5 ms·

Why wouldn't you be able to see your feet? Your head is also falling through the horizon (hopefully - otherwise you are going to be very unhappy), so the light

by ldunn 1y ago

Why wouldn't you be able to see your feet? Your head is also falling through the horizon (hopefully - otherwise you are going to be very unhappy), so the light from your feet doesn't need to escape the horizon for you to see it.

mr_toad 1y ago
The horizon is the distance at which escape velocity is c. Closer to the centre, including your feet, the escape velocity is higher. Electrical impulses wouldn’t be able to travel from the bottom of your brain to the top, so you’d be unconscious anyway.
- JoeAltmaier 1y ago
  ...and tidal forces may have reduced your body to red goo, so there's that
- ldunn 1y ago
  It is absolutely untrue that GR predicts that you would be knocked unconscious crossing the horizon. In fact one of the most fundamental aspects of GR (equivalence) predicts the exact opposite - there is no local experiment you can do as a freely falling observer to detect the horizon.
  - bencyoung 1y ago
    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
    - ldunn 1y ago
      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.
      - bencyoung 1y ago
        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-no-escape-3.svg 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. Happy not to discuss further!
        mr_toad 1y ago
        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.
        ldunn 1y ago
        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!
        bencyoung 1y ago
        Apart from my MSci in Physics... Perhaps you could post some links to the spacetime diagrams you are talking about?
        ldunn 1y 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_coordinates 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

by ldunn 1y ago

mr_toad 1y ago

The horizon is the distance at which escape velocity is c. Closer to the centre, including your feet, the escape velocity is higher. Electrical impulses wouldn’t be able to travel from the bottom of your brain to the top, so you’d be unconscious anyway.

JoeAltmaier 1y ago
...and tidal forces may have reduced your body to red goo, so there's that
ldunn 1y ago
It is absolutely untrue that GR predicts that you would be knocked unconscious crossing the horizon. In fact one of the most fundamental aspects of GR (equivalence) predicts the exact opposite - there is no local experiment you can do as a freely falling observer to detect the horizon.
- bencyoung 1y ago
  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
  - ldunn 1y ago
    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.
    - bencyoung 1y ago
      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-no-escape-3.svg 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. Happy not to discuss further!
      - mr_toad 1y ago
        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.
      - ldunn 1y ago
        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!
        bencyoung 1y ago
        Apart from my MSci in Physics... Perhaps you could post some links to the spacetime diagrams you are talking about?
        ldunn 1y 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_coordinates 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