Maxwell-Boltzmann distribution

Maxwell-Boltzmann distribution

parameter	${\displaystyle a>0\,}$
Domain of definition	${\displaystyle x\in [0;\infty )}$
Probability density	${\displaystyle {\sqrt {\frac {2}{\pi }}}{\frac {x^{2}e^{-x^{2}/(2a^{2})}}{a^{3}}}}$
Cumulative distribution function	${\displaystyle \operatorname {erf} \left({\frac {x}{{\sqrt {2}}a}}\right)-{\sqrt {\frac {2}{\pi }}}{\frac {xe^{-x^{2}/(2a^{2})}}{a}}}$
Expected value	${\displaystyle \mu =2a{\sqrt {\frac {2}{\pi }}}}$
mode	${\displaystyle {\sqrt {2}}a}$
Variance	${\displaystyle \sigma ^{2}={\frac {a^{2}(3\pi -8)}{\pi }}}$
Crookedness	${\displaystyle \gamma _{1}={\frac {2{\sqrt {2}}(16-5\pi )}{(3\pi -8)^{3/2}}}}$
Bulge	${\displaystyle \gamma _{2}=-4{\frac {96-40\pi +3\pi ^{2}}{(3\pi -8)^{2}}}}$
Entropy (in nats )	${\displaystyle \ln \left(a{\sqrt {2\pi }}\right)+\gamma -{\frac {1}{2}}}$ ( : Euler-Mascheroni constant ) ${\displaystyle \gamma }$

The Maxwell-Boltzmann distribution or Maxwell's velocity distribution is a probability density of statistical physics and plays an important role in thermodynamics , especially in kinetic gas theory . It describes the statistical distribution of the magnitude of the particle velocities in an ideal gas . It is named after James Clerk Maxwell and Ludwig Boltzmann , who first derived it in 1860. It results from the Boltzmann statistics . ${\displaystyle v=|{\vec {v}}|}$

Because of the simplifying assumption of an ideal gas, the velocity distribution of the particles in a real gas shows deviations. However, at low density and high temperature, the Maxwell-Boltzmann distribution is sufficient for most considerations.

Derivation of the velocity distribution in the kinetic gas theory

Derivation with the help of the Boltzmann factor

The kinetic energy of a particle state in the ideal gas is through

${\displaystyle E_{\text{kin}}={\frac {mv^{2}}{2}}}$

given, and the probability that it is occupied by a particle in the thermodynamic state of equilibrium of the particle system is given by the Boltzmann factor

${\displaystyle W(E)\propto e^{-{\frac {E_{\text{kin}}}{k_{\mathrm {B} }T}}}}$.

This contains the mass of the particle, the Boltzmann constant and the absolute temperature . The question is about the proportion of molecules with the amount of speed in an interval . The corresponding density of states is to be determined from the basic assumption that the density of states in the three-dimensional space of the velocity components is constant. According to all states of the same kinetic energy have the distance from the origin, so here they fill a spherical surface of the size . The volume element then belongs to the interval . Consequently, the sought-after proportion of molecules is equal to the product of the volume element, the Boltzmann factor, which is constant for the entire volume element, and a constant normalization factor : ${\displaystyle m}$ ${\displaystyle k_{\mathrm {B} }}$ ${\displaystyle T}$ ${\displaystyle [v,\,v{\mathord {+}}dv]}$ ${\displaystyle v_{x},\,v_{y},\,v_{z}}$ ${\displaystyle v^{2}{\mathord {=}}v_{x}^{2}{\mathord {+}}v_{y}^{2}{\mathord {+}}v_{z}^{2}}$ ${\displaystyle v}$ ${\displaystyle 4\pi v^{2}}$ ${\displaystyle [v,\,v{\mathord {+}}dv]}$ ${\displaystyle 4\pi v^{2}dv}$ ${\displaystyle c}$

${\displaystyle p(v)\,\mathrm {d} v=c\cdot 4\pi v^{2}\cdot e^{-{\frac {mv^{2}}{2\,k_{\mathrm {B} }T}}}\mathrm {d} v}$

The normalization factor results from the fact that the integral of the probability density has the value 1. ${\displaystyle c}$ ${\displaystyle \int _{0}^{\infty }p(v)\,\mathrm {d} v}$

