Calculating average of a function of molecule's orientation (Euler angles)

Question

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 $\sin(\theta)$.

Answer 1

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)

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

Answer 2

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.