You can do plenty of experiments to see if you are falling, e.g. hitting the surface of a planet you are falling towards. The event horizon is a surface like any other with a location in space and you can definitely see when you hit it (it's the bit where no light is coming out). And once you've crossed it, literally no EM radiation can move further from the singularity
I agree that if you are freely falling and then you are suddenly not freely falling because you hit the surface of a planet and experienced a huge acceleration, you will notice. That doesn't have anything to do with anything I said, but it is undeniably true.
An event horizon is not like the surface of a planet - you will not be accelerated as you pass through it.
It is, once again, irrelevant that light cannot propagate outward once you're behind the horizon because, again, you are falling towards the center, and in particular you are falling through the future light cone of your feet. Please look at some spacetime diagrams if you do not believe me, preferably ones in Kruskal-Szekeres coordinates.
In GR spacetime is locally flat and for an inertial observer special relativity applies, up to tidal corrections which can be made arbitrarily small at the horizon by considering a suitably large black hole. This is a deep and important fact about GR. The idea that falling through the horizon causes you to suddenly not be able to see your feet anymore appears to obviously violate this basic principle, so if you think your assertion is true you should be able to explain why either this principle of GR is actually not true, or why your assertion does not actually violate this principle.
Well, I disagree. Light literally can't move in a direction that makes it further from the singularity once inside the event horizon. I don't see what space being flat or not locally has to do with that. Check https://en.wikipedia.org/wiki/Event_horizon#/media/File:BH-n... for an example.
If your head is further from the singularity than your feet then you can't see them.
It doesn't have to move in such a direction! Look at a spacetime diagram and think about the trajectory of your head and feet! Read a book on GR! Do literally anything except have strong opinions about GR when you don't know any GR!
So when will we be able to just run general relativity numerical simulations on our desktop machines? So that you could set up Observer A at some point, and Observer B at some other point and and mass distribution, etc, then just crunch the numbers to see what each observer could see/measure as time evolves for each observer. Seems like the differential equations are straight forward enough(?). Is the possibility of singularities at the center of a black hole the hard part? What if you just simulated something that was 99.99% of the density needed to get a black hole? I suppose that you'd need a 4 dimensional matrix to hold the simulation (three space coordinates plus a time coordinate)? Is it that we just don't have enough RAM and storage yet in consumer machines? If your simulation did a 1,000 points in ever dimension, that would be 1e12 points. If there are 10 components of the tensor at each point and we are using 64-bit doubles per parameter, that means our simulation takes up ~80 TB. Or is it a that we don't have enough processing speed? Or are there still some philosophical issues that need to be decided when you program up the simulator? How many lines of code is a numerical general relativity solver using something like Euler's method? Is the core of a naive version less than, say 500 lines of C? I can see an optimized CUDA version being significantly larger of course.
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.
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.
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
I was considering a stationary observer inside the event horizon, but that’s not possible. ldunn is correct that a free-falling observer will catch up with the photons reflected from their feet.
Space being flat locally is important because if the gravitational gradient is too high (i.e. you get too close to the singularity) your feet will be accelerated much faster than your head.
