# Homework Help: Average force between electric dipoles

1. Aug 8, 2014

### WannabeNewton

1. The problem statement, all variables and given/known data

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.

3. 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 $$Z^{(2)} = \frac{1}{2}\beta^2 \frac{(\mu\mu')^2}{R^6}\int_{0}^{2\pi} d\phi' \int_{0}^{2\pi}d\phi \int_{0}^{\pi}d\theta' \int_{0}^{\pi}d\theta [4\cos^2\theta \cos^2\theta' + \sin^2\theta \sin^2\theta' \cos^2(\phi - \phi')] \\ = 8\pi^2 \beta^2 \frac{(\mu\mu')^2}{R^6}\int_{0}^{\pi}d\theta' \cos^2\theta'\int_{0}^{\pi}d\theta \cos^2\theta + \pi^2\beta^2 \frac{(\mu\mu')^2}{R^6} \int_{0}^{\pi}d\theta' \sin^2\theta'\int_{0}^{\pi}d\theta \sin^2\theta \\ = 2\pi^4 \beta^2 \frac{(\mu\mu')^2}{R^6} + \frac{\pi^4}{4}\beta^2 \frac{(\mu\mu')^2}{R^6} = \frac{9}{4} \pi^4\beta^2 \frac{(\mu\mu')^2}{R^6}$$

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!

2. Aug 8, 2014

### Orodruin

Staff Emeritus
So, statistical mechanics was never my forte, but: configuration space for each dipole is a unit sphere. What is the area element on a sphere?

3. Aug 8, 2014

### WannabeNewton

Hi, thanks for the reply Orodruin. With the area elements $\sin\theta' \sin\theta$ we get $$Z^{(0)} = 16\pi^2,Z^{(1)} = 0,\\ Z^{(2)} = 8\pi^2 \beta^2 \frac{(\mu\mu')^2}{R^6}\int_{0}^{\pi}d\theta' \sin\theta'\cos^2\theta'\int_{0}^{\pi}d\theta \sin\theta\cos^2\theta + \pi^2\beta^2 \frac{(\mu\mu')^2}{R^6} \int_{0}^{\pi}d\theta' \sin^3\theta'\int_{0}^{\pi}d\theta \sin^3\theta \\ = \frac{16}{3} \pi^2\beta^2 \frac{(\mu\mu')^2}{R^6}$$
so $\ln Z = \ln 16\pi^2 + \frac{1}{3}\beta^2 \frac{(\mu\mu')^2}{R^6} + O(\beta^3)$ which yields $\langle F \rangle = \hat{R}\partial_R \partial_{\beta}\ln Z = -4\frac{(\mu\mu')^2}{k_B T}\frac{\hat{R}}{R^7} + O(\beta^2)$ which is still off by a factor of $2$ so I don't know if I made an arithmetic mistake along the way.

But more importantly, why do we include the spherical area elements here? The partition function for this dipole-dipole system isn't a Cartesian integral that is transformed to spherical coordinates so as to acquire the usual Jacobian. Rather it's just an integral over all possible orientations of the two dipoles i.e. over all possible values of the respective polar and azimuthal angles $\theta,\theta', \phi, \phi'$, with respective polar and azimuthal conjugate momenta $p_{\theta}, p_{\theta'},p_{\phi}, p_{\phi'}$ corresponding to internal rotational degrees of freedom which I don't think the dipoles in this problem possess.

See for example the solution to problem 2(a): https://physics.ucsd.edu/students/courses/spring2013/physics210a/HOMEWORK/SOL03.pdf [Broken]

EDIT: Ok it would seem that whenever we include the degrees of freedom corresponding to the internal rotational kinetic energies of the dipoles, we need not include the spherical area elements whereas if we don't include them then we do need to include the spherical area elements, as can be seen by comparing the regular and alternative solutions to problem 2 (p.3): http://www.physics.oregonstate.edu/~tgiebult/COURSES/ph441/hmwk/441hw4s.pdf...but why is this?

Last edited by a moderator: May 6, 2017
4. Aug 9, 2014

### Orodruin

Staff Emeritus
Again, not my strongest subject, but my reasoning would be something along the following lines:

Look at the symmetries of phase space in the absence of a potential. In this case each spin direction should be equally probable so the density of states should be proportional to the area element on the unit sphere.

