The Arrow of Time: The Laws of Physics and the Concept of Time Reversal

[A-wal]

The example was supposed to illustrate that there is no objective 'mapping' of times in different coordinate systems. Only in nearly flat times you can use 'time dilation', which is like a low order correction. In BH you can'tuse that notion because it fails (you get infinities).

I've noticed. What does that tell you?

But if you draw light cones everything is simple and logical.

But they go past 90 degrees after the horizon, meaning anything on the inside is traveling back in time relative to anything outside.

In GR, speeds (for distant objects) faster than c are well know. Just an example - areas behind the cosmological horizon.

I don’t see how it can ever be possible to literally reach c relative to any object, no matter how distant.

That thread is quite long, please let me know: do you deny GR or do you deny some consequences of GR?

I'm not saying GR isn't true. I'm saying that I don't think it's the whole truth. I think the equivalence principle can be taken a step further, gravity is relative, and the radius of an event horizon changes depending on how close you get to it.

Sure, you can do it at the horizon. Take a freely falling observer and pick the event on that observer's worldline where he is just crossing the horizon. Make that event the origin of the local freely falling reference frame in which the observer's "acceleration" towards the black hole disappears. Call the (local) coordinates of that frame X and T; X increases in the outgoing radial direction, and T increases in the future time direction. Then the line X = T in that frame is the event horizon--more precisely, it's the little piece of the event horizon that lies within the range of the local freely falling frame.

I said AT the horizon. How can a bit of it lie within range if you're already there? The horizon is equivalent to traveling at c so the observer should experience zero proper time at the horizon and negative/minus/anti proper time inside the horizon. What does that mean? Nothing. It can’t happen, just liking crossing an event horizon.

Which it isn't.

If you compare the time dilation/length contraction from tidal acceleration and the time dilation/length contraction from acceleration in flat space-time don’t they look the same?

The properties of spacetime don't change depending on direction of travel. Outgoing and ingoing observers both see the same curvature of spacetime--meaning tidal gravity. They also both experience the same "acceleration due to gravity", inward towards the hole. The very fact that that "acceleration" is inward for all observers, regardless of their direction of travel, is *why* outgoing signals can't escape from any point on or inside the horizon, while ingoing signals can pass inward.

Why would having an inward velocity to start with make any difference? You’re treating the pull of gravity as absolute motion rather than relative.

No, "acceleration" is the rate of change of *velocity*, *not* the rate of change of acceleration, which is what "tidal force" is.

Using the river model again; you don’t feel the movement of the water because you’re in it. Tidal force is the acceleration of the water relative to the river bed?

No, they aren't. There is no "force of gravity". A body that is moving solely under the influence of "gravity" feels no force at all; it's weightless, in free fall.

Free-fall? We were talking about to different sized objects pushing against each other under (accelerating) under the influence of the "force of gravity" which you now seem to be insisting doesn't even exist anymore. You think when objects are in free-fall there is in fact no force of gravity despite the fact that they're obviously being pulled towards each other, which can even be felt slightly.

Once again, there is no "pull of gravity". It's simply that spacetime around the Earth is such that all the freely falling worldlines move inward, towards the Earth's center. You're correct that what prevents an object on the Earth's surface from following such a worldline is the fact that the object and the Earth are solid bodies; the Earth therefore pushes up on the object and keeps it from falling freely towards the center, so the object feels weight. The object also pushes down on the Earth--more precisely, it pushes down on the piece of the Earth's surface directly underneath it. But there's more Earth underneath that piece pushing back on it, and under that, and so on, so the Earth can't move inward in response to the object's push; its surface stays the same.

There is no pull of gravity? The fact that freely falling world-lines move inward towards the centre of the Earth proves that there is a force at work. Without gravity they wouldn’t feel their weight. If you explain how a force works it doesn't stop being a force. You could claim the other forces don't actually exist if they were defined well enough. Stop treating gravity as if it's special. It isn't. At least it's not until you show me why rather than how it's different. Unless you can tell me something that shows how they're different in practice. The event horizon of a black hole doesn't count as practice, it's still theory.

You're wrong. The rock will never be crushed under its own weight; it and the Moon can sit there in equilibrium indefinitely.

Again, you're wrong. You're basically claiming that every object will eventually collapse into a black hole, regardless of its initial state. That's wrong, and has been known to be wrong since the 1950's, when John Wheeler and some students of his studied the possible end states of matter. Kip Thorne talks about it in Black Holes and Time Warps.

That would mean the energy tied up in matter becomes infinite if it has an infinite lifespan. I’m not really at all even close to anything resembling sure about this. It seems to me that objects have to do work just to remain solid objects and to produce gravity, and the energy that allows this obviously can’t last forever. I'm sure I heard somewhere that matter doesn't have an infinite lifespan?

No, I listed four different manifestations of one cause, the stress-energy tensor. "Matter" and "energy" are just different units for expressing the stress-energy tensor; "matter-energy" vs. "pressure" and "stress" are just different components of the tensor, and how the tensor breaks up into components like that is frame-dependent; different observers in different states of motion will break up the tensor into "energy" vs. "pressure" or "stress" in different ways, but they will find the same physical laws, which depend only on the tensor as a whole, a single geometric object. It's all one cause.

But you said matter curves space-time completely differently to energy and the two processes were distinct and not equivalent? But you also said “In standard GR, curvature created by matter *is* indistinguishable from curvature generated by energy; in fact, "matter" and "energy" are really the same thing, just measured in different units, and the speed of light squared is just a conversion factor between the different units”.

No, you wouldn't. A freely falling object is always weightless, regardless of how curved spacetime is (how strong "gravity" is). This is one of the most basic ideas in general relativity; if you don't understand that, then it's no wonder you're having trouble with the rest of it.

I meant through tidal force. Maybe I shouldn't have said weight. You feel more tidal force the more you weigh though, so it's sort of your weight. I don’t think I’m not the one having trouble.

Only if you mean "curved" in the frame-invariant sense I gave. In that sense, the worldline of an object "hovering" at a constant radius over a black hole is curved. So is your worldline when you are standing motionless on the surface of the Earth. If you agree with both those statements, then we're OK.

Yes, and also your world-line is curved when you expend energy to accelerate because it's the same thing.

I'm not insisting on any such thing.

Yes you are! You keep claiming that you're not suggesting these things then go on to describe them anyway. If you're not even sure what you think then I'm not surprised you keep misinterpreting my words.

No, it isn't. Again, this (the difference between curvature of spacetime itself and curvature of a path in spacetime) is one of the most basic concepts in relativity. If you don't understand that, it's no wonder you're having trouble.

The main thing I'm having trouble with is getting through to you. It's like having a conversation with a God worshipper. They've already made their minds up and use backwards logic from there to explain away any inconsistency that anyone raises. The only difference is you've got slightly more to work with. There is no difference between saying that space-time is curved and saying that objects paths through space-time are curved. If every object were affected by a uniform force then everything would be accelerated depending on their distance from the sources, and you could just as easily use this say that space-time is curved. If you can't get that then I'm not sure how much help you can be to me to be honest. Still, I appreciate the effort you're making.

Not according to general relativity. In GR, spacetime is a dynamical entity of physics, right alongside matter-energy; its dynamics are contained in the Einstein Field Equation. That equation also includes matter-energy, so spacetime and matter-energy can affect each other.

If there's no matter then there can be no space-time to separate them, and without space-time there can be nowhere for the matter to exist in the first place. A non-informal change in the amount of space between matter is called acceleration. You can use energy to do this, and matter does it as well.

And I keep on telling you that you've got this backwards, and you keep on saying it anyway. The event horizon marks the point where nothing *outgoing* can catch up with *accelerating* objects outside the horizon. It does not mean that nothing from outside can catch an object free-falling inward.

And I keep on telling you that it doesn't matter. The point is that some coordinate systems show that you can reach the horizon while others that show the entire external space-time don't.

Have you not been reading all the previous posts where I explained this? Objects can experience "tidal force" in free fall. Acceleration in flat spacetime (or in curved spacetime, for that matter) means that an object is not in free fall. They're different concepts.

Yes, objects can experience acceleration in free-fall. It’s called tidal force. The fact that you're referring to them using different words does not make them different concepts. I haven't read a single word from you that suggests that there's any real difference.

 

[A-wal]

No. Acceleration, in the invariant sense (an object is accelerating if it feels weight), has nothing to do with the orientation of the light cones, either in flat or in curved spacetime.

I thought someone said you could do the same thing with acceleration in flat space-time. Why wouldn’t you be able to? I couldn’t find where that was mentioned, but I did find these:

Ingoing light rays behave differently from outgoing ones because they are ingoing, not outgoing. There's no physical requirement that the two *must* behave the same. In flat spacetime it so happens that light rays going in any direction behave the same, but that's because the spacetime is flat, not because of any physical requirement that light rays behave the same regardless of the direction they're traveling, in any spacetime whatsoever.

You just said there that the same-point in space-time behaves differently depending on the direction of the observer. Seems very wrong.

So basically, at the event of final evaporation of the hole, a flash is emitted that contains images of all the free-falling observers that crossed the horizon, just as they were crossing it. (Unfortunately, the free-falling observers themselves have already fallen into the singularity, since that is still to the future of any timelike worldline that crosses the horizon.) That single flash therefore gets assigned a single coordinate time in the "approximately Schwarzschild" coordinate system used by a distant observer, which will be the coordinate time of the final evaporation *and* of all the "crossing events" whose images are contained in the flash.

How could the free-falling observers have already fallen in? I don’t understand where this separation of an object and its light comes from. The objects can cross before the light from them? They fall in when they’re seen to fall in (we’re ignoring Doppler shift). If all the crossing events share a single coordinate time then that’s when they cross, or would do if it wasn’t too already too late. If they reach the singularity at the same time then they reach the event horizon at the same time because the black hole and the singularity are the same thing when you get close. Length contraction and time dilation make it appear to have a non-zero size and lifespan.

Read what I said again, carefully. I said radially *outgoing* light paths are bent to vertical at the horizon. I didn't say anything about *ingoing* light paths being vertical. They are still ingoing, and there is still "room" between the ingoing and outgoing radial light paths for ingoing timelike worldlines.

Let’s pretend for a moment that ingoing light paths are also vertical. What would happen then?

Wrong. The "angle" of the light cones at a given event is an invariant geometric feature of the spacetime. As such, it's the same for all observers, regardless of their state of motion.

