6 ms·

Black holes already have a time component since they exist in a universe with 3-dimensional space and 1-dimensional time. The equations include both the spatia

Black holes already have a time component since they exist in a universe with 3-dimensional space and 1-dimensional time. The equations include both the spatial components and the temporal components. One of the curious features of black holes is that they reverse the roles of time and space in a sense. In ordinary space we have freedom to move around in space, but are constrained to only move forward in time. Different reference frames may move forward at different rates relative to us, but they always move forward. By contrast, once you pass the event horizon of a black hole, these properties swap. It's possible to "move freely" in time in the sense that you can find reference frames that appear to move backwards in time. But all these reference frames are constrained to move forward in space.

Sniffnoy 3y ago
The article explicitly says that it's talking about number of spatial dimensions, it's not including time. It says the new theorem shows that black holes can form into up to 7 spatial dimensions (plus 1 for time). (This isn't to say they can't form beyond that, just this new theorem doesn't show that.)
Zamicol 3y ago
According to GR, time stops at the event horizon. It would take infinite time to cross the event horizon. Everything past the event horizon is speculative.
- magicalhippo 3y ago
  Not really. It just appears to take an infinite amount of time for an outside observer. For someone falling into a black hole, it takes a finite amount of proper time[1] to reach the event horizon. For that infalling observer, the horizon is a boundry where, once beyond, the singularity is always in their finite future. No matter what you do inside, you will reach it at some point. [1]: https://en.wikipedia.org/wiki/Proper_time https://en.wikipedia.org/wiki/Proper_time
  - antognini 3y ago
    Yes, this is correct. If you look at the Schwarzschild solution to the Einstein field equations you'll find that there are two apparent singularities: one at the center of the black hole, and another at the event horizon. However, the apparent singularity at the event horizon is not a true singularity because you can perform a coordinate transformation in which it disappears. Mathematically, this is doing what you describe: going from the reference frame of an external observer to one who is falling into the black hole. To the outside observer you appear to take an infinite amount of time to fall in, but from the perspective of the person falling in, it happens in a finite amount of time. The other singularity at r = 0 is different, though. It is a true singularity because there is no coordinate transformation you can make in which it disappears.
    - angiosperm 3y ago
      If the escape velocity from the event horizon is (defined to be) light speed, that seems to mean falling in from indefinitely far away, you would get indefinitely close to light speed (as perceived from at rest outside) at the point when you cross it,.
  - Zamicol 3y ago
    How long does it take a infalling observer to cross the event horizon of a black hole?
    - EA-3167 3y ago
      From the perspective of the person falling, it happens from one moment to the next. You as the person being accelerated don't notice any change in your own proper time, only distant observers will see that.
    - magicalhippo 3y ago
      If you were to fall radially inwards towards a non-rotating black hole, that is not spiraling around it but "straight in", it would be the same as with plain Newton's law[1]. There's a nice graph in the midde of this[2] page that shows the difference between the proper time and the aparent time observed by the outside observer. At r = 2m you can see the aparent time goes to infinity and the quickly back again. [1]: https://physics.stackexchange.com/questions/718222/proper-time-of-fall-in-schwarschild-metric https://physics.stackexchange.com/questions/718222/proper-ti... [2]: https://www.mathpages.com/rr/s6-04/6-04.htm https://www.mathpages.com/rr/s6-04/6-04.htm
qubex 3y ago
Aside: the fact that the radial spatial dimension towards the centre of the black hole where the singularity lies becomes time inside the event horizon is why all the drivel about “what you’d see if you crossed the event horizon” is drivel. You’d die instantly since circulating your blood or sending nerve impulses would essentially become impossible because it would involve reverse time travel — i.e., blood moving toward the centre would never be able to move out again because that would involve it travelling back in time. Greg Egan mentioned this in a short story of his.
- antognini 3y ago
  This is not true. It is possible for two observers inside the event horizon to send signals to each other. It's just that those signals cannot be sent to any radius larger than the radius they originated at. The only danger to a human would be the tidal forces. For a supermassive black hole, the tidal forces at the event horizon are quite modest, so you really would not notice it when you passed across the event horizon. It's only when you get close to the singularity that the tidal forces tear you apart.
  - angiosperm 3y ago
