- #1
Dox
- 26
- 1
Hello everybody!
I'm interested in a statistical system with N particles which possible energies are, [tex] 0,\epsilon>0[/tex].
a) Find the entropy of the system.
b) Find the most probable [tex]n_0[/tex] and [tex]n_1[/tex], and find the mean square fluctuations of these quantities.
c) Find the temperature as a function of [tex]U[/tex], and show it can be negative.
d) What happens when a system of negative temperature is allowed to exchange heat with a system of positive temperature?
[tex]W(n_0,n_1)=\frac{N!}{n_0!n_1!}[/tex]
[tex]S=k\ln \Omega, [/tex] with [tex]\Omega=\sum'_{\{n\}} W(n_0,n_1).[/tex]
[tex]\frac{1}{T}=\frac{\partial S}{\partial U}.[/tex]
I try the following,
[tex]N= n_0+n_1[/tex] and [tex]U=n_1 \epsilon_1-n_0\epsilon_0= n_1\epsilon_1.[/tex], so
[tex]W=\frac{N!}{(N-n_1)! n_1!}\;\Rightarrow \; n_1^*=\frac{N}{2}.[/tex]
That should answer the question (b).
Meanwhile, by using the Stirling's formula,
[tex]S=k\left(N\ln N -(N-n_1)\ln (N-n_1) - n_1\ln n_1,[/tex]
I used [tex]N=n_1+n_2[/tex] for deriving the temperature, but I got
[tex]\frac{1}{T}= \frac{k}{\epsilon_1}\ln\left(\frac{n_1+n_0}{n_1}\right),[/tex] which is not negative,
Finally, I guess that if I allow a sut up with negative temperature interact with a standard thermodynamical system the second law breaks... Am I right??
Homework Statement
I'm interested in a statistical system with N particles which possible energies are, [tex] 0,\epsilon>0[/tex].
a) Find the entropy of the system.
b) Find the most probable [tex]n_0[/tex] and [tex]n_1[/tex], and find the mean square fluctuations of these quantities.
c) Find the temperature as a function of [tex]U[/tex], and show it can be negative.
d) What happens when a system of negative temperature is allowed to exchange heat with a system of positive temperature?
Homework Equations
[tex]W(n_0,n_1)=\frac{N!}{n_0!n_1!}[/tex]
[tex]S=k\ln \Omega, [/tex] with [tex]\Omega=\sum'_{\{n\}} W(n_0,n_1).[/tex]
[tex]\frac{1}{T}=\frac{\partial S}{\partial U}.[/tex]
The Attempt at a Solution
I try the following,
[tex]N= n_0+n_1[/tex] and [tex]U=n_1 \epsilon_1-n_0\epsilon_0= n_1\epsilon_1.[/tex], so
[tex]W=\frac{N!}{(N-n_1)! n_1!}\;\Rightarrow \; n_1^*=\frac{N}{2}.[/tex]
That should answer the question (b).
Meanwhile, by using the Stirling's formula,
[tex]S=k\left(N\ln N -(N-n_1)\ln (N-n_1) - n_1\ln n_1,[/tex]
I used [tex]N=n_1+n_2[/tex] for deriving the temperature, but I got
[tex]\frac{1}{T}= \frac{k}{\epsilon_1}\ln\left(\frac{n_1+n_0}{n_1}\right),[/tex] which is not negative,
Finally, I guess that if I allow a sut up with negative temperature interact with a standard thermodynamical system the second law breaks... Am I right??