View Full Version : A few questions about black holes.

LorentzFactor

2017-Feb-10, 04:47 AM

Hypothetically were the mass of a black hole to double, at whichtime speed would the schwarschild radius be observed to expand? I'd expect it to expand at light speed at the same speed gravity changes can be observed.

If two objects, one a black hole, another a solid mass such as a planet or asteroid, were headed towards each other at above .5 light speed which as a combined total would exceed the escape velocity of the schwarschild radius, what would happen were the object to enter the event horizon. Would this allow it to escape? Why not?

ShinAce

2017-Feb-10, 01:06 PM

It isn't possible. It's like asking what would happen if the sun disappeared.

Hypothetically though, the event horizon would expand instantaneously if the mass were instantaneously doubled. It wouldn't be a black hole if it didn't.

Anything that crosses an event horizon is not coming back out.

Cougar

2017-Feb-10, 01:38 PM

Hypothetically were the mass of a black hole to double, at whichtime speed would the schwarschild radius be observed to expand? I'd expect it to expand at light speed at the same speed gravity changes can be observed.

Welcome to the boards, Lorentz.

First of all, I don't think the Schwarzschild radius is anything you can observe. But generally, I'd expect change to the black hole to occur at the same rate that mass is added.

mkline55

2017-Feb-10, 01:58 PM

To begin with, the Schwarzchild radius isn't a surface like you think of the surface of the Earth. It's a calculation. It's like calculating the radius around the Earth where gravitational acceleration is 5.0 m/s/s. If you doubled the mass then that changes the distance, but there's nothing to see.

So how would you double the mass of a black hole? The additional mass exists before it is merged with the black hole. If you took the total mass before the combination and determined the center of mass, you could calculate the Schwarzchild radius for that total mass before all of it is contained within that radius around the center of mass. It's just a number. Putting the mass inside the radius doesn't change that number. So, doubling the mass does not change the Schwarzchild radius of the total mass.

Similarly, take two equal-sized black holes. Say their individual Schwarzchild radius was 3 units each. Doubling the mass doubles the Schwarzchild radius, so their combined Schwarzchild radius is 6 units. That radius is around their combined center of mass. It doesn't matter whether they are within that radius or not. Their combined Schwarzchild radius is still 6 units. Their center of mass is halfway between the two masses, and it is also the center from which their combined Schwarzchild radius is calculated.

Strange

2017-Feb-10, 02:10 PM

If two objects, one a black hole, another a solid mass such as a planet or asteroid, were headed towards each other at above .5 light speed which as a combined total would exceed the escape velocity of the schwarschild radius, what would happen were the object to enter the event horizon. Would this allow it to escape? Why not?

It is a popular simplification to say that the reason nothing can escape is because escape velocity is the speed. of light. It is actually more complex than that: it is because space-time is so curved that there are no paths that lead out of the event horizon.

grant hutchison

2017-Feb-10, 02:24 PM

Yes, the event horizon is a set of coordinates, rather than a physical structure, so it's not constrained to move at light speed. If we imagine a spherical shell of matter falling towards a Schwarzschild black hole, there will come a point at which the mass of the spherical shell is outside the original event horizon, but inside the event horizon that corresponds to the mass of black hole + spherical shell. At that point, the horizon of the black hole (defined as the surface at which outwardly directed photons fall inwards) flicks outwards to encompass the shell of matter.

As to the issue of combined velocity, special relativity dictates that the two velocities >0.5c sum to <c for either of the moving observers. So although you (an outside observer) measure the velocities to be more than half of light speed, an observer travelling with either the black hole or the asteroid will measure their closing velocity to be less than light speed. So the asteroid cannot escape the event horizon once inside.

Grant Hutchison

Hornblower

2017-Feb-10, 03:55 PM

To begin with, the Schwarzchild radius isn't a surface like you think of the surface of the Earth. It's a calculation. It's like calculating the radius around the Earth where gravitational acceleration is 5.0 m/s/s. If you doubled the mass then that changes the distance, but there's nothing to see.

So how would you double the mass of a black hole? The additional mass exists before it is merged with the black hole. If you took the total mass before the combination and determined the center of mass, you could calculate the Schwarzchild radius for that total mass before all of it is contained within that radius around the center of mass. It's just a number. Putting the mass inside the radius doesn't change that number. So, doubling the mass does not change the Schwarzchild radius of the total mass.

Similarly, take two equal-sized black holes. Say their individual Schwarzchild radius was 3 units each. Doubling the mass doubles the Schwarzchild radius, so their combined Schwarzchild radius is 6 units. That radius is around their combined center of mass. It doesn't matter whether they are within that radius or not. Their combined Schwarzchild radius is still 6 units. Their center of mass is halfway between the two masses, and it is also the center from which their combined Schwarzchild radius is calculated.

