6 ms·

What about the center of a black hole? Is it younger than the surface?

by p1esk 7y ago

zamadatix 7y ago
Black holes are a singularity. They have an even horizon but it's not really a surface of an object it's just a place in space where certain conditions occur. What happens underneath that horizon (and in the singularity) isn't solidly known.
colechristensen 7y ago
That is a bit more difficult to reckon. A distant observer watching their unfortunate friend falling into a black hole will see their friends clock tick slower and slower such that they will never, from the outside, be seen to cross the event horizon, the passage of time will slow towards zero. From the outside, time does not appear to pass at all inside a black hole. The unfortunate traveller, however, won't even notice the crossing on a bigger black hole.
- raattgift 7y ago
  Our universe is more complicated than e.g. a vacuum Schwarzschild black hole spacetime. For starters, there are probably lots of black holes in our universe (and even in our galaxy). What happens if the unfortunate friend falling into the black hole is being watched from an orbit close to another black hole? (General Relativity has the somewhat frustrating property that adding a black hole to a black hole spacetime does not obey the principle of linear superposition, so things can get quite messy if the two black holes in the paragraph above are close together. One can get even kinkier, for instance by considering a stellar-mass black hole orbiting a supermassive black hole, and kinkier still if both of them rotate with unaligned rotational axes and/or with opposite spins.) Next, the metric expansion of space means that an isolated black hole would be more like a Schwarzschild-de Sitter solution. There we have the problem that two widely-separated observers can be highly cosmologically redshifted with respect to each other even if neither is anywhere near a black hole. One can of course also consider a collapsing universe in which two observers are blueshifted with respect to one another, and one could consider what that blueshift would do to observations by A of B if B were falling into a black hole in the collapsing universe. Moreover, the presence of matter can make a mess of things. One can contrive an arrangement of matter near to a black hole which is sufficiently dense as to offset the gravitational time dilation of someone near the horizon. Finally, there's kinematics: we can subject the unfortunate friend and the observer to a relative ultraboost wherein the doppler blueshift undoes the gravitational redshift. A relatively ultraboosted Schwarzschild black hole would look a little odd to the relatively ultraboosted observer: https://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultraboost https://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultrab... Things get weirder when you add other black holes, stars, dust, and so on to the Aichelburg-Sexl picture. > From the outside, time does not appear to pass at all inside a black hole From outside the event horizon of a static black hole there is no way to tell, even if one uses tricks like the above to do away with the objection that in general time appears to come to an effective standstill infinitesimally outside the horizon. However, if the infaller brings in significant mass or angular momentum, we would expect the horizon to reflect the internal configuration change, since it's the internals that generate the horizon in the first place. This is a dynamical rather than a static (up to tiny perturbation) black hole. This is extremely fiendishly difficult, and is run into in practice in terms of trying to work out exactly what the full theory says about the final merger of two black holes. Roughly speaking, the gravitational interactions between two merging black holes is best described by a metric wherein there is an effective third (and sometimes fourth) body.
Filligree 7y ago
The question is slightly ill-formed. I understand why you'd ask it, but there are a couple of possible answers, none of which work directly. - No, because a black hole has nothing underneath the event horizon. (This is true from the perspective of anyone outside it, at least to a first approximation. Gravity is affected by curvature, and whatever might be in the centre cannot be the cause of the event horizon, except historically, but the event horizon is stable in itself. It's a self-sustaining cascade of space-time.) - Mu, because once you pass the event horizon the extreme warping of space-time turns "inwards" into "future", making the centre of the hole a point in time, not space. This makes it difficult to determine what the question is asking. - Maybe, because singularities probably don't exist. There should be something in the centre, which might evolve over time. It might just be a faster-than-light cascade of space-time, though. We still can't answer your question, because we don't know how to accurately model the 'singularity'. (As anything other than a singularity, and that's essentially just the math giving up.)
- raattgift 7y ago
  > the extreme warping of space-time turns "inwards" into "future", making the centre of the hole a point in time, not space. Assuming an exact black hole solution to the Einstein Field Equations (EFEs), what's happening is that any observer sufficiently close to the singularity will inevitably collide with the singularity. One can then define the horizon the outer boundary at which the future light cone attached at every point ends in the singularity. In order to avoid the singularity in the future one would need a causal cone broader than that of the light cone. The boundary is only a surface in the abstract; it's not material, you can't touch it, nothing bounces off it, and the no-drama conjecture says that if you were to float through the horizon you won't immediately notice you've done so. Outside the horizon as defined above, some part of the future light cone will have the singularity in it, however as one choose an observer at a point further and further away from the black hole, the singularity occupies a smaller and smaller part of the future light cone for that observer at that point, and the observer would have to aim ever more carefully to collide with it. (The observer simply has to reach a point at the horizon). > a black hole has nothing underneath the event horizon That's true of vacuum solutions of the EFEs which contain black holes, but it's not true in general. The Vaiyda metric is an exact solution which equips a spherically symmetrical central mass with spherically symmetric non-interacting dust of incoming and/or outgoing radiation: a Vaiyda black hole can have some incoming radiation within the horizon. Astrophysical black