Find Energy from canonical function

[boddhisattva]

Homework Statement

Let Z=2T/s + 1/3 + s/(30T) be the partition function of a quantum rotor at s/T->0. Show that

$U=nk(T-s/6-s^2/(180T)$

Homework Equations

1/(1+x) = 1 -x

The Attempt at a Solution

$U=-kT^2(\partial _{T} ln(Z) )$

Hence
$U=-kT^2 [2/s - s/(30T^2)] / [ 2T/s + 1/3 + s/(30T) ]$

I tried dividing upper and lower equation by 2T/s and use 1/(1+x) = 1 -x but cannot find the result.

 

[TSny]

$U=-kT^2(\partial _{T} ln(Z) )$

Are you sure the negative sign on the right side is correct?

$U=-kT^2 [2/s - s/(30T^2)] / [ 2T/s + 1/3 + s/(30T) ]$

Try factoring out 2T/s in the denominator.

Note that you want to get a result for U that is good to second order in the small quantity s/T.

So, you might need to include the next higher order in the approximation 1/(1+x) = 1 - x

 

[nrqed]

Homework Statement

Let Z=2T/s + 1/3 + s/(30T) be the partition function of a quantum rotor at s/T->0. Show that

$U=nk(T-s/6-s^2/(180T)$

Homework Equations

1/(1+x) = 1 -x

The Attempt at a Solution

$U=-kT^2(\partial _{T} ln(Z) )$

I think that it should be $U=nkT^2(\partial _{T} ln(Z) )$

Hence
$U=-kT^2 [2/s - s/(30T^2)] / [ 2T/s + 1/3 + s/(30T) ]$

I tried dividing upper and lower equation by 2T/s and use 1/(1+x) = 1 -x but cannot find the result.

I got their answer. You are doing it right but, as already pointed out, maybe you forgot to keep the expansion to second order, which you have to do here,
$1/(1-x) = 1 + x + x^2 + \ldots$