Derivation using the normal distribution of the components of the velocity

According to the kinetic gas theory , not all gas particles in an ideal gas move at temperature (in Kelvin ) at the same speed , but randomly distributed at different speeds. No spatial direction is preferred here. Mathematically this can be formulated in such a way that the components of the velocity vector of the gas particles of the mass are independent of each other and normally distributed , with the parameters ${\displaystyle T}$ ${\displaystyle {\vec {v}}}$ ${\displaystyle m}$

Average speed: and the spread of the speeds ${\displaystyle \mu =0,\,}$ ${\displaystyle \sigma ={\sqrt {\frac {k_{\mathrm {B} }T}{m}}}}$

The density of the distribution of in the three-dimensional velocity space , here denoted by, results as the product of the distributions of the three components: ${\displaystyle {\vec {v}}}$ ${\displaystyle {\tilde {p}}({\vec {v}})}$

${\displaystyle {\tilde {p}}({\vec {v}})=\left({\frac {m}{2\pi k_{\mathrm {B} }T}}\right)^{3/2}\mathrm {e} ^{-{\frac {m|{\vec {v}}|^{2}}{2k_{\mathrm {B} }T}}}}$

To derive the Maxwell-Boltzmann distribution has to all particles with the same velocity magnitude integrate (or clearly "add up" this). These lie on a spherical shell with a radius and infinitesimal thickness around the speed 0: ${\displaystyle p(v)}$ ${\displaystyle v}$ ${\displaystyle \mathrm {d} v\rightarrow 0}$

${\displaystyle p(v)\mathrm {d} v=\iiint _{v<|{\vec {v}}'|<v+\mathrm {d} v}{\tilde {p}}({\vec {v}}')\;\mathrm {d} ^{3}{\vec {v}}'}$

The above-mentioned integral denotes all vectors with amounts in the interval . Since the definition of only includes the squared amount of the velocities (see definition above), which does not change in the infinitesimal interval , the integral can easily be transformed: ${\displaystyle \iiint _{v<|{\vec {v}}'|<v+\mathrm {d} v}d^{3}{\vec {v}}'}$ ${\displaystyle {\vec {v}}'}$ ${\displaystyle (v,v+\mathrm {d} v)}$ ${\displaystyle {\tilde {p}}({\vec {v}}')}$ ${\displaystyle (v,v+\mathrm {d} v)}$

${\displaystyle p(v)\mathrm {d} v=\left({\frac {m}{2\pi k_{\mathrm {B} }T}}\right)^{3/2}\mathrm {e} ^{-{\frac {mv^{2}}{2k_{\mathrm {B} }T}}}\iiint _{v<|{\vec {v}}'|<v+\mathrm {d} v}\;\mathrm {d} ^{3}{\vec {v}}'}$

Here only the simple volume integral remains to be solved. It just gives the volume of the infinitesimal spherical shell and we get the Maxwell-Boltzmann distribution we are looking for: ${\displaystyle 4\pi v^{2}\cdot \mathrm {d} v}$

${\displaystyle p(v)=4\pi \left({\frac {m}{2\pi k_{\mathrm {B} }T}}\right)^{3/2}v^{2}\exp \left(-{\frac {mv^{2}}{2k_{\mathrm {B} }T}}\right)}$

Significance and scope

Conclusions from the equations

Substance dependence of the velocity distribution at 0 ° C for hydrogen (H ₂ ), helium (He) and nitrogen (N ₂ )

