Could High speeds create a black hole?

ilikescience94

As you guys probably know, as an objects speed increases, so does its mass. I was wondering if a black hole could be induced by speeding up a neutron star or some other massive object with high density to high speeds? Also, what is the equation that explains this phenomena? Also, if this is true, then because a black hole has the highest entropy of any object of that volume(or surface area really) does velocity increase entropy, and would a black hole with the same radius and spin as another black hole, but a higher velocity have more entropy than a similar black hole moving at a slower velocity?

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phinds

Gold Member
You seem to have some serious misunderstandings here. Speed is relative, not absolute, and since the actual mass of an object, in its own reference frame, does not change just because it happens to be traveling fast in some other reference frame, your scenario does not make sense.

For example, you, right now as you are reading this, are traveling at almost the speed of light relative to an accelerated particle in the LHC. Do you feel any more massive? Do you think you are about to turn into a black hole?

ShayanJ

Gold Member
I agree with phinds' post and have nothing to add to it.
But I now see that SR is so confusing for newcomers when it reaches to mass increase.I once was asked about this and,when answering,realized that its in fact better to interpret the mass increase formula only to mean the increase of kinetic energy which is already relative in Newtonian mechanics.I think it is OK to hear different ideas on this.But because I don't want to hijack the thread,let's wait till the OP says they're OK with the answer.

Quantum Immortal

From what i understand, the source of the gravitational field is not really energy.
Energy is ill defined in relativity.
That explains that you can't create a black hole by simply going fast enough.

Mentor

ilikescience94

Ok, thanks guys, glad you could clear this all up for me, what is the equation that I would use to explain how much more massive an object appears then?

pervect

Staff Emeritus
Ok, thanks guys, glad you could clear this all up for me, what is the equation that I would use to explain how much more massive an object appears then?
I would suggest Olson, D.W.; Guarino, R. C. (1985). "Measuring the active gravitational mass of a moving object". http://dx.doi.org/10.1119/1.14280

If a heavy object with rest mass M moves past you with a velocity comparable to the speed of light, you will be attracted gravitationally towards its path as though it had an increased mass. If the relativistic increase in active gravitational mass is measured by the transverse (and longitudinal) velocities which such a moving mass induces in test particles initially at rest near its path, then we find, with this definition, that M rel=γ(1+β2)M. Therefore, in the ultrarelativistic limit, the active gravitational mass of a moving body, measured in this way, is not γM but is approximately 2γM.
Note that "active gravitational mass" more or less implies that the author has a Newtonian viewpoint. This is most likely a good thing, as the average reader usually has a Newtonian viewpoint as well.

The advantage of the formulation in this paper, "active gravitational mass", is that it is readily formulated and understood from a quasi-Newtonian viewpoint without a deep knowledge of GR. Furthermore the notion of velocities induced by a relativistic flyby is well defined and easy to describe. One should, however, take note that there is a lot more to say about the topic of mass in GR, this paper is more of a starting point than an ending point.

If you find and read the entire paper you'll see that GR has multiple definitions of mass, which the paper also discusses. The existence of multiple definitions should serve as a warning that the topic of mass in General Relativity is not trivial. The wiki article http://en.wikipedia.org/wiki/Mass_in_general_relativity has some more info on this as well.

PAllen

To add to this discussion, the way energy and pressure contribute to curvature (and thus any tendency to collapse into a black hole) is via the stress energy tensor, and this does not involve simply adding up terms as if computing a scalar sum. A specific example is as follows:

Consider two baseballs with initial state as comoving, with initial separation of .1 meter orthogonal to their direction of motion, with speed almost c (in some coordinates, or relative to some reference body). They would be predicted to slowly converge, but the result would be two comoving baseballs with no separation but no collapse or high pressure.

Consider two baseballs as above, but moving near c towards each other, initially aimed such that at low relative speeds, they would bypass each other with .1 meter closest approach. As speed goes to c, they will not bypass each other. Not only will they self capture, but for a relative speed insanely close to c they would be predicted to form a BH (at least if strong forms of the hoop conjecture are true).

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ilikescience94

OK, so I manipulated the e=mc^2+pc eqaution to find that m=√(E^2-(pc)^2)/c^2, would this tell me what the relativistic mass is? I substituted e for mc^2+pc, and plugged 5kg in for mass, and .95c for v, and came out with an m value of 8.51kg, would this be right, or am I thinking of this equation in the wrong sense? If I'm right, would p be determined by p=√(e^2-(mc^2)^2)/c?

