Relativistic motion of rigid body One friend of mine, who just entered university asked me about one problem on relativistic motion.
Problem sounds like this: System $K'$ moves with velocity $V$ parallel to $x$-axis relative to system K. In system $K'$ cylinder oriented parallel to $x'$-axis travels with velocity $V'_{y}$ parallel to $y'$-axis. Find angle $\theta$ in system $K$ that cylinder makes with $x$-axis. My initial take was just to take Lorentz transformations and than find $\arctan(V_{y}/V_{x})=\theta$. With $V'_{x}=0$ Lorentz transformation becomes:
$$V_{x}=V$$
$$V_{y}=V'_{y}\sqrt{1-\frac{V^{2}}{c^{2}}}$$
$$\theta = \arctan\left(\frac{V}{V_{y}^{'}}\gamma\right)$$
But this is angle between velocity and $x$-axis. And in my assumption cylinder is parallel to $x$-axis in $K$ system. But answer is different.
$$\theta = \arctan\left(\frac{V'_{y}V}{c^{2}}\gamma\right)$$
After that I found solution which begins with phrase: "Imagine lamp in the middle of cylinder. After light reaches edges of cylinder it starts to move with $V'_{y}$ in $K'$ system parallel to $x'$-axis." And then they say that in $K$ system edges will start to move in different time that's why angle should be like this. I understand all the formulas but I don't get physics in first sentence. I tried to imagine it like there is a force applied to the center of cylinder, and because of atomic structure this force then travels through this bonds pulling everything. Is this correct idea? What if force is applied uniformly through out cylinder? Will there still be a delay in $K$-system?
 A: Your question is somewhat ambiguous, but I believe the problem is that you have overlooked the relativity of simultaneity.
In frame K' the cylinder is oriented parallel to the x'axis and is moving along the y'axis. In other words, at a given time t' in K', both ends of the cylinder have moved the same distance in the y' direction.
However, in K, because of the relativity of simultaneity, a given time t corresponds to two different times at the cylinder ends in K'. In K, at a given time t, the leading edge of the cylinder is at one time in K' and the trailing edge is at a later time. In other words, at a given time in K, the trailing edge will have travelled further along the y'-axis than the leading edge, so the cylinder will no longer be parallel to the x'-axis.
To solve the problem, you need to consider the relativity of simultaneity, which will cause the two ends of the cylinder to have moved different distances from the x'axis, and will have caused the cylinder to be length contracted too.
As for the final part of your question, no real forces have acted on the cylinder- in its own rest frame its orientation and length have not changed.
A: The speed composition formulas give: $\theta =\arctan \left(\frac{V'_{y}}{\gamma V}\right)$
If we replace the cylinder with the light coming from a star, i.e. $V' = -c$, we can calculate the aberration angle of the light from this formula, i.e. $\theta'=-\pi/2$
The aberration angle is: $\alpha = -\pi/2-\theta=-(\pi/2+\theta)$
$\tan(\alpha)=-\tan(\pi/2+\theta)= \cot(\theta)$
$=- \frac{\beta}{\sqrt{1-\beta^{2}}},\ \beta=V/c$
The two formulas, relativistic and classical, coincide up to $\beta^{2}$