Temperature dependence of the velocity distribution for nitrogen

• From the above equations it follows that the proportion of particles in the speed interval is directly proportional to itself as long as it remains constant. So if you increase slightly or include more speeds in the interval, under the additional assumption that temperature and molar mass are constant, then the number of particles in it increases proportionally with the exception of small deviations . In other words: the distribution function is differentiable. ${\displaystyle f}$ ${\displaystyle \Delta \nu }$ ${\displaystyle \Delta \nu }$ ${\displaystyle F(\nu )}$ ${\displaystyle \Delta \nu }$ ${\displaystyle \Delta \nu }$
• The distribution function has a decreasing exponential function of the form with . Since the expression is directly proportional to the square of the particle speed at constant temperature and constant molar mass , it can be concluded from this that the exponential function and thus to a limited extent also the proportion of molecules becomes very small for high speeds and accordingly very large for low speeds (For the exact connection, see the images on the right). ${\displaystyle e^{-x}}$ ${\displaystyle x=Mv^{2}/2RT}$ ${\displaystyle x}$ ${\displaystyle v^{2}}$
• For gases with a large molar mass , assuming a constant temperature, the expression also becomes very large and the exponential function consequently decreases faster. This means that the probability of encountering heavy molecules at high speeds is very small and accordingly very high for lighter molecules with a low molar mass (see figure above right). ${\displaystyle M}$ ${\displaystyle x}$
• In the opposite case of a high temperature and a constant molar mass, the expression becomes very small and the exponential function accordingly approaches zero more slowly with increasing speed. At a very high temperature, the proportion of fast particles is therefore greater than at a lower temperature (see figure below right). ${\displaystyle x}$
• The lower the speed, the more the quadratic expression decreases outside of the exponential function. This means that the proportion of faster molecules also decreases faster at low speeds than the speed itself, but in return that it also increases quadratically with an increase in speed. ${\displaystyle v^{2}}$

All other quantities mean that the proportion of particles at a certain speed always moves in the interval between zero and one ( ). The two figures on the right illustrate the dependence of the Maxwell-Boltzmann distribution on particle mass and temperature of the gas. As the temperature rises , the average speed increases and the distribution becomes wider at the same time. With increasing particle mass, however, the average speed decreases and the speed distribution becomes narrower at the same time. This relationship between particle speed and temperature or particle speed and particle mass / molar mass can here also be described quantitatively. See also the section square average speed . ${\displaystyle [0,1]}$ ${\displaystyle T}$ ${\displaystyle m_{M}}$

Significance for thermodynamics

The Maxwell-Boltzmann distribution explains, for example, the process of evaporation. For example, damp laundry can dry at temperatures of 20 ° C, since there is a small proportion of molecules with the required high speed in this distribution curve that can detach from the liquid structure. So there will always be some molecules even at low temperatures that are fast enough to overcome the forces of attraction by their neighbors and from the liquid or solid aggregate state transition to the gaseous state of what is known as evaporation or sublimation referred. Conversely, among the comparatively fast particles of the gas there are always some that do not have sufficient speed and therefore change again from the gaseous to the liquid or solid state, which is called condensation or resublimation . These processes are summarized under the term phase change , whereby a dynamic equilibrium is established between particles that enter the gas phase and particles that exit the gas phase, provided there are no external disturbances . This is the subject of investigation in equilibrium thermodynamics , which is why it is also called thermodynamic equilibrium . The particles of the gaseous phase exert a pressure in the state of equilibrium , which is called the saturation vapor pressure . The phase behavior of substances is shown graphically in their phase diagram .

See also equation of state , fundamental equation , thermodynamic potential , ideal gas , real gas , triple , Critical Point

Particle velocities

With all distributions it is assumed that a reference point is chosen that does not move, otherwise there would be no symmetry of the velocity distribution and the gas mass moves as a whole.

Most likely speed

The most likely speed

${\displaystyle {\hat {v}}={\sqrt {\frac {2k_{\mathrm {B} }T}{m}}}={\sqrt {\frac {2RT}{M}}}}$

is the speed at which the density function has its maximum value. It can be calculated from the claim . is the particle mass and is the molar mass of the substance. ${\displaystyle p(v)}$ ${\displaystyle {\frac {{\text{d}}p(v)}{{\text{d}}v}}=0}$ ${\displaystyle m}$ ${\displaystyle M}$

Average speed

The mean speed is the average ${\displaystyle {\bar {v}}}$

${\displaystyle {\bar {v}}:={\frac {v_{1}+v_{2}+v_{3}+\ldots +v_{N}}{N}}}$

Here is the total number of particles and the ( ) their individual speeds. If one summarizes the particles with the same speed in each case, the result is ${\displaystyle N}$ ${\displaystyle v_{n}}$ ${\displaystyle n=1,2,3,\ldots ,N}$

${\displaystyle {\bar {v}}=\int _{0}^{\infty }v\,p(v)\,{\text{d}}v}$

The solution of the integral is:

${\displaystyle {\bar {v}}={\sqrt {\frac {8k_{\mathrm {B} }T}{\pi m}}}={\sqrt {\frac {8RT}{\pi M}}}}$

