# A On the formation of a black hole due to high kinetic energy

#### Joker93

Hello!
If energy bends spacetime, then an object moving at high velocity will bend spacetime a lot around it due to its really big kinetic energy. It follows, that an object can become a black hole at extremely high enough velocities.
But, since velocity is relative, we can find an observer for whom the object is not moving.
My question is, will some observers observe a black hole while others won't?

Related Special and General Relativity News on Phys.org

#### Ibix

No. If it's not a black hole, it's not a black hole.

See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html

Edit: by the way, you've marked this as an A thread, which means you were expecting an answer assuming graduate-level knowledge of GR. I guessed from the question that you didn't actually want me to start throwing tensors around, but more complex answers are available if I've guessed wrong.

Last edited:

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
To add to what @Ibix said, your motivation for asking this is likely that you read somewhere that energy is what curves space-time. This is only partially true. While energy does appear as a source term in Einstein's field equations, it is not the only source term. Momentum and stress also contribute and the final result is a result of the combined effects. The space-time itself will be the same and have the same properties regardless of what coordinates you happen to use.

#### Joker93

No. If it's not a black hole, it's not a black hole.

See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html

Edit: by the way, you've marked this as an A thread, which means you were expecting an answer assuming graduate-level knowledge of GR. I guessed from the question that you didn't actually want me to start throwing tensors around, but more complex answers are available if I've guessed wrong.
While it is true that a qualitative answer answers my question, it would be great if you could also add some mathematics to back it all up. But, I don't really have a graduate-level knowledge on GR. My level of GR is undergrad to grad. So, if there is a partial or full answer at that level, it would be great!

#### Geometry_dude

Hello!
If energy bends spacetime, then an object moving at high velocity will bend spacetime a lot around it due to its really big kinetic energy. It follows, that an object can become a black hole at extremely high enough velocities.
But, since velocity is relative, we can find an observer for whom the object is not moving.
My question is, will some observers observe a black hole while others won't?
Well, the existence of an object cannot depend on who's looking, right? So you have a contradiction. So you know your reasoning or assumptions must be flawed.

In standard GR, you have to specify what you mean by 'energy'. My suggestion would be the energy-momentum-tensorfield $T$ of a perfect fluid. Its components are:
$$T^{ij} = \left(\rho + \frac{p}{c^2}\right) X^i \, X^j - p g^{ij }$$
$\rho$ is the density, a positive scalar field, $p$ the internal pressure, also a scalar field, and $X$ is the fluids (relativistic) velocity field. Using Einstein summation, $X$ satisfies
$$g_{ij} \, X^i X ^j = c^2 \, ,$$
which makes sure that its integral curves are parametrized by proper time. I would argue that, up to a factor of $\rho/2$, this is also (pointwise) the (relativistic) kinetic energy - a constant. Can you see why I would say that?

Do you have an idea where the flaw in your reasoning might be?

Last edited:

Staff Emeritus
Do you really want an answer at that level? Because it sounds like you might not be quite at that level yet.

#### Geometry_dude

Why would you not leave that to him- or herself to decide?

On a side note, you might want to put a minus sign in front of what I termed "relativistic kinetic energy" above, but that's, of course, convention.

#### PAllen

While it is true that a qualitative answer answers my question, it would be great if you could also add some mathematics to back it all up. But, I don't really have a graduate-level knowledge on GR. My level of GR is undergrad to grad. So, if there is a partial or full answer at that level, it would be great!
You don' need detailed math to see that what you propose is impossible. You just need to have studied what are geometric invariants. An event horizon is geometric invariant feature of a Lorentzian manifold, so it cannot possbly depend on coordinates or observer.

#### Ibix

The source term in Newtonian gravity is the mass. The source term in general relativity is the stress-energy tensor. For a spherical lump of matter at rest, the stress-energy tensor inside the matter is $$T_i {}^j=\left (\begin {array}{cccc} \rho c^2&0&0&0\\ 0&p&0&0\\ 0&0&p&0\\ 0&0&0&p \end {array}\right)$$where $\rho$ is the density and p the internal pressure (note that this form isn't quite the same as geometry_dude's version, even allowing for his more compact notation). This is an idealised description of a planet or star. Just increase $\rho$ and $p$ as you get closer to something that has to collapse into a black hole.

