Can You Become a Black Hole from Approaching Light Speed?

[bbbl67]

TL;DR Summary

As an object gets close to speed of light, its relativistic mass increases, and its relativistic length contracts. Could this result in turning into a black hole?

If you're in a relativistic starship, approaching the speed of light, then if you get too close to it, do you end up becoming a black hole? Relativistic length decreases as you get closer to light speed. Relativistic mass increases as you do the same. Will your relativistic mass and relativistic length conspire to make you smaller than your own event horizon? I've got two opinions about this. (1) You do not reach your own event horizon from your own frame of reference, so you don't turn into a black hole. (2) You do reach the size of your own event horizon from an outside frame of reference. Which interpretation prevails?

 

[Ibix]

No. "Relativistic mass" has nothing to do with gravity. Confusion like this is one major reason everybody stopped using "relativistic mass" decades ago - if only popsci would catch up...

 

[PeterDonis]

If you're in a relativistic starship, approaching the speed of light, then if you get too close to it, do you end up becoming a black hole?

You are moving close to the speed of light right now relative to cosmic ray particles reaching the Earth. Are you turning into a black hole?

The point being that "speed" is frame-dependent, and no actual physical invariant, like whether or not an object is a black hole, can be affected by frame-dependent quantities.

 

[PeterDonis]

(1) You do not reach your own event horizon from your own frame of reference, so you don't turn into a black hole. (2) You do reach the size of your own event horizon from an outside frame of reference. Which interpretation prevails?

Neither; both are, in Wolfgang Pauli's phrase, not even wrong.

 

[Nugatory]

Remember that right now you are moving at 99% of the speed of light relative to a cosmic ray particle zipping past the earth... are you any heavier because of that? For that matter, you are also moving at 50% of of the speed of light relative to a different cosmic ray particle, and .000005% of the speed of light relative to the jetliner flying over your head. So which speed do we use to calculate your length contraction, time dilation, and relativistic mass?
What's actually going on here is that all of these relativistic effects are not something that happens to you because you are moving, they are something observed by someone at rest when you are moving relative to them. Your movement relative to someone else can have no physical effect on you so cannot turn you into a black hole or anything else.

Another issue, specific to the black hole question, is that the entire idea of relativistic mass was abandoned some decades ago; indeed we have an Insight article explaining why. It was invented as a sort of fudge to preserve the classical $F=ma$ and physicists eventually realized that this was a mistake: the correct classical equation to preserve is $F=\frac{dp}{dt}$ (with some subtleties that I'm glossing over) and with the correct definition of $p$ (not the classical $p=mv$ which is just an approximation good at speeds much less than lightspeed) $m$ does not change.

 

[anorlunda]

You are moving close to the speed of light right now relative to cosmic ray particles reaching the Earth. Are you turning into a black hole?

I am. Inch by inch, year by year, my Schwarzschild radius increases.

 

[bbbl67]

Remember that right now you are moving at 99% of the speed of light relative to a cosmic ray particle zipping past the earth... are you any heavier because of that? For that matter, you are also moving at 50% of of the speed of light relative to a different cosmic ray particle, and .000005% of the speed of light relative to the jetliner flying over your head. So which speed do we use to calculate your length contraction, time dilation, and relativistic mass?
What's actually going on here is that all of these relativistic effects are not something that happens to you because you are moving, they are something observed by someone at rest when you are moving relative to them. Your movement relative to someone else can have no physical effect on you so cannot turn you into a black hole or anything else.

Another issue, specific to the black hole question, is that the entire idea of relativistic mass was abandoned some decades ago; indeed we have an Insight article explaining why. It was invented as a sort of fudge to preserve the classical $F=ma$ and physicists eventually realized that this was a mistake: the correct classical equation to preserve is $F=\frac{dp}{dt}$ (with some subtleties that I'm glossing over) and with the correct definition of $p$ (not the classical $p=mv$ which is just an approximation good at speeds much less than lightspeed) $m$ does not change.

Yeah, I know that relativistic mass is problematic, but it is a useful way of representing kinetic energy.

 

[weirdoguy]

but it is a useful way of representing kinetic energy.

Kinetic energy is $mc^2(\gamma-1)$. I see no need for using relativistic mass in this formula.

 

[Orodruin]

I am. Inch by inch, year by year, my Schwarzschild radius increases.

As long as your actual radius also keeps increasing to keep you from collapsing ...

 

[Orodruin]

Yeah, I know that relativistic mass is problematic, but it is a useful way of representing kinetic energy.

No. It is just another name for total energy.

See also https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[Nugatory]

Yeah, I know that relativistic mass is problematic, but it is a useful way of representing kinetic energy.

How do you figure that? The kinetic energy in terms of relativistic mass is $E_k=(m_R-m_0)c^2$, more complicated and less informative than $mc^2(\gamma-1)$ - and at the risk of stating the obvious, the kinetic energy is not $\frac{1}{2}m_Rv^2$.

Expressing the energy in terms of relativistic mass also loses the energy-momentum relationship $E^2=(mc^2)^2+(pc)^2$; this is essential to the four-vector formalism of general relativity and also the relativistic quantum field theories that reconcile quantum mechanics and relativity.

