Boltzmann Factor Example

boltzmann-factor.md

 

Written by Andy, 2015/4/22 本來想寫Boltzmann Factor是怎麼導出來的，但是...太複雜了而且沒必要，就...。

What(TF) is Boltzmann Factor?

Boltzmann Factor, $e^{-\frac{E}{k_B T}}$, is

1. An energy distribution function that cause maximized entropy.
2. The un-normalized probability that you find between energy $E$ and $E+dE$ is the Boltzmann Factor times $dE$ when you pick up a particles randomly in a system of particles at equilibrium.

Average Energy

To calculate the average energy of a system, one should use the concept of expectation. Or you can think it of a normalize process. The general expression is $$ \left\langle E \right\rangle = \frac{ \int E(\tau) e^{-\frac{E(\tau)}{k_B T}} d\tau }{ \int e^{-\frac{E(\tau)}{k_B T}} d\tau }. $$ The involved parameter, $\tau$, is a $\mathbb{R}^n$ vector that indicate the property of a particle with interest Integrate in all microstates.

Example: Helium

A system composed of Helium, for example, is a simple system. The energy of a single atom is determined with $$E(\tau)=\frac{1}{2m} \left(p_x^2+p_y^2+p_z^2\right).$$ Adding its position, there are $6$ parameters in total, thus, $$d \tau = dx,dy,dz,dp_x,dp_y,dp_z$$

Integrate

As you can see later, the average of each parameter is symmetric, which means them can be separately calculated. Surprisingly, they are all equal. Here we just calculate it altogether: $$ \left\langle E \right\rangle = \frac{ \iiint!!\iiint \frac{1}{2m} \left(p_x^2+p_y^2+p_z^2\right) e^{-\frac{p_x^2+p_y^2+p_z^2}{2mk_BT}},dx,dy,dz,dp_x,dp_y,dp_z }{ \iiint!!\iiint e^{-\frac{p_x^2+p_y^2+p_z^2}{2mk_BT}},dx,dy,dz,dp_x,dp_y,dp_z } $$ To start with, use the following fact to split the tedious integrals,

$\iint f(x)g(y),dy,dx=\left(\int f(x),dx\right) \left(\int g(y),dy\right)$

And we end up with three integrals eliminated, $$ \frac{ \int \frac{p_x^2}{2m} e^{-\frac{p_x^2}{2mk_B T}}dp_x, \int e^{-\frac{p_y^2}{2mk_B T}}dp_y, \int e^{-\frac{p_z^2}{2mk_B T}}dp_z+\ \int e^{-\frac{p_x^2}{2mk_B T}}dp_x, \int \frac{p_y^2}{2m} e^{-\frac{p_y^2}{2mk_B T}}dp_y, \int e^{-\frac{p_z^2}{2mk_B T}}dp_z+\ \int e^{-\frac{p_x^2}{2mk_B T}}dp_x, \int e^{-\frac{p_y^2}{2mk_B T}}dp_y, \int \frac{p_z^2}{2m} e^{-\frac{p_z^2}{2mk_B T}}dp_z }{ e^{-\frac{p_x^2}{2mk_B T}}dp_x, e^{-\frac{p_y^2}{2mk_B T}}dp_y, e^{-\frac{p_z^2}{2mk_B T}}dp_z } $$

Simplify

This is a huge fraction! Hopefully we find more terms to eliminate. We write $$\begin{split} \mathcal{E}{1}&=\int e^{-\frac{u^2}{2mk_B T}},du & \mathrm{and}\ \mathcal{E}{u^2}&=\int u^2e^{-\frac{u^2}{2mk_B T}},du. \end{split}$$ (Better notation is welcome)

Continue, keeping in mind that which variable inside integral are independent of its value, $$ \frac{ \frac{1}{2m} \left(\mathcal{E}{u^2},\mathcal{E}{1},\mathcal{E}{1} + \mathcal{E}{1},\mathcal{E}{u^2},\mathcal{E}{1} + \mathcal{E}{1},\mathcal{E}{1},\mathcal{E}{u^2} \right) }{ \left( \mathcal{E}{1} \right)^3 } \ = \frac{3}{2m},\frac{\mathcal{E}{u^2}}{\mathcal{E}{1}} $$

$$ \begin{split} & A>0, \\ & \int_0^\infty e^{-Au^2},du & = \sqrt{\frac{\pi}{4A}}, \\ & \int_0^\infty u^2 e^{-Au^2},du & = \sqrt{\frac{\pi}{16A^3}}. \end{split} $$

So, eventually, $$ \frac{\mathcal{E}{u^2}}{\mathcal{E}{1}} = \frac{1}{2\frac{1}{2mk_B T}}=m k_B T, \ \left\langle E \right\rangle=\frac{3}{2m},\frac{\mathcal{E}{u^2}}{\mathcal{E}{1}}=\frac{3}{2}k_B T. $$

FIN {.text-center}

Written with StackEdit.