Boltzmann Distribution: Exploring Energy in High Density Reservoirs

[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 a lot 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?

 

[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.

 

[Rap]

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?

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

 

[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 :)

 

[Rap]

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 :)

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.

 

[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 $f_n=Ce^{-n/kT}$
$\sum{Ce^{-n} n}=1$
$C=\frac{(e-1)^2}{e}$

 

[aaaa202]

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.

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.

 

[Rap]

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 $f_n=Ce^{-n/kT}$
$\sum{Ce^{-n} n}=1$
$C=\frac{(e-1)^2}{e}$

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.

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.

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.

 

[aaaa202]

I'm just assuming we are counting energy in units of kT, and any particle can have 1,2,3... units of energy.

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..

 

[Rap]

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..

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.

 

[aaaa202]

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

 

[Rap]

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

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.

 

[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..

 

[f95toli]

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..

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?

 

[Rap]

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..

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

 

[aaaa202]

I don't 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.

 

[Rap]

I don't 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.

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

 

[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?

 

[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 $\beta=1/kT$?

 

[aaaa202]

Do you need to know how Boltzmann came up with his no-temperature equation? Do you now need to know how $\beta=1/kT$?

I would like to know both things actually! :)

 

[Rap]

Check out the Wikipedia article at http://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics

Basically, you have the number of particles N and a bunch of energy levels with energy, say 0,1,2,3... for example. Then you have the total energy E. Now you want to know how many ways you can fill those energy levels with N particles to get that total energy. Suppose you have 3 particles and the total energy is 4. There are 4 ways to do this:

0 1 2 3 4 = energy of level
---------
2 0 0 0 1 = number of particles in each level
1 1 0 1 0 "
1 0 2 0 0 "
0 2 1 0 0 "

4 3 3 1 1 = 4 times the average number in each level

You can see the distribution is high at level 0, dropping off for higher energy levels. Boltzmann (and Gibbs) carried out this analysis for N particles with total energy E, and figured out that if you have $N_i$ particles in energy level i with energy $\epsilon_i$ and total energy E, the number of ways ($W$) this could be done is:

$$W=\prod_i \frac{1}{N_i!}$$

for large N. Now we want to find the $N_i$ such that the sum of all the $N_i$ equals N and the sum of all the $\epsilon_i N_i$ equals E. In the table above, it was done for N=3 and E=4, now we want to do it for the general case. Since N is large, we can use Stirlings approximation for the factorial $x!\approx x^xe^{-x}$. Its better to work in the log of W, so we can do sums instead of products.

$$\ln(W)=\sum_i N_i-N_i\ln N_i$$

Now we want to find the $N_i$ where W is maximum. It turns out that that maximum is a HUGE maximum. The set of $N_i$ that gives the largest number of ways gives a number of ways that is MUCH larger than any other configuration. The way we find this maximum is to form a function:

$$f=\sum_i N_i-N_i\ln N_i +(N-\alpha \sum_iN_i) +(E-\sum_i\beta \epsilon_i N_i)$$

You can see that if you find the maximum of this function, it will give you the W you are looking for, that also gives you the right total N and E. So now take the derivative and set to zero:

$$\frac{d \ln(W)}{dN_i}=0=\sum_i -\ln N_i-\alpha-\beta \epsilon_i$$

or

$$N_i=e^{-\alpha-\beta\epsilon_i}$$

and that's Boltzmann's equation. Now you solve for $\alpha$ using the fact that $\sum_i N_i=N$ to get

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

and you can solve for $\beta$ knowing E. PLEASE NOTE - there is a lot more things that go into the derivation. The above derivation leaves a lot out, but if you can follow the above derivation, then you will be very ready for the real derivation.

As far a showing that $\beta=1/kT$, you have to use Boltzmann's famous equation for entropy $S=k\ln(W)$. Using the $N_i$ that you found above and substituting it into the expression for $\ln W$ you get

$$S/k=N\alpha+\beta E$$

differentiating and rearranging, you get

$$dE=\frac{1}{k\beta}\,dS-\frac{\alpha}{\beta}\,dN$$

which is just the fundamental equation of thermodynamics at constant volume:

$$dE=T\,dS+\mu dN$$

which shows that $T=1/k\beta$ and $\mu=-\alpha/\beta$ is the chemical potential.

 

[aaaa202]

neat! had actually looked up the derivation on wikipedia, where they use lagrangian multipliers to determine the maximum.

I'm ashamed to say that I'm however still a but confused. In your derivation β is a function of the CHANGE in total energy - not the total energy. So how is temperature determined from total energy with that argument?

 

[Rap]

neat! had actually looked up the derivation on wikipedia, where they use lagrangian multipliers to determine the maximum.

I'm ashamed to say that I'm however still a but confused. In your derivation β is a function of the CHANGE in total energy - not the total energy. So how is temperature determined from total energy with that argument?

You're right - strictly speaking the temperature is generally a complicated function of total energy, expressed in an equation of state, and its definition is in terms of small changes in total energy, not the total energy itself. In simple equations of state, far away from quantum effects, or for cases where the total energy doesn't change a lot, you can come up with approximate equations of state, like the ideal gas law, or the van der Waals equation of state, where temperature is proportional to total energy, but one good little example where its not is when you have a particle that can spin, or not. (not like the spin of an electron). At high energies, this spin angular momentum and energy adds an extra degree of freedom. Generally, total energy is U=C N (kT/2) where C is the dimensionless specific heat capacity, or the "effective degrees of freedom". At high temperatures, the spin energy is spread out over many spin energy levels and you can say that C=5 (instead of 3 for a particle which does not spin). But as you lower the temperature, the angular energy gets concentrated in just the low angular energy levels, and then C lies somewhere between 3 and 5. Finally, at lower temperature, the angular energy gets concentrated in the ground angular energy state, and the spin degree of freedom gets "frozen out" and C =3. Energy is now spread out over the three translational degrees of freedom, but not the angular degree of freedom. Now the total energy is U=3 N (kT/2). If you cool things WAY down to almost absolute zero, even the translational degrees of freedom start to freeze out, and C=3 is wrong. If the particles happen to be Bosons, you start getting a Bose-Einstein condensate, and things get quantum-complicated.

