# Collision and simultaneity in special relativity

An almost rigid rectangle with length $$D$$ can rotate around its center. Assume that rectangle is horizontal and does not rotate before collision. According to observer $$S$$ two balls collide at the same time at two ends points of rectangle $$R$$ & $$L$$, in $$t=0$$ and before collision two balls have same speed $$u$$ in $$y$$ axis therefore there is no rotation for rectangle at all after clash. For these collisions we can assign two events in spacetime, that is $$E_1=(-ct,r)=(0,-D/2,0,0)$$ and $$E_2=(0,+D/2,0,0)$$. Note that $$S$$ is at rest relative to rectangle.

Now let's consider another observer $$S'$$ who moves at speed $$-v$$ relative to rectangle in $$x$$ axis. By using Lorentz transformation we can see that $$E_1'=(-\gamma v D/2c,\gamma (-D/2+ vt),0,0)$$ and $$E_2'=(\gamma v D/2c,\gamma (+D/2+ vt),0,0)$$. Because two events $$E_1'$$ and $$E_2'$$ are not simultaneous in this frame ($$\Delta t'=\gamma vD/c^2$$) it's logical to deduce that the rectangle will rotate after first collision. While it's not possible for obvious reasons (one can put a bomb under rectangle for example). Where am i mistaken? Does system wait for a signal that is coming from two end points to the center?

PS: I am fully aware that 100% rigid bodies are not possible in SR, by almost rigid i meant that we have an elastic collision and rectangle tends to rotates than bends around it's center.

• It looks like you’ve answered your own question. The rectangle is not rigid, so it deforms slightly all along it’s length to transmit the momentum of each collision. In the rectangle’s frame, these deformations propagate simultaneously from each end. In other reference frames the deformation occur in a different sequence, but still all occur and transmit momentum. – Duncan Harris Apr 4 '19 at 16:17
• My concern is angular momentum, if these deformations does not propagate simultaneously to center, they won't arrive at the same time to the center necessarily (unless they are transmitted with speed of light i guess) hence they won't cancel each other out after that so there'd be a rotation, or at least a standing wave in the rectangle probably. – Paradoxy Apr 4 '19 at 16:27
• The event of the deformations reaching the center is localized in both space and time, so it is a proper relativistic “event”, so the deformations will cancel simultaneously at the center in all frames (assuming they cancel rather than producing a standing wave in any frame, which is dubious but perhaps possible). – Duncan Harris Apr 4 '19 at 16:33

It looks like you’ve almost answered your own question with your postscript. The rectangle is not rigid, so it deforms slightly all along it’s length to transmit the momentum of each collision. In the rectangle’s frame, these deformations propagate simultaneously from both ends to the center. In other reference frames the deformations occur in a different sequence, but still all occur and transmit momentum equivalently.

Perhaps of special interest is the event of the deformations reaching the center. This event is localized in both space and time, so it is a proper relativistic “event”, so the deformations will cancel simultaneously at the center in all frames (assuming they cancel rather than producing a standing wave in any frame, which is dubious but perhaps possible).

• Well you are right, but what if i put infinite number of small balls above rectangle? In that case, i would have a born rigid body, which eliminate the need of considering information transmutation between points of rectangle. – Paradoxy Apr 4 '19 at 17:17
• That case does not eliminate the need for deformations and momentum transmission unless you also eliminate the pivot constraint. Without the pivot the rectangle will accelerate. – Duncan Harris Apr 4 '19 at 17:24