Maxwell–Boltzmann distribution

For a system containing a large number of identical non-interacting, non-relativistic classical particles in themodynamic equilibrium, the fraction of the particles within an infinitesimal element of the three-dimensional velocity space ${\displaystyle d^{3}v}$, centered on a velocity vector of magnitude ${\displaystyle v}$, is given by

$${\displaystyle f(v)~d^{3}v=\left({\frac {m}{2\pi kT}}\right)^{3/2}\,e^{-{\frac {mv^{2}}{2kT}}}~d^{3}v,}$$

where ${\displaystyle m}$ is the particle mass, ${\displaystyle k}$ is the Boltzmann constant, and ${\displaystyle T}$ is thermodynamic temperature. ${\displaystyle f(v)}$ is a probability distribution function, properly normalized so that ${\textstyle \int f(v)\,d^{3}v}$ over all velocities is unity.

The speed probability density functions of the speeds of a few noble gases at a temperature of 298.15 K (25 °C). The y-axis is in s/m so that the area under any section of the curve (which represents the probability of the speed being in that range) is dimensionless.

One can write the element of velocity space as ${\displaystyle d^{3}v=dv_{x}\,dv_{y}\,dv_{z}}$, for velocities in a standard Cartesian coordinate system, or as ${\displaystyle d^{3}v=v^{2}\,dv\,d\Omega }$ in a standard spherical coordinate system, where ${\displaystyle d\Omega }$ is an element of solid angle.

The Maxwellian distribution function for particles moving in only one direction, if this direction is ${\displaystyle x}$, is

$${\displaystyle f(v_{x})~dv_{x}=\left({\frac {m}{2\pi kT}}\right)^{1/2}\,e^{-{\frac {mv_{x}^{2}}{2kT}}}~dv_{x},}$$

which can be obtained by integrating the three-dimensional form given above over ${\displaystyle v_{y}}$ and ${\displaystyle v_{z}}$.

Recognizing the symmetry of ${\displaystyle f(v)}$, one can integrate over solid angle and write a probability distribution of speeds as the function[6]

$${\displaystyle f(v)=\left({\frac {m}{2\pi kT}}\right)^{3/2}\,4\pi v^{2}e^{-{\frac {mv^{2}}{2kT}}}.}$$

This probability density function gives the probability, per unit speed, of finding the particle with a speed near ${\displaystyle v}$. This equation is simply the Maxwell–Boltzmann distribution (given in the infobox) with distribution parameter ${\textstyle a={\sqrt {kT/m}}}$. The Maxwell–Boltzmann distribution is equivalent to the chi distribution with three degrees of freedom and scale parameter ${\textstyle a={\sqrt {kT/m}}}$.

The simplest ordinary differential equation satisfied by the distribution is:

$${\displaystyle kTvf'(v)+f(v)\left(mv^{2}-2kT\right)=0,}$$

$${\displaystyle f(1)={\sqrt {\frac {2}{\pi }}}e^{-{\frac {m}{2kT}}}\left({\frac {m}{kT}}\right)^{3/2}}$$

or in unitless presentation:

$${\displaystyle a^{2}xf'(x)+\left(x^{2}-2a^{2}\right)f(x)=0,}$$

$${\displaystyle f(1)={\frac {{\sqrt {\frac {2}{\pi }}}e^{-{1}/{2a^{2}}}}{a^{3}}}.}$$

With the Darwin–Fowler method of mean values, the Maxwell–Boltzmann distribution is obtained as an exact result.

