Do black holes determine time's arrow?

[PeterDonis]

The time-reversal of someone falling into a black hole and yelling "Help, I'm falling into a black hole!" is not someone rising from a white hole yelling "Wow! A white hole, and I'm rising out of it!". The time reversal of the first scenario is someone yelling "Help, I'm falling into a black hole!" (speaking backwards).

On reading this over again, it occurs to me that we might be talking past each other because we are using the phrase "time reversal" to mean different things.

In the quote above, you are using "time reversal" to mean reversing the sign of the time coordinate. (You also showed this explicitly in an earlier post but the penny didn't drop for me then, unfortunately.) You were correct earlier when you said this cannot change the actual physical situation, since of course no coordinate transformation can do that.

However, throughout this discussion, I have been using "time reversal" to mean reversing the sign of proper time along the congruence of worldlines describing observers comoving with the matter. This is a coordinate-free operation that does change the actual physical situation (for the reasons I have given in previous posts), but of course it is not the same as a coordinate transformation and it can be described without any coordinate change at all.

It does happen that the effect of both of the above operations on the relative signs of the time coordinate and proper time along the congruence of worldlines is the same (both operations flip that relative sign), but they have that effect for different reasons. That means we shouldn't be using the same term for both operations. Unfortunately, I don't have a good alternate term for either one.

 

[PeterDonis]

Is this argument essentially about whether those two diagrams represent distinct physical models, and/or exactly what changes are needed to make them become distinct or not distinct?

It might be, but after realizing what I described in my previous post just now (that @stevendaryl and I have been using the phrase "time reversal" to mean different things), I'm not sure.

In addition to the issue I mentioned in that post just now (that I don't know of a good alternate term for either operation I described, reversing the time coordinate or reversing the proper time along worldlines), there is also the issue that, in the literature, it is not clear whether, for example, the labeling of the infinities, $i^+$, $i^-$, $I^+$, $I^-$, is supposed to be relative to the labeling of the time coordinate or the labeling of proper time along worldlines. As you have drawn the diagrams, you are reversing the proper time along the red worldline (since the events labeled 1, 2, and 3 come in reverse order), which is how I would do it, and you have relabeled the infinities relative to that. That makes your second diagram a distinct physical situation, since the observer whose worldline is the red worldline has the singularity in his past instead of in his future. But as @stevendaryl has described his understanding of "time reversal", the ordering of events along the red worldline would not change (so your second diagram would not be describing a "time reversal" as he has been using the term); all that would change is the sign of the time coordinate, so the events 1, 2, 3 along the red worldline would occur at a decreasing series of time coordinates instead of an increasing series of them. (I'm not sure whether @stevendaryl would relabel the infinities as you have done as part of such a transformation.)

Another issue in this discussion, I think, has been whether what I have just described (the order in which the observer experiences events along his worldline) requires thermodynamics. The diagrams you have drawn don't, at least IMO; proper time along a worldline is a geometric parameter and does not require any thermodynamics in the model. (In the idealized situations we have been discussing, including your diagrams, there is no meaningful thermodynamics since the microstate of the entire universe is known everywhere. But of course that is a physically unrealistic idealization. Part of the issue in this discussion might be what the impact of that physical unrealism actually is or ought to be.)

 

[stevendaryl]

However, throughout this discussion, I have been using "time reversal" to mean reversing the sign of proper time along the congruence of worldlines describing observers comoving with the matter. This is a coordinate-free operation that does change the actual physical situation (for the reasons I have given in previous posts), but of course it is not the same as a coordinate transformation and it can be described without any coordinate change at all.

My claim is that choosing a sign for proper time is as much of a modeling choice as choosing a coordinate system.

The definition of proper time along a parametrized path $\mathcal{P}(s)$ is:

$\tau = \int \sqrt{g_{\mu \nu} (\frac{d \mathcal{P}}{ds})^\mu (\frac{d \mathcal{P}}{ds}})^\nu ds$

The choice of the parameter $s$ is a modeling choice; you can choose any smooth, monotonically increasing parameter you want. The choice of $s$ does not affect the magnitude of $\tau$. But the sign of $\tau$ is doubly ambiguous. First, a square-root has two possible values, so that is source of ambiguity. But even if we adopt the convention that we always take the positive square root of any positive number, the choice of $s$ affects the sign of $\tau$. Choosing $\lambda = - s$ will flip the sign of $\tau$.

If you have already made a choice for an arrow of time everywhere along the path, then you can choose $s$ so that it increases towards the future.

So I don't see how going from talking about reversing the time coordinate to talking about reversing $\tau$ makes the issues any different. The sign of $t$ and the sign of the path parameter $s$ are both modeling choices, it seems to me.

 

[stevendaryl]

So I don't see how going from talking about reversing the time coordinate to talking about reversing $\tau$ makes the issues any different. The sign of $t$ and the sign of the path parameter $s$ are both modeling choices, it seems to me.

