6 ms·

I've always thought the event horizon for a black hole has to be spherical. But my physics intuition tells me that as two of them merge, the resulting BH shoul

I've always thought the event horizon for a black hole has to be spherical. But my physics intuition tells me that as two of them merge, the resulting BH should have a "peanut" shape, at least initially. And maybe it can keep having an irregular shape, depending on the mass distribution inside it?

itishappy 1y ago
It's only spherical in a Schwarzschild (non-rotating) black hole. A rotating black hole is called a Kerr black hole, and stuff gets weird, such as there being an oblate event horizon, a weird outer horizon called an ergosphere where spacetime gets dragged along such that it's impossible to stand still and you can accelerate objects using the black hole, a weirder inner horizon called the Cauchy horizon where time travel is possible, and a singularity in the shape of a ring. Your intuition is correct that during a merger it would be weirder still. https://en.wikipedia.org/wiki/Kerr_metric https://en.wikipedia.org/wiki/Kerr_metric https://arxiv.org/pdf/0706.0622 https://arxiv.org/pdf/0706.0622 https://en.wikipedia.org/wiki/Ergosphere https://en.wikipedia.org/wiki/Ergosphere https://en.wikipedia.org/wiki/Cauchy_horizon https://en.wikipedia.org/wiki/Cauchy_horizon Edit: Updated the bit about about horizons as I research a bit more. It's complicated, and I'm still not positive I have it exactly right, but I think it's now as good as I can get it.
- AnimalMuppet 1y ago
  Could you (or anyone) tell what the radius of the ring singularity is, in terms of mass and angular momentum? I haven't been able to find that.
  - itishappy 1y ago
    The math seems to suggest R=a, or simply the spin in terms of length. It's certainly an oversimplification, as the answer will depend on the choice of metric. Here's the best resources I've been able to find on the question. Roy Kerr himself responded to the Quora question: > There is no Newtonian singularity at the Center of the earth and there is no singularity inside a rotating black hole. The ring singularity is imaginary. It only exists in my solution because it contains no actual matter. When a star collapses into a black hole it keeps shrinking until the centrifugal force stabilizes it. The event shell forms between the star and the outside. In 57 years no one has actually proved that a singularity forms inside, and that includes Penrose. instead, he proved that there is a light ray of finite affine length. This follows from the “hairy ball theorem”. The stack overflow answer seems to describe the problem in terms I can better understand: > It seems unlikely to me that you're going to be able to formulate a notion of diameter that makes sense here. Putting aside all questions of the metric's misbehavior at the ring singularity, there is the question of what spacelike path you want to integrate along. For the notion of a diameter to make sense, there would have to be some preferred path. Outside the horizon of a Schwarzschild black hole, we have a preferred stationary observer at any given point, and therefore there is a preferred radial direction that is orthogonal to that observer's world-line. But this doesn't work here. https://physics.stackexchange.com/questions/471419/metric-diameter-of-a-ring-singularity https://physics.stackexchange.com/questions/471419/metric-di... https://www.quora.com/What-is-the-typical-diameter-roughly-of-the-ring-singularity-at-the-center-of-a-rotating-supermassive-black-hole https://www.quora.com/What-is-the-typical-diameter-roughly-o...
- CGMthrowaway 1y ago
  Simulation: https://www.youtube.com/watch?v=Tr1zDVbSjTM https://www.youtube.com/watch?v=Tr1zDVbSjTM
  - theGnuMe 1y ago
    Everything is a fluid in the end and at scale right??
  - grues-dinner 1y ago
    What kind of timeframe does that animation happen over?
- TMEHpodcast 1y ago
  No matter how chaotic the merger looks, the event horizon must asymptotically become either spherical (Schwarzschild) or oblate (Kerr). The mass distribution inside doesn’t change this, general relativity doesn’t allow static “lumpy” horizons. It’s wild how much happens in those milliseconds though. Numerical relativity papers like the one you shared from arxiv.org show the horizon “sloshing” before it stabilizes.
