Can kinetic energy lead to black hole?

[anorlunda]

I thought that the OP was seeking confirmation that total energy, not just rest mass, gravitates and thus should be included in calculating the Chandrasekhar limit. (Even if his example and expression of the question was flawed.)

A recent PF Insights, https://www.physicsforums.com/insights/do-photons-have-mass/, used the definition "The invariant mass of a particle is defined as the total energy of the particle measured in the particle’s rest frame divided by the speed of light squared." The definition says particle, but I take it to apply anybody or system including a black hole.

Physicists have stopped using the term relativistic mass because it causes confusion. Isn't the situation the same with the term rest mass? Why not abandon the word mass in all contexts and talk only about energy instead? If we did so, then then OP might not have needed to ask his question in the first place.

I see that particle physics with Feynman diagrams already uses elecron-volts for the energy of particles entering and leaving, eschewing use of mass.

 

[PeterDonis]

I thought that the OP was seeking confirmation that total energy, not just rest mass, gravitates

But that is not correct. What gravitates is the stress-energy tensor; "total energy" is not a good heuristic because it is frame-dependent. Failure to recognize that is what leads to the common misconception that you can turn yourself into a black hole if you move fast enough.

Physicists have stopped using the term relativistic mass because it causes confusion. Isn't the situation the same with the term rest mass?

No, because rest mass is an invariant. (Many relativity texts now use the term "invariant mass" instead of "rest mass" to underscore this point.) If you want to do a quick estimate of how much a given system will gravitate, the best quantity to use is the system's invariant mass. If the system is composed of multiple objects that are moving with respect to each other, the kinetic energy of those objects in the system's center of mass frame will contribute to the invariant mass of the system, and therefore will affect how much it gravitates. But you can't increase how much the system gravitates by transforming to a frame in which the system's center of mass is moving at relativistic speed and looking at the total energy in that frame.

I see that particle physics with Feynman diagrams already uses elecron-volts for the energy of particles entering and leaving, eschewing use of mass.

This has nothing to do with using energy instead of mass; it has to do with using energy units instead of mass units, because energy units are easier to compare to actual measurements.

 

[anorlunda]

What gravitates is the stress-energy tensor; "total energy" is not a good heuristic because it is frame-dependent.

But you can't increase how much the system gravitates by transforming to a frame in which the system's center of mass is moving at relativistic speed and looking at the total energy in that frame.

Isn't that handled in ZapperZ's definition by referring to the object's rest frame? That eliminates kinetic energy seen by other frames. Once again, "The invariant mass of a particle is defined as the total energy of the particle measured in the particle’s rest frame divided by the speed of light squared."

We don't know what happens inside a BH. All the original mass could be converted to massless particles or not. The point should be that we don't need to know. It is irrelevant with respect to the gravity external to the event horizon. E/c^2 gravitates the same as the m. I say that without reference to GR.

Doesn't the same thing arise in the "radiation dominated" version of the FRW equation in cosmology? It describes the gravitational potential energy in a universe with no matter. Again without reference to GR.

 

[Stephanus]

I thought that the OP was seeking confirmation that total energy, not just rest mass, gravitates and thus should be included in calculating the Chandrasekhar limit. (Even if his example and expression of the question was flawed.)

"Total energy"? is that the term. Yes, I think it's simpler. I should have rephrased my question. But I didn't think of that until your post came up.
Chandrasekhar limit?

Actually, I meant the Tolman–Oppenheimer–Volkoff limit which is not well known, but is believed to be around 3 solar masses.

PAllen post stated another limit. Tolman-Oppenheimer-Volkoff. I'll look it up in wiki.

[..]Failure to recognize that is what leads to the common misconception that you can turn yourself into a black hole if you move fast enough.

I often read that phrase.
"If you move fast enough, your clock slows down."
But I think the correct statement is, "If you move fast enough then your clock looks slower wrt a rest observer", is this right?
Should the statement be changed to this?
"You look like black hole according to the rest observer if you move fast enough according to the rest observer"?
Because as some mentors/advisors wrote in previous post.
"We are moving at 99.99%c according to particle at LHC. Do we see our clock slow down?"
Thanks for the responds guys.

