Non-sensical negative entropy? (grand canonical ensemble)

[maverick_starstrider]

Uhm, why do you divide by N! in your second post and you had not mentioned that in the OP?

We covered this, it's an approximation for indistinguishable particles since you double count states otherwise.

 

[Dickfore]

We covered this, it's an approximation for indistinguishable particles since you double count states otherwise.

I was trying to follow a pedagogical approach so that the OP will see their mistake.

 

[atyy]

I don't know if you read the post above your head, but I think the problem might be resolved :) It seems the reason is that the N! that one usual divides by is not exact but rather an approximation, and a horribly bad approximation in my system.

I guess the basic issue is to count correctly as Ken G pointed out. However, I did find an example calculation where counting could involve exactly an N! (explicitly) as well as not (ie. it's hidden in the correct counting).

In http://ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2007/lecture-notes/lec23.pdf the first method of counting introduces an N! between Eq VII.11 & VII.12, while there isn't such a factor in Eq VII.24 & VII.25.

Again in http://farside.ph.utexas.edu/teaching/sm1/lectures/node80.html there is no N! explicitly introduced to count, yet http://farside.ph.utexas.edu/teaching/sm1/lectures/node82.html shows that the N! is in fact not an approximation, but exact. So it's not so much that the N! is an approximation, but it's already secretly included in some methods of counting.

 

[DrDu]

I did not read the whole thread and do not know for sure whether this argument had been brought up: usually the N ln N term and some term in Z which goes like N ln V yield N ln V/N . Now in the thermodynamic limit V/N, the density is hold constant so that the combined term is extensive.

 

[vanhees71]

Yes, one should use the Sackur-Tetrode formula, but make sure that arguments of logarithms must be a dimensionless quantity, i.e., for a monatomic ideal gas with g=2s+1 spin-degrees of freedom, you have

S=k N \left \{\frac{5}{2}+ \ln \left [\frac{g V}{\hbar^3 N} \left (\frac{m U}{3 \pi N} \right) \right ] \right \}.

The important point is that you can get this formula of an extensive entropy (solving Gibbs's paradox) only as the semi-classical limit of the quantum distributions for indistinguishable particles. In this lowest-order approximation there's no difference concerning bosons or fermions. That's why \hbar enters in this formula. It cannot be eliminated here precisely because the argument of the logarithm must be dimensionless.

 

[Dickfore]

I think the OP should have used the canonical ensemble if the # of particles had been fixed. Then, because we are talking about indistinguishable particles, the multiplicity factor is 1. Then, the partition function should read:
<br /> Z = \sum_{n = 0}^{N}{\exp \left[ -\beta \, (n \, \epsilon + (N - n)(-\epsilon)) \right]}<br />
<br /> Z = e^{\beta \, N \, \epsilon} \, \sum_{n = 0}^{N}{e^{-2 \, \beta \, \epsilon \, n}}<br />
This is a sum of a finite geometric sequence. The result is:
<br /> Z = e^{\beta \, \epsilon \, N} \, \frac{1 - e^{-2 \, \beta \, \epsilon \, (N + 1)}}{1 - e^{-2 \, \beta \, \epsilon}}<br />
<br /> Z = e^{\beta \, \epsilon \, N} \, \frac{e^{-\beta \, \epsilon (N + 1)} \, \left[ e^{\beta \, \epsilon \, (N + 1)} - e^{-\beta \, \epsilon (N + 1)} \right]}{e^{-\beta \, \epsilon} \, \left[ e^{\beta \, \epsilon} - e^{-\beta \, \epsilon} \right]}<br />
<br /> Z = \frac{\sinh \left[\beta \, \epsilon \, (N + 1) \right] }{ \sinh (\beta \, \epsilon) }<br />
Try finding the entropy associated with this partition function!

 

[Ken G]

I agree-- see post #28! But you advanced the calculation another step, I was lazy!

 

[Dickfore]

I agree-- see post #28! But you advanced the calculation another step, I was lazy!

Yes, you had proposed the same idea.