holes are more complicated since the matter outside them is generally far from spherically symmetrical. However, within the horizon will be some of the cosmic microwave background's (CMB's) photon gas, and whatever else reaches the horizon. But, you're right that something must be deep inside an astrophysical black hole, and the something is perhaps not a classical singularity, but rather some configuration that preserves the quantum mechanical properties of the field-content that reached the horizon. > might evolve over time If there's infalling matter, it must evolve over time, as the mass will increase, and likely so will the angular momentum, and charges are likely to go back-and-forth as one cannot guarantee that only chargless matter (e.g. photons) will fall in, nor that if we let charged matter fall in, it will always be charge-neutralized (e.g., no ions allowed!). That's true of a classical singularity, and it will be even more true for any ultradense non-singularity structure that preserves QFT information. > faster-than-light cascade of space-time What's that? > Gravity is affected by curvature In General Relativity, gravity is curvature. > whatever might be in the centre cannot be the cause of the event horizon Mass-energy generates the metric. In the case of the Schwarzschild black hole solution to the EFEs, the central mass (which is all the mass and energy in the vacuum Schwarzschild spacetime) is static and concentrated at one point. However, a non-rotating star, or a slowly-rotating object like the moon, generates a very close approximation of the Schwarzschild solution. Astrophysical black holes generate an approximation of the vacuum Schwarzschild black hole spacetime, but the Kerr metric is a better approximation for rotating black holes, and there is plenty of matter in the universe outside black holes. The true metric of the entire universe is much much more complicated than our (useful!) approximations, but nevertheless it is determined by the scatter of mass-energy throughout the spacetime. The true metric almost certianly describes regions which behave almost indistinguishably from the interiors of Schwarzschild (and Kerr, etc.) black holes. Another way of looking at this: as one throws matter into a black hole, the mass of the black hole increases, and so therefore does the diameter of the black hole. If you allow matter to cross into the black hole and collide with the singularity (or whatever replaces it), how can you maintain the italicized part of the quote above? > the event horizon is stable in itself If you allow mass to increase by letting dust and gas and cosmic microwaves reach the horizon, how do you keep the horizon stable? Moreover, if you merge two black holes, how do you keep both black holes' horizons stable? > It's a self-sustaining Given Hawking radiation then a black hole that's a solution to the Einstien Field Equations and equipped with one or more quantum fields filling the whole spacetime where the energy-density of the quantum fields is below a critical threshold will shrink. In the limit of quantum vacuum outside, this shrinking becomes complete evaporation: the horizon simply is not found in the future, because the mass in the whole spacetime has reconfigured from concentration at a central point to diffusion away from the central point. Since mass-energy generates the metric, the metric ceases to be Schwarzschild (or whatever black hole metric had existed prior to final evaporation). In our universe, there is lots of mass-energy outside black holes, and even though the metric expansion of space will make the CMB energy-density very small, it will still be higher than the critical threshold for stellar mass and larger black holes for a lonnnnnng time. But this isn't the black holes self-sustaining: they would evaporate, except for the CMB dust falling onto them, so it's the CMB (and its neutrino equivalent) that's doing the sustaining.
pdonis 7y ago
The "center" of a black hole is not a place in space. It's a moment of time (that is to the future of all other moments inside the horizon).
- raattgift 7y ago
  Not disagreeing that there's a point of no return wherein an infaller will inevitably collide with the singularity, just whether it's really correct to say that makes the singularity a moment of time. For starters, different infallers can arrive at the singularity at different times (even ones that arrive at the horizon at the same time can be relatively accelerated, resulting in having different interval lengths). Also not sure it makes sense for non-eternal BHs. For a BH that forms by gravitational collapse and eventually completely evaporates, surely the singularity has a timelike worldline? Some "lucky" too-early infallers (a cosmic neutrino, say) might pass through the centre collapsing progenitor before the horizon forms, while too-late infallers might pass through the region after final evaporation (assuming no horizon-equipped remnant). The "just-miss" is depends on where in spacetime the coincidence happens, and any reasonable slicing or threading would attribute the miss to being at the wrong moment in time. Maybe easiest to think about that if the singularity is not always at the spatial origin of a system of coordinates. For instance, if SN1987a's remnant contains a black hole, that black hole is surely moving around the galaxy with the luminous matter "now", but before the final collapse there was neither horizon nor singularity. I'm also perhaps unreasonably superstitious about the slogan's survival of arbitrary parameterizations, e.g., one can do an affine parameter on even an eternal BH's singularity and have what looks like a decent proper time for it. Perhaps another way of putting it is that the usual Carter-Penrose BH diagrams showing long horizontal wavy line segments for the singularity are misleading because of exaggerations caused by the particular conformal chart it uses, kinda like how the Earth's very-near-polar regions are not that big even though they look that way on Mercator charts (those are also conformal, and you get really different results in the near-polar regions in other conformal projections even as closely related as a transverse Mercator/Gauss conformal projection, https://en.wikipedia.org/wiki/Map_projection#/media/File:Usgs_map_mercator.svg https://en.wikipedia.org/wiki/Map_projection#/media/File:Usg... versus https://commons.wikimedia.org/wiki/File:Usgs_map_traverse_mercator.PNG https://commons.wikimedia.org/wiki/File:Usgs_map_traverse_me... ). Just thinking aloud in the wee hours, and maybe having given too little weight (pardon the pun) to the qualification "inside the horizon" just before your close paren. There's a lot of rigour that can be hidden in those three words.