Simulation of a 2D gas relaxing towards a Maxwell–Boltzmann speed distribution

For particles confined to move in a plane, the speed distribution is given by

$${\displaystyle P(s<|{\vec {v}}|<s+ds)={\frac {ms}{kT}}\exp \left(-{\frac {ms^{2}}{2kT}}\right)ds}$$

This distribution is used for describing systems in equilibrium. However, most systems do not start out in their equilibrium state. The evolution of a system towards its equilibrium state is governed by the Boltzmann equation. The equation predicts that for short range interactions, the equilibrium velocity distribution will follow a Maxwell–Boltzmann distribution. To the right is a molecular dynamics (MD) simulation in which 900 hard sphere particles are constrained to move in a rectangle. They interact via perfectly elastic collisions. The system is initialized out of equilibrium, but the velocity distribution (in blue) quickly converges to the 2D Maxwell–Boltzmann distribution (in orange).

The Maxwell–Boltzmann distribution corresponding to the solar atmosphere. Particle masses are one proton mass, ${\displaystyle m=1{\text{ amu}}=1.67\times 10^{-24}{\text{ g }}}$, and the temperature is the effective temperature of the sun's photosphere, ${\displaystyle T=5800{\text{ K }}}$. ${\displaystyle {\tilde {V}},{\bar {V}},{\text{ and }}V_{\text{rms}}}$ mark the most probable, mean, and root mean square velocities, respectively. Their values are ${\displaystyle {\tilde {V}}\approx 9.79{\text{ km/s, }}}$ ${\displaystyle {\bar {V}}\approx 11.05{\text{ km/s, }}}$ and ${\displaystyle V_{\text{rms}}\approx 12.00{\text{ km/s }}}$.

The mean speed ${\displaystyle \langle v\rangle }$, most probable speed (mode) v_p, and root-mean-square speed ${\textstyle {\sqrt {\langle v^{2}\rangle }}}$ can be obtained from properties of the Maxwell distribution.

This works well for nearly ideal, monatomic gases like helium, but also for molecular gases like diatomic oxygen. This is because despite the larger heat capacity (larger internal energy at the same temperature) due to their larger number of degrees of freedom, their translational kinetic energy (and thus their speed) is unchanged.[7]

• The most probable speed, v_p, is the speed most likely to be possessed by any molecule (of the same mass m) in the system and corresponds to the maximum value or the mode of f(v). To find it, we calculate the derivative df/dv, set it to zero and solve for v:
  $${\displaystyle {\frac {df(v)}{dv}}=-8\pi \left({\frac {m}{2\pi kT}}\right)^{3/2}\ v\ e^{-{\frac {mv^{2}}{2kT}}}\left({\frac {mv^{2}}{2kT}}-1\right)=0}$$

  
  with the solution:
  $${\displaystyle {\frac {mv_{p}^{2}}{2kT}}=1}$$

  
  $${\displaystyle v_{p}={\sqrt {\frac {2kT}{m}}}={\sqrt {\frac {2RT}{M}}}}$$

  
  R is the gas constant and M is molar mass of the substance, and thus may be calculated as a product of particle mass, m, and Avogadro constant, N_A:
  $${\displaystyle M=mN_{A}}$$

  
  For diatomic nitrogen (N₂, the primary component of air)[8] at room temperature (300 K), this gives
  $${\displaystyle v_{p}\approx {\sqrt {\frac {2\cdot 8.31\ {\text{J}}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\ 300\ {\text{K}}}{0.028\ {\text{kg}}\cdot {\text{mol}}^{-1}}}}\approx 422\ {\text{m/s}}.}$$

  
• The mean speed is the expected value of the speed distribution, setting ${\textstyle b={\frac {1}{2a^{2}}}={\frac {m}{2kT}}}$:
  $${\displaystyle {\begin{aligned}\langle v\rangle &=\int _{0}^{\infty }v\,f(v)\,dv\\&=4\pi \left({\frac {b}{\pi }}\right)^{\frac {3}{2}}\int _{0}^{\infty }v^{3}e^{-bv^{2}}dv\\&=4\pi \left({\frac {b}{\pi }}\right)^{\frac {3}{2}}{\frac {1}{2b^{2}}}={\sqrt {\frac {4}{\pi b}}}\\&={\sqrt {\frac {8kT}{\pi m}}}={\sqrt {\frac {8RT}{\pi M}}}={\frac {2}{\sqrt {\pi }}}v_{p}\end{aligned}}}$$