On the other hand, if you move relative to a mass, your description of the stress-energy tensor becomes:
$$T_i {}^j=\pmatrix{\left(\rho c^2-p\beta^2\right)\gamma^2&\left(\rho c^2-p\right)\beta \gamma^2&0&0\cr \left(p-\rho c^2\right)\beta\gamma^2&\left(p-\rho v^2 \right)\gamma^2&0&0\cr 0&0&p&0\cr 0&0&0&p\cr }$$(assuming I didn't mess anything up there). Note that this is what a probe moving through the star at speed v($=\beta c$) would measure. You can see that this has a completely different form from the original. There are off-diagonal elements and the second element on the leading diagonal is no longer the same as the third and fouryth elements. This is a rather different behaviour from my description of $\rho$ and $p$ just getting bigger as you get closer to something that will collapse into a black hole. This is basically because "being about to collapse into a black hole" and "moving fast" are completely different things.

Staff Emeritus
Why would you not leave that to him- or herself to decide?
Did I not ask it as a question?

That said, someone who has taken a class in GR is unlikely to pose this question, since the natural first step will be "what does this look like in its rest frame?" If that is too tough, an A-level answer is unlikely to be helpful.

#### sweet springs

Hi. A different case along your idea would be a birth of BH by collision of two bodies with high speed wrt each other. Best.

#### PAllen

Hi. A different case along your idea would be a birth of BH by collision of two bodies with high speed wrt each other. Best.
That works, because the high energy remains in the COM frame. Potentially, an exact collision is not needed.

#### Joker93

Did I not ask it as a question?

That said, someone who has taken a class in GR is unlikely to pose this question, since the natural first step will be "what does this look like in its rest frame?" If that is too tough, an A-level answer is unlikely to be helpful.
The reason that I asked the particular question is because it seems nonsensical that an observer would see a black hole being formed while another would not.
Maybe I should have said that when I originally posted the question, but I think that it is only logical that if I thought that the reasoning that I described in my question was right and didn't find anything fishy about it, I wouldn't have posted it.
Also, my level is between undergrad and grad level. I have taken a course in undergrad GR and have self-studied some things by myself. That does not mean that I won't occasionally misunderstand some things or I won't make mistakes in my reasoning.

I was thinking of it in terms of the Schwarzschild radius, but, as other answers have pointed out, that corresponds to the Schwarzschild solution.
So, I made a mistake in my assumption that the usual criteria for the formation of a black hole that have to do with the Schwarzschild solution are generalizable. That does not mean that I won't understand a nearly A-level answer.

Thank you.

#### Joker93

Well, the existence of an object cannot depend on who's looking, right? So you have a contradiction. So you know your reasoning or assumptions must be flawed.

In standard GR, you have to specify what you mean by 'energy'. My suggestion would be the energy-momentum-tensorfield $T$ of a perfect fluid. Its components are:
$$T^{ij} = \left(\rho + \frac{p}{c^2}\right) X^i \, X^j - p g^{ij }$$
$\rho$ is the density, a positive scalar field, $p$ the internal pressure, also a scalar field, and $X$ is the fluids (relativistic) velocity field. Using Einstein summation, $X$ satisfies
$$g_{ij} \, X^i X ^j = c^2 \, ,$$
which makes sure that its integral curves are parametrized by proper time. I would argue that, up to a factor of $\rho/2$, this is also (pointwise) the (relativistic) kinetic energy - a constant. Can you see why I would say that?

Do you have an idea where the flaw in your reasoning might be?
Do you mean that the kinetic energy is $$T^{00}$$?
If so, wouldn't every component of the energy-momentum tensor contribute to the formation of the black hole?

As for your question if I can see how the energy is contant, just multiply both sides by $$g_{ij}$$, right?

#### Joker93

The source term in Newtonian gravity is the mass. The source term in general relativity is the stress-energy tensor. For a spherical lump of matter at rest, the stress-energy tensor inside the matter is $$T_i {}^j=\left (\begin {array}{cccc} \rho c^2&0&0&0\\ 0&p&0&0\\ 0&0&p&0\\ 0&0&0&p \end {array}\right)$$where $\rho$ is the density and p the internal pressure (note that this form isn't quite the same as geometry_dude's version, even allowing for his more compact notation). This is an idealised description of a planet or star. Just increase $\rho$ and $p$ as you get closer to something that has to collapse into a black hole.

On the other hand, if you move relative to a mass, your description of the stress-energy tensor becomes:
$$T_i {}^j=\pmatrix{\left(\rho c^2-p\beta^2\right)\gamma^2&\left(\rho c^2-p\right)\beta \gamma^2&0&0\cr \left(p-\rho c^2\right)\beta\gamma^2&\left(p-\rho v^2 \right)\gamma^2&0&0\cr 0&0&p&0\cr 0&0&0&p\cr }$$(assuming I didn't mess anything up there). Note that this is what a probe moving through the star at speed v($=\beta c$) would measure. You can see that this has a completely different form from the original. There are off-diagonal elements and the second element on the leading diagonal is no longer the same as the third and fouryth elements. This is a rather different behaviour from my description of $\rho$ and $p$ just getting bigger as you get closer to something that will collapse into a black hole. This is basically because "being about to collapse into a black hole" and "moving fast" are completely different things.
Sorry, I am having a bit of a trouble following you. I would appreciate if you could somehow state your last paragraph a bit differently.