That's what I thought. It wasn't a question. So all I need to do is show that they can't tilt all the way to 90 degrees because it's relative and it would be the same as reaching c.

You can't get out (from inside the horizon) at all, no matter how you move. You can be accelerated instead of in free fall and still get in; just accelerate inward, or accelerate outward but not enough to "hover".

I know what the standard description says. I was making the point that it doesn't make sense to say it doesn't have to move to approach the horizon but has to move to get away. You’re viewing the river as absolute motion again aren’t you.

No, you can't. They're different things. The "curvature" you're talking about using energy to "replicate" is curvature of a path--an object follows a curved path if it feels weight. The "curvature" of a geodesic path is curvature of spacetime itself; an object following a geodesic path feels no weight, and the *path* itself is as "straight" as it can be, in the spacetime it's in.

No they're not. They're the same. The "curvature" I'm talking about using energy to "replicate" is curvature of a path through flat space-time or a straight line through curved space-time. There's absolutely no distinction I far as I can see.

No, it isn't. In GR, the "direction of gravity" (meaning the direction of the effective "force" an object feels) also depends on the motion of the object. For example, a planet in an elliptical orbit does not feel a force that points directly towards the Sun. That's why the perihelion of the orbit precesses; this has been measured for the planet Mercury and the measurement agrees with the GR prediction. In all the discussion in this thread, I've been talking only about purely radial motion, where this issue doesn't arise.

Yea okay, but that's not really the issue. All I meant was that gravity/velocity is relative so all that's left is tidal force/acceleration. I think an elliptical orbit as being a stable out of balance orbit. Gravity starts to win, but this increases the relative velocity, so the object escapes, briefly. And so on.

Then what is the "obvious" way in a relativity forum?

Having an edge/event horizon that's relative and not in a fixed position.

None of these claims are true. The multiple coordinate systems are consistent with each other, there is no acceleration past c (objects always move within the light cones), the arrow of time is the same throughout the spacetime, and the "value of gravity" does not depend on your direction of travel (the curvature of spacetime is the same for all observers whatever their state of motion). This has been explained multiple times in this thread. If all you can do is keep repeating these false claims, there's not much point in continuing to discuss them, so I won't respond unless you actually can offer some substantive arguments. I'll only respond to correct outright false statements, or when I'm not sure I'm understanding something correctly.

Some coordinates show that the horizon can't be crossed while others show that it can. You can keep saying they cover different areas or are compatible if you understand the maths, but the fact is that they remain incompatible because they make contradictory predictions.

The light cone properties depend on whether you're inside or out, or the whatever equivalent you want to use. There’s acceleration past c as soon as the event horizon is crossed.

The arrow of time is broken because in your version an object would have to escape from within the event horizon if the arrow were reversed.

Gravity does have different strengths depending on your direction in your version because if time dilation went up to infinity at the horizon from the outside then you wouldn’t be able to reach the horizon and if it didn’t reach infinity inside then you would be able to escape. You could say they are different areas of space-time but if you were to move to one Plank length away from the horizon then the energy required to move away would jump up to infinity in an instant, and time dilation and length contraction either go to infinity at the horizon itself or they don’t. It can’t depend on your direction.

No object ever moves outside the light cones. That's all that "can't travel faster than light" means.

That seems backwards. I think it makes more sense say that no object can move faster than light. That's all “can't move outside the light cones" means.

Length is a quantity in the time dimension?

You know exactly what I meant. I still have no idea why length in one dimension would be any different than the others, or how it could possibly be infinite.

Wrong; that is not a necessary feature of a singularity. It happens to be true of the Big Bang singularity in the spacetimes used in cosmology, but it is *not* true of the singularity at the center of a black hole. That singularity is a spacelike line.

What? A space-like line? Maybe you're referring to the fact that it lives for a certain amount of time, so it's life-span make it a one dimensional line? But it doesn't exist for any amount of time at the horizon, in just the same way as it doesn't cover any amount space at the horizon. It just looks that way from a distance.

No.

Oh!

Because you don't understand the mathematics and logic that underlie the translation. It's perfectly straightforward and consistent. The description of Kruskal coordinates on the Wikipedia page shows how the translation is done. If you want more detail, all the major relativity textbooks describe the translation.

I know you can't move smoothly to infinity. I don't care what mathematical tricks you try to use.

Um, because the event of the crossing of the horizon is not in the spacetime external to the black hole, but another part of spacetime that those coordinates don't cover?

The Schwarzschild coordinates cover the entire space-time external to the black hole but you can't even reach the horizon because that represents an impossible situation, in the same way that a graph would show the energy required to accelerate increasing to infinity at c. Space-time is dynamic and it would change relative to you as you got closer, in exactly the same way that space-time appears to change as you accelerate in flat space-time.

If it's not consistent, then it's not just me that doesn't understand. It's the entire relativity physics community, that has been using the theory to make correct predictions for almost a century now (starting from Einstein in 1915).

Why wouldn't it still make correct predictions even if I was right?

You're describing the "river model" pretty well for the portion of spacetime outside the horizon; but adding the "river bed" to the model is gratuitous unless you can come up with a physical reason why it's necessary. The model predicts all the physics outside the horizon perfectly well without including the river bed at all, so Occam's razor says eliminate the river bed unless you can give a physical reason (some physical prediction that the model makes differently) why it's needed. And no, the "prediction" that nothing can cross the horizon is *not* a good physical reason, because that's precisely the point at issue. You have to find a reason *outside* the horizon why you need the river bed to make correct predictions. Otherwise you can just eliminate it and have the standard river model, which deals perfectly well with objects crossing the horizon and going inside.

You’re allowing objects to move with no limit other than relative to the river. Neither the river or anything in it can move at c relative to the riverbed, which is what would have to happen for an object to reach the event horizon. The speed of the river relative to the riverbed reaches c at the horizon. I've heard loads on how you could stretch the physical laws to breaking point to come up with a very shaky, highly suspect and over elaborate model. What I haven't heard is a single good reason why you need to do that. Why is gravity treated as though it's somehow special? You don't even need the riverbed for this. Why would anything in the river be able to move at c relative to the singularity, which is exactly what would have to happen for it to reach the horizon? What's up with these light cones? Inside the horizon they can tilt past 45 degrees despite the fact that this means they'll be moving faster than c and back in time relative to any objects on the outside? There is no space-time inside the horizon. That's why there's an event horizon in the first place. The whole concept of an object crossing an event horizon doesn't make any more sense than accelerating up to c in flat space-time. In flat space-time the event horizon is c. You're trying to use the acceleration of a black hole to bring the event horizon to you, but that can’t work anymore than trying to use energy to accelerate to c. You can accelerate towards it but never ever reach it. You would move into more length contracted/time dilated space-time as you accelerate through tidal force, which would stretch out the space-time from the perspective of your previous frame keeping the horizon out of reach. You use point-like objects to eliminate tidal force then carry on looking at it as curved space-time rather than as a force which blatantly doesn't work because tidal force replaces acceleration towards it. The energy required to pull an object from beyond the horizon is infinite, meaning the force pulling them inwards has to have infinite strength, no matter how you try to justify it. You use contradictory coordinate systems and claim it's okay because some only cover certain areas. What is that supposed to mean? What happened to those other areas from this perspective? Presumably they don't exist. Having it both ways is the main beauty of relativity. You can look at it in various ways and they're all right. You keep mixing and matching but you're treating them as though they're diferent.

 

[PeterDonis]

A-wal: Most of your post just re-states your position without making any new arguments, so I don't see much point in responding. I'll just focus on particular items where there may still be some useful clarification to be done.

I thought someone said you could do the same thing with acceleration in flat space-time. Why wouldn't you be able to? I couldn't find where that was mentioned

Probably because nobody has ever said that. Certainly I haven't, and I doubt anyone else in this thread has either (except you).

You just said there that the same-point in space-time behaves differently depending on the direction of the observer.

No, I said that observers moving on different worldlines (ingoing vs. outgoing) that both pass through the same point in spacetime may behave differently. The different behavior is due to the different worldlines the observers follow.

I don't understand where this separation of an object and its light comes from. The objects can cross before the light from them?

I'm not sure what you are trying to say here. If you mean that you think I'm saying that an ingoing free-falling object somehow reaches the horizon before ingoing light that it emits when it's outside the horizon, I'm not saying that (and it's false--ingoing light emitted by the object will, of course, reach the horizon before the object does).

Let's pretend for a moment that ingoing light paths are also vertical. What would happen then?

You might as well say let's pretend that two plus two equals five, what would happen then? If you think you can construct a consistent model where the ingoing and outgoing light paths are both vertical at the horizon, by all means post a description of your model. Otherwise I can't answer this question because it presupposes a situation which is mathematically inconsistent.

You're viewing the river as absolute motion again aren't you.

You're the one who claims there is an absolute "river bed", not me.

No they're not. They're the same. The "curvature" I'm talking about using energy to "replicate" is curvature of a path through flat space-time or a straight line through curved space-time. There's absolutely no distinction I far as I can see.

You don't see a distinction between a curved path through a flat spacetime and a straight line through a curved spacetime? No wonder you're having trouble. One path is curved and the other is straight. How can there possibly be no distinction between them?

All I meant was that gravity/velocity is relative so all that's left is tidal force/acceleration.

But gravity is not velocity, nor usefully analogous to it in any way that I can see, so this whole argument breaks down.

Having an edge/event horizon that's relative and not in a fixed position.

Meaning what, exactly?

The arrow of time is broken because in your version an object would have to escape from within the event horizon if the arrow were reversed.

The "time reverse" of a black hole is a "white hole", and it's also a perfectly valid mathematical solution to the Einstein Field Equation. In a white hole, yes, you're correct, objects can escape from inside the horizon to outside, but can't go from outside to inside. The white hole horizon is an *ingoing* null surface instead of an outgoing one, because of the time reversal.

Physically, I don't think anyone believes that white holes are realistic because there's no way for one to be formed. But mathematically, they're perfectly consistent. Remember that "time reversal" doesn't mean everything has to look identical when time's reversed; it only means that the physical laws have to be the same when time is reversed. Since the black hole and its time reverse (the white hole) are both valid solutions of the Einstein Field Equation (which is the relevant physical law), this condition is met.

I still have no idea why length in one dimension would be any different than the others, or how it could possibly be infinite.

I assume, then, that you believe the universe is closed? Otherwise its "length in future time" would be infinite (because it would keep on expanding forever).