On the other hand, rotational degrees of freedom are depending on the angular momentum which is rather related to the tangent space of the sphere, which is flat.

5. Aug 9, 2014

### vanhees71

I checked your integrals (of course with the correct area element of the unit spheres $\mathrm{d}^2 \Omega_j=\mathrm{d} \vartheta_j \mathrm{d} \varphi_j \sin \vartheta_{j}$, which must be there because the Hilbert space of a rotator is $L^2(\Omega)$ with this integration measure. You cannot use "canonical quantization" for non-Cartesian coordinates! See the textbook on pathintegrals by Kleinert, where he discusses this for the rigid rotator (quantization of the "spinning top", which can be used as rough model for the rotational modes of molecules).

Your result is too large by a factor of two, because you forgot the 1/2 from the virial expansion, which you apply here to the two-body order obviously. See Landau-Lifshits vol. V (Statistical Physics). In my German edition it's paragraph 75, Eq. (75,3).

I don't know the book by Pathria. That seems to be pretty tough. Do you think it's a good book? I'm still looking for a good graduate-level statistics textbook. My favorites so far are, Landau-Lifshits V, IX, X; Vol. V of Sommerfeld's Lectures on Theoretical Physics, and Lebellac, Equilibrium and non-equilibrium statistical mechanics (in this order), but all of them are either pretty tough or somewhat outdated (particularly Sommerfeld's lectures, but these are still the best books ever written on classical physics, if you ask me).

6. Aug 9, 2014

### WannabeNewton

I found my mistake; it was a silly one as usual. I wrote $\langle \vec{F} \rangle = -\partial_R \langle E \rangle = \partial_R \partial_{\beta}\ln Z$ but really it should be $$\langle \vec{F}\rangle = -\hat{R} \langle \partial_R E \rangle = -\hat{R} \frac{\int \partial_R E e^{-\beta E}}{\int e^{-\beta E}} = \hat{R}\frac{1}{\beta} \frac{\partial_R\int e^{-\beta E}}{\int e^{-\beta E}} = \hat{R}\frac{1}{\beta}\partial_R \ln Z$$
which makes more sense since $\langle \vec{F} \rangle$ is conjugate to $R$. This yields $\langle \vec{F} \rangle = -2\frac{(\mu\mu')^2}{k_B T}\frac{\hat{R}}{R^7} + O(\beta^2)$ as desired.

I forgot that thermodynamically we actually have $\langle F \rangle = -\partial_R \langle E \rangle |_S = -\partial_R \langle A \rangle |_T$ where $\langle A \rangle$ is the Helmholtz free energy. I mistakenly calculated $\langle F \rangle = -\partial_R \langle E \rangle |_T$ which of course does not follow from averaging the fundamental relation $F_n = -\partial_R E_n$ over all states $n$ in the canonical ensemble.

However I'm afraid I still don't understand the spherical area element vs. no spherical area element dichotomy even with your explanations vanhees and Orodruin. Why is it $Z = \int_{0}^{2\pi} d\phi\int _{0}^{\pi}d\theta \int_{-\infty}^{\infty} dp_{\theta}\int_{-\infty}^{\infty} dp_{\phi} e^{-\frac{\beta}{2I}(p_{\theta}^2 + \frac{p_{\phi}^2}{\sin^2\theta})}e^{-\mu \mathcal{E} \cos\theta}$ doesn't require a solid angle $d\Omega$ but $Z = \int_{0}^{2\pi} \int _{0}^{\pi}d\Omega e^{-\mu \mathcal{E} \cos\theta}$ does require a solid angle $d\Omega$? Thanks.

@vanhees, yes I think Pathria is a very good book! I've been using it in conjunction with Landau and Kardar.

Last edited: Aug 9, 2014
7. Aug 9, 2014

### vanhees71

Hm, but what's then with this factor 1/2 from the virial expansion?

Of course you are right, the average of the force is not the gradient of the average potential.

8. Aug 9, 2014

### Orodruin

Staff Emeritus
How much differential geometry do you know? Are you familiar with tangent bundles?

