📂Thermal Physics

Boltzmann Distribution

Theorem¹

The probability that a system with a temperature of $T$ has energy $\varepsilon$ is as follows.

$$ P(\varepsilon) \propto e^{ - \frac{\varepsilon}{k_{B} T} } $$

This distribution is called the Boltzmann distribution.

Derivation

An ensemble is simply ‘a situation made up of systems’.

Among them, a canonical ensemble is a situation with a large reservoir and a very small system as shown above.

The reservoir is assumed to have a temperature of $T$ and a very large thermal energy $E$, and is also called a heat bath. It is assumed to be so large that it can provide a lot of energy to the system and still maintain the same temperature afterward. It’s like going to the sea and scooping up a cup of seawater, after which the total amount of seawater is essentially unchanged.

The system can be a very small unit, like ‘one molecule’, for an extreme case. It is assumed that there is only one microstate for all the energy that the system can have. Therefore, $\Omega=1$. If there is no specific condition on how this system should be given, the same discussion would proceed for another system in the reservoir. Therefore, ’exploring the canonical ensemble’ is directly linked to ’exploring all molecules of a given system’.

Let’s assume that the system has gained a very small amount of energy $\varepsilon$ by coming into contact with the reservoir as shown above. Since the system is assumed to be very small, it must be viewed from a microscopic perspective, and the energy $\varepsilon$ will follow some distribution. And the probability that the energy of the given system is $\varepsilon$ is proportional to the number of microstates of the energy $E$ of the reservoir. In other words, $P(\varepsilon) \propto \Omega (E)$, and since $\Omega (E) = \Omega (E - \varepsilon) \Omega ( \varepsilon)$, we get the following equation.

$$ P(\varepsilon) \propto \Omega (E - \varepsilon) \Omega ( \varepsilon) $$

Since the system is assumed to be very small, $\Omega (\varepsilon ) = 1$, the above equation is as follows.

$$ P(\varepsilon) \propto \Omega (E - \varepsilon ) $$

Taylor’s Theorem

If the function $f(x)$ is continuous at $[a,b]$, and differentiable up to $n$ times at $(a,b)$, then for $x_{0} \in (a,b)$

$$ f(x) = \sum_{k=0}^{n-1} {{( x - x_{0} )^{k}\over{ k! }}{f^{(k)}( x_{0} )}} + {(x - x_{0} )^{n}\over{ n! }}{f^{(n)}(\xi)} $$

there exists $\xi \in (a,b)$.

Meanwhile, since the system is assumed to be very small, $\varepsilon \ll E$, and the Taylor expansion for $\ln \Omega (E - \varepsilon )$ around $E$ is as follows.

$$ \begin{align*} \ln \Omega (E - \varepsilon ) =& {{1} \over {0!}} \ln \Omega ( E ) + {{ \left[ ( E - \varepsilon) - E \right] } \over {1!}} \left( \ln \Omega (E) \right)^{\prime} + \cdots \\ =& \ln \Omega (E) - {{ d \ln \Omega (E) } \over { d E }} \varepsilon + \cdots \end{align*} $$

Definition of Temperature

$$ \dfrac{1}{k_{B} T} : = \dfrac{d \ln (\Omega)}{dE } $$

Then, based on the definition of temperature, it can be organized as follows.

$$ \ln \Omega (E - \varepsilon ) = \ln \Omega (E) - {{ 1 } \over {k_{B} T}} \varepsilon + \cdots $$

Since $\varepsilon$ is sufficiently small, the terms of $2$ or higher, $\varepsilon^{n}$, are considered almost equivalent to $0$. Then, we obtain the following equation.

$$ \begin{align*} \ln \Omega (E - \varepsilon ) =& \ln \Omega (E) - \dfrac{\varepsilon}{ k_{B} T} \\ =& \ln \Omega (E) + \ln e^{-\frac{\varepsilon}{k_{B}T}} \\ =& \ln \left( \Omega (E) e^{-\frac{\varepsilon}{k_{B}T}} \right) \end{align*} $$

Solving the log yields:

$$ \Omega (E - \varepsilon ) = \Omega ( E) e^{ - {{\varepsilon } \over {k_{B} T}} } $$

Therefore, $P(\varepsilon) \propto e^{ - {{\varepsilon } \over {k_{B} T}} }$, and such distribution is called the Boltzmann distribution. It is also called the Canonical Distribution due to its derivation from the canonical ensemble.