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

[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/

 

[Passionflower]

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/

I am in no way questioning the validity of GR, but in my opinion the H-K experiment could not have concluded it with the available test results, I have a similar opinion on the experiment by Eddington on Principe. In both cases the margins of error where too large in my opinion to make a conclusion.

 

[PeterDonis]

I am in no way questioning the validity of GR, but in my opinion the H-K experiment could not have concluded it with the available test results, I have a similar opinion on the experiment by Eddington on Principe. In both cases the margins of error where too large in my opinion to make a conclusion.

I definitely agree in the case of Eddington's eclipse observations in 1919. In that case, IIRC, the error bars were large enough that the results could not rule out the possibility that there was *zero* light bending by the Sun.

I'm not sure I agree in the case of the original H-K experiment. It was consistent with the GR predictions to within the accuracy of both the theoretical calculations and the experimental measurements, and it was definitely *not* consistent with no effect being present. (That is, the differences in clock readings were *not* zero to a very high degree of statistical significance.) It was also definitely *not* consistent with the known SR time dilation due to relative velocity being present (I say "known" because it had been verified by many other experiments by that time), but the predicted GR time dilation due to altitude difference not being present; only both effects together gave a prediction consistent with the experimental results. The Wikipedia page gives the values for the SR and GR predictions separately:

http://en.wikipedia.org/wiki/Hafele–Keating_experiment

I believe similar numbers are given in MTW (don't have my copy handy right now to check exactly where). In view of all this, I would say the H-K experiment gave significantly better evidence for GR than Eddington's eclipse observations did.

 

[A-wal]

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?

What? That’s not what I’m saying. I meant that it would be the equivalent of c because it’s the point when c wouldn’t be fast enough to move away. It’s the point when gravity would have accelerated you all the way to c. Mass can’t do that any more than energy can. The speed that the event horizon moves inwards is c at the horizon and slows down as an inverse square of the objects distance, so you would have to accelerate past c to catch it. The reason it moves inwards is because there’s a limit to how fast objects can move relative to other objects but there’s no limit to mass or energy, so it can keep on accelerating you towards it but you’ll never reach a relative velocity of c, so you get an event horizon.

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?

That’s not what I said. In that situation it’s two objects pushing against each other and the difference in mass determines how the energy is spread out between them and which one moves fastest relative to a third object. What I meant was simply that if you could turn off gravity it would take no more effort to lift Mount Everest than it would to lift your arms, as long as you had the Earth to push against. That’s why objects with different masses fall at the same rate. Objects with very different masses will move at the same speed if pushed in 0g and objects with very different masses will move (fall) at the same speed if there’s no energy.

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.

It wouldn’t fall faster if it was longer? Do you mean that it gets ‘weighted’ so that B moves closer to a keeping the difference in gravitational strength between A and C the same as the difference in gravitational strength between C and B?

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.

Yes I know. Tidal gravity would separate the parts and if they had something that was holding them together then they’d feel tidal force. Tidal gravity causes any extended object (all of them) to feel tidal force. What’s the problem?

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

I’m trying to show that I think the proper acceleration could be smoothed out to match gravity, making proper acceleration and tidal force equivalent.

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.

I still don’t really see how a point-like object fells anything. It’s not just a technicality.

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?

I wasn’t talking about a Rindler horizon. I was talking about an event horizon and its flat space-time equivalent, c!

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

What? Now I’m really confused. There is a point in flat space-time when no signal sent from an object will reach an accelerator, when they pass the accelerators Riindler horizon. There’s also an equivalent in curved space-time when an object has passed the free-fallers equivalent of the Rindler horizon.

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.)

In curved space-time the hoverer is equivalent to the free-faller/inertial observer because their acceleration from energy and their acceleration from mass are balanced. That’s why they hover. An accelerator free-falling towards the black hole is equivalent to an accelerator in flat space-time.

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.

I think the presence of the singularity is telling us that matter has collapsed enough to prevent any other matter reaching a certain distance of that point in space-time in a finite amount of proper time because they’d have to move faster than c, creating an event horizon.

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?)

The dimensions in flat space-time are at 90 degree angles to each other. The curvature of gravity/acceleration decreases the angle causing distance shortening. At c or an event horizon the angle is 0, so it’s changed by 90 degrees.

