# Boltzmann distribution

by aaaa202
Tags: boltzmann, distribution
 P: 789 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 $\beta=1/kT$?
P: 1,005
 Quote by Rap 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! :)
 P: 789 Check out the Wikipedia article at http://en.wikipedia.org/wiki/Maxwell...ann_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.
 P: 1,005 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?
P: 789
 Quote by 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?
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.
 P: 1,005 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 :)
P: 789
 Quote by 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.
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.

 Quote by aaaa202 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.

 Quote by aaaa202 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.
P: 1,005
 Quote by Rap 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..
P: 789
 Quote by aaaa202 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.

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