If you do believe the universe is closed, then a real, physical black hole does not have an infinite "length in time", because it will cease to exist when the universe recollapses to a final singularity. But that will not prevent objects from falling through the black hole's horizon before that happens.

What? A space-like line? Maybe you're referring to the fact that it lives for a certain amount of time, so it's life-span make it a one dimensional line?

No, I mean exactly what I say: the r = 0 singularity of a black hole is a one-dimensional line, infinitely long, in a spacelike direction. Inside the horizon, the Schwarzschild r and t coordinates "switch roles"; r becomes timelike (with decreasing r being the "future" direction of time) and t becomes spacelike. The r = 0 singularity goes from t = minus infinity to t = plus infinity in these coordinates (the Schwarzschild "interior" coordinates, which are not the same as the Schwarzschild "exterior" coordinates you've been using), so it's a line at one point of "time" (r = 0) that extends through all of "space" (the full range of t).

This also means that, since the decreasing r direction is the future time direction inside the horizon, the singularity at r = 0 is to the future of all events inside the horizon. That's why the singularity is unavoidable (and why ideas like "speed relative to the singularity" make no sense--see below).

I know you can't move smoothly to infinity. I don't care what mathematical tricks you try to use.

Then I guess you aren't familiar with the mathematical theory of limits and calculus, that puts things like "move smoothly to infinity" on a consistent, rigorous mathematical footing. If you consider things like calculus to be "mathematical tricks", then I'm not sure why you're bothering to even have a discussion in this forum, since calculus is a fundamental part of GR, and we tend to assume here that it's, well, valid.

Why wouldn't it still make correct predictions even if I was right?

Because you can't change just one thing. You can't just wave your hands and say, "my theory is exactly like GR and makes all the same predictions, except that nothing can ever reach a black hole's horizon". The predictions of GR are derived from an underlying model, and you can't change any single prediction without changing the entire model, meaning you will change lots of other predictions as well, including ones that have already been experimentally verified to high precision. The only way you could be right and still have all the other predictions come out the same would be if there were a consistent model that *only* differed from standard GR in that one single prediction, not anywhere else. If you think you have such a model, by all means post it--but you have to post the *model* itself, the whole logical structure, not just your prediction from it.

Neither the river or anything in it can move at c relative to the riverbed

Why not? That's the question I was asking before, and you haven't answered it. The "river bed" is not a physical thing; it's just a conceptual crutch that can be used to visualize what's going on. So there's no physical reason why the speed of anything relative to the riverbed needs to be limited.

What I haven't heard is a single good reason why you need to do that.

Um, to make correct predictions? See my comment on making correct predictions above.

You don't even need the riverbed for this. Why would anything in the river be able to move at c relative to the singularity, which is exactly what would have to happen for it to reach the horizon?

This is false; an object does not need to move at c "relative to the singularity" to reach the horizon, nor does it need to move faster than light "relative to the singularity" once it is inside the horizon. The idea of "speed relative to the singularity" doesn't even make sense; the singularity is in the future, not at a different place (see my comments above about the singularity being a spacelike line), so "speed relative to the singularity" makes no more sense than "speed relative to next Tuesday".

What's up with these light cones? Inside the horizon they can tilt past 45 degrees despite the fact that this means they'll be moving faster than c and back in time relative to any objects on the outside?

False. The light cones don't move at all. They are geometric features of spacetime. And, as I've already noted, nothing needs to move back in time inside the horizon.

You're trying to use the acceleration of a black hole to bring the event horizon to you

Also false. As I've said before, if you're freely falling, you don't have to do *anything* to reach and pass through the horizon, any more than you have to do anything to hit the surface of the Earth if you freely fall from a high altitude.

 

[DrGreg]

I know you can't move smoothly to infinity. I don't care what mathematical tricks you try to use.

Consider flat space time with standard Minkowski coordinates (t, x, y, z). Now consider a second coordinate system (T, X, Y, Z) defined by

T = \frac{t}{v-x}
X = x
Y = y
Z = z​

Consider a particle moving at constant velocity in Minkowski coordinates x = vt. As t increases from 0 to 1, x(=X) increases from 0 to v and T increases from 0 to ∞.

Now, according to your argument, as it takes an infinite amount of T-time to reach x=v, the particle never gets any further. Yet it's pretty obvious this is wrong when you use t-time instead, and there's nothing to stop the particle getting past x=v.

Moral: if you use the "wrong" coordinates, you can be misled to the wrong conclusion. Schwarzschild coordinates are the "wrong" coordinates for examining what happens at the event horizon.

 

[A-wal]

Probably because nobody has ever said that. Certainly I haven't, and I doubt anyone else in this thread has either (except you).

Probably. I thought was something like "in general relativity we use light cones which translates to special relativity just as well", or words to that effect. Did you edit it out?

No, I said that observers moving on different worldlines (ingoing vs. outgoing) that both pass through the same point in spacetime may behave differently. The different behavior is due to the different worldlines the observers follow.

You've lost me. If you want talking about the space-time they occupy then what? Behave differently how?

I'm not sure what you are trying to say here. If you mean that you think I'm saying that an ingoing free-falling object somehow reaches the horizon before ingoing light that it emits when it's outside the horizon, I'm not saying that (and it's false--ingoing light emitted by the object will, of course, reach the horizon before the object does).

No. I mean that saying an object has crossed the horizon but its light is still visible doesn't make sense when at any time the object could turn round and come back. If the light can't reach the horizon then the object can't.

You might as well say let's pretend that two plus two equals five, what would happen then? If you think you can construct a consistent model where the ingoing and outgoing light paths are both vertical at the horizon, by all means post a description of your model. Otherwise I can't answer this question because it presupposes a situation which is mathematically inconsistent.

Now you know how I feel every time I pretend you're version of it makes sense.

You're the one who claims there is an absolute "river bed", not me.

Yes then. If you refuse to even try to see it my way then of course it's not going to make any sense to you. And I'm not claiming that. I said river, not riverbed. The riverbed is just there to make the metaphor easier. It represents the frame of something at rest relative to the singularity and the objects starting frame in the example I used. If you move at c relative to the riverbed then you move at c relative to an object that stayed in our starting position. The river and anything in it move in the same way as anything else, because that's the only way you can do it. At the horizon the riverbed and anything in it would reach c relative to anything outside, but you can't move at c.

You don't see a distinction between a curved path through a flat spacetime and a straight line through a curved spacetime? No wonder you're having trouble. One path is curved and the other is straight. How can there possibly be no distinction between them?

You sound like someone who's just been told the concept of relative frames. Because whether you look at movement as objects moving through an unchangeable medium or as immoveable objects in a dynamic medium makes absolutely no difference. If two objects accelerate towards each other then the space between them has decreased. Was it the objects moving through flat space-time or did the space-time between them curve? Same thing! It's easier to view energy as creating real movement through space-time that's curved by gravity because normally matter lasts and energy doesn't.

But gravity is not velocity, nor usefully analogous to it in any way that I can see, so this whole argument breaks down.

This whole argument breaks down because you don't get the metaphor? Velocity is relative right? So there's no difference between constant velocity and being at rest right, so if you were going really, REALLY fast it would seem to you as though you're not moving at all. Gravity is relative right? So there's no difference between being in a higher gravitational field and a lower one from the perspective of a free-faller. You can feel acceleration because that isn't relative, well actually it is but not in quite the same way. You can feel acceleration from using energy or from moving into a stronger part of the gravitational field as you accelerate relative to anything far enough away to ignore the gravitational field. They are the same until you show me something, anything that even suggests there's any difference other than their direction or strength.

Meaning what, exactly?

Meaning that the event horizon in flat space-time is c. When there's gravity it means that you get time dilation and length contraction without relative movement. When gravity is strong enough to overpower the forces that hold up matter it creates the situation where a four-dimensional bubble is created over a certain volume of space-time where the amount of length contraction and time dilation go beyond infinity within that volume, in the same way that traveling at c would create infinite length contraction and time dilation. The edge of this volume of space-time is called an event horizon, which is strangely appropriate because no event can possibly happen inside the horizon. It means that you can't reach the horizon even if you accelerate towards it because you'd always run out of time as you travel through ever more and more length contracted space.

The "time reverse" of a black hole is a "white hole", and it's also a perfectly valid mathematical solution to the Einstein Field Equation. In a white hole, yes, you're correct, objects can escape from inside the horizon to outside, but can't go from outside to inside. The white hole horizon is an *ingoing* null surface instead of an outgoing one, because of the time reversal.

Physically, I don't think anyone believes that white holes are realistic because there's no way for one to be formed. But mathematically, they're perfectly consistent. Remember that "time reversal" doesn't mean everything has to look identical when time's reversed; it only means that the physical laws have to be the same when time is reversed. Since the black hole and its time reverse (the white hole) are both valid solutions of the Einstein Field Equation (which is the relevant physical law), this condition is met.

That's cheating! You call that a valid solution? You're going to have to do a lot better then that. Oh it just goes backward. Gravity doesn't go backwards! It doesn't care about the arrow of time. If it did then it would push up if the arrow were reversed and everything in the past would have been up in the air and it wasn't as far as I can remember. And you say I'm the one being hand-wavy. If they can't form when time is moving one way then they can't form when time is moving the other way either because the laws remain the same, so it's not a valid solution. Did you not read the beginning of this thread? What is the point of doing something that sneaky and underhanded if you don't even bloody notice? This is just like changing coordinate systems when one doesn't suit you or claiming objects reach c at the horizon, but not really. Stop cheating!

I assume, then, that you believe the universe is closed? Otherwise its "length in future time" would be infinite (because it would keep on expanding forever).

If you do believe the universe is closed, then a real, physical black hole does not have an infinite "length in time", because it will cease to exist when the universe recollapses to a final singularity. But that will not prevent objects from falling through the black hole's horizon before that happens.

I do believe it's closed. I don't believe it will "re"collapse. I don't think black holes last forever. I think that a black hole is an area that can't be reached because gravity has time-dilated and length contracted everything up to the horizon beyond the point where you could ever reach it in time.

No, I mean exactly what I say: the r = 0 singularity of a black hole is a one-dimensional line, infinitely long, in a spacelike direction. Inside the horizon, the Schwarzschild r and t coordinates "switch roles"; r becomes timelike (with decreasing r being the "future" direction of time) and t becomes spacelike. The r = 0 singularity goes from t = minus infinity to t = plus infinity in these coordinates (the Schwarzschild "interior" coordinates, which are not the same as the Schwarzschild "exterior" coordinates you've been using), so it's a line at one point of "time" (r = 0) that extends through all of "space" (the full range of t).

