The Boltzmann factor, e^-E/kT

E is energy of a molecule, k is the Boltzmann constant and T is absolute temperature. The probability that a molecule (in matter in thermal equilibrium) has energy E is proportional to the Boltzmann factor, and it turns out that kT is related to the average energy of a molecule. (In a monatomic gas, E_AV = ³/₂kT.) To see why the exponential is the appropriate probability function, consider picking two molecules: The probability of the first molecule having energy E₁ and the 2^nd having E₂ would be the product, P(E₁)P(E₂), just as in flipping two coins the probability of getting two heads is ¹/₄. But if there are N molecules, total energy E_T , then the probability is the same as the probability that all the other molecules have energy = E_T – (E₁+E₂), which is some function of E₁+E_2. So we are looking for a function such that the product P(E₁)P(E₂) = P(E₁+E₂).

The only function that behaves this way is the exponential (and by the way, e is not required, you can replace
e ^-E/kT with a ^-(E/kTlna)_, where a is any constant). The above argument about the P function is bogus so far, because they could be two different functions, P and P’. The reason P = P’ is that the equation must be valid for all molecules, and furthermore, P(E₁)P(E₂)P(E₃) = P’(E₁+E₂+E₃), etc., etc.

If we look at a simple example of 4 interacting bodies with integral energies, you will see why it is sort of exponential. (At temperature near absolute zero, the quantization of energy shows up, so this example is not absurd.) Say the total energy is 5 units. One could have E=5 and the others zero, and there are 4 ways that could occur. Next possibility: One could have E=4 and another with E=1, two with E=0. There are 12 ways for this to occur: 4 ways to pick the 4, then for each choice there are 3 arrangements for the others. Here are the possibilities:

Energies Number of ways
5,0,0,0 4
4,1,0,0 12
3,2,0,0 12
3,1,1,0 12
2,2,1,0 12
2,1,1,1 4
The total number of possible arrangements is 56. Now the probability of one having zero energy is computed as follows: ³/₄∙⁴/₅₆ + ¹/₂∙¹²/₅₆+ ¹/₂∙¹²/₅₆+ ¹/₄∙¹²/₅₆+ ¹/₄∙¹²/₅₆ = 21/56, because in the first line in the table above, ¾ are zeros, and that occurs 4/56 of the time, etc., etc. Probability of having energy E is P(E). We see that
P(0) = 21/56; similarly for the others:
P(1) = 15/56
P(2) = 10/56
P(3) = 6/56
P(4) = 3/56
P(5) = 1/56 Graph this, and you will see that it looks like a negative exponential curve. If the above procedure is done for a large number of bodies, it would agree with the theory very well.

In the case of an ideal monatomic gas, all the energy is kinetic, so P µ exp(-mv²/2kT), where exp means e to the power of. Let n(v) = the number of molecules in the small range v to v+Dv divided by Dv. Clearly this is proportional to P, and we will show that it is also proportional to v². We need to imagine velocity space. Regular xyz space has a corresponding v_x v_y v_z space. Any velocity is a vector from the origin. The set of molecules with speed v is a sphere of “radius” v and surface “area” 4pv² . Picture all molecular velocities in the range v +Dv with their arrow tips in the shell of “volume” 4pv²Dv. So n(v) is proportional to v² as well as P, so
n(v) µ v² exp(-mv²/2kT), called the Maxwell-Boltzmann velocity distribution. The graph of it looks sort of bell-shaped, but with a longer tail at high velocities. The most probable v can be found by setting the derivative to zero, and you can show (see below) that it is v_p = (2kT/m)^1/2. The average energy is (³/₂)kT, and from this we get v_rms = (3kT/m)^1/2, larger than v_p because of the asymmetry.

n(v) = Cv²exp(-mv²/2kT)

dn/dv = 2Cvexp( ) + Cv²exp( )(-2mv/2kT). Set this equal to zero for maximum n, factor out Cvexp( ), set the other factor equal to zero, and get 2 – v_p²m/kT = 0, from which we find v_p = (2kT/m)^1/2.

My main pages: on mechanics fluids, heat, e&m, vibrations& waves, quantum, index. fredrick.gram@ tri-c.edu