
$$N_0= 100\,e^{-0/kT}/Z = 100/3$$ or about 33 particles with energy 0. The number in the first energy state is ($\epsilon=1$)

$$N_1= 100\,e^{-1/kT}/Z = 200/9$$ or about 22 particles with total energy 22. The number in the second energy state is ($\epsilon=2$)

$$N_2= 100\,e^{-2/kT}/Z = 400/27$$ or about 15 particles with total energy 30 and so on.

If you lower the temperature, then the energy will not be 200 any more, it will be lower.

6. Mar 16, 2012

### Khashishi

To answer your example, we simply need to find the temperature such that the average energy per atom is 2. We can use the equipartition theorem here. Now, the answer depends on how many degrees of freedom are available to the system at these energies. I'll assume that there are 4. Then
$E=2=\frac{4}{2}kT$
$kT=1$

Then you can simply find the population distribution in each degree of freedom using Boltzmann distribution. Let us assume that one of the degrees of freedom is simply a harmonic oscillator. The ground state has energy 0, and excited states have energy 1, 2, 3 ...
$\sum{f_n E_n}=1$
We can use this constraint to find the normalization constant in front of the Boltzmann distribution $f_n=Ce^{-n/kT}$
$\sum{Ce^{-n} n}=1$
$C=\frac{(e-1)^2}{e}$

7. Mar 16, 2012

### aaaa202

Problem is as I see it, that temperature is not the same as energy, and you somehow assume that. Temperature is related to energy with the equation:

∂S/∂U = 1/T

So I don't see how you can say that energy is directly the same as temperature. A two state paramagnet can have zero energy but infinite temperature, as far as I recall. But I'm probably wrong about all this.

8. Mar 17, 2012

### Rap

This is valid, but it conceptually complicates things. The example I gave does not correspond to any physical situation (that I know of) but it illustrates the idea. The above analysis applies to an actual gas.

Well, in the example I gave, I'm not saying energy is the same as temperature. I'm just assuming we are counting energy in units of kT, and any particle can have 1,2,3... units of energy. Not a very physical situation, but the Boltzmann analysis will give an answer, nevertheless, and it shows how the energy is distributed. It's not overwhelmingly in the ground state as the OP suggested, and this is also true with a more realistic analysis.

9. Mar 17, 2012

### aaaa202

Exactly! But what do you base this assumption on? If it's the equipartition theorem then I can't really use it, since my book derives that theorem from the boltzmann factors, so that would make the whole proof circular..

10. Mar 17, 2012

### Rap

It's not an assumption, it's just a way of counting energy. You can count in Joules, ergs, or kT's. The point of my derivation was to show that you can't adjust the temperature and the total energy independently (keeping the same number of particles and the same volume). I tried to keep it as simple as possible, within your framework of 100 particles, 200 units of energy, but its not a real situation. I thought that was answering your question, which seemed to make the wrong assumption that you could set a high total energy and a low temperature at the same time.

If you want to do a real situation, then the energy isn't $\epsilon=n$ with n=0,1,2... the way I did it, its $\epsilon=n^2$ with n=1,2,3... in one dimension (for one degree of freedom). For three degrees of freedom its $\epsilon=n_1^2+n_2^2+n_3^2$ where each of the $n_i$ run from 1 to infinity. And even that is only approximate, because its wrong when the states with low $n_i$ contribute a lot. (That's a degenerate or quantum-dominated gas.) They don't contribute a lot for high temperature, so for high temperature, its a good approximation.

What Khashishi did was assume 4 degrees of freedom to simplify the math, and that the low-energy states don't contribute a lot.

11. Mar 17, 2012

### aaaa202

hmm okay, but aren't you assuming that energy and temperature are proportional?

12. Mar 17, 2012

### Rap

Yes. For an ideal monatomic gas, U= 3 N (kT/2) where U= total energy, N=total particles. For d degrees of freedom, U=d N (kT/2).

For the general situation, the "degrees of freedom" become a function of temperature, and then it may not be proportional.

13. Mar 17, 2012

### aaaa202

hmm okay, but all that is based on the equipartition theorem. And that is derived with the boltzmann factors, so the assumption is kind of circular from my point of view..

14. Mar 17, 2012

### f95toli

There are -as far as I remember- several ways to derive the equipartition theorem, and only the simplest of those uses the Boltzmann factors.

Have you had a look at the wiki for the equipartition theorem?

15. Mar 17, 2012

### Rap

Yes, it is, but I was trying to answer the original question. What really is the question?

16. Mar 17, 2012

### aaaa202

I dont know if there's an original question. Merely all my questions were just an example of my general confusion about the boltzmann distribution.
Overall I want to understand the whole thing based on my books assumptions. And my book certainly doesn't invoke the fact that energy and temperature should be proportional (they rather use the boltzmann distribution to prove that!).
They use the definition of temperature as stated in a previous post, and that's all.

17. Mar 17, 2012

### Rap

Is it confusion about how to derive the distribution, or is it confusion about what it means?

18. Mar 18, 2012

### aaaa202

How to derive it. I don't see how you can understand my example with 5 atoms and 10 atoms to be boltzmann distributed without invoking that E and T are proportional..
Overall it is just not intuitive for me. Because invoking the argument that multiplicity is largest you obviously if our specific atom is in the ground state, it is obvious that the most probable state is the ground state. But you could apply this procedure of treating one of the atoms as a single atom in contact with a resevoir to all of our 10 atoms, and then you'd get that the ground state is most probable for all.. Is that really true, when on average there are more than one energy unit per atom?

Last edited: Mar 18, 2012
19. Mar 18, 2012

### Rap

The bottom line for the Boltzmann distribution is there are many ways to distribute energy among the particles while keeping a particular value of the total energy. The Boltzmann distribution just says that when you have lots of particles and a total energy high enough, almost all of these many ways look about the same. That has nothing to do with temperature.

Boltzmann just says that the number in level i ($N_i$) is proportional to $N e^{-\beta \epsilon_i}$ where $\epsilon_i$ is the energy of the i-th level, N is the total number of particles and $\beta$ is some number that depends on the total energy. That makes no reference to temperature.

To be exact, we have to have the sum of the $N_i$ add up to $N$, so that means

$$N_i = N\, \frac{e^{-\beta \epsilon_i}}{Z(\beta)}$$

where

$$Z(\beta)=\sum_i e^{-\beta \epsilon_i}$$

We also have to have the total energy equal to some fixed value, call it $U$

$$U=\sum_i \epsilon_i N_i$$

If you know the energy of all the energy levels, the total energy, and the total number of particles, this will let you solve for the value of $\beta$.

Still no mention of temperature, only that $\beta$ is a function of the total energy, and we can solve for its value if we know $\epsilon_i,\,N$ and $U$. Now you have to get into thermodynamics, because that is where temperature is defined. It turns out, by using thermodynamics, you can show that $\beta=1/kT$.

That at least separates things into non-temperature ideas and temperature ideas. Do you need to know how Boltzmann came up with his no-temperature equation? Do you now need to know how $\beta=1/kT$?

Last edited: Mar 18, 2012
20. Mar 18, 2012

### aaaa202

I would like to know both things actually! :)