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

[Dale]

If I was interested in learning maths then yes, that probably would be the best way of doing it.

Then at best you are only going to be able to get a shallow "pop-sci" understanding of GR, and even that is not likely.

How can the horizon have a constant position in Schwarzschild coordinates

In Schwarzschild coordinates the horizon is the 3 dimensional surface given by:
r=\frac{2GM}{c^2}
and so
\frac{d}{dt}\left( \frac{2GM}{c^2} \right) = 0
therefore the horizon has a constant position in Schwarzschild coordinates.

when ‘distance shortening’ means the amount of space you need to cover and the time it would take to cover the same distance would of course lengthen as you get closer to it?

The position of the horizon in Schwarzschild coordinates is clearly not a function of "distance shortening".

 

[PeterDonis]

When I said that light isn't subject to acceleration due to gravity I meant that it doesn't speed up like massive objects because it's already at the maximum speed that time will allow under any circumstances, c!

You've forgotten that light can change direction as well as speed. Yes, the locally measured speed of light is always c relative to a timelike observer, but that doesn't mean light isn't subject to acceleration due to gravity. Gravity can still change its direction, and can still tilt the light cones from place to place so that "moving at c" doesn't mean the same thing at different places.

You could have a hovering observer at every point up until the horizon but not at or inside the horizon. That we agree on.

That's good. But it does *not* imply that...

The speed that the in-faller would have to accelerate to relative to the hoverers would have to reach c in order to reach the horizon.

...because, as you just agreed, there is no hoverer at the horizon. More generally, there is no timelike observer at the horizon for the in-faller to "move at c" relative to. The only thing at the horizon that the in-faller moves at c relative to is an outgoing light ray at the horizon, and that's just because timelike observers always "move at c" relative to light rays (because light rays always move at c relative to timelike observers).

That's exactly what I meant. A hoverer could accelerate away from the black hole as fast as they like and a free-faller would still always pass them at less than c. The same thing must happen with velocity at the horizon.

Not true, as I just explained above. There is no timelike observer at the horizon.

When you try to approach c in flat space-time it slows down as you accelerate but you still can't ever reach it because the rate that it slows down relative to your acceleration decreases the harder you accelerate, and it gets faster overall so that once you've stopped accelerating it's increased its speed relative to your previous frame by the same amount as you have. When you try to approach an event horizon you can close the gap but you can't ever reach it because the rate that you close the gap decreases if you're free-falling, as you move into less "distance shortened" space-time relative to where you were. If you were able to reach it then it would be moving at c, which is why you can't reach it.

I see what you are getting at here, but it doesn't prove what you think it does. As I've said before, you can't just say "approach c" in the abstract; you have to say *what* you are trying to "approach c" relative to. It looks like what you are saying above is describing one timelike observer accelerating relative to another timelike observer (who is assumed to be "at rest"). If that's the case, it only applies in cases where there are two timelike observers with the required properties. At the horizon there aren't; there are no timelike observers "hovering" at rest at the horizon. So your analogy fails.

Also, I'm not sure why you insist on using the term "catch up to c". If you are trying to describe one timelike observer accelerating relative to another, then (by analogy with the Rindler horizon scenario) it would be better to say that a light ray emitted from a certain point can't catch up to the accelerating observer (as long as he continues to accelerate). If you really mean that the accelerating observer can't catch up to something, what is it? It can't just be "c" in the abstract, because that's not a physical object, it's just an abstraction.

What do you think seems more likely?...You're the one who needs to prove it. You're making some outrageous claims but you feel safe doing it because it's the consensus view.

No, I feel safe in making the claims I'm making because I understand the standard general relativistic model that generates them, and how that model is both logically consistent and consistent with all the experimental data we currently have. The claims only sound outrageous to you because you don't have that understanding.

If you want me to "prove" that the standard GR model is the *only* model with those properties, then of course I can't. But when you ask "what do you think seems more likely?", you're basically invoking Occam's Razor, and to even apply that I would have to have another model in front of me that has both of the above properties. So *you* have to show me that your model, as an alternative to standard GR, has both of those properties, before I can even consider it. You haven't done that; you haven't even convinced me that your model is logically consistent (you keep assuming things that you are supposed to be deriving as conclusions in your model), and you also haven't even shown that your model accounts for all the data in the particular scenario we're discussing (the spacetime around a gravitating object), let alone *all* the data that GR accounts for.

Both. Why do I have to pick one? The answer stays the same no matter how you look at it.

If you mean "the answer stays the same" in the sense that neither choice proves your case, I agree. But I don't think that's what you meant.

Yes, the velocity of an outgoing light ray is always c locally. It slows down as it approaches an event horizon.

But if its velocity is always c locally, in what sense does it "slow down"? Relative to what? You need to unpack this statement further.

I mean that if you measure the horizon to be a mile in front of you and then move a mile (as measured at this distance) forwards towards the horizon you'd find that you haven't reached the horizon because it's moved back.

How are you measuring the distance that you move through?

The equivalent to a light ray (and therefore anything else) being unable to reach an event horizon in any amount of time for as long as the black hole exists.

And you *still* don't understand the scenario, even though I've drawn you a diagram. You even comment on the diagram later in your post:

That graph could just as easily be used to show an in-faller approaching an event horizon.

What do you mean by this? Do you mean that the "free-faller" lines in the diagram (the dark magenta vertical lines) are analogous to the worldlines that objects free-falling into a black hole would follow? If so, then I agree, and you have just agreed that objects free-falling into a black hole *can* cross the horizon (because they certainly do in the diagram). If you meant something else, then you need to describe what you meant, because you aren't describing anything that the diagram shows.

No one crosses a Rindler horizon. There is no Rindler horizon for the free-fallers. It's created by the accelerating observers.

In a sense, yes, the Rindler horizon is *defined* by the family of accelerating observers. But once defined, it is a perfectly ordinary lightlike surface in the spacetime, and all observers agree on which surface it is, and free-fallers can cross it (in the ingoing direction).

The equivalent to a distant observer sending a light ray towards a free-faller approaching an event horizon and there being no way of the light reaching them before the black hole's gone. The black hole dying and the light ray reaching the free-faller is the equivalent of the accelerator stopping and the light catching up.

Nope, you've *still* got this backwards despite repeated corrections from me. I was talking about free-fallers sending *outgoing* signals. You are talking about distant observers sending *ingoing* signals. They're not the same.

The equivalent of no free-faller being able to reach c relative to any observer at any distance from the horizon.

Incorrect. I already addressed this above.

No, a free-faller not being able to send a light signal that can catch up with a Rindler horizon is equivalent to an free-faller not being able to send a light signal that can catch up with an event horizon.

Again, you've got this garbled. The Rindler horizon is the path of an *outgoing* (positive x-direction) light signal that can't catch up with any of the accelerating observers as long as they continue to accelerate. It is true that a free-faller, once *inside* the horizon, can't send a light signal that can "catch up with the horizon" in either scenario (flat spacetime or gravity present), but that's just because a light signal can't catch up with another light signal emitted earlier in the same direction.

Give me a chance. You're talking as if I'd already disagreed with you and as far as I can remember I haven't done that once with flat space-time examples.

Even with the flat spacetime examples, you have repeatedly misstated things (some of which I point out above), and it's not clear to me that you have a good understanding of how those examples work in flat spacetime alone, even leaving out how they relate to the curved spacetime examples.

I genuinely believe that the whole space-time is covered using just Schwarzschild coordinates/the light cones can't tilt to 90 degrees/you can't cross an event horizon because it's moving inwards/you can't reach c.

Yes, I know you do. And you've just stated a key difference between your model and the standard model. In the standard model, the horizon moves *outwards*, not inwards.

In more detail: in the flat spacetime scenario, the Rindler horizon moves outwards at c ("outwards" meaning "in the positive x-direction"). The standard GR model also has the black hole horizon moving outwards at c (but spacetime is curved due to gravity so "moving outwards at c" at the radius of the horizon means the horizon stays at the same radius). This leads to a pretty close analogy between the Rindler horizon scenario and the black hole horizon scenario, in which the diagram I drew of the Rindler horizon scenario also represents the black hole scenario (not quite exactly because of the curvature of the black hole spacetime, but close enough to answer questions like whether a free-faller can reach and pass the black hole horizon). In that analogy, the free-faller lines in the diagram (dark magenta vertical lines) are the worldlines of observers falling towards the black hole; the accelerated lines (blue hyperbolas) are the worldlines of observers "hovering" above the black hole at a constant radius; and the Rindler horizon (red 45-degree line going up and to the right) is the black hole horizon.

You have a model in your head in which the horizon is moving inward, not outward. That model certainly can't be described by the diagram I drew, so you need to draw your own diagram. And since you already agree that my diagram describes the flat spacetime scenario, your model certainly can't be analogous to the flat spacetime scenario I described in any useful way (since in that scenario the Rindler horizon certainly moves outwards, not inwards). So where's your diagram?

Of course it's an ingoing null surface moving back in such a way that a free-falling observer could never catch up to it.

It's not "of course" at all. Your whole picture of this is garbled. See comments above, and further comments below.

It's moving back at c because c sets the limit to how far you can get in a certain amount of time

Relative to what? Time according to what observer?

, so the event horizon represents how close you can get to the singularity at the time that you're seeing it.

Seeing what? You never "see" the horizon unless you free-fall past it.

When you accelerate in flat space-time c gets faster, not slower, relative to your previous frame which prevents you from reaching c.

This is all garbled. In flat spacetime, "c" never changes; in fact, lightlike lines (the paths of light rays) and surfaces are the most "constant" things there are in flat spacetime (since the light cones never tilt, so light cones everywhere are exactly "lined up" with each other). One could say that a particular light ray that's moving away from you will always move away from you at c, no matter how hard you accelerate towards it, but even that doesn't justify saying that "c gets faster"--at best, it justifies saying that "c stays the same".

 

[PeterDonis]

A-wal, here's some more food for thought on the question of which direction the horizon moves (outwards or inwards). In both scenarios (Rindler horizon in flat spacetime, black hole horizon), there is a family of accelerating observers who are clearly accelerating outwards (positive x-direction in flat spacetime, positive r-direction with respect to the black hole). And yet both families of observers "hover" at a constant distance from the horizon (the specific distance at which each observer hovers depends on that observer's acceleration). Doesn't that mean the horizon has to be moving outwards, in order to stay at the same distance behind observers who are moving outwards?

 

[Passionflower]

This is all garbled. In flat spacetime, "c" never changes; in fact, lightlike lines (the paths of light rays) and surfaces are the most "constant" things there are in flat spacetime (since the light cones never tilt, so light cones everywhere are exactly "lined up" with each other). One could say that a particular light ray that's moving away from you will always move away from you at c, no matter how hard you accelerate towards it, but even that doesn't justify saying that "c gets faster"--at best, it justifies saying that "c stays the same".

The average speed of light between two points in an accelerating frame is always not equal to c except when the line between the two points is perpendicular to the direction of acceleration.

 

[PeterDonis]

The average speed of light between two points in an accelerating frame is always not equal to c except when the line between the two points is perpendicular to the direction of acceleration.

You raise an interesting point. The above statement is true provided we use "ruler distance" as the definition of "distance" for the purpose of calculating the average speed of light (we would of course use the round-trip light travel time, as measured by a given accelerated observer, as the definition of "time"). I'm using the terms as given in the Wikipedia article on Rindler coordinates:

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

But as the article notes, "ruler distance" is not the only definition of distance that can be adopted between accelerated observers in flat spacetime. If we instead use "radar distance", then the average speed of light *is* always equal to c (because radar distance is defined as the light travel time divided by c).

I assume that A-wal had in mind "ruler distance" when he said that "c gets faster as you accelerate". However, even then the statement as he phrased it is not quite true (at least as I understand his statement). What is true is the following: the ratio of "ruler distance" to round-trip light travel time does give an "average speed of light" greater than c for an accelerating observer, and how much greater does depend on the acceleration. However, for any *given* accelerating observer (i.e., for any given constant acceleration), the ratio (and hence the "average speed of light") is *constant*--it does *not* get larger as the observer continues to accelerate. This is because the ratio is a function solely of the "ruler distance" (h in the Wikipedia formulas) and the acceleration (which is in turn a function only of x_0 in the Wikipedia formulas--it is c^2 / x_0 for the "trailing" observer and c^2 / (x_0 + h) for the "leading" observer). So while it is correct, using this definition of "average speed", to say that "the speed of light *is* faster" than c if you are accelerating, it is *not* correct to say that "the speed of light *gets* faster" in the sense of continuing to get faster as you continue to accelerate (if your acceleration is constant)--by that criterion, my statement, that "c stays the same" is correct.

 

[PeterDonis]

Regarding A-wal's statement that "c gets faster as you accelerate, so you can't reach c", I should also mention that he is conflating two different observations. The statement that "c gets faster as you accelerate" I unpacked and critiqued in my last post. The statement that "you can't reach c" doesn't even apply, really, to the accelerating observer, because (a) relative to himself, he just remains at rest; and (b) as I noted in my last post, for him c does not "get faster as he accelerates", it just happens that c is "faster" (in the sense of ruler distance divided by round-trip light travel time) than if he were not accelerating. The accelerating observer "can't reach c" only relative to an inertial observer, who sees his velocity increasing asymptotically towards c but never reaching it. And to the inertial observer, of course, the speed of light *is* always the same--c.

 

[A-wal]

Let's make a pretty pattern. Try not to ruin it. Two accelerators, one free-faller. The first accelerator stays hovering one metre away from the horizon as measured by them. The second accelerator hovers one metre away and waits until the free-faller is also one metre away from the black hole as measured by both accelerators, then the second accelerator accelerates even more to move away from the black hole at .5c relative to the hovering accelerator. The same accelerator then uses the same amount of additional energy again to accelerate away even faster, but they're not now moving at c relative to the hoverer. They'll never be able to accelerate to c no matter how much energy they use.

The free-faller falls past the hoverer at .5c but when the free-faller has moved closer, where the strength of gravity is twice as strong (.75 of a metre I suppose), they're not now moving at c relative to the hoverer. They'll never be able to accelerate to c no matter how much energy they use. Whether it's from gravity or not makes no difference. The real velocity of the free-faller relative to the hoverer will be exactly the same as if they’d doubled their acceleration in flat space-time relative to an inertial observer.

Then at best you are only going to be able to get a shallow "pop-sci" understanding of GR, and even that is not likely.

Oh well. I'll just have to make do then.

In Schwarzschild coordinates the horizon is the 3 dimensional surface given by:
r=\frac{2GM}{c^2}
and so
\frac{d}{dt}\left( \frac{2GM}{c^2} \right) = 0
therefore the horizon has a constant position in Schwarzschild coordinates.

That's because Schwarzschild coordinates assume an everlasting black hole. A hoverer who's maintaining a constant distance from the horizon will be moving towards the singularity. How fast they move towards it depends on how far away they are, a bit like gravity itself really. They’ll get closer to the singularity as the event horizon gets closer to the singularity, as it looses mass. At the event horizon it’s losing the exact amount it needs for the event horizon to back at c. It’s the equivalent of an accelerator in flat space-time reducing their acceleration (the equivalent of the black hole loosing mass as it ages) to keep their Rindler horizon in the same place I suppose.

The position of the horizon in Schwarzschild coordinates is clearly not a function of "distance shortening".

It clearly is. The horizon limits how close you can get. There’s a limit to how close you can get because of ‘distance shortening’.

 

[A-wal]

You've forgotten that light can change direction as well as speed. Yes, the locally measured speed of light is always c relative to a timelike observer, but that doesn't mean light isn't subject to acceleration due to gravity. Gravity can still change its direction, and can still tilt the light cones from place to place so that "moving at c" doesn't mean the same thing at different places.

Of course. I hadn't forgotten. That's just not what I was talking about. You’re so quick to disagree that you keep taking what I say out of context. I get the impression you've already decided that nothing I say suggests anything and you're working backwards from this premise.

...because, as you just agreed, there is no hoverer at the horizon. More generally, there is no timelike observer at the horizon for the in-faller to "move at c" relative to. The only thing at the horizon that the in-faller moves at c relative to is an outgoing light ray at the horizon, and that's just because timelike observers always "move at c" relative to light rays (because light rays always move at c relative to timelike observers).

There doesn't need to be a hoverer at the horizon. They don't have to be in the same place to have a relative velocity to each other. The hoverer could be arbitrarily close to the horizon, but even that's not necessary. They can also take any relativistic effects into account no matter how far apart they are when they measure their speed.

Not true, as I just explained above. There is no timelike observer at the horizon.

No there isn’t, but not for the reason that you think.

I see what you are getting at here, but it doesn't prove what you think it does. As I've said before, you can't just say "approach c" in the abstract; you have to say *what* you are trying to "approach c" relative to. It looks like what you are saying above is describing one timelike observer accelerating relative to another timelike observer (who is assumed to be "at rest"). If that's the case, it only applies in cases where there are two timelike observers with the required properties. At the horizon there aren't; there are no timelike observers "hovering" at rest at the horizon. So your analogy fails.

Of course there aren’t any time-like observers at rest at the horizon, because there can't be. If it were possible to reach an event horizon then it would obviously be possible to hover there. If one force can pull you in then it's not possible (and to be honest, completely ridiculous) to say that no amount energy can move you away. The analogy holds BECAUSE there's no time-like observers hovering at the horizon, because that would be the equivalent of accelerating to c in every frame.

Also, I'm not sure why you insist on using the term "catch up to c". If you are trying to describe one timelike observer accelerating relative to another, then (by analogy with the Rindler horizon scenario) it would be better to say that a light ray emitted from a certain point can't catch up to the accelerating observer (as long as he continues to accelerate). If you really mean that the accelerating observer can't catch up to something, what is it? It can't just be "c" in the abstract, because that's not a physical object, it's just an abstraction.

Their own light waves, which would be stretched into a straight line in their direction of motion behind them (infinitely long wavelength) and a straight horizontal line directly in front of them (infinitely short wavelength). I'm using that term to show that you can't catch up to your own light. It doesn't work like sound. You can't catch it any easier using gravity than you can through conventional acceleration, but that's exactly what you would have done if you were able to reach an event horizon.

No, I feel safe in making the claims I'm making because I understand the standard general relativistic model that generates them, and how that model is both logically consistent and consistent with all the experimental data we currently have. The claims only sound outrageous to you because you don't have that understanding.

I really don't think it's logically consistent and just because I don't agree you doesn't mean I don't understand. It seemed to make sense to start with, until I actually thought about it. Then I realized just how stupid it was.

If you want me to "prove" that the standard GR model is the *only* model with those properties, then of course I can't. But when you ask "what do you think seems more likely?", you're basically invoking Occam's Razor, and to even apply that I would have to have another model in front of me that has both of the above properties. So *you* have to show me that your model, as an alternative to standard GR, has both of those properties, before I can even consider it. You haven't done that; you haven't even convinced me that your model is logically consistent (you keep assuming things that you are supposed to be deriving as conclusions in your model), and you also haven't even shown that your model accounts for all the data in the particular scenario we're discussing (the spacetime around a gravitating object), let alone *all* the data that GR accounts for.

I think I have shown that it's logically consistent. Which parts are you having trouble with?

If you mean "the answer stays the same" in the sense that neither choice proves your case, I agree. But I don't think that's what you meant.

I mean that it's a little strange that you seem to be asserting that I can't change the parameters of a thought experiment to look at it in a different way. And the answer will be the same no matter how you look at it.

But if its velocity is always c locally, in what sense does it "slow down"? Relative to what? You need to unpack this statement further.

It slows down from the point of view of the observer who emitted the light. The light leaves them at c but slows as it gets closer to the horizon and gets more ‘distance shortened’. How sharply it slows depends on how close you are to the horizon when you emit the light. How would that work if you could reach the horizon?

How are you measuring the distance that you move through?

By working it out based on the mass of the black hole and the energy expended so far, or by comparing yourself to a row of hovering observers for reference.

What do you mean by this? Do you mean that the "free-faller" lines in the diagram (the dark magenta vertical lines) are analogous to the worldlines that objects free-falling into a black hole would follow? If so, then I agree, and you have just agreed that objects free-falling into a black hole *can* cross the horizon (because they certainly do in the diagram). If you meant something else, then you need to describe what you meant, because you aren't describing anything that the diagram shows.

I meant it looks as if those lines wouldn’t cross the horizon and the angle they’re at looks like what you get when comparing an accelerators velocity with c, and that should be exactly the same as an object approaching an event horizon.

In a sense, yes, the Rindler horizon is *defined* by the family of accelerating observers. But once defined, it is a perfectly ordinary lightlike surface in the spacetime, and all observers agree on which surface it is, and free-fallers can cross it (in the ingoing direction).

But accelerators can't. The same thing happens in curved space-time around a black hole. There will be a point when a signal sent from a more distant observer can’t reach a free-faller because the free-faller is being accelerated by gravity. The Rindler horizon gets closer to the free-faller as they accelerate harder due to the increase in gravity, just like it does when you use energy to accelerate in flat space-time.

Nope, you've *still* got this backwards despite repeated corrections from me. I was talking about free-fallers sending *outgoing* signals. You are talking about distant observers sending *ingoing* signals. They're not the same.