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.

It is absolutely identical. I can’t stress that enough. When I use the analogy between the two pairs of horizons it’s meant to mean a lot more than that.

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.

The four horizons:

The Rindler horizon in flat space-time is the point when no signal sent will be able to catch an accelerator. It’s starts far away and follows the accelerator at a constant distance if they keep their acceleration constant. If they decrease their acceleration the gap between them increases and it decreases if they increase their acceleration. The Rindler horizon can never catch up to them. It gets harder to close the gap the harder they accelerate.

It’s the equivalent of c. C is a horizon in front of the accelerator that starts far away, which the accelerator follows at a constant distance if they keep their acceleration constant. If they decrease their acceleration the gap between them increases and it decreases if they increase their acceleration. They can never catch up it. It gets harder to close the gap the harder they accelerate. The rate they gain on the light is the same in every frame, even accelerated ones. The two horizons would meet at the same place as the accelerator if they were able to reach c.

It’s the exact equivalent of curved space-time. The Rindler horizon in curved space-time is the point when no signal sent will be able to catch a free-faller. It’s starts far away and follows the free-faller, getting closer to them as the strength of gravity increases. If they decrease their acceleration the gap between them increases and it decreases if they increase their acceleration. The Rindler horizon equivalent can never catch up to them. It gets harder to close the gap the harder they free-fall.

It’s the equivalent of the event horizon. The event horizon starts far away in front of the free-faller, which the free-faller approaches at an increasing rate to start with as the strength of gravity increases, then at a decreasing rate as it gets harder to close the gap the harder the free-fall. They can never catch up it. It gets harder to close the gap the harder they free-fall. The rate they gain on the light is the same in every frame, even accelerated ones. The two horizons would meet at the same place as the free-faller if they were able to reach c.

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.

All of it applies to an observer that’s in free-fall the whole time. Just replace energy with gravity:

An object can keep on free-falling towards an event horizon for as long as the black hole lasts, 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 mass that was accelerating them stopped.

Tidal gravity can be present even though it is not felt by any observer. I have repeatedly described such scenarios.

How can tidal gravity be present but not felt? There’s no such thing as an actual point-like object except a singularity, and even they don’t exist in for any amount of time so they can’t feel anything.

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.

How are they different things? Describing them differently doesn’t make them different.

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.

That’s not what I meant. They accelerate away before they reach the event horizon. If they can accelerate “all they want” then they can escape. That’s what accelerate all they want means! I don’t know why you are okay with this mystical attitude towards gravity? It accelerates objects in exactly the same way that energy does. Neither can accelerate you to c!

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

Easy. They know the mass of the object pulling in the free-faller, so they know exactly how fast they accelerate needs to move relative to the hoverer to keep it symmetric.

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.

That IS the Rindler horizon! The diagram would look identical to a Rindler in flat space-time because a Rindler horizon is equivalent to THIS horizon and NOT an event horizon.

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.

Z never crosses the event horizon. Z is the hoverer. I’ll call them F and H.

Hovering observer H emits a light beam *towards* the black hole. The light beam moves away from the free-falling observer in exactly the same way it would in flat space-time, then slows as it approaches the event horizon and from the perspective of H the light beam has a slower velocity relative to the light beam. 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.

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?

A diagram of what?

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.

You’ve completely lost me. Why cube? All I meant was if the acceleration was distributed through the object to match exactly with the way gravity is distributed through a free-faller then proper acceleration and tidal force would be exactly equivalent.

 

[A-wal]

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.

I wish you’d stop calling it my model. It’s worked the way it has forever whether I’m right or wrong. I didn’t make it. I found it, but apparently I’m the only sod who can see it so I’ve got to try to explain exactly what it looks like.

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.

No, catchING. They can never actually match its speed. It will always pull away. As I’ve said repeatedly, it gets harder to close the gap the harder you accelerate. From the free-faller/accelerators perspective it’s because the ‘distance shortening’ would have to be infinite to reach c. From a hoverer/inertial observers perspective light moves at a constant speed and it’s because the free-faller/accelerators mass increases.

I wish you wouldn’t call an inertial observer a free-faller. It’s very misleading. It implies an object at rest is equivalent to an object being accelerated by gravity. Giving them the same name doesn’t make them equivalent.