Anyway, from the partition function:

<br /> Z = \frac{\sinh \left[\beta \, \epsilon \, (N + 1) \right] }{ \sinh (\beta \, \epsilon) }<br />
Try finding the entropy associated with this partition function!

one can evaluate the free energy:
<br /> F = -\frac{1}{\beta} \, \ln{Z}<br />
and the entropy (\partial/\partial T = d \beta/d T \, \partial/\partial \beta = -\beta^2 \, \partial/\partial \beta) is:
<br /> S = -\frac{\partial F}{\partial T} = \beta^2 \, \frac{\partial F}{\partial \beta}<br />
Doing the derivatives, I get:
<br /> S = -\beta \, \epsilon \, \left[ (N + 1) \, \coth \left( \beta \, \epsilon \, (N + 1) \right) - \coth \left( \beta \, \epsilon \right) \right] + \ln \frac{\sinh \left( \beta \epsilon (N + 1) \right)}{\sinh \left( \beta \epsilon \right)}<br />
This has a finite limit when N \rightarrow \infty:
<br /> \lim_{N \rightarrow \infty} S = \beta \, \epsilon \, \coth \left( \beta \, \epsilon \right) - \ln \left[ 2 \, \sinh \left( \beta \, \epsilon \right) \right]<br />
Taking T = \frac{1}{\beta}, and plotting the above as a function of x \equiv T/\epsilon, we have the following plot:
http://www.wolframalpha.com/input/?i=Plot[Coth[1/x]/x+-+Log[2+Sinh[1/x]],{x,0,10}]

As can be seen, the entropy is non-negative.

 

[Ken G]

Thanks for carrying the calculation to its conclusion, but isn't it odd that the N dependence in S is not extensive? Or did you already calculate S/N here, the limits are tricky.

 

[Dickfore]

Thanks for carrying the calculation to its conclusion, but isn't it odd that the N dependence in S is not extensive? Or did you already calculate S/N here, the limits are tricky.

No, it is S, not S/N. The last expression is the limit of S when N \rightarrow \infty.

The "non-additivity" of the partition function comes from the fact that the partition function is not of the form:
<br /> Z \neq z^{N}_1(\beta, \epsilon)<br />

 

[Ken G]

Good point, the entropy here shouldn't be expected to be extensive. That answers mr vodka's question about that! We shouldn't have expected the entropy to be extensive, because indistinguishable particles don't get access to more states when you add more particles if they all go into the ground state. That's what must be happening here, a Bose-Einstein condensation.

 

[Dickfore]

The chemical potential of the system is:

<br /> \mu = \left( \frac{\partial F}{\partial N} \right)_{\beta} = -\frac{1}{\beta} \beta \, \epsilon \, \coth \left[ \beta \, <br /> \epsilon \, (N + 1) \right] = -\epsilon \, \coth \left[ \beta \, \epsilon \, (N + 1) \right]<br />

As long as \mu &lt; -\epsilon (lower than the lowest possible energy level), there is no macroscopic population of the ground state (Bose-Einstein condensation).

This corresponds to:
<br /> \coth \left[ \beta \, \epsilon \, (N + 1) \right] &gt; 1<br />
which is always true!

So, I don't think this system can undergo Bose-Einstein condensation.

 

[Ken G]

Yes, I was replacing the level energies with 0 and E rather than -epsilon and +epsilon. It's problematic to have a negative energy state with an unfixed N value-- clearly N will go to infinity because the reservoir will gain energy by making particles, and eventually the reservoir will gain infinite energy, violating the capacity of the reservoir itself. I don't think we really want negative energy states in a grand canonical distribution.

 

[Dickfore]

The choice of an energy reference level is irrelevant in statistical mechanics. The criterion for BEC is:
<br /> \mu = \epsilon_{\min}<br />
instead of:
<br /> \mu = 0<br />

 

[Ken G]

The reference level can't be irrelevant. If N is not fixed, and we have thermal contact with a reservoir, then there must be a very important physical difference between a positive and a negative energy level. The presence of a negative energy level will clearly create an energy divergence, and N will increase without bound, with no possibility of any equilibrium. Equilibrium occurs when the cost to the reservoir, in the currency of accessible configurations, of losing energy to the system is balanced by the number of configurations accessible to the system for having gained that energy. But if there is a negative energy state, then there is no such tradeoff, and it will populate infinitely in a grand canonical distribution.