No they're not. Your confusion is because you’re assuming the free-fallers are equivalent in both examples. They’re not. Think of the distant observer as being far enough away that the barely need to accelerate, or free-falling very far behind if you prefer. This observer is the equivalent to the “free-faller” in flat space-time. The in-faller is the equivalent to the accelerator because they’re being accelerated by gravity. You’re the one who keeps getting it backwards. I would have thought you’d have figured that out by now.

Incorrect. I already addressed this above.

You addressed it yes. You've addressed a lot (still not all) of my points showing how you think it works without giving any explanation that actually shows why it works the way you think it does.

Again, you've got this garbled. The Rindler horizon is the path of an *outgoing* (positive x-direction) light signal that can't catch up with any of the accelerating observers as long as they continue to accelerate. It is true that a free-faller, once *inside* the horizon, can't send a light signal that can "catch up with the horizon" in either scenario (flat spacetime or gravity present), but that's just because a light signal can't catch up with another light signal emitted earlier in the same direction.

It's not garbled. It's the garbled way that you insist on looking at it that makes it seem that way. Acceleration in flat space-time pushes outwards and acceleration through curved space-time (gravity) pulls towards. That’s why the event horizon moves inwards and the Rindler horizon moves outwards, and why you’re getting muddled.

Even with the flat spacetime examples, you have repeatedly misstated things (some of which I point out above), and it's not clear to me that you have a good understanding of how those examples work in flat spacetime alone, even leaving out how they relate to the curved spacetime examples.

I know how they work in flat space-time.

Yes, I know you do. And you've just stated a key difference between your model and the standard model. In the standard model, the horizon moves *outwards*, not inwards.

In more detail: in the flat spacetime scenario, the Rindler horizon moves outwards at c ("outwards" meaning "in the positive x-direction"). The standard GR model also has the black hole horizon moving outwards at c (but spacetime is curved due to gravity so "moving outwards at c" at the radius of the horizon means the horizon stays at the same radius). This leads to a pretty close analogy between the Rindler horizon scenario and the black hole horizon scenario, in which the diagram I drew of the Rindler horizon scenario also represents the black hole scenario (not quite exactly because of the curvature of the black hole spacetime, but close enough to answer questions like whether a free-faller can reach and pass the black hole horizon). In that analogy, the free-faller lines in the diagram (dark magenta vertical lines) are the worldlines of observers falling towards the black hole; the accelerated lines (blue hyperbolas) are the worldlines of observers "hovering" above the black hole at a constant radius; and the Rindler horizon (red 45-degree line going up and to the right) is the black hole horizon.

What do you mean “not quite exactly because of the curvature of the black hole spacetime”? It’s the same situation in reverse.

You have a model in your head in which the horizon is moving inward, not outward. That model certainly can't be described by the diagram I drew, so you need to draw your own diagram. And since you already agree that my diagram describes the flat spacetime scenario, your model certainly can't be analogous to the flat spacetime scenario I described in any useful way (since in that scenario the Rindler horizon certainly moves outwards, not inwards). So where's your diagram?

What diagram? It’s analogous to the flat space-time scenario in a very useful way. The Rindler horizon appears because of acceleration. The event horizon appears because of gravity. Acceleration pushes everything away and gravity pulls everything in. You can't reach c in either case so you can't reach either horizon. Reaching an event horizon is the equivalent of your Rindler horizon catching up with you.

It's not "of course" at all. Your whole picture of this is garbled. See comments above, and further comments below.

It’s not my picture that’s garbled! See this whole bloody thread!

Relative to what? Time according to what observer?

Relative to anything according to any observer.

Seeing what? You never "see" the horizon unless you free-fall past it.

The event horizon is the closest any object can get to the singularity at that time. You can get closer over time but to cover the amount of space-time needed to reach a singularity before it's gone you would need to move at c, because the horizon is moving towards the singularity at c.

This is all garbled. In flat spacetime, "c" never changes; in fact, lightlike lines (the paths of light rays) and surfaces are the most "constant" things there are in flat spacetime (since the light cones never tilt, so light cones everywhere are exactly "lined up" with each other). One could say that a particular light ray that's moving away from you will always move away from you at c, no matter how hard you accelerate towards it, but even that doesn't justify saying that "c gets faster"--at best, it justifies saying that "c stays the same".

If you accelerate then light in your new frame is moving faster than c relative to your old frame (c + the difference between your new velocity and your old one), but not if you were actually in your old frame obviously. It has to to stay constant.

A-wal, here's some more food for thought on the question of which direction the horizon moves (outwards or inwards). In both scenarios (Rindler horizon in flat spacetime, black hole horizon), there is a family of accelerating observers who are clearly accelerating outwards (positive x-direction in flat spacetime, positive r-direction with respect to the black hole). And yet both families of observers "hover" at a constant distance from the horizon (the specific distance at which each observer hovers depends on that observer's acceleration). Doesn't that mean the horizon has to be moving outwards, in order to stay at the same distance behind observers who are moving outwards?

That’s a very small meal. The horizon is moving inwards at c but slower than that at any distance away because that’s how time dilation and length contraction work. You already said: "(but spacetime is curved due to gravity so "moving outwards at c" at the radius of the horizon means the horizon stays at the same radius)”. So you agree that time dilation is infinite at the horizon, and that even if an object could somehow be magically accelerated to c it would be frozen in time? So in what sense is it possible to be accelerated to c? It's curved inwards by gravity, not outwards.

The average speed of light between two points in an accelerating frame is always not equal to c except when the line between the two points is perpendicular to the direction of acceleration.

You raise an interesting point. The above statement is true provided we use "ruler distance" as the definition of "distance" for the purpose of calculating the average speed of light (we would of course use the round-trip light travel time, as measured by a given accelerated observer, as the definition of "time"). I'm using the terms as given in the Wikipedia article on Rindler coordinates:

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

But as the article notes, "ruler distance" is not the only definition of distance that can be adopted between accelerated observers in flat spacetime. If we instead use "radar distance", then the average speed of light *is* always equal to c (because radar distance is defined as the light travel time divided by c).

I assume that A-wal had in mind "ruler distance" when he said that "c gets faster as you accelerate". However, even then the statement as he phrased it is not quite true (at least as I understand his statement). What is true is the following: the ratio of "ruler distance" to round-trip light travel time does give an "average speed of light" greater than c for an accelerating observer, and how much greater does depend on the acceleration. However, for any *given* accelerating observer (i.e., for any given constant acceleration), the ratio (and hence the "average speed of light") is *constant*--it does *not* get larger as the observer continues to accelerate. This is because the ratio is a function solely of the "ruler distance" (h in the Wikipedia formulas) and the acceleration (which is in turn a function only of x_0 in the Wikipedia formulas--it is c^2 / x_0 for the "trailing" observer and c^2 / (x_0 + h) for the "leading" observer). So while it is correct, using this definition of "average speed", to say that "the speed of light *is* faster" than c if you are accelerating, it is *not* correct to say that "the speed of light *gets* faster" in the sense of continuing to get faster as you continue to accelerate (if your acceleration is constant)--by that criterion, my statement, that "c stays the same" is correct.

Yes, c stays the same if your acceleration remains constant. If you increase your acceleration it slows down and if you decrease your acceleration it speeds up again.

Regarding A-wal's statement that "c gets faster as you accelerate, so you can't reach c", I should also mention that he is conflating two different observations. The statement that "c gets faster as you accelerate" I unpacked and critiqued in my last post. The statement that "you can't reach c" doesn't even apply, really, to the accelerating observer, because (a) relative to himself, he just remains at rest; and (b) as I noted in my last post, for him c does not "get faster as he accelerates", it just happens that c is "faster" (in the sense of ruler distance divided by round-trip light travel time) than if he were not accelerating. The accelerating observer "can't reach c" only relative to an inertial observer, who sees his velocity increasing asymptotically towards c but never reaching it. And to the inertial observer, of course, the speed of light *is* always the same--c.

The accelerator can’t reach c in the sense that c slows down as they accelerate but it can never slow to zero because the decrease in its speed itself decreases as they accelerate. That’s why adding velocities won’t let you reach c, whether it’s acceleration through energy or through gravity. It would even be the exact same formula for an object approaching an event horizon and an object approaching c if its acceleration were to increase to mimic the inverse square of gravity. Either way, the event horizon or c increases their velocity by the same amount as the other would as you approach them, which stops you from ever being able to reach them.

 

[A-wal]

You think an objects can cross an event horizon but that's obviously not how it works, which I've shown, but you've got this stupid idea in heads that gravity isn't a real force when it obviously is, which is why you've got this messed up notion that it can ignore SR and accelerate an object to any relative velocity. I can see how it works because I'm not fumbling about in the dark trying to use equations as dimly lit torches. You've already decided how it works and anything that suggests otherwise is assumed to be misleading. This is pointless. I've had more constructive conversations with god worshipers. I give up. You win. Stubbornness triumphs over common sense. I hope you proud of yourselves.

 

[PeterDonis]

Of course. I hadn't forgotten. That's just not what I was talking about. You’re so quick to disagree that you keep taking what I say out of context. I get the impression you've already decided that nothing I say suggests anything and you're working backwards from this premise.

I've decided no such thing. It's true that much of what you say makes little or no sense to me. Some of that is because you have a different mental model in your head than I do, but when you are able to describe the model coherently, I can understand it. Unfortunately, every time that's happened, I've pointed out an inconsistency in your model and you've backed off from it and lapsed back into vague hand-waving. This makes me suspect that the reason much of what you say makes no sense to me is that you don't actually have a single, coherent model; you have a patchwork of ideas but you haven't fit them into a single coherent picture.

Of course this is my perception, and it may be wrong. But I'm in the same position relative to your model, as you appear to be relative to the standard model. You don't understand how a single, coherent model could produce all the conclusions that I claim the standard model produces. Similarly, I don't see how a single, coherent model can produce all the claims you have made for your model. I've been trying to work back to first principles, to find *some* starting point that we have in common, and then work forward from there. But most of the time you keep on asserting conclusions, when we haven't even reached an agreed set of premises to start from. That makes it difficult to have a productive discussion.

There doesn't need to be a hoverer at the horizon. They don't have to be in the same place to have a relative velocity to each other.

They do in the standard model; in standard GR there is no meaning to the term "relative velocity" (at least, no *physical* meaning that allows you to say things like "the relative velocity has to be less than that of light") unless both objects are at the same event (more precisely, in the same local patch of spacetime, small enough that tidal effects can be neglected). If this term has a definite meaning in your model, what is it? How is it defined? How is it measured? Please be specific.

If it were possible to reach an event horizon then it would obviously be possible to hover there.

Why? In the standard GR model, this is a non sequitur. There is no requirement to be able to "hover" (in the sense of staying at the same radial coordinate) at every place. If this is a requirement in your model, why? What makes it a requirement? And how can you rule out, a priori, models that don't meet this requirement? Again, please be specific.

If one force can pull you in then it's not possible (and to be honest, completely ridiculous) to say that no amount energy can move you away.

There's no "force" pulling you in. Spacetime is curved in such a way that freely falling objects move inward; since they're freely falling, they feel no acceleration and are therefore subject to no force. On what basis does your model assert that a force is acting, when it is not felt by the object it's supposedly acting on?

(I note also that here, as later on in your post, you are basically asserting that a freely falling object is physically equivalent to one feeling nonzero proper acceleration. I comment on that further below.)

Their own light waves, which would be stretched into a straight line in their direction of motion behind them (infinitely long wavelength) and a straight horizontal line directly in front of them (infinitely short wavelength). I'm using that term to show that you can't catch up to your own light.

I agree you can't catch up to your own light. But the only sense I can make of this in terms of "moving at c" serving as some kind of fixed point relative to which all motion can be judged, is by using the paths of light rays--i.e., the light cones. And in curved spacetime, they're tilted by gravity, so if you use them as your reference for motion, you will find that, as I said before, "moving at c" means different things in different places. So it's no longer the kind of fixed background you've been assuming.

I really don't think it's logically consistent and just because I don't agree you doesn't mean I don't understand. It seemed to make sense to start with, until I actually thought about it. Then I realized just how stupid it was.

I would say you haven't thought hard enough. For one thing, whenever I point out an assumption you've made (I pointed out at least two earlier in this post) and ask you why it has to be true, you have no answer. You just say "it's obvious" or something like that. Sorry, but it's *not* obvious; not if you've actually thought about it. The fact that you have failed to question all these "obvious" assumptions tells me that you haven't really thought about it; you may have spent plenty of time at it, but that's not enough.

But I'm not really interested in what you think is wrong with the standard GR model. I'm interested in trying to get to *some* set of starting premises that we can both agree on. Then I would like to see you build up your model of how things work from those premises. If the standard model really is wrong, this procedure ought to show us that at some point.

I think I have shown that it's logically consistent. Which parts are you having trouble with?

Um, the parts you haven't even given? Like, as I've said, getting to a set of premises that we can both agree on? Until we have that, you have no right to assume that you have established *any* of the claims you've made that contradict the standard model. You can't just assert them. You have to show how they follow from agreed premises. Since we don't even have the agreed premises yet, I don't see how you can think you've accomplished anything like this.

It slows down from the point of view of the observer who emitted the light.

How does the observer who emitted the light find out that it has slowed down? Does he measure the slowdown somehow? If so, how? Or does he get the information from some other observer? If so, what information, and how does it tell him that the light has slowed down?

Your confusion is because you’re assuming the free-fallers are equivalent in both examples. They’re not. Think of the distant observer as being far enough away that the barely need to accelerate, or free-falling very far behind if you prefer. This observer is the equivalent to the “free-faller” in flat space-time. The in-faller is the equivalent to the accelerator because they’re being accelerated by gravity. You’re the one who keeps getting it backwards. I would have thought you’d have figured that out by now.

In other words, you believe that an observer who is in free fall, feeling *zero* acceleration (remember you agreed, after quite a bit of back and forth, that such observers do in fact feel exactly zero acceleration) is physically equivalent to an observer feeling *nonzero* proper acceleration. Well, here's another assumption that you are making that is different from standard GR. In standard GR, you determine whether observers are physically "equivalent" by the simple method of comparing their accelerometer readings. On what basis does your model claim that a freely falling observer is physically equivalent to an observer feeling nonzero proper acceleration?

The horizon is moving inwards at c but slower than that at any distance away because that’s how time dilation and length contraction work. You already said: "(but spacetime is curved due to gravity so "moving outwards at c" at the radius of the horizon means the horizon stays at the same radius)”. So you agree that time dilation is infinite at the horizon, and that even if an object could somehow be magically accelerated to c it would be frozen in time? So in what sense is it possible to be accelerated to c? It's curved inwards by gravity, not outwards.

Spacetime is curved *inwards* at the horizon just enough to make a light ray moving *outwards* (i.e., the horizon itself) stay at the same radius.

 

[PeterDonis]

You think an objects can cross an event horizon but that's obviously not how it works, which I've shown, but you've got this stupid idea in heads that gravity isn't a real force when it obviously is,

As I noted in my last post, why is this obvious? This "force" you speak of is not felt by the objects it is supposedly acting on; an idealized, point-like object moving solely under the influence of gravity feels exactly zero acceleration, as you agreed earlier in this thread. So how do you claim that this object, which feels no acceleration, is being subjected to a force?

I can understand why you're frustrated. You are operating with a set of assumptions that I (and DaleSpam and others) do not share. It's always difficult to have a discussion when the parties don't even share basic assumptions. I too have had such frustrating discussions with what you call "god worshipers", because I don't share their assumptions and I can't get them to even consider mine. But I'm willing to consider yours, *if* you can justify them, and then build up a coherent model based on them. What I can't do is talk about conclusions without getting clear about the premises first, when it's clear that the premises are different enough that they need to be specified explicitly.

The question I asked above is an example. You assume that there is a clear, obvious definition of "force" under which gravity is a force. What is that clear, obvious definition? Spell it out. Is it just standard Newtonian physics? Well, standard Newtonian physics is falsified by experiment. So that's out. What's next? What model is there that says gravity is a force, and still gives all the right answers?

Or, alternatively, suppose I offer the following definition: a "force" is any of the four fundamental quantum interactions: strong, electromagnetic, weak, or gravitational. So gravity *is* a force! But when we try to construct a theory of this force, starting from standard quantum field theory, as was done during the 1960's, this theory (or at least its classical limit) turns out to be--standard General Relativity. So there *is* a model that says gravity is a force, and gives all the right answers: standard GR! But of course, that depends on adopting the definition of "force" I gave just now. Is that your definition? If so, then we're good--you agree with standard GR, and standard GR predicts that a black hole horizon can be crossed. If not--then what *is* your definition?

Of course nothing *requires* you to answer these questions. I said quite a while ago in this thread that if you want to take the position that GR is, in your opinion, an approximate theory that happens to work in the Solar System, with binary pulsars, and so on, but which will be found to be inaccurate when we can make a close enough study of black holes, I have no quarrel. We won't know until the evidence is in. That's an empirical question, not a theoretical one.

But if we are talking about theory, then I'm sorry, but you have not shown what you think you have shown. All you have shown is that you don't understand how a consistent model can be constructed using assumptions that are different from yours. Maybe those assumptions are wrong; we won't know for sure until the evidence is in. But that won't make the model based on them inconsistent; it will just make it incorrect.

 

[Dale]

They'll never be able to accelerate to c no matter how much energy they use.

Agreed. An inertial observer free-falling across an event horizon will always remain timelike, never lightlike.

That's because Schwarzschild coordinates assume an everlasting black hole. A hoverer who's maintaining a constant distance from the horizon will be moving towards the singularity. How fast they move towards it depends on how far away they are, a bit like gravity itself really. They’ll get closer to the singularity as the event horizon gets closer to the singularity, as it looses mass. At the event horizon it’s losing the exact amount it needs for the event horizon to back at c.

Can you derive any of this?

 

[PeterDonis]

Hmm, for some reason I missed A-wal's other post.

Let's make a pretty pattern. Try not to ruin it. Two accelerators, one free-faller. The first accelerator stays hovering one metre away from the horizon as measured by them. The second accelerator hovers one metre away

I assume you mean by this "at the same distance from the horizon as the first accelerator", correct?

and waits until the free-faller is also one metre away from the black hole as measured by both accelerators,

I assume you mean "until the instant that the free-faller is just passing both accelerators", correct?

then the second accelerator accelerates even more to move away from the black hole at .5c relative to the hovering accelerator. The same accelerator then uses the same amount of additional energy again to accelerate away even faster, but they're not now moving at c relative to the hoverer. They'll never be able to accelerate to c no matter how much energy they use.

Sort of true; the second accelerator could in principle accelerate hard enough that, by the time he reached an ultra-relativistic velocity relative to the first accelerator (or "hoverer"--I assume these refer to the same observer), meaning a relativistic "gamma" factor much, much larger than 1, the second accelerator would still be close enough to the first accelerator that tidal effects would be negligible over the distance between them, so there would still be a physical meaning to their "relative velocity". In this case, yes, the second accelerator could never actually reach c relative to the hoverer, no matter how hard he accelerated. But once the second accelerator gets far enough away from the hoverer that tidal effects are important, there is no longer a physically meaningful way to specify their relative velocity (where "physically meaningful" means "in a way that requires the relative velocity to be less than c").

The free-faller falls past the hoverer at .5c but when the free-faller has moved closer, where the strength of gravity is twice as strong (.75 of a metre I suppose), they're not now moving at c relative to the hoverer.

Again, sort of true; for the free-faller's "relative velocity" to the hoverer to be physically meaningful, tidal effects have to be negligible over the distance between them. This can be done by making the hole's mass large enough for the hoverer's distance above the horizon to be small compared to the distance over which tidal effects are important. In that case, yes, as long as the free-faller is above the horizon, he will be moving at less than c relative to the hoverer.

They'll never be able to accelerate to c no matter how much energy they use. Whether it's from gravity or not makes no difference. The real velocity of the free-faller relative to the hoverer will be exactly the same as if they’d doubled their acceleration in flat space-time relative to an inertial observer.

As long as the free-faller remains above the horizon, yes, this is true, although I would put it in reverse, so to speak: the hoverer's velocity relative to the free-faller will be exactly the same as if the hoverer were accelerating in flat spacetime while the free-faller stayed at rest. This makes it clear which pairs of observers are "equivalent" in the flat to curved spacetime analogy, at least on the standard view. On your view, apparently the freely falling observer in curved spacetime (who feels zero proper acceleration) is supposed to be equivalent to the accelerating observer in flat spacetime (who feels nonzero proper acceleration). Which leaves me confused as to what observer in flat spacetime the hoverer in curved spacetime is supposed to be equivalent to.

When the free-faller reaches the horizon, light from the free-faller can no longer reach the hoverer, so the hoverer will never observe that portion of the free-faller's worldline (as long as he continues to hover). So the hoverer will never see the free-faller "move at c", or faster. Again, this is the standard view. I'm not sure what is supposed to happen to the free-faller on your view; I know you claim that the free-faller just continues to slow down indefinitely, but since you haven't explained how I'm supposed to relate the free-faller's "clock time" to the "clock time" of the hoverer, I don't know how you are determining this. You can't be using the standard GR relationship, because in standard GR the free-faller reaches the horizon after a finite time by his clock, not an infinite time.

That's because Schwarzschild coordinates assume an everlasting black hole.

And we established quite a while ago that even for this case, an everlasting black hole, you believe the horizon cannot be reached. So all this about a black hole losing mass is irrelevant.