This also means that, since the decreasing r direction is the future time direction inside the horizon, the singularity at r = 0 is to the future of all events inside the horizon. That's why the singularity is unavoidable (and why ideas like "speed relative to the singularity" make no sense--see below).

I wasn't talking about an interior view point because I don't believe such a stupid situation exists. It sounds just like you're explaining what would happen if you moved faster than c.

Then I guess you aren't familiar with the mathematical theory of limits and calculus, that puts things like "move smoothly to infinity" on a consistent, rigorous mathematical footing. If you consider things like calculus to be "mathematical tricks", then I'm not sure why you're bothering to even have a discussion in this forum, since calculus is a fundamental part of GR, and we tend to assume here that it's, well, valid.

You're right, I'm not familiar with the mathematical theory of limits and calculus. Finite number - larger finite number - even larger finite number - infinity = Jump. I don't see how you could do it without some kind of trick. I don't see how you could reach infinity no matter how fast the acceleration increases?

Because you can't change just one thing. You can't just wave your hands and say, "my theory is exactly like GR and makes all the same predictions, except that nothing can ever reach a black hole's horizon". The predictions of GR are derived from an underlying model, and you can't change any single prediction without changing the entire model, meaning you will change lots of other predictions as well, including ones that have already been experimentally verified to high precision. The only way you could be right and still have all the other predictions come out the same would be if there were a consistent model that *only* differed from standard GR in that one single prediction, not anywhere else. If you think you have such a model, by all means post it--but you have to post the *model* itself, the whole logical structure, not just your prediction from it.

I don't have to do anything. I have explained the logical structure.

Why not? That's the question I was asking before, and you haven't answered it. The "river bed" is not a physical thing; it's just a conceptual crutch that can be used to visualize what's going on. So there's no physical reason why the speed of anything relative to the riverbed needs to be limited.

Wow! Apart from special relativity, but who cares about that? The river needs to accelerate relative to something, everything in fact. That's what the riverbed's for. It represents one of an infinite number of frames and you can't reach c relative to any of them, so you can't reach the horizon.

Um, to make correct predictions? See my comment on making correct predictions above.

You don't need to do that to make correct predictions.

This is false; an object does not need to move at c "relative to the singularity" to reach the horizon, nor does it need to move faster than light "relative to the singularity" once it is inside the horizon. The idea of "speed relative to the singularity" doesn't even make sense; the singularity is in the future, not at a different place (see my comments above about the singularity being a spacelike line), so "speed relative to the singularity" makes no more sense than "speed relative to next Tuesday".

That makes less sense than anything else you've said in this thread. Quite an accomplishment. You can't move relative to a singularity because it doesn't exist in a specific place? You've got some very strange ideas about how the universe works. I'll tell you what, let's just assume it's right in the centre of the black hole mkay, unless of course that doesn't exist in space either. What else doesn't exist in space? You're viewing it from the inside again aren't you? That's why you're thinking of it in the future rather than in a place.

False. The light cones don't move at all. They are geometric features of spacetime. And, as I've already noted, nothing needs to move back in time inside the horizon.

They don't move? What? They move whenever you accelerate for whatever reason. And they do need to move back in time if their light cone goes beyond 90 degrees relative to anything.

Also false. As I've said before, if you're freely falling, you don't have to do *anything* to reach and pass through the horizon, any more than you have to do anything to hit the surface of the Earth if you freely fall from a high altitude.

Imagine there was infinite length contraction and time dilation at the surface because it would be the equivalent of traveling faster than c. Now you can't reach it no matter what you do! It's pushed to the limit at the horizon, literally. And guess what - It holds. There's no need to describe a trip to Never-Ever Land.

Consider flat space time with standard Minkowski coordinates (t, x, y, z). Now consider a second coordinate system (T, X, Y, Z) defined by

T = \frac{t}{v-x}
X = x
Y = y
Z = z​

Consider a particle moving at constant velocity in Minkowski coordinates x = vt. As t increases from 0 to 1, x(=X) increases from 0 to v and T increases from 0 to ∞.

Now, according to your argument, as it takes an infinite amount of T-time to reach x=v, the particle never gets any further. Yet it's pretty obvious this is wrong when you use t-time instead, and there's nothing to stop the particle getting past x=v.

Moral: if you use the "wrong" coordinates, you can be misled to the wrong conclusion. Schwarzschild coordinates are the "wrong" coordinates for examining what happens at the event horizon.

If you use two different “times” then you can do whatever you want. T and t aren't the same. You can't say this timeline doesn't support how I need it to work so I'll just use another one. Schwarzschild coordinates show everything up to the horizon and show that the time need to reach the horizon is infinite. End of. Unless you can give me an actual non-hand-wavy reason why it's okay to switch to a contradictory coordinate system and claim they're both right. You lot claim I'm being hand wavy because I don't use equations but when it comes to the logical structure of this stuff you're so much worse than me.

Sorry for the rant. It wasn't directed at you specifically, I'm just losing patience with close-minded people who've already made their minds up. Again, not directed at you. At least I'm trying to see it the other way, even if I don't think it works or even makes sense.

 

[A-wal]

Okay let's run a little thought experiment with my version and you can tell me what would happen in your version. An observer approaches a black hole which evaporates just before the object reaches the horizon. Now we reverse the arrow of time. The black hole forms again. The observer gets pushed away from the black hole. Gravity can't do that. It's the gamma ray burst of the newly formed singularity that pushes the observer away. Gravity and electro-magnetism switch over when you reverse the arrow of time. Has the penny dropped now?

 

[PeterDonis]

Probably. I thought was something like "in general relativity we use light cones which translates to special relativity just as well", or words to that effect. Did you edit it out?

I think I did say at some point that the concept of light-cones applies equally well to GR and SR (which it does), and that the difference is that in SR, since spacetime is globally flat, all the light-cones everywhere "line up" with each other, where in GR, since spacetime is curved, the light-cones at different events can be "tilted" with respect to each other. See comments further below on the light-cones.

You've lost me. If you want talking about the space-time they occupy then what?

The worldlines of the objects *are* the "spacetime they occupy" (in so far as that phrase means anything).

Behave differently how?

I was just referring to your perplexity at the fact that ingoing objects can cross the horizon from outside to inside, while outgoing objects cannot cross the horizon from inside to outside.

I mean that saying an object has crossed the horizon but its light is still visible doesn't make sense when at any time the object could turn round and come back.

But this argument breaks down if, as is actually the case, objects inside the horizon *cannot* "turn around and come back" (i.e., get outside the horizon again).

The riverbed is just there to make the metaphor easier. It represents the frame of something at rest relative to the singularity and the objects starting frame in the example I used.

Which is not a physical thing and has no physical effects, so it can't stop things from moving "faster than light" relative to it (since "relative to it" has no physical meaning).

If you move at c relative to the riverbed then you move at c relative to an object that stayed in our starting position.

Which is perfectly possible because the light cones may be tilted where we are now, relative to where they were at our starting position (see my earlier comment). It's the variable tilting of the light-cones that gives rise to the "appearance" of objects moving "faster than light" relative to distant locations. But objects are still moving inside the light cones *where they are* at any event.

Was it the objects moving through flat space-time or did the space-time between them curve? Same thing!

No, they're not. There is a real, physical difference between objects moving on "straight" paths (geodesics) and objects moving on "curved" paths. Objects moving on straight paths are in free fall--they feel no weight. Objects moving on curved paths feel weight. That is the crucial distinction, physically. In flat spacetime, it just so happens that the "straight" paths (the freely falling, weightless ones) are also straight in the everyday, Euclidean sense we're used to, while the accelerated paths along which objects feel weight are curved. But in curved spacetime, that correspondence no longer holds, so you have to actually look at the physics--what observers traveling on the paths actually feel, physically--to determine what kind of path it is.

Velocity is relative right? So there's no difference between constant velocity and being at rest right, so if you were going really, REALLY fast it would seem to you as though you're not moving at all.

Right so far.

Gravity is relative right? So there's no difference between being in a higher gravitational field and a lower one from the perspective of a free-faller.

Still ok.

You can feel acceleration because that isn't relative, well actually it is but not in quite the same way. You can feel acceleration from using energy or from moving into a stronger part of the gravitational field as you accelerate relative to anything far enough away to ignore the gravitational field.

Nope, here's where you go wrong. You can "move into a stronger part of the gravitational field" by freely falling--you will feel *no* acceleration by doing that and will see no local difference (as you yourself just said a moment before).

But if you "use energy" (in the sense of applying a force), you *will* feel acceleration--you will *not* be freely falling any more. The analogous situation to *that*, gravitationally, is something like standing at rest on the surface of a planet (like Earth)--you are "at rest" in the gravitational field, but you are *not* freely falling, and you *can* tell "where you are" in the field by how much weight you feel--how much force it takes to hold you at rest (the closer you are to the center of the field, the more force it takes).

Meaning that the event horizon in flat space-time is c. <rest of paragraph snipped>

What does any of this have to do with the "horizon" of the universe as a whole (which was what I was trying to get clarification on your views of)?

That's cheating! ... If they can't form when time is moving one way then they can't form when time is moving the other way either because the laws remain the same, so it's not a valid solution.

I guess I need to clarify a little bit. The term "white hole" as used to describe the "time reverse" of a black hole is usually meant only to apply to an "eternal" black hole, meaning the black hole never "formed" in the first place because it was always there (so the white hole would also be "eternal"). The time reverse of a real, physical black hole forming out of a collapsing star, say, would be a spacetime that started out looking like a white hole but at some point suddenly exploded into an expanding shell of matter. That is also a valid solution of the Einstein Field Equation (so "time reversibility" still applies), but it still has the same problem of trying to explain where the white hole originally came from. However, that is a problem not because it postulates something logically impossible but because we have strong evidence that our universe did not exist for an infinite time in the past (which is what would be required for it to contain white holes of either sort), but had a beginning a finite time ago (the Big Bang). In theories like a steady-state type of model where the universe has always existed, the problem I cited with white holes would not exist.

I do believe it's closed. I don't believe it will "re"collapse.

I believe these two statements are only consistent if there is a positive cosmological constant that is quite a bit larger than what we actually observe. I'll have to check to be sure, though. (With a zero or negative cosmological constant, the universe can be closed only if it recollapses. With a positive constant, it can be closed and still expand forever, but I believe the constant can't be too small for that to happen.)