Square mean speed

The root mean square speed is defined by: ${\displaystyle {\sqrt {\overline {v^{2}}}}}$

${\displaystyle {\sqrt {\overline {v^{2}}}}:={\sqrt {\frac {v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+\ldots +v_{N}^{2}}{N}}}}$

The following equation of state results from the kinetic gas theory :

${\displaystyle pV={\frac {1}{3}}Nm{\overline {v^{2}}}}$

The empirically determined equation of state of ideal gases is:

${\displaystyle pV=Nk_{\mathrm {B} }T}$

If you equate the expression , you get: ${\displaystyle pV}$

${\displaystyle {\frac {1}{3}}Nm{\overline {v^{2}}}=Nk_{\mathrm {B} }T}$

When rearranged according to the root , you finally get: ${\displaystyle {\overline {v^{2}}}}$

${\displaystyle {\sqrt {\overline {v^{2}}}}={\sqrt {\frac {3k_{\mathrm {B} }T}{m}}}={\sqrt {\frac {3RT}{M}}}}$

The root mean square of the velocity of the gas particles is thus directly proportional to the square root of the temperature, provided that the molecular mass does not change (e.g. due to a chemical reaction). A doubling of the temperature on the Kelvin scale leads to an increase in the square mean speed by the factor . Conversely, in this way the temperature can be defined by the kinetic gas theory. ${\displaystyle {\sqrt {2}}\approx 1{,}4}$

The same result is obtained also when substituted in the following equation, and then from 0 to integrated : ${\displaystyle p(v)}$ ${\displaystyle \infty }$

${\displaystyle {\sqrt {\overline {v^{2}}}}={\sqrt {\int _{0}^{\infty }v^{2}\,p(v)\,dv}}}$

The root mean square speed is also a measure of the mean kinetic energy ( E _kin ) of the molecules:

${\displaystyle {\overline {E_{\mathrm {kin} }}}={\frac {1}{2}}m{\overline {v^{2}}}={\frac {3}{2}}k_{\mathrm {B} }T}$

This statement can also be obtained using the uniform distribution theorem , since it is an ensemble mean value for a gas particle with three degrees of freedom. ${\displaystyle N=3}$

Harmonic mean

For the purposes of peak times, etc., you need a further mean, called the harmonic mean . The harmonic mean is defined by: ${\displaystyle {\breve {v}}}$

${\displaystyle {\frac {1}{\breve {v}}}:={\frac {1}{N}}\cdot \left({\frac {1}{v_{1}}}+{\frac {1}{v_{2}}}+{\frac {1}{v_{3}}}+\ldots +{\frac {1}{v_{N}}}\right)}$

Here ( ) are the individual velocities of the particles and their total number . ${\displaystyle v_{n}}$ ${\displaystyle n=1,2,3,\ldots ,N}$ ${\displaystyle N}$

${\displaystyle {\frac {1}{\breve {v}}}=\int _{0}^{\infty }{\frac {p(v)}{v}}\,{\text{d}}v}$

By substituting and and integrating one gets: ${\displaystyle {\frac {mv^{2}}{2k_{\mathrm {B} }T}}=z}$ ${\displaystyle v~dv={\frac {k_{\mathrm {B} }T}{m}}dz}$

${\displaystyle {\frac {1}{\breve {v}}}={\sqrt {\frac {2}{\pi }}}\cdot {\sqrt {\frac {m}{k_{\mathrm {B} }T}}}={\sqrt {\frac {2m}{\pi k_{\mathrm {B} }T}}}}$

or

${\displaystyle {\breve {v}}={\sqrt {\frac {\pi k_{\mathrm {B} }T}{2m}}}={\sqrt {\frac {\pi RT}{2{\text{M}}}}}}$

Relationships between speeds

Maxwell-Boltzmann velocity distribution for nitrogen

The picture on the right shows the Maxwell-Boltzmann speed distribution for nitrogen (N ₂ ) at three different temperatures. The most likely speed and the mean speed are also shown. It always applies that the most likely speed is lower than the mean speed. In general: ${\displaystyle {\hat {v}}}$ ${\displaystyle {\bar {v}}}$

