Maxwell Boltzmann Distribution Derivation

Consider an ideal system having n particles occupying a volume V, whose total energy is E. Here, the value of E is constant because no energy is being added or taken away from the system. So the total energy of the system is equivalent to the sum of total energies of the individual particles.

Since the system has multiple states, if we consider an ith state of the system, then, the probability at this state will be:

\[P_{i} \alpha E^{-\frac{\epsilon i}{kT}}\]

Where εi is the energy of a system in ith state.

k  = Boltzmann constant (= \[1.38 \times 10^{-23} JK^{-1}\])

T =Temperature

Now,

\[P_{i} = \frac{1}{z} E^{-\frac{\epsilon i}{kT}}\]

Here z is the partition function, which is the sum of the energies of all the states in the system.

Maxwell Boltzmann Distribution Derivation

The molecules inside the system travel at varying speeds so two persons named James Maxwell and Ludwig Boltzmann came up with a theory to demonstrate how the speeds of the molecule are distributed for an ideal gas which is Maxwell-Boltzmann distribution theory.

Consider a system having n particles occupying a volume V, whose total energy is E.

So, \[E = \sum_{i=0}^{\infty} n_{i} \epsilon_{i}\]

Where n is the number of particles having energy εi.

The number of particles N is also constant, i.e.,

\[N = \sum_{i=0}^{\infty} n_{i}\]

The number of microstates for the energy levels of molecules of a system can be expressed as:

\[W = \frac{N}{n_{0}! n_{1}! n_{2}}!\] .... (1)

Stirling's equation of eq(1) is,

ln N!. ≅ N ln N - N + 1/2 ln(2πN)...(2)

And, for large N it is:

N!. ≅ N ln N - N

Thus, in W it can be expressed as:

ln W ≅ N ln \[N - \sum_{i} n_{i} ln n_{i} - (N - \sum_{i} n_{i})\]

The term inside the parentheses is zero because, \[N = \sum_{i=0}^{\infty} n_{i}\] is a constant value.

Thus, ln W ≅ \[N ln N - \sum_{i} n_{i} ln  n_{i}\] ..(3)

Taking differential of eq(3):

\[d(ln W) = -\sum_{i} (1 + ln n_{i})dn_{i} = - \sum ln n_{i} d n_{i}\] …(4)

Since dni = dN = 0

Now using the Lagrange multipliers here, it is known that:

\[dE = 0   = \sum_{i} n_{i} \epsilon_{i}\] , and

\[dN = 0 = \sum_{i} dn_{i}\] …(5)

Adding (4) to (3), for any constant α and β, following should be true:

\[d(ln W) = - \sum_{i} ln n_{i} d n_{i} - \alpha \sum_{i} d n_{i} - \beta \sum_{i}d n_{i}\]

At maximum, W = 0

0 = - Σ (α + βεi + ln ni )dni 

Or, 0 = (α + βεi + ln ni )

ln ni  = - α - βεi 

Removing the logarithmic, we get:

\[n_{i} =e^{-\alpha} e^{-\beta \epsilon i}\]

Evaluating N using the distribution law here:

\[e^{\alpha} = \frac{N}{e^{-\beta \epsilon i}}\] and z, the partition function = \[\sum_{i} e^{-\beta \epsilon i}\]

Therefore, ni  = N/Q e-βεi

Applying zeroth law of thermodynamics, we get

β = 1/kT

Maxwell Boltzmann Distribution Equation Derivation

Maxwell distribution of velocities states that the gaseous molecules inside the system travel at different velocities.

Fraction F(v) = \[4 \pi N(\frac{m}{2 \pi k T})^{3/2} v^{2} e^{-mv^{2/2kT}}\]

The Maxwell distribution of velocities can be derived from Boltzmann’s equation:

\[f (E) = Ae^{-kT}\]

This equation tells us the probability that a molecule will be found with energy E that decreases exponentially with energy; i.e., any molecule is highly unlikely to capture much more than its average part of the total energy available to all the molecules.

(image will be uploaded soon)

If Maxwell-Boltzmann distribution is applied in one dimension of velocity for a molecule in an ideal gas is vz.

Where the factor F = dN/dV

This F states that if we consider a small rectangular inside the above graph, then the probability of the gas molecules traveling at velocity can be found.

Where Fraction F(v) = \[4 \pi N(\frac{m}{2 \pi k T})^{3/2} v^{2} e^{-mv^{2/2kT}}\]

Maxwell Boltzmann Equation Derivation

The probability of finding a molecule at some velocity is one. Then, by using the definite integral form here from minus-infinity to plus-infinity as:

\[\int_{-\infty}^{\infty} e^{-x^{2}} dx = \sqrt{\pi}\] and substituting \[x = \sqrt{\frac{m}{2kTv_{z}}}\]

\[A \sqrt{\frac{m}{2kT} v_{z}} \int_{-\infty}^{\infty} e^{-\frac{m}{2 \pi k T}} dx = \sqrt{\pi}\]

\[F(v_{z}) =\sqrt{\frac{m}{2kT}} v^{2} e^{-mvz^{2/2kT}}\]     \[\sqrt{\frac{m}{2kT}}  dv_{z} = 1\]

It gives \[A = \sqrt{\frac{m}{2kT}}\]

This normalizes the function to:  \[F(v_{z}) =\sqrt{\frac{m}{2kT}} v^{2} e^{-mvz^{2/2kT}}\]

In terms of three dimensions, it becomes:

\[F(v) =4 \pi (\frac{m}{2kT})^{3/2} v^{2} e^{-mvz^{2/2kT}}\]

Maxwell Boltzmann Derivation

Consider the fraction of molecules in a three-dimensional box having the translation energy ε, then, as a function, it will be:

ε  = h² / 8m [nx2 + ny2+ nz2]/[Lx2 + Ly2 + Lz2]

ni2   = nx2 + ny2+ nz2  ,and

 εi  = n²h²/8mL²

The length of each side of the box is equal.

So, we would consider the density of translational states as a sphere in space.

So the total volume of the sphere is:

\[\epsilon = \frac{4 \pi}{8} \int_{0}^{n} n^{2} dn\]

and  dε = π/2 n² dn

Also n²h²/8mL² = ½ mu², and

dn  = 2mV1/3/hdu

So,  π/2 n² dn = dNi/N = dε/Q \[e^{-\beta \epsilon i}\] 

Substituting in for thermal wavelength, we get:

\[\frac{1}{N} \frac{du}{dN} = 4 \pi u^{2} (m/2 \pi k T)^{3/2} e^{-mu^{2/2kT}}\]

This is the Maxwell-Boltzmann distribution.