You're right, I'm not familiar with the mathematical theory of limits and calculus. Finite number - larger finite number - even larger finite number - infinity = Jump.

From your point of view it's even worse than that; there is a whole region of spacetime (inside the horizon) where the "acceleration" is *greater* than "infinity". Of course, there's no actual problem, mathematically, but if you aren't familiar with the theory of limits and calculus I can see how it would be hard to visualize how that can possibly happen. Which doesn't change the fact that it *does* happen, and that it's all perfectly consistent.

I don't have to do anything. I have explained the logical structure.

No, you've continued to make claims that are based on premises which I claim are false, and you have not given a cogent logical argument for any of those premises; you've just continued to assert them.

The river needs to accelerate relative to something, everything in fact.

The "river" isn't a physical thing either; it's just another conceptual crutch to help with visualization. So it doesn't "need to accelerate" relative to anything.

You're viewing it from the inside again aren't you? That's why you're thinking of it in the future rather than in a place.

Yes, of course. The singularity is inside the horizon, so you have to "view from the inside" to talk about "where" it is and how to "get to" it.

They don't move? What? They move whenever you accelerate for whatever reason.

No, the *light cones* don't move at all. At every event in spacetime, the light cones are fixed; they never change at that event. But light-cones at *different events* can be "tilted" relative to one another. So as *you* "move" (meaning you travel along your worldline), you can encounter light cones that point in different directions. The light cones don't "belong" to you; they "belong" to each event you pass through.

Sorry for the rant. It wasn't directed at you specifically, I'm just losing patience with close-minded people who've already made their minds up. Again, not directed at you.

No offense taken. But I should admit that, from your point of view, I *am* one of those "close-minded people who've already made their minds up". Nothing you have said has had the slightest effect on my picture of how black holes work, because I understand the mistake you are making and how it's a very easy mistake to make (I made it myself back when I was first learning about this stuff, and I expect most people who try to learn GR make it at some point). You are assuming that there is some flat "background" spacetime that constrains what can happen (the "river bed" is one term you use to describe that background). There isn't, and there doesn't need to be. That's as simple as I can put it. But I understand that it's hard to see *how* things can possibly work without that background.

 

[PeterDonis]

Okay let's run a little thought experiment with my version and you can tell me what would happen in your version. An observer approaches a black hole which evaporates just before the object reaches the horizon. Now we reverse the arrow of time. The black hole forms again. The observer gets pushed away from the black hole. Gravity can't do that. It's the gamma ray burst of the newly formed singularity that pushes the observer away. Gravity and electro-magnetism switch over when you reverse the arrow of time. Has the penny dropped now?

But if you reverse the direction of time, there is no gamma ray burst emanating from the singularity--because while time was moving forward, there was no shell of gamma rays converging on the singularity. The gamma ray burst goes into the *future*, not the *past*; when you reverse time at the moment of the black hole evaporating, you're moving into the past, so you don't see the burst.

Also, remember that the principle of "time reversal" is *not* that things have to look the same when time is reversed, but that the laws of physics have to be the same when time is reversed. The time reverse of an observer falling towards the black hole, and accelerating towards it, is an observer rising away from the black hole and *decelerating* as it moves away. (This works the same as the time reverse of a ball dropped from a height; it's a ball rising and decelerating until it reaches the point where it was dropped.) The physical law is that the acceleration is towards the hole (that's highly oversimplified since we're talking about full-blown GR now, but it will do for purely radial motion), and that's true in the time-reversed case as well as the time-forward case.

(By the way, this principle also applies to white holes, which as I noted previously are the time reverses of black holes. An object coming out of a white hole horizon *decelerates* as it rises away from the horizon. It does *not* get "pushed away".)

 

[A-wal]

I think I did say at some point that the concept of light-cones applies equally well to GR and SR (which it does), and that the difference is that in SR, since spacetime is globally flat, all the light-cones everywhere "line up" with each other, where in GR, since spacetime is curved, the light-cones at different events can be "tilted" with respect to each other. See comments further below on the light-cones.

Right. Now presumably acceleration in flat space-time also causes the light-cones to be tilted? At c they would be tilted to 90 degrees. That can't actually happen and I don't think gravity can do it either. Even if you put the two together it would be no different than accelerating harder.

The worldlines of the objects *are* the "spacetime they occupy" (in so far as that phrase means anything).

Then how can they possibly be different depending on what direction you're traveling in? Surely if they share the same point in space-time then they behave in the same way. What possible difference could their direction make? Take just one moment when they're right next to each other just outside the horizon. Now we've removed direction so they should behave the same.

I was just referring to your perplexity at the fact that ingoing objects can cross the horizon from outside to inside, while outgoing objects cannot cross the horizon from inside to outside.

I can't believe it doesn't seem stupid to you.

But this argument breaks down if, as is actually the case, objects inside the horizon *cannot* "turn around and come back" (i.e., get outside the horizon again).

How can they be inside the horizon? This is like what I said when you asked me if I though an object could cross the horizon of an ever-lasting black hole and I said that it wouldn't make sense if you could because what if the black hole lasted longer than the time it would take to cross an ever-lasting horizon? From the perspective of anyone on the outside it's not possible for anything to reach the horizon if it's always possible for any object to move away from the horizon. I would have thought that was obvious.

Which is not a physical thing and has no physical effects, so it can't stop things from moving "faster than light" relative to it (since "relative to it" has no physical meaning).

Don't think of it as one entity. Think of all the pebbles on the riverbed as individual 'hovering' observers. Now you can't move at c relative to them.

Which is perfectly possible because the light cones may be tilted where we are now, relative to where they were at our starting position (see my earlier comment). It's the variable tilting of the light-cones that gives rise to the "appearance" of objects moving "faster than light" relative to distant locations. But objects are still moving inside the light cones *where they are* at any event.

You mean when they tilt past 90 degrees? That's just another way of saying they've broken the light barrier.

No, they're not. There is a real, physical difference between objects moving on "straight" paths (geodesics) and objects moving on "curved" paths. Objects moving on straight paths are in free fall--they feel no weight. Objects moving on curved paths feel weight. That is the crucial distinction, physically. In flat spacetime, it just so happens that the "straight" paths (the freely falling, weightless ones) are also straight in the everyday, Euclidean sense we're used to, while the accelerated paths along which objects feel weight are curved. But in curved spacetime, that correspondence no longer holds, so you have to actually look at the physics--what observers traveling on the paths actually feel, physically--to determine what kind of path it is.

Objects in free-fall are the equivalent to inertially moving objects in flat space-time. Objects moving on curved paths are the equivalent of objects accelerating in flat space-time. What's the difference?

Nope, here's where you go wrong. You can "move into a stronger part of the gravitational field" by freely falling--you will feel *no* acceleration by doing that and will see no local difference (as you yourself just said a moment before).

But if you "use energy" (in the sense of applying a force), you *will* feel acceleration--you will *not* be freely falling any more. The analogous situation to *that*, gravitationally, is something like standing at rest on the surface of a planet (like Earth)--you are "at rest" in the gravitational field, but you are *not* freely falling, and you *can* tell "where you are" in the field by how much weight you feel--how much force it takes to hold you at rest (the closer you are to the center of the field, the more force it takes).

Tidal force is movement into a stronger gravitational field!

What does any of this have to do with the "horizon" of the universe as a whole (which was what I was trying to get clarification on your views of)?

The horizon of the universe as a whole in flat space-time is c, sort of.

I guess I need to clarify a little bit. The term "white hole" as used to describe the "time reverse" of a black hole is usually meant only to apply to an "eternal" black hole, meaning the black hole never "formed" in the first place because it was always there (so the white hole would also be "eternal"). The time reverse of a real, physical black hole forming out of a collapsing star, say, would be a spacetime that started out looking like a white hole but at some point suddenly exploded into an expanding shell of matter. That is also a valid solution of the Einstein Field Equation (so "time reversibility" still applies), but it still has the same problem of trying to explain where the white hole originally came from. However, that is a problem not because it postulates something logically impossible but because we have strong evidence that our universe did not exist for an infinite time in the past (which is what would be required for it to contain white holes of either sort), but had a beginning a finite time ago (the Big Bang). In theories like a steady-state type of model where the universe has always existed, the problem I cited with white holes would not exist.

The problem would still exist because they would still have to form because black holes don't exist forever. Even if you think they last forever they're not eternal because they form at a certain point in time, no matter which direction you point the arrow.

Gravity doesn't push when you reverse the arrow of time. A time-reversed black hole is...a black hole.

I believe these two statements are only consistent if there is a positive cosmological constant that is quite a bit larger than what we actually observe. I'll have to check to be sure, though. (With a zero or negative cosmological constant, the universe can be closed only if it recollapses. With a positive constant, it can be closed and still expand forever, but I believe the constant can't be too small for that to happen.)

What is the cosmological constant? I've heard that phrase before but never knew what it meant. Is it the strength of gravity? That would make sense with what you and others have said.

From your point of view it's even worse than that; there is a whole region of spacetime (inside the horizon) where the "acceleration" is *greater* than "infinity". Of course, there's no actual problem, mathematically, but if you aren't familiar with the theory of limits and calculus I can see how it would be hard to visualize how that can possibly happen. Which doesn't change the fact that it *does* happen, and that it's all perfectly consistent.

Don't be silly.

No, you've continued to make claims that are based on premises which I claim are false, and you have not given a cogent logical argument for any of those premises; you've just continued to assert them.

That's not true. I've explained how I think it works and why I think that. I've also tried to explain why I think the standard description contradicts itself and that I don't see the need for approaching an event horizon to be treated any differently than acceleration in flat space-time. I'm not sure what else I can do.

The "river" isn't a physical thing either; it's just another conceptual crutch to help with visualization. So it doesn't "need to accelerate" relative to anything.

Okay, technically anything in it needs to accelerate relative to anything not in it. I'm aware there is no actual dividing line between in and out of the river, but you know what I mean.

Yes, of course. The singularity is inside the horizon, so you have to "view from the inside" to talk about "where" it is and how to "get to" it.

It's in the middle and you can't get to it.

No, the *light cones* don't move at all. At every event in spacetime, the light cones are fixed; they never change at that event. But light-cones at *different events* can be "tilted" relative to one another. So as *you* "move" (meaning you travel along your worldline), you can encounter light cones that point in different directions. The light cones don't "belong" to you; they "belong" to each event you pass through.

Change rather than move then.

