Derivation of Boltzmann Factor via Reservoir Method (canonical ensemble)

[yucheng]

Homework Statement

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Relevant Equations

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https://scholar.harvard.edu/files/schwartz/files/7-ensembles.pdf
https://mcgreevy.physics.ucsd.edu/s12/lecture-notes/chapter06.pdf

On page 3 of both the notes above, the author merely claims that $$P \propto \Omega_{\text{reservoir}}$$

But isn't $$P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}}$$?

We have

$$\mathrm{ln} \ \Omega_{\text{res}} (E_{\text{tot}} - E_{\text{sys}})= \mathrm{ln} \ \Omega_{\text{res}} (E_{\text{tot}} ) - E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{res}}}{\partial E_{\text{sys}}} + \text{higher order terms}$$

Why are higher order terms negligible? We know that they are small compared to the total energy, and how large they are in numerical terms, depends on the unit of energy. Hence we can scale the unit such that $E_{\text{sys}}<<1$ then the higher order terms obviously disappear. Is reasonable?

$$\mathrm{ln} \ \Omega_{\text{sys}} (E_{\text{sys}})= \mathrm{ln} \ \Omega_{\text{sys}} (0) + E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} + \text{higher order terms} = E_{\text{sys}} \frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} + \text{higher order terms}$$

so... in equilibrium, both the reservoir and system have the same temperature, but the partial derivatives are evaluated at different energies, but since they are in equilibrium,

Thus $$\frac{\partial \mathrm{ln} \ \Omega_{\text{sys}}}{\partial E_{\text{sys}}} \bigg \rvert_{0} = \beta_{sys}$$
$$\frac{\partial \mathrm{ln} \ \Omega_{\text{res}}}{\partial E_{\text{sys}}} \bigg \rvert_{E_{\text{tot}}} = \beta_{res}$$

But aren't $$\beta = 1/kT$$? So... aren't they equal?

Which just means that
$$P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}} \propto e^{\beta({E_{\text{sys}} - E_{\text{sys}}})} \propto 1$$?

hence $P \propto \Omega_{\text{system}}\Omega_{\text{reservoir}}$ cannot be right!?

Thanks in advance!

 

[Lord Jestocost]

The first equation holds under the following conditions:

"Since the $system + reservoir$ is a closed system, the total energy of the $system + reservoir$ is fixed at $E_{tot}$. Since we have fixed the microstate $k$ of the system, the total number of states is determined only by properties of the reservoir. More precisely, the probability of finding the system in microstate $k$ is proportional to the number of ways of configuring the $system + reservoir$ with the system in microstate $k$."

see page 3 in https://scholar.harvard.edu/files/schwartz/files/7-ensembles.pdf

 

[yucheng]

the total number of states is determined only by properties of the reservoir. More precisely, the probability of finding the system in microstate $k$ is proportional to the number of ways of configuring the $system + reservoir$ with the system in microstate $k$."

Why is the total number of states is determined only by properties of the reservoir?
Is it because the reservoir is so large that it swamps out the system i.e. we can neglect the properties of the system itself when enumerating the total microstates of the composition of system and reservoir?

 

[Lord Jestocost]

When one fixes the system to be in state $k$ with energy $E_k$, this states can be realized $\Omega_{reservoir}(E_{tot}-E_k)$ times.

 

[yucheng]

When one fixes the system to be in state $k$ with energy $E_k$, this states can be realized $\Omega_{reservoir}(E_{tot}-E_k)$ times.

I think I understand where I am confused.

My intuition: the probability of the system having a particular energy is $\propto$ to the number of microstates the system takes

So where was this fact incorporated? In the summation at the end... where the summation in the partition function is taken over all the microstates a system can take (not over the distinct energies)!

P.S. in fact, I was very careless. You are right, one fixes the microstate, not the energy, hence the author derived the probability of being in that microstate, not the probability of being in that energy!

P.S.S. if one were to take the summation in the partition function over distinct energies, one would need to associate a weight factor with each distinct energy; this serves as the motivation for section 3.4 of Pathria & Beale! And that was what confused me even more when the author uses $P$ for both the probability of being in a microstate and of having a definite energy!

Thanks!