Mass velocity at a black hole event horizon.

[Shaw]

SR implies that mass can't reach c, but mass does reach a black hole's event horizon. How is this reconciled with SR?

 

[wabbit]

One thing a massive object cannot do is stay at the horizon, which would indeed imply moving at light speed - but crossing it is no issue.

 

[PeterDonis]

SR implies that mass can't reach c, but mass does reach a black hole's event horizon. How is this reconciled with SR?

There's nothing to reconcile. A massive object crossing a black hole's horizon is not moving at c. The horizon itself is moving at c; it's an outgoing null surface.

 

[Shaw]

Thanks for the explanations. I know that GR sits on top of SR, so to speak, when it comes to gravitation, so I expected there would be a reason for it not to be taken literally.

 

[PeterDonis]

I know that GR sits on top of SR, so to speak, when it comes to gravitation, so I expected there would be a reason for it not to be taken literally.

What, exactly, do you think is "not to be taken literally"? If what you are referring to are statements like "objects crossing the horizon are moving at the speed of light", that kind of statement is just wrong; it's not a matter of not taking it literally.

 

[Shaw]

If we just consider SR, acceleration to the event horizon implies acceleration to the speed of light, but you've explained that it's the event horizon itself that's moving at the speed of light, so SR calculations don't apply here, therefore it can't be taken" literally."

 

[PeterDonis]

If we just consider SR, acceleration to the event horizon implies acceleration to the speed of light

It does no such thing. First of all, in SR there is no such thing as an "event horizon", because spacetime is flat. The best we can do in SR is to consider analogues of the black hole event horizon, which can share some of its properties, but not all.

Second, if we consider the closest analogue, a Rindler horizon, a free-falling observer who appears, to a Rindler observer, to be "accelerating towards the horizon" is not accelerating "to the speed of light"--again, the horizon is moving at the speed of light, not the inertial observer.

you've explained that it's the event horizon itself that's moving at the speed of light, so SR calculations don't apply here

Incorrect; in the SR analogue of the black hole horizon, the horizon is indeed moving at the speed of light and the inertial observer is not. See above.

 

[Shaw]

Many thanks for this expansion on your original explanations. I can't access authoritative explanations so quickly anywhere else. This is why I post on PhysicForums. I've reached a mathematically provable conclusion from all my previous posts, and I will present it shortly.

 

[Nugatory]

I've reached a mathematically provable conclusion from all my previous posts, and I will present it shortly.

Before you do, please review the Physics Forums rules about personal theories:

Physics Forums is not intended as an alternative to the usual professional venues for discussion and review of new ideas, e.g. personal contacts, conferences, and peer review before publication. If you have a new theory or idea, this is not the place to look for feedback on it or help in developing it.

 

[Shaw]

Thanks for the reminder. I'll run it past an advisor first for review. I'll only post it if he's happy with the math.

 

[nitsuj]

If we just consider SR, acceleration to the event horizon implies acceleration to the speed of light, but you've explained that it's the event horizon itself that's moving at the speed of light, so SR calculations don't apply here, therefore it can't be taken" literally."

If we consider acceleration is what an accelerometer measures and Newton's first law.

It does no such thing. First of all, in SR there is no such thing as an "event horizon", because spacetime is flat. The best we can do in SR is to consider analogues of the black hole event horizon, which can share some of its properties, but not all.

Second, if we consider the closest analogue, a Rindler horizon, a free-falling observer who appears, to a Rindler observer, to be "accelerating towards the horizon" is not accelerating "to the speed of light"--again, the horizon is moving at the speed of light, not the inertial observer.
Incorrect; in the SR analogue of the black hole horizon, the horizon is indeed moving at the speed of light and the inertial observer is not. See above.

Would an analogy be the expansion of space. For example, I'd suspect an accelerometer would measure zero for "stuff" crossing the EH (distant observer). there is no "force" acting on the object, in turn is not subject to the restrictions of acceleration approaching c. And again similar to expanding space, for outside observers is an "elsewhere" region of spacetime as it's accelerating faster than c from us. is that right-ish?

Maybe asked different, idealizing a constant (infinite/ gravitational potential and an object local to it and experiencing no acceleration, ..this is tough to word, but could the object "be" faster than light compared to observers not "materially" affected by the gravitational potential.

lol as I think it through the wording gets better. what is the max "rate" for gravity or is there even one? I suppose that is asking if the metric can flip spacetime to timespace lol (joking aside it's symmetric if no slower than light!)

 

[PeterDonis]

I'd suspect an accelerometer would measure zero for "stuff" crossing the EH

If it was in free fall, yes. But an object crossing the horizon doesn't necessarily have to be in free fall.

(distant observer)

Not sure why you put this in.

there is no "force" acting on the object, in turn is not subject to the restrictions of acceleration approaching c

What "restrictions"? "Acceleration approaching c" is relative to some particular reference frame, which in curved spacetime would have to be local anyway. There is no absolute sense in which any object is "accelerating approaching c".

idealizing a constant (infinite/ gravitational potential

The concept of gravitational potential does not make sense at or inside the horizon, so it's not helpful in trying to understand what happens when objects cross the horizon.

what is the max "rate" for gravity or is there even one?

I'm not sure what "rate for gravity" means, but I suspect the answer is "there isn't one".

I suppose that is asking if the metric can flip spacetime to timespace

I'm not sure what this means either, but I suspect the answer is "no, it can't, since the idea doesn't make sense".