Why can’t you apply that logic to an event horizon? Look:

The free-falling ship *never* "catches up" to its own light from *either* perspective, its own or that of the hoverer that sees the free-fallers velocity approaching c but never quite reaching it. From *both* perspectives, a light ray emitted by the free-falling 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 free-falling ship to the light beam *decreases*, or when the free-falling ship appears to move faster than the light (which is what reaching the event horizon would mean), from either perspective.


The accelerator almost reaches the speed of their own light from the perspective of the inertial observer, and from their own perspective. As they accelerate they send a series of signals, one every second of the accelerators proper time. The signals received by the inertial observer get progressively less frequent, but they never stop coming because the accelerator can never reach the speed of their own light.

The free-faller almost reaches the speed of their own light from the perspective of the hoverer, and from their own perspective. As they free-fall they send a series of signals, one every second of the free-fallers proper time. The signals received by the hoverer get progressively less frequent, but they never stop coming because the free-faller can never reach the event horizon.

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.

Doesn’t work you run an object crossing an event horizon backwards though does it?

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 is the point when the even the speed of light isn’t enough to move you towards a hoverer. It is exactly equivalent. Working out acceleration relative to a hoverer in free-fall is no different from acceleration relative to an inertial observer in flat space-time. You know how acceleration works, the fact that it gets harder to accelerate the faster your relative velocity. Why do you think gravity can accelerate an object to a relative velocity of c? That would take a huge jump, an infinite jump in fact, of energy. It couldn’t happen smoothly because up until the event horizon acceleration and gravity are equivalent because you can always move away using energy.

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.

The event horizon moves inwards because gravity pulls inwards, and c moves outwards because energy pushes outwards. They both move away.

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.

What I’ve been saying is consistent with Schwarzschild space-time and it's GR predictions.


There’s still some important things I just don’t get. How does your model explain these:

(1) What happens if you free-fall towards the black hole with your light in front of you crossing the event horizon, then pull away before you reach the horizon? Your light would have to turn round and come back across the event horizon! WFT!

(2) How can two coordinate systems that say completely different things (one saying an object crosses a black hole and one saying it doesn’t) possibly be consistent with each other? I still haven’t had an explain of how they could both be right. This is the most glaring problem.

(3) What is the difference between flat and curved space-time? The reason you’re getting these ridiculous predictions about gravity is because GR distinguishes between the movement of an object and a change in the amount of space-time between them. What’s the difference? The difference between flat and curved space-time is in flat space-time it's the falling object that moves and gravity is a force, whereas in curved space-time the faller is at rest. It's relative. There is no distinction! Gravity works just like acceleration does!

(4) You know that an object in free-fall outside the event horizon can’t reach a velocity of c relative to an object accelerating away from the black hole at almost c. Gravity can keep on accelerating them but it gets harder as their relative velocity increases. The way ‘distance shortening’ works is to create less space-time for the accelerator to travel through from their perspective and more from the perspective of a hoverer/inertial observer. So the free-faller is constantly traveling through a smaller and smaller amount of space-time from their perspective compared to the perspective of a distant observer, just like accelerating in flat space-time. Do you really believe it makes sense if the tiniest amount more gravity could accelerate a free-faller to c relative to a hoverer when the energy of a hydrillion cyanoid big bangs couldn’t?

 

[PeterDonis]

What? That’s not what I’m saying. I meant that it would be the equivalent of c because it’s the point when c wouldn’t be fast enough to move away.

So far, so good.

It’s the point when gravity would have accelerated you all the way to c.

Nope. As DaleSpam and I have both said repeatedly, an object free-falling through the horizon is timelike, not lightlike. It never moves faster than c (it always moves within the light cones).

The speed that the event horizon moves inwards is c at the horizon

The event horizon moves outwards, not inwards.

The reason it moves inwards is because there’s a limit to how fast objects can move relative to other objects but there’s no limit to mass or energy, so it can keep on accelerating you towards it but you’ll never reach a relative velocity of c, so you get an event horizon.

I have no idea what this means. Why can't you draw a diagram?

What I meant was simply that if you could turn off gravity it would take no more effort to lift Mount Everest than it would to lift your arms, as long as you had the Earth to push against.

