Material Behavior at the Event Horizon

[MattRob]

In classical mechanics, to raise from some height $h_{0}$ to infinity over a gravitational body, takes a certain amount of energy, the energy associated with escape velocity, let's just call it $ε$.

$ε = \lim_{t\rightarrow +\infty} \int_{h_{0}}^t ƒ(h)dh$

Likewise, it's time-reversible, so dropping something from stationary at an infinite distance, then when it reaches $h_{0}$, because of potential energy becoming kinetic, it will have that same amount of energy in kinetic.

So, if the energy required to escape from the event horizon is infinite, then what keeps something falling into a black hole from achieving an infinite amount of energy as it reaches the event horizon, thus contributing an infinite amount of mass to the black hole?

I guess time-reversal doesn't really apply in the same way here, since a time-reversal would mean reversing the direction of gravity as well (since its a curve in spacetime), creating a white hole. But I'm still wondering how it can be that something can require an infinite amount of energy to escape from a certain $h_{0}$, yet not achieve an infinite energy when dropping from higher up down to $h_{0}$. So different equations would be used to describe something falling as opposed to something attempting to rise out of the gravity well?

 

[Orodruin]

You are using energy arguments, which do not apply to GR. Sufficient to say, the event horizon is characterised by light not being able to increase the $r$ coordinate.

 

[PeterDonis]

if the energy required to escape from the event horizon is infinite

That's not correct. The correct statement is that "the energy required to escape from the event horizon" is not well-defined, because the spacetime is not static at (or inside) the event horizon, and the "energy" you are talking about is only defined in the region where the spacetime is static.

what keeps something falling into a black hole from achieving an infinite amount of energy as it reaches the event horizon, thus contributing an infinite amount of mass to the black hole?

The mass that the object falling in from rest at infinity adds to the hole is not determined by the kinetic energy gained as it falls. It is determined by the object's original rest mass, when it was sitting at infinity. (This assumes that the object free-falls through the horizon.) The reason for this is similar to the reason why an object going very, very fast does not turn into a black hole: kinetic energy is frame-dependent, but whether or not an object is a black hole, or how much mass an infalling object adds to a black hole, is not frame-dependent; it's an invariant, the same for all observers.

 

[bcrowell]

You are using energy arguments, which do not apply to GR.

I guess this is a funny case where the correctness of your statement depends on the punctuation. With the comma, it sounds like you're making a general statement that energy arguments don't apply to GR. That would be wrong. A test particle in a static spacetime does have a conserved energy. But without the comma, maybe you mean that the particular energy arguments used by the OP are wrong in GR, and then I would agree with you.

 

[bcrowell]

In classical mechanics, to raise from some height $h_{0}$ to infinity over a gravitational body, takes a certain amount of energy [...]

$ε = \lim_{t\rightarrow +\infty} \int_{h_{0}}^t ƒ(h)dh$

[...]So, if the energy required to escape from the event horizon is infinite

This description makes it sound as though a finite amount of energy would be enough to make the particle come out through the event horizon and rise to some height, but not to infinity. Actually, you can't get the particle to come out through the event horizon at all.

Likewise, it's time-reversible

What you've really stated is a paradox involving time-reversal, not energy. The energy part isn't needed in order to create the paradox, which is fundamentally just this: if we have trajectories for particles falling into a black hole, then by time-reversal, why don't we have trajectories for particles coming out?

If you construct the maximal extension of the Schwarzschild spacetime, you get a spacetime that is time-reversal symmetric, and in which there is a white hole as well as a black hole. There are indeed outgoing trajectories for test particles, but they are emerging from the white hole.

In a black hole that forms by gravitational collapse, the spacetime doesn't have this symmetry. This total lack of symmetry isn't so obvious when you just look at the expression for the Schwarzschild metric, which applies after the black hole has settled down. It's very obvious, though, if you look at the Penrose diagram. Keep in mind that inside the event horizon, it's the Schwarzschild r coordinate that's timelike, not t, so the transformation $t\rightarrow -t$ isn't a time-reversal.

 

[Orodruin]

I guess this is a funny case where the correctness of your statement depends on the punctuation. With the comma, it sounds like you're making a general statement that energy arguments don't apply to GR. That would be wrong. A test particle in a static spacetime does have a conserved energy. But without the comma, maybe you mean that the particular energy arguments used by the OP are wrong in GR, and then I would agree with you.

I would agree with removing the comma, yes. Thank you for clarifying.

 

[MattRob]

That's not correct. The correct statement is that "the energy required to escape from the event horizon" is not well-defined, because the spacetime is not static at (or inside) the event horizon, and the "energy" you are talking about is only defined in the region where the spacetime is static.
The mass that the object falling in from rest at infinity adds to the hole is not determined by the kinetic energy gained as it falls. It is determined by the object's original rest mass, when it was sitting at infinity. (This assumes that the object free-falls through the horizon.) The reason for this is similar to the reason why an object going very, very fast does not turn into a black hole: kinetic energy is frame-dependent, but whether or not an object is a black hole, or how much mass an infalling object adds to a black hole, is not frame-dependent; it's an invariant, the same for all observers.