9. Aug 9, 2014

### WannabeNewton

Yes. Most of my differential geometry knowledge comes from GR.

10. Aug 9, 2014

### Orodruin

Staff Emeritus
Ok, so the phase space of a classical system is simply the cotangent bundle to the configuration space (a point in phase space is a point in configuration space together with a cotangent vector). If the configuration space has a particular symmetry, the density of states should be invariant under this symmetry. This implies taking the invariant measure on the configuration space (the invariant surface element in the case of the sphere). However, the cotangent space at each point will get a flat measure. Integration over the states will imply integrating with this invariant measure over the cotangent bundle (phase space).

11. Aug 9, 2014

### vanhees71

Now, that I've read the entire thread again, I realize that we have misunderstood each other, because I'm used to the grand-canonical potential from many-body QFT, where this is the quantity you use. You worked in the canoncial ensemble, and your analysis of the factor of 2 is completely right.

Now to your other question concerning the integration measure. I guess in your example, where you claim not to have a "solid-angle measure" you are referring to the two-atomic mulecule's rotation degrees in freedom.

You must keep in mind that you have to integrate over phase space. The integration measure is always
$$\mathrm{d}^N q \mathrm{d}^N p/(2 \pi \hbar)^N,$$
where $q$ and $p$ are arbitrary generalized coordinates and their momenta, which have to be determined from generalized coordinates going via the Lagrangian (holonomous generalized velocities and the associated canonical momenta).

Since the center-mass coordinates and momenta totally separate, we can treat the rotational part separately since then $Z=Z_{\text{trans}} Z_{\text{rot}}$. The translational part is just as for the monatomic ideal gas.

For the two-atomic molecule the kinetic energy (which is its Lagrangian) reads
$$L_{\text{rot}}=\frac{I}{2} \left (\dot{\vartheta}^2+\sin^2 \vartheta \dot{\psi}^2 \right).$$
The rotation around the line connecting the two atoms can be neglected, because the moment of inertia is much smaller compared to the ones around axes perpendicular to it.

The canonical momenta are
$$p_{\vartheta}=I \dot{\vartheta}, \quad p_{\varphi}=I \sin^2 \vartheta \dot{\psi}^2.$$
The Hamiltonian is
$$H_{\text{rot}}=\frac{1}{2I} \left (p_{\vartheta}^2+\frac{p_{\varphi}^2}{\sin^2 \vartheta} \right ).$$
The single-particle partition sum thus is
$$Z_{\text{rot}}=\frac{1}{(2 \pi \hbar)^2} \int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \int_{\mathbb{R}^2} \mathrm{d} p_{\vartheta} \mathrm{d} p_{\varphi} \exp(-\beta H_{\text{rot}}).$$
This is pretty tedious to evaluate.

Thus you remember that you can express the kinetic energy also in terms of the coordinates in the principle-axes rigid-body system, where the Hamiltonian becomes much simpler. Here we can use a shortcut by just realizing that a substitution
$$J_1=p_{\vartheta}, \quad J_2=\frac{p_{\varphi}}{\sin \vartheta}$$
leads to a much simpler integral, but then you need to include the corresponding Jacobian
$$\mathrm{d} p_{\vartheta} \mathrm{d} p_{\varphi}=\mathrm{d} J_1 \mathrm{d} J_2 \sin \vartheta.$$
The $\sin \vartheta$ comes from the orientation of the axis of the molecule relative to an arbitrarily fixed $z$ axis. The derivation shows that the new angular-momentum components $J_1$ and $J_2$ are not canonically conjugated momenta to the corresponding angle variables, and thus you must be careful with the integration measure. Only for pairs of conjugate coordinates and momenta you have the simple phase-space measure $$\mathrm{d}^N q \mathrm{d}^N p/(2 \pi\hbar)^3$$
to naively count the number of quantum states in the classical limit.