No offense taken. But I should admit that, from your point of view, I *am* one of those "close-minded people who've already made their minds up". Nothing you have said has had the slightest effect on my picture of how black holes work, because I understand the mistake you are making and how it's a very easy mistake to make (I made it myself back when I was first learning about this stuff, and I expect most people who try to learn GR make it at some point). You are assuming that there is some flat "background" spacetime that constrains what can happen (the "river bed" is one term you use to describe that background). There isn't, and there doesn't need to be. That's as simple as I can put it. But I understand that it's hard to see *how* things can possibly work without that background.

Yes you are one of those people, but from your point of view you're just trying to help someone who doesn't get it.

The background is what everything would move relative to if every point in space had an inertial observer for you to compare yourself with. The fact that there isn't an inertial observer at every point in space-time is irrelevant. The background radiation could do this job in reality, but I prefer to think of lots of individual observers because you can do more with that.

But if you reverse the direction of time, there is no gamma ray burst emanating from the singularity--because while time was moving forward, there was no shell of gamma rays converging on the singularity. The gamma ray burst goes into the *future*, not the *past*; when you reverse time at the moment of the black hole evaporating, you're moving into the past, so you don't see the burst.

The grb would have to move outwards no matter which way the arrow is pointing, so what does happen to it when time is reversed?

Also, remember that the principle of "time reversal" is *not* that things have to look the same when time is reversed, but that the laws of physics have to be the same when time is reversed. The time reverse of an observer falling towards the black hole, and accelerating towards it, is an observer rising away from the black hole and *decelerating* as it moves away. (This works the same as the time reverse of a ball dropped from a height; it's a ball rising and decelerating until it reaches the point where it was dropped.) The physical law is that the acceleration is towards the hole (that's highly oversimplified since we're talking about full-blown GR now, but it will do for purely radial motion), and that's true in the time-reversed case as well as the time-forward case.

I meant that the source of an objects acceleration switches between gravity and electro-magnetic when the arrow is reversed. If gravity still pulls and electro-magnetism still pushes then when the arrow is reversed it means that the two should be equivalent and interchangeable. The ball using electro-magnetism to rise until gravity overpowers it and pulls it back down. When time is reversed what was it's upwards acceleration is now downwards and it's downwards acceleration is now upwards, and gravity only pulls and electro-magnetism only pushes, so it means they've switched over.

(By the way, this principle also applies to white holes, which as I noted previously are the time reverses of black holes. An object coming out of a white hole horizon *decelerates* as it rises away from the horizon. It does *not* get "pushed away".)

Then white holes do pull? They would have to if the object decelerates. But then how could the object escape from inside the horizon in the first place?

 

[PeterDonis]

Right. Now presumably acceleration in flat space-time also causes the light-cones to be tilted?

As I have said several times now, *NO*, it does *NOT*. Acceleration in flat space-time has *NO* effect on the light cones.

Take just one moment when they're right next to each other just outside the horizon. Now we've removed direction so they should behave the same.

No, you haven't removed direction. They are right next to each other, but one is moving inward and the other is moving outward. They have different velocities. So they follow different worldlines, and experience different things.

Don't think of it as one entity. Think of all the pebbles on the riverbed as individual 'hovering' observers. Now you can't move at c relative to them.

Doesn't make any difference. You are still implicitly assuming that all the pebbles, sitting at rest with respect to each other, *must* cover the entire spacetime. That assumption is simply false in the standard GR model. You have not proven that that assumption being false leads to any logical inconsistency; you've simply kept on asserting the assumption, and saying that the standard GR model is inconsistent with the assumption being true. Of course it is, since in the standard GR model the assumption is false. That doesn't make the standard GR model inconsistent; it just means your assumption doesn't hold in that model.

Objects in free-fall are the equivalent to inertially moving objects in flat space-time. Objects moving on curved paths are the equivalent of objects accelerating in flat space-time. What's the difference?

Once again, the difference is that objects that are accelerating feel weight, and objects in free fall don't, and that distinction is crucial in GR. Do you honestly not see any physical difference between a freely falling object and an object that feels weight? Bear in mind that that distinction is also fundamental in *special* relativity.

The problem would still exist because they would still have to form because black holes don't exist forever. Even if you think they last forever they're not eternal because they form at a certain point in time, no matter which direction you point the arrow.

Just to clarify terms: the "eternal" black hole in standard GR lasts forever in both directions of time, past and future. A black hole that forms from a collapsing star does form at a certain finite time, but (in standard GR, without bringing quantum effects in--black hole evaporation is a quantum phenomenon) it lasts forever in the future. The time reverses of those two cases are an "eternal" white hole, which also lasts forever in both directions of time, and a white hole that existed for an infinite time into the past, but at some finite time suddenly "explodes" into an expanding shell of matter (the time reverse of the collapsing star). See below, though, for more on that case.

What is the cosmological constant? I've heard that phrase before but never knew what it meant.

It's also sometimes referred to as "dark energy" in cosmology. It's basically a form of energy that is possessed by what we normally think of as "empty space" or "vacuum". The Wikipedia page (link below) is an OK starting point for learning about it.

http://en.wikipedia.org/wiki/Cosmological_constant

I'm not sure what else I can do.

You can give some kind of logical argument, starting from premises we all accept, that shows why the underlying assumptions of your model, which are false in standard GR, somehow *must* be true. You haven't done that. You've simply continued to assert those assumptions, without argument, even though I and others have kept on telling you those assumptions are false in GR, so you can't just help yourself to them; you have to argue for them. Just saying "it seems obvious to me" or "doesn't it seem silly to you?" isn't an argument.

The background is what everything would move relative to if every point in space had an inertial observer for you to compare yourself with.

But this, by itself, isn't enough to support the claims you're making. You also have to require that *all* of those inertial observers, at *every* point of space, can be at rest relative to each other. Once again, that assumption is false in the standard GR model of a black hole spacetime.

The grb would have to move outwards no matter which way the arrow is pointing, so what does happen to it when time is reversed?

The grb moves outward only in the *forward* direction of time. Call the instant of black hole evaporation and the emission of the grb time T. Then at time T + dT, say (where dT is some positive number), the grb has spread outward from the point of evaporation. So if we start at time T + dT and look at the time-reversed picture, we see a contracting shell of radiation converging inward on the point of evaporation, and reaching it at time T--and then *disappearing* into the time-reversed black hole, so at times earlier than T, the grb is *nowhere*--it doesn't exist.

Then white holes do pull? They would have to if the object decelerates. But then how could the object escape from inside the horizon in the first place?

Yes, white holes do pull. Objects escape from inside the white hole horizon because the horizon is an *ingoing* null surface, not an *outgoing* null surface (remember, the white hole is the time reverse of the black hole--the time reverse of an outgoing null surface is an ingoing null surface). Objects emerge from the white hole singularity, which is in the past for all objects inside the white hole horizon (the time reverse of the black hole singularity, which is in the future for all objects inside the black hole horizon), and when they emerge they are moving outward; they decelerate because of the hole's gravity, but the horizon moves inward at the speed of light, so no matter how much the objects decelerate, they can't possibly "catch up" to the ingoing horizon.

 

[PeterDonis]

You also have to require that *all* of those inertial observers, at *every* point of space, can be at rest relative to each other. Once again, that assumption is false in the standard GR model of a black hole spacetime.

On re-reading this and the question it was in response to, I realized I should clarify a couple of things:

(1) The family of observers that are "hovering" outside the black hole horizon, at rest relative to each other, are *not* inertial observers; they are accelerated and feel weight. No two inertial observers in a black hole spacetime can be at rest relative to each other for more than a single instant unless they are at rest relative to each other at the same radial coordinate r (which r coordinate that is will change with time, as the observers free fall towards the hole, but if they both start free falling at the same r, they will always be at the same r, so they will always be at rest relative to each other--at least, as long as they are close enough together that the tangential tidal gravity is negligible).

(2) Even if we allow the family of "hovering" observers, at rest relative to each other, to define a "background" coordinate system for the spacetime, that family of observers does not cover the entire spacetime around the black hole. Inside the horizon, there are *no* "hovering" observers *at all*--*all* observers, no matter how they move or how hard they accelerate, *must* decrease their radial coordinate r with time. So inside the horizon, there are *no* observers, inertial *or* accelerated, that are at rest even for an instant relative to the family of "hovering" observers outside the horizon.

Item (2) is what I was referring to in the quote above: there is a region of spacetime around a black hole (the region inside the horizon) where it is impossible for any observer, even for an instant, to be at rest relative to what A-wal calls the "river bed" (or "pebbles" or "background" or whatever term you want to use). Such observers can only exist outside the horizon. But as item (1) shows, even outside the horizon, observers at rest relative to the "river bed" are *not* inertial observers. That's what I wanted to clarify.

 

[A-wal]

As I have said several times now, *NO*, it does *NOT*. Acceleration in flat space-time has *NO* effect on the light cones.

I thought it would make sense if they did in special relativity as well. You mean light cones aren't used in this way or they can't be? Seems like they could be.

I've just got the number of a really hot Columbian bird. I'm taking her out Friday. I love public libraries.

No, you haven't removed direction. They are right next to each other, but one is moving inward and the other is moving outward. They have different velocities. So they follow different worldlines, and experience different things.

That's special relativity. I don't think their velocity should make any difference. If they're in the same area of space-time then although their different velocities would of course mean they experience different things, the effect of gravity would be the same for both of them if they're in the same place.

Doesn't make any difference. You are still implicitly assuming that all the pebbles, sitting at rest with respect to each other, *must* cover the entire spacetime. That assumption is simply false in the standard GR model. You have not proven that that assumption being false leads to any logical inconsistency; you've simply kept on asserting the assumption, and saying that the standard GR model is inconsistent with the assumption being true. Of course it is, since in the standard GR model the assumption is false. That doesn't make the standard GR model inconsistent; it just means your assumption doesn't hold in that model.

No, it's not that they must. It's just that they do in this particular thought experiment. If you want to reach the horizon then you're going to have to break the light barrier relative to them and that's not possible.

Once again, the difference is that objects that are accelerating feel weight, and objects in free fall don't, and that distinction is crucial in GR. Do you honestly not see any physical difference between a freely falling object and an object that feels weight? Bear in mind that that distinction is also fundamental in *special* relativity.

No I don't see a difference. Objects in free-fall are always accelerating because the river is always taking them closer to the gravitational source. An object in free-fall is the equivalent of an object with a different relative velocity undergoing no acceleration only if we ignore tidal force. Tidal force is the equivalent of acceleration in flat space-time.

