# Thermodynamics (Reif)

1. Nov 6, 2006

### quasar987

I find the mathematical manipulation of thermodynamics are often dubious and this one is doubly so. It is the derivation of the Boltzman probability distribution made by Reif (Section 6.2).

He considers an isolated system $A^{(0)}$ of energy $E^{(0)}$ in thermodynamical equilibrium composed of two system A and A' in thermal contact, where A' has much more degrees of freedom that A (A'>>A). He also supposes that the energy of interaction of A and A' are negligible so that $E^{(0)}=E+E'$.

These initial conditions set, he tries to find the probability of finding system A in state r of corresponding energy $E_r$. It follows from the postulate of equiprobability that the probability of finding A in state r is just the ratio of all the possible states of $A^{(0)}$ given that A is in state r to the ratio of all possible states of $A^{(0)}$*:

$$P_r=\frac{\Omega'(E^{(0)}-E_r)}{\Omega^{(0)}}$$

He then says, let us expand $\Omega'(E^{(0)}-E_r)$ in a Taylor series about $E^{(0)}$. But he actually expands the log of that in order to get a faster convergence:

$$\ln(\Omega'(E^{(0)}-E_r))\approx \ln(\Omega'(E^{(0)}))-\left[\frac{\partial \Omega'}{\partial E'} \right]_{E^{(0)}}E_r$$

And he justifies the dropping of the higher order terms by the fact that $E_r <<E^{(0)}$.

But I have three problems with this justification. (problem #3 is closely linked to #2)

1) $E_r=E^{(0)}$, E'=0 is a possible state! And in principle, we do not know just how unlikely it is until we've effectively found the formula for $P_r$!

2) It seems to me that $E_r <<E^{(0)}$ justifies nothing! We want $E_r<<1$ so that the higher the power n, the smaller $(E_r)^n$ is.

3) And wheter $E_r$ is <1 actually depends on the units of energy used! In, say, a relativistic taylor expansion, the argument is v/c which is dimensionless, so the problem of dimensionality does not arise. But in this case?!

I would really appreciate it if someone could nullify these two objections for me.

Many thanks!

*Reif uses $\Omega'(E')$ to denote the cardinality of the set of all states for wich the energy of A' is E' and $\Omega^{(0)}$ to denote the cardinality of the set of all possible states of the system $A^{(0)}$ as a whole.

Last edited: Nov 6, 2006
2. Nov 7, 2006

### lalbatros

Concerning your point (2) and (3).

Its is Er/E° that must be << 1 , not Er.
There is no problem in the dimensions, because the dimension of dW/dE * dEr is right.
There is small mistake in your formula for the Taylor expansion of ln(W).

For point (1), for large enough system, the probability distribution of the total energy of the (sub-) system is a very sharp peak around the mean value. You should try to calculate that sometimes: try a perfect gas and the M-B distribution. Also, have a look back at statistics: how does the sum of many (gaussian) distribution shape as a function of the number of items in the sum? Nice games, with generating functions, etc ...

3. Nov 7, 2006

### quasar987

My point exactly.

I miswrote the expansion; I forgot an "ln". So if I redo the expansion, expliciting every step, I want to expand $\ln(\Omega'(E'))$ about $E^{(0)}$, so

$$\ln(\Omega'(E'))=\ln(\Omega'(E^{(0)}))+\left[ \frac{\partial \ln(\Omega'(E'))}{\partial E'} \right]_{E^{(0)}}(E'-E^{(0)})+...$$

But $E'-E^{(0)}=-E_r$, so it becomes

$$\ln(\Omega'(E^{(0)}-E_r))=\ln(\Omega'(E^{(0)}))-\left[ \frac{\partial \ln(\Omega'(E'))}{\partial E'} \right]_{E^{(0)}}E_r+...$$

As you can see, it is E_r and not Er/E° that needs to be <<1 in order for the approximation to be justified, but this 1) is not true for all r, 2) depends on the units in which we measure the energy.

I am aware of this result. But if we restrict our study to the states r of energies near the average value, how can we later on assert that the distribution holds for all r?

4. Nov 7, 2006

### Galileo

The point is that E0 is waaaay larger than Er. So you're still very much in the neighbourhood of E0 which justifies the expansion. It's always the relative magnitude that is important.

If you like you can write $$E^0-E_r=E^0(1-\frac{E_r}{E_0})$$.

5. Nov 7, 2006

### Tomsk

But
$$\frac{\partial \ln(\Omega'(E'))}{\partial E'} = 1/k_{B}T$$

Right? Which is basically just the energy of the reservoir, so the units work, and $E_r << E^{(0)}$ by supposition, so any higher terms are going to cancel.

6. Nov 7, 2006

### quasar987

Not necessarily. We're looking for the probability of r where r is any accesible state. An accesible state is any state that is compatible with the macroscopic state. In particular, Er=E0 and E'=0 is an accesible state. It is unlikely because there are not many ways in which this can be realized, and the probability that it be realized is the ratio of this number of alll the accesible states. But we don't know that yet. That's what we're supposed to find out!

What does this change in the context of the expansion above?

Last edited: Nov 7, 2006
7. Nov 7, 2006

### quasar987

It has never been hypothesized that E_r << E^{(0)}. But if we suppose that it is, in order to make the approximation valid, i.e. if we choose to only consider this particular band of energies, then how can we say that the answer we are going to find is valid for all r even for those for which Er=E0?