The intuition about proper time for a path is that it is the time shown on an idealized clock traveling that path (or is related to it through a linear transformation). But because the laws of physics are time-symmetric, for any clock whose elapsed time increases along a path, there exists an initial condition for that clock so that its elapsed time decreases along the same path. "The time that clocks show" does not uniquely determine the sign of proper time, unless you assume that all clocks agree on an arrow of time.

 

[stevendaryl]

No, it works against the point you are making. You are basically claiming that I can't define the proper time parameter along an observer's worldline without introducing thermodynamics. The paper, correctly, recognizes, in the quote I gave, that that is not the case; proper time is a geometric parameter and can be included in a geometric model without any thermodynamics at all.

There are two different issues that perhaps are getting mixed up. You can certainly define "the future" to be the timelike direction away from the Big Bang, or in the case of a black hole, you can define "the future" to be toward the central singularity. But once you do that, a Big Bang and a Big Crunch are not different cosmologies. A black hole and a white hole are not different solutions. If you use geometry to determine the arrow of time, then the future is by definition the direction in which the universe is colder and less dense, and the future is the direction in which the singularity lies in a freefalling worldline. Using geometry implies that there is only one solution, not two. The author makes that point explicitly.

Alternatively, you can use thermodynamics to determine the direction of time. This is actually a more relevant and a richer definition, because it allows you to distinguish between a black hole and a white hole. A black hole is one in which the thermodynamic arrow of time points towards the singularity, and a white hole is one in which the thermodynamic arrow of time points away from it. The benefit to me of this approach is that it makes it an empirical question whether you are falling into a black hole or rising from a white hole. It's an empirical question whether we live in a Big Bang cosmology or a Big Crunch cosmology.

The third option is just to say that what you mean by a "solution of GR" is a solution to the field equations, plus an assignment of future/past direction to each timelike path. My complaint about that approach is just that unless the thermodynamic arrow of time agrees with the assignment, it's a physically meaningless choice. Observations and experiments can't determine such an arrow of time.

 

[PeterDonis]

My claim is that choosing a sign for proper time is as much of a modeling choice as choosing a coordinate system.

If by "modeling choice" you mean that we choose the model to match the actual physical behavior of a clock following the worldline, yes, I agree. I'm not sure that's what you mean, but if it's not, then I disagree. The whole point of the "modeling choice" you refer to is to make the proper time along the worldline, as a mathematical parameter in the model, match the clock time along the worldline, as a physical observable. You can't reverse the sign of the parameter in the math and keep the physical interpretation the same.

because the laws of physics are time-symmetric, for any clock whose elapsed time increases along a path, there exists an initial condition for that clock so that its elapsed time decreases along the same path.

And the way you model this changed initial condition is to change the sign of the proper time, the mathematical parameter in the model, to match the behavior of the physical clock. You can't change the clock's physical behavior and keep the mathematical parameter in the model the same; then the model no longer matches the physics.

"The time that clocks show" does not uniquely determine the sign of proper time,

Yes, it does, because that's the rule of physical interpretation for models in GR. I'm not going to budge from this position unless you can show me references from GR textbooks that say that the rule of physical interpretation of proper time is something different from what I've said.

There are two different issues that perhaps are getting mixed up.

No, that's not the problem. The problem is that you are not adopting the rule of physical interpretation that I have described. Plus you are making up your own rule of interpretation that comes from nowhere, as far as I can see:

If you use geometry to determine the arrow of time, then the future is by definition the direction in which the universe is colder and less dense

Nonsense. The Oppenheimer-Snyder model of a star collapsing to a black hole has the matter more dense towards the future. So does a collapsing FRW universe (which is what the matter region of the O-S model is a portion of). You are basically claiming that gravitational collapse cannot occur. That makes no sense to me.

Using geometry implies that there is only one solution, not two. The author makes that point explicitly.

I've already stated repeatedly that I don't buy this point in the paper you referenced, and explained why. Not to mention that, as I've already said, it's a paper stating a proposal by the author; the portion you describe is certainly not stating the standard viewpoint of physicists in GR, it's stating the author's opinions.

The third option is just to say that what you mean by a "solution of GR" is a solution to the field equations, plus an assignment of future/past direction to each timelike path. My complaint about that approach is just that unless the thermodynamic arrow of time agrees with the assignment, it's a physically meaningless choice.

In other words, you refuse to accept any idealized model that doesn't have any thermodynamics--which includes all of the idealized models we have discussed in this thread. I can't stop you from having this opinion, but I don't think it reflects any standard viewpoint among relativity physicists. It certainly doesn't reflect my viewpoint.