 

[ergospherical]

It's probably not what you had in mind, but there's still some interesting avenues to think about. In the rest frame of the mass the spacetime is Schwarzschild; you can boost coordinates by $v$ and take $v \rightarrow 1$ (there's a little subtlety required in taking this limit because you need to fiddle around with the coordinates to make the limit exist - see "Aichelburg-Sexl ultraboost"). The resulting metric has a delta-function containing term which spikes on the hyperplane $x=t$. The strong localised curvature constitutes a so-called pp-plane-wave pulse, which is a measurable gravitational pulse.

 

[Vanadium 50]

Yeah, I know that relativistic mass is problematic, but it is a useful way of representing kinetic energy.

No, it is not.
It is foolish to latch on to an outdated concept that is used primarily in popularizations and unused by practicing physicists.

see "Aichelburg-Sexl ultraboost"

B-level.

 

[Nugatory]

In the rest frame of the mass the spacetime is Schwarzschild; you can boost coordinates by $v$ and take $v \rightarrow 1$ (there's a little subtlety required in taking this limit because you need to fiddle around with the coordinates to make the limit exist - see "Aichelburg-Sexl ultraboost"). The resulting metric has a delta-function containing term which spikes on the hyperplane x=t. The strong localised curvature constitutes a so-called pp-plane-wave pulse, which is a measurable gravitational pulse.

Or translated into B-level speak... A star zipping past us at near lightspeed is exactly the same problem as us zipping past the star at near lightspeed. We don't expect anything unusual to happen to us in the first case, so the same has to be true for the second case as well; the basic premise in the initial post about "an object gets close to speed of light" is mistaken.

This mistake is more fundamental than the one about relativistic mass.

 

[ergospherical]

no comprendo señor, I just wanted to say that a really really fast moving particle can in fact produce gravitational effects, to whit a gravitational plane wave pulse along the (arbitrarily) $x$ direction (see my posting #12)

have a mad friday! love u

 

[PeterDonis]

I just wanted to say that a really really fast moving particle can in fact produce gravitational effects

The spacetime geometry produced by the particle is invariant; it does not depend on how fast you are moving relative to it.

a gravitational plane wave pulse along the (arbitrarily) $x$ direction (see my posting #12)

A better (and closer to B-level) way to say it would be that this is what the invariant spacetime geometry produced by the source "looks like" if you and it have a relative speed along the $x$ direction that is close to the speed of light. It does not mean that, just by moving relative to you at close to the speed of light, the source magically acquires gravitational effects that it would not have otherwise.

 

[ergospherical]

yeah man that's so true, right on sir
there's a close analogy to electromagnetism, I can elucidate if OP would like (what's your background mate?)

 

[hmmm27]

It's probably not what you had in mind, but there's still some interesting avenues to think about. In the rest frame of the mass the spacetime is Schwarzschild; you can boost coordinates by $v$ and take $v \rightarrow 1$ (there's a little subtlety required in taking this limit because you need to fiddle around with the coordinates to make the limit exist - see "Aichelburg-Sexl ultraboost"). The resulting metric has a delta-function containing term which spikes on the hyperplane $x=t$. The strong localised curvature constitutes a so-called pp-plane-wave pulse, which is a measurable gravitational pulse.

(realizing I may regret asking this, but...)
That any relation to frame-dragging ? (a gravitational or probably "simply" spacetime phenomenon)

 

[Ibix]

(realizing I may regret asking this, but...)
That any relation to frame-dragging ?

It's a rigorous version of the only-in-some-senses-accurate claim that the gravitational field of a massive body will length contract in its direction of motion. In the limit as its speed relative to you approaches $c$ it becomes an impulsive change in your course.

(a gravitational or probably "simply" spacetime phenomenon)

Not sure what distinction you are trying to make here.

 

[ergospherical]

Not really, no. Frame-dragging crops up in various places where things are rotating, e.g. in Kerr particles with zero angular momentum $l = u \cdot (\partial / \partial \phi)$ exhibit non-zero angular velocities $d\phi / dt$.

 

[hmmm27]

Not sure what distinction you are trying to make here.

the possibility of gravitons distinct from spacetime ; just leaving some options open.

Not really, no. Frame-dragging crops up in various places where things are rotating, e.g. in Kerr particles with zero angular momentum $l = u \cdot (\partial / \partial \phi)$ exhibit non-zero angular velocities $d\phi / dt$.

What about two massy objects on hyperbolic "orbits" to each other: there is a relative rotation as they approach pass and depart.

 

[ergospherical]

That's the 2-body problem, and it's quite complicated relativistically. You approximate the solutions with the so-called $k$PN post-Newtonian approximations, where you keep $k$ post-Newtonian terms in the expansion of the metric. I've come across some work to this effect regarding binary star systems, but not the situation you're describing. (In any case, not particularly any relation to what is known as frame dragging.)

 

[PeterDonis]

any relation to frame-dragging ?

No. As @ergospherical says, frame dragging is a feature of spacetimes in which there are rotating sources. It has nothing to do with the relative velocity of the source and a test object.

 

[PeterDonis]

the possibility of gravitons distinct from spacetime

I'm not sure what this means, but whatever it is, it's personal speculation.

just leaving some options open.

Personal speculations are off limits here. Please take note.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

The OP's question has been answered, and the thread will remain closed. Thanks to everyone who responded.