July 17, 2011 12:50 AM Subscribe

Does a black hole that decelerates from relativistic speeds lose enough mass to cease to be a black hole?

A while ago on an episode of Astronomy Cast I heard a question that stumped the presenters. I still haven't found a good answer, but perhaps metafilter can come to the rescue.

Imagine you have a stationary 1kg sphere. As you begin to accelerate the sphere to relativistic speeds its mass increases. Eventually, when the sphere is travelling at a high enough percentage of c, the mass (and therefore density) of the sphere will be high enough to form a black hole. Now imagine we decelerate this black hole. It will lose mass (and therefore density) until... will it remain a black hole? Will it turn back into normal matter? Will it explode it a puff of logic?

Taking this question forwards, for the purposes of science fiction, could a spacecraft trapped inside a black hole be rescued if the entire black hole is decelerated externally?

A while ago on an episode of Astronomy Cast I heard a question that stumped the presenters. I still haven't found a good answer, but perhaps metafilter can come to the rescue.

Imagine you have a stationary 1kg sphere. As you begin to accelerate the sphere to relativistic speeds its mass increases. Eventually, when the sphere is travelling at a high enough percentage of c, the mass (and therefore density) of the sphere will be high enough to form a black hole. Now imagine we decelerate this black hole. It will lose mass (and therefore density) until... will it remain a black hole? Will it turn back into normal matter? Will it explode it a puff of logic?

Taking this question forwards, for the purposes of science fiction, could a spacecraft trapped inside a black hole be rescued if the entire black hole is decelerated externally?

In short, no. One of the cardinal rules of relativity is that there are no preferred reference frames where the physics is "true." In the frame of your sphere, the sphere itself is always stationary and if it emitted light (say it is a disco sphere) that light would necessarily zip away at the speed of light (another cardinal rule: the speed of light is the same in all reference frames). Since those trajectories are the same as what any other observer in any other reference frame would see, adjusted only for length and time contractions, your sphere can't be a black hole anywhere. Basically, you can't have closed light trajectories in some reference frames and not others, and if thus if you aren't a black hole when stationary, you can't ever be.

posted by Schismatic at 1:25 AM on July 17, 2011 [3 favorites]

posted by Schismatic at 1:25 AM on July 17, 2011 [3 favorites]

Sure it could! Physicists who read the story would snicker. But that might not matter.

posted by flabdablet at 2:45 AM on July 17, 2011

There's no such thing as deceleration, btw. Only acceleration in the opposite direction.

posted by empath at 7:18 AM on July 17, 2011 [1 favorite]

It's correct that the density of the sphere will increase as you speed it up, though this is just due to its Lorentz contraction rather than any change in the mass itself. However, it's not just mass density that affects the gravitational field; so do the presence of momentum density and pressure. All three of these are combined into a big mathematical entity called the stress-energy tensor, and it's this entire entity that determines the curvature of spacetime. You need to look at all parts of this mathematical entity to determine how strong the gravitational effects are, and thus whether a black hole will form or not. Most of the time, the mass density is much larger than the other components of the stress-energy tensor, and you can get away with ignoring things like momentum density and pressure. When you have a massive sphere moving at a significant fraction of the speed of light, though, these other effects become non-negligible.

posted by Johnny Assay at 7:32 AM on July 17, 2011

Time dilation, length contraction, etc, are all consequences of observing a distant object moving at the speed of light in relation to your inertial reference frame. If you watched a distant spaceship moving near the speed of light relative to you, it would *appear* to become a black hole to you at some point. But if you're in the object itself, nothing would have appeared to change whatsoever -- in fact, you'd appear to become a black hole to it. (Since relative to itself, you'd be moving near the speed of light relative to it).

The difference between that and a*real* black hole is that there is no reference frame in which a black hole is not a black hole.

posted by empath at 9:23 AM on July 17, 2011

The difference between that and a

posted by empath at 9:23 AM on July 17, 2011

Looks like everyone else has covered the main points, but in case you're still bothered by the idea that eventually the apparent local energy density of your 1 kg sphere would exceed that required to form a black hole; I'll add that the Schwartzschild solution for a black hole (the one that says how densely packed non-rotating, non-charged matter has to be in order to form a black hole) only works in a very simple set of assumptions. Among those is that the matter is stationary with respect to the coordinates you've chosen. You can't extrapolate the result simply from that frame to another and expect it to still apply.