#### Ibix

Sorry, I am having a bit of a trouble following you. I would appreciate if you could somehow state your last paragraph a bit differently.
I'm not sure what to tell you. The stress energy tensor of matter inside a star at rest is diagonal with the first term different from the other three. The values of the terms change as the mass of the star increases. Simply moving with respect to the matter changes the stress-energy tensor, giving it off-diagonal elements and changing the relationship between the on-diagonal elements. This makes it look different from matter that's part of a star about to collapse into a black hole. A star accreting mass and about to collapse will have increasing $\rho$ and $p$ terms. That's not the effect increasing velocity has on the stress-energy tensor, so why would you expect it to behave as if it were accreting mass?

#### PeterDonis

Mentor
it seems nonsensical that an observer would see a black hole being formed while another would not.
That's right; it is nonsensical, and it doesn't happen.

I was thinking of it in terms of the Schwarzschild radius, but, as other answers have pointed out, that corresponds to the Schwarzschild solution.
Yes, but you can also, for example, look at a similar solution for an ordinary planet or star, where there is no black hole, and then see what it looks like from the viewpoint of someone flying past the planet or star at close to the speed of light. You will see that there is still no black hole, even though the object's energy is large enough that if you converted it into a Schwarzschild radius, it would be larger than the radius of the object.

I made a mistake in my assumption that the usual criteria for the formation of a black hole that have to do with the Schwarzschild solution are generalizable.
That's not the issue. The issue is that you are using the wrong criterion, even with respect to the Schwarzschild solution. More precisely, you are taking a criterion that only works in the rest frame of the object, and treating it as if it were frame-independent.

Do you mean that the kinetic energy is
$$T^{00}$$
?
$T^{00}$ is total energy (density), not just kinetic energy. It counts rest energy as well (and chemical energy, internal energy stored as vibrations of atoms, etc., etc., but we can ignore all those complications here).

#### PeterDonis

Mentor
wouldn't every component of the energy-momentum tensor contribute to the formation of the black hole?
To properly answer this question, we have to look at the actual correct criterion for the presence of a black hole, in frame-invariant terms. That criterion is that there is an event horizon, i.e., a null surface that bounds a region of spacetime that cannot send light signals out to infinity. That criterion is frame-invariant because the causal structure of spacetime--which events can send light signals to which other events--is frame-invariant. (So, for example, two observers, one at rest relative to an object and one flying past it at close to the speed of light, will both agree on whether or not there is an event horizon.)

Notice, however, that this criterion says nothing about the energy density, or indeed about any component of the stress-energy tensor. That's because the SET is local, but the criterion of there being an event horizon is not; it's global. That is, the presence or absence of an event horizon is a global property of the entire spacetime, not a local property that you can figure out by looking at the metric, curvature, stress-energy, etc. in a local region. And it turns out that there is no local property that exactly corresponds to this global one, so there is no local way of knowing for sure whether a particular object is or is not a black hole.

Another thing to consider when trying to relate the SET components to the presence or absence of a black hole is that the "density" of a black hole isn't really well-defined anyway, because the spatial "volume" of a black hole isn't well-defined. In fact, for some choices of coordinates, the spatial volume of the hole is infinite! So we shouldn't expect there to be a criterion for something being a black hole based on energy density, or more generally on the SET components.

#### Joker93

I'm not sure what to tell you. The stress energy tensor of matter inside a star at rest is diagonal with the first term different from the other three. The values of the terms change as the mass of the star increases. Simply moving with respect to the matter changes the stress-energy tensor, giving it off-diagonal elements and changing the relationship between the on-diagonal elements. This makes it look different from matter that's part of a star about to collapse into a black hole. A star accreting mass and about to collapse will have increasing $\rho$ and $p$ terms. That's not the effect increasing velocity has on the stress-energy tensor, so why would you expect it to behave as if it were accreting mass?
Yes, I understand now.
Thanks for the patience and your time.

