# Black hole traveling near the speed of light

## Main Question or Discussion Point

Suppose a black hole travels at something like v = 0.999999999c relative to some observer. Does the black hole's event horizon becomes length contracted, thus appearing to turn into a black disk?

## Answers and Replies

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Drakkith
Staff Emeritus
Ohh, interesting question. I don't know, but I would think so.

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Simon Bridge
Homework Helper
That would be the same as asking what the geometry of the event horizon would be from the POV of an observer travelling relativistically wrt it.

Note: an observer stationary wrt a large sphere would see the sphere as a disk.
Observe: Sun, Moon, Planets, near starts etc: all look like disks.

Measure the diameter, in different directions re travel direction, as you pass one at speed, and they look like ellipses.
Very fast and you get a thin ellipse.

Aside: artistic interpretations aside, BHs are unlikely to look black.

Stephanus
PeterDonis
Mentor
2019 Award
That would be the same as asking what the geometry of the event horizon would be from the POV of an observer travelling relativistically wrt it.

Note: an observer stationary wrt a large sphere would see the sphere as a disk.
Observe: Sun, Moon, Planets, near starts etc: all look like disks.

Measure the diameter, in different directions re travel direction, as you pass one at speed, and they look like ellipses.
Very fast and you get a thin ellipse.
But this reasoning doesn't apply to a black hole's horizon, because there's no way to measure the diameter; it doesn't have one.

Also, the usual interpretation of length contraction depends on the "world tube" of the object being composed of timelike curves; different frames take "slices" out of the world tube at different angles, so the slices have different geometries. But the "world tube" of the event horizon is composed of null curves, not timelike curves; so I'm not sure you can take "slices" of the tube the same way you would with an ordinary object.

Finally, there's no way to actually observe the "geometry" of the horizon itself, since light can't escape from it; see below.

artistic interpretations aside, BHs are unlikely to look black.
What doesn't look black in these scenarios isn't the hole's horizon; it's stuff falling into the hole but still outside the horizon, which can radiate intensely as it falls in. The horizon itself can't be seen, since outgoing light at the horizon stays at the horizon forever, never escaping.

Simon Bridge
Homework Helper
Yep, which was the reason for including the link - in the event there was nothing falling in, the hole would probably occlude stars ... but there would also be significant lensing.

All this means that we have to be careful talking about the geometry of a black hole even horizon.
I suspect that OP is imagining that you can see it... that it's like a black ball, and that passing a length-contracted ball is like passing a flattened disk.

But need feedback to clarify the question.

I'd rate it as asking what would be a sensible way of treating relativistic length contraction for a black hole.
Presumably from an observer far enough away for the terms to make sense.

WannabeNewton
Suppose a black hole travels at something like v = 0.999999999c relative to some observer. Does the black hole's event horizon becomes length contracted, thus appearing to turn into a black disk?
In classical GR, a black hole is quite literally space-time geometry with an event horizon generated by null curves. It isn't a time-like object i.e. the concept of length contraction doesn't apply. That being said there are a myriad of optical effects that can occur due to black holes so that might be of interest to you.

martinbn
Probably I am confused about something very basic but it seems perfectly ok to take different time slices of the event horizon. The fact that it is null generated is not a problem. This would imply that the cross section area will be the same (for a stationary black hole) for all observers. So it seems natural to ask about geometric properties other than area as well.

When I thought of this possibility I was more interested in the ramifications for the gravity of the black hole combined with the relativistic mass increase of the black hole as it approaches c. What effect would this have on the imaginary time which would be experienced by an object falling into a black hole itself moving at relativistic speed?. Would there be two imaginary times to consider or would they cancel out?

The black hole is approaching a speed of c in relation to what?

PeterDonis
Mentor
2019 Award
I was more interested in the ramifications for the gravity of the black hole combined with the relativistic mass increase of the black hole as it approaches c
Relativistic mass is not the source of gravity in GR. The mass of the black hole, as far as its gravity is concerned, is an invariant. The motion of a test object relative to the black hole does affect how the hole's gravity curves its trajectory, but there's no useful way to view that as an effect of "relativistic mass increase" of the black hole.

What effect would this have on the imaginary time which would be experienced by an object falling into a black hole itself moving at relativistic speed?
An object falling into a black hole doesn't experience "imaginary time"; it experiences perfectly normal time flow. The "imaginary time" you are thinking of is not something physical; it's an artifact of choosing particular coordinates.

A.T.
But this reasoning doesn't apply to a black hole's horizon, because there's no way to measure the diameter; it doesn't have one.
But how does the Schwarzschild geometry outside the horizon look like in a frame where the BH is moving fast? Or, alternatively the Schwarzschild geometry of an massive star, that isn't a BH yet. Is it similar to a contracted version of the usual Schwarzschild geometry?

If the black hole were accelerating toward earth near the speed of light and a device were dropped into the front of it which attempted to send out a signal; is it possible that the red shifting of the signal might be cancelled by the blue shifting of the same signal as it hurled toward us at 99.9999% of c.? In other words: within the dark star's inertial frame it would be red shifted but what about outside the accelerating frame? My thinking was that general relativity does not distinguish between the body in accelerated motion and the body hovering outside a gravitationla well. If so, and if the moving black hole system were taken as an inertial frame, then could an obserrver at rest with respect to the black hole see deeper into the black hole than if it were stationary?