Okay; so I've been reading the book recommended in this thread: https://www.physicsforums.com/threads/looking-for-books-on-gr.821003/#post-5154506

And I came across this; I think this is what you were referring to?
$m^{2} = E^{2} - ρ^{2}$,
where $E$ and $ρ$ will differ for different observers, but in such a way that $m$ remains constant? Wouldn't that kind of make this equation a metric, too, then?

That's really fascinating, then. I can see how they derived this, but it's still somewhat baffling that you'd subtract the momentum from the energy like that to get mass. Fascinating!

So mass really is a constant, then. Doesn't that mean that whole thing about "mass increase as your approach the speed of light" is kind of a misconception, then?

I've seen that explanation used before - that if you watch a rocket (accelerating with a constant proper acceleration) accelerate indefinitely (ignoring fuel constraints for simplicity), then its approach to $c$ will be asymptotic because its mass increases by a factor of $γ = \frac{1}{1-β^{2}}$ as it gains velocity relative to the coordinate frame; thus will never reach $c$ in the coordinate frame (ie; mass in coordinate frame = $m_{0}γ$, with $m_{0}$ being rest mass).

But, while this would yield the same result for coordinate position as a function of coordinate time(?), this is a misconception that's inconsistent with other physics - that a more accurate approach to take is to describe the rocket as experiencing a constant acceleration in its proper frame, but because its "proper clock" appears to tick more and more slowly to the coordinate frame, its acceleration likewise decreases in the coordinate frame. When $\frac{d\tau}{dt} = 1$, then $\frac{dv}{dt} = a_{0}$ (where $a_{0}$ is its proper acceleration), but as it increases in velocity, $\frac{d\tau}{dt} = i > 1$, $\frac{dv}{dt} = \frac{a_{0}}{i}$, and as $v→c$, $\frac{d\tau}{dt} → ∞$, and thus $\frac{dv}{dt} → \frac{a_{0}}{∝} = 0$. So in other words, as it approaches $c$, its acceleration in the coordinate frame approaches zero in such a way that its velocity approaches $c$ asymptotically, not because of an increase in mass, but due to time dilation slowing its acceleration.

Thus, the idea of an object gaining mass due to relativistic velocities is simply incorrect - an artifact of time dilation. Is this understanding correct?

But then what of mass-energy equivalence; where is that kinetic energy's "mass"?

 

[Orodruin]

So mass really is a constant, then. Doesn't that mean that whole thing about "mass increase as your approach the speed of light" is kind of a misconception, then?

I suggest you have a look at our relativity FAQ section: https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-it-is-not-used-much.796527/

 

[PeterDonis]

I think this is what you were referring to?

$m^{2} = E^{2} - ρ^{2}$

where $E$ and $ρ$ will differ for different observers, but in such a way that mm remains constant? Wouldn't that kind of make this equation a metric, too, then?

No. It's just the relativistic energy-momentum relation. It has nothing to do with the metric; it's the same in any spacetime.

a more accurate approach to take is to describe the rocket as experiencing a constant acceleration in its proper frame

This is equivalent to saying it has constant proper acceleration, yes. But this has nothing to do with an object freely falling into a black hole; such an object has zero proper acceleration. There isn't any useful analogy between these two cases.

 

[MattRob]

No. It's just the relativistic energy-momentum relation. It has nothing to do with the metric; it's the same in any spacetime.
This is equivalent to saying it has constant proper acceleration, yes. But this has nothing to do with an object freely falling into a black hole; such an object has zero proper acceleration. There isn't any useful analogy between these two cases.

It's only tangentially related to something freely falling into a black hole, since as you said, something freely falling into a black hole experiences no proper acceleration.

The point, though, is that the asymptotic nature of accelerating to $c$ can be explained without saying that objects increase in mass as they accelerate in any particular frame of reference; this effect can instead be attributed to time dilation.

The relation to the original question, is that mass is not frame-dependent.

EDIT:

Which statement I find somewhat confusing, though, due to mass-energy equivalence and that energy is frame dependent.

I suggest you have a look at our relativity FAQ section: https://www.physicsforums.com/threads/what-is-relativistic-mass-and-why-it-is-not-used-much.796527/

I'm still taking this in. But in the course of it, I found this: https://www.physicsforums.com/threads/what-is-the-massenergy-equivalence.763067/ that cleared that up. I'll continue to mull these over.

 

[PeterDonis]

the asymptotic nature of accelerating to $c$ can be explained without saying that objects increase in mass as they accelerate in any particular frame of reference; this effect can instead be attributed to time dilation.

Neither of these is an explanation; they are just different ways of looking at the side effects of proper acceleration. The explanation is the causal structure of spacetime: timelike vectors and null vectors are fundamentally different things, and an object's 4-momentum vector can only be of one type or the other; it can't change type. Proper acceleration changes the "direction in spacetime" of an object's 4-momentum, but it can't change a timelike 4-momentum into a null 4-momentum, which is what "accelerating to c" non-asymptotically would require.

The relation to the original question, is that mass is not frame-dependent.

This is just another way of saying that the norm of an object's 4-momentum (as with the norm of any 4-vector) is not frame-dependent. But that's not the same as saying the norm can't change from timelike to null. That's a different statement.