 

[Nugatory]

But I think the correct statement is, "If you move fast enough then your clock looks slower wrt a rest observer", is this right?

Although that statement can be read in a way that makes it not flat-out wrong, it is so easily misunderstood and so misleading that I cannot bring myself to say that it is "right". The problem is that to speak in terms of you "moving fast enough" as opposed to being a "rest observer" is to invite the demon of absolute motion back into the discussion.

You look like black hole according to the rest observer if you move fast enough according to the rest observer

And that's an example of the sort of misunderstanding that results from accepting the previous statement.

Perhaps "Clocks moving relative to you read slower than clocks at rest relative to you" would be better? However, there's only so precise a natural-language statement can be.

 

[PeterDonis]

"Total energy"? is that the term.

No. See my post #32.

Chandrasekhar limit?

That limit is not a limit on total energy, which, as I pointed out, is frame-dependent. It is best viewed as a limit on invariant mass.

Should the statement be changed to this?
"You look like black hole according to the rest observer if you move fast enough according to the rest observer"?

No. Whether or not an object is a black hole is an invariant; it doesn't change if you change frames. An object cannot look like a black hole to some observers but not others. This is exactly the sort of confusion I was warning against in post #32.

 

[PeterDonis]

Isn't that handled in ZapperZ's definition by referring to the object's rest frame?

Obviously not, since your phrasing has led Stephanus into exactly the sort of confusion I warned against. The term "total energy" suggests something frame-dependent; the term "invariant mass" does not. That's why you use the latter when you are talking about the conditions for an object becoming a black hole, since being a black hole is an invariant.

We don't know what happens inside a BH.

We can't observe what happens inside from the outside, true. But whether or not an object is a black hole is still an invariant, so we should use invariant language when talking about it.

Doesn't the same thing arise in the "radiation dominated" version of the FRW equation in cosmology? It describes the gravitational potential energy in a universe with no matter.

No, it doesn't. The concept of gravitational potential energy is not well-defined in an expanding universe. It's only well-defined in stationary spacetimes.

Also, the radiation dominated cosmology has no "matter" in the sense of "stuff moving at non-relativistic speeds". But it certainly has a nonzero stress-energy tensor, and, as I've already pointed out, the stress-energy tensor is the source of gravity.

Again without reference to GR.

Um, what? You do know that the FRW spacetimes (all of them, not just the radiation-dominated one) are GR solutions, right?

 

[Stephanus]

"Total energy"
What if a 3 solar mass neutron star collides with an 3 solar mass anti matter neutron star?
(I know this is very unlikely. Even for a non scientist, I have read about baryongenesis and something like baryon asymetry. And again, let's dispense with technical difficulties, that there is somehow an anti matter neutron star)
As you know, this will create enormouse energy. What is the result? There's just pure energy, no mass in that so packed diameter.
Also I have read a "wrong question". What if a black hole from matter collide with anti matter black hole. And not to be confused with our discussion, there's no matter (or anti matter, for that matter ) black hole.
So if a huge energy is packed in a small radius, smaller than the Schwarzschild radius for matter equivalent, will it become a black hole?
@PeterDonis
I see that you edit your post before I reply.
Yes, I knew, it's in your latter post rather in your first post.

 

[anorlunda]

Obviously not, since your phrasing has led Stephanus into exactly the sort of confusion I warned against. The term "total energy" suggests something frame-dependent; the term "invariant mass" does not.

The prasing was not mine. It is ZapperZ's phrasing in the PF Insights article linked above. The phrasing does not dodge use of invariant mass it is the definition of invariant mass.

Um, what? You do know that the FRW spacetimes (all of them, not just the radiation-dominated one) are GR solutions, right?

See the Sussikind video below. He derives the FRW equations, including radiation dominated, from Newtonian mechanics. FRW may be a solution of GR, but GR is not necessary to derive it.