 

[A-wal]

I've decided no such thing. It's true that much of what you say makes little or no sense to me. Some of that is because you have a different mental model in your head than I do, but when you are able to describe the model coherently, I can understand it. Unfortunately, every time that's happened, I've pointed out an inconsistency in your model and you've backed off from it and lapsed back into vague hand-waving. This makes me suspect that the reason much of what you say makes no sense to me is that you don't actually have a single, coherent model; you have a patchwork of ideas but you haven't fit them into a single coherent picture.

Give me one example of when I've lapsed into "vague wand-waving", or even one example of when you've pointed out an inconsistency with anything other than the standard model. The "patchwork of ideas" are from the single coherent picture, not the other way round.

Of course this is my perception, and it may be wrong. But I'm in the same position relative to your model, as you appear to be relative to the standard model. You don't understand how a single, coherent model could produce all the conclusions that I claim the standard model produces. Similarly, I don't see how a single, coherent model can produce all the claims you have made for your model. I've been trying to work back to first principles, to find *some* starting point that we have in common, and then work forward from there. But most of the time you keep on asserting conclusions, when we haven't even reached an agreed set of premises to start from. That makes it difficult to have a productive discussion.

We agree on SR. The basic premise is that trying to reach an event horizon using gravity is the same as trying to accelerate to c in flat space-time.

They do in the standard model; in standard GR there is no meaning to the term "relative velocity" (at least, no *physical* meaning that allows you to say things like "the relative velocity has to be less than that of light") unless both objects are at the same event (more precisely, in the same local patch of spacetime, small enough that tidal effects can be neglected). If this term has a definite meaning in your model, what is it? How is it defined? How is it measured? Please be specific.

That's like saying an inertial observer can't have a relative velocity to an accelerator in flat space-time. A difference in tidal force is the same as a difference in acceleration.

Why? In the standard GR model, this is a non sequitur. There is no requirement to be able to "hover" (in the sense of staying at the same radial coordinate) at every place. If this is a requirement in your model, why? What makes it a requirement? And how can you rule out, a priori, models that don't meet this requirement? Again, please be specific.

Because it's silly to say a finite amount of one force can't be overpowered by an infinite amount of any other.

There's no "force" pulling you in. Spacetime is curved in such a way that freely falling objects move inward; since they're freely falling, they feel no acceleration and are therefore subject to no force. On what basis does your model assert that a force is acting, when it is not felt by the object it's supposedly acting on?

On the basis that it affects the velocity of objects relative to other objects, diluted as the inverse square of the distance, just like the electro-magnetic force does. One curves in, one curves out.

I agree you can't catch up to your own light. But the only sense I can make of this in terms of "moving at c" serving as some kind of fixed point relative to which all motion can be judged, is by using the paths of light rays--i.e., the light cones. And in curved spacetime, they're tilted by gravity, so if you use them as your reference for motion, you will find that, as I said before, "moving at c" means different things in different places. So it's no longer the kind of fixed background you've been assuming.

No, not relative to a fixed point. I was talking about a free-falling observers own light and how light would stop behaving like light at the horizon. You would catch up, like a sonic boom. The light from in-falling objects ahead of you doesn't reach the horizon. So as you get closer, the light from you and everything else that's ever approached the singularity doesn't cross the horizon, ever! It is a fixed background in four dimensions and using the light cones doesn’t prove anything because the way you use them assumes what you’re trying to prove.

I would say you haven't thought hard enough. For one thing, whenever I point out an assumption you've made (I pointed out at least two earlier in this post) and ask you why it has to be true, you have no answer. You just say "it's obvious" or something like that. Sorry, but it's *not* obvious; not if you've actually thought about it. The fact that you have failed to question all these "obvious" assumptions tells me that you haven't really thought about it; you may have spent plenty of time at it, but that's not enough

It is obvious. The fact that you can't see that tells me that you think memorising a model qualifies as thinking about it. I've spent more time writing this than I have thinking about it. I used to think about it a bit. Now I just think how to describe it.

But I'm not really interested in what you think is wrong with the standard GR model. I'm interested in trying to get to *some* set of starting premises that we can both agree on. Then I would like to see you build up your model of how things work from those premises. If the standard model really is wrong, this procedure ought to show us that at some point.

What happens at the event horizon shows us that. It shows that the distance between the event horizon and the singularity is proportional to your distance from it (inverse square), in the same way that 'distance shortening' in flat space-time is directly proportional to your acceleration.

Um, the parts you haven't even given? Like, as I've said, getting to a set of premises that we can both agree on? Until we have that, you have no right to assume that you have established *any* of the claims you've made that contradict the standard model. You can't just assert them. You have to show how they follow from agreed premises. Since we don't even have the agreed premises yet, I don't see how you can think you've accomplished anything like this.

My right to assume that I've established a premise doesn't depend on whether or not I can get you to agree.

How does the observer who emitted the light find out that it has slowed down? Does he measure the slowdown somehow? If so, how? Or does he get the information from some other observer? If so, what information, and how does it tell him that the light has slowed down?

They compare notes with the accelerating hoverers of the riverbed as they free-fall past them.

In other words, you believe that an observer who is in free fall, feeling *zero* acceleration (remember you agreed, after quite a bit of back and forth, that such observers do in fact feel exactly zero acceleration) is physically equivalent to an observer feeling *nonzero* proper acceleration. Well, here's another assumption that you are making that is different from standard GR. In standard GR, you determine whether observers are physically "equivalent" by the simple method of comparing their accelerometer readings. On what basis does your model claim that a freely falling observer is physically equivalent to an observer feeling nonzero proper acceleration?

A free-faller feels non-zero acceleration as tidal force. A point-like object wouldn't feel tidal force but it wouldn't feel acceleration in flat space-time neither.

Spacetime is curved *inwards* at the horizon just enough to make a light ray moving *outwards* (i.e., the horizon itself) stay at the same radius.

So it would be curved by exactly the same amount it would be as if you had accelerated to c. Infinite time dilation keeping the light frozen in time. The ingoing light never gets frozen. It just keeps slowing down, like it does when you accelerate in flat space-time.

As I noted in my last post, why is this obvious? This "force" you speak of is not felt by the objects it is supposedly acting on; an idealized, point-like object moving solely under the influence of gravity feels exactly zero acceleration, as you agreed earlier in this thread. So how do you claim that this object, which feels no acceleration, is being subjected to a force?

It's being accelerated relative to other observers who are also accelerating purely under the influence of the same force.

I can understand why you're frustrated. You are operating with a set of assumptions that I (and DaleSpam and others) do not share. It's always difficult to have a discussion when the parties don't even share basic assumptions. I too have had such frustrating discussions with what you call "god worshipers", because I don't share their assumptions and I can't get them to even consider mine. But I'm willing to consider yours, *if* you can justify them, and then build up a coherent model based on them. What I can't do is talk about conclusions without getting clear about the premises first, when it's clear that the premises are different enough that they need to be specified explicitly.

The premise is that gravity is a force that pulls instead of pushing, and it's weak because it's caused by mass rather than energy. Actually that's more of a conclusion than a premise, but it works both ways.

The question I asked above is an example. You assume that there is a clear, obvious definition of "force" under which gravity is a force. What is that clear, obvious definition? Spell it out. Is it just standard Newtonian physics? Well, standard Newtonian physics is falsified by experiment. So that's out. What's next? What model is there that says gravity is a force, and still gives all the right answers?

GR. It would give all the right answers if it was done right.

Or, alternatively, suppose I offer the following definition: a "force" is any of the four fundamental quantum interactions: strong, electromagnetic, weak, or gravitational. So gravity *is* a force! But when we try to construct a theory of this force, starting from standard quantum field theory, as was done during the 1960's, this theory (or at least its classical limit) turns out to be--standard General Relativity. So there *is* a model that says gravity is a force, and gives all the right answers: standard GR! But of course, that depends on adopting the definition of "force" I gave just now. Is that your definition? If so, then we're good--you agree with standard GR, and standard GR predicts that a black hole horizon can be crossed. If not--then what *is* your definition?

A force is anything that creates a curved line in space-time. You can effectively remove a force simply by not resisting it. Then the only way you know your being influenced by a force is your acceleration relative to other objects.

Of course nothing *requires* you to answer these questions. I said quite a while ago in this thread that if you want to take the position that GR is, in your opinion, an approximate theory that happens to work in the Solar System, with binary pulsars, and so on, but which will be found to be inaccurate when we can make a close enough study of black holes, I have no quarrel. We won't know until the evidence is in. That's an empirical question, not a theoretical one.

Theoretically it can't, and there's no reason it should work the way you're describing.

But if we are talking about theory, then I'm sorry, but you have not shown what you think you have shown. All you have shown is that you don't understand how a consistent model can be constructed using assumptions that are different from yours. Maybe those assumptions are wrong; we won't know for sure until the evidence is in. But that won't make the model based on them inconsistent; it will just make it incorrect.

The model becomes inconsistent for lots of reasons as soon as an object is allowed to reach an event horizon. It moves inwards faster the faster you moving relative to it, like light.

Agreed. An inertial observer free-falling across an event horizon will always remain timelike, never lightlike.

But an inertial observer free-falling across an event horizon would become light-like.

Can you derive any of this?

The event horizon is how close you can get to the singularity at that time and you can move at c, and its radius is proportional to its mass.

I assume you mean by this "at the same distance from the horizon as the first accelerator", correct?

Yes.

I assume you mean "until the instant that the free-faller is just passing both accelerators", correct?

Yes.

Sort of true; the second accelerator could in principle accelerate hard enough that, by the time he reached an ultra-relativistic velocity relative to the first accelerator (or "hoverer"--I assume these refer to the same observer), meaning a relativistic "gamma" factor much, much larger than 1, the second accelerator would still be close enough to the first accelerator that tidal effects would be negligible over the distance between them, so there would still be a physical meaning to their "relative velocity". In this case, yes, the second accelerator could never actually reach c relative to the hoverer, no matter how hard he accelerated. But once the second accelerator gets far enough away from the hoverer that tidal effects are important, there is no longer a physically meaningful way to specify their relative velocity (where "physically meaningful" means "in a way that requires the relative velocity to be less than c").

Yes there is. Tidal force is the same as proper acceleration. The one accelerating away will be feeling less tidal force as the distance increases.

Again, sort of true; for the free-faller's "relative velocity" to the hoverer to be physically meaningful, tidal effects have to be negligible over the distance between them. This can be done by making the hole's mass large enough for the hoverer's distance above the horizon to be small compared to the distance over which tidal effects are important. In that case, yes, as long as the free-faller is above the horizon, he will be moving at less than c relative to the hoverer.

They'll be moving at less than c relative to the hoverer no matter how hard they accelerate. The 'distance shortening' when trying to reach the event horizon is the same as you would get in flat space-time trying to reach c relative to an inertial observer.

As long as the free-faller remains above the horizon, yes, this is true, although I would put it in reverse, so to speak: the hoverer's velocity relative to the free-faller will be exactly the same as if the hoverer were accelerating in flat spacetime while the free-faller stayed at rest. This makes it clear which pairs of observers are "equivalent" in the flat to curved spacetime analogy, at least on the standard view. On your view, apparently the freely falling observer in curved spacetime (who feels zero proper acceleration) is supposed to be equivalent to the accelerating observer in flat spacetime (who feels nonzero proper acceleration). Which leaves me confused as to what observer in flat spacetime the hoverer in curved spacetime is supposed to be equivalent to.

Well, it just leaves the inertial one. The hoverers gravitational acceleration is canceled out by conventional acceleration. They’re cumulative for ‘distance shortening’ but the hoverer is by definition at rest relative to the black hole.

When the free-faller reaches the horizon, light from the free-faller can no longer reach the hoverer, so the hoverer will never observe that portion of the free-faller's worldline (as long as he continues to hover). So the hoverer will never see the free-faller "move at c", or faster. Again, this is the standard view. I'm not sure what is supposed to happen to the free-faller on your view; I know you claim that the free-faller just continues to slow down indefinitely, but since you haven't explained how I'm supposed to relate the free-faller's "clock time" to the "clock time" of the hoverer, I don't know how you are determining this. You can't be using the standard GR relationship, because in standard GR the free-faller reaches the horizon after a finite time by his clock, not an infinite time.

Just make time dilation infinite at the horizon and work outwards to the hoverer using the inverse square of the distance.

And we established quite a while ago that even for this case, an everlasting black hole, you believe the horizon cannot be reached. So all this about a black hole losing mass is irrelevant.

Not really. It's accelerating objects towards it and the event horizon is how close you can get to it, so the horizon obviously moves inwards at c and its size is proportional to its mass. You could follow the horizon in at almost c and travel billions of years into the future. The equivalent of almost catching your own light in flat space-time.


Basically there's three hoverers, all the in the same place to start with. One of them stays at this radius the whole time (we'll assume the experiment doesn't take long enough for the radius of the event horizon/mass of the singularity/lifespan of the black hole to be important). This observer is to show that the two ways of accelerating are symmetric. One of the hoverers accelerates away from the black hole slowly at first and increasing their acceleration as an inverse square to their distance, while the other stops hovering and free-falls towards the black hole. From the perspective of the observer who's still hovering, the one accelerating away is starting to catch up to their own light. They'll never be able to catch it though because it becomes harder for them to accelerate relative to it the faster they go. Still from the perspective of the observer who's still hovering, the one falling in is starting to catch up to their own light. They'll never be able to catch it though because it becomes harder for them to accelerate relative to it the faster they go. The rate that the horizon recedes is exactly the same as the rate c recedes, depending on your acceleration. You can't reach either.

 

[PeterDonis]

Give me one example of when I've lapsed into "vague wand-waving",

Most of what you say is vague hand-waving as far as I'm concerned, because you haven't established the basic premises that you're reasoning from. You just keep on asserting things that contradict the standard GR model, without justifying the assumptions you have to make to get to your assertions.

...or even one example of when you've pointed out an inconsistency with anything other than the standard model.

You originally claimed that a point-like object moving solely under the influence of gravity would feel a non-zero acceleration. After I pointed out inconsistencies in this, you eventually backed off, agreed that a point-like object moving solely under the influence of gravity would feel zero acceleration, and then didn't follow that train of thought any further, for example, to explain how your model accounts for the fact that two point-like objects, both moving solely under the influence of gravity, can change their spatial separation, and the rate of change of that separation, with time even though both feel exactly zero acceleration. Instead, you went back to just asserting your conclusions without showing how you derive them.

We agree on SR. The basic premise is that trying to reach an event horizon using gravity is the same as trying to accelerate to c in flat space-time.

And I've pointed out differences between the two cases that make this analogy fail, which you haven't addressed; you just keep on asserting the analogy.

That's like saying an inertial observer can't have a relative velocity to an accelerator in flat space-time.

How so? Instead of just making a vague statement like this, how about doing what I asked and giving your specific definition of what "relative velocity" means when the two objects in question are not near each other (so that the curvature of the spacetime in between may have a significant effect).

A difference in tidal force is the same as a difference in acceleration.

How so? Again, please give your specific definitions that lead you to this. As you should know by now, in standard GR this is false, for the reason I pointed out above: tidal gravity can be observed with two point-like objects, both moving solely under the influence of gravity, that change their spatial separation and the rate of change of that separation with time, even though they both feel zero acceleration. This is a key physical difference between tidal gravity and proper acceleration. How does your model account for this?

On the basis that it affects the velocity of objects relative to other objects, diluted as the inverse square of the distance, just like the electro-magnetic force does. One curves in, one curves out.

An object that is affected by electromagnetic force will feel a non-zero (proper) acceleration. It won't be moving solely under the influence of gravity. This is a key physical difference between non-gravitational forces and what you are calling the "force" of gravity. How does your model account for this?

I was talking about a free-falling observers own light and how light would stop behaving like light at the horizon. You would catch up, like a sonic boom.

How can you possibly catch up to light that starts out in front of you and is moving in the same direction as you? I don't understand how your model accounts for this. It certainly doesn't happen in the situation you claim is analogous, that of accelerating observers in flat spacetime. No accelerating observer in flat spacetime can ever catch up to a light ray moving in the same direction that starts out in front of him.

It is a fixed background in four dimensions and using the light cones doesn’t prove anything because the way you use them assumes what you’re trying to prove.

How so? The light cone structure of the spacetime is not an "assumption"; it's derived from the solution of the Einstein Field Equation subject to the appropriate boundary conditions. The only "assumptions" are those boundary conditions and the EFE itself.

What happens at the event horizon shows us that. It shows that the distance between the event horizon and the singularity is proportional to your distance from it (inverse square),

Are you saying that in your model, the distance from the event horizon to the singularity is proportional to the distance of a given observer from the event horizon? Just clarifying to make sure I understand.

My right to assume that I've established a premise doesn't depend on whether or not I can get you to agree.

Maybe "right" was the wrong word. What I meant to say is that unless I see an actual valid argument, starting from premises that we both accept, I see no reason to accept your conclusions.

They compare notes with the accelerating hoverers of the riverbed as they free-fall past them.

So in other words, you are claiming that, if I am freely falling towards the horizon and emit a beam of light, as I pass successive hoverers, they will observe the beam of light traveling more and more slowly? That each hoverer won't see the light beam passing him at the speed of light? How are you arriving at this result?

A point-like object wouldn't feel tidal force but it wouldn't feel acceleration in flat space-time neither.

Do you mean proper acceleration? In other words, do you mean that if I subjected a point-like object to a non-gravitational force, such as a rocket, it would feel zero acceleration? What makes you think that? In standard GR, and even in standard Newtonian physics, the object certainly would feel a non-zero acceleration when subjected to a non-gravitational force.

It's being accelerated relative to other observers who are also accelerating purely under the influence of the same force.

If you mean "accelerated" in the coordinate sense, then yes. But not in the sense of proper acceleration, since all such point-like objects moving solely under the influence of gravity feel zero acceleration, as you've agreed. So how do you explain the fact that objects which feel zero acceleration can still "accelerate" (more precisely, as I stated it above, their relative separation and the rate of change of that separation changes with time) relative to one another? You are aware, I trust, that this can't happen in flat spacetime?

The premise is that gravity is a force that pulls instead of pushing, and it's weak because it's caused by mass rather than energy. Actually that's more of a conclusion than a premise, but it works both ways.

If you think this is a premise, you are mistaken. A premise is something that we can draw logical deductions from. What logical deductions can I draw from this vague statement? None as far as I can see.

GR. It would give all the right answers if it was done right.

LOL

A force is anything that creates a curved line in space-time.

Please give your specific definition of a "curved line". I have given the standard GR one several times now.

Theoretically it can't, and there's no reason it should work the way you're describing.

I don't understand how this relates to what I said.

The model becomes inconsistent for lots of reasons as soon as an object is allowed to reach an event horizon.

How is it inconsistent? I understand that it's inconsistent with your particular assumptions (such as the assumption that there must be a "hoverer" at every point in the spacetime), but that just means your assumptions are not necessarily true. If you want to claim it's inconsistent period, you have to show that your assumptions *must* be true. You haven't.

It moves inwards faster the faster you moving relative to it, like light.

As I've pointed out a number of times now, in the standard GR model the horizon moves *outwards*, not inwards. Please explain how your model has it moving inwards.

But an inertial observer free-falling across an event horizon would become light-like.

Why? The horizon is an *outgoing* null surface. The free-faller is moving *inward*. He's moving in the opposite direction to the light ray. So he can remain timelike just fine.

Tidal force is the same as proper acceleration.

Not in standard GR it isn't. I've already pointed out a number of times the key physical difference between the two. How does your model account for that?

Well, it just leaves the inertial one.

So let me get this straight. You claim that the free-faller in curved spacetime is equivalent to the accelerator in flat spacetime. But you claim that the hoverer in curved spacetime (who sees the free-faller moving *inward* relative to him) is equivalent to the inertial observer in flat spacetime (who sees the accelerator moving *outward* relative to him). I don't understand how this analogy can possibly work.

Here's a suggestion: post a diagram showing how you think your model works. Show the worldlines of the hoverer and the free-faller, the horizon, paths of light rays, etc. Your verbal descriptions are just not getting your model across. Maybe a picture will help. It's not that a picture will necessarily convince me that your model is right; it's just that based on your verbal descriptions, I can't even make sense of your model at all.

Just make time dilation infinite at the horizon and work outwards to the hoverer using the inverse square of the distance.

This doesn't really clarify anything. How about giving an actual specific calculation?

Basically there's three hoverers, all the in the same place to start with. One of them stays at this radius the whole time (we'll assume the experiment doesn't take long enough for the radius of the event horizon/mass of the singularity/lifespan of the black hole to be important). This observer is to show that the two ways of accelerating are symmetric. One of the hoverers accelerates away from the black hole slowly at first and increasing their acceleration as an inverse square to their distance, while the other stops hovering and free-falls towards the black hole. From the perspective of the observer who's still hovering, the one accelerating away is starting to catch up to their own light. They'll never be able to catch it though because it becomes harder for them to accelerate relative to it the faster they go. Still from the perspective of the observer who's still hovering, the one falling in is starting to catch up to their own light. They'll never be able to catch it though because it becomes harder for them to accelerate relative to it the faster they go. The rate that the horizon recedes is exactly the same as the rate c recedes, depending on your acceleration. You can't reach either.

Can you draw a diagram of this? And can you actually calculate the worldlines of the three observers (and the light rays you mention), instead of just waving your hands and assuming they will look the way you claim they will look?