    How could a nerve impulse propagate in any direction but inward, then? Thinking involves signals going in all directions.
    - raattgift 3y ago
      Thanks for asking this good (and pretty difficult) question, rather than trying to make an assertion; cf. a sibling comment of mine. tl;dr: that the body is all attached for some time after crossing the horizon is [a] likely and [b] sufficent to justify the notion of stimulation in the feet being processed in the brain, when falling feet-first within a black hole. Firstly and basically the whole of [b], I'm just gonna assume that nerve impulses are the result of chemicals spat within neurons and between them, all very slowly compared to the speed of light, and all quite local (one chemical change or protein-conformation change at a time, step by step). I don't know the full details, but the relevant thing is that the speeds are slow, and that's the basis for the question ("how does something not faster than light rise within a black hole?"). Now on to what happens to a self-connected body inside a black hole, i.e., a justification for [a]. My nth iteration of an answer to this (abandoned a few because they got too technical or were unsatisfying; really fatigue drove me to a decision to submit as-is or to just walk away from an attempt at a good answer) is that the "rules" of a black hole are that your body's centre of mass cannot climb relative to the centre of mass of the black hole. More particularly, acceleration of the body's centre of mass is a vector (i.e., with magnitude, direction and sense) which is constrained to have the sense be at least slightly inwards at every point inside the horizon. However, there is no rule about where exactly your centre of mass is found relative to the bits of you stuck to your skeleton and encased in your skin. Equivalently, the sense constraint does not apply to very small parts of your body so long as those small parts do not break the rule that applies to the bulk's centre of mass. (This is another way of stating the universality of free fall version of the weak equilvalence principle, where the vacuum worldline is that of the bulk, the microscopic components of the bulk (atoms in haemoglobin in blood in the aorta etc) are obviously not in vacuum themselves.) Internal adhesive forces (and other contact/surface forces) win for a while in a battle to keep your extremities from breaking off and taking their own free-falling trajectories, and so (if you are falling feet-first) you could think of your head supporting the weight of your body. As you fall inwards, your feet will feel heavier and heavier and ultimately you become what medical people like to call "disarticulated". This is a result of three rules: * A spinning spherical mass becomes oblate. Earth is very slightly narrower between the poles than between two opposite points on the Equator. * Objects falling onto a spinning spherical mass become prolate, that is longer in a vertical direction and narrower in the directions perpendicular to that. * Objects dropped above different points on a spinning spherical mass will follow converging trajectories. Handy little diagram: <https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fmedia.wired.com%2Fphotos%2F5ac50754a54678516963cd74%2Fmaster%2Fw_1644%2Ch_1148%2Cc_limit%2Fweightontheearth.png&f=1&nofb=1&ipt=ea263f5d4140bc37212bca27d467a3de645d5baa2accc99d09327b0f47fd3346&ipo=images https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2F...> If we put the stick figure at the equator on the tropic of Cancer instead, the fat arrows pointing towards the centre still converge, just with a smaller angle. These are all taken to exxtreem in a spinning black hole. The hole itself is slightly oblate. This induces different notions of "straight down" that get weirder as one gets deeper within. The vertical stretch and inward squashed induced on an infaller becomes strong as one gets close to the black hole's centre. And how much inducement there is for parts of a body to find their own notion of "straight down" (the gravitational "pseudo force") becomes stronger than the inducement on them to stay attached to the rest of the body. If you are falling feet-first, the skin cells on your shoulders want to take their own notion of "straight to the centre", but they are dragged onto your centre of mass's "straight to the centre" because of the forces that stick your skin cells to the layers underneath. Your feet want to fall faster than your hips, but your connective tissue and skeleton keep them from doing that. Your head wants to lag behind, but it too is prevented. Until you split apart (e.g. your femoral artery ruptures, your spine separates, whatever) this structural integrity supports nervous impulses and blood rising from your feet towards your brain. You can still think on your way down. You could consider this as your internal forces changing the local spacetime geometry to allow nerve impulses and blood to rise upwards; or alternatively you could think of it as your upper bits suspending your lower bits at a higher gravitational potential. The real physics is a complex set of tensors, and with enough initial data you could grind out a full picture with mathematical rigour. You'd then cut down the tensorful field equations into