  
• The mean square speed ${\displaystyle \langle v^{2}\rangle }$ is the second-order raw moment of the speed distribution. The "root mean square speed" ${\displaystyle v_{\mathrm {rms} }}$ is the square root of the mean square speed, corresponding to the speed of a particle with median kinetic energy, setting ${\textstyle b={\frac {1}{2a^{2}}}={\frac {m}{2kT}}}$:
  $${\displaystyle {\begin{aligned}v_{\mathrm {rms} }&={\sqrt {\langle v^{2}\rangle }}=\left(\int _{0}^{\infty }v^{2}\,f(v)\,dv\right)^{1/2}\\&=\left(4\pi \left({\frac {b}{\pi }}\right)^{3/2}\int _{0}^{\infty }v^{4}e^{-bv^{2}}dv\right)^{1/2}\\&=\left(4\pi \left({\frac {b}{\pi }}\right)^{3/2}{\frac {3}{8}}{\sqrt {\frac {\pi }{b^{5}}}}\right)^{1/2}=\left({\frac {3}{2b}}\right)^{1/2}\\[4pt]&={\sqrt {\frac {3kT}{m}}}={\sqrt {\frac {3RT}{M}}}={\sqrt {\frac {3}{2}}}v_{p}\end{aligned}}}$$

  

In summary, the typical speeds are related as follows:

$${\displaystyle v_{p}\approx 88.6\%\ \langle v\rangle <\langle v\rangle <108.5\%\ \langle v\rangle \approx v_{\mathrm {rms} }.}$$

The root mean square speed is directly related to the speed of sound c in the gas, by

$${\displaystyle c={\sqrt {\frac {\gamma }{3}}}\ v_{\mathrm {rms} }={\sqrt {\frac {f+2}{3f}}}\ v_{\mathrm {rms} }={\sqrt {\frac {f+2}{2f}}}\ v_{p},}$$

where ${\textstyle \gamma =1+{\frac {2}{f}}}$ is the adiabatic index, f is the number of degrees of freedom of the individual gas molecule. For the example above, diatomic nitrogen (approximating air) at 300 K, ${\displaystyle f=5}$[9] and

$${\displaystyle c={\sqrt {\frac {7}{15}}}v_{\mathrm {rms} }\approx 68\%\ v_{\mathrm {rms} }\approx 84\%\ v_{p}\approx 353\ \mathrm {m/s} ,}$$

the true value for air can be approximated by using the average molar weight of air (29 g/mol), yielding 347 m/s at 300 K (corrections for variable humidity are of the order of 0.1% to 0.6%).

The average relative velocity

$${\displaystyle v_{\rm {rel}}\equiv \langle |{\vec {v}}_{1}-{\vec {v}}_{2}|\rangle =\int \!d^{3}v_{1}\,d^{3}v_{2}\left|{\vec {v}}_{1}-{\vec {v}}_{2}\right|f({\vec {v}}_{1})f({\vec {v}}_{2})={\frac {4}{\sqrt {\pi }}}{\sqrt {\frac {kT}{m}}}={\sqrt {2}}\langle v\rangle }$$

where the three-dimensional velocity distribution is

$${\displaystyle f({\vec {v}})\equiv {\frac {1}{\left(2\pi kT/m\right)^{3/2}}}e^{-{\frac {1}{2}}m{\vec {v}}^{2}/kT}.}$$

The integral can easily be done by changing to coordinates ${\displaystyle {\vec {u}}={\vec {v}}_{1}-{\vec {v}}_{2}}$ and ${\displaystyle {\vec {U}}={\frac {{\vec {v}}_{1}+{\vec {v}}_{2}}{2}}.}$