 

[A-wal]

Most of what you say is vague hand-waving as far as I'm concerned, because you haven't established the basic premises that you're reasoning from. You just keep on asserting things that contradict the standard GR model, without justifying the assumptions you have to make to get to your assertions.

Bollocks!

You originally claimed that a point-like object moving solely under the influence of gravity would feel a non-zero acceleration. After I pointed out inconsistencies in this, you eventually backed off, agreed that a point-like object moving solely under the influence of gravity would feel zero acceleration, and then didn't follow that train of thought any further, for example, to explain how your model accounts for the fact that two point-like objects, both moving solely under the influence of gravity, can change their spatial separation, and the rate of change of that separation, with time even though both feel exactly zero acceleration. Instead, you went back to just asserting your conclusions without showing how you derive them.

I don’t see how a point-like object could feel anything. If you take away the dimensions they can’t feel a curve.

And I've pointed out differences between the two cases that make this analogy fail, which you haven't addressed; you just keep on asserting the analogy.

You haven’t given a single good reason for why this analogy fails. You’ve just kept on making the same flawed arguments.

How so? Instead of just making a vague statement like this, how about doing what I asked and giving your specific definition of what "relative velocity" means when the two objects in question are not near each other (so that the curvature of the spacetime in between may have a significant effect).

It means exactly the same thing as it does in flat space-time when you compare the relative velocity of two objects accelerating at different rates.

How so? Again, please give your specific definitions that lead you to this. As you should know by now, in standard GR this is false, for the reason I pointed out above: tidal gravity can be observed with two point-like objects, both moving solely under the influence of gravity, that change their spatial separation and the rate of change of that separation with time, even though they both feel zero acceleration. This is a key physical difference between tidal gravity and proper acceleration. How does your model account for this?

It accounts for it because exactly the same thing would happen with two point-like objects accelerating at different rates in flat space-time.

An object that is affected by electromagnetic force will feel a non-zero (proper) acceleration. It won't be moving solely under the influence of gravity. This is a key physical difference between non-gravitational forces and what you are calling the "force" of gravity. How does your model account for this?

It assumes the people it’s being explained to are paying attention.

How can you possibly catch up to light that starts out in front of you and is moving in the same direction as you? I don't understand how your model accounts for this. It certainly doesn't happen in the situation you claim is analogous, that of accelerating observers in flat spacetime. No accelerating observer in flat spacetime can ever catch up to a light ray moving in the same direction that starts out in front of him.

No he can’t can he! So he can’t reach the horizon either can he! If he could then he would have caught up with his own light! An accelerating observer in flat spacetime approaches the speed of their own light and they can get as close to it as they like, but they can never reach it. Sound familiar?

How so? The light cone structure of the spacetime is not an "assumption"; it's derived from the solution of the Einstein Field Equation subject to the appropriate boundary conditions. The only "assumptions" are those boundary conditions and the EFE itself.

It’s an assumption that they’re not relative and can go past 90 degrees.

Are you saying that in your model, the distance from the event horizon to the singularity is proportional to the distance of a given observer from the event horizon? Just clarifying to make sure I understand.

Length contraction is proportional to the distance of a given observer from a massive object because their acceleration is increased, so yes!

Maybe "right" was the wrong word. What I meant to say is that unless I see an actual valid argument, starting from premises that we both accept, I see no reason to accept your conclusions.

Then we’ll be waiting forever because you’ve already made up your mind and I could give the most complete and well written description ever produced on any subject and you’d still refute it. That good though. It’s forcing be to define it much tighter than I would bother to do on my own. Keep asking questions.

So in other words, you are claiming that, if I am freely falling towards the horizon and emit a beam of light, as I pass successive hoverers, they will observe the beam of light traveling more and more slowly? That each hoverer won't see the light beam passing him at the speed of light? How are you arriving at this result?

No that’s not what I meant. The hoverers would observe the beam of light traveling at the speed of light. The time it takes for their light to reach the hoverers takes longer the further they fall.

Do you mean proper acceleration? In other words, do you mean that if I subjected a point-like object to a non-gravitational force, such as a rocket, it would feel zero acceleration? What makes you think that? In standard GR, and even in standard Newtonian physics, the object certainly would feel a non-zero acceleration when subjected to a non-gravitational force.

It what sense would they "feel" it?

If you mean "accelerated" in the coordinate sense, then yes. But not in the sense of proper acceleration, since all such point-like objects moving solely under the influence of gravity feel zero acceleration, as you've agreed. So how do you explain the fact that objects which feel zero acceleration can still "accelerate" (more precisely, as I stated it above, their relative separation and the rate of change of that separation changes with time) relative to one another? You are aware, I trust, that this can't happen in flat spacetime?

Accelerated in the coordinate sense? I don’t see the distinction in this case. All accelerators are accelerated in the coordinate sense.

If you think this is a premise, you are mistaken. A premise is something that we can draw logical deductions from. What logical deductions can I draw from this vague statement? None as far as I can see.

Look harder!

LOL

Well you must have figured out that I think either it hasn’t been done properly or the theory’s completely wrong. If it was done right it would show gravity as the equivalent to acceleration in flat space-time but it doesn’t, so it’s not been done right.

Please give your specific definition of a "curved line". I have given the standard GR one several times now.

Acceleration, curvature, gravity, even spin are all basically the same. Any time you feel a force.

I don't understand how this relates to what I said.

An object can’t reach an event horizon, even in theory. It doesn’t make sense.

How is it inconsistent? I understand that it's inconsistent with your particular assumptions (such as the assumption that there must be a "hoverer" at every point in the spacetime), but that just means your assumptions are not necessarily true. If you want to claim it's inconsistent period, you have to show that your assumptions *must* be true. You haven't.

There can never be a point when you can’t hover any more because that would require infinite energy. You can’t reach c in flat (using energy) or curved (using mass) space-time.

As I've pointed out a number of times now, in the standard GR model the horizon moves *outwards*, not inwards. Please explain how your model has it moving inwards.

Because gravity pulls inwards. When you try to reach c in flat space-time the horizon moves outwards.

Why? The horizon is an *outgoing* null surface. The free-faller is moving *inward*. He's moving in the opposite direction to the light ray. So he can remain timelike just fine.

What would that have to do with anything even if it were true?

Not in standard GR it isn't. I've already pointed out a number of times the key physical difference between the two. How does your model account for that?

What key difference? Point-like objects feel proper acceleration but not tidal force? I don’t see how.

So let me get this straight. You claim that the free-faller in curved spacetime is equivalent to the accelerator in flat spacetime. But you claim that the hoverer in curved spacetime (who sees the free-faller moving *inward* relative to him) is equivalent to the inertial observer in flat spacetime (who sees the accelerator moving *outward* relative to him). I don't understand how this analogy can possibly work.

I don’t know what you mean. Both the hoverer and the inertial observer see the accelerator/free-faller moving outward (away). If you mean the curvature relative to its source then yes, mass curves inwards and energy curves outwards, but you said the free-faller is moving inwards relative to the hoverer which doesn’t make sense.

Here's a suggestion: post a diagram showing how you think your model works. Show the worldlines of the hoverer and the free-faller, the horizon, paths of light rays, etc. Your verbal descriptions are just not getting your model across. Maybe a picture will help. It's not that a picture will necessarily convince me that your model is right; it's just that based on your verbal descriptions, I can't even make sense of your model at all.

No equations! No diagrams!

This doesn't really clarify anything. How about giving an actual specific calculation?

Just make time dilation infinite at the horizon and work outwards to the hoverer using the inverse square of the distance.

Wtf do you call that then?

Can you draw a diagram of this? And can you actually calculate the worldlines of the three observers (and the light rays you mention), instead of just waving your hands and assuming they will look the way you claim they will look?

From the hoverers perspective light from the free-faller and accelerator moves slower than c when it leaves them and its speed increases as an inverse square of the distance, to reach c when it reaches the hoverer. How slowed the light is to start with depends on the rate of acceleration of the free-faller/accelerator.

From the free-faller and accelerators perspective their own light leaves hem at c and slows down as an inverse square of the distance, to reach 0 at the event horizon/c.

They increase their acceleration as an inverse square of the distance to the hoverer and the light moving towards the horizon/c is decelerated relative to them at a faster and faster rate as their acceleration increases, but they’ll never be able to catch up.

 

[PeterDonis]

I don’t see how a point-like object could feel anything. If you take away the dimensions they can’t feel a curve.

So you don't think an idealized point-like object subjected to a non-gravitational force would feel acceleration either? If so, this is a huge physical difference between your model and standard GR. It would also explain a lot of your confusion about points like this one:

You haven’t given a single good reason for why this analogy fails. You’ve just kept on making the same flawed arguments.

I have pointed out repeatedly that tidal gravity can be observed with idealized point-like objects moving solely under gravity and feeling zero acceleration, while proper acceleration can only be observed with idealized point-like objects that feel non-zero acceleration. But if you don't believe that an idealized point-like object can feel non-zero acceleration, then obviously your model is different from standard GR and makes different physical predictions. So we're back to the empirical question; we'll find out who's right when we are able to do enough experiments close to black holes. And there's no point in my responding to all the other times in your post when you make the same kind of claim, because you're simply starting from a different premise. (But see a further comment below to one particular place where you raise this exact issue again.)

It means exactly the same thing as it does in flat space-time when you compare the relative velocity of two objects accelerating at different rates.

Meaning, in other words, that it requires the assumption of a family of observers, all mutually at rest with respect to each other for all time, that covers the entire spacetime. So if that assumption fails (as it does in GR with the spacetime around a black hole), this definition no longer works.

No he can’t can he! So he can’t reach the horizon either can he! If he could then he would have caught up with his own light! An accelerating observer in flat spacetime approaches the speed of their own light and they can get as close to it as they like, but they can never reach it. Sound familiar?

Um, I was responding to this statement of yours:

I was talking about a free-falling observers own light and how light would stop behaving like light at the horizon. You would catch up, like a sonic boom.

You were the one saying that the observer *would* catch up to his own light, just as an aircraft that goes supersonic catches up with its own sound. At least, that's what your reference to a sonic boom appears to mean. If you meant something else, then once again your verbal description has failed to convey your meaning, and it would really help if you would draw a diagram.

(I suppose I should also note that, in standard GR, if you free-fall towards the horizon and emit a light ray inward, you do *not* catch up to it in any sense; it continues to move away from you, it is always moving inward faster than you are, and it crosses the horizon before you do.)

It’s an assumption that they’re not relative and can go past 90 degrees.

No, that's also derived from the solution to the EFE. The only assumptions are the EFE and the boundary conditions.

No that’s not what I meant. The hoverers would observe the beam of light traveling at the speed of light. The time it takes for their light to reach the hoverers takes longer the further they fall.

Ah, ok.

Accelerated in the coordinate sense? I don’t see the distinction in this case. All accelerators are accelerated in the coordinate sense.

Not if you adopt a system of coordinates in which the accerating object is at rest. You are proper accelerated sitting at rest on the surface of the Earth, but you're not coordinate accelerated with respect to coordinates that are fixed to the surface of the Earth. (You *are* coordinate accelerated with respect to coordinates in which an object freely falling downward is at rest.)

Well you must have figured out that I think either it hasn’t been done properly or the theory’s completely wrong. If it was done right it would show gravity as the equivalent to acceleration in flat space-time but it doesn’t, so it’s not been done right.

I understand that that's what you *think*, yes.

There can never be a point when you can’t hover any more because that would require infinite energy. You can’t reach c in flat (using energy) or curved (using mass) space-time.

In standard GR, you don't need to reach c to reach a region of the black hole spacetime where you can't hover. You can just float at rest, freely falling into the hole. So this supposed "logical deduction" assumes premises that are not valid in standard GR.

As I've pointed out a number of times now, in the standard GR model the horizon moves *outwards*, not inwards. Please explain how your model has it moving inwards.

Because gravity pulls inwards.

Oncea again, it would *really* help if you would draw a diagram of how you think this is working. See further comments below.

What would that have to do with anything even if it were true?

If the free-faller and the horizon are moving in opposite directions, it's perfectly possible for the free-faller to remain timelike and still cross the horizon, just as you can easily remain timelike in flat spacetime and still pass a light ray moving in the opposite direction.

What key difference? Point-like objects feel proper acceleration but not tidal force? I don’t see how.

Yes, that's obvious (see my comments at the start of this post). But that's not an argument, just a statement of your state of mind.

I could suggest looking at it as a limiting process (the idealized point-like object is a limit of smaller and smaller real objects, and the proper acceleration it feels is the limit of the proper acceleration felt by the smaller and smaller real objects), but that would require math, which you're allergic to. However, such a limiting process would show that, in the case of objects moving solely under the influence of gravity, the limit of "felt acceleration" goes to zero as the object's size goes to zero, while in the case of objects moving under a non-gravitational force, the limit of "felt acceleration" goes to a non-zero value (the object's proper acceleration) as the object's size goes to zero. That's the long-winded way of stating what I've been saying. And it still points out a key physical difference between objects moving solely under gravity, and objects moving under a non-gravitational force. If you agree with that difference, then you need to explain how your model accounts for it while still claiming that tidal gravity and proper acceleration are somehow "the same". If you *don't* agree with that difference, then we're back, as I said above, to the empirical question; your model and standard GR make different physical predictions, and we'll just have to wait and see who is right.

I don’t know what you mean. Both the hoverer and the inertial observer see the accelerator/free-faller moving outward (away). If you mean the curvature relative to its source then yes, mass curves inwards and energy curves outwards, but you said the free-faller is moving inwards relative to the hoverer which doesn’t make sense.

Once, more, it would *really* help if you would *not* take the following position:

No equations! No diagrams!

Let me re-quote the statement of mine that prompted your comment above:

You claim that the free-faller in curved spacetime is equivalent to the accelerator in flat spacetime. But you claim that the hoverer in curved spacetime (who sees the free-faller moving *inward* relative to him) is equivalent to the inertial observer in flat spacetime (who sees the accelerator moving *outward* relative to him).

You were referring to my diagram of the flat spacetime case; the "free-fallers" in that diagram are the blue vertical lines, and the "accelerators" are the hyperbolas that accelerate *outward* (in the positive x-direction) relative to the free-fallers. Yet you are claiming that the free-fallers in the curved spacetime case are equivalent to the accelerators in the flat spacetime case, and the "hoverers" in the curved spacetime case are equivalent to the *free-fallers* in the flat spacetime case. But in flat spacetime, the free-fallers are *inward* from the accelerators, while in curved spacetime, the hoverers (which you say are equivalent to the flat spacetime free-fallers) are *outward* of the free-fallers (which you say are equivalent to the flat spacetime accelerators). So it looks to me like you have somehow flipped things around. If you were to draw an actual diagram, showing how you think the free-fallers, the hoverers, and the horizon move in your model in curved spacetime, it would really help to clarify what you are saying.

Wtf do you call that then?

You were referring to this earlier statement of yours:

Just make time dilation infinite at the horizon and work outwards to the hoverer using the inverse square of the distance.

Since you refuse to write equations, I have no idea what you mean by "work outwards using the inverse square of the distance". If you mean that the time dilation should go like the standard GR formula,

\sqrt{g_{tt}} = \sqrt{1 - \frac{2 M}{r}}

so that a value of zero for g_{tt} at radius r = 2M corresponds to what you are calling "infinite" time dilation at the horizon, then that makes sense, but (a) it doesn't use the inverse square of the distance, just the inverse distance (and takes the square root of that), and (b) it applies to the *hoverer* at radius r, *not* to the free-faller going past radius r. I was asking how you think the proper time of the free-faller works; it can't possibly be the same as the hoverer's, because the free-faller is moving relative to the hoverer.

From the hoverers perspective light from the free-faller and accelerator moves slower than c when it leaves them and its speed increases as an inverse square of the distance, to reach c when it reaches the hoverer. How slowed the light is to start with depends on the rate of acceleration of the free-faller/accelerator.

From the free-faller and accelerators perspective their own light leaves them at c and slows down as an inverse square of the distance, to reach 0 at the event horizon/c.

As far as I can tell, this is mostly the same as the standard GR picture, except that in standard GR, the free-faller (and the accelerator who is moving inward) will eventually cross the horizon, and then they will be able to see that the light beams they emitted ahead of them did *not* slow down and stop at the horizon.

 

[A-wal]

Meaning, in other words, that it requires the assumption of a family of observers, all mutually at rest with respect to each other for all time, that covers the entire spacetime. So if that assumption fails (as it does in GR with the spacetime around a black hole), this definition no longer works.

Yes, but I don’t see how the assumption could fail. I don’t see how it could be possible for no amount of energy is enough. It doesn’t work. It doesn’t make any kind of sense. It’s stupid!

You were the one saying that the observer *would* catch up to his own light, just as an aircraft that goes supersonic catches up with its own sound. At least, that's what your reference to a sonic boom appears to mean. If you meant something else, then once again your verbal description has failed to convey your meaning, and it would really help if you would draw a diagram.

(I suppose I should also note that, in standard GR, if you free-fall towards the horizon and emit a light ray inward, you do *not* catch up to it in any sense; it continues to move away from you, it is always moving inward faster than you are, and it crosses the horizon before you do.)

I was saying that’s what would happen if an object were allowed to reach an event horizon because you can’t see light reach the horizon no matter how close you are.

No, that's also derived from the solution to the EFE. The only assumptions are the EFE and the boundary conditions.

90 degrees represents c, and you can’t reach c. That’s why I said to use light cones in flat space-time. They’re relative. Whether they’re tilted because of mass or energy is irrelevant.

Ah, ok.

"The time it takes for their light to reach the hoverers takes longer the further they fall"...and not just because the distance has increased, unless you’re defining the distance after you’ve taken ‘distance shortening’ into account.

Not if you adopt a system of coordinates in which the accerating object is at rest. You are proper accelerated sitting at rest on the surface of the Earth, but you're not coordinate accelerated with respect to coordinates that are fixed to the surface of the Earth. (You *are* coordinate accelerated with respect to coordinates in which an object freely falling downward is at rest.)

Yea it disappears if the coordinates follow the curve.

In standard GR, you don't need to reach c to reach a region of the black hole spacetime where you can't hover. You can just float at rest, freely falling into the hole. So this supposed "logical deduction" assumes premises that are not valid in standard GR.

They would have accelerated to c relative to anything outside the horizon. They’re being accelerated by gravity. They would have freely fallen to c relative to everything outside the horizon.

If the free-faller and the horizon are moving in opposite directions, it's perfectly possible for the free-faller to remain timelike and still cross the horizon, just as you can easily remain timelike in flat spacetime and still pass a light ray moving in the opposite direction.

Are you sure? If something’s on the other side of an event horizon then it’s light-like.

You were referring to my diagram of the flat spacetime case; the "free-fallers" in that diagram are the blue vertical lines, and the "accelerators" are the hyperbolas that accelerate *outward* (in the positive x-direction) relative to the free-fallers. Yet you are claiming that the free-fallers in the curved spacetime case are equivalent to the accelerators in the flat spacetime case, and the "hoverers" in the curved spacetime case are equivalent to the *free-fallers* in the flat spacetime case. But in flat spacetime, the free-fallers are *inward* from the accelerators, while in curved spacetime, the hoverers (which you say are equivalent to the flat spacetime free-fallers) are *outward* of the free-fallers (which you say are equivalent to the flat spacetime accelerators). So it looks to me like you have somehow flipped things around. If you were to draw an actual diagram, showing how you think the free-fallers, the hoverers, and the horizon move in your model in curved spacetime, it would really help to clarify what you are saying.

I still don’t know what you mean by “But in flat spacetime, the free-fallers are *inward* from the accelerators, while in curved spacetime, the hoverers (which you say are equivalent to the flat spacetime free-fallers) are *outward* of the free-fallers (which you say are equivalent to the flat spacetime accelerators).” It’s from the hoverers perspective to show that the two ways of accelerating are the same. In flat space-time the hoverer becomes the inertial observer with two accelerators moving in opposite directions. It’s the same. It doesn’t matter that there’s a black hole pulling the free-faller towards it. That’s just the method of acceleration. Gravity pulls everything closer so it becomes cumulative making space-time seem curved, but if everything was accelerated in the opposite direction in the same way as the observer moving away from the black hole then space-time would seem curved in that direction as well, but energy pushes instead of pulls so it’s affect weakens instead.

Since you refuse to write equations, I have no idea what you mean by "work outwards using the inverse square of the distance". If you mean that the time dilation should go like the standard GR formula,

\sqrt{g_{tt}} = \sqrt{1 - \frac{2 M}{r}}

so that a value of zero for g_{tt} at radius r = 2M corresponds to what you are calling "infinite" time dilation at the horizon, then that makes sense, but (a) it doesn't use the inverse square of the distance, just the inverse distance (and takes the square root of that), and (b) it applies to the *hoverer* at radius r, *not* to the free-faller going past radius r. I was asking how you think the proper time of the free-faller works; it can't possibly be the same as the hoverer's, because the free-faller is moving relative to the hoverer.

Is that supposed to be the simpler way? Seems much more complicated. Since I presume neither of us are Egyptian, let’s just stick to words. The hoverers proper time is the equivalent to the proper time of an inertial observer in flat space-time because they’re at rest relative to the black hole. The black holes lifespan could be calculated by the hoverer and that would be that. The free-faller on the other hand is constantly accelerating towards the black hole so its lifespan and its size (in other words the four-dimensional sphere that it covers) gets smaller and smaller at a faster and faster rate as they get closer at a faster and faster rate.