vectors (which way do your various bits get accelerated) by choosing a suitable set of coordinates; probably you'd want to use ones where your centre of mass is always at the origin, or where each of your bits and pieces have their own set of coordinates ("Fermi coordinates") and you do some coarse-graining like averaging. Either way, your body-bits' mutual accelerations overwhelm the tidal forces (and the tidal forces are themselves vectors which we extract from the Weyl tidal tensor in applying the same suitable coordinates). In support of this line of thinking about how you can still think when you're in a black hole is the "no-drama" conjecture. The radius of curvature just outside the horizon of a supermassive black hole is much greater than that at Earth. Our bodies tend not to be disassembled by Earth's gravitation (in the "tidal" sense; that's that sense which is captured in the three rules above -- a Nasa-produced animation exaggerating the prolation induced by the moon on the Earth's oceans <https://www.youtube.com/watch?v=l37ofe9haMU https://www.youtube.com/watch?v=l37ofe9haMU>), so it would be surprising if they were disassembled by the gentler gravitation just outside a big black hole. You should just free-fall through a big black hole's horizon without feeling much in the "tidal" sense (your view of distant stars would give things away though, even before you are at the point of no return). For a supermassive black hole you should continue not to notice much for some time (minutes on your wristwatch). Eventually (possibly an hour or so on your wristwatch) you notice and it hurts and your head probably pops off ~minutes later, and your remains "spaghettify". Unfortunately we don't have enough evidence to select an low-curvature-radius-but-still-near-horizon effective theory, so there are certainly plausible descriptions very different from mine above. For example, certain physicists reject the no-drama conjecture quite strongly, often because their preferred theories of quantum fields (or strings) predict that the horizon is a tangible surface which supports very high energy radiation. Black hole horizons in general relativity do nothing of the sort; they're just a notion of a point of no return[*], and any particle crossing that point falls inwards, be they super high energy or super low energy or something in between. (The curvature singularity isn't a surface that supports anything either; infallers' remains don't really collide with it or pile on top of it, or rather we can't say one way or another what the case is because the standard equations become divergent. They do not diverge at the horizon though.) In some of those theories you are fried by high intensity gamma radiation before you get to the horizon; in others there is a "firewall" just inside that blows you apart with ~Planck-length radiation. If those are right, there's drama at or very near the horizon, and nothing will be able to spaghettified because it will be disintegrated by radiation. [*] yes yes, but you (complaining expert) don't know which horizon I mean, and there are many to choose from (Visser, <https://arxiv.org/abs/1407.7295v2 https://arxiv.org/abs/1407.7295v2>, 2nd paragraph)
cyberax 3y ago
> By contrast, once you pass the event horizon of a black hole, these properties swap. Sigh. No. They do not. Moreover, nothing special at all happens to you when you pass the event horizon. In fact, if you are free-falling then you should not even be able to detect the crossing using only local experiments. > It's possible to "move freely" in time in the sense that you can find reference frames that appear to move backwards in time. Nope. What happens is that your spatial directions become more and more constrained, until they collapse into a single point (the singularity). And then you'll just exist in this single point forever, according to GR. From your viewpoint, it'll look like the singularity becomes an infinite plane that cuts off most of your field of vision. You'll be falling towards this plane, but until the last moment you'll be able to receive signals from outside of the black hole. If someone puts a stationary clock outside the black hole's event horizon, you won't see it going faster or slower. And for any realistic black hole, your trip to the singularity will consume only a short time according to that clock.
- Spivak 3y ago
  And to add to this a singularity is predicted by GR, we don't actually know if such a thing exists and there are models where it doesn't like LQG where black holes are just a different kind of star whose collapse is opposed by a repulsive force that is predicted to exist by the uncertainty principle.
- dvt 3y ago
  > Sigh. No. They do not. Technically, GP is right that they (mathematically) swap, but yeah, it has a meaningless physical effect (the geodesic is always smooth). It's akin to describing the rotation of a kicked ball with imaginary exponentiation and thinking something spooky is going on.
  - antognini 3y ago
    So I would agree with this for the Schwarzschild metric. The swapping of signs in the space and time components of the metric does not have any real physical consequences. But when you move to a Kerr metric it absolutely does because it permits the construction of closed timelike curves.

