Statistical physics: counting states, entropy and temperature

In summary, the system has N=E/N\epsilon=g excited states, and the entropy is S(E,N)=k \ln \Omega(E,N). The temperature is \frac{1}{T}=(\frac{\partial{S}}{\partial{E}})_{V,N}, and the region where the temperature is negative is at ν=-T/k.
  • #1
EricTheWizard
14
0
Hi everyone, I've hit a bit of a snag with part c of this problem (can't figure out how to invert a function T(ν)), so I'm starting to question whether I have the previous parts correct.

Homework Statement


Consider a system of N identical but distinguishable particles, each of which has a
nondegenerate ground state with energy zero, and a g-fold degenerate excited state with
energy ε > 0.

(a) Let the total energy of the system be fixed at E = Mε, where M is the number of
particles in an excited state. What is the total number of states (E,N)?
(b) What is the entropy S(E,N)? Assume the system is thermodynamically large. You
may find it convenient to define ν ≡ M/N, which is the fraction of particles in an
excited state.
(c) Find the temperature T(ν). Invert this relation to find ν(T).
(d) Show that there is a region where the temperature is negative.
(e) What happens when a system at negative temperature is placed in thermal contact
with a heat bath at positive temperature?

Homework Equations


entropy: [itex]S(E,N) = k \ln \Omega (E,N) [/itex], where [itex]\Omega[/itex] is the number of states with energy E and particle number N.

temperature: [itex]\frac{1}{T} = (\frac{\partial{S}}{\partial{E}})_{V,N}[/itex]

Stirling's approximation: [itex]k! ≈ \sqrt{2\pi k}e^{-k}k^{k}[/itex]

The Attempt at a Solution


a) A particle can be in either the ground state or one of g excited states, so the number of states with excited particle number M should be [itex]\Omega(M,N) = {{N}\choose{M}} g^M[/itex] because there are [itex]{N}\choose{M}[/itex] ways to arrange which particles in an ensemble are excited, and then g choices for each particle of which degenerate state they're in. This seems to make sense to me, anyways.

b) To find the entropy, simply take the natural logarithm of Ω (which requires you to make Stirling's approximation so that you can pull parts of the factorials through the log)
[tex]S = k \ln\Omega ≈ k \ln{(\frac{N^{N+1/2}g^M}{\sqrt{2\pi}(N-M)^{N-M+1/2}M^{M+1/2}})} = k[(N+1/2)\ln N + \frac{E}{\epsilon}\ln g - \frac{1}{2}\ln(2\pi)-(N-\frac{E}{\epsilon}+\frac{1}{2})\ln{(N-\frac{E}{\epsilon})}-(\frac{E}{\epsilon}+\frac{1}{2})\ln{\frac{E}{ε}} ][/tex]

c) Now find the temperature distribution [itex]T(\nu)[/itex] making the substitution [itex]\nu = M/N = E/N\epsilon [/itex], the fraction of particles in the excited state and using the formula for temperature given above.
[tex]\frac{1}{T} = \frac{\partial S}{\partial E} = \frac{k}{\epsilon}[\ln g + \ln(N - E/\epsilon) + \frac{N-E/\epsilon+1/2}{N-E/\epsilon}-\ln\frac{E}{\epsilon}-(1-\frac{\epsilon}{2E})] = \frac{k}{\epsilon}[\ln\frac{g(1-\nu)}{\nu}+\frac{1}{2N}\frac{2\nu-1}{\nu (1-\nu)}] = \frac{1}{T(\nu)}[/tex]
Now this is the part I'm stuck at; I simply can't invert T(ν) because it's not a 1-to-1 function. If you plot it in Mathematica or similar, you get something like this:
JFkiV.jpg

which (and you can't see this very well) turns downward near ν=1 very sharply and makes the distribution non-invertible (at least in the negative temperature zone). Have I made a mistake somewhere?

d) Set T<0 and solve for the pertinent conditions (haven't tried this yet)

e) Heat will flow from a negative temperature system to a positive temperature one if they come in contact.
 
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  • #2
EricTheWizard said:
[tex]\frac{k}{\epsilon}[\ln\frac{g(1-\nu)}{\nu}+\frac{1}{2N}\frac{2\nu-1}{\nu (1-\nu)}] = \frac{1}{T(\nu)}[/tex]

Just a thought: Can you ignore the second term on the left if you assume N is large?
 
  • #3
TSny said:
Just a thought: Can you ignore the second term on the left if you assume N is large?

You're right; the problem does say to assume the system is thermodynamically large (N order of 10^23 or whatever). Sounds like a valid strategy (and makes the answer [itex]\nu(T)=\frac{g}{e^{\epsilon/kT}+g}[/itex]).

Thanks for the inspiration!
 

1. What is statistical physics?

Statistical physics is a branch of physics that uses statistical methods and concepts to explain and predict the behavior of physical systems, particularly at the microscopic level. It is based on the principles of thermodynamics and statistical mechanics.

2. What is the concept of counting states in statistical physics?

Counting states refers to the process of determining the number of possible microscopic configurations or arrangements of particles in a physical system. This allows us to calculate the probability of a certain configuration occurring and ultimately understand the behavior of the system as a whole.

3. What is entropy in statistical physics?

Entropy is a measure of the disorder or randomness in a system. In statistical physics, it is often described as the number of ways in which a system can be arranged at the microscopic level. It is closely related to the concept of probability and is used to understand the behavior of systems that are in thermal equilibrium.

4. How is temperature related to statistical physics?

Temperature is a measure of the average kinetic energy of particles in a system. In statistical physics, it is related to the entropy of a system through the Boltzmann constant. As the temperature of a system increases, the entropy also increases, leading to a more disordered state.

5. What are some practical applications of statistical physics?

Statistical physics has many practical applications in various fields, such as thermodynamics, materials science, and even economics. It is used to understand and predict the behavior of gases, liquids, and solids, and has also been applied to complex systems like biological networks and social systems.

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