Maxwell-Boltzmann distribution formula

Hi, I'm sure that all of you study in school thermodynamics. And you might have heard at least once about Maxwell-Boltzmann distribution, which looks like....this:


Have you ever wondered where this huge formula came from? Well, I did. In the post I'll try to explain it as clearly as possible! Enjoy!

First of all I will introduce a few statistical terms, like microstate, macrostate and multiplicity.

A microstate is simply a possible distribution of the elements( particles, energy...) of a system.
A macrostate, on the other hand is a more general characteristic of a system, focusing on the quantity of the elements in the system  distributions.

For example, suppose we drop 3 coins. I'll note one side of a coin with 1 and the other one with 0.

Now, after I drop them, the possible states are:

1 1 1                1 0 0
1 1 0                0 1 0
1 0 1                0 0 1
0 1 1                0 0 0

Each of the 8 occurences is called a microstate!

What about quantities? Well we have:

-3 1 and no 0
-2 1 and 1 0
- 1 1 and 2 0
- no 1 and 3 0

Each of the 4 descriptions of the system is called macrostate!

I think it that this new terms are clear enough now. The multiplicity (W) of a system is the ratio between the number of the microstates and the number of the macrostates. So,in our example the multiplicity is 2. The bigger the multiplicity is , the greater the disorder grade of a system is. For example, if we have one ball and ten baskets, there is one macrostate( the number of the empty baskets stays the same), but the number of microstates is ten( the ball can be placed in 10 different places), so the multiplicity is 10. However, if we have one ball and 20 baskets, the multiplicity is 20! It is  harder to guess where the ball is in the second case, so that system is more chaotic, or more disordered! So, the multiplicity measures the disorder of a system.
*The fundamental assumption of thermodynamics: In an isolated system in thermal  equilibrium, all accessible microstates are equally probable. It is obvious that the macrostates of a system have different multiplicities. In our example, the macrostate "2 one and 1 zero" had a multiplicity equal to 3.

Thus, we can introduce the second law of thermodynamics: A system evolves to the macrostate with the highest multiplicity. It is quite simple to understand why. It is matter of probability. The one with more lottery tickets has the greatest chances to win!  In other words, systems then to flow to the highest degree of chaos. Look at our Universe, which is constantly expanding!

In order to exemplify this better, I will discuss the interaction between two bodies, A and B, that can interact by exchanging energy( they form a closed system). Their total energy is U,so UA + UB = U. The energy of each body can obviously vary from 0 to U. The graph between the multiplicity of the system and the energy of one body, let's say UA looks like this:

 You can observe that the curve is sharper if the energy and the number of oscillators of the system is higher. So, systems with very high numbers of particles( 10^23) will be found almost certainly in a specific state.( the   apex of the curve). Now you should already understand what the thermal equilibrium between two bodies actually mean. The equilibrium temperature established between bodies in contact you learn in school is actually the temperature for which the system is found in the macrostate with the highest grade of disorder. So, temperature is a matter of statistic after all!

Since the equilibrium between the bodies is met at the apex, it is obvious that

dSTOTAL / dUA = 0 at equilibrium.

STOTAL = SA +SB

 so (dSA +dSB)/ dUA =0

the system is isolated, so dUA=dUB

In the end,we get that 

dSA/ dUA = dSB/ dUB= T

thus, I can now introduce the definition of temperature,

T=(dS/ dU)V.N ( at constant volume and number of particles)




The multiplicity of the large systems like gases or other solid or liquid bodies, which are formed of about 10^24 atoms is....a big number. So, scientists introduced a new size, called entropy, noted with S.

           

where k is the Boltzmann constant.
As multiplicity, entropy measures the disorder grade of a system.`

Of course, a graph between entropy and energy of a body can be sketched as well.

Now, we'll get serious. It's time to demonstrate Boltzmann distribution formula. Boltzmann, not Maxwell-Boltzmann! not yet...

As I stated before,if  a system is isolated, all microstates are equally probable.So if we connect the system to a reservoir with temperature T held constant, microstates with different energies will have different probabilities.
 Let's take to states of the system, s1 and s2, with energies E1 and E2 and probabilities P1 and P2

The probabilities are proportional to the multiplicity of the system, so

P1 / P2 = W1/ W2

or P1/ P2 = e^S1/k /e^S2/k, since     , S1 and S2 are quantities for the reservoir

dER= TdSR for the reservoir( we write this relation only for this reservoir because its size is so big that its temperature remains constant during the process).

dES=-dER( energy exchanges between the system and the reservoir are complementary), so dES= -TdS so ES= TS( we can integrate like this, since T remains the same)

thus, we obtain:

P1/P2=e^-E1/kT / e^-E2/kT

if the system consist of a large number of particles, the ratio between the number Ni having the energy Ei and the total number of particles N is given by the formula:



where  is the degeneracy (meaning, the number of levels having energy ; sometimes, the more general 'states' are used instead of levels, to avoid using degeneracy in the equation) and Z(T) is the partition function.
Considering  

 
we get that the partition function equals:
\

The ideal gas is a particular system where Boltzmann distribution applies!

For the ideal gas gi is 0 and  the potential energy between the particles are neglected and the energy is only kinetic.
These are the representations of the Boltzmann distribution for gases at different temperatures!

Next, I'll show you how to derive Maxwell-Boltzmann distribution using Boltzmann distribution.

First, you should understand that the Boltzmann distribution is dealing with energies. You can deduce that according to it, the most probable energy in a system of bodies is 0. However, since the energy in an ideal gas is only kinetic energy, you might infer that the most probable speed is 0, which seems illogical( it's weird to think of perfectly still molecules in the middle of the gas). What's the answer to the riddle? Remember that I said that Boltzmann distribution is only dealing with the probabilities of energies, not probabilities of velocities!

Now, imagine a momentum (or velocity space, since p=mv), which has the axes px, py, pz




Imagine now that all energies in the gas are equally probable. If that were true, the probability of a specific speed is proportional to the area of the velocity sphere , 4πv^2( actually is proportional with the volume of the velocity layer of an infinitely small thickness, 4πv^2 dv),, where

That means that 0 velocity has 0 probability and the greatest velocities are the most probable. But that's true only if all energies were equally probable....That's where Boltzmann distribution interfere and offer a model for energy distribution. Now the probability of a specific velocity has the form

,dN/ Ntotal= C 4πv^2  *e^-E/kT dv, 

where dN is the fraction of molecules having velocity of v. After integrating the relation, for the all the domain of speeds and number of molecules, we get the constant C. Finally, I am able to write the Maxwell-Boltzmann distribution formula:


,where f(v) is dN/(Ntotal *dv)