High speeds - becoming a black hole

[dhi]

Recently someone I know made this claim: accelerating a spaceship until it (almost) reaches the speed of light it will transform the spaceship into a black hole.

Though I believe the contrary, I want to see other opinions. Is it so, is it not? And why?

Thank you.

 

[Physics Monkey]

I'm afraid your acquaintance was mistaken, though it is a common misconception. See here for a nice discussion: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html

 

[russ_watters]

It will not change into a black hole, because relativistic mass is frame of reference dependant (since the ship's speed is frame dependant), and in the only frame that matters - the frame of the ship itself - the ship's mass has not changed.

 

[aaroman]

The claim was argumented this way: in order to accelerate it to the (almost) light speed, you need to add energy. (Almost) an infinite energy. Considering the equivalence between mass and energy, this will result in making a black hole, after enough acceleration.
If the ship would have to carry its own fuel (that is energy) needed for acceleration, the mass would be enough for that black hole from the start, but one that likes this kind of argument can "escape" with a solution to feed the ship with energy during acceleration. Perhaps even with energy extraction from void :)

Ok, I've seen that there are other posts already: please do not forget the acceleration thing: it does not come from nothing.

By the way, that article cannot convince with such arguments:

This would be paradoxical since we would expect things to appear very differently to an observer who is stationary relative to the star.

Things do look different for different observers. Imagine for example the ship going with a speed -> c away from the star. That means if I'm not mistaken, that redshift -> infinity. According with the equivalence principle, that is the same as a gravitational redshift -> infinity, and it looks like a black hole.

 

[Physics Monkey]

aaroman, a black hole is a frame independent feature of spacetime, a black hole in one frame is a black hole in every frame. If there isn't a black hole in the rest frame of the ship then there can't be a black hole in any other frame. Things do look different to different observers, but not that different.

Also, if you add other objects into spacetime (like accelerators supplied with lots of energy) then you complicate the situation. If you accelerate the ship very slowly, then it can still reach high speeds without ever requiring a preponderance of energy located in a small region. The ship is going fast and it isn't a black hole.

 

[aaroman]

Why? Saying it is not enough for me. I don't see why an object cannot be seen as a black hole from one frame of referrence, and not a black hole from another.

It seems that there is a somewhat related thread here:
https://www.physicsforums.com/showthread.php?t=35884
From there I extracted this:

This is what Hawking said about creating black holes in particle accelerators.
.
"If one had a particle with an energy above what is called the Planck energy, 10 million million million Gev (1 followed by 19 zeros), its mass would be so concentrated that it would cut itself off from the rest of the universe and form a little black hole."

It looks like the one that is considered one of the best experts in the world on this subject, did say something else.

PS I'm far from being an expert in the field, but I know at least that: relativistic mass do have gravitational influence. It's enough to look at the equations, to see there the energy and the flow of momentum, not only the rest mass. So it might happen that if a ship does travel near an object which such a high speed, its gravitational effects on the object would be felt like it would be a black hole, if, of course, the speed is fast enough. So, if it looks like a black hole, and it behaves like a black hole, it is a black hole. For that observer, that is.

 

[Physics Monkey]

A traditional black hole is a place in spacetime where the curvature diverges. Just calculate the curvature invariant I = R^{\alpha \beta \gamma \delta} R_{\alpha \beta \gamma \delta}, and note that this object is scalar which is invariant between frames. A black hole in one frame means I diverges somewhere, but since it is a scalar it must diverge at that point in all frames.

I don't know the context of Hawking statement, but I can assure you he wasn't saying that a particle moving really fast all by itself would become a black hole. Any GR textbook will affirm this fact, so has russ, and so does the Usenet FAQ which is maintained by reputable physicists who know what they're talking about. I think it wise that I not speculate on exactly what Hawking meant until you can provide me a reference. Classically, any point particle should form a black i.e. some mass in zero volume. But if you are more careful and couple a quantum field theory to classical GR then you find that the particle like excitations are not surrounded by black holes.

Also, be careful about declaring the "relativistic mass" really does contribute. Imagine a massive ball with some uniform density. You are right that the energy density goes up in in a moving frame, but the energy and momentum flux are now also present. You implicitly assume that these new terms will add to field, but the field doesn't actually get any stronger (say as measured by the curvature invariant), only its components look different.

