Time of collision of two relativistic speed particles

Question

Suppose I have two particles, one moving at $0.9c$ to the right, starting at $(-0.9c,0,0)$ in the lab frame at $t=0$ and the second one moving at $0.9c$ to the left, starting at $(0.9c,0,0)$. In the lab frame at $t=1$ these particles are going to collide at the origin $(0,0,0)$ where we've placed a marker and annihilate. At the beginning the marker is $0.9c$ from the first particle and the second particle is $1.8c$ away from it in the lab frame.

In the lab frame the annihilation will happen at the point of the marker exactly.

We know that in the frame of the first particle, the marker is moving towards the particle at speed $0.9c$ and by the speed addition formula the second particle is moving towards the first particle at speed $\frac{0.9c+0.9c}{1+\frac{0.81c^2}{c^2}} = 0.995c$. We also have length contraction in the frame of the first particle which sees the distance between it's starting point and the marker as $0.9c\sqrt{1-0.81c^2/c^2} = 0.39c$. It also sees the distance between its starting point and the starting point of the second particle as twice this, so $0.78c$.

The marker is moving towards it at 0.9c, so from the reference frame of the particle the marker reaches it in $0.39c/0.9c = 0.43$ seconds. The second particle is moving towards the first particle at $0.995c$ and has a distance of $0.78c$ to cover which takes $0.78c/0.995c = 0.785$ seconds to reach the first particle and annihilate both of them.However when $0.785$ seconds will have elapsed the marker will be well behind the particle and so in this frame the annihilation event will not happen at the space location of the marker.

To me this seems weird and my intuition isn't helping me see why it's possible for the location of the annihilation event to differ from the marker just based on the reference frame we're using. What's the right way to be thinking about this situation?

EDIT: Here is the spacetime diagram as requested:

This actually confuses me more if anything, it seems to show that the time of the collision in the reference frame of the first particle is at $t' > 1$ which means from the point of view of the particle it takes longer to collide than from the lab frame view, which is weird since distances are contracted for the particle frame compared to the lab frame, so the collision should occur in less time.

Answer 1

In the lab frame, the first particle passes through 0.9 at the same time that the second particle passes through -0.9. But these events will not necessarily be simultaneous in other reference frames.

Suppose you start in the location of the target, equidistant from each particle. You will receive a signal when each particle is at +/- 0.9. If you are in the lab frame, you will receive each signal at the same time and can conclude that these events were simultaneous.

But if you are in the frame of one of the particles, then the signal from the other (approaching) particle will reach you first and therefore you will conclude that it reached 0.9 before the particle in your reference frame.

Equivalently, suppose you will trigger emitters at +/- 0.9 with a signal when you are equidistant from them. Depending on your reference frame, your signal could reach the emitters at the same time or one could be triggered before the other.

Either way, you will need to account for this head start when determining when the particles will strike the target.

Answer 2

The invalid step is in assuming that the length measured by the original frame will only be contracted. i.e. it is not $.39c$

The issue is that once you changed the frame of reference, what was simultaneous in the original frame acquires a time difference in the new frame, and that part is being neglected by your calculation.