Derivation of Maxwell Speed Distribution

Introduction

Maxwell derived the exponential atomic speed distributions without any reference to Boltzmann's "partition" functions.  To me his derivation is much easier to understand than the latter.

Setting Mono-Energetic Speeds

To "randomly" choose the components while requiring that the energy be E we could do the following:  First choose vi such that

           

            $\begin{array}{l}v{x}_0=random\left(\right)-0.5\\ v{y}_0=random\left(\right)-0.5\\ v{z}_0=random\left(\right)-0.5\end{array}$             (1.1)

where random() results in a floating point random number between 0 and 1.  Then compute

            $r=\sqrt{v{x}_0{}^2+v{y}_0^2+v{z}_0^2}$   (1.2)

then actual velocity components are:

            $\begin{array}{l}{v}_x=\frac{v_{\max }v{y}_0}{r}\\ v_y=\frac{v_{\max }v{y}_0}{r}\\ v_z=\frac{v_{\max }v{y}_0}{r}\end{array}$     (1.3)

where

            $v_{\max }=\sqrt{\frac{2E}{m}}$     (1.4)

and E is the energy of all of the atoms.

           

Maxwell's Derivation without Energy Constraints

Maxwell claims that the (v_x,v_y,v_z) components are independent of each other so he can write the three dimensional probability function as the product of three one-dimensional probability functions:

            $F(v_x,v_y,v_z)=f(v_x)f(v_y)f(v_z)$          (1.6)

 Now, since F or the fs are not expected to be zero or negative, it is acceptable to take the natural log of both sides of this equation:

            $\ln (F)=\ln [f(v_x)]+\ln [f(v_y)]+\ln [f(v_z)]$         (1.7)

Now we can take the derivative of this equation with respect to vx:

            $\frac{\partial \ln (F)}{\partial v_x}=\frac{d\ln [f(v_x)]}{d{v}_x}$  (1.8)

Note that the total speed is:

            $v=\sqrt{v_x^2+v_y^2+v_z^2}$          (1.9)

and therefore we can use the chain rule to express the derivative:

            $\frac{\partial \ln (F)}{\partial v_x}=\frac{\partial \ln [F]}{\partial v}\frac{\partial v}{\partial v_x}$  (1.10)

Now

            $\frac{\partial v}{\partial v_x}=\frac{v_x}{v}$           (1.11)

so that:

            $\frac{\partial \ln (F)}{\partial v_x}=\frac{\partial \ln [F]}{\partial v}\frac{v_x}{v}$    (1.12)

Then using a previous equation we can write:

 

            $\frac{\partial \ln (F)}{\partial v_x}=\frac{\partial \ln [F]}{\partial v}\frac{v_x}{v}=\frac{d\ln [f(v_x)]}{d{v}_x}$      (1.13)

which can be written:

            $\frac{\partial \ln [F]}{\partial v}\frac{1}{v}=\frac{1}{v_x}\frac{d\ln [f(v_x)]}{d{v}_x}$       (1.14)

The next ansatz is that all of the (v_x,v_y,v_z) versions of the right hand side of this equation must be equal to the same constant.  We will call this constant -b and let the initial value of f(v_x)=a.  Then we can write:

            $\begin{array}{l}\frac{1}{v_x}\frac{d\ln [f(v_x)]}{d{v}_x}=-b\\ {\displaystyle \int d\ln [f(v_x)]=\ln [f(v_x)]-\ln (a)=-b{\displaystyle \int v_{x}d{v}_x=-\frac{b{v}_x^2}{2}}}\end{array}$  (1.15)

or integrating and taking the exponential of both sides:

            $f(v_x)=a\exp \left(\frac{-b{v}_x^2}{2}\right)$  (1.16)

To evaluate a we must compute the integral of f(v_x) from negative infinity to + infinity:

            $a{\displaystyle {\int }_{-\infty }^{\infty }\exp \left(-\frac{b{v}_x^2}{2}\right)d{v}_x=a\sqrt{\frac{2\pi }{b}}=1}$        (1.17)

so that

            $a=\sqrt{\frac{b}{2\pi }}$          (1.18)

To determine b we have to integrate the energy associated with the v_x component and (and for 3D) set it equal to 1/3 of the total kinetic energy

           

            $\frac{m}{2}\sqrt{\frac{b}{2\pi }}{\displaystyle {\int }_{-\infty }^{\infty }{v}_x^2}\exp \left(-\frac{b{v}_x^2}{2}\right)d{v}_x=\frac{1}{3}{E}_{kinetic}$          (1.19)

Evaluating the integral in equation 1.3 we obtain:

 

            $\frac{m}{2}\sqrt{\frac{b}{2\pi }}\left(\frac{1}{2}\sqrt{\frac{8\pi }{b^3}}\right)=\frac{1}{3}\frac{m{v}^2}{2}$      (1.20)

Solving for b in equation 1.5 we have:

            $\begin{array}{l}\frac{1}{b}=\frac{1}{3}\frac{v^2}{2}\\ b=\frac{3}{\frac{v^2}{2}}\end{array}$           (1.21)

Therefore the equation for f(vx) is actually:

            $f(v_x)=\sqrt{\frac{3}{\pi v^2}}\exp \left(\frac{-3v_x^2}{v^2}\right)$      (1.22)

In 3D the standard definition of the temperature associated with a particle with total translational kinetic energy mv²/2 is

            $\frac{m{v}^2}{2}=\frac{3}{2}kT$     (1.23)

and therefore converting the v² to temperature we have:

            $f(v_x)=\sqrt{\frac{m}{\pi kT}}\exp \left(\frac{-m{v}_x^2}{kT}\right)$    (1.24)