 

[cd27]

this is interesting...i have always assumed that a black hole would be created. thank you for posting this here.

cd

 

[aaroman]

Also, be careful about declaring the "relativistic mass" really does contribute

Is the photon mass relativistic? It is not rest mass, I'm pretty sure.
Its path is curved when it travels near a big mass. I assume that if the mass has gravitational effects on it, the photon has gravitational effects on the mass. Or else we could build nice thing that accelerates by itself. Just put a box full of a lot of photons in a box at the end of a rod, and a big mas at the other end. The mass will drag the photons in one way, but the photons will not drag the mass in the other way :)

PS Hawking's quote is from "A brief history of time". I just checked and it's accurate.

 

[Physics Monkey]

The photon doesn't have mass in the modern parlance, it has energy. I choose not to use the notion of relativistic mass for the confusion it can bring. A light beam is deflected by massive objects but only because it follows the null geodesics of spacetime which are determined by those massive objects. As for whether the light beam generates any curvature, what do you think?

Also, I'm not suggesting the quote isn't accurate, but one needs to know the context to understand exactly what Hawking is saying. Otherwise, you end up arguing, just as you are now doing, over something that is known to have simple answer. I hate to just appeal to authority on this one, but do you really think the well known Usenet FAQ is just flat wrong, that there is some massive misunderstanding amongst physicists about what would happen?

 

[aaroman]

As for whether the light beam generates any curvature, what do you think?

Well, I think that it does. It has energy and momentum, you said it. Or else I just invented a perpetuum mobile :)
But so it has a moving ship. It has energy and momentum. Apart from its rest mass.

About that curvature invariant, it looks like it is. I'm far from knowing general relativity, I promise to learn it well next year when I'll take the course, but isn't that curvature the curvature of spacetime? So since space and time looks different for the two observers, so it should look the gravity.
The gravity does look different for different observers, even the curvature is invariant, that can be sayed by anybody that experienced a free fall in a place where in other circumstances it feels a "force". So at least this is confirmed experimentally.

 

[pervect]

I hate to just appeal to authority on this one, but do you really think the well known Usenet FAQ is just flat wrong, that there is some massive misunderstanding amongst physicists about what would happen?

Not only the "appeal to authority" has been presented, but a very simple argument based on the principles of general covariance was already given that shows that an object cannot become a black hole just by moving.

It's really a very simple argument. The key issue as to whether an object is a black hole is whether or not the light-like geodesics (null geodesics) originating at the surface of the object reach infinity. This is the definition of a black hole.

By far the easiest way to calculate the null geodesics (the paths of light beams) is to calculate them in the rest frame of the black hole.

General covariance says that if you have the equations for a beam of light (null geodesic) in one coordinate system, you get the equations for the same beam of light in a different coordinate system by just by mapping the old coordinates to the new coordinates.

The result of this is that because the coordinate transforms aren't singular, a beam of light that escapes to infinity in the rest frame of the object still escapes to infinity in "the" coordinate system in which the object is moving.

This means that the object cannot be a black hole just because it moves.

There is one issue I've glossed over, that's the issue of how to do the coordinate transform. Actually it's miselading to say that there is just one coordinate system associated with a moving observer in curved space-time. Because GR has the ability to work with arbitrary coordinate systems, an ability that is needed in curved space-time, it's hard to single out a single preferred coordinate system to describe the space around a moving mass. I suspect that for the problem at hand, the easiest requirement to "pin down" the coordinate system would be to ask that radially outgoing null geodesics appear to be "straight lines" in the new coordinate system, but actually the details don't matter. Any coordinate system choice that approaches the flat-space Lorentz transform is going to be non-singular, and because GR can happily deal with arbitrary coordinate systems, the choice doesn't really matter.

So we've quoted the appropriate authorities, and we've presented a detailed argument. There's not much more we can do except ask aaroman to listen.

 

[Peterdevis]

to set things right:

As for whether the light beam generates any curvature, what do you think?

In fact it does! The energy-momentum tensor contains al energy (except gravitational) en influence the curvature.

