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

[PeterDonis]

Things just keep on falling till they hit something. The only limit to how far an object can fall towards an event horizon is c, which is constant. The only way an event horizon could be reached is if gravity had accelerated you past c, but it's constant. "Distance shortening" is relative to c, so traveling any distance in any amount of time won't be enough to reach it.

You keep on throwing around hand-waving terms and descriptions without answering the question I asked. How do you figure out what acceleration a "freely falling" object should feel (since in your model it isn't zero)? Statements like " it's the increase in the rate of acceleration relative to riverbed that should be felt as proper acceleration" are meaningless unless you can give a rule for how to figure out "the increase in the rate of acceleration relative to the riverbed"; otherwise you're just substituting one meaningless phrase for another. As for your latest post, none of it even talks about what acceleration an object should feel at all. Are you going to actually answer the question or not?

 

[kmarinas86]

[O]bjects in free fall, including radially infalling ones, feel zero acceleration to within the limits of accuracy of our current measuring instruments, which are certainly accurate enough to detect a 1 g acceleration (which is, of course, what this formula gives for objects falling in vacuum near the surface of the Earth). So I ask again: what is the rule in your model for determining what acceleration a "freely falling" object feels?

1 g of acceleration is "felt" only because there is compression on the body being accelerated. The difference between 1 g of acceleration in a car vs freefall is that the forces are concentrated at one side in the former case, while being more or less equally distributed in the latter case. Theoretically, one could accelerate at 10 g, 100 g, or even 1000 g and not feel a thing if all the forces were distributed evenly throughout the body. However, we lack technology to produce such an effect. With a more massive and compact gravitational body, such as a "black hole", tidal forces (which result in an uneven force in a body) only become significant at even higher g-forces, allowing achievement of such accelerations without the full acceleration being "felt". (See http://www.astronomycafe.net/qadir/q357.html)

 

[PeterDonis]

The difference between 1 g of acceleration in a car vs freefall is that the forces are concentrated at one side in the former case, while being more or less equally distributed in the latter case.

In the case of an extended object, with internal forces between its parts, there is a way of interpreting what you're saying that makes it correct (by interpreting "forces" as meaning "internal forces between the parts of the body"), although for an object to be in free fall it must also be the case that the net resultant of the "forces" on the body as a whole is zero (perhaps that's what you're getting at by "equally distributed", though if so the qualifier "more or less" shouldn't be there; if the net resultant is not exactly zero, the body is not in free fall--see below).

If you read my previous posts you'll note that I qualified the case of an extended body with internal structure by talking about the (net) acceleration felt at the object's center of mass, which is still zero if the object as a whole is in free fall. But right now in this thread we're talking about an idealized, point-like object ("point-like" meaning "small enough that there is no detectable tidal gravity over its length); for such an object, the acceleration is exactly zero, and so are the "forces" (there are no "internal" forces because the object has no internal structure).

OTOH, if by "forces" above you mean actual "external" forces that have a non-zero net resultant on the body as a whole, as I noted above, in standard GR a body in free fall has *no* such forces acting on it. In particular, "gravity" is *not* such a force; an (idealized, "point-like"--see below) object moving solely under the influence of "gravity" has zero force on it and feels zero acceleration. "Gravity" in standard GR is the same as "fictitious forces" like centrifugal force; it only appears if you adopt a certain coordinate system and insist on referring "forces" to that system instead of adopting a coordinate-free, physical definition of what a "force" is (a force is "something you can feel", something that causes a non-zero net proper acceleration of a body, which gravity does not do).

(I put "gravity" in quotes above because once you've removed the "Newtonian" part of gravity, which is not a force in GR, you still have tidal gravity left over, which in Newtonian terms is the difference in the Newtonian "acceleration due to gravity" as you move in space--in GR it's somewhat more complicated than that, but the Newtonian language will do for right now. Even tidal gravity, however, only appears as a "force" in an extended object which has non-gravitational internal forces between its parts; those internal forces can be non-zero if an object is large enough for the "natural" motions of its parts relative to one another to be affected by tidal gravity, so that the internal forces have to be non-zero to keep the parts at the same relative distances. Again, we're idealizing that away right now in this thread by considering the motion of "point-like" objects which are too small to "see" any tidal gravity and have no internal forces between parts because they have no parts.)

