ldunn 7 months ago

The diagram on the Wikipedia page for Kruskal-Szekeres coordinates[1] does the job. There you see the trajectory of some infalling observer along with some future light cones[2] of points along that trajectory and the event horizon marked as the dashed line. The usual Schwarzschild r and t coordinates are also shown as the pale hyperbolas.

Say the trajectory that's drawn on the diagram is the trajectory of your feet. Now consider a second trajectory which begins slightly displaced "outwards" (that is, rightwards at t=0 on the diagram) from this first one - that's your head. Hopefully you agree that the head-trajectory would have to do something pretty strange to avoid crossing through the future lightcone of your feet, even behind the horizon. This doesn't require signals from your feet to travel "outward" - it's just that your head is travelling "inward".

K-S coordinates make it pretty clear that nothing drastic happens to the structure of spacetime at the event horizon - everything is perfectly regular. It's just that once you cross the horizon, the singularity (the thick hyperbola at the top of the diagram) is inevitably in your future: there is no trajectory within any future lightcone behind the horizon that doesn't run into the singularity. You're doomed to run into it in finite time, and all your future lightcones lie entirely behind the horizon.

[1]: https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coord...

[2]: A useful feature of K-S coordinates is that lightcones are always at +-45 degrees

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  • bencyoung 7 months ago

    Thanks! Hmm, I think we're talking about slightly different things but it's been too long since I studied it to put it in the right words :)

    I completely agree that spacetime can be "flatish" for a large block hole, but the event horizon does still represent a boundary right?

    Consider the edge case of crossing the event horizon itself at some speed <<c (because you've got magic thrusters fighting the pull). At some point your feet will be through the event horizon and your head won't be. Do you agree that at that point you won't be able to see your feet?

    I agree that your head will pass through the future light cone of your feet, and so could do somethign to affect your head (by emitting something falling slower than your head), but I'm not sure any light rays could follow that path.

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    • bencyoung 7 months ago

      Ok I drilled down a bit and looks like you are right, although I'm still not sure I've built up a clear understanding! (https://physics.stackexchange.com/questions/187917/thought-e...). In fact that question (series of onions) is exactly how I visualised it...

      I'm not quite sure from that discussion why an event horizon is equivalent to a body moving outwards at the speed of light but it does make some sense. GR is always fun!

      I still don't have a good idea of the "slow moving crossing the event horizon" case" but I'll read around it some more

      Maybe the difference is between "free fall through an event horizon" vs "hover" (as much as is possible) at an event horizon

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