As far as I can tell, this is mostly the same as the standard GR picture, except that in standard GR, the free-faller (and the accelerator who is moving inward) will eventually cross the horizon, and then they will be able to see that the light beams they emitted ahead of them did *not* slow down and stop at the horizon.

How would that work? You’re approaching the horizon and no light reaches it until you do, and then it suddenly jumps ahead of you as if it had never been slowed down? This is what I meant about the contradiction. There’s two DIFFERENT versions that you keep switching between, but they can’t possibly both be right. In one version the free-faller notices nothing special and merrily passes the event horizon. This is the wrong one. In the other version the free-faller starts to catch up to their own light as the gap between them and the horizon shortens, just like an observer accelerating in flat space-time because you can’t see light cross the horizon no matter how close you are to it. This is the right version.

What inwards accelerating observer? There’s a hoverer, a free-faller, and an accelerator who matches their velocity relative to the hoverer with the free-fallers. You could have an observer accelerating towards the black hole using the same amount of energy as the one accelerating away and an equivalent observer accelerating away from the black hole using twice the energy of the first one. That shows that using energy or mass to accelerate is equivalent and you can even mix them together on the same observer. It makes no difference. They’re equivalent.

 

[Dale]

But an inertial observer free-falling across an event horizon would become light-like.

No it doesn't. A free-falling observer remains timelike as it crosses the event horizon. A free-falling observer follows a geodesic, geodesics parallel transport their tangent vector, and parallel transport preserves dot products.

The event horizon is how close you can get to the singularity at that time and you can move at c, and its radius is proportional to its mass.

So, if its mass remains constant then this doesn't apply. You have asserted earlier that it is not possible to cross the event horizon even of a constant-mass "eternal" black hole. Are you now revising that opinion? Do you now agree that an observer could cross the event horizon of a constant mass black hole in a finite amount of proper time?

 

[PeterDonis]

Yes, but I don’t see how the assumption could fail. I don’t see how it could be possible for no amount of energy is enough. It doesn’t work. It doesn’t make any kind of sense. It’s stupid!

Again, I understand that this is what you think, but it's not an argument, it's just a description of your state of mind. Can you think of an actual logical *reason* why the assumption *cannot* fail?

90 degrees represents c, and you can’t reach c. That’s why I said to use light cones in flat space-time. They’re relative. Whether they’re tilted because of mass or energy is irrelevant.

In flat spacetime light cones don't tilt at all. So "use lightcones in flat spacetime" really means "no tilting" period. But gravity *has* to tilt the light cones, because it bends light; so in the presence of gravity light cones *do* tilt, and 90 degrees no longer "represents c".

Yea it disappears if the coordinates follow the curve.

I'm not sure what you think this means. We agreed many, many posts ago that you can't change the physics by changing coordinate systems.

They would have accelerated to c relative to anything outside the horizon. They’re being accelerated by gravity. They would have freely fallen to c relative to everything outside the horizon.

Your definition of "relative velocity" doesn't work, so this statement has no meaning. In terms of standard GR, the free-faller continues to move inside the light-cones, so he certainly hasn't "freely fallen to c". Can you find a definition of "relative velocity" for objects at different radial coordinates that implies a valid constraint on that relative velocity to be less than c?

Are you sure? If something’s on the other side of an event horizon then it’s light-like.

Nope, you stay timelike if you free-fall inside the horizon, just as you stay timelike after you've passed a light ray going in the opposite direction.

I still don’t know what you mean by “But in flat spacetime, the free-fallers are *inward* from the accelerators, while in curved spacetime, the hoverers (which you say are equivalent to the flat spacetime free-fallers) are *outward* of the free-fallers (which you say are equivalent to the flat spacetime accelerators).” It’s from the hoverers perspective to show that the two ways of accelerating are the same. In flat space-time the hoverer becomes the inertial observer with two accelerators moving in opposite directions.

Well, the second accelerator isn't anywhere on my diagram; I'm just trying to understand how you think the hoverer plus the free-faller falling inward (or the inward "accelerator" in your terminology) correspond to worldlines on my diagram, since you said those two were the two that had a correspondence. If you would just draw a diagram of your own we wouldn't have to go through all these gyrations.

Also, if you say that "the hoverer becomes the inertial observer" then you are saying that the hoverer crosses the horizon (because the inertial observer in flat spacetime does). I don't think you mean to say that, which means that your diagram can't possibly correspond to my diagram. So again, why can't you just draw a diagram of your own instead of saying that there's a correspondence to my diagram, and then basically refuting any such correspondence when you try to describe it?

Is that supposed to be the simpler way? Seems much more complicated.

LOL.

Since I presume neither of us are Egyptian, let’s just stick to words. The hoverers proper time is the equivalent to the proper time of an inertial observer in flat space-time because they’re at rest relative to the black hole.

But the hoverer's proper time varies with distance from the black hole. The inertial observer's proper time in flat spacetime (for inertial observers all at rest relative to each other, which I assume is what you mean) is the same everywhere; it doesn't vary in space. So again, you are asserting an analogy that simply can't work. So much for words.

How would that work? You’re approaching the horizon and no light reaches it until you do,

Where did I say that? I said the opposite: I said the light reaches the horizon *before* you do.

There’s two DIFFERENT versions that you keep switching between, but they can’t possibly both be right. In one version the free-faller notices nothing special and merrily passes the event horizon. This is the wrong one.

Nope, this is the right one, and it's the *only* one I have been asserting. Anything else is just you not understanding, or refusing to accept that the assumptions you are making can be violated.

What inwards accelerating observer? There’s a hoverer, a free-faller, and an accelerator who matches their velocity relative to the hoverer with the free-fallers.

Not sure what part of my post you were responding to here, but you were the one who called the free-faller in curved spacetime an inward accelerating observer (accelerating relative to the hoverer). You said that the hoverer, not the free-faller, was analogous to an inertial observer in flat spacetime, and the free-faller was analogous to an accelerator in flat spacetime. (Though as I've pointed out, this analogy does not work when I actually ask for more details.)

 

[A-wal]

No it doesn't. A free-falling observer remains timelike as it crosses the event horizon. A free-falling observer follows a geodesic, geodesics parallel transport their tangent vector, and parallel transport preserves dot products.

I’ll take your word for it on this occasion.

So, if its mass remains constant then this doesn't apply. You have asserted earlier that it is not possible to cross the event horizon even of a constant-mass "eternal" black hole. Are you now revising that opinion? Do you now agree that an observer could cross the event horizon of a constant mass black hole in a finite amount of proper time?

NO NO NO! ANY given amount of time won't be enough to reach the horizon, just as no amount of time in flat space-time is enough to reach c no matter how much energy you use. If no amount of energy is enough then it doesn’t matter how much time there is to apply that energy. They’d just keep on accelerating. It doesn’t make sense for an object to reach an event horizon!

Again, I understand that this is what you think, but it's not an argument, it's just a description of your state of mind. Can you think of an actual logical *reason* why the assumption *cannot* fail?

Because the methods of acceleration are equivalent. It’s very self-evident that they’re equivalent. You haven’t given a good enough reason to show this premise is wrong. Not even close.

In flat spacetime light cones don't tilt at all. So "use lightcones in flat spacetime" really means "no tilting" period. But gravity *has* to tilt the light cones, because it bends light; so in the presence of gravity light cones *do* tilt, and 90 degrees no longer "represents c".

Acceleration bends light in the same way. Just treat it in the same way you would for the curvature of gravity. I’m not sure why you think you can’t compare velocities properly when it’s gravity that causing the acceleration, as if that makes a difference.

I'm not sure what you think this means. We agreed many, many posts ago that you can't change the physics by changing coordinate systems.

Of course. When you follow an object with a different relative velocity, that velocity disappears. When you free-fall you’re following the curve, so it disappears. You don’t river the river current because you’re moving with it. Etc.

Your definition of "relative velocity" doesn't work, so this statement has no meaning. In terms of standard GR, the free-faller continues to move inside the light-cones, so he certainly hasn't "freely fallen to c". Can you find a definition of "relative velocity" for objects at different radial coordinates that implies a valid constraint on that relative velocity to be less than c?

If the gravitational source was accelerated to match the current velocity of the free-falling object then the strength of gravity would remain constant for the free-faller, so you could compare their relative velocity with a more distant observer. Or you could just do what you do for a continuously accelerating observer in flat space-time.

Nope, you stay timelike if you free-fall inside the horizon, just as you stay timelike after you've passed a light ray going in the opposite direction.

Hmm. If it made sense for the horizon to move outwards I suppose that would make sense.

Well, the second accelerator isn't anywhere on my diagram; I'm just trying to understand how you think the hoverer plus the free-faller falling inward (or the inward "accelerator" in your terminology) correspond to worldlines on my diagram, since you said those two were the two that had a correspondence. If you would just draw a diagram of your own we wouldn't have to go through all these gyrations.

No the free-faller is equivalent to the hoverer. I never said that. The hoverer is equivalent to an inertial observer in flat space-time. The free-faller is equivalent to an accelerator in flat space-time. All three start off as hoverers so that the two types of acceleration cancel each other out and they become the equivalent of inertial observers in flat space-time. One of them shows this by accelerating away at a faster and faster rate but never reaching c relative to the hoverer. The free-faller does exactly the same thing.

Also, if you say that "the hoverer becomes the inertial observer" then you are saying that the hoverer crosses the horizon (because the inertial observer in flat spacetime does). I don't think you mean to say that, which means that your diagram can't possibly correspond to my diagram. So again, why can't you just draw a diagram of your own instead of saying that there's a correspondence to my diagram, and then basically refuting any such correspondence when you try to describe it?

You’re diagram has the event horizon moving outwards, so if an observer were to hover relative to the singularity then they would cross the horizon.

But the hoverer's proper time varies with distance from the black hole. The inertial observer's proper time in flat spacetime (for inertial observers all at rest relative to each other, which I assume is what you mean) is the same everywhere; it doesn't vary in space. So again, you are asserting an analogy that simply can't work. So much for words.

It’s not the words thought that you’re going out of your way to misrepresent them to prove a point. The hoverers distance from the black hole doesn’t vary. They’re hovering.

Where did I say that? I said the opposite: I said the light reaches the horizon *before* you do.

But it can’t. No light can reach the horizon until you do?

Nope, this is the right one, and it's the *only* one I have been asserting. Anything else is just you not understanding, or refusing to accept that the assumptions you are making can be violated.

But this isn’t what someone free-falling across an event horizon would experience, if it was possible. Light can’t reach the horizon until you do. So if you’re right it must suddenly jump in front of you when you reach the horizon?

Not sure what part of my post you were responding to here, but you were the one who called the free-faller in curved spacetime an inward accelerating observer (accelerating relative to the hoverer). You said that the hoverer, not the free-faller, was analogous to an inertial observer in flat spacetime, and the free-faller was analogous to an accelerator in flat spacetime. (Though as I've pointed out, this analogy does not work when I actually ask for more details.)

Yes it does. Ask away. The free faller is accelerating inwards towards the black hole yes, but from the hoverers perspective they’re both accelerating away. You could remove the black hole and the acceleration of the hoverer to make the hoverer an inertial observer and the free-faller now just accelerates away at the same rate as before and it’s the same.

 

[PeterDonis]

Because the methods of acceleration are equivalent. It’s very self-evident that they’re equivalent. You haven’t given a good enough reason to show this premise is wrong. Not even close.

I have repeatedly pointed out that objects being "accelerated" by gravity feel zero acceleration (as measured by an accelerometer), while objects being accelerated by "energy" (in your terminology) feel nonzero acceleration (as measured by an accelerometer). I even clarified how a limiting process makes sense of these statements when applied to idealized, point-like objects. I'm not trying to prove that your premise is wrong; I'm simply showing that it is false in GR, whatever its status is in your theory, and therefore it does not *have* to be right. I've even said that if you want to maintain that these statements, as physical facts, are *not* true, that's your prerogative: you simply have a different empirical theory than GR, and the decision which is right will be made by experiment, not argument.

I’m not sure why you think you can’t compare velocities properly when it’s gravity that causing the acceleration, as if that makes a difference.

Because acceleration due to "energy" does not invalidate the premise of SR that objects which feel no acceleration must have constant relative velocity to each other for all time. "Acceleration" due to gravity does. That's the key difference.

Of course. When you follow an object with a different relative velocity, that velocity disappears. When you free-fall you’re following the curve, so it disappears. You don’t river the river current because you’re moving with it. Etc.

So basically you are denying that the "curvature" in question is a physical invariant; you are saying that "curvature" can always be transformed away by changing coordinates. This is not true for the GR definition of curvature, either of a worldline (what you are calling "acceleration due to energy" and which I am calling proper acceleration) or of spacetime (tidal gravity). Both types of curvature are physical invariants and are present regardless of the coordinate system. (They are *different* physical invariants, but they are both physical invariants.) Can you give a specific definition of curvature in your model such that it always *can* be transformed away?

If the gravitational source was accelerated to match the current velocity of the free-falling object then the strength of gravity would remain constant for the free-faller, so you could compare their relative velocity with a more distant observer.

If by accelerating the gravitational source, you just mean changing coordinates, then this is simply false; the two objects (the source and the free-faller) will move towards each other regardless of what coordinates you use. If by accelerating the source you mean adding a new source of energy to the problem (like attaching a huge rocket to the Earth, say, so that it moves away from an object falling towards it), then you are changing the problem and we'll need to start from scratch in analyzing it. I'd rather not do that since we don't even have agreement on the simpler problem where the Earth doesn't have a rocket attached to it.

The hoverer is equivalent to an inertial observer in flat space-time.

And if this is supposed to be an analogy to the diagram I drew, the inertial observer in flat spacetime crosses the horizon, so by your analogy the hoverer should cross the horizon. So the above statement simply can't be a proper description of whatever analogy you think there is between the hoverer in curved spacetime and the diagram I drew for flat spacetime.

You’re diagram has the event horizon moving outwards, so if an observer were to hover relative to the singularity then they would cross the horizon.

The hoverer is hovering relative to the horizon, not the singularity. My diagram is drawn in flat spacetime, remember; in flat spacetime there is no singularity, so there is no such thing as "hovering relative to the singularity".

In the curved spacetime equivalent to my diagram, which would be a Kruskal diagram, the singularity is a hyperbola up in region II (the region where t > 0 and x < t, on the opposite side of the horizon to the hoverers), and it makes no sense to ask "how far away" it is from the hoverers since it's not a timelike surface, it's a spacelike surface--i.e., a surface of constant "time", not a surface of constant "position", so asking how far away it is would be like asking how far away next Tuesday is. The question doesn't make sense. It does make sense to hover relative to the horizon, and that's what the hoverers are doing in curved spacetime. (In the standard GR analogy, of course, hoverers are analogous to, um, hoverers.)

It’s not the words thought that you’re going out of your way to misrepresent them to prove a point. The hoverers distance from the black hole doesn’t vary. They’re hovering.

I'm not misrepresenting your words; I'm simply pointing out that they can't possibly express a valid analogy between anything in curved spacetime and the diagram I drew for flat spacetime. If you stop asserting such an analogy in terms which are obviously not valid, I won't have to comment on such assertions. And if you would just draw a diagram of how things look in your model we wouldn't have to go through all these gyrations of you trying to describe how you think your model relates to my diagram and me pointing out that what you're saying makes no sense if I try to interpret it as expressing any such relationship.

But this isn’t what someone free-falling across an event horizon would experience, if it was possible. Light can’t reach the horizon until you do. So if you’re right it must suddenly jump in front of you when you reach the horizon?

Nope. The free-faller sees the light moving away from him in much the same way he would if he were floating at rest in flat spacetime. Locally, there is no way he can even tell when either the light or he himself crosses the horizon; he has to use observations of distant objects to tell that.

 

[Dale]

NO NO NO! ANY given amount of time won't be enough to reach the horizon, just as no amount of time in flat space-time is enough to reach c no matter how much energy you use. If no amount of energy is enough then it doesn’t matter how much time there is to apply that energy. They’d just keep on accelerating. It doesn’t make sense for an object to reach an event horizon!

So if an observer cannot cross the event horizon of a constant-mass black hole then your previous responses are not relevant.:

In Schwarzschild coordinates the horizon is the 3 dimensional surface given by:
r=\frac{2GM}{c^2}
and so
\frac{d}{dt}\left( \frac{2GM}{c^2} \right) = 0
therefore the horizon has a constant position in Schwarzschild coordinates.

That's because Schwarzschild coordinates assume an everlasting black hole. A hoverer who's maintaining a constant distance from the horizon will be moving towards the singularity. How fast they move towards it depends on how far away they are, a bit like gravity itself really. They’ll get closer to the singularity as the event horizon gets closer to the singularity, as it looses mass. At the event horizon it’s losing the exact amount it needs for the event horizon to back at c. It’s the equivalent of an accelerator in flat space-time reducing their acceleration (the equivalent of the black hole loosing mass as it ages) to keep their Rindler horizon in the same place I suppose.

Can you derive any of this?

The event horizon is how close you can get to the singularity at that time and you can move at c, and its radius is proportional to its mass.

So again, the event horizon has a constant position in Schwarzschild coordinates, as I proved above.

 

[A-wal]

I have repeatedly pointed out that objects being "accelerated" by gravity feel zero acceleration (as measured by an accelerometer), while objects being accelerated by "energy" (in your terminology) feel nonzero acceleration (as measured by an accelerometer). I even clarified how a limiting process makes sense of these statements when applied to idealized, point-like objects. I'm not trying to prove that your premise is wrong; I'm simply showing that it is false in GR, whatever its status is in your theory, and therefore it does not *have* to be right. I've even said that if you want to maintain that these statements, as physical facts, are *not* true, that's your prerogative: you simply have a different empirical theory than GR, and the decision which is right will be made by experiment, not argument.

Okay, you haven't given a good enough reason to show why the standard GR version is the right one. There’s plenty to suggest that it’s not right.

Because acceleration due to "energy" does not invalidate the premise of SR that objects which feel no acceleration must have constant relative velocity to each other for all time. "Acceleration" due to gravity does. That's the key difference.

Well it shouldn’t.

So basically you are denying that the "curvature" in question is a physical invariant; you are saying that "curvature" can always be transformed away by changing coordinates. This is not true for the GR definition of curvature, either of a worldline (what you are calling "acceleration due to energy" and which I am calling proper acceleration) or of spacetime (tidal gravity). Both types of curvature are physical invariants and are present regardless of the coordinate system. (They are *different* physical invariants, but they are both physical invariants.) Can you give a specific definition of curvature in your model such that it always *can* be transformed away?

On Earth we’re accelerating up, but it doesn’t really seem like it (although it feels like it) because the coordinate system is defined by us. We’re at rest and we’re being pulled down.

If by accelerating the gravitational source, you just mean changing coordinates, then this is simply false; the two objects (the source and the free-faller) will move towards each other regardless of what coordinates you use. If by accelerating the source you mean adding a new source of energy to the problem (like attaching a huge rocket to the Earth, say, so that it moves away from an object falling towards it), then you are changing the problem and we'll need to start from scratch in analyzing it. I'd rather not do that since we don't even have agreement on the simpler problem where the Earth doesn't have a rocket attached to it.

(: Fair enough. I basically just don’t see the problem with comparing velocities in curved space-time because it’s the equivalent of comparing observers with different amounts of acceleration in flat space-time. In fact it’s easier because you can always predict exactly how much gravity they’ll feel as they fall.

And if this is supposed to be an analogy to the diagram I drew, the inertial observer in flat spacetime crosses the horizon, so by your analogy the hoverer should cross the horizon. So the above statement simply can't be a proper description of whatever analogy you think there is between the hoverer in curved spacetime and the diagram I drew for flat spacetime.

If the inertial observer crosses any horizon then it’s not from the inertial observers perspective. It must be from the accelerators. From the free-fallers (accelerators) perspective the hoverer (inertial observer) doesn’t cross the event horizon. What you’re describing is not equivalent to a free-faller crossing an event horizon.

The hoverer is hovering relative to the horizon, not the singularity. My diagram is drawn in flat spacetime, remember; in flat spacetime there is no singularity, so there is no such thing as "hovering relative to the singularity".

Congratulations. You’ve just managed to "transform away" a curve by changing coordinates.

In the curved spacetime equivalent to my diagram, which would be a Kruskal diagram, the singularity is a hyperbola up in region II (the region where t > 0 and x < t, on the opposite side of the horizon to the hoverers), and it makes no sense to ask "how far away" it is from the hoverers since it's not a timelike surface, it's a spacelike surface--i.e., a surface of constant "time", not a surface of constant "position", so asking how far away it is would be like asking how far away next Tuesday is. The question doesn't make sense. It does make sense to hover relative to the horizon, and that's what the hoverers are doing in curved spacetime. (In the standard GR analogy, of course, hoverers are analogous to, um, hoverers.)

The distance to the singularity would be the distance to the horizon plus the diameter/radius (whatever, I can never remember) of the black hole.

It’s not the words thought that you’re going out of your way to misrepresent them to prove a point. The hoverers distance from the black hole doesn’t vary. They’re hovering.

Thought? Fault! There not even similar, they just sound similar. I'm definitely a bit dyslexic, or a bit something.