 

[russ_watters]

An interpretation of the Hawking quote from that thread:

This is what Hawking said about creating black holes in particle accelerators.
.
"If one had a particle with an energy above what is called the Planck energy, 10 million million million Gev (1 followed by 19 zeros), its mass would be so concentrated that it would cut itself off from the rest of the universe and form a little black hole."

"Of course, the Planck energy is a very long way from the energies of around a hundred GeV, which are the most we can produce in the laboratory at the present time. We shall not bridge that gap with particle accelerators in the foreseeable future!"

p225 The Illustrated A Brief History of Time [1996 edition].

I do not interpret this as meaning you can accelerate a particle less massive than the Planck energy into a black hole. Rather, that if you could accelerate and collide less massive particles at high enough energies, a Planck energy size particle could be created [i.e., a tiny black hole result]. This is the same process CERN used to discover the W+, W- and Zo particles. They did not convert less massive particles to these highly massive particles through acceleration, it was through the energy-matter conversion resulting from collisions between highly accelerated less massive particles.

 

[aaroman]

Still I don't see any difference between "rest mass" energy, that is the that is the mass that you gain by colliding the particles, and the energy that is not rest mass, that is just before the particles "touch" each other (but they are under the Schwartzchild radius). I don't see why the particles would become a black hole only after they touch each other and form a whole, whatever that means :)

PS About this:

The result of this is that because the coordinate transforms aren't singular, a beam of light that escapes to infinity in the rest frame of the object still escapes to infinity in "the" coordinate system in which the object is moving.

If a black hole is moving, the beam of light escapes to infinity along with the black hole, in "the" coordinate system in which it moves. Does it mean that the black hole isn't a black hole if it's moving? Or if somebody is moving with respect to it? :)

 

[Physics Monkey]

The relevant energy for particle production is the energy of the system in the center of mass frame. It is this energy, if you like, that could give you a black hole in principle. When physicists do collider experiments at CERN, they don't have to worry about the fact that the Earth is clipping along at a nice pace around the Sun or that the Earth is spinning. The only energy that matters for particle production is the energy in the center of mass frame. It is not a coincidence that the center of mass frame of a single particle system is just the rest frame of the particle. Thus a single fast moving particle won't become a black hole, but if you collide two fast moving particles with sufficient center of mass energy then you could in principle form a black hole, at least classically.

 

[aaroman]

So it's the same thing if you colide two particles that go against each other with 0.9c and the center of mass is standing still, compared with a system where the center of the mass of the two particles go with 0.9999999c compared with the observer (but still they go against each other with 0.9c from the "center of mass" point of view)? I really doubt that you obtain the same things. Let's suppose that in collision some photons are released. Do they have the same frequency for the two scenarios?

 

[Physics Monkey]

aaroman, I'm sorry you don't believe me, but I can assure it is only the center of mass energy that is relevant for particle production. Of course any light emitted will be Doppler shifted in a moving frame, but so what? It isn't as if more massive particles can be created or different particles will appear if you observe the system moving fast relative to it. Look it up in any particle physics book, or any relativity book, or probably just about anywhere.

I don't understand the combative attitude, can't we all be friends?

 

[Pippo]

Recently someone I know made this claim: accelerating a spaceship until it (almost) reaches the speed of light it will transform the spaceship into a black hole.
Though I believe the contrary, I want to see other opinions. Is it so, is it not? And why?
Thank you.

My question is what is the spaceship ? is it just the space occupied by the ship or is it the relativistic space time all around distorted by ship in movement ?
I would say that such a ship creates a wormhole, creates and use it.

 

[aaroman]

Of course any light emitted will be Doppler shifted in a moving frame, but so what? It isn't as if more massive particles can be created or different particles will appear if you observe the system moving fast relative to it. Look it up in any particle physics book, or any relativity book, or probably just about anywhere.

Ok, so the light will be Doppler shifted. That means more energetic photons. Does that mean stronger gravity from them? If that is true, just accelerating enough we could obtain gravity as strong as we want (in theory).

PS I am your friend :) But when somebody tells me that things look the same in two frames of refference, when obviously do not, I don't believe.

 

[Hans de Vries]

PS I am your friend :) But when somebody tells me that things look the same in two frames of refference, when obviously do not, I don't believe.

He was absolutely right when he said that something can not be a black hole
in one reference frame while at the same time it isn’t a black hole in another
reference frame!

