# Derivation of Boltzmann Factor via Reservoir Method (canonical ensemble)

• yucheng
In summary, the author claims that the partial derivative of the temperature of a system and reservoir is equal to beta, which can only be right if P = Omega system-reservoir.
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!?

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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

Lord Jestocost said:
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?

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
Lord Jestocost said:
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!

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vanhees71 and Lord Jestocost

## 1. What is the Boltzmann factor?

The Boltzmann factor is a mathematical expression used in statistical mechanics to describe the probability of a particle occupying a particular energy state. It is defined as e^(-E/kT), where E is the energy of the state, k is the Boltzmann constant, and T is the temperature of the system.

## 2. What is the reservoir method in the derivation of the Boltzmann factor?

The reservoir method is a technique used to derive the Boltzmann factor in the canonical ensemble. It involves considering a large external system, or "reservoir", that is in thermal equilibrium with the smaller system of interest. This allows for the use of the principles of thermodynamics to derive the Boltzmann factor.

## 3. How is the Boltzmann factor derived using the reservoir method?

In the reservoir method, the total energy of the system is divided into two parts: the energy of the reservoir and the energy of the system. By considering the energy exchange between the two, the Boltzmann factor can be derived using the principles of thermodynamics, such as the conservation of energy and the second law of thermodynamics.

## 4. What is the significance of the Boltzmann factor in statistical mechanics?

The Boltzmann factor is a fundamental concept in statistical mechanics as it relates the macroscopic properties of a system, such as temperature, to the microscopic behavior of its particles. It allows for the calculation of the probability of a particle occupying a particular energy state, which is crucial in understanding the behavior of systems at the atomic and molecular level.

## 5. Can the Boltzmann factor be applied to all systems?

The Boltzmann factor can be applied to systems that are in thermal equilibrium, meaning that the temperature is constant throughout the system. It is commonly used in the canonical ensemble, which describes systems with a fixed number of particles, volume, and temperature. However, it may not be applicable to systems that are out of equilibrium or have varying temperatures.

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