# Relating temperature to energy in microstates of a system (Blundell)

1. Mar 31, 2012

### yeracxam

Hi there,

I am working through the preliminary chapters of "Concepts in Thermal Physics" by Blundell. I have not really studied statistics before, so have a lot to get to grips with, but I thought I was making headway until this question cast questions over my understanding again!!

1. The problem statement, all variables and given/known data

a) A system comprises N states which can have energy 0 or Δ. Show that the number of ways Ω(E) of arranging the total system to have energy E = rΔ (where r is an integer) is given by:

Ω(E) = N! / (r! * (N-r)! )

b) Now remove a small amount of energy sΔ from the system, where s << r. Show that:

$\Omega(E - \epsilon) \approx \Omega(E) \frac{r^{s}}{(N-r)^{s}}$

c) And hence show that the system has temperature T given by

$\frac{1}{k_{B}T} = \frac{1}{\Delta} ln(\frac{N-r}{r})$

d) Sketch kBT as a function of r from r=0 to r=N and explain the result.

2. Relevant equations

$\frac{1}{k_{B}T} = \frac{dln(\Omega)}{dE}$

3. The attempt at a solution

a) Okay, no problems here.
b)
$\Omega(E-\epsilon) = \frac{N!}{(r-s)!(N-r+s)!}$

from Stirling approx ln(N!) ≈ Nln(N) - N :

$ln\Omega(E-\epsilon) \approx Nln(N) - (r-s)ln(r-s) - (N-r+s)ln(N-r+s)$

$= Nln(N) - rln(r-s) - (N-r)ln(N-r+s) +s(ln(r-s) - ln(N-r+s))$

If I make the approximation ln(r-s) ≈ ln(r), ln(N-r+s) ≈ ln(N-r) because s<<r then I can rearrange to say:

$ln\Omega(E-\epsilon) \approx ln\Omega(E) + ln(r^{s}) - ln((N-r)^{s})$

$\Rightarrow \Omega(E-\epsilon) \approx \Omega(E)\frac{r^{s}}{(N-r)^{s}}$

Which gets me to the solution, but I am not sure why I am justified in saying ln(r-s) ≈ ln(r) but not (r-s) ≈ r?

c) I can get to this by simple differentiation of the stirling approximation, but this does not involve using my solution to b) to get there, so I am not sure if this is the correct approach?

d) Now comes the bulk of my confusion: Plotting the function seems to result in kBT → ∞ as r → N/2, and negative kBT for N/2 < r < N.

This seems to imply infinite and negative values for T, which I can't fathom!

My insights and questions:
- r = N/2 is clearly the macrostate which has the most microstates able to accommodate it
- When r changes, the energy in the system changes linearly... where does this energy come from / go?
- Shouldn't temperature have a linear relationship with the total energy in the system?
- Suppose this was a system connected to a heat bath / reservoir at temperature T. It will reach equilibrium with the reservoir and will therefore have temperature T and we can say something about the total energy the system is likely to contain. Does the result from this question suggest that it will, in fact, never contain more energy than NΔ/2, half the energy it is "capable" of containing?
- This does actually seem to fit with the shape of the Boltzmann distribution, with higher energies being exponentially less likely...

I think I am on the verge of understanding this, but at the moment my mind is going around in circles a bit, so I would appreciate some pointing in the right direction. Sorry for not posing a clearer question, but it's more help with understanding I am looking for if anybody would be able to oblige?

Many thanks!
Max

2. Mar 31, 2012

### nonequilibrium

Hello there. It seems you're starting an adventure in statistical mechanics. Be sure to enjoy the ride, it's a beautiful subject :)

Because ln(x) increases less rapidly than x. Or more exactly, in first order of s, we have that (by Taylor expansion): $\ln(r-s) \approx \ln(r) - \boxed{ \frac{s}{r} }$ whereas (also in first order of s) we have $r-s = r - \boxed{ s }$ (kind of silly to write this, but it's to show the analogy), and obviously $\frac{s}{r} \ll s$, justifying your approximations.