If you believe physicists about black holes and relativistic mass why don't you
believe the same physicists that your acquaintance is telling you nonsense?

The Earth is moving at near light speed for some observers. Would you expect
them to see us all crushed to dead in a black hole? Regards, Hans

 

[aaroman]

Well, I think that is already established that what one sees from one reference system is not necessarily what one sees from another.
I don't expect "to see us crushed", I expect to not see us at all.

 

[Hans de Vries]

Well, I think that is already established that what one sees from one reference system is not necessarily what one sees from another.

Yes, it has been established for over a century now: Lorentz contraction, Time dilation.

I don't expect "to see us crushed", I expect to not see us at all.

But not this... This goes against all the textbooks on Special Relativity.


Regards, Hans.

 

[aaroman]

Probably a lot of things go against special relativity when there are things that have gravitation, isn't it?

 

[Hans de Vries]

Probably a lot of things go against special relativity when there are things that have gravitation, isn't it?

No, in this case it's not!

You can either study some textbook to find things out yourself or ask
questions here. I can guarantee you that the answers you've got here
are conform the textbooks.

But I doubt if this will help considering your past posts...

You should be aware that this is all pretty basic physics.Regards, Hans.

 

[aaroman]

Do you see that actually you are not responding to some questions I ask, but instead you do something else?

 

[Hans de Vries]

Do you see that actually you are not responding to some questions I ask, but instead you do something else?

My interpretation of:

Probably a lot of things go against special relativity when there are things that have gravitation, isn't it?

Was that your were just casting doubt on my reply. If you instead want to
change the subject of the discussion then maybe you should make that clearer.


Regards, Hans

 

[aaroman]

Ok, here was the last question with no reply:

Ok, so the light will be Doppler shifted. That means more energetic photons. Does that mean stronger gravity from them? If that is true, just accelerating enough we could obtain gravity as strong as we want (in theory).

 

[pervect]

Well, I think that is already established that what one sees from one reference system is not necessarily what one sees from another.
I don't expect "to see us crushed", I expect to not see us at all.

You are apparently grossly misunderstanding the answer that several different people (and the written FAQ written specifically to answer the question you just asked) are all trying to give you.

Physical reality is independent of the coordinates used to describe it. If person A shoots person B on a moving train, person B is dead in all frames of reference.

He is not alive in some frames, and dead in others. He is dead in all frames.

It does not matter if person C is watching on the train, or person D is watching from the station - person B is dead. (C and D may disagree on the time of death, but they won't disagree that B dies).

 

[aaroman]

Physical reality is independent of the coordinates used to describe it. If person A shoots person B on a moving train, person B is dead in all frames of reference.

Not at the same time, though. Fore some is dead, for some is still living. For some is still unborn, so no, person B is not dead in all frames.
Going back to gravity from this analogy, in one frame the "force" felt could be very weak, in another one could be very strong. One frame just has to accelerate enough :)
I'm not questioning physical reality, but the appearance. If I'm not mistaken, for an outside observer, a falling object will never pass through the event horizon of a black hole. That is not true for the object that falls in. It passes through the event horizon in a finite time. So in some sort of way, for one observer B never "dies", for the one that falls though it's completely another matter.

PS I just looked over some papers: either they are wrong, or the affirmation that a black hole is a black hole in all reference frames which was presented on this thread is wrong. Talking abaout the Schwartzchild solution, is says something like that:

one can discover different coordinate frames in terms of which no singularity is seen. ... Introduce new coordinates ("Kruskal coordinates")

Well, I think I gave enough. It's from "Introduction to General Relativity" by G. Hooft. So what's the truth here? Is it a black hole or not in all coordinates?

 

[Physics Monkey]

PS I am your friend :) But when somebody tells me that things look the same in two frames of refference, when obviously do not, I don't believe.

Wonderful! However, you have taken the mantra "everything is relative" too far. Black holes are not relative (at least not in the sense that I think you want them to be). No matter how fast you move, the components of the curvature are always finite if there is no black hole, and the components are always divergent at some point if there is a black hole. Also, like prevect said, no matter how fast you move, light beams that escape in one frame will still escape in every other frame hence no black hole in one frame means no black hole in every frame.