To state my viewpoint briefly: we have idealized models in which we model things like the increasing readings on clocks and the experience of observers of time flowing from past to future, using proper time along timelike worldlines. The relationship between proper time along timelike worldlines and other geometric parameters (such as where singularities are or the density and temperature of matter) is part of the physical interpretation of the model. Again, I'm not making this viewpoint up; I'm taking it from the GR textbooks I've read.

If we want to make a more complicated model that includes thermodynamics, then of course I agree that the thermodynamic arrow of time in the more complicated model must agree with the geometric arrow of time defined by proper time along timelike worldlines. So, for example, if we want our model to obey the second law of thermodynamics, we can't have proper time increasing along timelike worldlines towards a singularity that has zero entropy, like the singularity in the idealized Big Bang model (without inflation), which has zero Weyl curvature, or a hot, dense state with very low entropy, like the "Big Bang" state in more realistic models (where we might have an inflation epoch prior to the hot, dense "Big Bang" state, and where that state is not exactly uniform but has density and temperature fluctuations that will later on cause lead to gravitational clumping). This implies, for example, that a black hole singularity in a more realistic model that included thermodynamics would not be the idealized spherically symmetric one in the idealized Schwarzschild model, but something more like a BKL singularity, which has increasingly chaotic fluctuations of curvature and hence increasing entropy towards the singularity.

I don't have a problem with any of this. I just have a problem with claiming that idealized models that don't include thermodynamics either have no physical interpretation at all, or have a physical interpretation that can be adjusted at will.

 

[Demystifier]

Does black holes determines time's arrow?

No, it's the other way around. The statistical time arrow (of which the thermodynamic time arrow is a special case) determines the black hole.

Otherwise how to explain with time reversed objects been pushed out from black holes?

That's indeed explained by the time arrow, but it's important to understand that the statistical time arrow is a more general principle, while black hole is just one of its manifestations/consequences.

 

[Aanta]

No.
The idea that a black hole would become start to spew out matter under time reversal is a very odd one.
Since gravitation is invariant under time reversal.

Will all events be the same under time reversal then?
No alternative current will flip so that transformer coils will have opposite magnetic polarity.
But they will still work fine of course.

 

[PeterDonis]

The idea that a black hole would become start to spew out matter under time reversal is a very odd one.
Since gravitation is invariant under time reversal.

"Gravitation" is not the same as "black hole spacetime". The time reverse of an object made of matter that collapses to a black hole is a white hole expanding into an object made of matter. That means that in the time reversed solution, the matter does come out of the white hole.

What is invariant under time reversal is, in Newtonian terms, the "acceleration due to gravity". In the black hole case, the matter accelerates as it falls down into the black hole. In the white hole case, the matter decelerates as it comes upwards out of the white hole. In both cases the "acceleration due to gravity" points towards the hole.

 

[Aanta]

Hello PeterDonis
I have a hunch the OP had a quite good reason for asking the question here.
Lets flip the time arrow according to the question.

In the reversed time, the black hole event horizon is still supposed to exist.
But nothing can pass out from inside an event horizon.
According to your reasoning matter is supposedly now able to do so, but that is not the case.
Since the black hole would then by necessity not be a black hole with an event horizon.

I have a hunch this is why the OP asked about this matter.

But gravity IS invariant when the time arrow point the other direction.

If gravity suddenly would be repulsive, all objects would start to rip apart.
Would also the electroweak interaction have opposite polarity and matter would flow apart and make a diffuse mist of atoms? And the W and Z bosons rip even individual nuclei to shreds?
The answer is still no.

I will provide a most simple example.
The filament in a light bulb does not have the properties that make it possible for it to work as a solar cell, even less is it able to suck up photons from an increasingly darkening room while it convert those to electricity. So it will of course not do so if the time arrow is flipped.

 

[stevendaryl]

If you actually study the trajectory of a free-falling particle near an eternal black hole, you will see that it is completely time-symmetric. What's true is that the radial component of the 4-velocity of a particle cannot change sign while inside the event horizon. That means that if a particle falls through the event horizon (so the radial component of its 4-velocity is negative, meaning that the Schwarzschild coordinate r is decreasing with proper time), then it can never turn around inside the event horizon to have a positive radial velocity.

In order for a particle to rise out of a black hole (having a positive radial velocity), it must have ALWAYS had a positive radial velocity. This means it must have come into existence inside the event horizon, arising out of the central singularity.

A realistic black hole is not eternal. In the far past, there was no black hole, there was a cloud of gas that collapsed into a black hole. So the black hole grows by particles falling through the event horizon, and so all the particles inside the event horizon are in the first category, they can only have a negative radial velocity when inside the event horizon.

So ultimately, the time asymmetry of black holes---the fact that things fall in, but don't come out---follows from a time asymmetry in our assumed initial conditions. We assume that the black holes were formed from collapsing balls of gas. The time-reversed initial condition would assume that the black holes were always there in the far past, and the particles inside the black hole all had positive radial velocity. That's the white hole solution to Einstein's equations.