So total energy is not proportional to temperature in the big picture, the "constant of proportionality" (C) is not constant at all, it is itself a function of temperature.

 

[aaaa202]

I think that all in all by looking through your posts I have learned what I came for. That I can't just say suppose you have this and this energy and this and this temperature, since the temperature is something that comes from the combinatorics and energy. Indeed I'm almost tempted to say that the Boltzmann distribution merely defines what the temperature is.

So with high energy density - which concerned my original question - you for instance get a very high temperature, which means the the exponential curve approaches a very flat curve meaning equal probability for all states.

I hope what I said so far was correct, because now I want to ask you a final question (you have been immensely helpful so far):
Can you make it intuitive for me, that the probabilities for each energy level always approach the probability for the lowest state, but never exceed it?
I can't quite make it intuitive by own arguments, but maybe you could put up and example like: suppose our atom acquires one unit of energy, then the probability for acquiring another one must always be a little less bla bla bla..

Hope you understand what I mean :)

 

[Rap]

I think that all in all by looking through your posts I have learned what I came for. That I can't just say suppose you have this and this energy and this and this temperature, since the temperature is something that comes from the combinatorics and energy. Indeed I'm almost tempted to say that the Boltzmann distribution merely defines what the temperature is.

Well, no, classical thermodynamics defines temperature, Boltzmann just explains it in microscopic terms. Classical thermodynamics defines all of the thermodynamic parameters, and the laws of thermodynamics tell you how they inter-relate. Statistical mechanics then tells you why they inter-relate the way they do.

So with high energy density - which concerned my original question - you for instance get a very high temperature, which means the the exponential curve approaches a very flat curve meaning equal probability for all states.

No, when the temperature is high, its still an exponential curve, but the "width" of the curve is much greater than the distance between energy levels. The probability is never equal for all states, its higher for the low energy states, less for the high energy states.

I hope what I said so far was correct, because now I want to ask you a final question (you have been immensely helpful so far):
Can you make it intuitive for me, that the probabilities for each energy level always approach the probability for the lowest state, but never exceed it?
I can't quite make it intuitive by own arguments, but maybe you could put up and example like: suppose our atom acquires one unit of energy, then the probability for acquiring another one must always be a little less bla bla bla..

Hope you understand what I mean :)

I think so - you mean that the probability of the ground state is always higher than any other state - the exponential curve is highest at zero energy. I think a better way of saying it is that the probability of any state is always less than that for any state below it in energy. If you say that, then its easy to show that the ground state is most likely.

About the intuitive explanation, I'm not sure. If you have 100 particles and 200 energy units, you could have all but one at zero, and one at 200. That one at 200 has a higher population number than all the levels below it except the ground state, but there is only one way this can happen. You can see that there are many more ways if you move some particles out of the ground state, and drop that 200 particle down in energy. By the same token, if you have any energy level that has more particles in it than some state below it, you can always find many more ways to distribute that energy by removing some of its particles, some up and more down than you can by just leaving things more or less the same.

Whatever intuition you come up with, it must have the idea that the equilibrium distribution has the most ways of being realized - its the most likely distribution. If you come up with a better intuition, let me know.

 

[aaaa202]

Well, no, classical thermodynamics defines temperature, Boltzmann just explains it in microscopic terms. Classical thermodynamics defines all of the thermodynamic parameters, and the laws of thermodynamics tell you how they inter-relate. Statistical mechanics then tells you why they inter-relate the way they do.

But really the way temperature is defined is rather randomly. Why does it have to exactly be:

1/T = ∂S/∂U, where S = kln(W)

isn't it exactly this because, this is the definition that makes sense in for instance the Boltzmann distribution? And thus the distribution more or less defines what temperature is..

 

[Rap]

But really the way temperature is defined is rather randomly. Why does it have to exactly be:

1/T = ∂S/∂U, where S = kln(W)

isn't it exactly this because, this is the definition that makes sense in for instance the Boltzmann distribution? And thus the distribution more or less defines what temperature is..

That is not a definition of temperature. Temperature is defined (to within a scale factor) by the second law of thermodynamics, with help from the zeroth and first. That's classical thermodynamics, not statistical mechanics. Its also not a definition of temperature because you cannot measure changes in entropy directly. There is no "entrometer". You can measure temperature, pressure, volume, and mass directly, not entropy and not chemical potential. You can control the change in internal energy, so in some cases, you can "measure" a change in internal energy. But there's no way to measure ∂S/∂U and so T= ∂S/∂U cannot be a definition of temperature. It is a relationship that follows from the laws of thermodynamics.

The definition of temperature is an experimental definition, as all classical thermodynamic definitions are. The definition of temperature tells you how to measure temperature, and makes no reference to atoms or molecules or statistical mechanics. The laws of classical thermodynamics puts constraints on the results of measurements. Then Boltzmann comes along and assumes that classical thermodynamics is explained by atoms and molecules and their statistics, and develops (along with others) the explanation of classical thermodynamics called statistical mechanics. What falls out of this explanation is an explanation of many things that are just an unexplained set of measurements in classical thermodynamics, like the specific heat.