The Hamiltonian for a particle is given by:

$$H_1 = \frac{P^2}{2m} + \frac{p^2_\theta}{2I} + \frac{p^2_\phi}{2Isin^2 \theta}$$

$I$ is the moment of inertia

$$I = \frac{mR^2}{4}$$

To get the partition function for one particle $Z_1(P, Q)$ the following integral has to be solved:

$$Z_1(P, Q, p_ \theta, \theta, p_\phi, \phi) = \int_ {-\infty}^{+\infty}\frac{dPdQ}{\lambda^3} dp_\theta d\theta dp_\phi d\phi e^{-\beta H_1}$$

Where $\lambda$ is the thermal wavelength:

$$\lambda = h \sqrt{\frac{\beta}{2 \pi m}}$$

My issue here is that I do not know how to solve this 6 dimensional integral.

I am asked for the average energy < E >

$$\langle E \rangle = -\frac{ \partial \log(Z_1)}{\partial \beta}$$


$$\langle E \rangle = \frac{5}{2\beta}$$

  • 3
    $\begingroup$ Z factorizes into 5 one dimensional Gaussian integral, the ensemble average over E becomes a sum of the 5 contributions, each one contributes $\dfrac{1}{2} k T$ $\endgroup$ – Count Iblis Oct 3 '18 at 17:26

With $$H = \frac{P^2}{2m} + \frac{P_{\theta}^{2}}{2 m r} + \frac{P_{\phi}^{2}}{2 m r \sin^2\theta}$$ then $$I = \int e^{- \beta H} \, dP \, dP_{\theta} \, dP_{\phi} \, dr \, d\theta \, d\phi$$ becomes: \begin{align} I &= \int d^{3}q \, \int e^{-\beta P^{2}/2m} \, dP \cdot \int e^{- \beta P_{\theta}^{2}/(2 m r)} \, dP_{\theta} \, \cdot \int e^{- \beta P_{\phi}^{2}/(2 m r \sin \theta)} \, dP_{\phi} \\ &= \int d^{3}q \, \left(\frac{2 m \pi}{\beta}\right)^{1/2} \, \left(\frac{2 m \pi r}{\beta}\right)^{1/2} \, \left(\frac{2 m \pi r \sin^2\theta}{\beta}\right)^{1/2} \\ &= \left( \frac{2 \pi m}{\beta}\right)^{3/2} \, \int r^2 \sin\theta \, dr \, d\theta \, d\phi \\ &= 4 \pi \left( \frac{2 \pi m}{\beta}\right)^{3/2}. \end{align}

This is how the integral is computed. The remainder is to add in the various constants and compute the average energy.

  • $\begingroup$ What I do not see is why you changed $R$ by $r$ with this equation: $R = 2 r$. Did you do it just for convenience (make the integral easier)? Note that without any change you get $\frac{2 p^2_\theta}{mR^2}$ in the second term (for instance) $\endgroup$ – JD_PM Oct 4 '18 at 8:57
  • $\begingroup$ The result of this integral $\int r^2 \sin\theta \, dr \, d\theta \, d\phi$ would not be $\frac{4\pi}{3}$ ? $\endgroup$ – JD_PM Oct 4 '18 at 9:49
  • $\begingroup$ More details if you are interested: physics.qandaexchange.com/?qa=3122/… $\endgroup$ – JD_PM Oct 11 '18 at 18:57

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