Sniffnoy 3y ago

The article explicitly says that it's talking about number of spatial dimensions, it's not including time. It says the new theorem shows that black holes can form into up to 7 spatial dimensions (plus 1 for time). (This isn't to say they can't form beyond that, just this new theorem doesn't show that.)

Zamicol 3y ago

According to GR, time stops at the event horizon. It would take infinite time to cross the event horizon. Everything past the event horizon is speculative.

magicalhippo 3y ago
Not really. It just appears to take an infinite amount of time for an outside observer. For someone falling into a black hole, it takes a finite amount of proper time[1] to reach the event horizon. For that infalling observer, the horizon is a boundry where, once beyond, the singularity is always in their finite future. No matter what you do inside, you will reach it at some point. [1]: https://en.wikipedia.org/wiki/Proper_time https://en.wikipedia.org/wiki/Proper_time
- antognini 3y ago
  Yes, this is correct. If you look at the Schwarzschild solution to the Einstein field equations you'll find that there are two apparent singularities: one at the center of the black hole, and another at the event horizon. However, the apparent singularity at the event horizon is not a true singularity because you can perform a coordinate transformation in which it disappears. Mathematically, this is doing what you describe: going from the reference frame of an external observer to one who is falling into the black hole. To the outside observer you appear to take an infinite amount of time to fall in, but from the perspective of the person falling in, it happens in a finite amount of time. The other singularity at r = 0 is different, though. It is a true singularity because there is no coordinate transformation you can make in which it disappears.
  - angiosperm 3y ago
    If the escape velocity from the event horizon is (defined to be) light speed, that seems to mean falling in from indefinitely far away, you would get indefinitely close to light speed (as perceived from at rest outside) at the point when you cross it,.
- Zamicol 3y ago
  How long does it take a infalling observer to cross the event horizon of a black hole?
  - EA-3167 3y ago
    From the perspective of the person falling, it happens from one moment to the next. You as the person being accelerated don't notice any change in your own proper time, only distant observers will see that.
  - magicalhippo 3y ago
    If you were to fall radially inwards towards a non-rotating black hole, that is not spiraling around it but "straight in", it would be the same as with plain Newton's law[1]. There's a nice graph in the midde of this[2] page that shows the difference between the proper time and the aparent time observed by the outside observer. At r = 2m you can see the aparent time goes to infinity and the quickly back again. [1]: https://physics.stackexchange.com/questions/718222/proper-time-of-fall-in-schwarschild-metric https://physics.stackexchange.com/questions/718222/proper-ti... [2]: https://www.mathpages.com/rr/s6-04/6-04.htm https://www.mathpages.com/rr/s6-04/6-04.htm

qubex 3y ago

Aside: the fact that the radial spatial dimension towards the centre of the black hole where the singularity lies becomes time inside the event horizon is why all the drivel about “what you’d see if you crossed the event horizon” is drivel. You’d die instantly since circulating your blood or sending nerve impulses would essentially become impossible because it would involve reverse time travel — i.e., blood moving toward the centre would never be able to move out again because that would involve it travelling back in time. Greg Egan mentioned this in a short story of his.

