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Homework Statement
Consider a pair of electric dipoles ##\mu## and ##\mu'##, oriented in the directions ##(\theta,\phi)## and ##(\theta',\phi')## respectively; the distance ##R## between their centers is assumed to be fixed. The potential energy in this orientation is given by ##\varepsilon = -\frac{\mu\mu'}{R^3}[2\cos\theta \cos\theta' - \sin\theta \sin\theta' \cos(\phi - \phi')]##.
Now, consider this pair of dipoles to be in thermal equilibrium, their orientations being governed by a canonical distribution. Show that the mean force between these dipoles, at high temperatures, is given by ##\langle F \rangle = -2\frac{(\mu\mu')^2}{k_B T}\frac{\hat{R}}{R^7}##, with ##\hat{R}## being the unit vector in the direction of the line of centers.
This is Pathria problem 3.36 3rd edition.
The Attempt at a Solution
The problem itself is straightforward. Letting ##(\theta,\theta', \phi,\phi')## be the configuration space coordinates, we calculate the canonical partition function ##Z = \int_{0}^{2\pi} d\phi' \int_{0}^{2\pi}d\phi \int_{0}^{\pi}d\theta' \int_{0}^{\pi}d\theta e^{\beta\frac{\mu\mu'}{R^3}[2\cos\theta \cos\theta' - \sin\theta \sin\theta' \cos(\phi - \phi')] }## and from this we get ##\langle E \rangle = -\partial_{\beta}\ln Z ## and finally we calculate ##\langle \vec{F} \rangle = - \vec{\nabla} \langle E \rangle ## and we're done.
The issue of course is in calculating the partition function itself. Since we are working with very high temperatures, so that ##\beta \ll 1##, I expanded the integrand to second order in ##\beta##: ##e^{\beta\frac{\mu\mu'}{R^3}[2\cos\theta \cos\theta' - \sin\theta \sin\theta' \cos(\phi - \phi')] } \\= 1 + \beta\frac{\mu\mu'}{R^3}[2\cos\theta \cos\theta' - \sin\theta \sin\theta' \cos(\phi - \phi')] \\+ \frac{1}{2}\beta^2 \frac{(\mu\mu')^2}{R^6}[2\cos\theta \cos\theta' - \sin\theta \sin\theta' \cos(\phi - \phi')]^2 + O(\beta^3)##.
I expanded to second order in ##\beta## because it's the lowest non-vanishing and non-trivial term in the expansion that contributes to the partition function; it's clear that the first order term does not contribute to ##Z##. The zeroth order term just gives ##4\pi^4##; noting that the cross term in the expansion of the second order term does not contribute to the partition function , we are left with for the second order term [tex]Z^{(2)} = \frac{1}{2}\beta^2 \frac{(\mu\mu')^2}{R^6}\int_{0}^{2\pi} d\phi' \int_{0}^{2\pi}d\phi \int_{0}^{\pi}d\theta' \int_{0}^{\pi}d\theta [4\cos^2\theta \cos^2\theta' + \sin^2\theta \sin^2\theta' \cos^2(\phi - \phi')] \\ = 8\pi^2 \beta^2 \frac{(\mu\mu')^2}{R^6}\int_{0}^{\pi}d\theta' \cos^2\theta'\int_{0}^{\pi}d\theta \cos^2\theta + \pi^2\beta^2 \frac{(\mu\mu')^2}{R^6} \int_{0}^{\pi}d\theta' \sin^2\theta'\int_{0}^{\pi}d\theta \sin^2\theta \\ = 2\pi^4 \beta^2 \frac{(\mu\mu')^2}{R^6} + \frac{\pi^4}{4}\beta^2 \frac{(\mu\mu')^2}{R^6} = \frac{9}{4} \pi^4\beta^2 \frac{(\mu\mu')^2}{R^6} [/tex]
Then ##\ln Z = \ln (4\pi^4 + \frac{9}{4} \pi^4\beta^2 \frac{(\mu\mu')^2}{R^6} + O(\beta^3)) \\= \ln 4\pi ^4+ \ln (1 + \frac{9}{16} \beta^2 \frac{(\mu\mu')^2}{R^6} + O(\beta^3)) =\ln 4\pi ^4 + \frac{9}{16} \beta^2 \frac{(\mu\mu')^2}{R^6} + O(\beta^3)##
where I've expanded the logarithm to second order in ##\beta##. Thus ##\langle E \rangle = -\partial_{\beta}\ln Z = -\frac{9}{8}\beta \frac{(\mu\mu')^2}{R^6} + O(\beta^2)##. Finally, choosing the origin of spherical coordinate system to be on the center of one of the dipoles we have ##\langle \vec{F} \rangle = -\partial_R \langle E \rangle \hat{R} = -\frac{27}{4}\frac{(\mu\mu')^2}{k_B T}\frac{\hat{R}}{R^7} + O(\beta^2)##.
As you can see everything is correct except for that factor of ##\frac{27}{4}## which should really be a ##2##. I've backtracked over and over and I can't figure out why I'm getting the wrong multiplicative factor. Either I've not been careful in retaining all second order terms in ##\beta## throughout the calculation, or the approximations just don't work, or I've just made a silly arithmetic error somewhere that I can't spot. Any help is appreciated, thanks in advance!