Regarding your question about the doppler shifted photons, the gravitational field is the same! What do I mean by this? If you make a transformation of coordinates corresponding to a Lorentz transformation then all that happens is that the components of R_{\alpha \beta \gamma \delta} (or any other tensor) get scrambled up. The curvature invariant that I defined earlier is the same in any frame. Geodesics still deviate in the same way as before, so no true new gravitational effects have been produced (although, of course, the coordinates you assign to geodesic and its tangent vector change).

 

[aaroman]

no black hole in one frame means no black hole in every frame

Well, I just edited previous post, I found a paper that says that a singularity can be no singularity in another coordinate frame, so they seem relative if that is true.

Regarding your question about the doppler shifted photons, the gravitational field is the same!

Energy and momentum is different. Why is the gravitational result the same?

The curvature invariant that I defined earlier is the same in any frame

The curvature is invariant, but the force one feels might be different, or else no matter of the reference frame, all would feel the same gravity, which is not true. I tested this experimentally :)

 

[Physics Monkey]

Well, I just edited previous post, I found a paper that says that a singularity can be no singularity in another coordinate frame, so they seem relative if that is true.

You have confused a so called coordinate singularity with a true space time singularity. The traditional Schwarzschild metric is
<br /> ds^2 = - \left(1 - \frac{2 M}{r}\right) dt^2 + \frac{1}{\left(1 - \frac{2 M}{r}\right)}dr^2 + r^2 d\Omega^2 <br />
and there is clearly a coordinate singularity at r = 2 M, the event horizon. However, the components of the curvature are perfectly well defined here so there is no actual singularity. The apparent singularity is an artifact of the coordinate system. The Kruskal-Szekeres coordinate system is one way of removing this artifical coordinate singularity. However, the singularity at r = 0 is a true spacetime singularity and cannot be removed by a clever choice of coordinates. The curvature diverges at r = 0.

 

[aaroman]

Now I might have a problem with points in physical reality, but for the rest I get it.

 

[pervect]

Not at the same time, though. Fore some is dead, for some is still living. For some is still unborn, so no, person B is not dead in all frames.
Going back to gravity from this analogy, in one frame the "force" felt could be very weak, in another one could be very strong. One frame just has to accelerate enough :)

I'm not questioning physical reality, but the appearance. If I'm not mistaken, for an outside observer, a falling object will never pass through the event horizon of a black hole. That is not true for the object that falls in. It passes through the event horizon in a finite time. So in some sort of way, for one observer B never "dies", for the one that falls though it's completely another matter.

You are correct that a falling object will pass through the event horizon at t=infinity in the coordinate system of the outside observer, and that it will pass through the event horizon in a finite proper time as measured by its own clock. This happens because the coordinate transforms are singular - specifically, it is the Schwarzschild coordinates that are singular in this cae.

If you look at the more detailed post I gave earlier, response #12

https://www.physicsforums.com/showpost.php?p=825259&postcount=12

I mentioned that it was important that the coordinate transforms were not singular.

The result of this is that because the coordinate transforms aren't singular, a beam of light that escapes to infinity in the rest frame of the object still escapes to infinity in "the" coordinate system in which the object is moving.

In flat space-time, you are probably familiar with time dilation for a moving object. The time dilation factor can be very large, but will always be finite. This finite time dilation means that there are no singularities in the coordinate transform for a moving observer - the Lorentz transform - any event t occurring on the moving train at a finite time occurs at a finite time t' on the non-moving station, and vica-versa.

PS I just looked over some papers: either they are wrong, or the affirmation that a black hole is a black hole in all reference frames which was presented on this thread is wrong. Talking abaout the Schwartzchild solution, is says something like that:

one can discover different coordinate frames in terms of which no singularity is seen. ... Introduce new coordinates ("Kruskal coordinates")

Well, I think I gave enough. It's from "Introduction to General Relativity" by G. Hooft. So what's the truth here? Is it a black hole or not in all coordinates?

This quote is about the "bad behavior" of the Schwarzschild coordinates at the event horizon that was mentioned earlier. A black hole in Kruskal coordinates remains a black hole - the Kruskal coordinates simply remove the "bad behavior" that the Schwarzschild coordinates exhibit at the event horizon.