Maxwell-Boltzmann distribution

The Maxwell-Boltzmann distribution describes the most probable energy distribution for the molecules in a system. It can be derived using Statistical Mechanics (see Derivation of the partition function) and forms the basis for the [Kinetic theory of gases]?.

The Maxwell-Boltzmann distribution can be expressed as:

     Ni/N = exp(-Ei/kT)/∑exp(-βEj/kT)  (1)

where N_i/N is the number fraction of molecules at equilibrium temperature T, having energy level E_i and k is Boltzmanns constant.

Maxwell-Boltzmann Velocity Distribution

For the case of an "ideal gas" consisting of non-interacting atoms in the ground state, the energy is only kinetic energy. From the Particle in a box problem in Quantum mechanics we know that the energy levels for a gas in a rectangular box with sides of lengths a_x, a_y, a_z are given by:

   Ei = (nx2/ax2 + ny2/ay2 + nz2/az2)(h2/8m)   (2)

where, n_x, n_y, and n_z are the quantum numbers for x,y, and z motion, respectively. However, for a macroscopic sized box, the energy levels are very closely spaced, so the energy levels can be considered continuous and we can replace the sum with an integral. Furthermore, we can recognize that (h²n_i²/4a_i²) corresponds to the square of the ith component of momentum, p_i² giving:

     Ni/N = q-1 exp[-1/2mkT(px2 + py2 + pz2)]  (3)

where q corresponds to the denomenator in Equation 1. This distribution of N_i/N is proportional to the probability distribution function f_p for finding a molecule with these values of of momentum components, so:

    fp(px, py, pz) = cq-1 exp[-1/2mkT(px2 + py2 + pz2)]   (4)

The constant of proportionality, c, can be determined by recognizing that the probability of a molecule having any momentum must be 1. Therefore the integral of equation 4 over all p_x, p_y, and p_z must be 1.

It can be shown that:

   ∫∫∫dpxdpydpz q-1 exp[-1/2mkT(px2 + py2 + pz2)] = q-1<sup>(2mπ/kT)<sup>3/2   (5)

so in order for the integral of equation 4 to be 1,

    c = q(2πmkT)-3/2     (6)

Substituting Equation 6 into Equation 4 and using p_i=mv_i for each component of momentum gives:

    fp(px, py, pz) = (1/2πmkT)3/2 exp[-m/2kT(vx2 + vy2 + vz2)]   (7)

Finally recognizing that the velocity probability distribution, f_v is proportional to the momentum probability distribution function as f_v = m³f_p, we get:

   fv(vx, vy, vz) = (m/2πkT)3/2 exp[-m/2kT(vx2 + vy2 + vz2)]   (8)

Which is the Maxwell-Boltzmann velocity distribution.

Velocity Distribution in One Direction

For the case of a single direction Equation 8 can be reduced to:

   fv(vx) = (m/2πkT)1/2 exp[-mvx2/2kT]   (9)

This distribution has the form of a Gaussian error curve. As expected for a gas at rest, the average velocity in any particular direction is zero.

Distribution of Speeds

Usually, we are more interested in the speed of molecules rather than the component velocities, where speed, v is defined such that:

     v2 = vx2 + vy2 + vz2  (10)

The corresponding speed distribution is:

   F(v) = 4πv2(m/2&pikT)3/2exp(-mv2/2kT)   (11)

Someone with time, please add the following!!!

Average Speed

Most Probable Speed

Mean Speed

Root-mean-square Speed