PAllen

OK, so I manipulated the e=mc^2+pc eqaution to find that m=√(E^2-(pc)^2)/c^2, would this tell me what the relativistic mass is?
This tells you the invariant mass. For a single body, it is the rest mass. For a system, it is the mass as measured in the COM frame.
I substituted e for mc^2+pc, and plugged 5kg in for mass, and .95c for v, and came out with an m value of 8.51kg, would this be right, or am I thinking of this equation in the wrong sense?
No, if you solve for m and make no mistakes, you get 5 kg back. For E you use mγc^2, for p you use mγv. If you plug those in, you find that this equation gives back the rest mass you started with.
If I'm right, would p be determined by p=√(e^2-(mc^2)^2)/c?
Yes, you can do t his.

ilikescience94

This tells you the invariant mass. For a single body, it is the rest mass. For a system, it is the mass as measured in the COM frame.

No, if you solve for m and make no mistakes, you get 5 kg back. For E you use mγc^2, for p you use mγv. If you plug those in, you find that this equation gives back the rest mass you started with.

.
Ok, so what would mγv be then?

Also can the changing of mass then not be found using the e=mc^2+pc equation?

pervect

Staff Emeritus
OK, so I manipulated the e=mc^2+pc eqaution to find that m=√(E^2-(pc)^2)/c^2, would this tell me what the relativistic mass is?
No, that gives you the invariant mass.
I substituted e for mc^2+pc, and plugged 5kg in for mass, and .95c for v, and came out with an m value of 8.51kg, would this be right, or am I thinking of this equation in the wrong sense? If I'm right, would p be determined by p=√(e^2-(mc^2)^2)/c?
When you get a value for "mass", how do you intend to apply it? I'm concerned that you might just substitute it into Newtonian equations (like F=ma, or F=gmM/r^2) which would most likely mislead you.

Do you have some specic problem in mind that you're trying to solve, or are you just learning about "mass"?

PAllen

Ok, so what would mγv be then?

Also can the changing of mass then not be found using the e=mc^2+pc equation?
In all these formulas, m is rest mass, period. They don't use or define 'relativistic mass'. In modern practice, there is no changing of mass with speed; there is changing of energy.

ilikescience94

In all these formulas, m is rest mass, period. They don't use or define 'relativistic mass'. In modern practice, there is no changing of mass with speed; there is changing of energy.
Ok cool, and then energy changes as momentum changes damn I love relativity haha this stuff is so mind blowing, I just found the equation for relativistic mass m=(m0/√(1-v^2/c^2)) this is a lot like the time dilation equation except doesn't t0 equals relativistic time and t=rest time, whereas m= relativistic mass, and m0=rest mass, or is there something I'm missing?

PAllen

Ok cool, and then energy changes as momentum changes damn I love relativity haha this stuff is so mind blowing, I just found the equation for relativistic mass m=(m0/√(1-v^2/c^2)) this is a lot like the time dilation equation except doesn't t0 equals relativistic time and t=rest time, whereas m= relativistic mass, and m0=rest mass, or is there something I'm missing?
so we don't need to repeat everything here.

Briefly, relativistic mass is an older term that has fallen out of favor due to causing more confusion than utility (did you know, that when it was in use, you had to define the concept of longitudinal versus transverse relativistic mass? all of this goes away with newer terminology). Note, physics didn't change, just abandoning confusing names for particular expressions that appeared in certain formulas.

In particular, in all the formulas you quoted, m is simply 'mass' which means rest mass in the newer terminology. If you were going to write, for example, the E^2 = m^2 c^4 + p^2 c^2 formula using the terminology of m and m0 you might write:

m^2 c^4 = m0^2 c^4 + p^2 c^2

This is a correct equation using the old definition of m versus m0; but nobody writes it or uses it. When you see it just stated with m, the m means rest mass, which is the only kind used nowadays in SR.

ilikescience94

so we don't need to repeat everything here.

Briefly, relativistic mass is an older term that has fallen out of favor due to causing more confusion than utility (did you know, that when it was in use, you had to define the concept of longitudinal versus transverse relativistic mass? all of this goes away with newer terminology). Note, physics didn't change, just abandoning confusing names for particular expressions that appeared in certain formulas.

In particular, in all the formulas you quoted, m is simply 'mass' which means rest mass in the newer terminology. If you were going to write, for example, the E^2 = m^2 c^4 + p^2 c^2 formula using the terminology of m and m0 you might write:

m^2 c^4 = m0^2 c^4 + p^2 c^2

This is a correct equation using the old definition of m versus m0; but nobody writes it or uses it. When you see it just stated with m, the m means rest mass, which is the only kind used nowadays in SR.
That's been very helpful, thanks, but I still don't know what γ equals.

ShayanJ

Gold Member
$\gamma=\Large{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}$

ilikescience94

$\gamma=\Large{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}$
Oh duh haha gamma of course, not my finest hour, thanks

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