Expressions for energy & entropy from free energy (discrete distribution)

[baseballfan_ny]

Homework Statement

(a) Find an expression for the free energy as a function of $\tau$ (temperature) of a system with two states, one at energy 0 and one at energy $\epsilon$. (b) From the free energy, find expressions for energy and entropy of the system. The entropy is plotted in Figure 3.11 (see below).

Relevant Equations

$$ Z = \sum_i e^{ frac {-E_i} {k_BT} } $$
$$ F = -k_B T \ln{Z} $$
$$ F = U - TS $$

So just by by using the definition of the partition function...
$$ Z = \sum_i e^{ \frac {-E_i} {k_BT} } = e^{ \frac {-0} {k_BT} } + e^{ \frac {-\epsilon} {k_BT} } = 1 + e^{ \frac {-\epsilon} {k_BT} } $$

And then, a result we obtained in class by using the Boltzmann H factor to solve for $S$ and finding an expression for $U - TS$ gave us $F = -k_B T \ln{Z}$. So applying that gives part (a)...

$$ F = -k_B T \ln{ 1 + e^{ \frac {-\epsilon} {k_BT} } } $$

Now for (b), I'm confused. I'm not sure if I have an equation to get S and U out of just this free energy expression. All I have is the free energy definition, $F = U - TS$, and if I try to solve using that and my above expression for free energy, I'm going to get something like $U = U$ or $S = S$. So I'm not sure which equation/principle to use. I have thought about using differential with constant volume but I don't think that would help and the question seems pretty explicit about using the free energy expression from (a) to obtain both U and S.

 

[TSny]

Think of $U$ as $\langle E \rangle$. For the two-state system, how would you calculate $\langle E \rangle$? Hint: What is the probability that the system has energy 0? What is the probability the system has energy $\epsilon$?

 

[baseballfan_ny]

Think of $U$ as $\langle E \rangle$. For the two-state system, how would you calculate $\langle E \rangle$? Hint: What is the probability that the system has energy 0? What is the probability the system has energy $\epsilon$?

Oh ok. So ...
$$ \langle E \rangle = \sum_{E = 0, E = \epsilon} E*P(E) = \frac {e^{\frac {0} {k_BT}}} {Z}*0 + \frac {e^{\frac {-\epsilon} {k_BT}}} {Z} *\epsilon $$

$$ \langle E \rangle = \frac {e^{\frac {-\epsilon} {k_BT} } } {1 + {e^{\frac {-\epsilon} {k_BT} } } *\epsilon $$

Is this the expression for "the energy of the system" that (b) is asking me for? So then I plug this into $F = U - TS$ to solve for entropy?

 

[TSny]

Yes

 

[baseballfan_ny]

That's interesting, I would have that by "energy of the system," they mean some sort of general expression, like energy as a function of the state. But it actually refers to an ensemble average of energy. Is there a reason for that? My guess is it has something to do with the system being discrete.

 

[TSny]

That's interesting, I would have that by "energy of the system," they mean some sort of general expression, like energy as a function of the state. But it actually refers to an ensemble average of energy. Is there a reason for that? My guess is it has something to do with the system being discrete.

I'm not sure I understand what you are asking here. Taking the thermodynamic $U$ to correspond to the statistical mechanics $\langle E \rangle$ is standard, I believe.

Another approach to the problem is to first find $S$ using the thermodynamic relation $S = -\frac {\partial F}{\partial T}$. Then you can find $U$ from $F = U - TS$.

Or, maybe you have seen the relation $U= - \large \frac {\partial \ln Z}{\partial \beta}$, where $\beta = \large \frac 1 {kT}$. This can be used to find $U$.

 

[baseballfan_ny]

Taking the thermodynamic U to correspond to the statistical mechanics ⟨E⟩ is standard, I believe.

That's exactly what I was asking :) Now that I think about, I think our instructor had mentioned it once briefly during class. Also got the same answer from trying the other two methods. Thanks for the help.