 

[nitsuj]

lol my post were questions...not sure your intent here but it wasn't to answer.

But nice reply, mentor

 

[PeterDonis]

my post were questions...not sure your intent here but it wasn't to answer

My intent was to answer. If you'd rather someone else did, that's fine; but you'll get more or less the same response. If parts of your question, or the assumptions underlying it, don't make sense, I'm going to say so. But I'll refrain from responding to you further in this thread.

 

[thedaybefore]

SR implies that mass can't reach c, but mass does reach a black hole's event horizon. How is this reconciled with SR?

According to GR expert Lev Okun, the rest energy of an object/particle recduces when the object goes deeper into a gravity well. The reduction in rest mass is determined by the G00 element of the metric tensor. Since the G00 element of the metric tensor goes to zero at the event horizon, the rest energy of a paticle goes to zero at the event horizon.

See http://www.itep.ru/theor/persons/lab180/okun/em_12.pdf

Especially equation 8.

 

[PeterDonis]

the rest energy of an object/particle recduces when the object goes deeper into a gravity well

Not as measured locally; only as "measured" (using gravitational red shift) by an observer far outside the gravity well. Locally, the object's rest mass is unchanged.

Also, this reasoning applies to an object that is static in the gravitational field; but there are no such objects at or inside the horizon. So this isn't relevant to the OP's question anyway.

 

[thedaybefore]

Not as measured locally; only as "measured" (using gravitational red shift) by an observer far outside the gravity well. Locally, the object's rest mass is unchanged.

Also, this reasoning applies to an object that is static in the gravitational field; but there are no such objects at or inside the horizon. So this isn't relevant to the OP's question anyway.

You should re-read the Okun article. The reduction in rest energy is not limited to zero relative velocity situations.

As the object aproaches the event horizon, more and more of its enrgy will be shifted from internal energy to momentum/kinetic energy. Talking about what happens when an object aproaches an event horizon doesn't make sense unless you talk about it from some observer's perspective. All observations require an observer and an observed object.

 

[PeterDonis]

As the object aproaches the event horizon, more and more of its enrgy will be shifted from internal energy to momentum/kinetic energy.

Relative to a particular viewpoint, things can be interpreted this way, yes. Although even there I would question interpreting what is happening as "transferring" energy from internal (rest) energy to kinetic energy. A static observer at a given finite altitude, measuring the free-falling object passing it, will not measure its total energy to be the "reduced" amount; he will measure it to be the object's full rest energy plus the kinetic energy it has accumulated while falling. The "reduced" amount of energy is an amount calculated by a distant observer by taking into account the gravitational redshift from the finite altitude to infinity.

Talking about what happens when an object aproaches an event horizon doesn't make sense unless you talk about it from some observer's perspective.

The object itself counts as an "observer's perspective". And measurements made in the object's local inertial frame will show its rest mass to be unchanged.

 

[nitsuj]

Not sure why you put this in.

So "he" wasn't affected by gravity in any significant way

 

[thedaybefore]

Relative to a particular viewpoint, things can be interpreted this way, yes. Although even there I would question interpreting what is happening as "transferring" energy from internal (rest) energy to kinetic energy. A static observer at a given finite altitude, measuring the free-falling object passing it, will not measure its total energy to be the "reduced" amount; he will measure it to be the object's full rest energy plus the kinetic energy it has accumulated while falling. The "reduced" amount of energy is an amount calculated by a distant observer by taking into account the gravitational redshift from the finite altitude to infinity.



The object itself counts as an "observer's perspective". And measurements made in the object's local inertial frame will show its rest mass to be unchanged.

If you are measuring the object from a local inertial frame, then the object will have no velocity because the observer will be in the same frame as the object, which means there is no worry about light speed.

Unless, you mean that a local inertial frame is a co-moving frame outside the gravity well (an infintly far location with a g00 value of 1). By that definition of a local inertial frame, the rest enrgy of the object goes to zero as it aproaches the event horizon.

 

[PeterDonis]

If you are measuring the object from a local inertial frame, then the object will have no velocity

Yes, if you choose a local inertial frame in which the object is at rest. But you actually don't have to; if you choose, for example, a local inertial frame in which a static observer just outside the horizon is momentarily at rest, then the infalling object will certainly have a velocity (close to c) in this frame. And in this frame, the object's total energy will be, as I said, its full rest energy (with no "redshift factor" applied) plus all the kinetic energy it has gained in falling (i.e., the kinetic energy associated with moving at close to c).

Unless, you mean that a local inertial frame is a co-moving frame outside the gravity well (an infintly far location with a g00 value of 1).

There are local inertial frames out there, yes; but they do not contain the infalling object's worldline, because they are local. A global frame in which $g_{00} \rightarrow 1$ as $r \rightarrow \infty$ can be viewed as "inertial" at infinity, but it is not a local inertial frame; it is global, and the metric is not Minkowski everywhere (only at infinity).

By that definition of a local inertial frame, the rest enrgy of the object goes to zero as it aproaches the event horizon.

No, in a global frame such as I just described, a quantity which can (charitably) be interpreted as "rest energy" goes to zero as the object approaches the event horizon. But, as I've said, this interpretation is somewhat strained, since local measurements of the object's rest energy are unchanged (they do not go to zero).

 

[PeterDonis]

Thread closed for moderation.

 

[Drakkith]

There's been some 'behind the scenes' problems with this thread. Since the original question has been answered, the thread will stay locked.