Really? How do you figure that? Suppose I'm in zero g (way out in deep space somewhere), and I run the following experiment: I pick a certain force, F, and I exert that force both on Mount Everest and on my arms. (How I exert the force is immaterial; I can exert it by pushing against the Earth, or by using a small rocket engine, or whatever.) Since Mount Everest's mass is much greater than the mass of my arms, the acceleration induced on Mount Everest by force F will be much smaller than that induced on my arms. So I don't see how "it would take no more effort to lift Mount Everest than it would to lift your arms" in zero g, unless by "no more effort" you simply mean I can exert the same force on both objects without caring that the accelerations induced are vastly different. But I don't think that's what you meant, because you say:

Objects with very different masses will move at the same speed if pushed in 0g

If by "pushed" you mean "pushed with the same force", this is false.

Do you mean that it gets ‘weighted’ so that B moves closer to a keeping the difference in gravitational strength between A and C the same as the difference in gravitational strength between C and B?

More or less, yes. But I am not trying to say that this will continue to hold regardless of how long the object is; I'm limiting discussion to objects which are very short compared to the size of the gravitational source.

Tidal gravity would separate the parts and if they had something that was holding them together then they’d feel tidal force. Tidal gravity causes any extended object (all of them) to feel tidal force. What’s the problem?

You just stated that the objects only feel tidal force when something is holding them together. So clearly the tidal *force* is not caused by tidal *gravity* (which would separate the parts), but by the something that's holding them together.

I’m trying to show that I think the proper acceleration could be smoothed out to match gravity, making proper acceleration and tidal force equivalent.

And I'm saying you haven't succeeded.

I still don’t really see how a point-like object fells anything. It’s not just a technicality.

I'll comment on this near the end when I discuss the definition of tidal gravity.

I wasn’t talking about a Rindler horizon. I was talking about an event horizon and its flat space-time equivalent, c!

No, the flat spacetime equivalent of the event horizon *is* the Rindler horizon. It's not "c"--saying the event horizon is "equivalent to c" doesn't even make sense.

What? Now I’m really confused. There is a point in flat space-time when no signal sent from an object will reach an accelerator, when they pass the accelerators Riindler horizon. There’s also an equivalent in curved space-time when an object has passed the free-fallers equivalent of the Rindler horizon.

You keep on mixing up the case of *outgoing* motion with the case of *ingoing* motion. The original statement of yours that I responded to was about *ingoing* motion--you were saying that there was a point where the hoverer couldn't send signals to the free-faller. Now you're talking about *outgoing* motion, about the free-faller sending signals to the hoverer. Those are two different cases, and if you would stop mixing them up it would help you to avoid confusion.

In curved space-time the hoverer is equivalent to the free-faller/inertial observer because their acceleration from energy and their acceleration from mass are balanced. That’s why they hover. An accelerator free-falling towards the black hole is equivalent to an accelerator in flat space-time.

Continuing to repeat this doesn't make it right. You still have not explained how an observer that feels acceleration (the hoverer) can possibly be physically equivalent to an observer that doesn't (the free-faller). Unless you can address that issue, what you are saying here is simply not valid physics, and no amount of repetition will make it so.

The dimensions in flat space-time are at 90 degree angles to each other. The curvature of gravity/acceleration decreases the angle causing distance shortening.

How do you figure this? I think you're confusing the "angles of the dimensions" (not very good terminology, but I think I understand what you mean by it), which do *not* change with curvature, with the angle between the outgoing and ingoing sides of the light cones, which *does* change with curvature. But even the latter does not collapse to zero at the horizon.

It is absolutely identical. I can’t stress that enough. When I use the analogy between the two pairs of horizons it’s meant to mean a lot more than that.

So you are claming that flat spacetime is absolutely identical to curved spacetime? Then how do you explain the fact that tidal gravity, an observable physical phenomenon, is present in curved spacetime but not in flat spacetime?

The Rindler horizon in flat space-time is the point when no signal sent will be able to catch an accelerator.

This is ok, although I'm not sure I understand the rest of what you say about it, with the acceleration changing. I think you're putting in too many complications at once, which makes it difficult to focus on particular issues.

C is a horizon in front of the accelerator that starts far away, which the accelerator follows at a constant distance if they keep their acceleration constant.