antognini 3y ago
This is not true. It is possible for two observers inside the event horizon to send signals to each other. It's just that those signals cannot be sent to any radius larger than the radius they originated at. The only danger to a human would be the tidal forces. For a supermassive black hole, the tidal forces at the event horizon are quite modest, so you really would not notice it when you passed across the event horizon. It's only when you get close to the singularity that the tidal forces tear you apart.
- angiosperm 3y ago
  How could a nerve impulse propagate in any direction but inward, then? Thinking involves signals going in all directions.
  - raattgift 3y ago
    Thanks for asking this good (and pretty difficult) question, rather than trying to make an assertion; cf. a sibling comment of mine. tl;dr: that the body is all attached for some time after crossing the horizon is [a] likely and [b] sufficent to justify the notion of stimulation in the feet being processed in the brain, when falling feet-first within a black hole. Firstly and basically the whole of [b], I'm just gonna assume that nerve impulses are the result of chemicals spat within neurons and between them, all very slowly compared to the speed of light, and all quite local (one chemical change or protein-conformation change at a time, step by step). I don't know the full details, but the relevant thing is that the speeds are slow, and that's the basis for the question ("how does something not faster than light rise within a black hole?"). Now on to what happens to a self-connected body inside a black hole, i.e., a justification for [a]. My nth iteration of an answer to this (abandoned a few because they got too technical or were unsatisfying; really fatigue drove me to a decision to submit as-is or to just walk away from an attempt at a good answer) is that the "rules" of a black hole are that your body's centre of mass cannot climb relative to the centre of mass of the black hole. More particularly, acceleration of the body's centre of mass is a vector (i.e., with magnitude, direction and sense) which is constrained to have the sense be at least slightly inwards at every point inside the horizon. However, there is no rule about where exactly your centre of mass is found relative to the bits of you stuck to your skeleton and encased in your skin. Equivalently, the sense constraint does not apply to very small parts of your body so long as those small parts do not break the rule that applies to the bulk's centre of mass. (This is another way of stating the universality of free fall version of the weak equilvalence principle, where the vacuum worldline is that of the bulk, the microscopic components of the bulk (atoms in haemoglobin in blood in the aorta etc) are obviously not in vacuum themselves.) Internal adhesive forces (and other contact/surface forces) win for a while in a battle to keep your extremities from breaking off and taking their own free-falling trajectories, and so (if you are falling feet-first) you could think of your head supporting the weight of your body. As you fall inwards, your feet will feel heavier and heavier and ultimately you become what medical people like to call "disarticulated". This is a result of three rules: * A spinning spherical mass becomes oblate. Earth is very slightly narrower between the poles than between two opposite points on the Equator. * Objects falling onto a spinning spherical mass become prolate, that is longer in a vertical direction and narrower in the directions perpendicular to that. * Objects dropped above different points on a spinning spherical mass will follow converging trajectories. Handy little diagram: <https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fmedia.wired.com%2Fphotos%2F5ac50754a54678516963cd74%2Fmaster%2Fw_1644%2Ch_1148%2Cc_limit%2Fweightontheearth.png&f=1&nofb=1&ipt=ea263f5d4140bc37212bca27d467a3de645d5baa2accc99d09327b0f47fd3346&ipo=images https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2F...> If we put the stick figure at the equator on the tropic of Cancer instead, the fat arrows pointing towards the centre still converge, just with a smaller angle. These are all taken to exxtreem in a spinning black hole. The hole itself is slightly oblate. This induces different notions of "straight down" that get weirder as one gets deeper within. The vertical stretch and inward squashed induced on an infaller becomes strong as one gets close to the black hole's centre. And how much inducement there is for parts of a body to find their own notion of "straight down" (the gravitational "pseudo force") becomes stronger than the inducement on them to stay attached to the rest of the body. If you are falling feet-first, the skin cells on your shoulders want to take their own notion of "straight to the centre", but they are dragged onto your centre of mass's "straight to the centre" because of the forces that stick your skin cells to the layers underneath. Your feet want to fall faster than your hips, but your connective tissue and skeleton keep them from doing that. Your head wants to lag behind, but it too is prevented. Until you split apart (e.g. your femoral artery ruptures, your spine separates, whatever) this structural integrity supports nervous impulses and blood rising from your feet towards your brain. You can still think on your way down. You could consider this as your internal forces changing the local spacetime geometry to allow nerve impulses and blood to rise upwards; or alternatively you could think of it as your upper bits suspending your lower bits at a higher gravitational potential. The real physics is a complex set of tensors, and with enough initial data you could grind out a full picture with mathematical rigour. You'd then cut down the tensorful field equations into vectors (which way do your various bits get accelerated) by choosing a suitable set of coordinates; probably you'd want to use ones where your centre of mass is always at the origin, or where each of your bits and pieces have their own set of coordinates ("Fermi coordinates") and you do some coarse-graining like averaging. Either way, your body-bits' mutual accelerations overwhelm the tidal forces (and the tidal forces are themselves vectors which we extract from the Weyl tidal tensor in applying the same suitable coordinates). In support of this line of thinking about how you can still think when you're in a black hole is the "no-drama" conjecture. The radius of curvature just outside the horizon of a supermassive black hole is much greater than that at Earth. Our bodies tend not to be disassembled by Earth's gravitation (in the "tidal" sense; that's that sense which is captured in the three rules above -- a Nasa-produced animation exaggerating the prolation induced by the moon on the Earth's oceans <https://www.youtube.com/watch?v=l37ofe9haMU https://www.youtube.com/watch?v=l37ofe9haMU>), so it would be surprising if they were disassembled by the gentler gravitation just outside a big black hole. You should just free-fall through a big black hole's horizon without feeling much in the "tidal" sense (your view of distant stars would give things away though, even before you are at the point of no return). For a supermassive black hole you should continue not to notice much for some time (minutes on your wristwatch). Eventually (possibly an hour or so on your wristwatch) you notice and it hurts and your head probably pops off ~minutes later, and your remains "spaghettify". Unfortunately we don't have enough evidence to select an low-curvature-radius-but-still-near-horizon effective theory, so there are certainly plausible descriptions very different from mine above. For example, certain physicists reject the no-drama conjecture quite strongly, often because their preferred theories of quantum fields (or strings) predict that the horizon is a tangible surface which supports very high energy radiation. Black hole horizons in general relativity do nothing of the sort; they're just a notion of a point of no return[*], and any particle crossing that point falls inwards, be they super high energy or super low energy or something in between. (The curvature singularity isn't a surface that supports anything either; infallers' remains don't really collide with it or pile on top of it, or rather we can't say one way or another what the case is because the standard equations become divergent. They do not diverge at the horizon though.) In some of those theories you are fried by high intensity gamma radiation before you get to the horizon; in others there is a "firewall" just inside that blows you apart with ~Planck-length radiation. If those are right, there's drama at or very near the horizon, and nothing will be able to spaghettified because it will be disintegrated by radiation. [*] yes yes, but you (complaining expert) don't know which horizon I mean, and there are many to choose from (Visser, <https://arxiv.org/abs/1407.7295v2 https://arxiv.org/abs/1407.7295v2>, 2nd paragraph)