#### PeterDonis

Mentor
there is no local property that exactly corresponds to this global one, so there is no local way of knowing for sure whether a particular object is or is not a black hole.
To somewhat counteract this rather gloomy-sounding statement, note that there is a conjectured criterion that, while not precisely "local", is still a lot better than having to know the entire global causal structure of the spacetime. This is called the "hoop conjecture", and is originally due to Kip Thorne. The conjecture goes like this: measure the mass of the object (by, for example, putting objects in orbit around it and measuring the orbital parameters); calculate the Schwarzschild radius corresponding to that mass; then imagine a hoop with that radius being passed around the object in all possible orientations (in the object's rest frame). If the hoop encloses the object in all possible orientations, then the object is a black hole; if not, not. In other words, if the object's radius in all directions in its rest frame is less than the Schwarzschild radius corresponding to its mass, then it's a black hole.

Note, though, that this criterion still doesn't look at the SET components, and it isn't frame-invariant (you have to do it in the object's rest frame). But it more or less corresponds to the intuition that an object "smaller than its Schwarzschild radius" should be a black hole. (Note that this criterion does not require that the object be spherically symmetric, or described by the Schwarzschild solution.) However, it's only a conjecture and has not been proven (although AFAIK practically all physicists believe it to be valid).

#### Joker93

To somewhat counteract this rather gloomy-sounding statement, note that there is a conjectured criterion that, while not precisely "local", is still a lot better than having to know the entire global causal structure of the spacetime. This is called the "hoop conjecture", and is originally due to Kip Thorne. The conjecture goes like this: measure the mass of the object (by, for example, putting objects in orbit around it and measuring the orbital parameters); calculate the Schwarzschild radius corresponding to that mass; then imagine a hoop with that radius being passed around the object in all possible orientations (in the object's rest frame). If the hoop encloses the object in all possible orientations, then the object is a black hole; if not, not. In other words, if the object's radius in all directions in its rest frame is less than the Schwarzschild radius corresponding to its mass, then it's a black hole.

Note, though, that this criterion still doesn't look at the SET components, and it isn't frame-invariant (you have to do it in the object's rest frame). But it more or less corresponds to the intuition that an object "smaller than its Schwarzschild radius" should be a black hole. (Note that this criterion does not require that the object be spherically symmetric, or described by the Schwarzschild solution.) However, it's only a conjecture and has not been proven (although AFAIK practically all physicists believe it to be valid).
Are there people working(researching) on finding a more rigorous criterion?

#### PeterDonis

Mentor
Are there people working(researching) on finding a more rigorous criterion?
Do you mean, working on trying to prove the hoop conjecture? That's been an open problem in GR since it was first conjectured, in the late 1960s, but nobody has been able to prove it. I don't know how much active effort there is in this area.

If you mean, trying to find some other criterion besides the hoop conjecture, I don't think anyone has come up with any promising candidates. There is another approximate criterion, the formation of a trapped surface (basically a surface at which radially outgoing light stays in the same place), but that is known to be only approximate--that is, in a general spacetime with an event horizon, the event horizon will not exactly coincide everywhere with a trapped surface. (Roughly speaking, if the black hole is gaining mass, the event horizon will be outside the trapped surface, while if it is losing mass, such as via Hawking radiation, the EH will be inside the trapped surface. But the exact relationship between them can't be known locally.)

#### Joker93

Do you mean, working on trying to prove the hoop conjecture? That's been an open problem in GR since it was first conjectured, in the late 1960s, but nobody has been able to prove it. I don't know how much active effort there is in this area.

If you mean, trying to find some other criterion besides the hoop conjecture, I don't think anyone has come up with any promising candidates. There is another approximate criterion, the formation of a trapped surface (basically a surface at which radially outgoing light stays in the same place), but that is known to be only approximate--that is, in a general spacetime with an event horizon, the event horizon will not exactly coincide everywhere with a trapped surface. (Roughly speaking, if the black hole is gaining mass, the event horizon will be outside the trapped surface, while if it is losing mass, such as via Hawking radiation, the EH will be inside the trapped surface. But the exact relationship between them can't be known locally.)
Great! Thank you.
Lastly, would you recommend me any textbook on grad-level GR?

#### PeterDonis

Mentor
would you recommend me any textbook on grad-level GR?
If you've read the classics (MTW and Wald), you've read as much as I have at that level. I know there are other more recent texts out there, but I haven't read them so I can't make any recommendations.

#### Joker93

If you've read the classics (MTW and Wald), you've read as much as I have at that level. I know there are other more recent texts out there, but I haven't read them so I can't make any recommendations.
BTW, I think that MTW is getting re-released!
In any way, thank you for your time! I appreciate it.

"On the formation of a black hole due to high kinetic energy"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving