0
$\begingroup$

Suppose $N$ fair coin tosses. It is easy to represent this system in the micro-canonical ensemble as follows. There are 2^N possible results. Taking the entropy as the total number of micro state, we get

$S=N k_B \ln(2)$

$E=N k_B T \ln(2)$

and each bit contribute $k_B \ln(2)$ to the entropy.

I am trying to recover this results from the canonical ensemble, but it seems to fail. I try to obtain the partition function as follows: 1. Each bit (0 or 1) has the same energy level $E_0$. Hence the partition function is degenerate.

$Z=\sum_{i=0}^N{e^{-2\beta E_0}}=Ne^{-2\beta E_0}$

But when I take the energy using $\overline{E}=-\frac{\partial \ln{Z}}{\partial \beta}$

$\overline{E}=-\frac{1}{Ne^{-2\beta E_0}}(Ne^{-2\beta E_0})(-2 E_0)$

$\overline{E}=2E_0$

Why do I not recover the same results as the micro canonical case? Here, the average energy does not even depend on $T$.

Reverse engineering the process, to recover the micro canonical results, the partition function must be

$Z=\sum_{i=0}^N{e^{- E_0 \ln{\beta} \ln{2}}}$

then taking $\overline{E}=-\frac{\partial \ln{Z}}{\partial \beta}$ will give $\overline{E}=Nk_BT\ln{2}$. But a partition function with $\ln{\beta}$ instead of $\beta$ as a conjugate variable is hardly legit.

What am I doing wrong?

$\endgroup$
1
$\begingroup$

Let's start with the microcanonical case. Your expression for the entropy is correct. As far as the energy goes, in the microcanonical ensemble, the energy is a specified, not a derived, quantity, and by the description of the system you have given, $$ E = N E_0 $$ You seem to have assumed $S = \frac{\partial E}{\partial T}$, but this is not the case. You're looking for the formula $$T = \left( \frac{\partial S}{\partial E} \right)^{-1}$$ Then $dE = E_0 dN$, and so $$ T = \left ( \frac{1}{E_0} \frac{\partial}{\partial N} \left [ k_B N \ln 2 \right ] \right )^{-1} = \frac{E_0}{k_B \ln 2} $$ although just what the "temperature" is in the MCE is a little fuzzy.

Next, let's look at the canonical ensemble. The exponents in the partition function are not correct; they should be $\beta$ times the energy of the whole microstate. Every microstate has the same energy $$ E = \overline E = N E_0 $$ This is the reason that the energy does not depend on the temperature. Then $$ Z = \sum_{n=0}^{2^N} e^{-\beta N E_0} = 2^N e^{-\beta N E_0} $$ (Note that this correctly reproduces $\overline E$ via $E = -\frac{\partial \ln Z}{\partial \beta}$.) We can calculate the entropy via $$ S = k_B \left ( \ln Z + \beta \overline E \right ) = k_B \left ( N \ln 2 - \beta N E_0 + \beta \overline E \right ) = k_B N \ln 2 $$ and the temperature via $$ T = \frac{\partial \overline E}{\partial S} = \frac{1}{k_B \ln 2}\frac{\partial N E_0}{\partial N} = \frac{E_0}{k_B \ln 2} $$

These results agree, but please note that the example is kind of janky. The microcanonical and canonical ensembles describe systems in different situations. In the microcanonical ensemble, $E$ is given and $T$ is derived (if ontologically dubious); in the canonical ensemble, $T$ is given by thermodynamic contact with a heat bath of infinite heat capacity, and $E$ is derived. There's an artificial character that arises in this case because you've specified a system with only one energetic macrostate. A much more interesting example is that in which the "heads" and "tails" states for individual coins have different energies; for the sake of simplicity, say "heads" is $0$ and "tails" is $E_0 > 0$. I encourage you to work it out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.