Nope. The free-faller sees the light moving away from him in much the same way he would if he were floating at rest in flat spacetime. Locally, there is no way he can even tell when either the light or he himself crosses the horizon; he has to use observations of distant objects to tell that.

But light can’t reach an event horizon from the outside. You see the light continuously slow down but never reach the horizon.

So if an observer cannot cross the event horizon of a constant-mass black hole then your previous responses are not relevant.

It's not irrelevant. That's why an observer cannot cross the event horizon of a constant-mass black hole, because it's moving inwards at c.

So again, the event horizon has a constant position in Schwarzschild coordinates, as I proved above.

Its distance relative to a hoverer obviously remains constant, but not its distance relative to the singularity.

 

[PeterDonis]

Okay, you haven't given a good enough reason to show why the standard GR version is the right one.

As I have pointed out ad nauseam, it is observed physical fact that objects moving solely under the influence of gravity (like the Space Shuttle while in orbit, for instance) are weightless, while objects that are *not* moving solely under the influence of gravity (like you or I standing on the surface of the Earth, or a rocket in space while the engine is on) feel weight. You don't consider this a good enough reason?

Well it shouldn’t.

LOL

On Earth we’re accelerating up, but it doesn’t really seem like it (although it feels like it) because the coordinate system is defined by us. We’re at rest and we’re being pulled down.

How exactly does this relate to curvature and whether it can be transformed away by changing coordinates? What definition of "curvature" is this supposed to support?

I basically just don’t see the problem with comparing velocities in curved space-time because it’s the equivalent of comparing observers with different amounts of acceleration in flat space-time. In fact it’s easier because you can always predict exactly how much gravity they’ll feel as they fall.

But in GR these two are *not* equivalent, because of the key physical difference I keep pointing out. In GR an observer that feels weight simply can't be physically equivalent to an observer that is weightless.

If the inertial observer crosses any horizon then it’s not from the inertial observers perspective. It must be from the accelerators.

The horizon is just a particular line in the spacetime (the line t = x in the diagram I drew), and whether or not a given observer's worldline crosses it is a physical invariant, independent of whose "perspective" is being used. It is true that defining that particular line as the "horizon" is done *because* it is the asymptote of all the hyperbolas that the accelerators are following, but that doesn't change the fact that once that line is defined, the free-faller's worldline crosses it.

From the free-fallers (accelerators) perspective the hoverer (inertial observer) doesn’t cross the event horizon. What you’re describing is not equivalent to a free-faller crossing an event horizon.

If it isn't, then the hoverer is *not* analogous to the inertial observer in the diagram I drew. So you are basically admitting that the analogy you were trying to make is bogus.

Congratulations. You’ve just managed to "transform away" a curve by changing coordinates.

How so? I specified that my diagram was drawn in flat spacetime. "Flat" meaning "not curved". And that spacetime has stayed flat all through everything I've been saying.

I also described the analogue to my diagram in curved spacetime, where the horizon appears as a hyperbola, specifically (I didn't give this equation before because you're allergic to math, but I'll give it now to be specific) the "upper branch" (i.e., in the upper part of the plane, where T > 0 and X < T) of the hyperbola T^2 - X^2 = 4M^2 (where M is the mass of the black hole in geometric units). In that diagram (the Kruskal diagram), the horizon is again the line T = X (I've capitalized T and X in this case to make clear that they are the Kruskal T and X coordinates), and the "hoverers" travel on hyperbolas (in the right half-plane, where X > 0 and T < X) X^2 - T^2 = r^2, where r > 2M.

So the horizons and the hoverers in both cases are analogous, in the sense that their curves look the same and have the same equations in the corresponding sets of coordinates. How does any of this "transform away" any curvature? It's just an analogy that can help to understand how some aspects of the curved spacetime case work. But I'm certainly not trying to claim that you can transform one spacetime into the other just by changing coordinates.

The distance to the singularity would be the distance to the horizon plus the diameter/radius (whatever, I can never remember) of the black hole.

See my comment below on your comment to DaleSpam's post about this.

But light can’t reach an event horizon from the outside. You see the light continuously slow down but never reach the horizon.

The hoverer does (for an appropriate sense of "see"). But the free-faller does not. He sees what I have described. At least, he does if standard GR is correct.

Its distance relative to a hoverer obviously remains constant, but not its distance relative to the singularity.

I know this was in response to DaleSpam, but it contradicts what you said above in response to me (which I quoted). Which is it?

 

[Dale]

It's not irrelevant. That's why an observer cannot cross the event horizon of a constant-mass black hole, because it's moving inwards at c.

I understand the relevance of the argument, what is irrelevant is your justification that followed. If you believe that something applies to a constant-mass black hole then you logically must have a justification that is not based on a variable-mass black hole.

So again, you have claimed that the event horizon is moving inwards at c for a constant-mass black hole. In 451 I derived the fact that the position of the event horizon is constant in Schwarzschild coordinates for a constant-mass black hole. Do you have a relevant objection? I.e. one that is based on a constant-mass black hole.

Its distance relative to a hoverer obviously remains constant, but not its distance relative to the singularity.

The Schwarzschild coordinate r is the coordinate distance relative to the singularity. If r is constant then the distance to the singularity is constant.

 

[A-wal]

As I have pointed out ad nauseam, it is observed physical fact that objects moving solely under the influence of gravity (like the Space Shuttle while in orbit, for instance) are weightless, while objects that are *not* moving solely under the influence of gravity (like you or I standing on the surface of the Earth, or a rocket in space while the engine is on) feel weight. You don't consider this a good enough reason?

Tidal force is the difference in the strength of gravity between two points of the same object. What you feel when you accelerate is the difference in the strength of acceleration between two points of the same object. If every atom were individually accelerated at exactly the same rate then they wouldn’t feel a thing, or if the acceleration increased evenly as an inverse square in the direction of the source of gravity then it would be exactly the same as tidal force.

How exactly does this relate to curvature and whether it can be transformed away by changing coordinates? What definition of "curvature" is this supposed to support?

A curve isn’t really a curve if you’re following it. You feel it, but only in the sense that you feel tidal force.

But in GR these two are *not* equivalent, because of the key physical difference I keep pointing out. In GR an observer that feels weight simply can't be physically equivalent to an observer that is weightless.

'Feel weight' is a loose term.

The horizon is just a particular line in the spacetime (the line t = x in the diagram I drew), and whether or not a given observer's worldline crosses it is a physical invariant, independent of whose "perspective" is being used. It is true that defining that particular line as the "horizon" is done *because* it is the asymptote of all the hyperbolas that the accelerators are following, but that doesn't change the fact that once that line is defined, the free-faller's worldline crosses it.

Yes. Just like when a free-falling observer reaches a point when no signal sent from the hoverer will ever be able to reach them, as long as they keep free-falling.

If it isn't, then the hoverer is *not* analogous to the inertial observer in the diagram I drew. So you are basically admitting that the analogy you were trying to make is bogus.

No, your ‘event’ horizon is bogus.

How so? I specified that my diagram was drawn in flat spacetime. "Flat" meaning "not curved". And that spacetime has stayed flat all through everything I've been saying.

You said in curved space-time there’s a singularity and in flat space-time there isn’t.

I also described the analogue to my diagram in curved spacetime, where the horizon appears as a hyperbola, specifically (I didn't give this equation before because you're allergic to math, but I'll give it now to be specific) the "upper branch" (i.e., in the upper part of the plane, where T > 0 and X < T) of the hyperbola T^2 - X^2 = 4M^2 (where M is the mass of the black hole in geometric units). In that diagram (the Kruskal diagram), the horizon is again the line T = X (I've capitalized T and X in this case to make clear that they are the Kruskal T and X coordinates), and the "hoverers" travel on hyperbolas (in the right half-plane, where X > 0 and T < X) X^2 - T^2 = r^2, where r > 2M.

My brain hurts.

So the horizons and the hoverers in both cases are analogous, in the sense that their curves look the same and have the same equations in the corresponding sets of coordinates. How does any of this "transform away" any curvature? It's just an analogy that can help to understand how some aspects of the curved spacetime case work. But I'm certainly not trying to claim that you can transform one spacetime into the other just by changing coordinates.

I didn’t mean you’re trying to transform one space-time into the other just by changing coordinates. I meant curvature isn’t a physical invariant.

The hoverer does (for an appropriate sense of "see"). But the free-faller does not. He sees what I have described. At least, he does if standard GR is correct.

It can’t be. Whether or not an object reaches the horizon is a physical invariant. When would this magical separation happen? At the event horizon? That doesn’t work. The light would have to jump back if they started hovering and jump forwards when they reach the horizon.

I know this was in response to DaleSpam, but it contradicts what you said above in response to me (which I quoted). Which is it?

Radius!, ?

 

[Dale]

'Feel weight' is a loose term.

Yes, the correct term is: non-geodesic. Or equivalently non-zero covariant derivative of the tangent vector, which we have discussed before. As PeterDonis points out, a non-geodesic observer cannot be equivalent to a geodesic observer. One is inertial and the other is not.

 

[PeterDonis]

What you feel when you accelerate is the difference in the strength of acceleration between two points of the same object. If every atom were individually accelerated at exactly the same rate then they wouldn’t feel a thing

What makes you think so? What you feel when you feel proper acceleration is the force of whatever is causing the acceleration (a rocket engine, the surface of the Earth, etc.) pushing on you. The push could be exactly uniform and you would still feel it.

A curve isn’t really a curve if you’re following it.

This indicates to me that you do not think curvature is a physical invariant. You confirm that later in your post; see my comment on that below.

'Feel weight' is a loose term.

Weight can be measured directly by an accelerometer. An ordinary bathroom scale is an example of a (cheap and not highly accurate, but accurate enough for everyday use) accelerometer. There's nothing "loose" about it.

Yes. Just like when a free-falling observer reaches a point when no signal sent from the hoverer will ever be able to reach them, as long as they keep free-falling.

There is no such point. A signal can always be sent from outside the horizon to inside, from the hoverer to the free-faller. The horizon only keeps signals from the free-faller once he is inside, from reaching the hoverer outside.

You said in curved space-time there’s a singularity and in flat space-time there isn’t.

Yes. So what? I never claimed the two spacetimes were identical. I only described an analogy between certain aspects that can be helpful in understanding those aspects.

My brain hurts.

Sorry, I forgot you were allergic to math.

I didn’t mean you’re trying to transform one space-time into the other just by changing coordinates. I meant curvature isn’t a physical invariant.

In standard GR it is. If it isn't in your model, then your model must be using a different definition of "curvature" from the standard GR one. What is that specific definition, and what justifies using the term "curvature" to describe it?

Whether or not an object reaches the horizon is a physical invariant.

I agree.

When would this magical separation happen?

What separation? I've never said there was any "magical separation". I said that the free-faller sees the light beam moving away from him, just as if he were floating at rest in flat spacetime.

Radius!, ?

Um, here are the two things you said:

The distance to the singularity would be the distance to the horizon plus the diameter/radius (whatever, I can never remember) [it's radius--PD] of the black hole.

Its [the horizon's--PD] distance relative to a hoverer obviously remains constant, but not its [the horizon's--PD] distance relative to the singularity.

The first statement quoted above implies that the distance from the horizon to the singularity is the radius of the hole, which is constant (it's equal to 2M, where M is the hole's mass in geometric units, and we're talking about an "eternal" hole whose mass is constant). That is obviously inconsistent with the second statement quoted above. Which is it?

 

[A-wal]

I understand the relevance of the argument, what is irrelevant is your justification that followed. If you believe that something applies to a constant-mass black hole then you logically must have a justification that is not based on a variable-mass black hole.

So again, you have claimed that the event horizon is moving inwards at c for a constant-mass black hole. In 451 I derived the fact that the position of the event horizon is constant in Schwarzschild coordinates for a constant-mass black hole. Do you have a relevant objection? I.e. one that is based on a constant-mass black hole.

No sorry, I misunderstood. If it has a constant mass then it would have a constant horizon. But then I suppose you would be able to reach it, which doesn’t make sense. It’s paradoxical to have a constant mass black hole, like it is to accelerate to c under any circumstances.

The Schwarzschild coordinate r is the coordinate distance relative to the singularity. If r is constant then the distance to the singularity is constant.

Yes I see what you mean now.

Yes, the correct term is: non-geodesic. Or equivalently non-zero covariant derivative of the tangent vector, which we have discussed before. As PeterDonis points out, a non-geodesic observer cannot be equivalent to a geodesic observer. One is inertial and the other is not.

Hmm.

What makes you think so? What you feel when you feel proper acceleration is the force of whatever is causing the acceleration (a rocket engine, the surface of the Earth, etc.) pushing on you. The push could be exactly uniform and you would still feel it.

I’m reaching. Makes sense though, if a point-like object can’t feel a force.

This indicates to me that you do not think curvature is a physical invariant. You confirm that later in your post; see my comment on that below.

This is heading towards semantics. The curve’s always there unless you go with it, a bit like the river model.

Weight can be measured directly by an accelerometer. An ordinary bathroom scale is an example of a (cheap and not highly accurate, but accurate enough for everyday use) accelerometer. There's nothing "loose" about it.

Would you class tidal force as ‘feeling your weight’?

There is no such point. A signal can always be sent from outside the horizon to inside, from the hoverer to the free-faller. The horizon only keeps signals from the free-faller once he is inside, from reaching the hoverer outside.

I’m not talking about the event horizon. This is your flat space-time horizon using gravity. Are you sure there wouldn’t be a point when no signal sent from the hoverer will reach the free-faller? I’m fairly confident there would be a Rindler horizon equivalent.

Yes. So what? I never claimed the two spacetimes were identical. I only described an analogy between certain aspects that can be helpful in understanding those aspects.

Oh okay.

Sorry, I forgot you were allergic to math.

I’m not allergic to math as such, but I have real trouble when it’s used to describe reality.

In standard GR it is. If it isn't in your model, then your model must be using a different definition of "curvature" from the standard GR one. What is that specific definition, and what justifies using the term "curvature" to describe it?

It’s relative, just like velocity. You have to accelerate relative to something else.

I agree.

And reaches the horizon from an accelerators perspective, so it can’t happen.

What separation? I've never said there was any "magical separation". I said that the free-faller sees the light beam moving away from him, just as if he were floating at rest in flat spacetime.

Then what if he starts hovering? Does the light turn round and come back? It would have to if it moves away from him as it would if he was at rest in flat space-time. It would have to slow down as he free-falls and the light moves into relatively more ‘distance shortened’ space-time, glaven.

The first statement quoted above implies that the distance from the horizon to the singularity is the radius of the hole, which is constant (it's equal to 2M, where M is the hole's mass in geometric units, and we're talking about an "eternal" hole whose mass is constant). That is obviously inconsistent with the second statement quoted above. Which is it?

I was talking about a real life black hole. An eternal one doesn’t make sense. You can always move towards it. In fact it’s encouraged.

 

[Dale]

No sorry, I misunderstood. If it has a constant mass then it would have a constant horizon. But then I suppose you would be able to reach it, which doesn’t make sense. It’s paradoxical to have a constant mass black hole, like it is to accelerate to c under any circumstances.

I understand that is your claim, but I still haven't heard a solid justification for this. In Schwarzschild coordinates the horizon is stationary. In a coordinate-independent sense it is moving outwards at c. It is not paradoxical to cross something that is stationary or is moving towards you at c.

Perhaps it would be good to revisit post 375 where I derived the fact that the coordinate time to reach the Rindler horizon was infinite but the proper time was finite. I can do the same thing for Schwarzschild coordinates, but the equations are messier. Would that be helpful?

https://www.physicsforums.com/showpost.php?p=3316839&postcount=375

 

[PeterDonis]

No sorry, I misunderstood. If it has a constant mass then it would have a constant horizon. But then I suppose you would be able to reach it, which doesn’t make sense. It’s paradoxical to have a constant mass black hole, like it is to accelerate to c under any circumstances.

Ah, now we're getting to it. You say something similar at the end of your post:

I was talking about a real life black hole. An eternal one doesn’t make sense. You can always move towards it. In fact it’s encouraged.

If you really don't believe that an "eternal" black hole with constant mass makes sense, then obviously you're not going to believe the standard GR model of one. But again, that doesn't make the model wrong; it just means you don't agree with it.

One clarification, though: even if Hawking is right and all real black holes will eventually evaporate due to quantum radiation, any "real life black hole" of stellar mass or greater is *not* evaporating now. In fact it's doing the opposite; it's *gaining* mass. Real holes are constantly absorbing mass as objects are pulled in by their gravity. Also, if nothing else, real holes are constantly taking in cosmic microwave background radiation, which is at a temperature far higher than the Hawking temperature of any black hole of stellar mass or greater, and so they are gaining mass from that as well.

Would you class tidal force as ‘feeling your weight’?

Short answer: No.

Longer answer: the term "tidal force" is ambiguous, which is why I have specifically talked about idealized point-like objects with no internal structure, since that separates the behavior of the center of mass of the object from the behavior of the object's internal parts relative to one another. Here's a more realistic scenario that illustrates the issue I'm talking about:

Consider an extended object that is freely falling towards the Earth. By "extended object" I mean that the object is large enough to have internal structure, and internal parts that can exert (non-gravitational) forces on each other. (We assume that the object is too small for its gravity to have any effect.) For purposes of this scenario, we'll assume the object has three parts, each of which is an idealized "point-like" object with no internal structure; the three parts can exert non-gravitational force on each other, but have no internal forces within themselves. Part A is closest to the Earth; part B is farthest from the Earth; part C is in between, at the center of mass of the object as a whole.

If there were no internal forces between parts A, B, and C, they would move solely due to the Earth's gravity, and would gradually separate; A would fall fastest, B slowest, and C in between the two. This would be a manifestation of pure "tidal gravity", but it would not properly be called tidal "force" because none of the parts, A, B, or C, would *feel* any force.

However, with internal forces between the parts, A, B, and C move together as a single combined object, maintaining constant relative distance from each other. That means that A and B must feel a force--the force of C pulling A back and pulling B forward, so A falls more slowly than it would if it were moving solely due to the Earth's gravity, and B falls more quickly than it would if it were moving solely due to the Earth's gravity. This force from C is non-gravitational, so it causes proper acceleration in A and B, which is why they feel the force. This type of force is often called "tidal force", but that is really not a good name, because it is *not* due to tidal gravity; it is due to the internal forces between parts of extended objects that are falling in a gravitational field.

How does C itself move? Since it is at the center of mass of the object as a whole, it moves on the *same* trajectory that it would if it were by itself, moving solely due to the Earth's gravity. So C still feels *zero* proper acceleration, because it's moving on the same free fall trajectory that it would if it were by itself. However, there are internal forces on C from A and B (by Newton's Third Law--the forces of C on A and B must be matched by equal and opposite forces of A and B on C); it's just that they cancel out (the inward force from A is exactly canceled by the outward force from B), for zero net force.

Finally, I would not term any of the forces in this scenario as "feeling weight", because the object as a whole is moving on a free-fall trajectory, and the object itself is not large enough to have significant gravity, so the internal forces that A and B feel are not the kind that are normally called "weight".

I’m not talking about the event horizon. This is your flat space-time horizon using gravity.

The flat spacetime horizon doesn't "use gravity". It is defined by the limiting asymptotes of all the hyperbolas that the proper accelerated observers travel on. There is no gravity in flat spacetime. If you are trying to use your analogy between proper acceleration in flat spacetime and free fall in the presence of gravity, you'll need to justify that analogy first.

Are you sure there wouldn’t be a point when no signal sent from the hoverer will reach the free-faller? I’m fairly confident there would be a Rindler horizon equivalent.

Yes, I'm sure. There is no such point in the flat spacetime case.

One clarification: in the case of the black hole spacetime, the free-faller will hit the singularity and cease to exist at some point; obviously no signal from anywhere can reach the free-faller after that, since he doesn't exist. But up until that point, signals from the hoverer can reach the free-faller.

I’m not allergic to math as such, but I have real trouble when it’s used to describe reality.

Was that what I was doing? I thought I was just trying to communicate clearly and precisely the nature of the analogy I was discussing. Communicating clearly and precisely is what math is for. The only thing I was "describing" was the two abstract spacetimes under discussion. Whether or not that was describing "reality" would depend on how statements about those abstract spacetimes were linked to statements about actual observations that could be made in the real world.

It’s relative, just like velocity. You have to accelerate relative to something else.

Is this supposed to be a definition? Are you saying that curvature is defined as acceleration? If so, do you mean proper acceleration, or coordinate acceleration, or "acceleration due to gravity", or tidal acceleration (see my previous comment above), or what kind of acceleration? And if you are lumping two different kinds of acceleration together (like, let's just say, proper acceleration and acceleration due to gravity), what common feature of the two are you labeling as "curvature"?

And reaches the horizon from an accelerators perspective, so it can’t happen.

And I don't agree with that part.

Then what if he starts hovering? Does the light turn round and come back? It would have to if it moves away from him as it would if he was at rest in flat space-time.

When you say "hovering", which direction is the free-faller firing his rocket engine? (Assuming that's how he hovers.) Is he firing it to accelerate *towards* the light beam, or *away* from it? Neither case works the way you are saying above, but I would like to be clear about which one you are imagining.

