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In this paper, orientational average of a function of Euler angles, $f(\phi,\theta,\psi)$, is defined as: $$\langle f\rangle=\frac{1}{8 \pi^2} \int_0^\pi \int_0^{2 \pi} \int_0^{2 \pi} f(\theta, \phi, \chi) \sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} \chi$$

My question is: Why do we need the $\sin(\theta)$? What's wrong with defining it as follows: $$\langle f\rangle=\frac{1}{4 \pi^3} \int_0^\pi \int_0^{2 \pi} \int_0^{2 \pi} f(\theta, \phi, \chi) \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} \chi$$ This still has the correct normalization without the $\sin(\theta)$.

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2 Answers 2

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The reason is the same why, in the case of the spherical coordinates, the surface area element contains the $\sin(\theta)$ factor: we would like to have a uniform measure on the unit sphere surface described by the coordinates $\theta$ and $\phi$.

Lines at constant longitude ($\phi$), the "meridians," have the same length. Lines at constant latitude ($\theta$), the "parallels," have an increasing size going from the poles (zero-length) to the "Equator" (same length as the meridians). It is a simple exercise to check that the length of the parallels is proportional to a $\sin (\theta)$ factor. Therefore, the surface area of two small elements on the surface of a unitary radius sphere, corresponding to the same coordinate variation $d\theta, d\phi$, at different latitudes (like the green and the red elements in the following figure) enter image description here

must be written as $$ \sin (\theta) d \theta d \phi.$$ Similarly, for the average on Euler's angles.

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Maybe this is just a silly comment, but it is too long to be a comment itself.

@GiorgioP 's answer explains why the "$\sin$" has to be in your expression, but is still interesting to notice that sometimes it can be "hidden" in the integrand:

$$\int_0^\pi \int_0^{2 \pi} \int_0^{2 \pi} f(\theta, \phi, \chi) \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} \chi=\int_0^\pi \int_0^{2 \pi} \int_0^{2 \pi} f'(\theta, \phi, \chi) \sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} \chi.$$

Where $f$ and $f'$ are different functions. This might not be as trivial as one would think. A nice example is the problem of computing the canonical ensemble of the diatomic molecule in the classical approximation.

Let the Hamiltonian of one diatomic molecule in cartesian coordinates be $H(\vec{r_1},\vec{r_2},\vec{p_2},\vec{p_2})=\frac{\vec{p_2}^2+\vec{p_1}^2}{2m}+V(|\vec{r_1}-\vec{r_2}|)$. If one compute the partition function: $$\mathcal{Z}\propto \int d\vec{p_1} \, d\vec{p_2}\int d\vec{r_1} \, d\vec{r_2}e^{-\beta H(\vec{r_1},\vec{r_2},\vec{p_2},\vec{p_2})}\propto \int dr \, d\theta \, d\phi \, r^2 \sin(\theta)e^{-\beta V(r)}. $$

Where I introduced the Jacobian of the change of coordinates to express the problem in spherical coordinates.

One can also face the problem by expressing the Hamiltonian in the coordinates of the center of mass directly, having $H(\vec{r}_{CM},r,\phi,\theta,\vec{p}_{CM},p_r,p_\phi,p\theta)=\frac{\vec{p}_{CM}^2}{2m}+V(r)+\frac{1}{2m}(p_r^2+\frac{p_\theta}{r^2}+\frac{p_\phi}{r^2\sin(\theta)}).$ Then, the partition function reads:

$$\mathcal{Z}\propto \int d\vec{p}_{CM} \, dp_r\, dp_\phi \, dp_\theta \int d\vec{r}_{CM} \, dr \, d\phi\, d\theta \, e^{-\beta H(\vec{r}_{CM},r,\phi,\theta,\vec{p}_{CM},p_r,p_\phi,p\theta)} $$.

So I am integrating now on spherical coordinates without the Jacobian of the transformation! Obviously, this is just an apparent contradiction, both integrals give the same result, but the "$\sin$" is "hidden" in the exponential in the second case.

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  • $\begingroup$ Using the center of mass coordinates does not imply necessarily the use of spherical coordinates. Since the integral over the relative position must be the same in spherical and cartesian coordinates, the Jacobian of the transformation must appear in any integral over spherical coordinates and also in the formula for the two-body partition function. $\endgroup$
    – GiorgioP
    Jun 28 at 11:34
  • $\begingroup$ The process is: Say the Hamiltonian is expressed in coordinates $X$-> $H_X(X,P_X)$. Say also that you have a different set of coordinates ($Y(X)$), so you also can express the Hamiltonian in these coordinates ($H_Y(Y,P_Y)$). Then, $\int dX\, dP_X e^{-\beta H_X(X,P_X) }=\int dY\, dP_X J(X/Y) H_X(Y,P_X)=\int dY\, dP_Y e^{-\beta H_Y(Y,P_Y) } $. Where $J(X/Y)$ is de Jacobian of the transformation. $\endgroup$
    – Javi
    Jun 28 at 11:45
  • $\begingroup$ If you were right, the second virial coefficient should not contain the $r^2$ factor and should be reduced by a factor 2 ( see en.wikipedia.org/wiki/Virial_coefficient ). $\endgroup$
    – GiorgioP
    Jun 28 at 12:05
  • $\begingroup$ @GiorgioP Sorry, I don't get your argument on the virial coefficient. I am just stating that the Jacobian of a canonical transformation equals 1. In my notation, $J(X,P_X/Y,P_Y)=1$. This is a well-known result in classical mechanics (e.g. en.wikipedia.org/wiki/… or physics.stackexchange.com/a/459673/252421). $\endgroup$
    – Javi
    Jun 28 at 15:25
  • $\begingroup$ The key point is that the transformation between cartesian and spherical coordinates in configuration space is not a canonical transformation. The canonical transformation from the old canonical coordinates (including the $p$'s and the new (including the new momenta) has Jacobian equal to 1. And the $\sin$ term is not hidden in the exponential but is canceled by a similar term in the momentum volume element. You can work out easily the simpler case of polar coordinates in the plane to realize that $dx dy \neq dr d\phi$, but $dp_x dp_y dx dy = dp_r dp_{\phi} dr d{\phi}$. $\endgroup$
    – GiorgioP
    Jun 28 at 17:50

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