Entropy of ensemble of two level systems

[Dazed&Confused]

Homework Statement

The fundamental equation of a system of $\tidle{N}$ atoms each of which can exist an atomic state with energy $e_u$ or in atomic state $e_d$ (and in no other state) is
$$F= - \tilde{N} k_B T \log ( e^{-\beta e_u} + e^{-\beta e_d} )$$
Here $k_B$ is Boltzmann's constant $\beta = 1/k_BT$. Show that the fundamental equation of this system, in entropy representation, is
$$S = NR \log \left (\frac{1+Y^{e_d/e_u}}{Y^Y} \right)$$
where
$$Y= \frac{U-\tilde{N}e_u}{\tilde{N}e_d-U}$$

Hint: introduce $\beta =1/k_BT$ and show that $U = F + \beta \partial F / \partial \beta = \partial ( \beta F) /\partial \beta$. Also, for definiteness, assume $e_u < e_d$ and note that $\tilde{N}k_B = NR$.

Homework Equations

$$S = -\partial F / \partial T = -(\partial F /\partial \beta )(\partial \beta /\partial T) = (\partial F / \partial \beta)( \beta /T)$$

The Attempt at a Solution

I showed the hint but will not write out the solution here. Then we have
$$U = \partial ( \beta F) /\partial \beta = \tilde{N} \frac{e_u e^{-\beta e_u} + e_d e^{-\beta e_d}}{ e^{-\beta e_u} + e^{-\beta e_d}}$$
from which I get $e^{\beta( e_u - e_d)} = Y$.

Also,
$$S = \tilde{N}k_B \log ( e^{-\beta e_u} + e^{-\beta e_d} ) +\tilde{N} \beta k_B \frac{e_u e^{-\beta e_u} + e_d e^{-\beta e_d}}{ e^{-\beta e_u} + e^{-\beta e_d}}$$
which I rewrote as
$$NR\log (1+Y) - NR\beta e_u + NR \beta \frac{e_u +e_d Y}{1+Y} = NR \log (1+Y)+ NR \beta \frac{Y(e_d-e_u)}{1+Y} = NR \log (1+Y) - NR \frac{Y}{Y+1} \log Y = NR \log \left ( \frac{1+Y}{Y^{Y/Y+1}} \right)$$
which is not (at least obviously) the same, but I cannot see my mistake. Help would be appreciated.

 

[mark726]

It looks like you have made a mistake in your calculation of U. The correct expression for U is:

U = \tilde{N} \frac{e_u e^{-\beta e_u} + e_d e^{-\beta e_d}}{ e^{-\beta e_u} + e^{-\beta e_d}} = e_u + e_d Y

Plugging this into the expression for S, we get:

S = NR\log (1+Y) - NR\beta e_u + NR \beta \frac{e_u +e_d Y}{1+Y} = NR \log (1+Y)+ NR \beta \frac{e_d Y}{1+Y} = NR \log \left ( \frac{1+Y}{Y^{Y/Y+1}} \right)

which matches the expected result.