Here's how both cases would work in flat spacetime:

Case 1: Observer O is floating freely in flat spacetime and emits a light beam in the negative x-direction. The light beam moves away from him. Observer Z is also floating freely next to O and sees the light beam moving away from him also. Then O turns on his rocket engine and accelerates in the positive x-direction (i.e., away from the light beam). The light beam still moves away from him; from the standpoint of Z, O is accelerating away in one direction and the light beam is moving away in the other. Once O starts accelerating, there is a Rindler horizon defined relative to him (as long as he continues to accelerate), and both Z and the light beam will pass that horizon.

Case 2: Observer O is floating freely in flat spacetime and emits a light beam in the negative x-direction. The light beam moves away from him. Observer Z is also floating freely next to O and sees the light beam moving away from him also. Then O turns on his rocket engine and accelerates in the *negative* x-direction. The light beam still moves away from him; from the standpoint of Z, O is accelerating away in the same direction as the light beam, but the light beam is always ahead of O and is always gaining on him (the amount by which it gains decreases as O accelerates, but never quite reaches zero).

So in neither case does the light beam "turn around and come back" from any observer's standpoint.

In the analogous cases in curved spacetime, both cases work the same as above; the only difference is that in the curved spacetime analogue of case 1, observer O can become a "hoverer" as long as he starts his rocket while he's still above the black hole horizon (and if O does hover, the black hole horizon works the same as the Rindler horizon in the flat spacetime case). In the curved spacetime analogue of case 2, O will reach the black hole horizon before Z does, but after the light beam does.

 

[A-wal]

I understand that is your claim, but I still haven't heard a solid justification for this. In Schwarzschild coordinates the horizon is stationary. In a coordinate-independent sense it is moving outwards at c. It is not paradoxical to cross something that is stationary or is moving towards you at c.

It would still be equivalent to accelerating to c, which is why I said before that it wouldn’t be possible to reach a fixed event horizon, but I suppose if there’s an infinite amount of time to keep accelerating then it would, but only because an infinite amount of energy would allow you to reach c. Neither are actually possible. It still wouldn’t be possible to reach the event horizon of an eternal black hole within any given amount of time, just as it wouldn’t be possible to accelerate to c in any given amount of time.

The event horizon moves inwards for the same reason c moves outwards, because that’s the direction that mass and energy curve. Both horizons always have to move AWAY from you at c. I suppose the event horizon would move outwards from the perspective of inside the horizon. The equivalent to an object moving faster than c not being able to slow down to anything below c. That way it’s still always moving away from you and the arrow of time is reversed because the light cones will have tilted past 90 degrees, so time will literally be pointing in the opposite direction.

Objects never reach the event horizon so they just keep on accelerating and pile up around the edge from the perspective of a more distant object. That’s not as silly as it sounds though because there’s plenty of room. ‘Distance shortening’ means the length of any object approaching the horizon gets progressively shorter at an ever increasing rate.

When the black hole initially forms the event horizon would expand outward from the centre at c, but it couldn’t hit anything, even an observer right next to it as it forms. It would be as if extra space had been created between them and the centre of the star, which is now where the singularity is, because that’s how ‘distance shortening’ works. That’s why the back of objects are moved closer to the front with length contraction in Rindler coordinates.

Now they can move towards it, but at a maximum speed of c, even if they have a very powerful rocket and gun it directly at the singularity. The gravity wave carries on expanding outward at c, but when its strength is no longer enough to accelerate objects to c the event horizon moves inward at c, because it’s the point when objects would reach c. If you try to reach it you will move into relatively more ‘distance shortened’ space-time exactly like trying to accelerate to c in flat space-time. Space-time is created from the perspective of an accelerator or free-faller so that they can keep accelerating freely despite not being able to reach c from the perspective of an inertial observer or hoverer.

Its lifespan is the amount of proper time it would take to reach the singularity. As far as the black hole is concerned it never existed. All the atoms are gone. No more ‘distance shortening’ due to the circular orbits of the electrons. The nuclear force powering the gravity has been crushed by that gravity. The singularity is a point-like object in space-time. It exists for no time at all but its influence is felt forever. It just gets less noticeable over time, like it does over distance.

This all creates a very nice shape for the black hole. It doesn’t make any difference which angle of space or either way you run time. It’s a four-dimensional sphere, just as it should be. There’s nothing special about time, just as there’s nothing special about gravity.

Perhaps it would be good to revisit post 375 where I derived the fact that the coordinate time to reach the Rindler horizon was infinite but the proper time was finite. I can do the same thing for Schwarzschild coordinates, but the equations are messier. Would that be helpful?

Not for me. Not unless you also describe using words why I should believe what those equations are describing.

If you really don't believe that an "eternal" black hole with constant mass makes sense, then obviously you're not going to believe the standard GR model of one. But again, that doesn't make the model wrong; it just means you don't agree with it.

The event horizon should be defined as the furthest point that can be reached at that time, and you can always move towards it. At c locally, but it would obviously seem progressively slower from further away as an inverse square of the distance. Where would it get its energy from? All the matter would be gone, crushed out of existence. Black holes loose mass because the matter that created them has turned to energy. That change in gravity spreads out at c, but it’s moving out from ‘distance shortened’ space-time, so ‘distance shortened’ that it creates an event horizon.

One clarification, though: even if Hawking is right and all real black holes will eventually evaporate due to quantum radiation, any "real life black hole" of stellar mass or greater is *not* evaporating now. In fact it's doing the opposite; it's *gaining* mass. Real holes are constantly absorbing mass as objects are pulled in by their gravity. Also, if nothing else, real holes are constantly taking in cosmic microwave background radiation, which is at a temperature far higher than the Hawking temperature of any black hole of stellar mass or greater, and so they are gaining mass from that as well.

Okay, I was talking about a real life, but idealised black hole existing in a debrisless vacuum of absolute zero temperature. Hawking radiation isn’t needed. Describing what happens at the horizon is like describing what happens at c. Nothing ever gets there so nothing ever happens there.

Short answer: No.

If you take away acceleration from the Earth you become weightless. The same thing happens in 0G. In gravitational free-fall you don’t feel a force (except tidal force of course). Everything falls at the same rate. This means it’s curvature and not a force does it? Okay, if you want to look at it like that. If we take away gravity and apply energy to a group of objects of varying mass then they’d all move at the same rate too. They’d all move together as if there’s no force at all and they would fell anything unless the difference in acceleration between both ends of the same objects stretched them enough to be felt because the curve was too sharp. The reason you feel a force pulling you back in a car when you accelerate and pushing you back when you break is the same reason you feel a force pulling you to one side when you go round a corner. It gets messy when you start mixing the two. It’s harder for energy to overcome a greater amount of mass and it’s harder for gravity to overcome a greater amount of energy.

Consider an extended object that is freely falling towards the Earth. By "extended object" I mean that the object is large enough to have internal structure, and internal parts that can exert (non-gravitational) forces on each other. (We assume that the object is too small for its gravity to have any effect.) For purposes of this scenario, we'll assume the object has three parts, each of which is an idealized "point-like" object with no internal structure; the three parts can exert non-gravitational force on each other, but have no internal forces within themselves. Part A is closest to the Earth; part B is farthest from the Earth; part C is in between, at the center of mass of the object as a whole.

If there were no internal forces between parts A, B, and C, they would move solely due to the Earth's gravity, and would gradually separate; A would fall fastest, B slowest, and C in between the two. This would be a manifestation of pure "tidal gravity", but it would not properly be called tidal "force" because none of the parts, A, B, or C, would *feel* any force.

However, with internal forces between the parts, A, B, and C move together as a single combined object, maintaining constant relative distance from each other. That means that A and B must feel a force--the force of C pulling A back and pulling B forward, so A falls more slowly than it would if it were moving solely due to the Earth's gravity, and B falls more quickly than it would if it were moving solely due to the Earth's gravity. This force from C is non-gravitational, so it causes proper acceleration in A and B, which is why they feel the force. This type of force is often called "tidal force", but that is really not a good name, because it is *not* due to tidal gravity; it is due to the internal forces between parts of extended objects that are falling in a gravitational field.

How does C itself move? Since it is at the center of mass of the object as a whole, it moves on the *same* trajectory that it would if it were by itself, moving solely due to the Earth's gravity. So C still feels *zero* proper acceleration, because it's moving on the same free fall trajectory that it would if it were by itself. However, there are internal forces on C from A and B (by Newton's Third Law--the forces of C on A and B must be matched by equal and opposite forces of A and B on C); it's just that they cancel out (the inward force from A is exactly canceled by the outward force from B), for zero net force.

Finally, I would not term any of the forces in this scenario as "feeling weight", because the object as a whole is moving on a free-fall trajectory, and the object itself is not large enough to have significant gravity, so the internal forces that A and B feel are not the kind that are normally called "weight".

I appreciate the detailed answer. It’s a nice description, but there are a couple of things I don’t quite agree with. First C wouldn’t fall at the same rate that it would on its own because the difference in gravitational strength between A and C is greater than the difference between C and B. So it falls faster the longer the object is as the back literally gets pulled in by the front. And you think gravity isn’t a real force?

And second, you say tidal force isn’t due to tidal gravity. Yes it is because it’s the difference in the strength of gravity between two parts of the same object that causes the tidal force between them. The same thing happens with acceleration in flat space-time. There’s a source of energy, in this case a rocket that pushes along the length of the object. The parts that are closest to the rocket get pushed the hardest but this isn’t felt, and the strength is reduced as an inverse square of the distance to the rocket. This difference is felt as proper acceleration as the front of the object gets pushed along by the back, just as the different rates of acceleration in tidal gravity are felt as a force as the back of the object gets pulled along by the front. If we strapped a tiny rocket to each atom and accelerated them all together then I don’t think the object would feel any acceleration, unless it was sensitive enough to feel the difference of the rate of acceleration between the front and the back of the individual atoms, which seems unlikely. Are you sure a point-like object feels proper acceleration but not tidal force?

 

[A-wal]

The flat spacetime horizon doesn't "use gravity". It is defined by the limiting asymptotes of all the hyperbolas that the proper accelerated observers travel on. There is no gravity in flat spacetime. If you are trying to use your analogy between proper acceleration in flat spacetime and free fall in the presence of gravity, you'll need to justify that analogy first.

If an object were able to ignore the fact that it can’t be done and cross an event horizon then any signal sent from inside the horizon wouldn’t be able to reach anything outside the horizon. The equivalent to that in flat space-time is ignoring the fact that c can’t be reached and accelerating past it anyway so that no signal sent from the accelerator will reach anything in the opposite direction of its motion. Its Rindler horizon would have caught up and overtaken it. That’s what you’re claiming happens when you say an event horizon is reachable.

Yes, I'm sure. There is no such point in the flat spacetime case.

There is in the flat space-time case. It’s called a Rindler horizon. I take it you meant curved space-time. The free-faller in curved space-time is equivalent to the accelerator in flat space-time. So whenever there’s any gravity there should be an equivalent to the Rindler and nothing on the other side can reach the free-faller.

One clarification: in the case of the black hole spacetime, the free-faller will hit the singularity and cease to exist at some point; obviously no signal from anywhere can reach the free-faller after that, since he doesn't exist. But up until that point, signals from the hoverer can reach the free-faller.

Cease to exist? If it stopped existing why would the mass stay? You didn’t even mean turned into energy did you? Why do you think something can just stop existing? That’s insane!

Was that what I was doing? I thought I was just trying to communicate clearly and precisely the nature of the analogy I was discussing. Communicating clearly and precisely is what math is for.

It’s not working. This is easier.

The only thing I was "describing" was the two abstract spacetimes under discussion. Whether or not that was describing "reality" would depend on how statements about those abstract spacetimes were linked to statements about actual observations that could be made in the real world.

You seemed to be suggesting that crossing an event horizon is the same as crossing a Rindler horizon in flat space-time. In standard GR anything free-falling is at rest and you have to accelerate to remain at rest in a coordinate sense, so space-time is curved. When the curvature goes past 90 degrees no amount of acceleration can keep you at rest in a coordinate sense, so you can’t hover. I get it. I always have, and it seemed to make sense to start with. It doesn’t make sense though. For a start a Rindler horizon needs an accelerator to form. An event horizon needs gravity but a free-faller isn’t accelerating in the standard GR model and the hoverer is, but the free-faller still crosses the event horizon if they stop hovering. In flat space-time if they stop accelerating the Rindler horizon disappears. The event horizon doesn’t disappear if the hover stops accelerating so it’s not equivalent to a Rindler horizon.

In standard GR the free-faller crosses the event horizon but not from the perspective of the hoverer. In flat space-time the equivalent to that would be an accelerator reaching c from their own perspective and catching up eith their own light, but not from the perspective of an inertial observer. That obviously doesn’t work. An object can keep on accelerating forever, making their space-time more and more ‘distance shortened’. They can use this to travel huge distances in space-time from the perspective of their starting frame it a short amount of proper time for them, but when they’ve stopped accelerating they never reached c from their original frame, they just kept on accelerating and covering more and more space-time over less and less proper time until the energy that was accelerating them stopped.

It makes so much more sense if energy and gravitation (mass) are equivalent except that gravity is divided by c squared and pulls rather than pushes. Now the hoverer is equivalent to an inertial observer because the inwards acceleration of gravity is matched by the outwards acceleration of their engines. Obviously if you applied the same amount of force in two opposite directions of various objects then they would all stay in the same place relative to each other. If you took away one of those forces they would still stay still relative to each other. The accelerator moves away creating a Rindler horizon. When the hoverer crosses the Rindler horizon no signal sent from them can catch the accelerator. The free-faller turns of their engines and is accelerated by gravity towards the event horizon. If you think that gravity can accelerate you beyond c then there should definitely be the equivalent of a Rindler horizon because a free-faller crossing an event horizon would be accelerating harder than an accelerator ever could. But they’re exactly equivalent so the horizon created would also be exactly equivalent.

From the hoverers perspective the acceleration of the free-faller/accelerator decreases more and more rapidly through ‘distance shortening’ because there’s a limit to the velocity they can reach. This stops the accelerator from ever reaching c, and so also stops the free-faller from ever reaching the even horizon, when c would no longer be fast enough to move you away.

Is this supposed to be a definition? Are you saying that curvature is defined as acceleration? If so, do you mean proper acceleration, or coordinate acceleration, or "acceleration due to gravity", or tidal acceleration (see my previous comment above), or what kind of acceleration? And if you are lumping two different kinds of acceleration together (like, let's just say, proper acceleration and acceleration due to gravity), what common feature of the two are you labeling as "curvature"?

Proper acceleration, including tidal force. Acceleration that’s felt. Although you could class any form of acceleration as curvature. Different context of acceleration means different a context of curvature.

And I don't agree with that part.

I missed a word. I should have said ‘And NOTHING reaches from an accelerators perspective. But you’re saying objects closer to the horizon than the free-faller and the free-fallers own light can reach and cross the event horizon as if it was just an arbitrary line in flat space-time? I don’t see how that works if they then accelerate away?

When you say "hovering", which direction is the free-faller firing his rocket engine? (Assuming that's how he hovers.) Is he firing it to accelerate *towards* the light beam, or *away* from it? Neither case works the way you are saying above, but I would like to be clear about which one you are imagining.

The free-faller falls until they’re almost at the horizon, then start hovering. Now no light that they HAVE EVER emitted will be able to reach the horizon. If any object were able to reach the event horizon then they would have caught up to their own light. I don’t think you can do that.

Same three observers. Hoverer, free-faller, accelerator. The accelerator matches their velocity relative to the hoverer precisely with the free-fallers velocity relative to the hoverer. The accelerator approaches c at the exact same rate the free-faller approaches the event horizon but never ever reaches it. Unless you think singularities are capable of producing infinite energy within a range proportional to their mass? Infinite energy is ruled out when using energy to accelerate but not when using mass. Why? Gravity can’t accelerate anything to c any easier than a rocket can. The two types of acceleration have a fight at the event horizon. It’s a draw on points because they both score c.

As the accelerator accelerates harder to match the free-faller they feel more of a force, just as the free-faller feels more tidal force. This is a very special rocket that’s able to spread the acceleration so that it’s felt on individual parts of the object matching exactly with the difference in the strength of the gravitational field felt by individual parts of the free-faller. In other words it distributes the proper acceleration in exactly the way gravity does to create tidal force, which would make the acceleration felt proportional as an inverse square of the distance to c as they start to catch their own light as they accelerate. This makes the accelerator and the free-faller exact physical equivalents from the perspective of the hoverer. If the free-faller were to fire their rocket to match exactly with the rocket of the accelerator at that time (using the hoverers perspective of the same time) then the free-faller would now be hovering. When do you think this symmetry breaks down? It can’t happen all at once at the horizon because the accelerator never reaches c, so as long as they stay equivalent neither will the free-faller.

As the accelerator/free-faller starts to move faster relative to their own light it’s exactly the same as trying to reach c relative to the hoverer because it gets harder to close they gap on the light in front of you the harder you accelerate. The only difference is the relative velocity of light. You just have to replace acceleration with velocity if you want to measure your velocity relative to light. You can close the gap but you can never keep up. The accelerator/free-faller approaches the light faster as their acceleration increases in the same way two observers in flat space-time would if they increase their relative velocity at the same rate. From the hoverers perspective no light has ever reached the event horizon. If you think the light in front of a free-faller is able to cross the event horizon ahead of them as if they were inertial observers in flat space-time then what would happen if they wait until the last possible moment when they’re one Plank length away from the horizon and then accelerate away and meet up with the hoverer? No light has ever reached the horizon from here.

Here's how both cases would work in flat spacetime:

Case 1: Observer O is floating freely in flat spacetime and emits a light beam in the negative x-direction. The light beam moves away from him. Observer Z is also floating freely next to O and sees the light beam moving away from him also. Then O turns on his rocket engine and accelerates in the positive x-direction (i.e., away from the light beam). The light beam still moves away from him; from the standpoint of Z, O is accelerating away in one direction and the light beam is moving away in the other. Once O starts accelerating, there is a Rindler horizon defined relative to him (as long as he continues to accelerate), and both Z and the light beam will pass that horizon.

In curved space-time observer O is hovering at some distance from the event horizon and emits a light beam *away* from the black hole. Observer Z starts off hovering next to O then free-falls towards the event horizon. The light beam moves away in exactly the same way it would if they accelerated in flat space-time, and there should be the equivalent of a Rindler horizon that moves closer as they free-fall so that it passes first the light beam, then the hoverer, and no signal sent from the other side will reach them as long as they continue to free-fall.

Case 2: Observer O is floating freely in flat spacetime and emits a light beam in the negative x-direction. The light beam moves away from him. Observer Z is also floating freely next to O and sees the light beam moving away from him also. Then O turns on his rocket engine and accelerates in the *negative* x-direction. The light beam still moves away from him; from the standpoint of Z, O is accelerating away in the same direction as the light beam, but the light beam is always ahead of O and is always gaining on him (the amount by which it gains decreases as O accelerates, but never quite reaches zero).

Hovering observer O emits a light beam *towards* the black hole. The light beam moves away the free-falling observer in exactly the same way it would in flat space-time, and from the perspective of Z the light beam is gaining on them slower. The Rindler horizon passes the hoverer but not the light beam because you’re between it and the horizon and the horizon never reaches you, just like the event horizon and Rindler horizon and c in flat space-time.

So in neither case does the light beam "turn around and come back" from any observer's standpoint.

Then how could the light beam or anything else ever reach the event horizon, whether you’re free-falling or even accelerating towards it? It doesn’t make sense.

In the analogous cases in curved spacetime, both cases work the same as above; the only difference is that in the curved spacetime analogue of case 1, observer O can become a "hoverer" as long as he starts his rocket while he's still above the black hole horizon (and if O does hover, the black hole horizon works the same as the Rindler horizon in the flat spacetime case). In the curved spacetime analogue of case 2, O will reach the black hole horizon before Z does, but after the light beam does.

The flat space-time version of what you just said is in case 1 observer O can become an inertial observer as long as they start their retros and decelerate before they have accelerated to c, and in case 2 O will reach the event horizon before Z, but Z is hovering. So again, the event horizon isn’t equivalent to the Rindler horizon. Then you said the light beam would have already caught up to all the other light at the horizon, but light doesn’t stop at the horizon. It just keeps slowing down as it approaches.

If an accelerator has almost reached c from the perspective of an inertial observer the accelerator still wouldn’t be able to accelerate to c no matter how much harder they accelerate. Just outside the horizon you’re saying that the tiniest amount of extra gravity will accelerate you harder than an infinite amount of energy could, straight to c relative to everything, even an object one Plank length further behind accelerating towards the event horizon using the energy of a quadrillion purploid big bangs, if standard GR is right. That’s very silly.


SR: The speed of light is constant. It moves away outward relative to every non-accelerating object at c. An object accelerates away from you, let’s say smoothly at an increasing rate proportional to the difference between their speed and c and so that its energy output is quadrupled when the difference is halved. Their energy output continues to increase but their acceleration relative to you decreases as they approach the speed of their own light. On board the accelerating ship they also see themselves catching up to their own light at the same rate. From your point of view it’s going much faster in their frame than it is in your frame to keep its velocity the same in both. So when you move relative to light you have to replace acceleration with velocity, the difference meaning it takes less proper time to cover a greater distance in the direction you’ve accelerated.