${\displaystyle {\hat {v}}<{\bar {v}}<{\sqrt {\overline {v^{2}}}}}$

The relationship between the speeds results from:

${\displaystyle {\bar {v}}={\sqrt {8 \over {3\pi }}}\cdot {\sqrt {\overline {v^{2}}}}\approx 0{,}921\cdot {\sqrt {\overline {v^{2}}}}}$

Conversion factors between the different particle speeds (rounded)

to ↓ \ from →	${\displaystyle {\hat {v}}}$	${\displaystyle {\bar {v}}}$	${\displaystyle {\sqrt {\overline {v^{2}}}}}$	${\displaystyle {\breve {v}}}$
${\displaystyle {\hat {v}}}$	1	0.886	0.816	1.128
${\displaystyle {\bar {v}}}$	1.128	1	0.921	1.273
${\displaystyle {\sqrt {\overline {v^{2}}}}}$	1.225	1.085	1	1,382
${\displaystyle {\breve {v}}}$	0.886	0.785	0.724	1

Example values ​​for the different particle velocities

T \ v	${\displaystyle {\hat {v}}}$	${\displaystyle {\bar {v}}}$	${\displaystyle {\sqrt {\overline {v^{2}}}}}$	${\displaystyle {\breve {v}}}$
100 K (−173.15 ° C)	243.15 m / s	274.36 m / s	297.79 m / s	215.43 m / s
300 K (26.85 ° C)	421.15 m / s	475.20 m / s	515.78 m / s	373.14 m / s
800 K (526.85 ° C)	687.74 m / s	776.02 m / s	842.29 m / s	609.34 m / s

Conversion factors between the different particle speeds (exactly)

to ↓ \ from →	${\displaystyle {\hat {v}}}$	${\displaystyle {\bar {v}}}$	${\displaystyle {\sqrt {\overline {v^{2}}}}}$	${\displaystyle {\breve {v}}}$
${\displaystyle {\hat {v}}}$	1	${\displaystyle {\frac {\sqrt {\pi }}{2}}}$	${\displaystyle {\sqrt {\frac {2}{3}}}}$	${\displaystyle {\frac {2}{\sqrt {\pi }}}}$
${\displaystyle {\bar {v}}}$	${\displaystyle {\frac {2}{\sqrt {\pi }}}}$	1	${\displaystyle {\sqrt {\frac {8}{3\pi }}}}$	${\displaystyle {\frac {4}{\pi }}}$
${\displaystyle {\sqrt {\overline {v^{2}}}}}$	${\displaystyle {\sqrt {\frac {3}{2}}}}$	${\displaystyle {\sqrt {\frac {3\pi }{8}}}}$	1	${\displaystyle {\sqrt {\frac {6}{\pi }}}}$
${\displaystyle {\breve {v}}}$	${\displaystyle {\frac {\sqrt {\pi }}{2}}}$	${\displaystyle {\frac {\pi }{4}}}$	${\displaystyle {\sqrt {\frac {\pi }{6}}}}$	1
${\displaystyle {\sqrt {\frac {k_{\mathrm {B} }T}{m}}}}$	${\displaystyle {\sqrt {2}}}$	${\displaystyle {\sqrt {\frac {8}{\pi }}}}$	${\displaystyle {\sqrt {3}}}$	${\displaystyle {\sqrt {\frac {\pi }{2}}}}$

Derivation in the canonical ensemble

The Maxwell-Boltzmann distribution can be derived using the methods of statistical physics. One considers a particle system with the Hamilton function ${\displaystyle N}$

${\displaystyle H=\sum _{i=1}^{N}{\frac {p_{i}^{2}}{2m}}+U({\vec {x}}_{1},\ldots ,{\vec {x}}_{N})}$

For the derivation only the assumption is made that the potential is conservative, i.e. independent of the. Therefore, the following derivation also applies to real gases . ${\displaystyle U}$ ${\displaystyle p_{i}}$