  - pavel_lishin 1y ago
    Is it even sensible to talk about a "mass distribution" inside of an event horizon?
    - kadoban 1y ago
      Sure, especially consider if singularities are not real. Then what's inside the event horizon is just some bunch of unknown material in some actual shape. Why wouldn't it be? If singularities are real...same thing but more boring answer maybe? (the distribution just being: in the center).
      - lisper 1y ago
        > Why wouldn't it be? Because the whole concept of "shape" assumes properties of space that might not apply inside an event horizon?
        db48x 1y ago
        There’s no reason to expect that the properties of space are different inside the event horizon than outside. Of course the direction of time turns sharply as you go inside, but otherwise space is just space. You only get an asymmetric black hole during the milliseconds of a merger. And that asymmetry is entirely due to the mass distribution inside the black hole. The black hole only becomes spherical again once the singularities have merged. Or in the more common case of rotating black holes, they only become properly oblate again once their ringularities have merged. Either way it happens quite quickly.
captainkrtek 1y ago
Here is an animation from MIT/CalTech of what a merger looks like: https://youtu.be/1agm33iEAuo https://youtu.be/1agm33iEAuo
chasil 1y ago
Does the black hole's spin deform the event horizon? I think so? https://archive.ph/VrzwW https://archive.ph/VrzwW Edit: "The Kerr metric also predicts the existence of an inner and outer event horizon, with the shape of these horizons being oblate rather than perfectly spherical due to the rotation."
- mr_toad 1y ago
  Kind of. Because the black hole drags space around with it you need to go faster near the ‘equator’ than the poles just to stand still. So the event horizon is fatter at the equator.
micw 1y ago
Is there a "mass distribution" inside? AFAIK a black hole is a singularity which means it's mass is in one (infinitely small) point.
BlackFly 1y ago
It is difficult to talk about the shape of the event horizon because the ordinary definition of a sphere is "surface where all points are equidistant from a given point" is already complex in a differentiable manifold, but even more so when the distance is infinite because of a singularity (or the point doesn't exist/isn't unique because of geodesic structure). So you switch to a definition of "surface of constant scalar curvature with the topology of a sphere", the topology being important to distinguish it from a plane and a hyperboloid. From there, I haven't personally done or seen the calculations of the shape of the horizon for Kerr or merging black holes, but my intuition is that it would be indeed peanut shaped for a merger (there are likely some saddle points). The coordinate shape certainly is but you can choose coordinates so that a Schwarzschild black hole is a coordinate peanut so coordinates aren't very meaningful.
Permik 1y ago
I'm no physicist but what I've learned on the internet through osmosis, I'd wager that black holes aren't spherical per-se, but they appear sphere-like to us dimensionally challenged beings. It's more like a manifold (mobius-gate? not a mobius surface), that changes your spatial directions to parallel temporal directions to spatial ones all leading to the singularity.
veunes 1y ago
But black holes are incredibly efficient at radiating away asymmetries via gravitational waves
deleted 1y ago
[deleted]
fpoling 1y ago
From our perspective there is no event horizon since the collapsing star has not reached the black hole state. In fact it takes infinite amount of time from the point of view of an external observer for the event horizon to form. In almost all situations it does matter as the collapsing star will behave as it is a black hole. But for the merge of black holes it is significant as it allows to release energy as there is no event horizon.