Just to clarify terms: the "eternal" black hole in standard GR lasts forever in both directions of time, past and future. A black hole that forms from a collapsing star does form at a certain finite time, but (in standard GR, without bringing quantum effects in--black hole evaporation is a quantum phenomenon) it lasts forever in the future. The time reverses of those two cases are an "eternal" white hole, which also lasts forever in both directions of time, and a white hole that existed for an infinite time into the past, but at some finite time suddenly "explodes" into an expanding shell of matter (the time reverse of the collapsing star). See below, though, for more on that case.

I'm not sure having an object that can't form is a valid solution.

It's also sometimes referred to as "dark energy" in cosmology. It's basically a form of energy that is possessed by what we normally think of as "empty space" or "vacuum". The Wikipedia page (link below) is an OK starting point for learning about it.

http://en.wikipedia.org/wiki/Cosmological_constant

Oh it's dark energy. Okay that makes sense.

You can give some kind of logical argument, starting from premises we all accept, that shows why the underlying assumptions of your model, which are false in standard GR, somehow *must* be true. You haven't done that. You've simply continued to assert those assumptions, without argument, even though I and others have kept on telling you those assumptions are false in GR, so you can't just help yourself to them; you have to argue for them. Just saying "it seems obvious to me" or "doesn't it seem silly to you?" isn't an argument.

No it's not an argument, but that's not all I've done. Look at it from my point of view. I've got a nice simple model in my head and there's no need for white holes, infinite energy, I don't need multiple coordinate systems to describe one thing, there's no acceleration up to c, it doesn't brake the arrow of time, gravity doesn't behave differently depending on your direction, and it makes intuitive sense rather than giving me the impression that it made to fill gaps that didn't even need filling. There's nothing you've said that makes me think that version of it is right because you haven't given me a single reason why all of those highly dubious things need to be brought in.

But this, by itself, isn't enough to support the claims you're making. You also have to require that *all* of those inertial observers, at *every* point of space, can be at rest relative to each other. Once again, that assumption is false in the standard GR model of a black hole spacetime.

You've completely lost me here. Why wouldn't they be able to be at rest relative to each other?

The grb moves outward only in the *forward* direction of time. Call the instant of black hole evaporation and the emission of the grb time T. Then at time T + dT, say (where dT is some positive number), the grb has spread outward from the point of evaporation. So if we start at time T + dT and look at the time-reversed picture, we see a contracting shell of radiation converging inward on the point of evaporation, and reaching it at time T--and then *disappearing* into the time-reversed black hole, so at times earlier than T, the grb is *nowhere*--it doesn't exist.

You mean call the instant of black hole formation and the emission of the grb time T. Yea that makes sense. I had the example of the ball being thrown in my head. I was thinking that the upwards movement of the ball would be the downwards movement caused by gravity but it doesn't quite work like that. Must have smoked one too many. Gravity pulls down on the object in the same way on the way up as it does on the way down. It a constant force so it doesn't change when the arrow is reversed.

Yes, white holes do pull. Objects escape from inside the white hole horizon because the horizon is an *ingoing* null surface, not an *outgoing* null surface (remember, the white hole is the time reverse of the black hole--the time reverse of an outgoing null surface is an ingoing null surface). Objects emerge from the white hole singularity, which is in the past for all objects inside the white hole horizon (the time reverse of the black hole singularity, which is in the future for all objects inside the black hole horizon), and when they emerge they are moving outward; they decelerate because of the hole's gravity, but the horizon moves inward at the speed of light, so no matter how much the objects decelerate, they can't possibly "catch up" to the ingoing horizon.

They pull and push then. Why would gravity push? Definitely not a valid solution. The horizon moving inward at the speed of light is what I think a black hole would do at the horizon. Maybe you need to switch it over to a white hole when you cross the horizon because everything gets flipped and that's why objects can't reach the horizon from the perspective of the inside. I still think it would be an imaginary rather than a literal inside though. In reality the black hole has zero size because it's just the singularity. Time dilation and length contraction make it appear bigger from a distance.


I haven’t got time to reply to your most recent post atm. I’ll do it Friday night when I’m back at ‘work’.

 

[PeterDonis]

If they're in the same area of space-time then although their different velocities would of course mean they experience different things, the effect of gravity would be the same for both of them if they're in the same place.

While they're in the same place, yes, the effect of gravity is the same--but they only remain in the same place for an instant because their velocities are different. Then they move apart and the curvature of spacetime is different at their two different locations, so they experience different things.

(I should also mention that for an ingoing and outgoing object to be at the same place, even for an instant, they both have to be outside the horizon. Inside the horizon there are no outgoing objects, not even light rays.)

No, it's not that they must. It's just that they do in this particular thought experiment. If you want to reach the horizon then you're going to have to break the light barrier relative to them and that's not possible.

Still doesn't make any difference. Your claim that an object would have to "break the light barrier" relative to the hovering objects implicitly assumes that the "frame" defined by the hovering objects covers the entire spacetime. Once again, that assumption is false in the standard GR model. (See below for more details on this.)

No I don't see a difference. Objects in free-fall are always accelerating because the river is always taking them closer to the gravitational source. An object in free-fall is the equivalent of an object with a different relative velocity undergoing no acceleration only if we ignore tidal force. Tidal force is the equivalent of acceleration in flat space-time.

Ok, at least I understand why you don't see a difference. I don't agree, but at least I understand how this particular claim of yours ties in with the rest of your claims.

I'm not sure having an object that can't form is a valid solution.

It's not that it "can't form", it's that the mathematical solution describes an object that has existed for an infinite time in the past. If you have good physical reasons to believe that the entire universe has only existed for a finite time in the past (which we do), then obviously any mathematical solution describing an object that would have had to exist for an infinite time in the past is not a good candidate for describing an actual, physical object. But it's still a perfectly valid solution mathematically.

Look at it from my point of view. I've got a nice simple model in my head and there's no need for white holes, infinite energy, I don't need multiple coordinate systems to describe one thing, there's no acceleration up to c, it doesn't brake the arrow of time, gravity doesn't behave differently depending on your direction, and it makes intuitive sense rather than giving me the impression that it made to fill gaps that didn't even need filling. There's nothing you've said that makes me think that version of it is right because you haven't given me a single reason why all of those highly dubious things need to be brought in.

I've mentioned making correct predictions before. What does your nice, simple model predict for the following:

(1) The precession of the perihelion of Mercury's orbit?

(2) The bending of light by the Sun?

(3) The changes in the orbits of binary pulsars due to the emission of gravitational waves?

(4) The precession of gyroscopes orbiting the Earth due to gravitomagnetism?

The standard GR model that predicts all these phenomena correctly *also* predicts that black holes will behave as I've been describing. That's why physicists believe in the standard GR model of black holes that I've been describing. If you can show how your model reproduces all these correct predictions *without* requiring black holes to behave as I've been describing, please do so. But you can't just wave your hands and say, "well, obviously my model looks just like GR outside the horizon", because the way GR arrives at all the above predictions is inseparably linked, mathematically, to the way it describes black holes and their horizons. So you have to start from scratch, and work through how your model would deal with the above phenomena, *without* making use of any of the machinery or results of GR.

(And no, you can't get any of the above results just by applying non-relativistic Newtonian gravitational theory. That's why I chose these examples.)

You've completely lost me here. Why wouldn't they be able to be at rest relative to each other?

Because you can't directly assign any physical meaning to the "relative velocity" of two objects at different places. Suppose I have observer A, well outside the horizon and hovering at a constant r. Then I have observer B, who has just freely fallen through the horizon. In order to make sense of the "relative velocity" of A and B, I have to implicitly assume a third observer, C, who is at the same radial coordinate r as B (i.e., r a little less than the radius of the horizon), but who is at rest relative to A, so that I can say that the relative velocity of A and B is equal to the relative velocity of C and B (which I can assign a direct physical meaning to because C and B are at the same place). But in the standard GR model, there can't be any such observer C; *no* observer inside the horizon can "hover" at a constant radius, not even for an instant. So the only way of physically assigning a meaning to the concept "relative velocity of A and B" breaks down if A is outside the horizon and B is inside.

(One clarification: by "physical meaning of relative velocity" I mean a meaning that would justify the requirement that the "relative velocity" of two objects can't be faster than light. If two observers are at the same place, then I can apply special relativity locally and impose that requirement. But I can't do it for observers that are separated, if the curvature of spacetime is significantly changed from one to the other. Of course, I can arbitrarily define the "relative velocity" of A and B by simply using, for example, dr/dt, the derivative of the radial coordinate r with respect to the "time" coordinate t. But this meaning of "relative velocity" does *not* require that the relative velocity can't be faster than light, because it's just an arbitrary number; it doesn't correspond to anything that any possible physical observer could ever observe.)

They pull and push then. Why would gravity push? Definitely not a valid solution.

Why do you think there's a push? There's no push anywhere. Objects emerge from the white hole singuarity, but that's not a "push" because it's not due to any "force" from the singularity; the objects just emerge. As soon as they emerge, they start decelerating, so the only "force" observed is a pull.

 

[PeterDonis]

On re-reading earlier posts I found some things in post #301 that I wanted to respond to, because they involve points that I haven't touched on, or because they reinforce points that I think are very important.

It seems to me that objects have to do work just to remain solid objects and to produce gravity, and the energy that allows this obviously can't last forever.

What makes you think this? And don't say "I heard it somewhere" (see next comment). Give me some sort of logical argument, based on premises we all accept, that makes this seem reasonable to you. (To me, as should be obvious from my previous posts, it's just wrong as it stands.)

I'm sure I heard somewhere that matter doesn't have an infinite lifespan?

You keep on saying "I heard somewhere" something, and give a vague description of it, but can't give any actual reference or explain what you mean beyond the vague description. That's not very helpful in understanding what you're talking about. Even the OP in this thread suffers from this problem.

If you are thinking of things like normal matter "quantum tunnelling" into other states (as described, for example, in the page linked to below), yes, according to QM that will eventually happen if nothing else does, but that process doesn't require any energy, and normal matter certainly doesn't do any work or "build up" energy while it's "waiting" for this to happen.

http://math.ucr.edu/home/baez/end.html

But you said matter curves space-time completely differently to energy and the two processes were distinct and not equivalent?

