Homework T06.1.c)
In the exercise, we must obtain for N particles with a given Hamiltonian three different potentials, for the case in c) we must find the average potential energy for a potential with the form:
This can be solved in a very similar way than the b), because we obtain almost the same integral for the partition function, in b) I found that this integral was:
for c) we can see that is going to be almost the same partition function but as we can see in b) is taken the absolute value of x, and in c) for x<0, the function goes to 0 and for x>0 it has a value, so, we won't multiply by 2 as we did in the integral for b) so our partition function is:
We already know that the solution is going to be also very similar than in b) that can be obtained with a very easy variable change, so we obtain:
We take the thermodynamic limit obtaining:
Finally, to get the average potential energy, I used the fact that the average energy shoud be given by:
and from the equipartition theorem we know that
and that
in that way, if we solve for the average potential energy, we find:
Doing the math I found that the average potential energy for the potential of c) is:







The idea should be correct in principle. However you have a few mistakes.
ResponderBorrar1. You write = - . There, obviously, the minus sign should be a plus
2. is not equal to k_B T/2, it is equal to N d k_B T/2
3. You did not put your result for . It is = Nd k_B T/2 (1 + 1/n).
4. Doing the math, one gets: = Nd k_B T/(2n)