cyberax 3y ago

> By contrast, once you pass the event horizon of a black hole, these properties swap. Sigh. No. They do not. Moreover, nothing special at all happens to you when you pass the event horizon. In fact, if you are free-falling then you should not even be able to detect the crossing using only local experiments. > It's possible to "move freely" in time in the sense that you can find reference frames that appear to move backwards in time. Nope. What happens is that your spatial directions become more and more constrained, until they collapse into a single point (the singularity). And then you'll just exist in this single point forever, according to GR. From your viewpoint, it'll look like the singularity becomes an infinite plane that cuts off most of your field of vision. You'll be falling towards this plane, but until the last moment you'll be able to receive signals from outside of the black hole. If someone puts a stationary clock outside the black hole's event horizon, you won't see it going faster or slower. And for any realistic black hole, your trip to the singularity will consume only a short time according to that clock.

Spivak 3y ago
And to add to this a singularity is predicted by GR, we don't actually know if such a thing exists and there are models where it doesn't like LQG where black holes are just a different kind of star whose collapse is opposed by a repulsive force that is predicted to exist by the uncertainty principle.
dvt 3y ago
> Sigh. No. They do not. Technically, GP is right that they (mathematically) swap, but yeah, it has a meaningless physical effect (the geodesic is always smooth). It's akin to describing the rotation of a kicked ball with imaginary exponentiation and thinking something spooky is going on.
- antognini 3y ago
  So I would agree with this for the Schwarzschild metric. The swapping of signs in the space and time components of the metric does not have any real physical consequences. But when you move to a Kerr metric it absolutely does because it permits the construction of closed timelike curves.