The system is in the canonical state with phase space density

${\displaystyle w={\frac {1}{Z(\beta )}}e^{-\beta H({\vec {x}}_{1},{\vec {p}}_{1},\,\ldots \,,{\vec {x}}_{N},{\vec {p}}_{N})}}$

and the canonical partition function

${\displaystyle Z(\beta )=\int _{\mathbb {R} ^{6N}}e^{-\beta H({\vec {x}}_{1},{\vec {p}}_{1},\,\ldots \,,{\vec {x}}_{N},{\vec {p}}_{N})}{\text{d}}\tau }$   With   ${\displaystyle {\text{d}}\tau ={\frac {1}{N!(2\pi \hbar )^{3N}}}\;{\text{d}}^{3}x_{1}{\text{d}}^{3}p_{1}\,\ldots \,{\text{d}}^{3}x_{N}{\text{d}}^{3}p_{N}}$

The parameter is proportional to the inverse temperature ${\displaystyle \beta }$

${\displaystyle \beta ={\frac {1}{k_{\rm {B}}T}}}$

The expectation value of a classical observable is given by

${\displaystyle \langle o\rangle =\int _{\mathbb {R} ^{6N}}o\,w\,{\text{d}}\tau }$

The following applies to the transformation of probability densities: A random variable and a probability density and a mapping are given . Then is the probability density of the random variable . ${\displaystyle X}$ ${\displaystyle {\mathcal {P}}_{X}:\;\mathbb {R} ^{n}\rightarrow \mathbb {R} }$ ${\displaystyle Y:\;\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}}$ ${\displaystyle {\mathcal {P}}_{Y}(y)=\int _{\mathbb {R} ^{n}}\delta (y-Y(x))\,{\mathcal {P}}_{X}(x)\,{\text{d}}x=\langle \delta (y-Y(x))\rangle }$ ${\displaystyle Y}$

Now we can calculate the probability density for the momentum of any particle in the system. According to the above transformation theorem applies: ${\displaystyle {\vec {p}}}$ ${\displaystyle j\in \left\{1,\ldots ,N\right\}}$

${\displaystyle {\begin{aligned}{\mathcal {P}}_{{\vec {p}}_{j}}({\vec {p}})&=\langle \delta ({\vec {p}}_{j}-{\vec {p}})\rangle =\int _{\mathbb {R} ^{6N}}\delta ({\vec {p}}_{j}-{\vec {p}})\,w\,{\text{d}}\tau ={\frac {1}{Z(\beta )}}\int _{\mathbb {R} ^{6N}}\delta ({\vec {p}}_{j}-{\vec {p}})\,e^{-\beta H}\,{\text{d}}\tau \\&={\frac {\int _{\mathbb {R} ^{6N}}{\delta }({\vec {p}}_{j}-{\vec {p}})\,e^{-\beta H}\,{\text{d}}\tau }{\int _{\mathbb {R} ^{6N}}{e^{-\beta H}}\,{\text{d}}\tau }}\\&={\frac {\int _{\mathbb {R} ^{3N}}{e^{-\beta U}\,{\text{d}}^{3}x_{1}\,\ldots \,{\text{d}}^{3}x_{N}\;}\ \int _{\mathbb {R} ^{3N}}{\delta ({\vec {p}}_{j}-{\vec {p}})\,e^{-{\frac {\beta }{2m}}\sum _{i=1}^{N}{p_{i}^{2}}}\,{\text{d}}p_{1}\,\ldots \,{\text{d}}^{3}p_{N}\;}}{\int _{\mathbb {R} ^{3N}}{e^{-\beta U}\,{\text{d}}^{3}x_{1}\,\ldots \,{\text{d}}^{3}x_{N}\;}\ \int _{\mathbb {R} ^{3N}}{e^{-{\frac {\beta }{2m}}\sum _{i=1}^{N}{p_{i}^{2}}}\,{\text{d}}^{3}p_{1}\,\ldots \,{\text{d}}^{3}p_{N}\;}}}\end{aligned}}}$

All location integrations can be shortened, as can all momentum integrations for . Thus only the integration remains. ${\displaystyle i\neq j}$ ${\displaystyle p_{j}}$

${\displaystyle {\mathcal {P}}_{{\vec {p}}_{j}}({\vec {p}})={\frac {\int _{\mathbb {R} ^{3}}\delta ({\vec {p}}_{j}-{\vec {p}})\,e^{-{\frac {\beta }{2m}}p_{j}^{2}}\,{\text{d}}^{3}p_{j}}{\int _{\mathbb {R} ^{3}}e^{-{\frac {\beta }{2m}}p_{j}^{2}}{\text{d}}^{3}p_{j}}}}$