What you're in effect asking is "how densely packed would matter have to be to form a black hole singularity while moving at X% of the speed of light relative to the coordinate system of some observer*according to that observer*" (the assumption here is that you've defined the coordinate system to be stationary relative to you). I've never done the calculation in that frame, but if you wanted to, the way to do it is "boost" from the stationary frame to the frame of the object, find out how much matter density you'd need there (exactly how much the Schwarzschild solution requires, since now you've boosted to the Schwarzschild metric) and then boost back to the stationary observer. What you would find is that the matter density required in your frame would be much larger than the naive Schwarzschild result would tell you. This is fine, since the two values are answering two different questions.

empath, I think you're incorrect. I don't think that anyone would see a singularity form in any frame of reference, for basically the reason Schismatic stated. I could be wrong though. As I said, I've never done this calculation explicitly, but I remember the paradox bothering the hell out of me in college and I think the solution is as I've written above. It would appear to me that having a black hole in one frame and not in the other will really screw with the 2nd law (not to mention causality and some other basic facts about the Universe I'm rather fond of), but I'd love to be proven wrong on this. Also, the distance to the observer has nothing to do with the reality of the Lorentz effects; there's no metric to worry about here. Atmospheric muons, for example, do really see the atmosphere as Lorentz contracted (alternatively, we see them as Lorentz dilated), and they are passing through you right now.

I'll also add that an accelerating observer (which is different from an observer moving with constant velocity relative to another observer) will see a event horizon form which the stationary (i.e. non-accelerating) observers don't see. This will have "Hawking radiation" (really Unruh radiation) associated with it, and can be attributed to the fact that, as long as you are accelerating, there are places in the Universe that the light emitted will never be able to reach you from.

posted by physicsmatt at 9:44 AM on July 17, 2011 [2 favorites]

What you're in effect asking is "how densely packed would matter have to be to form a black hole singularity while moving at X% of the speed of light relative to the coordinate system of some observer

empath, I think you're incorrect. I don't think that anyone would see a singularity form in any frame of reference, for basically the reason Schismatic stated. I could be wrong though. As I said, I've never done this calculation explicitly, but I remember the paradox bothering the hell out of me in college and I think the solution is as I've written above. It would appear to me that having a black hole in one frame and not in the other will really screw with the 2nd law (not to mention causality and some other basic facts about the Universe I'm rather fond of), but I'd love to be proven wrong on this. Also, the distance to the observer has nothing to do with the reality of the Lorentz effects; there's no metric to worry about here. Atmospheric muons, for example, do really see the atmosphere as Lorentz contracted (alternatively, we see them as Lorentz dilated), and they are passing through you right now.

I'll also add that an accelerating observer (which is different from an observer moving with constant velocity relative to another observer) will see a event horizon form which the stationary (i.e. non-accelerating) observers don't see. This will have "Hawking radiation" (really Unruh radiation) associated with it, and can be attributed to the fact that, as long as you are accelerating, there are places in the Universe that the light emitted will never be able to reach you from.

posted by physicsmatt at 9:44 AM on July 17, 2011 [2 favorites]

I'm probably wrong if you say so :) I guess I meant 'appear to meet the requirements of a black hole'.

posted by empath at 9:56 AM on July 17, 2011

posted by empath at 9:56 AM on July 17, 2011

Does relativistic mass effect gravity? That is, would star moving very fast relative to an observer appear to bend light passing through it more than an identical one that is stationary?

posted by empath at 10:05 AM on July 17, 2011

posted by empath at 10:05 AM on July 17, 2011

I'm pretty sure that a 10-solar-mass black hole at rest in your frame and a 10-solar-mass black hole moving at 0.9

posted by Johnny Assay at 10:51 AM on July 17, 2011

I'm pretty sure you wouldn't see a black hole moving towards you at near the speed of light until it was too late :)

posted by empath at 11:24 AM on July 17, 2011

posted by empath at 11:24 AM on July 17, 2011

Johnny has it (or, at least, I agree with his reasoning- again without having done the calculation).

posted by physicsmatt at 11:38 AM on July 17, 2011

posted by physicsmatt at 11:38 AM on July 17, 2011

Thanks a lot everyone! Now I'm left wondering why something which is apparently so simple would leave the presenters stumped.

Thanks again for some very interesting comments.

posted by MighstAllCruckingFighty at 12:00 PM on July 17, 2011

Thanks again for some very interesting comments.

posted by MighstAllCruckingFighty at 12:00 PM on July 17, 2011

It's certainly not simple (notice that I haven't done any of the math to prove this, partly because I'm pretty sure I know the result, partly because I have other physics to work on, but mostly because it's hard to calculate). The question requires one to combine what they know from general relativity to what they know from special relativity. The problem is that exact solutions in GR are hard to come by. The Schwarzschild solution is one of the known ones, but if you don't think about exactly how you got that solution, you'll just recall the result (that X amount of mass inside the Schwarzschild radius = a black hole) without remember what conditions apply to the result.

