5 ms·

For a proper head spin, consider that gravity propagates at the speed of light... And light is not fast enough to escape the black hole. https://www.forbes.com

For a proper head spin, consider that gravity propagates at the speed of light... And light is not fast enough to escape the black hole. https://www.forbes.com/sites/startswithabang/2019/10/24/this-is-why-the-speed-of-gravity-must-equal-the-speed-of-light/ https://www.forbes.com/sites/startswithabang/2019/10/24/this...

thehappypm 3y ago
Light is “fast enough” it just redshifts to nothing. Gravity has no equivalent
- pdonis 3y ago
  > Light is “fast enough” it just redshifts to nothing. Gravity has no equivalent This is wrong in two ways. First, light at the hole's horizon does not "redshift to nothing"; it just stays at the same radial coordinate because of the curvature of spacetime. The "redshift to nothing" view is an illusion, created by a bad choice of coordinates; that illusion was corrected in the late 1950s and early 1960s by the discovery of better coordinates. Second, gravity does have an equivalent to light: gravitational radiation. Gravitational waves travel on null geodesics, just like light, and any gravitational waves at the hole's horizon would stay at the same radial coordinate just as light does.
  - az09mugen 3y ago
    Thank you a lot pdonis for all your explanations along the comments !
    - pdonis 3y ago
      You're welcome! :-)
  - thehappypm 3y ago
    Let’s say there is a light bulb inside the event horizon. It emits a photon. From the light bulb’s perspective, it sees the photon speeding away at the speed of light. From an outside observer, we never see the photon. It can’t make it out. But — the photon does exist, and it is traveling at the speed of light over a finite (though deeply warped in spacetime) distance. How does the photon not reach us then? Is it not going at the speed of light? The redshift model applies here; the photon is redshifted until it has no energy from our frame.
pdonis 3y ago
> gravity propagates at the speed of light... And light is not fast enough to escape the black hole. Gravity doesn't have to get out of the hole. The gravity you feel outside a black hole is due to the global spacetime geometry, not to any "force" coming out of the hole. If you want to attribute it ultimately to the presence of matter, it is the matter in your past light cone, which originally collapsed to form the hole, before it fell below the event horizon. See this Usenet Physics FAQ article: https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles...
- pixl97 3y ago
  Ya if gravity had to get out of the hole the if a black hole happened in that universe it would just snap shut and kind of disappear.
BurningFrog 3y ago
My understanding is that the black hole curves space so much that there is no path out of it. This does trap light inside, but speed has nothing to do with it.
- pdonis 3y ago
  No timelike or null path out of it, yes. (That is, no path through spacetime that an ordinary object or a light ray can follow.)

thehappypm 3y ago

Light is “fast enough” it just redshifts to nothing. Gravity has no equivalent

pdonis 3y ago
> Light is “fast enough” it just redshifts to nothing. Gravity has no equivalent This is wrong in two ways. First, light at the hole's horizon does not "redshift to nothing"; it just stays at the same radial coordinate because of the curvature of spacetime. The "redshift to nothing" view is an illusion, created by a bad choice of coordinates; that illusion was corrected in the late 1950s and early 1960s by the discovery of better coordinates. Second, gravity does have an equivalent to light: gravitational radiation. Gravitational waves travel on null geodesics, just like light, and any gravitational waves at the hole's horizon would stay at the same radial coordinate just as light does.
- az09mugen 3y ago
  Thank you a lot pdonis for all your explanations along the comments !
  - pdonis 3y ago
    You're welcome! :-)
- thehappypm 3y ago
  Let’s say there is a light bulb inside the event horizon. It emits a photon. From the light bulb’s perspective, it sees the photon speeding away at the speed of light. From an outside observer, we never see the photon. It can’t make it out. But — the photon does exist, and it is traveling at the speed of light over a finite (though deeply warped in spacetime) distance. How does the photon not reach us then? Is it not going at the speed of light? The redshift model applies here; the photon is redshifted until it has no energy from our frame.

pdonis 3y ago

> gravity propagates at the speed of light... And light is not fast enough to escape the black hole. Gravity doesn't have to get out of the hole. The gravity you feel outside a black hole is due to the global spacetime geometry, not to any "force" coming out of the hole. If you want to attribute it ultimately to the presence of matter, it is the matter in your past light cone, which originally collapsed to form the hole, before it fell below the event horizon. See this Usenet Physics FAQ article: https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles...

pixl97 3y ago
Ya if gravity had to get out of the hole the if a black hole happened in that universe it would just snap shut and kind of disappear.

BurningFrog 3y ago

My understanding is that the black hole curves space so much that there is no path out of it. This does trap light inside, but speed has nothing to do with it.

pdonis 3y ago
No timelike or null path out of it, yes. (That is, no path through spacetime that an ordinary object or a light ray can follow.)