I realize it is often said that energy is not specified to within a fixed overall constant additive term, but what is meant here by a state of -epsilon energy is that creating a particle and putting it in that state releases energy epsilon to the reservoir. That is already a change in energy, so is not ambiguous to within an additive constant term. If there is an energy cost for creating the particle, that has to be included in the meaning of -epsilon-- the chemical potential is not hard-wired to know how much energy it takes to make a particle, the chemical potential has to be told that (in our description of the energies of the states), and then it responds by telling us how much energy is invested on average per particle. If this has the wrong sign, and we have a grand canonical distribution, it has to blow up.

 

[Dickfore]

The reference level can't be irrelevant. If N is not fixed, and we have thermal contact with a reservoir, then there must be a very important physical difference between a positive and a negative energy level. The presence of a negative energy level will clearly create an energy divergence, and N will increase without bound, with no possibility of any equilibrium. Equilibrium occurs when the cost to the reservoir, in the currency of accessible configurations, of losing energy to the system is balanced by the number of configurations accessible to the system for having gained that energy. But if there is a negative energy state, then there is no such tradeoff, and it will populate infinitely in a grand canonical distribution.

I realize it is often said that energy is not specified to within a fixed overall constant additive term, but what is meant here by a state of -epsilon energy is that creating a particle and putting it in that state releases energy epsilon to the reservoir. That is already a change in energy, so is not ambiguous to within an additive constant term. If there is an energy cost for creating the particle, that has to be included in the meaning of -epsilon-- the chemical potential is not hard-wired to know how much energy it takes to make a particle, the chemical potential has to be told that (in our description of the energies of the states), and then it responds by telling us how much energy is invested on average per particle. If this has the wrong sign, and we have a grand canonical distribution, it has to blow up.

If N is not fixed, then we are in contact with a "particle reservoir" at some chemical potential μ (similar to the case where if E is not fixed, then we are in contact to a thermal reservoir with some temperature T). However, μ itself, being an energy, is determined "up to a reference level".

Without going into details, perhaps in inquiry into thermodynamics can help. The fundamental equation of thermodynamics for a system with a variable number of particles is given by:
<br /> dE = T \, dS - \Lambda \, d\lambda + \mu \, dN<br />
where we used a generalized coordinate \lambda, and a corresponding generalized force \Lambda.

Next, suppose:
<br /> \mu = \mu&#039; + \epsilon_0<br />
where \epsilon_0 is some arbitrary reference level. It is straightforward to show that a simultaneous redefinition:
<br /> E = E&#039; - \epsilon_0 \, N<br />
satisfies the fundamental equation for thermodynamics with E&#039; and \mu&#039;.

Many textbooks (especially describing BE condensation and photon systems), use the condition:
<br /> \mu = \left( \frac{\partial F}{\partial N} \right)_{\lambda, T} = 0<br />
as one of the necessary conditions for a minimum of the free energy when the number of particles is not fixed. However, a careful analysis shows that, when in contact to a "particle reservoir" at chemical potential \mu_0, this condition is:
<br /> \mu = \mu_0<br />
instead.

EDIT:
As for your example, if the chemical potential of the particle reservoir is \mu &lt; -\epsilon, then, actually it costs energy to populate the lower lying level.

 

[Ken G]

Something here just doesn't make physical sense. The energy epsilon is not an arbitrary energy that could have a fixed number added to it with no physical consequences, because epsilon is not an energy level, it has to be an energy difference. It is the difference between the energy of having a particle in that level, and having no particle at all (this would not be necessary if N were fixed, then only the energy difference between the two levels would be physically significant). So we cannot say that adding something to epsilon will just add something to mu and have no other physical consequence at all, because the conservation of energy says that any time we end up with a particle in the level with energy epsilon, the reservoir at T must provide that energy, which comes at a cost in number of configurations. Or, if we end up with a particle at energy -epsilon, the reservoir gains that energy. That's what epsilon has to mean here, it can't mean anything else for the problem to be well posed. So it cannot have an unspecified constant term added to it arbitrarily, or else the whole problem is underdetermined. For example, we didn't say if the particle has mass or not, so we must assume that epsilon accounts for any mass. Or, we didn't say if the particle has to be extracted from some deep potential well, so again that has to be in epsilon. We aren't free to let epsilon be whatever we want, because if it doesn't mean the energy the reservoir must part with to end up with a particle in that state, then we do not have enough information to solve for the thermodynamics of this system.