The convolution property of the delta distribution is used in the numerator to evaluate this expression

${\displaystyle \int _{\mathbb {R} ^{3}}{\delta }({\vec {x}}-{\vec {x}}_{0})f({\vec {x}})\,{\text{d}}^{3}x=f({\vec {x}}_{0})}$

In the denominator one integrates via a Gaussian function ; the integration in three dimensions can be reduced to a one-dimensional integral with ${\displaystyle x^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$

${\displaystyle \int _{\mathbb {R} ^{3}}{e^{-ax^{2}}{\text{d}}^{3}x}=\int _{\mathbb {R} ^{3}}{e^{-ax_{1}^{2}}e^{-ax_{2}^{2}}e^{-ax_{3}^{2}}{\text{d}}x_{1}{\text{d}}x_{2}{\text{d}}x_{3}}=\left(\int _{-\infty }^{\infty }{e^{-ax^{2}}{\text{d}}x}\right)^{3}\,=\left({\frac {\pi }{a}}\right)^{\frac {3}{2}}}$

The probability density for the momentum of any particle is obtained:

${\displaystyle {\mathcal {P}}_{{\vec {p}}_{j}}({\vec {p}})=\left({\frac {\beta }{2m\pi }}\right)^{\frac {3}{2}}e^{-{\frac {\beta }{2m}}p^{2}}}$

The pre-factor essentially corresponds to the thermal De Broglie wavelength . ${\displaystyle ({\tfrac {\beta }{2m\pi }})^{3/2}=({\tfrac {\lambda }{h}})^{3}}$ ${\displaystyle \lambda }$

This allows the probability density for the velocity amount to be determined with the transformation set ${\displaystyle v=|{\vec {p}}|/m}$

${\displaystyle {\mathcal {P}}(v)=\int _{\mathbb {R} ^{3}}{\mathcal {P}}_{{\vec {p}}_{j}}({\vec {p}})\,\delta \left(v-{\frac {|{\vec {p}}|}{m}}\right)\,{\text{d}}^{3}p=\left({\frac {\beta }{2m\pi }}\right)^{\frac {3}{2}}\int _{\mathbb {R} ^{3}}e^{-{\frac {\beta }{2m}}p^{2}}\,\delta (v-{\frac {|{\vec {p}}|}{m}})\,{\text{d}}^{3}p}$

The integration is carried out in spherical coordinates and the relationship is used ${\displaystyle \delta (v-|{\vec {p}}|/m)=m\,\delta (p-mv)}$

${\displaystyle {\mathcal {P}}(v)=\left({\frac {\beta }{2m\pi }}\right)^{\frac {3}{2}}4\pi \int _{0}^{\infty }p^{2}e^{-{\frac {\beta }{2m}}p^{2}}m\,\delta (p-mv)\,{\text{d}}p}$

Now the convolution property of the delta distribution is to be used again

${\displaystyle {\mathcal {P}}(v)=\left({\frac {\beta }{2m\pi }}\right)^{\frac {3}{2}}4\pi (mv)^{2}m\,e^{-{\frac {\beta }{2m}}(mv)^{2}}\Theta (v)=\left({\frac {\beta m}{2\pi }}\right)^{\frac {3}{2}}4\pi v^{2}e^{-\beta {\frac {m}{2}}v^{2}}\Theta (v)}$

here is the Heaviside step function , which makes the probability of negative magnitude velocities disappear. ${\displaystyle \Theta (v)}$

If you set for , you get to the Maxwell-Boltzmann distribution ${\displaystyle \beta ={\frac {1}{k_{\rm {B}}T}}}$

${\displaystyle {\mathcal {P}}(v)={\sqrt {\frac {2}{\pi }}}\left({\frac {m}{k_{\mathrm {B} }T}}\right)^{\frac {3}{2}}v^{2}\,e^{-{\frac {mv^{2}}{2k_{\mathrm {B} }T}}}\,\Theta (v)}$

See also

• H theorem

Individual evidence

1. ↑ Klaus Stierstadt, Günther Fischer: Thermodynamics: From Microphysics to Macrophysics (Chapter 4.2) . Springer, Berlin, New York 2010, ISBN 978-3-642-05097-8 ( limited preview in Google book search).