Categories: Physics, Statistics, Thermodynamics.

Maxwell-Boltzmann distribution

The Maxwell-Boltzmann distributions are a set of closely related probability distributions with applications in classical statistical physics.

Velocity vector distribution

In the canonical ensemble (where a fixed-size system can exchange energy with its environment), the probability of a microstate with energy EE is given by the Boltzmann distribution:

f(E)exp ⁣( ⁣ ⁣βE)\begin{aligned} f(E) \:\propto\: \exp\!\big(\!-\! \beta E\big) \end{aligned}

Where β1/kBT\beta \equiv 1 / k_B T. We split E=K+UE = K + U, where KK and UU are the total contributions from the kinetic and potential energies of the system. For NN particles with positions r(r1,...,rN)\va{r} \equiv (\vb{r}_1, ..., \vb{r}_N) and momenta p=(p1,...,pN)\va{p} = (\vb{p}_1, ..., \vb{p}_N), then KK only depends on p\va{p} and UU only on r\va{r}, so the probability of a specific microstate (r,p)(\va{r}, \va{p}) is as follows:

f(r,p)exp ⁣( ⁣ ⁣β(K(p)+U(r)))\begin{aligned} f(\va{r}, \va{p}) \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\va{p}) + U(\va{r}) \big) \Big) \end{aligned}

Since this is classical physics, we can split the exponential (in quantum mechanics, the canonical commutation relation would prevent that):

f(r,p)exp ⁣( ⁣ ⁣βK(p))exp ⁣( ⁣ ⁣βU(r))\begin{aligned} f(\va{r}, \va{p}) \:\propto\: \exp\!\big(\!-\! \beta K(\va{p}) \big) \exp\!\big(\!-\! \beta U(\va{r}) \big) \end{aligned}

Classically, the probability distributions of the momenta and positions are independent:

fK(p)exp ⁣( ⁣ ⁣βK(p))fU(r)exp ⁣( ⁣ ⁣βU(r))\begin{aligned} f_K(\va{p}) \:\propto\: \exp\!\big(\!-\! \beta K(\va{p}) \big) \qquad \qquad f_U(\va{r}) \:\propto\: \exp\!\big(\!-\! \beta U(\va{r}) \big) \end{aligned}

We cannot evaluate fU(r)f_U(\va{r}) further without knowing U(r)U(\va{r}) for a system. We thus turn to fK(p)f_K(\va{p}), and see that the total kinetic energy K(p)K(\va{p}) is simply the sum of the particles’ individual kinetic energies Kn(pn)K_n(\vb{p}_n), which are well-known:

K(p)=n=1NKn(pn)whereKn(pn)=pn22m\begin{aligned} K(\va{p}) = \sum_{n = 1}^N K_n(\vb{p}_n) \qquad \mathrm{where} \qquad K_n(\vb{p}_n) = \frac{|\vb{p}_n|^2}{2 m} \end{aligned}

Consequently, the probability distribution f(px,py,pz)f(p_x, p_y, p_z) for the momentum vector of a single particle is as follows, after normalization:

f(px,py,pz)=(12πmkBT)3/2exp ⁣( ⁣ ⁣(px2+py2+pz2)2mkBT)\begin{aligned} f(p_x, p_y, p_z) = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \bigg) \end{aligned}

We now rewrite this using the velocities vx=px/mv_x = p_x / m, and update the normalization, giving:

f(vx,vy,vz)=(m2πkBT)3/2exp ⁣( ⁣ ⁣m(vx2+vy2+vz2)2kBT)\begin{aligned} \boxed{ f(v_x, v_y, v_z) = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \bigg) } \end{aligned}

This is the Maxwell-Boltzmann velocity vector distribution. Clearly, this is a product of three exponentials, so the velocity in each direction is independent of the others:

f(vx)=m2πkBTexp ⁣( ⁣ ⁣mvx22kBT)\begin{aligned} f(v_x) = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\bigg( \!-\!\frac{m v_x^2}{2 k_B T} \bigg) \end{aligned}

The distribution is thus an isotropic Gaussian with standard deviations given by:

σx=σy=σz=kBTm\begin{aligned} \sigma_x = \sigma_y = \sigma_z = \sqrt{\frac{k_B T}{m}} \end{aligned}

Speed distribution

That was the distribution of the velocities along each axis, but what about the speed v=vv = |\vb{v}|? Because we do not care about the direction of v\vb{v}, only its magnitude, the density of states g(v)g(v) is not constant: it is the rate-of-change of the volume of a sphere of radius vv:

g(v)=ddv(4π3v3)=4πv2\begin{aligned} g(v) = \dv{}{v} \bigg( \frac{4 \pi}{3} v^3 \bigg) = 4 \pi v^2 \end{aligned}

Multiplying the velocity vector distribution by g(v)g(v) and substituting v2=vx2+vy2+vz2v^2 = v_x^2 + v_y^2 + v_z^2 then gives us the Maxwell-Boltzmann speed distribution:

f(v)=4π(m2πkBT)3/2v2exp ⁣( ⁣ ⁣mv22kBT)\begin{aligned} \boxed{ f(v) = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\bigg( \!-\!\frac{m v^2}{2 k_B T} \bigg) } \end{aligned}

Some notable points on this distribution are the most probable speed vmodev_{\mathrm{mode}}, the mean average speed vmeanv_{\mathrm{mean}}, and the root-mean-square speed vrmsv_{\mathrm{rms}}:

f(vmode)=0vmean=0vf(v)dvvrms=(0v2f(v)dv)1/2\begin{aligned} f'(v_\mathrm{mode}) = 0 \qquad \quad v_\mathrm{mean} = \int_0^\infty v \: f(v) \dd{v} \qquad \quad v_\mathrm{rms} = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2} \end{aligned}

Which can be calculated to have the following exact expressions:

vmode=2kBTmvmean=8kBTπmvrms=3kBTm\begin{aligned} \boxed{ v_{\mathrm{mode}} = \sqrt{\frac{2 k_B T}{m}} } \qquad \quad \boxed{ v_{\mathrm{mean}} = \sqrt{\frac{8 k_B T}{\pi m}} } \qquad \quad \boxed{ v_{\mathrm{rms}} = \sqrt{\frac{3 k_B T}{m}} } \end{aligned}

Kinetic energy distribution

Using the speed distribution, we can work out the kinetic energy distribution. Because KK is not proportional to vv, we must do this by demanding that:

f(K)dK=f(v)dv    f(K)=f(v)dvdK\begin{aligned} f(K) \dd{K} = f(v) \dd{v} \quad \implies \quad f(K) = f(v) \dv{v}{K} \end{aligned}

We know that K=mv2/2K = m v^2 / 2, meaning dK=mvdv\dd{K} = m v \dd{v} so the energy distribution f(K)f(K) is:

f(K)=f(v)mv=2mπ(1kBT)3/2vexp ⁣( ⁣ ⁣mv22kBT)\begin{aligned} f(K) = \frac{f(v)}{m v} = \sqrt{\frac{2 m}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} v \exp\!\bigg( \!-\!\frac{m v^2}{2 k_B T} \bigg) \end{aligned}

Substituting v=2K/mv = \sqrt{2 K/m} leads to the Maxwell-Boltzmann kinetic energy distribution:

f(K)=2Kπ(1kBT)3/2exp ⁣( ⁣ ⁣KkBT)\begin{aligned} \boxed{ f(K) = 2 \sqrt{\frac{K}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{K}{k_B T} \bigg) } \end{aligned}


  1. H. Gould, J. Tobochnik, Statistical and thermal physics, 2nd edition, Princeton.