itishappy 1y ago

It's only spherical in a Schwarzschild (non-rotating) black hole. A rotating black hole is called a Kerr black hole, and stuff gets weird, such as there being an oblate event horizon, a weird outer horizon called an ergosphere where spacetime gets dragged along such that it's impossible to stand still and you can accelerate objects using the black hole, a weirder inner horizon called the Cauchy horizon where time travel is possible, and a singularity in the shape of a ring. Your intuition is correct that during a merger it would be weirder still. https://en.wikipedia.org/wiki/Kerr_metric https://en.wikipedia.org/wiki/Kerr_metric https://arxiv.org/pdf/0706.0622 https://arxiv.org/pdf/0706.0622 https://en.wikipedia.org/wiki/Ergosphere https://en.wikipedia.org/wiki/Ergosphere https://en.wikipedia.org/wiki/Cauchy_horizon https://en.wikipedia.org/wiki/Cauchy_horizon Edit: Updated the bit about about horizons as I research a bit more. It's complicated, and I'm still not positive I have it exactly right, but I think it's now as good as I can get it.

AnimalMuppet 1y ago
Could you (or anyone) tell what the radius of the ring singularity is, in terms of mass and angular momentum? I haven't been able to find that.
- itishappy 1y ago
  The math seems to suggest R=a, or simply the spin in terms of length. It's certainly an oversimplification, as the answer will depend on the choice of metric. Here's the best resources I've been able to find on the question. Roy Kerr himself responded to the Quora question: > There is no Newtonian singularity at the Center of the earth and there is no singularity inside a rotating black hole. The ring singularity is imaginary. It only exists in my solution because it contains no actual matter. When a star collapses into a black hole it keeps shrinking until the centrifugal force stabilizes it. The event shell forms between the star and the outside. In 57 years no one has actually proved that a singularity forms inside, and that includes Penrose. instead, he proved that there is a light ray of finite affine length. This follows from the “hairy ball theorem”. The stack overflow answer seems to describe the problem in terms I can better understand: > It seems unlikely to me that you're going to be able to formulate a notion of diameter that makes sense here. Putting aside all questions of the metric's misbehavior at the ring singularity, there is the question of what spacelike path you want to integrate along. For the notion of a diameter to make sense, there would have to be some preferred path. Outside the horizon of a Schwarzschild black hole, we have a preferred stationary observer at any given point, and therefore there is a preferred radial direction that is orthogonal to that observer's world-line. But this doesn't work here. https://physics.stackexchange.com/questions/471419/metric-diameter-of-a-ring-singularity https://physics.stackexchange.com/questions/471419/metric-di... https://www.quora.com/What-is-the-typical-diameter-roughly-of-the-ring-singularity-at-the-center-of-a-rotating-supermassive-black-hole https://www.quora.com/What-is-the-typical-diameter-roughly-o...
CGMthrowaway 1y ago
Simulation: https://www.youtube.com/watch?v=Tr1zDVbSjTM https://www.youtube.com/watch?v=Tr1zDVbSjTM
- theGnuMe 1y ago
  Everything is a fluid in the end and at scale right??
- grues-dinner 1y ago
  What kind of timeframe does that animation happen over?
TMEHpodcast 1y ago
No matter how chaotic the merger looks, the event horizon must asymptotically become either spherical (Schwarzschild) or oblate (Kerr). The mass distribution inside doesn’t change this, general relativity doesn’t allow static “lumpy” horizons. It’s wild how much happens in those milliseconds though. Numerical relativity papers like the one you shared from arxiv.org show the horizon “sloshing” before it stabilizes.
- pavel_lishin 1y ago
  Is it even sensible to talk about a "mass distribution" inside of an event horizon?
  - kadoban 1y ago
    Sure, especially consider if singularities are not real. Then what's inside the event horizon is just some bunch of unknown material in some actual shape. Why wouldn't it be? If singularities are real...same thing but more boring answer maybe? (the distribution just being: in the center).
    - lisper 1y ago
      > Why wouldn't it be? Because the whole concept of "shape" assumes properties of space that might not apply inside an event horizon?
      - db48x 1y ago
        There’s no reason to expect that the properties of space are different inside the event horizon than outside. Of course the direction of time turns sharply as you go inside, but otherwise space is just space. You only get an asymmetric black hole during the milliseconds of a merger. And that asymmetry is entirely due to the mass distribution inside the black hole. The black hole only becomes spherical again once the singularities have merged. Or in the more common case of rotating black holes, they only become properly oblate again once their ringularities have merged. Either way it happens quite quickly.

captainkrtek 1y ago

Here is an animation from MIT/CalTech of what a merger looks like: https://youtu.be/1agm33iEAuo https://youtu.be/1agm33iEAuo

chasil 1y ago

Does the black hole's spin deform the event horizon? I think so? https://archive.ph/VrzwW https://archive.ph/VrzwW Edit: "The Kerr metric also predicts the existence of an inner and outer event horizon, with the shape of these horizons being oblate rather than perfectly spherical due to the rotation."

mr_toad 1y ago
Kind of. Because the black hole drags space around with it you need to go faster near the ‘equator’ than the poles just to stand still. So the event horizon is fatter at the equator.

micw 1y ago

Is there a "mass distribution" inside? AFAIK a black hole is a singularity which means it's mass is in one (infinitely small) point.

BlackFly 1y ago

It is difficult to talk about the shape of the event horizon because the ordinary definition of a sphere is "surface where all points are equidistant from a given point" is already complex in a differentiable manifold, but even more so when the distance is infinite because of a singularity (or the point doesn't exist/isn't unique because of geodesic structure). So you switch to a definition of "surface of constant scalar curvature with the topology of a sphere", the topology being important to distinguish it from a plane and a hyperboloid. From there, I haven't personally done or seen the calculations of the shape of the horizon for Kerr or merging black holes, but my intuition is that it would be indeed peanut shaped for a merger (there are likely some saddle points). The coordinate shape certainly is but you can choose coordinates so that a Schwarzschild black hole is a coordinate peanut so coordinates aren't very meaningful.

Permik 1y ago

I'm no physicist but what I've learned on the internet through osmosis, I'd wager that black holes aren't spherical per-se, but they appear sphere-like to us dimensionally challenged beings. It's more like a manifold (mobius-gate? not a mobius surface), that changes your spatial directions to parallel temporal directions to spatial ones all leading to the singularity.

veunes 1y ago

But black holes are incredibly efficient at radiating away asymmetries via gravitational waves

deleted 1y ago

[deleted]

fpoling 1y ago

From our perspective there is no event horizon since the collapsing star has not reached the black hole state. In fact it takes infinite amount of time from the point of view of an external observer for the event horizon to form. In almost all situations it does matter as the collapsing star will behave as it is a black hole. But for the merge of black holes it is significant as it allows to release energy as there is no event horizon.