I have never said any such thing. I have said repeatedly that only one "thing" curves spacetime, and that's the stress-energy tensor (which includes what you are calling "matter", and "pressure", and also includes "energy" as standard physics uses the term, but you don't always use that term correctly).

I *have* said that what you sometimes refer to as "energy" (meaning something like firing a rocket engine to accelerate, and therefore feeling weight) *is* different from the stress-energy tensor curving spacetime, because acceleration (in the sense of feeling weight) curves your worldline, not spacetime, and you can have a curved worldline in a flat spacetime, so the two concepts are distinct. (See further comment below on this.)

Yes you are! You keep claiming that you're not suggesting these things then go on to describe them anyway. If you're not even sure what you think then I'm not surprised you keep misinterpreting my words.

I'm quite sure what I think. I'm also quite sure that you don't understand it, and that the reason you don't understand it is that your thinking is based on assumptions that you think are obviously true, whereas I have a consistent model in which they're false. I keep on asking you to give actual arguments for your assumptions, instead of just assuming they're true even though I've repeatedly said I don't accept them, but you never do, you just keep asserting them. It's like trying to explain how matrix multiplication works to a person who keeps insisting that multiplication *has* to be commutative, even though it keeps being pointed out that in fact, matrix multiplication is *not* commutative, just as a matter of mathematical fact.

The main thing I'm having trouble with is getting through to you. It's like having a conversation with a God worshipper. They've already made their minds up and use backwards logic from there to explain away any inconsistency that anyone raises. The only difference is you've got slightly more to work with. There is no difference between saying that space-time is curved and saying that objects paths through space-time are curved. If every object were affected by a uniform force then everything would be accelerated depending on their distance from the sources, and you could just as easily use this say that space-time is curved. If you can't get that then I'm not sure how much help you can be to me to be honest. Still, I appreciate the effort you're making.

You are ignoring one key difference between what I've been saying and what you've been saying. Every time you have made a statement I disagree with, I have given a physical reason why I disagree, whereas you have just kept on asserting your statements without ever responding to the physical reason for my disagreement. Take the statement in bold above. You have asserted it repeatedly, without ever responding to the physical reason I've given for why the two cases *are* different: because the curvature of an object's path (which is determined by whether or not it feels weight--objects on curved paths feel weight, objects on straight paths do not) is completely independent of the curvature of spacetime itself (which is determined by whether tidal gravity exists, or in more explicit terms, by whether two objects, both in free fall, both starting out close together and at rest with respect to each other at a given time, continue to remain at rest with respect to each other for all time or not--if they do, spacetime is flat, if they don't, spacetime is curved).

In the quote above, in the part that's underlined, you at least have given some sort of amplification of your statement, but it's still wrong, and the reason why it's wrong has been brought up repeatedly in this thread: a family of accelerated "Rindler observers" in flat spacetime has precisely the set of properties you describe, but the spacetime is still flat, as is easily seen by the test I gave above (*inertial* observers that start at rest with respect to each other remain at rest with respect to each other for all time). So your argument is still wrong. (And don't say that you've looked at Rindler observers and it doesn't change your mind, because every time you've described how you think Rindler observers and a Rindler horizon work, you've gotten it wrong, as I've pointed out repeatedly.)

So the reason why you have not been "getting through" to me is that, as I've said before, I already understand the mistakes you're making, so seeing you continue to make them does not change my mind.

 

[Buckleymanor]

Also, remember that the principle of "time reversal" is *not* that things have to look the same when time is reversed, but that the laws of physics have to be the same when time is reversed. The time reverse of an observer falling towards the black hole, and accelerating towards it, is an observer rising away from the black hole and *decelerating* as it moves away. (This works the same as the time reverse of a ball dropped from a height; it's a ball rising and decelerating until it reaches the point where it was dropped.)

There seems to be an inconsistency in this.How can this work for the time reversal of a ball dropped from a height.A ball rising and decelerating until it reaches the point where it was dropped is push gravity it just looks the same.The laws of physics won't be the same if this happens we don't see balls rising up from the floor.Which seems to contradict what you are explaining.

 

[PeterDonis]

There seems to be an inconsistency in this.How can this work for the time reversal of a ball dropped from a height.

Please go back and read my post #2 in this thread (yes, I know it was a *long* time ago...). That addresses the case you are asking about. If you still have questions after reading that post, by all means ask.

 

[Buckleymanor]

Please go back and read my post #2 in this thread (yes, I know it was a *long* time ago...). That addresses the case you are asking about. If you still have questions after reading that post, by all means ask.

Yes I have read the post and it is hypothetical and not based on what we have observed.You would have to ignore certain aspects like the ball jumping up from the ground this just does not happen.You can imagine it might but that does not make it real or a possibility.

 

[PeterDonis]

Yes I have read the post and it is hypothetical and not based on what we have observed.You would have to ignore certain aspects like the ball jumping up from the ground this just does not happen.You can imagine it might but that does not make it real or a possibility.

We don't observe balls jumping up from the ground by themselves, but that does not mean that behavior is inconsistent with the laws of Newtonian physics. It isn't. It's just very, very unlikely because of the second law of thermodynamics. My post #2 was not intended to take into account the second law, but there was discussion of it in later posts; see DaleSpam's post #33 for a good summation of where that topic ended up.

 

[Buckleymanor]

We don't observe balls jumping up from the ground by themselves, but that does not mean that behavior is inconsistent with the laws of Newtonian physics. It isn't. It's just very, very unlikely because of the second law of thermodynamics. My post #2 was not intended to take into account the second law, but there was discussion of it in later posts; see DaleSpam's post #33 for a good summation of where that topic ended up.

Thanks for the information.
Yes it's difficult to unscramble an egg, don't know if balls jumping up from the ground would be more consistent with QM than Newtonian mechanics either way I will keep a look out.

 

[Dale]

Balls jump up from the ground all of the time. Have you never seen a basketball game? Assuming no inelasticities or friction the time reverse of a ball bouncing is a ball bouncing, whether it is going up or down.

 

[Dmitry67]

PeterDonis, I watch this thread and just wanted to say that I admire your patience!

 

[Buckleymanor]

Balls jump up from the ground all of the time. Have you never seen a basketball game? Assuming no inelasticities or friction the time reverse of a ball bouncing is a ball bouncing, whether it is going up or down.

So do bricks jumping back to walls broken glass bottles reforming and jumping back to tables, and no need for there to be a motor industry or hospitals you can just wait for a crashed vehicle to mend itself and the occupants.

 

[Dale]

So do bricks jumping back to walls broken glass bottles reforming and jumping back to tables, and no need for there to be a motor industry or hospitals you can just wait for a crashed vehicle to mend itself and the occupants.

Obviously not. Those are all situations where the entropy increases, and the second law of thermodynamics is only symmetric if entropy does not change. That is why I specified no inelasticities or friction in my example above. In any classical situation where the entropy is unchanged the time reverse follows the same laws as the time forward.

In any case, I was responding to your incorrect statement that a ball going upwards from the ground and decelerating was an example of push gravity:

A ball rising and decelerating until it reaches the point where it was dropped is push gravity it just looks the same.

Do you see the error of this statement now?

 

[PeterDonis]

PeterDonis, I watch this thread and just wanted to say that I admire your patience!

It's probably stubbornness as much as patience, but thanks!

 

[Buckleymanor]

Balls jump up from the ground all of the time. Have you never seen a basketball game? Assuming no inelasticities or friction the time reverse of a ball bouncing is a ball bouncing, whether it is going up or down.

Actualy this was not the point that was being made.There was no bounce.

 

[Dale]

Actualy this was not the point that was being made.There was no bounce.

The (incorrect) point you made was:

A ball rising and decelerating until it reaches the point where it was dropped is push gravity it just looks the same.

A bouncing ball is a counter-example which clearly disproves the point you were making. A bouncing ball is an example of a ball rising and decelerating until it reaches the point where it was dropped and yet gravity is still pull gravity.

 

[Buckleymanor]

The (incorrect) point you made was: A bouncing ball is a counter-example which clearly disproves the point you were making. A bouncing ball is an example of a ball rising and decelerating until it reaches the point where it was dropped and yet gravity is still pull gravity.

No you are making the point about a bouncing ball I am trying to make the point about a ball riseing from the ground without a bounce two separate things not counter but totally different observations.

 

[Dale]

No you are making the point about a bouncing ball I am trying to make the point about a ball riseing from the ground without a bounce two separate things not counter but totally different observations.

That is not what you said above. Above you said nothing about the time prior to when the ball was rising and decelerating, you simply claimed that a ball rising and decelerating is an example of push gravity, which is wrong as I demonstrated.

Look, this is easy to show mathematically. For a ball falling in a gravitational field the law is:
x&#039;&#039;=-g
Which, if we are given initial conditions x(0)=x_0 and x&#039;(0)=v_0 leads to the equation of motion:
x(t)=-\frac{1}{2}g t^2 + v_0 t + x_0

Now, if we time reverse the equation by making the substitution t=-T we get:
x(T)=-\frac{1}{2}g T^2 - v T + x
Note, that this expression is the same as the above equation of motion except with the initial condition x&#039;(0)=-v_0. Taking the second derivative of this expression (wrt T) we recover the same law
x&#039;&#039;=-g
where gravity is still "pull" gravity. This is what is meant by the time reverse symmetry of gravity.

Peter Donis is correct in his statements about the time reversal of gravity. Your comments about it being inconsistent are wrong as demonstrated both by the counter example and by the math above. In fact, as the math shows, for any scenario of an object falling under pull gravity, the time reverse is also an object falling under pull gravity only with the opposite initial velocity.

 

[nitsuj]

Peter Donis is correct in his statements about the time reversal of gravity. Your comments about it being inconsistent are wrong as demonstrated both by the counter example and by the math above. In fact, as the math shows, for any scenario of an object falling under pull gravity, the time reverse is also an object falling under pull gravity only with the opposite initial velocity.

DaleSpam I admire your understanding, how you can reconcille that concept above with the idea that it is a description of "time running" backwords

(perhaps you are not discussing the idea of "time running" backwords, ie time reversal is not time running backwords).

 

[Dale]

DaleSpam I admire your understanding, how you can reconcille that concept above with the idea that it is a description of "time running" backwords

(perhaps you are not discussing the idea of "time running" backwords, ie time reversal is not time running backwords).

The usual term is "time reversal", or more explicitly "time reversal symmetry". It refers to a coordinate transformation like I described above (t=-T) and the invariance of the laws of physics under such a transformation.