And I am not saying that the units don't "work" in the sense that ln(W) doesn not turn out dimensionless. I am saying that since the Taylor expansion is in powers of E_r, if we measure the energy in units such that E_r is gigantic, the dropping of the terms is unjustified.

(or are you saying that k_B will be consequently adjusted so as to maintain the product Er/kT the same in any unit system? This is confusing.. :/)

Last edited: Nov 7, 2006
8. Nov 7, 2006

### Tomsk

The probability of the smaller system A being in a state $E_r$ is proportional to $\Omega(E')\Omega(E_r)$, as for every $\Omega(E')$, the small system can be in any one of $\Omega(E_r)$. But $\Omega(E_r)=1$ because the system is so small it only has one state, Er. So if you want ANY state up to E' ~ Er, then you need to Taylor expand $\Omega(E')\Omega(E_r)$.

9. Nov 7, 2006

### Tomsk

$E_r=E^{(0)}$ is not a possible state if the number of degrees of freedom of A' >> A, which is what has been assumed. It follows that $E_r << E^{(0)}$

10. Nov 7, 2006

### quasar987

I was imagining a large lattice and a small latice. Even though the large latice has more atoms, it is possible that at some instant, they all be at rest and the enegy be all in the small latice.

11. Nov 7, 2006

### Tomsk

If we use units that give Er to be very big, then E' will be even bigger. kT is the energy E' I think, so it becomes negligible for higher powers. I think that follows, not entirely sure. I think you'd get something like Er^2/E'^2, which is pretty small if Er<<E'

12. Nov 7, 2006

### Tomsk

Doesn't that seem to be very improbable though? There are lots more ways of arranging the energy so that it is spread around the larger lattice than the number of way all of that energy can be arranged in the small lattice. A small lattice is only going to keep a large amount of energy for a very small time, it wouldn't be stable with it. I see your point, but I think this approximation is really for when the energy of the system stays well below the energy of the reservoir. Since this is a much more likely observation, I think this approximation is valid.

13. Nov 7, 2006

### vanesch

Staff Emeritus
The assumption is that $$\Omega(E')$$ is a steeply rising function of E' and $$\Omega(E_r)$$ is a steeply rising function of Er.
In fact, they are essentially power functions with the power of the order of the number of degrees of freedom (see section 2.5).
If you then look at section 3.3, it is treated why the probability
$$P(E) = C \Omega(E)\Omega'(E^0 - E)$$ is such a sharp function.

Once you have this, this justifies the appoximations of working in a very small neighbourhood of the maximum.

As concerning to your "dimensional" argument, it is not necessary that in a series a_1 x + a_2 x^2 + ... a_n x^n...
x be smaller than 1 to have quick convergence. It depends also on the n-dependence of a_n.

Clearly, if you change energy units, you will change the numerical value of x, but also of all coefficients in the expansion, and the a_n coefficient will change with the unit conversion factor ^ n.

14. Nov 7, 2006

### Galileo

I don't have Reif's book, but in any thermobook I've read they assume Er<<E0.

You noted yourself that ln(W) doesn't depend on which units you use (to whatever order). If you measure in eV Er will be much bigger, but derivatives wrt E will drop in the same proportion giving the same entropy ln(W). So the problem is not the units, but whether the entropy S=ln(W) doesn't change much around S(E) when Er is small compared to E.

I think it can be shown, maybe from the convexity of S. Also, in the end it is only the relative magnitudes of $$\Omega(E^0-Er)$$ that will play a role, since it will be normalized such that the probabilities add to 1. I'm a bit fried at the moment but I`ll give it some thought.

15. Nov 7, 2006

### lalbatros

You could also write

ln(W) = ln(W°) + dln(W°)/dln(E°) * Er /E°

This shows you that Er/E° << 1 is the right condition.
And of course the coefficient in the taylor expansion which is

dln(W°)/dln(E°)

has no dimension.

16. Nov 8, 2006

### quasar987

And the logarithm is such a function? Do you have numbers? (ex: precise to k% regardless of the value of x when keeping only the linear term)

Eeeh? Would you mind explaining what you did here?

17. Nov 9, 2006

### quasar987

Here's what I think is a pretty convincing reasoning using bits and pieces from all the people who gave their opinion on this:

$\Omega'(E')\propto E^{f_A'}$ (where f_A' is the degree of freedom of system A', which is large by hypothesis). Now,

$$\frac{\partial \ln(\Omega'(E'))}{\partial E'}\propto \frac{\partial f\ln(E')}{\partial E'}=\frac{f}{E'}$$

So

$$\left[ \frac{\partial \ln(\Omega'(E'))}{\partial E'} \right]_{E^{(0)}}E_r\propto f\frac{E_r}{E^{(0)}}$$

In general,

$$\frac{1}{n!}\left[ \frac{\partial^n \ln(\Omega'(E'))}{\partial E'^n} \right]_{E^{(0)}}E_r\propto (-1)^{n-1}\frac{1}{n}f\left(\frac{E_r}{E^{(0)}}\right)^n$$

So at least now we have a power series in E_r/E°, which is always lesser or equal to unity.

Last edited: Nov 9, 2006