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The angle $\theta$ between two 3D-vectors with a uniform random orientation in space is distributed according to $sin(\theta)$. In Fig. 1 I have simulated random points on a sphere (like in How to find a random axis or unit vector in 3D?) and calculated the angle between the vectors to the points and the z-axis:

Fig 1: https://i.stack.imgur.com/TtkCx.jpg

NOW MY QUESTION: I am looking for a general expression if the vectors are not uniformly oriented in space anymore but rather look all in one direction with just small, gaussian distributed deviations in the angles (e.g. like molecules in a crystal lattice). I have the feeling that it should look something like $$ p_n(\theta)\cdot p_s(\theta), \text{ where } p_n \sim \mathcal{N}(\mu, \sigma), p_s \sim sin(\cdot)$$ (see simulation in Fig. 2):

Fig 2: https://i.stack.imgur.com/1usjb.jpg

Can that be true and how can I argue that from a probability theory (and also intuitive) point of view?

The expression $ p(\theta) = p_s(\theta)\cdot p_n(\theta) $ looks somewhat strange to me as I am multiplying two distributions of the same variable.

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magnetilo
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  • It's bit unclear, which are the random variables ("the objects" which behaves randomly) and what you want to find and what you exactly mean by orientation. I understand it this way: you want to determine the distribution of the angles given two randomly picked $v_1, v_2$, which are distributed according to $D$. In this case, what is $D$? – bjn Nov 18 '16 at 20:50
  • @bjn Yes, the formulation of the problem is part of my difficulties. My goal is to find an expression for the angle dist. between two molecules in a crystal lattice. If the molecules would vibrate in a 2D plane, the orientation could be expressed by one angle which would be gaussian distributed (my assumption). But in 3D I don't know how to express the orientation by just one variable and therefore I don't know how to express $D$. My idea was, that if a uniform dist. in 2D converts to a sin(.) dist. in 3D, a gaussian dist. in 2D converts to a "gauss*sin" in 3D. But I don't know how to justify. – magnetilo Nov 23 '16 at 15:52
  • Can you explain more what you mean by converting distribution of $\mathbb{R}^2$to $\mathbb{R}^3$? Although I must say this is rather strange, because $\mathbb{R}^2$ and $\mathbb{R}^3$ differs in dimension. I think one would need one more parameter to fix. – bjn Nov 27 '16 at 17:41
  • This may rather subject-specific related, hence my help is rather limited: On which basis/knowledge you choose a distribution of orientations $\mathbb{R}^3$? Can you not just define a distribution on $\mathbb{R}^3$(and so we can used the derived distributions of on set of orientations)? – bjn Nov 27 '16 at 17:49
  • I guess you mean by an orientation of a vector $v \in \mathbb{R}^3 \setminus {0}$ the vector $\frac{v}{| v |}$, which is an element of the unit sphere surface. In this case, an orientation cannot just described by one variable (as the orientations in $\mathbb{R}^2$), but by two variables, namely azimuthal and polar angle (see wikipedia article about spherical coordinate system). So you may define a distributions for azimuthal angle and polar angle to fix a distribution for orientations in $\mathbb{R}^3$. – bjn Nov 27 '16 at 18:17
  • You may use also other coordinate systems just as cylinderical coordiante system (see also same named wikipedia article), which is also used in your quoted post "How to find..." – bjn Nov 27 '16 at 18:17
  • Do you want to choose a specific distribution on orientations in $\mathbb{R}^3$, such as gaussian distribution or uniform distribution, and then you want to know the derived distribution of angles between such distributed orientations? In this case which distribution would like on the orientations in $\mathbb{R}^3$? – bjn Nov 27 '16 at 18:21

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In case anyone is once interested in this topic, I think I was now able to answer my question. I had especially problems with finding the right description for a "random orientation" when asking this question. The clue idea was that we can normalize, transform and turn two vectors in space, such that both vectors start from the origin and one of the two vectors is aligned with the z-axis, without affecting the probability distribution of the angle between the two vectors. The "random-orientation" of the second vector is then a point density on the unit sphere and the angle between the two vectors coincides with the polar angle of the spherical coordinate system.

If we now assume the two independent random angles $\Theta'$ and $\Phi'$, where $\Theta'$ is distributed in the yz-plane according to $f_{\Theta'}(\theta')$ and $\Phi'$ is distributed in the xy-plane according to $f_{\Phi'}(\phi')$, we find the probability of a point on the unit sphere contained in a differential area according to $$f(\theta', \phi') sin(\theta') d\theta' d\phi' = f_{\Theta'}(\theta') f_{\Phi'}(\phi') sin(\theta') d\theta' d\phi',$$ where $sin(\theta') d\theta' d\phi'$ is the surface area element of the spherical coordinates. Therefore we can get the probability distribution of $\theta$ from $$f_{\Theta}(\theta) d\theta = \int_{\phi} f_{\Theta'}(\theta) f_{\Phi'}(\phi') sin(\theta) d\theta' d\phi' = f_{\Theta'}(\theta) sin(\theta) d\theta$$ as $$f_{\Theta}(\theta) = f_{\Theta'}(\theta) sin(\theta).$$

In particular we can asume $f_{\Theta'}(\theta')$ to approximately be Gaussian in the interval $[0,\pi]$.

Please correct me if this is wrong..

magnetilo
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