What about the center of a black hole? Is it younger than the surface?

by p1esk 7y ago

What about the center of a black hole? Is it younger than the surface?

Black holes are a singularity. They have an even horizon but it's not really a surface of an object it's just a place in space where certain conditions occur. What happens underneath that horizon (and in the singularity) isn't solidly known.

colechristensen 7y ago

That is a bit more difficult to reckon. A distant observer watching their unfortunate friend falling into a black hole will see their friends clock tick slower and slower such that they will never, from the outside, be seen to cross the event horizon, the passage of time will slow towards zero. From the outside, time does not appear to pass at all inside a black hole. The unfortunate traveller, however, won't even notice the crossing on a bigger black hole.

raattgift 7y ago
Our universe is more complicated than e.g. a vacuum Schwarzschild black hole spacetime. For starters, there are probably lots of black holes in our universe (and even in our galaxy). What happens if the unfortunate friend falling into the black hole is being watched from an orbit close to another black hole? (General Relativity has the somewhat frustrating property that adding a black hole to a black hole spacetime does not obey the principle of linear superposition, so things can get quite messy if the two black holes in the paragraph above are close together. One can get even kinkier, for instance by considering a stellar-mass black hole orbiting a supermassive black hole, and kinkier still if both of them rotate with unaligned rotational axes and/or with opposite spins.) Next, the metric expansion of space means that an isolated black hole would be more like a Schwarzschild-de Sitter solution. There we have the problem that two widely-separated observers can be highly cosmologically redshifted with respect to each other even if neither is anywhere near a black hole. One can of course also consider a collapsing universe in which two observers are blueshifted with respect to one another, and one could consider what that blueshift would do to observations by A of B if B were falling into a black hole in the collapsing universe. Moreover, the presence of matter can make a mess of things. One can contrive an arrangement of matter near to a black hole which is sufficiently dense as to offset the gravitational time dilation of someone near the horizon. Finally, there's kinematics: we can subject the unfortunate friend and the observer to a relative ultraboost wherein the doppler blueshift undoes the gravitational redshift. A relatively ultraboosted Schwarzschild black hole would look a little odd to the relatively ultraboosted observer: https://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultraboost https://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultrab... Things get weirder when you add other black holes, stars, dust, and so on to the Aichelburg-Sexl picture. > From the outside, time does not appear to pass at all inside a black hole From outside the event horizon of a static black hole there is no way to tell, even if one uses tricks like the above to do away with the objection that in general time appears to come to an effective standstill infinitesimally outside the horizon. However, if the infaller brings in significant mass or angular momentum, we would expect the horizon to reflect the internal configuration change, since it's the internals that generate the horizon in the first place. This is a dynamical rather than a static (up to tiny perturbation) black hole. This is extremely fiendishly difficult, and is run into in practice in terms of trying to work out exactly what the full theory says about the final merger of two black holes. Roughly speaking, the gravitational interactions between two merging black holes is best described by a metric wherein there is an effective third (and sometimes fourth) body.

