Maxwell Boltzmann Distribution Derivation

Not all the air molecules surrounding us travel at the same speed. Some air molecules travel faster, some move at moderate speeds, and some air molecules will hardly move at all. Hence, instead of asking about the speed of any particular gas molecule, we ask about the distribution of speed in a gas at a certain temperature. James Maxwell and Ludwig Boltzmann came up with a theory to show how the speeds of the molecules are distributed for an ideal gas. The distribution is often represented using the following graph.

Maxwell Boltzmann Distribution Derivation

In the next section, let us look at the Maxwell Boltzmann distribution derivation.

Derivation of Maxwell-Boltzmann Distribution

Consider a system that consists of identical yet distinguishable particles.

  • Let the total number of particles in the system be ‘n’.
  • The total volume of the system is fixed and is given by ‘V’.
  • The total amount of the energy is fixed and is given by ‘U’.

We now want to know the number of particles at a given energy level.

The energy levels available within the system are given by ε0, ε1, ε2 ………., εr. The energy levels are fixed for the system.

The number of particles sitting at each energy level is variable, and the number of particles is given by n1, n2, n3 …….., nr.

The number of ways to attain a given microstate is given by the formula

\(\begin{array}{l}\omega =\frac{n!}{n_{0}!n_1!n_{2}!….n_{r}!}… (1)\end{array} \)
 
We need to know the combination of n1,n2,n3 up to nr that maximizes the value of ω. This combination results in the most probable microstate. If it is the most probable state, then it can be considered the equilibrium state. To find the combination, let us alter eq (1) as follows

\(\begin{array}{l}\ln \omega =\frac{\ln n!}{\coprod_{i=0}^{i=r}=n_{i}!}…(2)\end{array} \)
 
To maximise the value of ω, instead of dealing with ω let us deal with \lnω.

\(\begin{array}{l}\ln \omega=\ln n!-\sum_{i=0}^{i=r}\ln n_i!… (3)\end{array} \)
 
According to Sterling’s approximation, the log of factorial can be approximated as

\(\begin{array}{l}\ln x!=x\ln x-x\end{array} \)
 
Making use of the Stirling approximation in eq (3), we get

\(\begin{array}{l}\ln \omega =n\ln n-n-\sum_{i=0}^{i=r}[n_{i}\ln{n_i}-n_i]\end{array} \)
 
By taking the derivative of the above equation, we get

\(\begin{array}{l}\partial \ln \omega=-\sum_{i=0}^{i=r}\partial n_{i}\ln n_{i}+n_{i}\times \frac{1}{n_{i}}-\partial n_{i}=0\end{array} \)
 
The above equation implies that

\(\begin{array}{l}\sum_{i=0}^{i=r}[\partial n_i\ln (n_i)]=0… (4)\end{array} \)
 
As we know that n is a constant, therefore we can say that

\(\begin{array}{l}\sum_{i=0}^{i=r}n_i=n=constant\end{array} \)
 
This means that

\(\begin{array}{l}\sum_{i=0}^{i=r}\partial n_i=0…(5)\end{array} \)
 
The sum of all the changes in the system should sum up to zero.

The total energy of the system should also be constant.

\(\begin{array}{l}\sum_{i=0}^{i=r}\varepsilon _i n_i=U=constant\end{array} \)
 
Differentiating the above equation, we get

\(\begin{array}{l}\sum_{i=0}^{i=r}\varepsilon _i \partial n_i=U=0…(6)\end{array} \)
 
To maximise the function (4) subjected to constraints (5) and (6), let us use Lagrange’s method of undetermined multipliers.

Multiplying (5) by ɑ and multiplying (6) by 𝛽 and then adding 4, 5 and 6, we get

\(\begin{array}{l}\sum_{i=0}^{i=r}[\ln n_{i}+\alpha +\beta \varepsilon_i]\,\partial{n_i}=0…(7)\end{array} \)
 
From eq (7) to be zero, the term
\(\begin{array}{l}[\ln n_{i}+\alpha +\beta \varepsilon_i]\,\, \textup{has to be zero.}\end{array} \)
 
\(\begin{array}{l}\ln n_{i}+\alpha +\beta \varepsilon_i=0\end{array} \)
 
Simplifying the above equation, we get

\(\begin{array}{l}n_i=e^{-\alpha}\times e^{-\beta\epsilon_i}…(8)\end{array} \)
 
The above equation signifies the number of particles in i th level.
 
In eq (8), the only variable parameter is ω, therefore taking the sum, we get

\(\begin{array}{l}\sum_{i=0}^{i=r}n_i=e^{-\alpha}\sum_{i=0}^{i=r} e^{-\beta\epsilon_i}=n\end{array} \)
 
Simplifying, we get

\(\begin{array}{l}e^{-\alpha}=\frac{n}{\sum_{i=0}^{i=r} e^{-\beta\epsilon_i}}\end{array} \)
 
The above equation is the same as

\(\begin{array}{l}e^{-\alpha}=\frac{n}{P}…(9)\end{array} \)

where P is the partition function,

\(\begin{array}{l}P=\sum_{i=0}^{i=r} e^{-\beta\epsilon_i}\end{array} \)
 
Substituting (9) in (8), we get

\(\begin{array}{l}n_i=\frac{n}{P}e^{-\beta \epsilon _i}…(10)\end{array} \)
 
The value of 𝛽 in the above equation is
\(\begin{array}{l}\beta=-\frac{1}{k_{b}T}\end{array} \)
where kb is the Boltzmann constant and T is the temperature.

Substituting the value of 𝛽 in eq (10), we get

\(\begin{array}{l}n_i=\frac{n}{P}\,e^{\frac{-\epsilon_i}{k_{B}T}}\end{array} \)
 
The above expression helps find the number of particles in the most probable microstate. This expression is known as the Maxwell Boltzmann statistics expression.

Also, Read

Derivation of Phase Rule Stokes Law Derivation

 

Frequently Asked Questions – FAQs

Q1

How do you define the Maxwell-Boltzmann distribution law?

The Maxwell-Boltzmann distribution is concerned with the distribution of energy between identical but distinct particles. It reflects the probability of the distribution of states in a system with varying energies.
Q2

What is kb in Maxwell-Boltzmann?

kb = Boltzmann constant.
Q3

What characteristics distinguish Maxwell Boltzmann statistics?

In statistical mechanics, the distribution of particles from classical materials across different energy levels in thermal equilibrium is described by Maxwell-Boltzmann statistics.
Q4

What does the Boltzmann constant mean in terms of physics?

Boltzmann’s constant has a value of around 1.3807 x 10–23 joules per kelvin (J.K–1).
Q5

What is Maxwell plot?

According to their speed, the Maxwell distribution function is a function of the number of molecules moving at a particular speed. According to the article’s curve, the peak corresponds to the most likely speed.

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