 

[PeterDonis]

In the reversed time, the black hole event horizon is still supposed to exist.

No, the time reverse of a black hole is a white hole, which has no event horizon. It has a horizon, but it's a past horizon, not a future horizon. See below.

But nothing can pass out from inside an event horizon.

A white hole horizon, as above, is not an event horizon. Things can come out of a white hole horizon just fine; but they can't go into a white hole horizon.

According to your reasoning matter is supposedly now able to do so, but that is not the case.

Yes, it is. See above.

gravity IS invariant when the time arrow point the other direction.

The "acceleration due to gravity" points in the same direction, yes; gravity is attractive in the time reversed solution as well as the original solution. But the fact that gravity's direction is the same does not mean everything is the same.

If gravity suddenly would be repulsive

Nobody is claiming that it is. You are attacking a straw man.

 

[PeterDonis]

In order for a particle to rise out of a black hole (having a positive radial velocity), it must have ALWAYS had a positive radial velocity. This means it must have come into existence inside the event horizon, arising out of the central singularity.

This is impossible inside an actual black hole. But in the time reverse of a black hole, namely a white hole, this is indeed what happens: the particle comes from the white hole singularity (which is not the same as the black hole singularity) and emerges from the white hole horizon (which, per my post #117, is not an event horizon since it is a past horizon, not a future horizon).

In the maximally extended Kruskal-Szekeres geometry, both a black hole and a white hole are present, and they are not the same; they are different, distinct regions of the spacetime. That solution itself is fully time symmetric.

However, in the more realistic case of a black hole that forms by gravitational collapse of a massive object, that solution is time asymmetric, as you say, and its time reverse is another time asymmetric solution, a white hole that expands into a massive object.

Note that I am using "time reverse" here in the way I explained in earlier discussion in this thread.

 

[PeterDonis]

If you actually study the trajectory of a free-falling particle near an eternal black hole, you will see that it is completely time-symmetric.

Note that the only possible meaning for "eternal black hole" here is actually the maximally extended Kruskal-Szekeres geometry; no other "black hole" is eternal. And, as I noted in post #118 just now, in that geometry, there are two "hole" regions, not one: there is a black hole region and a white hole region, and they are not the same. So while it is possible to have a fully time symmetric trajectory for a free-falling particle in this geometry, any such trajectory will start on the white hole singularity, emerge from the white hole horizon, rise to some maximum altitude in the exterior region, fall back inside the black hole horizon, and end on the black hole singularity. It will not fall back into the same region of spacetime from which it emerged.

 

[Dale]

If gravity suddenly would be repulsive, all objects would start to rip apart.

As already mentioned, time reversed gravity is still attractive, but a time reversed black hole horizon is a white hole horizon.

 

[stevendaryl]

Note that the only possible meaning for "eternal black hole" here is actually the maximally extended Kruskal-Szekeres geometry; no other "black hole" is eternal. And, as I noted in post #118 just now, in that geometry, there are two "hole" regions, not one: there is a black hole region and a white hole region, and they are not the same. So while it is possible to have a fully time symmetric trajectory for a free-falling particle in this geometry, any such trajectory will start on the white hole singularity, emerge from the white hole horizon, rise to some maximum altitude in the exterior region, fall back inside the black hole horizon, and end on the black hole singularity. It will not fall back into the same region of spacetime from which it emerged.

Well, it’s always impossible to fall back to the same region of spacetime that you came from. Time will be different if nothing else.

 

[stevendaryl]

This is impossible inside an actual black hole.

If we take the Schwarzschild geometry as the background geometry, and consider test particles moving in that background, then there are solutions of the equations of motion where the particle is “falling” as a function of proper time and solutions where the particle is “rising” as a function of proper time. In the region with r less than the Schwarzschild radius, the sign of dr/ds can never change.

 

[PeterDonis]

If we take the Schwarzschild geometry as the background geometry, and consider test particles moving in that background, then there are solutions of the equations of motion where the particle is “falling” as a function of proper time and solutions where the particle is “rising” as a function of proper time.

Yes. The former solutions are inside the black hole region, and the latter solutions are inside the white hole region.

In the region with r less than the Schwarzschild radius, the sign of dr/ds can never change.

That's correct, but there are two such regions, one for each possible sign of dr/ds.

 

[PeterDonis]

it’s always impossible to fall back to the same region of spacetime that you came from.

By "region of spacetime" I meant one of the four regions in the maximally extended Schwarzschild geometry: the "right" exterior, the black hole, the "left" exterior, and the white hole.

 

[Dale]

The former solutions are inside the black hole region, and the latter solutions are inside the white hole region.

And in the region outside the horizons. Both types of solutions exist there.