 

[nonequilibrium]

But Ken G, isn't what you're saying based on the assumption that the ground energy state of the reservoir is zero? Why can it not also be negative, as the system's?

This ties into Dickfore's explanation: mu quantizes this, mu is the measure of how much energy changes if a particle is added. According to your assumption, namely that the reservoir has ground level 0 or higher, the mu of the reservoir is 0 or higher. In that case you indeed have a problem and the bosons come flooding in. But mu is taken to be negative in such a case.

 

[Ken G]

But Ken G, isn't what you're saying based on the assumption that the ground energy state of the reservoir is zero? Why can it not also be negative, as the system's?

I'm saying epsilon has to be relative to that ground state energy, or else we cannot solve the problem, we won't have enough information. How can we find the entropy of a system if we don't even know the expectation value of N? I guess the point is, we are not solving for that, we are only solving for the entropy as a function of that expectation value, and finding that it reaches a constant value for large N, independent of the expectation of N. But that still might not be meaningful if there is no equilibrium for the system-- as would be true if there really is a negative energy for every particle that appears from the particle reservoir.

 

[nonequilibrium]

I'm saying epsilon has to be relative to that ground state energy, or else we cannot solve the problem, we won't have enough information. How can we find the entropy of a system if we don't even know the expectation value of N?

I'm not sure what you're saying, why are you claiming "else we cannot solve the problem". mu determines the expectation value of N. Do you have any backing of the statement "I'm saying epsilon has to be relative to that ground state energy, or else we cannot solve the problem, we won't have enough information."? It seems wrong to me, but I don't know how to prove that before I know why you think it's true.

 

[Ken G]

I'm not sure what you're saying, why are you claiming "else we cannot solve the problem". mu determines the expectation value of N. Do you have any backing of the statement "I'm saying epsilon has to be relative to that ground state energy, or else we cannot solve the problem, we won't have enough information."? It seems wrong to me, but I don't know how to prove that before I know why you think it's true.

If there is a particle reservoir, it will keep cranking out particles as long as the number of configurations the system+reservoir can reach is increased by adding particles. That's what determines the expectation value of N, physically. The chemical potential may give us a clever mathematical way to calculate this, but we should not lose sight of the basic physics. If we have a negative energy state, meaning that every time a particle is created from the particle reservoir and placed into that state, it releases energy, then that's just what the particle reservoir is going to do, and that is an impossible situation-- it will never reach equilibrium. We cannot derive both mu and N, we need one to tell us the other, but presumably there is some physical truth that sets mu, and so N is what we cannot control unless we have access to the mechanism that controls mu.

So what I'm saying is, it makes sense to me that if we observe the expectation value of N, we can infer mu, and epsilon can be negative or can have anything added to it, and it will just change our meaning of mu. Or, if we know something about what goes into mu, we can infer N, but what we know about mu had better make sure that energy is not liberated by -epsilon (so the value of -epsilon becomes physically important in an absolute way). Finally, we can take the limit as N goes to infinity, as Dickfore did, and get a result that does not depend on N so would seem to be able to infer mu. But this means epsilon is placing a constraint on the physics of mu such that we'll end up with a large but not infinite N, and a different epsilon puts a different physically meaningful limit on mu. What is physically meaningful is the energy that appears in the system when a particle appears in the system, and that is how I interpreted the value of epsilon. This in turn means that Dickfore's analysis can be used, not to derive N, but to find the physically necessary mu, relative to the energy of the state, that creates a large number of particles without creating an infinite number of particles. In other words, there is a physical meaning to the energy that appears in the system when a particle does, and that had better not be a negative number, because the particle didn't exist before and there's no place but the reservoir to conserve that energy.