I have no idea what this means; I am not aware of any "horizon" in *front* of an accelerator in flat spacetime, much less one that somehow moves at c and yet remains at a constant distance in front of the accelerator. Why can't you draw a diagram?

The Rindler horizon in curved space-time is the point when no signal sent will be able to catch a free-faller.

And as I've repeatedly said, there is no such point, at least not in standard GR. Why can't you draw a diagram that shows how you think this works?

The event horizon starts far away in front of the free-faller, which the free-faller approaches at an increasing rate to start with as the strength of gravity increases,

Ok, more or less; it would be nice if you could define what "approaches at an increasing rate" actually means, because it might help you to see why this...

then at a decreasing rate as it gets harder to close the gap the harder the free-fall.

...is false.

All of it applies to an observer that’s in free-fall the whole time.

But an observer in flat spacetime who is accelerating due to "energy" is *not* in free fall! So how can he possibly be equivalent to a free-faller in curved spacetime (or anywhere else)?

How can tidal gravity be present but not felt?

I have repeatedly described how tidal gravity works. It causes freely falling trajectories to converge or diverge. That's the definition of tidal gravity. So by definition, it is not felt; it applies to objects in free fall. See next comment.

There’s no such thing as an actual point-like object

No, but there are certainly objects that are in free fall. Yes, "point-like object" is an idealization; it assumes that any internal structure of the object can be ignored, so it can be treated as though it were a single point located at its center of mass. The idealization works because the actual internal forces in objects in free fall are so small that they can be ignored for practical purposes in many problems; the motion of the object *is*, to within the precision of our measurements, the same as what we predict using the idealization of a "point-like object". If you weren't allergic to math you could actually work some problems and see this.

I suppose you could claim that, as long as there are *any* internal forces, then there is no such thing as true "free fall". However, even that would not get you off the hook; you would still have to make a quantitative prediction about what "tidal forces" your model says should be felt, and then actually measure the tiny internal forces that standard GR idealizes away in many cases, and show that those tiny internal forces are what your model predicts they should be. If you're able to do this, by all means go ahead.

How are they different things? Describing them differently doesn’t make them different.

Showing a physical observable that differs (whether or not a force is felt) does.

If they can accelerate “all they want” then they can escape. That’s what accelerate all they want means!

No it doesn't. "Accelerate all they want" means that the measured proper acceleration (measured by an accelerometer) is unbounded; it can be as large as you like. It does *not* mean that that unbounded proper acceleration can put you anywhere in spacetime that you want.

I don’t know why you are okay with this mystical attitude towards gravity? It accelerates objects in exactly the same way that energy does.

No, it doesn't. Acceleration due to energy causes you to feel acceleration (and you can measure it with an accelerometer). "Acceleration" due to gravity does not; an accelerometer reads zero. I don't know why you are okay with claiming that these two cases, with an obvious difference in an obvious physical observable, are somehow magically the same.

Easy. They know the mass of the object pulling in the free-faller, so they know exactly how fast they accelerate needs to move relative to the hoverer to keep it symmetric.

I don't see how this answers my question. We have one object that is feeling acceleration and one that is not. How can the two possibly match velocities for more than an instant?

That IS the Rindler horizon! The diagram would look identical to a Rindler in flat space-time because a Rindler horizon is equivalent to THIS horizon and NOT an event horizon.

No it isn't. Draw the diagram and you will see why.

Hovering observer H emits a light beam *towards* the black hole. The light beam moves away from the free-falling observer in exactly the same way it would in flat space-time, then slows as it approaches the event horizon and from the perspective of H the light beam has a slower velocity relative to the light beam. 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.

Again, draw the diagram.

A diagram of what?

Did you see the diagram I posted a while back? It was a spacetime diagram; time on the vertical axis and space (one dimension of it, but that's enough for what we're discussing) on the horizontal axis. Then a bunch of worldlines, plots of time vs. space for various objects and for any horizons that you claim are present. That's the kind of diagram I'm talking about. If you really have a clear mental picture of your model, you should be able to draw such diagrams for every case you've described.

You’ve completely lost me. Why cube? All I meant was if the acceleration was distributed through the object to match exactly with the way gravity is distributed through a free-faller then proper acceleration and tidal force would be exactly equivalent.