Not sure if there's a point here other than physics can be hard sometimes. Relativity in particular is filled with easily pose-able questions that require a lot of thought to see how to resolve the paradox.

posted by physicsmatt at 12:27 PM on July 17, 2011

Not sure if there's a point here other than physics can be hard sometimes. Relativity in particular is filled with easily pose-able questions that require a lot of thought to see how to resolve the paradox.

posted by physicsmatt at 12:27 PM on July 17, 2011

Great question, interesting answers.

Almost the identical question arises from a consideration of extremely energetic photons.

I've never heard of an upper limit on photon energy (not that that means much) but if there isn't one, then you could have a photon so energetic that the equivalent rest mass (via e=mc^2) would form a black hole of any given size, and with photons the higher the energy the shorter the wavelength and in some sense the 'smaller'-- as witness the failure of photons to pass through holes much smaller than their wavelength under ordinary circumstances-- which would intensify the paradox.

One of the interesting things black holes might seem to do is limit the strength of gravitational fields we can experience, because all stronger fields are within Schwarzschild radii, but if extremely energetic photons (EEPs) don't collapse into black holes, that limit would seem to be removed (as it would with the very fast-moving objects of Johnny Assay's answer too, I suppose).

I take the deflection of photons by gravitational fields to demonstrate that photons do have a gravitational field of their own, because conservation of momentum demands that the sources of the field be deflected by the photon in turn, and gravity is the only candidate for the force making such a compensating deflection.

But if photons are attracted by masses because of the gravitational fields of those masses, shouldn't photons attract each other gravitationally? And if they do, couldn't two photons orbit each other the way twin stars do? (Perhaps the geometry of the photon field would make this difficult.)

And if so, and the photons are EEPs, you might have a situation in which some sphere containing the orbits of the photons had enough rest mass equivalent inside it to make a black hole.

posted by jamjam at 1:35 PM on July 17, 2011

Almost the identical question arises from a consideration of extremely energetic photons.

I've never heard of an upper limit on photon energy (not that that means much) but if there isn't one, then you could have a photon so energetic that the equivalent rest mass (via e=mc^2) would form a black hole of any given size, and with photons the higher the energy the shorter the wavelength and in some sense the 'smaller'-- as witness the failure of photons to pass through holes much smaller than their wavelength under ordinary circumstances-- which would intensify the paradox.

One of the interesting things black holes might seem to do is limit the strength of gravitational fields we can experience, because all stronger fields are within Schwarzschild radii, but if extremely energetic photons (EEPs) don't collapse into black holes, that limit would seem to be removed (as it would with the very fast-moving objects of Johnny Assay's answer too, I suppose).

I take the deflection of photons by gravitational fields to demonstrate that photons do have a gravitational field of their own, because conservation of momentum demands that the sources of the field be deflected by the photon in turn, and gravity is the only candidate for the force making such a compensating deflection.

But if photons are attracted by masses because of the gravitational fields of those masses, shouldn't photons attract each other gravitationally? And if they do, couldn't two photons orbit each other the way twin stars do? (Perhaps the geometry of the photon field would make this difficult.)

And if so, and the photons are EEPs, you might have a situation in which some sphere containing the orbits of the photons had enough rest mass equivalent inside it to make a black hole.

posted by jamjam at 1:35 PM on July 17, 2011

The energy of a photon is related to the frequency, which is limited by quantization, no?

posted by empath at 1:48 PM on July 17, 2011

Photons have 0 mass and they always move at the speed of light (as would anything that has 0 mass), and they always travel in straight lines (or geodesics in curved space). The only reason they appear to be curved by gravity is that massive objects curve space time.

posted by empath at 1:52 PM on July 17, 2011

This thread is closed to new comments.

As you begin to accelerate the sphere to relativistic speeds its mass increases.No, it really doesn't. That is a pedagogical lie used for teaching SR that has fallen out of favor.

The real story is that in SR you have to correct the Newtonian definition of momentum from p = mv to p = š¯›¾mv, where š¯›¾ is the Lorentz factor. Now it's tempting to look at that and squint and say that the "effective mass" or something is š¯›¾m, so that you can continue to use the old familiar Newtonian equations like F = ma where m is rest mass plus some relative mass term. But that's not really what's happening. Mass is not really increasing, it's just that momentum has this Lorentz factor added, just like Ī”t' = š¯›¾Ī”t (time dilation) and Ī”x' = Ī”x/š¯›¾ (length contraction).

posted by Rhomboid at 1:24 AM on July 17, 2011 [5 favorites]