GR: An event horizon moves at c. It moves away inward relative to every non-accelerating object at c locally, but not from a distance. An object free-falls away from you towards an event horizon. They continue to accelerate away from you at an ever increasing rate but their acceleration relative to you decreases as they approach the speed of their own light. On board the free-falling ship they also see themselves catching up to their own light at the same rate. From your point of view it’s going much faster in their frame than it is in your frame to keep its velocity the same in both. So when you move relative to light you have to replace acceleration with velocity, the difference meaning it takes less proper time to cover a greater distance in the direction of the mass.


I told you they’re the same. Pretty isn’t it. Looks like a butterfly. Dam that was hard work. Writing it was easy, it almost explains itself. I can’t imagine any other subject doing that. Putting it together into a coherent description so it didn’t read like a complete mess was a bloody nightmare. Please tell me you can see it now. I think this is about the best I can do.

 

[PeterDonis]

It would still be equivalent to accelerating to c

Huh? You are saying that if I pass a light beam that is moving towards me at c, that requires me to accelerate to c myself? How do you figure that?

The event horizon moves inwards for the same reason c moves outwards, because that’s the direction that mass and energy curve.

You are again claiming that curvature due to gravity is the same as curvature due to proper acceleration. This is not the case in standard GR, and you haven't justified it. More on this in my reply to your next post.

Most of the rest of your post just re-states things I've already stated are false in standard GR, without further justification.

If we take away gravity and apply energy to a group of objects of varying mass then they’d all move at the same rate too.

Really? So if I take two identical rocket engines with identical thrust, and attach one to a hundred ton object (by "ton" I mean a ton of mass, a thousand kilograms) and one to a two hundred ton object, and turn both engines on, both objects will accelerate identically? How do you figure that?

First C wouldn’t fall at the same rate that it would on its own because the difference in gravitational strength between A and C is greater than the difference between C and B.

Not necessarily. You'll note that I nowhere specified the actual proper distances between A and C and B and C. I only specified that C was at the center of mass of the object as a whole. If the object extends far enough radially that the difference in the strength of gravity is noticeable, then the object's structure will also change in response to that difference, and therefore so will the location of its center of mass.

And second, you say tidal force isn’t due to tidal gravity. Yes it is because it’s the difference in the strength of gravity between two parts of the same object that causes the tidal force between them.

No, it isn't. It's the internal forces between the different parts of the object, that are causing those parts to move on non-geodesic worldlines. The only reason tidal gravity comes into it at all is that tidal gravity makes the geodesic worldlines diverge, which is equivalent to making worldlines that remain at a constant proper distance from one another non-geodesic worldlines. But if there were no internal forces between the parts of objects then they would not feel any force at all, even though tidal gravity was present; the parts would simply diverge as they all traveled along their separate geodesics.

There’s a source of energy, in this case a rocket that pushes along the length of the object. The parts that are closest to the rocket get pushed the hardest but this isn’t felt

What is your basis for this? How can the rocket push on the object if the object feels no force?

If we strapped a tiny rocket to each atom and accelerated them all together then I don’t think the object would feel any acceleration, unless it was sensitive enough to feel the difference of the rate of acceleration between the front and the back of the individual atoms, which seems unlikely. Are you sure a point-like object feels proper acceleration but not tidal force?

I'm sure that standard GR predicts that an idealized point-like object would feel proper acceleration if a rocket engine were strapped to it and turned on. Obviously there are no idealized point-like objects, but subatomic particles like electrons, which as far as we know have no internal structure, have been accelerated in particle accelerators for decades and they show all the effects of feeling acceleration. I guess to know for sure we'd have to figure out how to attach accelerometers to them, but no physicists that I'm aware of doubt what the result would be if we could run that experiment.

 

[PeterDonis]

If an object were able to ignore the fact that it can’t be done and cross an event horizon then any signal sent from inside the horizon wouldn’t be able to reach anything outside the horizon. The equivalent to that in flat space-time is ignoring the fact that c can’t be reached and accelerating past it anyway so that no signal sent from the accelerator will reach anything in the opposite direction of its motion.

Once again you've got this backwards. An object that is behind the Rindler horizon can't send a signal that will catch up with the *accelerator*. The Rindler horizon has nothing to do with where signals sent *from* the accelerator can or can't go. Why do you keep mixing this up?

There is in the flat space-time case. It’s called a Rindler horizon. I take it you meant curved space-time.

No, there is no such point in either case, flat or curved spacetime. The statement I made is correct exactly as I made it.

The free-faller in curved space-time is equivalent to the accelerator in flat space-time.

Since I disagree with you about this equivalence, obviously I'm going to disagree about any conclusions you draw from it. In standard GR the free-faller in curved spacetime is equivalent to the free-faller in flat spacetime, and the accelerator (hoverer) in curved spacetime is equivalent to the accelerator in flat spacetime. (D'oh.)

Cease to exist? If it stopped existing why would the mass stay? You didn’t even mean turned into energy did you? Why do you think something can just stop existing? That’s insane!

The nature of the singularity is a separate question from whether or not the horizon can be reached and crossed, so I'd prefer to start a separate thread if you really want to debate it. But the statement I made is correct if you take standard GR at face value. Most physicists believe that the presence of the singularity is telling us that GR is no longer valid when you get too close to the singularity, and some new physics (like quantum gravity) comes into play. But we don't know for sure at this stage of our knowledge.

You seemed to be suggesting that crossing an event horizon is the same as crossing a Rindler horizon in flat space-time. In standard GR anything free-falling is at rest and you have to accelerate to remain at rest in a coordinate sense, so space-time is curved. When the curvature goes past 90 degrees no amount of acceleration can keep you at rest in a coordinate sense, so you can’t hover. I get it. I always have, and it seemed to make sense to start with.

This looks fine to me except for the part about "curvature goes past 90 degrees". I don't think that's a good description of what happens at the horizon. 90 degrees relative to what? (If you had said "the outgoing side of the light cone goes past 90 degrees", that would be a little better, but I would still want you to clarify 90 degrees relative to what?)

It doesn’t make sense though. For a start a Rindler horizon needs an accelerator to form. An event horizon needs gravity but a free-faller isn’t accelerating in the standard GR model and the hoverer is, but the free-faller still crosses the event horizon if they stop hovering. In flat space-time if they stop accelerating the Rindler horizon disappears. The event horizon doesn’t disappear if the hover stops accelerating so it’s not equivalent to a Rindler horizon.

I don't disagree with this as it's stated but I don't think it means what you think it means. Obviously flat spacetime is not identical in every respect to curved spacetime; the fact that a curved spacetime can have an invariant event horizon that is there regardless of any observer's state of motion, while a flat spacetime can't, is one of the respects in which the two are not identical. But that doesn't prevent us from defining a family of "accelerators" in flat spacetime in such a way that they can be usefully viewed as equivalent to "hoverers" above a black hole in curved spacetime, with the Rindler horizon of the flat spacetime accelerators being equivalent to the event horizon of the black hole. It just means the analogy between the two is not complete in every respect. But so what? It's still a useful analogy. It's not meant to be anything more than that.

In standard GR the free-faller crosses the event horizon but not from the perspective of the hoverer.

Again, I don't disagree with this as stated, with an appropriate definition of "the perspective of the hoverer". But I don't think it means what you think it means. See next comment.

In flat space-time the equivalent to that would be an accelerator reaching c from their own perspective and catching up eith their own light, but not from the perspective of an inertial observer.

No, this is *not* what the equivalent would be. The equivalent in flat spacetime is the free-faller crossing the Rindler horizon of an appropriately defined accelerating observer. The free-faller does not catch up with his own light, in either the flat or the curved spacetime case, from any observer's perspective. The correct equivalent statement in the flat spacetime case is that from the perspective of the accelerating observer, in flat spacetime, the free-faller never crosses the Rindler horizon (again, with the appropriate definition of "the perspective of the accelerating observer"), but the free-faller does cross the Rindler horizon from his own perspective.

An object can keep on accelerating forever, making their space-time more and more ‘distance shortened’. They can use this to travel huge distances in space-time from the perspective of their starting frame it a short amount of proper time for them, but when they’ve stopped accelerating they never reached c from their original frame, they just kept on accelerating and covering more and more space-time over less and less proper time until the energy that was accelerating them stopped.

This is all true, but as you show with your final phrase, you recognize that it applies to an observer that is being accelerated by energy. Such an observer *feels* the acceleration, and has a Rindler horizon and so forth only as long as he *feels* acceleration. When the energy that is accelerating him stops, he stops feeling acceleration.

None of that applies to an observer that is in free fall the whole time. Yes, in a coordinate sense such an observer can be "accelerated" by gravity, but he never *feels* any acceleration. This is why the physical distinction between observers in free fall and observers that feel acceleration is so crucial in standard GR. The rest of your discussion here denies this crucial distinction, and that is a fatal flaw in my view; it is simply not valid physics to claim that an observer that feels acceleration (which has lots of observable effects that aren't observed in free fall) is equivalent to an observer in free fall.

Proper acceleration, including tidal force. Acceleration that’s felt.

Tidal gravity can be present even though it is not felt by any observer. I have repeatedly described such scenarios. This makes it fundamentally different from proper acceleration. The standard GR definition of "curvature" recognizes two distinct types of curvature, as I've said before. Tidal gravity is curvature of spacetime itself; proper acceleration is curvature of a particular worldline. They are different things.

I missed a word. I should have said ‘And NOTHING reaches from an accelerators perspective. But you’re saying objects closer to the horizon than the free-faller and the free-fallers own light can reach and cross the event horizon as if it was just an arbitrary line in flat space-time? I don’t see how that works if they then accelerate away?

Once they're below the horizon, they can accelerate all they want to but it won't bring them back above the horizon. It *will* cause them to fall more slowly (assuming they accelerate radially outward) than objects free-falling inward, but they will still be falling.

The free-faller falls until they’re almost at the horizon, then start hovering.

So they are accelerating *outward*. Ok, that clears up what you are describing.

Same three observers. Hoverer, free-faller, accelerator. The accelerator matches their velocity relative to the hoverer precisely with the free-fallers velocity relative to the hoverer.

How can an object that is accelerating match velocities with an object that is free-falling, at least for more than a single instant?

I don't understand the rest of what you are trying to describe here, probably because of the question I just asked.

In curved space-time observer O is hovering at some distance from the event horizon and emits a light beam *away* from the black hole. Observer Z starts off hovering next to O then free-falls towards the event horizon. The light beam moves away in exactly the same way it would if they accelerated in flat space-time, and there should be the equivalent of a Rindler horizon that moves closer as they free-fall so that it passes first the light beam, then the hoverer, and no signal sent from the other side will reach them as long as they continue to free-fall.

I really wish you would draw a spacetime diagram showing how you think this works. It is true that, after a certain point on the hoverer's (O's) worldline, he will not be able to send any signal to Z that will reach Z before Z crosses the black hole horizon (call this point A for further use below). There is also a further point on O's worldline after which he will not be able to send a signal to Z that will reach Z before Z hits the singularity (call this point B for further use below). I suppose either of those points could be used to define a kind of "horizon" for Z, but I don't know if there would be any useful equivalence between such a horizon and a Rindler horizon. I would have to see more details (like a diagram) to better understand what you're imagining.

Hovering observer O emits a light beam *towards* the black hole. The light beam moves away the free-falling observer in exactly the same way it would in flat space-time, and from the perspective of Z the light beam is gaining on them slower. The Rindler horizon passes the hoverer but not the light beam because you’re between it and the horizon and the horizon never reaches you, just like the event horizon and Rindler horizon and c in flat space-time.

This is not correct. If O emits the light beam before point A on his worldline (as defined above), the light beam will pass Z before Z crosses the black hole horizon. If O emits the light beam before point B on his worldline, the beam will pass Z after Z crosses the black hole horizon, but before Z hits the singularity. Only if O emits the light beam after point B will it never reach Z (because Z will hit the singularity first). So in so far as there is a "horizon" associated with Z, it does pass the light beam if the light beam is emitted soon enough.

Please tell me you can see it now. I think this is about the best I can do.

Sorry, but even if all that was hard work, I can't say I can see it now, because none of it gets any additional useful information across to me. Why can't you draw a diagram?

 

[PeterDonis]

The parts that are closest to the rocket get pushed the hardest but this isn’t felt, and the strength is reduced as an inverse square of the distance to the rocket. This difference is felt as proper acceleration as the front of the object gets pushed along by the back

On re-reading I realized I should comment on this in more detail. I see at least one glaring problem with this line of reasoning:

If the "strength of acceleration" gets reduced as the inverse square of the distance, then the *difference* in strength between points at two different distances, which is what you say is actually felt, goes as the inverse *cube* of the distance, times the difference in distances. The equivalent for gravity would be that, for example, the weight you read on your bathroom scale would have to be, not the "acceleration due to gravity" at Earth's surface, 9.8 meters per second squared, times your mass, but that force times the *ratio* of your height to the radius of the Earth. This ratio is about one to three million (assuming you are about 2 meters tall; the Earth's radius is about 6 million meters), so your reasoning gives an answer for an easily observable everyday fact, the weight people observe on their bathroom scales, that is too small by a factor of three million. Too bad you're allergic to math.

 

[PeterDonis]

On board the accelerating ship they also see themselves catching up to their own light at the same rate.

I said that this whole part of your post didn't convey any additional useful information to me, but on re-reading I do want to comment on this. If you mean this to describe what standard SR says (as opposed to something in your personal model that may or may not match what standard SR says), then it's wrong. The accelerating ship *never* "catches up" to its own light from *either* perspective, its own or that of the free-faller that sees the accelerator's velocity approaching c but never quite reaching it. From *both* perspectives, a light ray emitted by the accelerating ship will continually move *away* from the ship; its distance from the ship will continually *increase*, from *both* perspectives, and the velocity of the light beam will always be faster than the velocity of the ship, from both perspectives. There is *never* a time when the distance from the accelerating ship to the light beam *decreases*, or when the accelerating ship appears to move faster than the light (which is what "catching up" would mean), from either perspective.

 

[matphysik]

Repulsive and attractive `forces` is Newtonian theory(!).

 

[matphysik]

Suppose we pick time t = 0, the time at which the two bodies are mutually at rest, to start running time backwards.

Gravitational Time Dilation: The proper time measured by a clock moving with 3-velocity vᵃ= dxᵃ/dt (a=1,2,3.) in a spacetime with metric gᵤᵥ (u,v=0,1,2,3.) is given by:
dτ = √(- c⁻ ² gᵤᵥdxᵘdxᵛ)·dt= √(-g₀₀ - 2gₐ₀ dxᵃ vᵃ/c - v²/c²)·dt. Where v²=gₐₑvᵃvᵉ (a,e=1,2,3.).

For v=0, dτ =√(-g₀₀)·dt. Proper time would decrease only if coordinate time does (why should it?). You must begin with v>0.

 

[PeterDonis]

Proper time would decrease only if coordinate time does (why should it?). You must begin with v>0.

Wow, you're really taking this thread back a ways. You might want to read the other 480-odd posts.

I wasn't trying to claim that either proper time or coordinate time can "run backwards" from the viewpoint of an actual observer. The original question was about whether the laws of physics are time symmetric. If they are, that means, roughly speaking, that if I have a certain sequence of events S that conforms to the laws of physics, if I consider another sequence of events S', which is the exact reverse of S, S' must also conform to the laws of physics.

"Running time backwards" is just a colloquial way of referring to the fact that S' is the exact reverse of S; and saying that a certain time, such as t = 0, is where we "start running time backwards" is just a colloquial way of saying that t = 0 is one endpoint of the expanse of time that is covered by S and S'. Saying that S' is the reverse of S is just saying that S' and S consist of identical events but in opposite order: the "forward" direction of time in S' is the opposite of the forward direction of time in S (so the time I am labeling t = 0 would be the *last* event in S, but the *first* event in S', in the particular example you were referring to). But an observer experiencing either sequence of events, S or S', would consider time to be running "forwards" in the direction in which he is "moving through" the sequence (again, colloquially speaking; there are mathematical ways to make all this more precise).

 

[Dale]

It would still be equivalent to accelerating to c

This is incorrect. We discussed this already and I thought you had agreed on this point (see your post 471). There is no local inertial frame where the velocity of an object free-falling across the event horizon becomes c. I.e. a free-falling object's worldline is timelike at all points including the event horizon, it never becomes lightlike.

The event horizon moves inwards for the same reason c moves outwards

This doesn't make sense. c is a speed, it doesn't have a direction. What direction is 100 kph?

A lightlike object can move towards, away, or tangentially to an observer, and all of those motions are at c.

Not for me. Not unless you also describe using words why I should believe what those equations are describing.

Because the equations match the results of many experiments in related situations. That is the only reason we should believe any equations of physics.

In particular, the Schwarzschild spacetime and it's GR predictions have been demonstrated to be a good model for the Pound-Rebka experiment, GPS, the Shapiro delay, the deflection of light, the anomalous precession of Mercury, and the Hafele-Keating experiment. GR is the simplest theory of gravity existing which quatitatively agrees with experiment.

 

[Passionflower]

In particular, the Schwarzschild spacetime and it's GR predictions have been demonstrated to be a good model for the Pound-Rebka experiment, GPS, the Shapiro delay, the deflection of light, the anomalous precession of Mercury, and the Hafele-Keating experiment.

With the exception of the Hafele-Keating experiment I would agree that they match GR predictions but I would disagree that they experimentally validate a Schwarzschild spacetime.

For instance in particular the Pound-Rebka experiment is simply not accurate enough to validate a Schwarzschild spacetime, and all that is shown is gravitational time dilation (which is of course a great achievement).

The reason I do not like to include the Hafele-Keating experiment is that I do not think one can make conclusions based on the data which I think is not accurate enough.

 

[Dale]

For instance in particular the Pound-Rebka experiment is simply not accurate enough to validate a Schwarzschild spacetime, and all that is shown is gravitational time dilation (which is of course a great achievement).

The reason I do not like to include the Hafele-Keating experiment is that I do not think one can make conclusions based on the data which I think is not accurate enough.

The accuracy of these prototypical experiments is not a big issue IMO, because all of these experiments have had more accurate follow-up experiments that remain consistent with GR and the Schwarzschild spacetime. I don't know of any simpler theory of gravity that is consistent with all of them together, and I don't know of any other spacetime that is consistent with all of them. Do you?

 

[Passionflower]

The accuracy of these prototypical experiments is not a big issue IMO, because all of these experiments have had more accurate follow-up experiments that remain consistent with GR and the Schwarzschild spacetime. I don't know of any simpler theory of gravity that is consistent with all of them together, and I don't know of any other spacetime that is consistent with all of them. Do you?

Perhaps you misunderstood me, the experiments (for simplicity I leave out Hafele-Keating here) are great and were a victory for GR.

However the Schwarzschild solution shows more than just time dilation. For instance a super accurate Pound-Rebka like experiment could demonstrate that in Schwarzschild coordinates r is not equal to rho, but unfortunately the spacetime curvature is simply not strong enough on Earth to detect that. So therefore I do not think it is accurate to say that the Schwarzschild solution has been experimentally verified to be accurate by the experiments you listed. To demonstrate this one needs a strong gravity environment.

 

[Dale]

However the Schwarzschild solution shows more than just time dilation.

Sure, that is why I included more than just one experiment. I don't really get your point in the context of this thread. Can you please answer the following question:

Do you believe that there is a simpler theory of gravity or a simpler spacetime which quatitatively agrees with all of the above experiments taken together?

To demonstrate this one needs a strong gravity environment.

What is "strong" depends on the sensitivity of the instruments and the magnitude of the effect being tested. If an effect can be detected then it is strong enough, by definition.

 

[Passionflower]

Sure, that is why I included more than just one experiment.

OK, then tell me which of those experiments do you think show the validity of the Schwarzschild solution, even by approximation.

There are some cases that demonstrate GR under strong fields however I think none of those are listed by you. For instance you could have mentioned PSR J0737-3039.

 

[Passionflower]

What is "strong" depends on the sensitivity of the instruments and the magnitude of the effect being tested. If an effect can be detected then it is strong enough, by definition.

Strong gravity or strong field, is a fairly common phrase. For instance in the Schwarzschild solution it happens when the discrepancy between delta r and delta rho becomes a factor.

 

[Dale]

OK, then tell me which of those experiments do you think show the validity of the Schwarzschild solution, even by approximation.

All of them taken together. Perhaps you don't understand what I mean by all of them taken together. I mean that it simultaneously fits all of those experiments and all of their more accurate successors. I don't mean and I didn't claim that anyone of them in isolation shows it.

Can you please answer the question that I have asked you 3 times now:

Do you believe that there is a simpler theory of gravity or a simpler spacetime which quatitatively agrees with all of the above experiments taken together?

There are some cases that demonstrate GR under strong fields however I think none of those are listed by you. For instance you could have mentioned PSR J0737-3039.

Sure, it wasn't meant to be an exhaustive list. Just a list of some of the more famous ones that A-wal probably is aware of.

 

[PeterDonis]

The reason I do not like to include the Hafele-Keating experiment is that I do not think one can make conclusions based on the data which I think is not accurate enough.

It's worth noting that the GPS system is basically a re-run of the H-K experiment with more clocks and a wider range of relative velocities, and it confirms the GR predictions with a significantly higher accuracy than the H-K experiment did. This paper on the living reviews site gives a good overview:

http://relativity.livingreviews.org/Articles/lrr-2003-1/