Think Science .

If what you want is the actual proof, I suggest you look at any good book on statistical mechanics. One of my favorite books for that subject is

Fundamentals of Statistical and Thermal Physics, by Frederick Reif
http://www.amazon.com/Fundamentals-Statistical-Thermal-Physics-McGraw-Hill/dp/0070518009
The law of equipartition of energy is valid for a system (classical or quantum) (a) whose hamiltonian is a quadratic function of the generalized coordinates and momenta describing the system in question, and (b) in thermal contact with a reservoir at constant temperature.

Because of (b), the probability of any microstate of the system is proportional to exp(-ßH), where ß = 1/(kT) and H is the hamiltonian of the system. Thus, the average value of any dynamical quantity F(q, p), depending on the generalized coordinates q and/or momenta p, is given by

<F> = ( ? F(q, p) exp[ -ß H(q, p) ] dq dp ) / ( ? exp[ -ß H(q, p) ] dq dp )

So far, this is perfectly general. Now, if (a) is true, then you can actually compute the average above quite easily. To make things easier here (because of orkut’s limitations), let’s say that the Hamiltonian is H = q^2, that is, it’s quadratic on the generalized coordinate q.

What’s the average value of the energy of the system, then? It’s simply

<H> = <q^2> = ( ? q^2 exp[ -ß q^2 ] dq ) / ( ? exp[ -ß q^2 ] dq )

Now consider the integral Z(ß) = ? exp[ -ß q^2 ] dq as a function of ß. It’s the integral appearing as the denominator of <H>. Taking the derivative with respect to ß, you find:

dZ(ß)/dß = - ? q^2 exp[ -ß q^2 ] dq

which, except for the minus sign, is exactly the numerator appearing in <H>. Thus,

<q^2> = - (1/Z) dZ/dß = - (d/dß) [ ln Z(ß) ]

So, all we need is to compute Z(ß). The limits of integration (which I haven’t shown in the above) are from negative to positive infinity, since the coordinate can take any of those values. So, with u = sqrt(ß) q,

Z(ß) = ? exp[ -ß q^2 ] dq = ? exp[ -u^2 ] du / sqrt(ß) = ß^(-1/2) x some constant

lnZ(ß) = -(1/2) lnß + some constant

and

<q^2> = - (d/dß) [ ln Z(ß) ] = (1/2) (1/ß) = kT/2

So, the average value of any quadratic generalized coordinate is kT/2. The same is true for the generalized momenta. Hence, for a system described by N generalized coordinates and N generalized momenta, all of which appearing quadratically in the hamiltonian, the average energy is going to be given by

<E> = N kT/2 + N kT/2 = N kT

That’s true also if the generalized coordinates and/or the generalized momenta appear in H as general quadratic functions, that is, with linear terms as well.

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