Answer to the Question 04/97
The question was:A particle is placed in a one-dimensional potential
2 B
U(x)=A*x + ---
2
x
at temperature T. Its motion is restricted to
x>0. Find its mean energy.
The problem was solved correctly by Oded Farago from Tel Aviv U. (5/15/97).
The answer: The mean total energy is kT+2*sqrt(A*B)
The solution:
1. The mean kinetic energy of the particle is < p2/2m> = kT/2 (Equipartition theorem)
2. The mean potential energy can be split into two parts:
< A*x2+B/x2> = < A*x2-B/x2> + 2*B*< 1/x2>.
a. The first part can be evaluated using virial theorem:
< A*x2-B/x2> = 1/2*< x*dU/dx> = kT/2
b. The second part can be evaluated as follows: First perform a change in variable y=sqrt(sqrt(A/B))x. In this new variable the potential is brought to the form U=sqrt(AB)*(y2+1/y2). Thus the average
< 1/x2> =sqrt(A/B) int (1/y2) exp[-(sqrt(AB)/kT)(y2+1/y2)] dy / int exp[-(sqrt(AB)/kT)(y2+1/y2)] dy
Now in the first integral perform variable change z=1/y and it will become identical with the second integral, leaving us with the answer < 1/x2> =sqrt(A/B).
Collecting all the above results we finally get the total mean energy: kT+2*sqrt(A*B).
Y. Kantor: Note something interesting: the potential has a minimum at x=sqrt(sqrt(B/A)). The potential energy at that point is U=2*sqrt(A*B). At low T the potential can be treated as approximately harmonic (parabolic), i.e. the potential energy due to the temperature will be kT/2. If we add similar term for kinetic energy, we get the total energy kT+sqrt(A*B). The surprising fact is that this relation holds also when the temperatures are not low and the potential cannot be approximated as harmonic...
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