It is my understanding that if two black holes approach each other on a collision course, their respective event horizons will become elongated, roughly egg-shaped with the small ends toward each other. When they get close enough together they will merge into a temporary dumbbell shape before settling into a final spherical shape. It will not be spherical while the mass distribution is not yet spherically symmetrical.

I read about this in a Sky and Telescope article several years ago, and it would take a lot of searching to find it now. You may wish to "Google" something like "black hole merger".

grant hutchison

2017-Feb-10, 04:54 PM

It is my understanding that if two black holes approach each other on a collision course, their respective event horizons will become elongated, roughly egg-shaped with the small ends toward each other. When they get close enough together they will merge into a temporary dumbbell shape before settling into a final spherical shape. It will not be spherical while the mass distribution is not yet spherically symmetrical.That's right - the horizons reach out to each other. There's a nice diagram of a small black hole approaching the event horizon of a very large black hole here (http://iopscience.iop.org/0264-9381/33/15/155003/downloadFigure/figure/cqgaa2d8ff3), taken from Exact event horizon of a black hole merger (http://iopscience.iop.org/article/10.1088/0264-9381/33/15/155003) by Emparan & Martinez.

There used to be some nice animated gifs out there of a more equal merger, showing the "ring down" period when a dumbell-shaped event horizon rotates and radiates gravitational waves, but they're now difficult to find because the search results are now dominated by LIGOs rather beautiful animations of the visual appearance of merging black holes (https://www.youtube.com/watch?v=I_88S8DWbcU). It's interesting that we don't actually get to see the event horizons reaching out to touch, because that process is obscured by light taking other routes between the black holes.

The reason the event horizons reach outwards is because, as they approach each other, a region forms between the two black holes from which light can't escape because of the combined mass of the two black holes.

Grant Hutchison

Cougar

2017-Feb-10, 07:38 PM

That's right - the horizons reach out to each other....

Even though these are large, massive objects, I expect the whole merger, from initial "reaching" to end of ring-down, to go pretty quickly. Isn't this what amounts to a "chirp"?

Glom

2017-Feb-10, 09:19 PM

Yes, the event horizon is a set of coordinates, rather than a physical structure, so it's not constrained to move at light speed.

But, but, but, on Voyager they fell into a black hole and cracked the event horizon.

More seriously, if the event horizon were to move at faster than the speed of light, then something just outside the event horizon could find itself enveloped rather quickly. As this would potentially be observable from a distance, which could breach causality maybe?

Strange

2017-Feb-10, 10:20 PM

Even though these are large, massive objects, I expect the whole merger, from initial "reaching" to end of ring-down, to go pretty quickly. Isn't this what amounts to a "chirp"?

Indeed. And if you read about the frequencies (and hence orbital periods) of the chirp, it is truly amazing.

I have a vague recollection of someone (probably on this forum) saying that once the "reaching out" starts, the merger is inevitable and cannot be reversed. (But, like much of what I remember, that could be completely wrong!)

Noclevername

2017-Feb-10, 11:41 PM

More seriously, if the event horizon were to move at faster than the speed of light, then something just outside the event horizon could find itself enveloped rather quickly. As this would potentially be observable from a distance, which could breach causality maybe?

IIRC, the Event Horizon is an observational property of the space around a BH. It doesn't really move, the way an object with mass does.

grant hutchison

2017-Feb-11, 12:13 AM

More seriously, if the event horizon were to move at faster than the speed of light, then something just outside the event horizon could find itself enveloped rather quickly. As this would potentially be observable from a distance, which could breach causality maybe?I can't see how.

There are actually two horizons in slightly different places, when a black hole is accreting.

The absolute horizon contains the region of spacetime from which no light will escape - it actually starts to edge outwards as infalling matter approaches.

Below that is the apparent horizon, which encloses all the spacetime in which outward-directed photons fall inwards. It stays where it is until the infalling matter meets the outward moving absolute horizon, and then it leaps outwards to merge with the absolute horizon.

During the accretion, there's a zone outside the apparent horizon and inside the absolute horizon, in which photons are temporarily able to move outwards, but will never escape the black hole, because the infalling matter will increase its mass and pull them back inwards again.

So you have a choice between the absolute horizon, which starts moving in anticipation of infalling matter, or the apparent horizon which leaps out instantaneously when infalling material gets close enough.

Grant Hutchison

Strange

2017-Feb-11, 08:43 AM

I can't see how.

There are actually two horizons in slightly different places, when a black hole is accreting.

Presumably these don't often get talked about (in popular science articles) because the usual description of a Schwarzschild black hole is a static solution.

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