Once again, too bad you're allergic to math. If you weren't you would see that it's obvious that the derivative of 1/r^2 is proportional to 1/r^3. You are saying that "acceleration is distributed" as 1/r^2; that means the *change* in acceleration (its spatial derivative) is proportional to 1/r^3. So over a given spatial separation dr, the difference in acceleration, which is what you claim is actually felt, goes like 1/r^3 * dr.

If that's still too abstruse for you, think of it this way. I'm standing on the surface of the Earth, so the acceleration at my feet goes like 1/R^2, where R is the radius of the Earth. My height is h, which is much less than R. So at my head, the acceleration is 1/(R + h)^2. You are saying that it is the *difference* between these two that is what I actually feel as weight. Calculate the numbers and you'll see that the difference between those two numbers is a factor of about three million too small. All the bit about the cube was just a quick way of calculating that difference, roughly, without going through the full computation.

 

[PeterDonis]

I wish you’d stop calling it my model. It’s worked the way it has forever whether I’m right or wrong. I didn’t make it. I found it, but apparently I’m the only sod who can see it so I’ve got to try to explain exactly what it looks like.

Can you give a reference for where you found it? It doesn't resemble any model I've seen anywhere else, as far as I can tell.

No, catchING. They can never actually match its speed. It will always pull away.

If by "pull away" you mean "distance constantly increasing", then I have no issue, but that doesn't seem consistent with previous things you've said, which have strongly implied, at least to me, that you believe the accelerating ship can somehow *decrease* the distance between itself and the light beam if it accelerates long enough.

I wish you wouldn’t call an inertial observer a free-faller. It’s very misleading. It implies an object at rest is equivalent to an object being accelerated by gravity. Giving them the same name doesn’t make them equivalent.

I use the term free-faller as a simple physical description: a free-faller is an observer who doesn't feel any acceleration (accelerometer reads zero). Yes, that does imply that all such observers are equivalent in an important physical sense, even if one is at rest in flat spacetime and the other is moving solely under the influence of gravity in curved spacetime. That physical equivalence is one of the cornerstones of GR, so I don't have any intention of changing terminology. To me, it's you who are using misleading terminology, by saying that an object in free fall is being "accelerated by gravity" when it clearly feels zero acceleration.

The accelerator almost reaches the speed of their own light from the perspective of the inertial observer,

True.

and from their own perspective.

How so? From the accelerator's perspective, the light beam he emits always moves away from him at the same speed; his speed relative to it never changes. It's only from the inertial observer's perspective that the accelerator's speed changes relative to the light.

As they accelerate they send a series of signals, one every second of the accelerators proper time. The signals received by the inertial observer get progressively less frequent, but they never stop coming because the accelerator can never reach the speed of their own light.

You are mixing things up again. You're describing the accelerator sending signals to the free-faller, but as we'll see in a moment, that is *not* the analogous situation to the free-faller sending signals to the hoverer in curved spacetime.

If we turn this situation around and consider a free-faller sending signals to the accelerator, once every second of his proper time, we will find that the signals from the free-faller *do* stop coming; there is a "last signal" that the free-faller sends, the last second of his proper time before he crosses the accelerator's Rindler horizon and no signal he sends can reach the accelerator any more.

The free-faller almost reaches the speed of their own light from the perspective of the hoverer,

Again, true.

and from their own perspective.

Again, how so? From the free-faller's perspective, the light always moves away from him at c. It's only from the hoverer's perspective that the free-faller's speed appears to approach c. Furthermore, this similarity between the free-faller and the accelerator in flat spacetime is misleading, because this...

As they free-fall they send a series of signals, one every second of the free-fallers proper time. The signals received by the hoverer get progressively less frequent, but they never stop coming because the free-faller can never reach the event horizon.

...is *not* true! As with the free-faller crossing the Rindler horizon of an accelerator in flat spacetime, if a free-faller falling into a black hole sends signals outward to a hoverer, once every second of his proper time (remember, you specified *proper* time), then those signals *will* stop coming after a finite time by the hoverer's clock! There will be a "last signal" that the free-faller sends, the last second of his proper time before he crosses the horizon and no signal he sends can reach the hoverer any more.

Doesn’t work you run an object crossing an event horizon backwards though does it?

Sure it does. The time reverse of a black hole horizon is a "white hole" horizon, where objects can pass outward but not inward.