A way to do it using the result of b (although any approach is okay, even yours):
$\frac{1}{k_B T} = \frac{\mathrm d \ln \Omega}{\mathrm d E} = \lim_{\epsilon \to 0} \frac{\ln \Omega(E) - \ln \Omega(E-\epsilon)}{\epsilon} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \ln \left( \frac{\Omega(E)}{\Omega(E- \epsilon)} \right) = \lim_{s \to 0} \frac{1}{s \Delta} \ln \frac{(N-r)^s}{r^s} = \lim_{s \to 0} \frac{1}{s \Delta} \cdot s \cdot \ln \frac{N-r}{r} = \frac{1}{\Delta} \ln \frac{N-r}{r}$

That's because this concept of statistical temperature (not an official term) is quite more general than what you're used to (i.e. the thing you measure with a thermometer). What you measure with a thermometer is the motion of the atoms in an object, also called kinetic temperature. But it's easy to see that in such a system there is no upper limit to the energy, whereas in the system you're describing, there is! That explains why your intuitive feeling of temperature isn't valid, but okay that still doesn't explain this specific, quite weird behaviour. But the following does:

Let's take a look at the definition of temperature to understand what it means. $\frac{1}{k_B T} = \frac{\mathrm d \ln \Omega}{\mathrm d E}$ and this inverse temperature $\frac{1}{k_B T}$ is sometimes (in statistical mechanics) also called temperature (since it's actually the more logical concept, but hey we're stuck with the inverse of beta), but to avoid confusion I'll here call it the "beta temperature", hence defined as $\beta = \frac{\mathrm d \ln \Omega}{\mathrm d E}$. What does the beta temperature of a system tell us about a system? Well, it's a measure of how much the possible microscopic configurations (consistent with a certain energy) change when adding energy, after all that's what $\frac{\mathrm d \ln \Omega}{\mathrm d E}$ means!

So what would be expect for the beta temperature, qualitatively, without any calculations, just based on what we know of the system (i.e. that it has a minimum energy 0, a maximum energy $E_\textrm{max}$ and that the most microscopic configurations go along with an energy of $\frac{ E_\textrm{max} }{2}$ (Is this latter claim evident enough?))?
Well say the energy is pretty low, then almost all the particles have about same energy (probably zero), so there are not a lot of things you can swap around, i.e. there are only a small amount of possible microscopic configurations consistent with that energy, and it's also clear that if you add energy, you gain more possible configurations, hence the beta temperature should be positive. But if you keep on adding the same amount of energy to the system, the effect lessens each time: when nearly every particle has energy zero, adding just a little bit of energy increases Omega considerably (i.e. very large beta), and as you near $\frac{ E_\textrm{max} }{2}$ beta keeps dropping, but still staying positive, since adding energy also adds more possible configurations. Until you hit the $\frac{ E_\textrm{max} }{2}$ mark: this is the state of maximum disorder: the chance that a particle has either 0 or $\Delta$ energy is fifty percent, allowing the maximum number of possible microstates for a given energy. If you now add even more energy, you're causing more than half of the particles having $\Delta$ energy, and as you keep adding energy more and more particles will have to have a $\Delta$ energy, so there are less ways to mix things up: hence you lose microstates. What does this say about beta? Well that it is negative: adding energy decrease the number of microstates. And as you approach $E_\textrm{max}$ beta approaches minus infinity, since if you give it all the energy it can handle, every single particle is in a $\Delta$ state: there is only one microscopic configuration!

Conclusion: the graph of beta is that of a cotangent function: at zero it starts at +infinity, it approaches zero and then goes to -infinity. What about the "real" temperature T? Well you just have to take the inverse of beta, and that gets you that weird graph: you start at 0 (= $\frac{1}{+\infty}$) go to + infinity (= $\frac{1}{+0}$), then where beta is zero you switch to minus infinity (= $\frac{1}{-0}$) and then approach zero (= $\frac{1}{-\infty}$) once again.