Filligree 7y ago

The question is slightly ill-formed. I understand why you'd ask it, but there are a couple of possible answers, none of which work directly. - No, because a black hole has nothing underneath the event horizon. (This is true from the perspective of anyone outside it, at least to a first approximation. Gravity is affected by curvature, and whatever might be in the centre cannot be the cause of the event horizon, except historically, but the event horizon is stable in itself. It's a self-sustaining cascade of space-time.) - Mu, because once you pass the event horizon the extreme warping of space-time turns "inwards" into "future", making the centre of the hole a point in time, not space. This makes it difficult to determine what the question is asking. - Maybe, because singularities probably don't exist. There should be something in the centre, which might evolve over time. It might just be a faster-than-light cascade of space-time, though. We still can't answer your question, because we don't know how to accurately model the 'singularity'. (As anything other than a singularity, and that's essentially just the math giving up.)

raattgift 7y ago
> the extreme warping of space-time turns "inwards" into "future", making the centre of the hole a point in time, not space. Assuming an exact black hole solution to the Einstein Field Equations (EFEs), what's happening is that any observer sufficiently close to the singularity will inevitably collide with the singularity. One can then define the horizon the outer boundary at which the future light cone attached at every point ends in the singularity. In order to avoid the singularity in the future one would need a causal cone broader than that of the light cone. The boundary is only a surface in the abstract; it's not material, you can't touch it, nothing bounces off it, and the no-drama conjecture says that if you were to float through the horizon you won't immediately notice you've done so. Outside the horizon as defined above, some part of the future light cone will have the singularity in it, however as one choose an observer at a point further and further away from the black hole, the singularity occupies a smaller and smaller part of the future light cone for that observer at that point, and the observer would have to aim ever more carefully to collide with it. (The observer simply has to reach a point at the horizon). > a black hole has nothing underneath the event horizon That's true of vacuum solutions of the EFEs which contain black holes, but it's not true in general. The Vaiyda metric is an exact solution which equips a spherically symmetrical central mass with spherically symmetric non-interacting dust of incoming and/or outgoing radiation: a Vaiyda black hole can have some incoming radiation within the horizon. Astrophysical black holes are more complicated since the matter outside them is generally far from spherically symmetrical. However, within the horizon will be some of the cosmic microwave background's (CMB's) photon gas, and whatever else reaches the horizon. But, you're right that something must be deep inside an astrophysical black hole, and the something is perhaps not a classical singularity, but rather some configuration that preserves the quantum mechanical properties of the field-content that reached the horizon. > might evolve over time If there's infalling matter, it must evolve over time, as the mass will increase, and likely so will the angular momentum, and charges are likely to go back-and-forth as one cannot guarantee that only chargless matter (e.g. photons) will fall in, nor that if we let charged matter fall in, it will always be charge-neutralized (e.g., no ions allowed!). That's true of a classical singularity, and it will be even more true for any ultradense non-singularity structure that preserves QFT information. > faster-than-light cascade of space-time What's that? > Gravity is affected by curvature In General Relativity, gravity is curvature. > whatever might be in the centre cannot be the cause of the event horizon Mass-energy generates the metric. In the case of the Schwarzschild black hole solution to the EFEs, the central mass (which is all the mass and energy in the vacuum Schwarzschild spacetime) is static and concentrated at one point. However, a non-rotating star, or a slowly-rotating object like the moon, generates a very close approximation of the Schwarzschild solution. Astrophysical black holes generate an approximation of the vacuum Schwarzschild black hole spacetime, but the Kerr metric is a better approximation for rotating black holes, and there is plenty of matter in the universe outside black holes. The true metric of the entire universe is much much more complicated than our (useful!) approximations, but nevertheless it is determined by the scatter of mass-energy throughout the spacetime. The true metric almost certianly describes regions which behave almost indistinguishably from the interiors of Schwarzschild (and Kerr, etc.) black holes. Another way of looking at this: as one throws matter into a black hole, the mass of the black hole increases, and so therefore does the diameter of the black hole. If you allow matter to cross into the black hole and collide with the singularity (or whatever replaces it), how can you maintain the italicized part of the quote above? > the event horizon is stable in itself If you allow mass to increase by letting dust and gas and cosmic microwaves reach the horizon, how do you keep the horizon stable? Moreover, if you merge two black holes, how do you keep both black holes' horizons stable? > It's a self-sustaining Given Hawking radiation then a black hole that's a solution to the Einstien Field Equations and equipped with one or more quantum fields filling the whole spacetime where the energy-density of the quantum fields is below a critical threshold will shrink. In the limit of quantum vacuum outside, this shrinking becomes complete evaporation: the horizon simply is not found in the future, because the mass in the whole spacetime has reconfigured from concentration at a central point to diffusion away from the central point. Since mass-energy generates the metric, the metric ceases to be Schwarzschild (or whatever black hole metric had existed prior to final evaporation). In our universe, there is lots of mass-energy outside black holes, and even though the metric expansion of space will make the CMB energy-density very small, it will still be higher than the critical threshold for stellar mass and larger black holes for a lonnnnnng time. But this isn't the black holes self-sustaining: they would evaporate, except for the CMB dust falling onto them, so it's the CMB (and its neutrino equivalent) that's doing the sustaining.