 

[PeterDonis]

The prasing was not mine.

I realize you didn't write the article, but you linked to it in this thread, so you're the one claiming its phrasing is relevant to this discussion.

He derives the FRW equations, including radiation dominated, from Newtonian mechanics.

No, he shows how the FRW equations can make sense if you use an interpretation based on Newtonian mechanics. This is not news. But it is not the same as an actual derivation.

 

[Dale]

This particular form of the problem is unnecessarily difficult. Would you be satisfied with considering a simpler problem, two such rockets, colliding head on?

With this modification, the center-of-mass of the collision is stationary, and the center-of-mass energy density is what is immportant.

Since 1 solar mass is about 1.5 km, a rocket smaller than 1km would easily produce the energy densities required to create a black hole.

While you didn't ask this question, I don't believe it would even be necessary for the two rockets to actually collide, passing within, say, the previoiusly mentioned kilometer of each other should be easily good enough to create a black hole.

Don't you need spherical symmetry also? I believe that you are basically referencing Birkhoff's theorem, but I am not sure that there is an equivalent theorem for axisymmetric scenarios instead of spherically symmetric scenarios.

 

[PeterDonis]

Don't you need spherical symmetry also? I believe that you are basically referencing Birkhoff's theorem

I think he's actually referring to the hoop conjecture, which doesn't assume spherical symmetry.

 

[Stephanus]

Dear Mentors/Advisors,
I should have stopped watching this thread , once I got the answer from @PeterDonis that a one ton rocket hitting a neutron star at 99.9999...% the speed of light or just passing by inside the supposed Schwarzschild radius as stated by @pervect is enough to create a black hole. It's all the answer that fulfill my curiousity.
But then again I remember DrGreg post.

By the way. that's the wrong formula. In relativity it's $E_k = (\gamma - 1) mc^2$

Now I have another question confusion.
Motion is relative, right. What if we apply this equation to the neutron star. $E_k = (\gamma - 1)m_{star}c^2$, not $E_k = (\gamma - 1)m_{rocket}c^2$
The neutron star shouldn't move at 99.9999% the speed of light. Just 99% speed of light will do, then gamma is 7. And if the neutron star hits the rocket the neutron star and the rocket meet then could this process create a black hole?
What if we don't have to move the neutron star, it will require tremendous energy, right. What if we push the rocket just 99% the speed of light. Perhaps this process will take, say.. 10 years. And in 10 years + 1 day, all the other observer away from the star and the rocket will only see that these two objects are approaching each other.
And what should the observer calculate the process?
A: $E_k = (\gamma - 1) m_{star}c^2$, or
B: $E_k = (\gamma -1) m_{rocket}c^2$?
Perhaps the answer is a technical one, not mathematic. Still, I hope this can clear some confusion.

 

[Dale]

I think he's actually referring to the hoop conjecture, which doesn't assume spherical symmetry.

Oh, cool. I didn't know about that.

 

[pervect]

I think he's actually referring to the hoop conjecture, which doesn't assume spherical symmetry.

Yes, the hoop conjecture, mentioned by other posters, is mainly what I was thinking of. Using Google, there is also some other theoretical work that has been studying gravity wave collisions in higher-dimensional ADS space-times, see for instance http://arxiv.org/pdf/0902.4062.pdf, http://link.springer.com/article/10.1007/s11232-009-0152-x#page-1. These may not be strictly relevant to the results of near-collision of ultra-relativistic masse in Minkowskii space-time, though I find them highly suggestive, especially combined with the hoop conjecture. I am also assuming that the formation of an apparent horizon (a trapped null surface) would imply the formation of an absolute horizon in the classical context where there aren't any issues with black hole evaporation. I could be wrong on this point, too.

 

[PeterDonis]

I am also assuming that the formation of an apparent horizon (a trapped null surface) would imply the formation of an absolute horizon in the classical context where there aren't any issues with black hole evaporation.

This is correct; in fact, it's basically the content of the singularity theorems of Hawking and Penrose.