Proof of the Equipartition Theorem

[dRic2]

The proof for the ET I've found in some of the undergrad books for statistical physics (for example in Reif's "Statistical and Thermal Physics") assumes the form of the Hamiltonian of the system to be:
$$H = bp_i^2 + E'(q_1,...,p_f)$$
where $b$ is a constant.
My professor in his notes, says that $b$ can be a function of all the coordinates and momenta as long as it does not depend on the i-th term, i.e., $b = b(q_j, p_{j \neq i})$. I fail to understand the proof however. Since now $b$ is a function of the generalized coordinates and momenta the integral
$$<bp_i^2> = \int d^{3f}q d^{3f}p bp_i^2 e^{-\beta H}$$
no longer factorizes.
In his notes he just says that recalling that the integral in $p_i$ is the second moment of a Gaussian we immediately obtain $<bp_i^2> = \frac 1 2 k_B T$. But I am unable to perform the calculations.

Any help would be greatly appreciated.

thanks
Ric

 

[vanhees71]

But the integral factorizes. If $b$ doesn't depend on $p_i$ you have
$$\langle b p_i^2 \rangle=\frac{1}{Z} \int \mathrm{d}^{3f}q \mathrm{d}^{3f-3} p b \exp(-\beta H') \int_{\mathbb{R}} \mathrm{d}^3 p_i p_i^2 \exp[-\beta b p_i^2] ,$$
where the last integral leaves out the $p_i$ integration, which is done explicitly and $H'$ just omits the term $b p_i^2$. Also don't forget the normalization factor, i.e., the partition sum which cancels all the stuff you omit to get the equipartition theorem in the given form.

 

[dRic2]

don't forget the normalization factor

Sorry, my fault.

But the integral factorizes ...
...
...
the partition sum which cancels all the stuff you omit to get the equipartition theorem in the given form.

Still I don't get it. $b$ depends on all the other coordinates, so how can you simplify the remaining therms?