Special Relativity: 2-D Collisions for Alice & Bob

[MrBlank]

Alice and Bob are initially in the same inertial frame. There are 2 point test masses m1 and m2. Initially m1 is at the origin and m2 is on the positive x-axis. At time zero, m1 is instantaneously accelerated to velocity Vx in the positive x-direction. After some time, m1 collides with m2.

An instant before time zero, Bob is instantaneously accelerated to velocity Vy in the negative y-direction, from Alice’s point of view. From Bob’s point of view, m1 and m2 will have velocity Vy in the positive y-direction before m1 is accelerated. After m1 is accelerated in the positive x-direction, it will have a velocity less than Vy in the positive y-direction. See

https://www.physicsforums.com/threads/acceleration-in-special-relativity-in-2-dimensions.977658/
It will appear to Bob that m1 and m2 do not collide. However, in reality they do collide. Bob will see that m1 and m2 are affected by this collision, even though he observed no collision.

It’s possible to adjust the initial position of m2 so that Alice observes that m1 and m2 do not collide, but Bob observes that they do collide. However, in reality they do not collide. Bob will observe m1 and m2 pass through each other with no effect on either mass.

As far as I can tell, in order to know if a mass is involved in any collisions, you must be in the rest frame of the mass.

Is this correct?

 

[Dale]

All frames agree on whether or not they collided. If they collide then that means that both objects are at the same event. If they are at the same event in one frame then they are clearly at the same event in every frame

 

[Ibix]

As far as I can tell, in order to know if a mass is involved in any collisions, you must be in the rest frame of the mass.

Is this correct?

Obviously not. Either the objects collided or they did not, and all you need to do is watch them - no frames are needed in this approach.

Alice and Bob are initially in the same inertial frame.

Everything is "in" all inertial frames. A frame is simply a choice of coordinates, a way to describe reality. From your subsequent description, I think you mean that Alice and Bob are initially at rest in the same inertial frame.

After m1 is accelerated in the positive x-direction, it will have a velocity less than Vy in the positive y-direction.

No it won't. In the thread you linked, pervect did provide a circumstance in which this could happen, but as you've stated this setup that circumstance cannot apply because you specifically set things up so that the two objects do collide. All you need to do is write down the coordinates of the objects in one frame and Lorentz transform them. You'll see that they have equal y velocities.

 

[Ibix]

All you need to do is write down the coordinates of the objects in one frame and Lorentz transform them.

I have a few minutes free, so I thought I'd do this for you, @MrBlank. In Alice's frame, m1 has coordinates $(t,x,y)=(t,v_xt,0)$ and m2 has coordinates $(t,X,0)$, where $X$ is some constant and $v_x$ is what you called Vx.

Now we boost to Bob's frame, moving at $v_y$ in the $+y$ direction. To do this we use the Lorentz transforms$$\begin{eqnarray*}
t'&=&\gamma\left(t-\frac{v_y}{c^2}y\right)\\
x'&=&x\\
y'&=&\gamma(y-v_yt)
\end{eqnarray*}$$where $\gamma=\sqrt{1-v_y^2/c^2}$. Inserting the unprimed coordinates of m1 we get primed coordinates $(t',x',y')=(\gamma t,v_xt,-\gamma v_yt)$. Inserting the coordinates of m2 we get $(\gamma t,X,-\gamma v_yt)$. Clearly the objects always have the same $y'$ coordinate (you can write it as $y'=-v_yt'$ if you prefer), so this frame also describes the two objects colliding (unsurprisingly).

As Nugatory commented in your previous thread, velocity addition does not work as you seem to think it does. Do calculate the velocity of m1 in this frame. It remains below $c$, although it's a little tricky to see it (I recommend writing $v_x=c-\epsilon_x$ and $v_y=c-\epsilon_y$ and considering the case where the two $\epsilon$ are very small compared to $c$).

Handwaving arguments are no substitute for maths.