We have a cylinder of length $l$ (in units of its radius $d,$ as basic unit of length set to $d=1.$) in a box, and we consider an orthonormal Cartesian coordinate system with its origin placed at the centre of the box. We know the position of the cylinder in terms of the coordinates of its centre of mass, so $\vec{R}=(r_x,r_y,r_z)$ and in terms of its orientation vector (vector along its main/long axis) $\vec{O}=(o_x,o_y,o_z).$ Assuming at a later time the position and orientation of the cylinder have been randomized (still in the box), we'd like to estimate the displacement vector in terms of: how much the cylinder has moved parallel to its long axis ($\Delta \vec{R}_{||}$), how much it's moved in the plane perpendicular to its long axis ($\Delta \vec{R}_{\perp}$), how much it's been rotated in terms of $\Delta \theta$ and $\Delta \phi$ (usual angular definitions in spherical coordinates).

Knowing $\vec{R}$ and $\vec{O}$ both before and after displacement, I'm wondering how to estimate these correctly. For instance I guess for the displacement along the long axis, it's probably easy and given by: $\Delta \vec{R}_{||} = (\Delta \vec{R}\cdot \vec{O}_0)\vec{O}_0/l, $ right? How about the other ones?


You're correct, if the orientational vector is unitary. Otherwise you must calculate the unitary vector in the direction of $\vec{O}_1$ and then perform the projection (just divide by the norm of the orientational vector). The perpendicular component is just the rejection vector of $\Delta \vec{R}$, given by the definition: $\Delta\vec{R}_{\perp} = \Delta\vec{R} - \Delta\vec{R}_{\parallel} $.

If you draw the vectors you will be able to visualize that this is correct and then convince yourself. Take a look:

enter image description here

You can determine straight forward the rotation by the internal product between the two orientational vectors $\vec{O}_1$ and $\vec{O}_2$.

$\theta = \arccos{\left( \frac{\vec{O}_1 \cdot \vec{O}_2}{|\vec{O}_1||\vec{O}_2|}\right)} $

  • $\begingroup$ Thank you very much for posting this answer. Indeed this sounds very reasonable. On the other hand, is my approach for the displacement parallel to the long axis correct? If yes, I admit it is incredibly naive to ask, but can $\Delta \vec{R}_{\perp}$ be obtained from $\Delta \vec{R}$ and $\Delta \vec{R}_{||}?$ First guess one d think it is simply $\Delta \vec{R} - \Delta \vec{R}_{||}$ but I cannot convince myself :( thank you again $\endgroup$ – user929304 Dec 12 '17 at 23:10
  • $\begingroup$ I edited my answer. Take a look. :) $\endgroup$ – The Doctor Dec 13 '17 at 14:02
  • $\begingroup$ Because you want to know how much the cylinder was displaced and not only the direction. If you project a vector over another vector with norm greater than 1 you will have the correct direction but the components would be the result of the components multiplication. Remember, you want to project the components of your vector at the same time that you conserve the norm. The general definition of a projection $p$ of the vector $a$ over any given vector $b$ is: $\vec{p} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$ $\endgroup$ – The Doctor Dec 13 '17 at 15:37
  • $\begingroup$ I would recommend you to read the wikipedia article about vector projection: en.wikipedia.org/wiki/Vector_projection Please, if you think that I have answered all your questions, consider selecting my answer as the best one. :) $\endgroup$ – The Doctor Dec 13 '17 at 15:39
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    $\begingroup$ sure of course, thanks a lot again. Makes sense, I'll get back to you if i'm confused again :) $\endgroup$ – user929304 Dec 13 '17 at 15:42

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