Question

Prove 3/2kt from maxwell boltzmann distribution of energy

Answer 1

This comes from Statistical Mechanics. I'm not sure of your background, but I'll post the derivation I know. If you need to expand, pick one Statistical Mechanics book like Fundamentals of Statistical and Thermal Physics by Reif.

In order to be precise, the whole idea is: considering the classical phase space M endow it with a probability density function ρ:M→R. This probability density is undestood exactly as you might think:

P(A)=∫Aρ(x)dx

is the probability that the system is in a microstate contained in the region of position and momentum A.

The question one tries to answer then is: considering the system is in thermal equilibrium at temperature T, what is ρ?

The answer, whose derivation is found in Reif's book, is (considering M=R6N being N the number of particles)

ρ(p,q)=1Ze−βH(p,q)

with p=(p1,…,pN) the momenta of the particles, q=(q1,…,qN) their coordinates, H the Hamiltonian of the system, β=(kBT)−1 and Z being the normalization factor called the "partition function" given by

Z=∫e−βH(p,q)d3Npd3Nq

Notice also that pi=(pix,piy,piz) and qi=(qix,qiy,qiz).

Now what is the energy? The energy is the mean value of H, considering the probability density ρ, in other words:

E=⟨H⟩=∫ρ(p,q)H(p,q)d3Npd3Nq=1Z∫H(p,q)e−βH(p,q)d3Npd3Nq

but it is clear that

H(p,q)e−βH(p,q)=−∂∂βe−βH(p,q)

thus

E=−1Z∂∂β∫e−βH(p,q)d3Npd3Nq=−1Z∂Z∂β

Now think of N free particles of same mass - i.e., a gas. In that case the Hamiltonian is given by H(p,q)=∑p2i2m - i.e, the sum of their kinetic energies. Now in this case finding Z is simple because the integrals will factor:

Z=∫e−β∑p2i2md3Npd3Nq=∏i=1N∫e−βp2i2md3pi∫d3qi,

but each integral over the coordinates just gives the volume V of the box where the particles are, so we have

Z=VN∏i=1N∫e−βp2ix2md3pix∫e−βp2iy2md3piy∫e−βp2iz2md3piz

the three integrals are identical, so we compute just once using the well known Gaussian Integral

Z=VN(∫e−βp22mdp)3N=VN(2mπβ−−−−−√)3N=VN(2mπβ)3N/2

Now compute E

E=−1Z(VN(2mπ)3N/2)(−3N2β−3N/2−1)=3N2β3N/2VN(2mπ)3N/2(VN(2mπ)3N/2)β−3N/2−1

This is exactly E=3N2β−1 or else

E=3N/2kBT