pdonis 7y ago

The "center" of a black hole is not a place in space. It's a moment of time (that is to the future of all other moments inside the horizon).

raattgift 7y ago
Not disagreeing that there's a point of no return wherein an infaller will inevitably collide with the singularity, just whether it's really correct to say that makes the singularity a moment of time. For starters, different infallers can arrive at the singularity at different times (even ones that arrive at the horizon at the same time can be relatively accelerated, resulting in having different interval lengths). Also not sure it makes sense for non-eternal BHs. For a BH that forms by gravitational collapse and eventually completely evaporates, surely the singularity has a timelike worldline? Some "lucky" too-early infallers (a cosmic neutrino, say) might pass through the centre collapsing progenitor before the horizon forms, while too-late infallers might pass through the region after final evaporation (assuming no horizon-equipped remnant). The "just-miss" is depends on where in spacetime the coincidence happens, and any reasonable slicing or threading would attribute the miss to being at the wrong moment in time. Maybe easiest to think about that if the singularity is not always at the spatial origin of a system of coordinates. For instance, if SN1987a's remnant contains a black hole, that black hole is surely moving around the galaxy with the luminous matter "now", but before the final collapse there was neither horizon nor singularity. I'm also perhaps unreasonably superstitious about the slogan's survival of arbitrary parameterizations, e.g., one can do an affine parameter on even an eternal BH's singularity and have what looks like a decent proper time for it. Perhaps another way of putting it is that the usual Carter-Penrose BH diagrams showing long horizontal wavy line segments for the singularity are misleading because of exaggerations caused by the particular conformal chart it uses, kinda like how the Earth's very-near-polar regions are not that big even though they look that way on Mercator charts (those are also conformal, and you get really different results in the near-polar regions in other conformal projections even as closely related as a transverse Mercator/Gauss conformal projection, https://en.wikipedia.org/wiki/Map_projection#/media/File:Usgs_map_mercator.svg https://en.wikipedia.org/wiki/Map_projection#/media/File:Usg... versus https://commons.wikimedia.org/wiki/File:Usgs_map_traverse_mercator.PNG https://commons.wikimedia.org/wiki/File:Usgs_map_traverse_me... ). Just thinking aloud in the wee hours, and maybe having given too little weight (pardon the pun) to the qualification "inside the horizon" just before your close paren. There's a lot of rigour that can be hidden in those three words.