EDIT: Since I accidentally clicked "post" too soon, I'll leave this post as it is (my explanation is done), and I'll answer your specific questions in a new post.

EDIT2: On second thought let me first add two remarks on my explanation, based on the question I'll state myself: What does this have to do with "temperature" as we know it?
1) It turns out that if the system under inspection is for example a box of gas, and you calculate beta the same way, and again take the inverse, that you indeed get the temperature as you know it.
2) This general temperature that we're discussing still has the same properties a "temperature" should have: if you have two systems at different temperatures, heat will flow from the hot one to the cold one. But now you know this more general concept of temperature can be negative, it becomes a bit more subtle: you have to remember that a negative temperature is actually larger than any positive temperature (in the sense that a system with a negative temperature is hotter)!!! Why is this? Well that statement is equivalent with saying that if you have a system A with a negative temperature and a system B with a positive temperature, that heat will flow from A to B. That's just rephrasing my weird statement, not explainig it. But let's rephrase it again, coming closer to the explanation: if you have a system A with a negative beta temperature and a system B with a positive beta temperature, heat will flow from A to B. And this is actually logical! But perhaps not on first sight, because this is new to you... But I don't want to spoil everything, so give it some thought yourself, why this is true. To see why my last underlined statement is true, simply think back of what beta means (a measure of how the number of microstates changes when you change energy...). Once you understand that last underlined statement, you also understand why systems with negative temperature are actually hotter than any system with positive temperature. (Exercise: think of what happens if you have the above system, in your OP, with a negative temperature, and you bring it into contact with a box of gas (which always has a positive temperature, as you know))

Last edited: Mar 31, 2012
3. Mar 31, 2012

### nonequilibrium

(Note that I added an extra note to my previous post, cf "EDIT2")

It is not specified. r is simply a parameter that quantifies how much energy is in your system. r is just a parameter as E is a parameter. A bit like the number of particles N is a parameter: in principle it can change, but that is irrelevant sometimes.

Nope, why should it? It is the case, however, for the special case of a box of gas, where you get $E = \frac{3}{2} Nk_B T$ but this is not a general or deep truth.

Well, I presume you're intuitively giving your reservoir a positive temperature T. If that is the case, then yes your system will never contain more energy than NΔ/2. But again, this is not a deep truth, but it's just a consequence of you choosing a reservoir with positive temperature.

Indeed, it's not a Boltzmann distribution. The Boltzmann distribution applies to the speed of (possibly interacting) particles, but our system is quite different. First of all, nowhere does it mention the word "speed", and secondly such a system (of moving particles) wouldn't have an upper limit to the energy, but our system has.

On a more comforting note: I was once where you are now, totally flabbergasted by this weird new concept. But in the end you will understand it, and you will actually see how marvelously logical it all is, and how brilliant a man Boltzmann must have been :)

Last edited: Mar 31, 2012
4. Mar 31, 2012

### yeracxam

Thank you, mr. vodka, for such a great answer!

I think I have been bringing too many preconceptions to the table from the study of ideal gases, which your second post helped to highlight.

Thanks for taking the time to write such a thorough explanation, you have certainly given me a lot to digest, and also a much clearer idea of how I should be thinking about things. Your second edit, expanding on the nature of "negative" temperatures was particularly helpful.

I understand what you mean about it being a beautiful subject; it is not something I am directly covering in my course at the moment, but it has really captured my imagination and I am determined to spend some of my Easter break getting to grips with some of the basic concepts.

You have been a great help, and I am looking forward to learning more, so thanks again! :)

5. Mar 31, 2012

### nonequilibrium

Glad to be of service, and happy to hear you're enjoying the ride :)

The confusion statistical mechanics offers at first is quite tasty, as you can rest assured eventually it will all become crystal clear (unlike quantum mechanics, I'd say). It's a shame it hasn't got the same glamour as relativity or quantum, in my eyes it surely deserves it.

I'll try to keep my eyes open for more of your questions, but I don't often check the homework forums, so if you happen to post something new and nobody seems to be replying, feel free to PM me to check out your post.