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.

[*]: knowing $\vec{R}_0,$ $\vec{O}_0$ and $\vec{R},$$\vec{O}.$ Notation: $\Delta \vec{R} = \vec{R}-\vec{R}_0$

My attempts:

  1. To know the displacement vector along the long axis of the cylinder, I take the scalar product between $\Delta \vec{R}$ and $\vec{O}_0,$ i.e. $\Delta \vec{R}_{||} = (\Delta \vec{R}\cdot \vec{O}_0)\vec{O}_0/l $ normalized by $l.$ Is this correct?
  2. Now how can I get the components of the displacement vector in the plane perpendicular to the long axis of the cylinder? that is, the components of $\Delta \vec{R}_{\perp}.$ Could it simply be $\Delta \vec{R}_{\perp} = \Delta \vec{R}-\Delta \vec{R}_{||}?$
  3. Knowing the orientation vectors $\vec{O}_0$ and $\vec{O}$ before and after the displacement respectively, can we estimate the angular displacement? my guess would be something along: the angular displacement $\Delta \theta =\arccos{(\frac{\vec{O}\cdot \vec{O}_0}{l^2}}).$
  • $\begingroup$ Is $\|\vec O_0\|=l$? $\endgroup$ – amd Dec 13 '17 at 21:04
  • $\begingroup$ @amd indeed, also $||\vec{O}||=l$, sorry I forget to properly state that in the post. $\endgroup$ – user929304 Dec 13 '17 at 21:19

If I correctly understood what it is that you’re trying to compute, then it looks more or less right to me.

You’re resolving $\Delta \vec R$ into its components parallel and orthogonal to $\vec O_0$. The former is given by the orthogonal projection $$\Delta\vec R_\parallel = \left(\Delta \vec R \cdot {\vec O_0 \over \|O_0\|}\right){\vec O_0 \over \|O_0\|} = {\Delta\vec R \cdot \vec O_0 \over \vec O_0 \cdot \vec O_0}\vec O_0 = \frac1{l^2}(\Delta\vec R \cdot \vec O_0)\,\vec O_0.$$ Note that the denominator is $l^2$, not $l$—you have to use a unit vector in the dot product, too.

The perpendicular component is the orthogonal rejection from $\vec O_0$, which is indeed simply $\Delta \vec R_\perp = \Delta\vec R-\Delta\vec R_\parallel$, and the angle between two vectors can be extracted from their dot product, as you have it. If you want to get a sense of the direction of rotation, consider examining $\vec O_0 \times \vec O$. This will give you a vector normal to the plane of rotation with length proportional to the sine of the angle, and whose direction gives you an orientation for a clockwise rotation to align the two vectors.


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