NB: It's really quantum also in the classical statistical theory; I've never seen a convincing argument from classical mechanics, why the phase-space measure is to be chosen as was conjectured by Boltzmann. Only from quantum theory that's easily derived; see Landau vol. V; also all troubles with the Gibbs paradox are immediately gone. Of course, at low temperatures you must the use full quantum theory of the rigid top (see Landau-Lifshits vol. III, or the textbook by Kleinert, where an elegant method using group theory is given to get the correct integration measure in the Hilbert space of the spinning top in Euler-angle representation; the naive "canonical quantization" fails!).

In the case of the two dipoles you use their distance from each other and the the two spherical-coordinate angles for their orientation relative to a fixed axis in space. You can choose this axis as the direction of the vector connecting the locations of the two dipoles. Then you immediately get the formula for $\epsilon$ you started from. The phase space of this orientational degrees of freedom is thus the corresponding two unit spheres of these orientations and thus the phase-space elements are $\mathrm{d} \Omega_1 \mathrm{d} \Omega_2=\mathrm{d} \vartheta_1 \mathrm{d} \vartheta_2 \mathrm{d} \varphi_1 \mathrm{d} \varphi_2 \sin \vartheta_1 \sin \vartheta_2$.

12. Aug 9, 2014

### WannabeNewton

Thank you both very much. You explanations made perfect sense. I think I understand now.

vanhees just to recapitulate, if I have a set of generalized coordinates $(\theta,\phi)$ of dipole orientations, with associated generalized conjugate momenta corresponding to internal rotations, then the partition function measure is always $d\theta d\phi dp_{\theta}dp_{\phi}$ correct?

The partition function itself is $$\int_{0}^{2\pi} d\phi\int _{0}^{\pi}d\theta \int_{-\infty}^{\infty} dp_{\theta}\int_{-\infty}^{\infty} dp_{\phi} e^{-\frac{\beta}{2I}(p_{\theta}^2 + \frac{p_{\phi}^2}{\sin^2\theta})}e^{-\beta\mu \mathcal{E} \cos\theta}$$

We then define a new coordinates $\sin\theta J_{\phi} = p_{\phi}, J_{\theta} = p_{\theta}$ which will of course no longer be the conjugate momenta to the coordinates $\theta,\phi$. With this we then have $e^{-\frac{\beta}{2I}(p_{\theta}^2 + \frac{p_{\phi}^2}{\sin^2\theta})} = e^{-\frac{\beta}{2I}(J_{\theta}^2 + J_{\phi}^2)}$ and $dp_{\phi} = \frac{\partial p_{\phi}}{\partial J_i}dJ_i = \sin\theta, dp_{\theta} = dJ_{\theta}$ so $$Z = \int_{0}^{2\pi} d\phi\int _{0}^{\pi}d\theta \sin\theta e^{-\beta\mu \mathcal{E} \cos\theta} \int_{-\infty}^{\infty} dJ_{\theta}\int_{-\infty}^{\infty} dJ_{\phi} e^{-\frac{\beta}{2I}(J_{\theta}^2 + J_{\phi}^2)}$$

which separates the contribution of the internal rotation to the partition function from the contribution of the dipole potential and allows us to evaluate separately the dipole contribution $Z_p = \int_{0}^{2\pi} d\phi\int _{0}^{\pi}d\theta \sin\theta e^{-\beta\mu \mathcal{E} \cos\theta}$ without having to evaluate the rotational kinetic part since that has been separated off. Right?

13. Aug 9, 2014

### Staff: Mentor

I think the answer can be quite simple: Your angular coordinates are arbitrary - they have no physical meaning, you could choose any other coordinates. There is no special reason to expect $d\theta d\phi$ to be fundamental (this is different for $d\theta d\phi dp_{\theta}dp_{\phi}$). The relevant element is dΩ where you need the sin(θ) in your coordinates.

14. Aug 10, 2014

### vanhees71

It's all correct what you write, but now I don't understand, where the $-\beta \mu \mathcal{E} \cos \theta$ term comes from. The evaluation of the spatial part of the dipole-dipole interaction was correct already in your posting #6.

15. Aug 10, 2014

### WannabeNewton

Sorry I was just using the case of a single dipole in an external electric field for simplicity since the concept is the same for the dipole-dipole interaction.

Yes I see your point.

Thanks!

16. Aug 11, 2014

### vanhees71

I see, that's also correct then.