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Looking at the Kruskal diagram for black holes, it seems to me that at the horizon, all free falling objects cross the horizon at the same coordinate radius and time. Does this not mean that all objects crossing the horizon do so at the same time (i.e. two objects starting free fall at different locations must meet at the surface at the same instant because they cross at the same coordinate radius and time)? Since we can say 'two objects begin to fall at t=0', doesn't this mean that two objects at the same radius and coordinate time are 'the same event'

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It just means the Schwarzschild coordinate system is faulty at the horizon because it assigns the same coordinates $(t,r,\theta, \phi)$=$(\infty,2m,\theta,\phi)$ to multiple events that are actually distinct events.

If you look at the Kruskal-Szekeres coordinate system you can pick an event on the horizon and then draw the past light cone and those are events it can see before it crosses the event horizon. You can also pick and event at the singularity and draw the past light cone of that event. And the events in that past light cone are the are the events it can see before it hits the singularity.

So if you draw the past light cone of an event on the singularity where particle one hits it. Then pick a world line (for particle two) with a future pointing tangent that crosses the event horizon outside that past light cone. Then from the path the first particle sees it doesn't even know if the second particle crosses the horizon or not.

And that is because there are multiple future pointing curves that agree on that past light cone. And some of them go in and some of them do not. So it depends on whether that second particle fires its rockets or not.

So they definitely don't cross together because particles that cross together see each other.

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