DaveC426913
Gold Member
The black hole is approaching a speed of c in relation to what?
Read the OP more carefully:

"...relative to some observer..."

Lira&Mira
ShayanJ
Gold Member
But the "world tube" of the event horizon is composed of null curves, not timelike curves
I'm confused here. Its true that an EH is a null surface but that applies to non-moving BHs as well. So it seems to me its not related to the state of the motion of the BH.
Also when you say a curve is null, it means that the corresponding "thing" that the curve is describing the motion of, is moving at the speed of light. I seem to remember that you once said the EH moves radially outward at the speed of light or something like this. But this isn't the motion we're talking about because this applies to non-moving BHs as well. Also if a BH is moving, I think we prefer it moves slower than light. So it seems to me the worldtube of the EH can't be composed of null curves.
I think my problem is that, here we're talking about a feature of spacetime geometry moving, not an object. That just seems weird!

pervect
Staff Emeritus
Suppose a black hole travels at something like v = 0.999999999c relative to some observer. Does the black hole's event horizon becomes length contracted, thus appearing to turn into a black disk?
Something vaguely like that happens - as the black hole approaches "c", the space-time geometry approaches a plane wave solution called the Aichelberg-sexl ultraboost. See https://en.wikipedia.org/wiki/Aichelburg–Sexl_ultraboost Which is basically disc-like, in that it's basically a plane wave. It's definitely not spherically symmetrical.

Lorentz contraction isn't quite applicable though - the distance between a timelike observer and a null worldline (like a light ray or the event horizon of a black hole) one follows a different contraction law than the familiar Lorentz contraction. This later issue comes up a lot in questions like "what is the distance to the event horizon". Basically a space-like interval contracts with velocity by the factor ##\gamma##, the interval between a space-like and a time-like observer follows the relativistic doppler shift law ##k = \sqrt{\frac{1 + \beta}{1-\beta}}##.

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Nugatory
WannabeNewton
So it seems to me the worldtube of the EH can't be composed of null curves.
It's the center of mass of the black hole, relative to some asymptotic Lorentz frame, that moves at subluminal speeds; the center of mass is simply the conserved charge associated with boosts so it is easily defined in terms of the ADM 4-momentum. The worldtube of the event horizon is still generated by a null congruence.

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micromass
WannabeNewton
This would imply that the cross section area will be the same (for a stationary black hole) for all observers. So it seems natural to ask about geometric properties other than area as well.
There is no unique choice of space-like section so geometric quantities defined in this way would not be intrinsic to the black hole but instead would have some gauge freedom. Now this is all fine if all we're concerned with is geometric properties of the event horizon relative to some family of observers. However in general such space-like sections would not correspond to simultaneity surfaces of any family of observers so it isn't clear how one would measure geometric quantities calculated from them.

But if we go the other way, wherein you give me a family of observers and ask for geometric quantities derived from space-like sections which are simultaneity surfaces of this family, then we would have to restrict ourselves to those 4-velocity fields whose local Lorentz frames are non-singular on the horizon and whose vorticity vanishes e.g. the family of observers freely falling radially from infinity. In this case I do not see an issue with calculating geometric properties of the event horizon relative to said space-like sections.

Would we find black holes moving near the speed of light in nature? Orbiting very close together or blasted out of a violent galactic nucleous? Or is it just a thought experiment?

Drakkith
Staff Emeritus
Would we find black holes moving near the speed of light in nature? Orbiting very close together or blasted out of a violent galactic nucleous? Or is it just a thought experiment?
I think its a thought experiment. I cant imagine how a solar mass size object could be accelerated to such speeds.

Ibix
Would we find black holes moving near the speed of light in nature? Orbiting very close together or blasted out of a violent galactic nucleous? Or is it just a thought experiment?
Depends what you mean. From the perspective of a black hole, it's a cinch that there are particles passing it at very high fractions of c. From the point of view of the particles, it's the black hole that is moving at high speed, and the point of relativity is that either perspective is valid. So in that sense, yes, there are billions upon billions of particles encountering black holes moving at substantial fractions of c all the time. However, I tend to agree with Drakkith that we're unlikely to be able to see this directly ourselves.

It would be possible - at least in principle - to build a probe and send it whizzing past a black hole at high speed and see what happens. Don't hold your breath, though. That kind of thing is well beyond our technology, and it's a long way to the nearest black hole, anyway.

Thank you for clarifying about the the two perscpectives. BTW I just found dozens of serious articles about stars moving at near the speed of light. They have not yet been observed because they a far away and faint but astronomers believe that when two supermassive black holes collide, as will happen with the Milkyway and Andromeda, as they get locked in the proverbial death dance some stars will be hurled out at relativistic speeds. And if stars get accelerated that way then why not black holes?

Drakkith
Staff Emeritus
Thank you for clarifying about the the two perscpectives. BTW I just found dozens of serious articles about stars moving at near the speed of light. They have not yet been observed because they a far away and faint but astronomers believe that when two supermassive black holes collide, as will happen with the Milkyway and Andromeda, as they get locked in the proverbial death dance some stars will be hurled out at relativistic speeds. And if stars get accelerated that way then why not black holes?
I see. I don't see any reason black holes wouldn't have the same thing happen under similar circumstances. I guess I've just never read about relativistic stars.

We have not seen anything which would suggest the existence of black holes traveling at relativistic speed from the point of view here on Earth.
Until we do see such a thing I don't think it will become an issue.

Ibix