In n-dimensional space, Maxwell–Boltzmann distribution becomes:

$${\displaystyle f(v)~d^{n}v=\left({\frac {m}{2\pi kT}}\right)^{n/2}\,e^{-{\frac {m|v|^{2}}{2kT}}}~d^{n}v}$$

Speed distribution becomes:

$${\displaystyle f(v)~dv={\text{const.}}\times e^{-{\frac {mv^{2}}{2kT}}}\times v^{n-1}~dv}$$

The following integral result is useful:

$${\displaystyle {\begin{aligned}\int _{0}^{+\infty }v^{a}e^{-{\frac {mv^{2}}{2kT}}}dv&=\left[{\frac {2kT}{m}}\right]^{(a+1)/2}\int _{0}^{+\infty }e^{-x}x^{\frac {a}{2}}dx^{\frac {1}{2}}\\&=\left[{\frac {2kT}{m}}\right]^{(a+1)/2}\int _{0}^{+\infty }e^{-x}x^{\frac {a}{2}}{\frac {x^{-{\frac {1}{2}}}}{2}}dx\\&=\left[{\frac {2kT}{m}}\right]^{(a+1)/2}{\frac {\Gamma ({\frac {a+1}{2}})}{2}}\end{aligned}}}$$

where ${\displaystyle \Gamma (z)}$ is the Gamma function. This result can be used to calculate the moments of speed distribution function:

$${\displaystyle {\begin{aligned}\langle v\rangle &={\frac {\displaystyle \int _{0}^{+\infty }v\cdot v^{n-1}e^{-{\frac {mv^{2}}{2kT}}}dv}{\displaystyle \int _{0}^{+\infty }v^{n-1}e^{-{\frac {mv^{2}}{2kT}}}dv}}\\[4pt]&=\left[{\frac {2kT}{m}}\right]^{1/2}{\frac {\Gamma \left({\frac {n+1}{2}}\right)}{\Gamma \left({\frac {n}{2}}\right)}}\end{aligned}}}$$

which is the mean speed itself ${\textstyle v_{\text{avg}}=\langle v\rangle =\left[{\frac {2kT}{m}}\right]^{1/2}{\frac {\Gamma \left({\frac {n+1}{2}}\right)}{\Gamma \left({\frac {n}{2}}\right)}}}$.

$${\displaystyle {\begin{aligned}\langle v^{2}\rangle &={\frac {\displaystyle \int _{0}^{+\infty }v^{2}\cdot v^{n-1}e^{-{\frac {mv^{2}}{2kT}}}dv}{\displaystyle \int _{0}^{+\infty }v^{n-1}e^{-{\frac {mv^{2}}{2kT}}}dv}}\\&=\left[{\frac {2kT}{m}}\right]{\frac {\Gamma ({\frac {n+2}{2}})}{\Gamma ({\frac {n}{2}})}}\\&=\left[{\frac {2kT}{m}}\right]{\frac {n}{2}}={\frac {nkT}{m}}\end{aligned}}}$$

which gives root-mean-square speed ${\textstyle v_{\text{rms}}={\sqrt {\langle v^{2}\rangle }}=\left[{\frac {nkT}{m}}\right]^{1/2}}$.

The derivative of speed distribution function:

$${\displaystyle {\frac {df(v)}{dv}}={\text{const.}}\times \ e^{-{\frac {mv^{2}}{2kT}}}\left(-{\frac {mv}{kT}}v^{n-1}+(n-1)v^{n-2}\right)=0}$$

This yields the most probable speed (mode) ${\textstyle v_{\text{p}}=\left[{\frac {(n-1)kT}{m}}\right]^{1/2}}$.

• Quantum Boltzmann equation
• Maxwell–Boltzmann statistics
• Maxwell–Jüttner distribution
• Boltzmann distribution
• Boltzmann factor
• Rayleigh distribution
• Kinetic theory of gases

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Wikimedia Commons has media related to Maxwell–Boltzmann distributions.

• "The Maxwell Speed Distribution" from The Wolfram Demonstrations Project at Mathworld