📂Thermal Physics

Maxwell Distribution

Theorem¹

The random variable $V$ representing the speed of gas molecules follows the probability density function described by the Maxwell distribution.

$$ f(v) = \dfrac{4}{\sqrt{ \pi}} \left( \dfrac{m}{2 k_{B} T} \right)^{3/2} v^{2} e^{-mv^2 / 2k_{B}T } $$

Explanation

The Maxwell distribution is derived from the Boltzmann distribution and is also called the Maxwell-Boltzmann speed distribution. It is a distribution rarely seen in statistics, so much so that the name statistical mechanics might seem inappropriate. If anything, it is related to the skewness or kurtosis of the normal distribution.

Through the derivation of the Maxwell distribution, we probabilistically grasp and statistically understand the motion of gas molecules. We may not be able to verify the motion of each molecule microscopically, but macroscopically, it makes sense. For the derivation, it is assumed that the size of a molecule is sufficiently small compared to the distance between molecules and that the forces between molecules can be neglected.

Derivation

Part 1. Distribution of Speed

• Let’s say a certain gas molecule has a mass $m$, velocity $\mathbf{v} := ( v_{x} , v_{y} , v_{z} )$, and speed $v := | \mathbf{v} |$. Then the kinetic energy is as follows:

  $$ {{1} \over {2}} m v^2 = {{1} \over {2}} m v_{x}^2 + {{1} \over {2}} m v_{y}^2 + {{1} \over {2}} m v_{z}^2 $$

  Boltzmann distribution

  The probability that the energy of a system at temperature $T$ is $\epsilon$ is as follows:

  $$ P(\epsilon) \propto e^{ - \epsilon /k_{B} T } $$

  Then, the probability that the kinetic energy in the direction of $v_{x}$ axis is $\displaystyle E = {{1} \over {2}} m v_{x}^2$ is as follows:

  $$ g( E ) \propto e^{-mv_{x}^{2} / 2k_{B}T } \implies g( E ) = C e^{-mv_{x}^{2} / 2k_{B}T } $$

  Here $C$ is a constant. Now, let’s normalize $g$ to make it a probability density function. That means integrating $g$ over the entire region so that it equals $1$.

  Gaussian integral

  $$ \int_{-\infty}^{\infty} e^{-x^2} dx= \sqrt{\pi} $$

  Substituting with $v_{x} = \sqrt{\dfrac{2 k_{B} T}{m}} x$ and using the Gaussian integral gives the following:

  $$ \begin{align*} \int_{-\infty}^{\infty}g =& C \int_{-\infty}^{\infty} e^{ -m v_{x}^2 / 2 k_{B}T} dv_{x} \\ =& C \int_{-\infty}^{\infty} \sqrt{{2 k_{B} T} \over {m}} e^{ - x^2 } dx \\ =& C \sqrt{\dfrac{2 k_{B} T}{m}} \int_{-\infty}^{\infty} e^{ - x^2 } dx \\ =& C \sqrt{\dfrac{2 k_{B} T}{m}}\cdot \sqrt{\pi} \\ =& C \sqrt{\dfrac{2 \pi k_{B} T }{m}} \\ =& 1 \end{align*} $$

  Therefore, $C$ is as follows:

  $$ C = \sqrt{\dfrac{m}{2 \pi k_{B} T }} $$

  Thus, $g(v_{x})$ is as follows:

  $$ g(v_{x} ) = \sqrt{ {m} \over {2 \pi k_{B} T } } e^{ - {{m v_{x}^2 } \over {2 k_{B} T}} } $$

  Therefore, the proportion of gas molecules having speed between $(v_{x}, v_{y}, v_{z})$ and $(v_{x} + dv_{x}, v_{y} + dv_{y}, v_{z} + dv_{z})$ is as follows:

  $$ g(v_{x}) d v_{x} g(v_{y}) d v_{y} g(v_{z}) d v_{z} \propto e^{- {{m (v_{x}^{2} + v_{y}^{2} + v_{z}^{2})} \over {2 k_{B} T}} } d v_{x} d v_{y} d v_{z} = e^{- {{m v^2} \over {2 k_{B} T}} } d v_{x} d v_{y} d v_{z} $$

• Part 2. Distribution of Speed

  If the probability density function for the distribution of speed is denoted as $f(v)$, then it should be $\displaystyle \int_{0 } ^{\infty} f(v) dv = 1$. Since a gas molecule can move in any direction with speed $v$, consider a sphere with center $\mathbb{0}$ and radius $v$.

  

  The surface area of a sphere is $4 \pi v^2$, so if the thickness of the sphere’s outer shell is $dv$, it can be expressed as follows:

  $$ e^{- \frac{m v^2}{2 k_{B} T} } d v_{x} d v_{y} d v_{z} = 4 \pi v^2 e^{- \frac{m v^2}{2 k_{B} T} } dv $$

  $$ f(v) dv \propto v^2 e^{- \frac{m v^2}{2 k_{B} T} } dv $$

  If the explanation including the figure is not clear, it is acceptable to simply consider it mathematically and understand that a Jacobian has been multiplied. Finally, using the Gaussian integral and integration by parts for normalization gives the following.

  $$ f(v) = {{4} \over { \sqrt{ \pi } }} \left( {{m} \over {2 k_{B} T}} \right)^{{3} \over {2}} v^2 e^{- \frac{m v^2}{2 k_{B} T} } $$

■