2
$\begingroup$

Because of time dilation we cannot observe a black hole forming in a finite amount of time. For the same reason I suppose we also cannot observe the horizon moving: everything happening on the horizon takes an eternity to witness from the outside perspective.

Therefore, would a moving black hole result in new horizons (almost) forming according to an outside observer? The initial horizon would remain frozen in time, followed by the horizons around the moving singularity. Meaning, would moving black holes leave a trail of blackness behind, everywhere it passed?

An important issue here is the moving reference frames. Could one really claim that there is a perspective where the observer moved, rather than the black hole? The observer isn't curving spacetime to extremes, while spacetime is a medium: it's a fabric, it seems more than something described by coordinate systems.

Edit: I changed the question title, it used to be "Moving reference frame of a black hole" but the new title better suits my question. The issue with reference frames is more a follow-up question.

$\endgroup$
  • 1
    $\begingroup$ Think of your question this way. Say, I am stationary relative to a black hole, but you are moving toward it. Obviously I don't see any weird "trail of blackness" on the other side of the black hole from you. If I don't see it, it doesn't exist. Would you see it? Remember, in relativity, physics doesn't depend of the frame of reference. So, in your frame, you'd see me moving toward you. How about the black hole? Remember, it is stationary relative to me. And despite a typical answer of BC, you don't need to actually see the horizon to know a black hole is moving. $\endgroup$ – safesphere Oct 14 '18 at 2:30
  • $\begingroup$ @safesphere: agreed, therefore it seems to me that black holes could be an exception to the abitrary frames of reference. Think about spacetime as an actual fabric, perhaps like a quantized grid where nearby points are separated at the order of the Planck-length. Once the grid has been curved by a BH so that the time dilation becomes infinite, wouldn't it need to remain curved according to an outside observer? $\endgroup$ – Geert VS Oct 14 '18 at 8:00
  • $\begingroup$ I see your point now. Spacetime with an intrinsic curvature is equivalent to relative aether: "According to the general theory of relativity space without ether is unthinkable", - Albert Einstein. So you argue that a critically curved aether may cease to be relative and not be able to move. However, black holes move and even hit each other, so the intrinsic curvature concept is challenged. Sorry if this is not clear, but +1. You would need luck getting a good answer. $\endgroup$ – safesphere Oct 14 '18 at 8:30
  • $\begingroup$ I came to a new insight: Just like photons don't age but still move, black hole horizons don't age but still move. $\endgroup$ – Geert VS Oct 29 '18 at 23:08
  • $\begingroup$ If a critically curved aether would remain curved, we probably would have measured the huge build-up of curvature in clusters of galaxies, where supermassive black holes move around since they are probably in the center of a galaxy. $\endgroup$ – Geert VS Oct 29 '18 at 23:15
3
$\begingroup$

GR doesn't have global reference frames, only local ones. Therefore you can't have a frame of reference big enough to surround a black hole.

So a better way to phrase this question would be in terms of a moving observer. No, a moving observer cannot observe the horizon. The definition of the horizon is that causal curves from the horizon cannot reach outside events, and this definition precludes any external observer from observing the horizon. This definition is independent of the state of motion of an observer, and is independent of any choice of coordinates.

Because of time dilation we cannot observe a black hole forming in a finite amount of time.

It's not really because of time dilation, it's simply because the definition of a horizon is that it's something you can't observe (the boundary of an externally unobservable region of spacetime).

$\endgroup$
  • $\begingroup$ 1. The gravity and spacetime curvature far from a BH are the same as of any star. So sure I can have a reference frame, say, of the Milky Way galaxy with a black hole inside. 2. You don't need to see the horizon to know a black hole is moving. There are enough other indications like gravitational lensing and orbiting stuff. 3. There is no horizon. As the OP clearly stated, the horizon hasn't formed in a frame of reference of any observer. The horizon could hypothetically exist only for an observer crossing it, but a falling observer does not have a reference frame at the moment of crossing. $\endgroup$ – safesphere Oct 14 '18 at 2:44
  • $\begingroup$ As safesphere commented on my post: doesn't the lack of observing the horizon enable an observer to know where the horizon is? And with an horizon I mean an almost-horizon; since the BH almost forms but needs an infinite amount of time to actually form from an outside perspective. $\endgroup$ – Geert VS Oct 14 '18 at 7:48
  • $\begingroup$ @safesphere: 1. The frame of reference you describe is a frame of reference for the asymptotically flat part of the spacetime, which can't be extended to include the region of high curvature near the black hole. 2. You seem to have misunderstood what I was claiming. I gave a technical definition at "The definition of the horizon..." 3. This is moot because of your misunderstanding of frames of reference in GR. $\endgroup$ – Ben Crowell Oct 14 '18 at 14:32
  • $\begingroup$ @GeertVS: doesn't the lack of observing the horizon enable an observer to know where the horizon is? Yes. The unobservability of the horizon has an exact technical definition, which I gave at "The definition of the horizon..." It doesn't mean that we can't tell whether, e.g., Sag A* is a black hole. It is, and we can tell. $\endgroup$ – Ben Crowell Oct 14 '18 at 14:34
  • $\begingroup$ My understanding is fine, thank you. I am well aware of what I do or do not understand. See this, "$\mathcal{N}$ [the set of Schwarzschild observers] cannot be complemented in a continuous way by an observer meeting $\gamma$ [free falling observer] in $h$ [the horizon] (such an observer would have to move with the speed of light)" - arxiv.org/abs/0804.3619 - Your answer however is not helpful. The asymptotic frame doesn't need to be extended to the black hole to see it moving. We see galaxies move and know they contain black holes. This is sufficient in the context of this question. $\endgroup$ – safesphere Oct 14 '18 at 17:41
-1
$\begingroup$

Since I found an answer to my own question, which I first only wrote as a comment, I'll put it here to wrap things up:

Just like photons don't age but still move, black hole horizons don't age but still move.

$\endgroup$
  • $\begingroup$ (Hawking radiation aside) $\endgroup$ – Geert VS Dec 22 '18 at 13:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.