 

[Dale]

The event horizon is the point when the even the speed of light isn’t enough to move you towards a hoverer.

While this is true it does not imply that a free-faller becomes lightlike.

It is exactly equivalent. Working out acceleration relative to a hoverer in free-fall is no different from acceleration relative to an inertial observer in flat space-time.

You seem to have the analogy backwards. The free falling Schwarzschild observer is analogous to the inertial observer in flat spacetime, and the hovering Schwarzschild observer is analogous to the accelerating observer in flat spacetime.

The event horizon moves inwards because gravity pulls inwards

I don't understand your reasoning here, this is in fact why the event horizon moves outward locally.

What I’ve been saying is consistent with Schwarzschild space-time and it's GR predictions.

Then derive what you have been saying using GR.

(1) What happens if you free-fall towards the black hole with your light in front of you crossing the event horizon, then pull away before you reach the horizon? Your light would have to turn round and come back across the event horizon! WFT!

I don't know why you think this. Light isn't attached to you by some sort of rubber band that would make it turn around simply because you did. This is a silly idea and certainly is not a prediction of GR.

(2) How can two coordinate systems that say completely different things (one saying an object crosses a black hole and one saying it doesn’t) possibly be consistent with each other? I still haven’t had an explain of how they could both be right. This is the most glaring problem.

You have had many explanations of this, but just didn't understand any of them. Your lack of comprehension is not a failure of GR, it is a failure of you to even want to learn the required math. If you choose not to make the effort to learn the math that is fine, but then you have chosen not have the qualifications to assess the self-consistency of GR. You cannot have it both ways. The mathematical framework is precisely what ensures the self-consistency of GR, so if you avoid the math then you cannot claim that inconsistencies exist.

(3) What is the difference between flat and curved space-time?

The difference between flat and curved space-time is the presence of tidal forces.

(4) You know that an object in free-fall outside the event horizon can’t reach a velocity of c relative to an object accelerating away from the black hole at almost c. Gravity can keep on accelerating them but it gets harder as their relative velocity increases. The way ‘distance shortening’ works is to create less space-time for the accelerator to travel through from their perspective and more from the perspective of a hoverer/inertial observer. So the free-faller is constantly traveling through a smaller and smaller amount of space-time from their perspective compared to the perspective of a distant observer, just like accelerating in flat space-time. Do you really believe it makes sense if the tiniest amount more gravity could accelerate a free-faller to c relative to a hoverer when the energy of a hydrillion cyanoid big bangs couldn’t?

No, I don't believe it makes sense.

 

[PeterDonis]

You seem to have the analogy backwards.

He certainly does. I've been going back and forth with him about it for many posts now.

 

[A-wal]

I have got a half written reply but I’m not going to be finishing it any time soon so I’ll get straight to the point.

You claim that objects can reach even horizons in a finite amount of their own proper time. If a line of objects were continuously falling into a black hole like a conveyer belt then this would cause a few problems. None of the objects in front can reach the horizon before they themselves do. So they all have to cross the horizon at the same time. If everything that will have to reach a horizon has to do it at a specific time then at what point in the black holes life will that be? Just after it's gone!

If we do the same thing with a row of accelerating objects that accelerate harder the closer they are to a distant object and started them off equally spaced then they would start to separate, and at a quicker rate the closer they are to the object that they're heading towards. An objects velocity is always c / the distance squared to the destination object. None of these objects would be able to reach the stationary object that they're heading towards. They'd just get more and more time dilated and length contracted as they get closer at a slower and slower rate from any distance.

Using the river model; there will never be a point when the river is moving at or faster than c relative to the singularity. It's no different than adding relative velocities in flat space-time.

15-20 minutes that took. That's better.

 

[Dale]

Hi A-wal, welcome back!

You claim that objects can reach even horizons in a finite amount of their own proper time. If a line of objects were continuously falling into a black hole like a conveyer belt then this would cause a few problems. None of the objects in front can reach the horizon before they themselves do. So they all have to cross the horizon at the same time.

You are mixing up proper time and Schwarzschild coordinate time. Proper time alone cannot be used to establish an ordering of events on different worldlines. You need some synchronization convention for that.

Using the Schwarzschild synchronization convention they all reach the horizon in the limit as t->∞. If you want to think of that as "at the same time" then go ahead.