 

[A-wal]

You keep on throwing around hand-waving terms and descriptions without answering the question I asked. How do you figure out what acceleration a "freely falling" object should feel (since in your model it isn't zero)? Statements like " it's the increase in the rate of acceleration relative to riverbed that should be felt as proper acceleration" are meaningless unless you can give a rule for how to figure out "the increase in the rate of acceleration relative to the riverbed"; otherwise you're just substituting one meaningless phrase for another. As for your latest post, none of it even talks about what acceleration an object should feel at all. Are you going to actually answer the question or not?

I seem to have offended you for some reason, and I have answered...

Oops, just realized that "acceleration relative to the river bed" in your version of the model is not the same as proper acceleration in GR. A "hovering" observer's "acceleration relative to the river bed" is zero, since the hovering observer is at rest in Schwarzschild coordinates. It doesn't change the main point I was making, but I phrased it wrong. Here's what I should have said:

In the limit where gravity is weak (i.e., very far away from the hole), I believe you can say that tidal force is equal to the *rate of change* of the "acceleration of the river relative to the river bed" (*not* the acceleration itself). I say "I believe" because I haven't actually worked out the math to confirm that, when you compute the "acceleration of the river relative to the river bed" far away from the hole, you do in fact get the correct Newtonian formula, - GM/r^2, for "acceleration due to gravity". If you do, then the (radial) tidal gravity, in the Newtonian approximation, is equal to the (radial) rate of change of that acceleration, i.e., 2GM/r^3.

If all that is the case, then the equality would continue to hold approaching, at, and inside the horizon, because, as I've already pointed out, the formula for tidal gravity remains the same as the Newtonian formula all the way into r = 0, even though the "acceleration due to gravity" does not (it acquires an extra sqrt(1 - 2GM/c^2r) term in the denominator, so the acceleration diverges as the horizon is approached). The formula for "acceleration of the river relative to the river bed" should also remain the same as the Newtonian formula all the way into r = 0 (since the formula for the velocity of the river remains the same, and the acceleration is just the radial rate of change of that velocity), so radial tidal gravity should continue to equal the radial rate of change of that acceleration. (This means, of course, that as the horizon is approached, reached, and passed, the "acceleration of the river relative to the river bed" is no longer equal to the correct relativistic "acceleration due to gravity", since that diverges as the horizon is approached.)

All this depends, however, on the formula for "acceleration of the river relative to the river bed" working out as I said it needs to above. I'll have to check that when I get a chance.

...and so have you.

1 g of acceleration is "felt" only because there is compression on the body being accelerated. The difference between 1 g of acceleration in a car vs freefall is that the forces are concentrated at one side in the former case, while being more or less equally distributed in the latter case. Theoretically, one could accelerate at 10 g, 100 g, or even 1000 g and not feel a thing if all the forces were distributed evenly throughout the body. However, we lack technology to produce such an effect. With a more massive and compact gravitational body, such as a "black hole", tidal forces (which result in an uneven force in a body) only become significant at even higher g-forces, allowing achievement of such accelerations without the full acceleration being "felt". (See http://www.astronomycafe.net/qadir/q357.html)

In the case of an extended object, with internal forces between its parts, there is a way of interpreting what you're saying that makes it correct (by interpreting "forces" as meaning "internal forces between the parts of the body"), although for an object to be in free fall it must also be the case that the net resultant of the "forces" on the body as a whole is zero (perhaps that's what you're getting at by "equally distributed", though if so the qualifier "more or less" shouldn't be there; if the net resultant is not exactly zero, the body is not in free fall--see below).

If you read my previous posts you'll note that I qualified the case of an extended body with internal structure by talking about the (net) acceleration felt at the object's center of mass, which is still zero if the object as a whole is in free fall. But right now in this thread we're talking about an idealized, point-like object ("point-like" meaning "small enough that there is no detectable tidal gravity over its length); for such an object, the acceleration is exactly zero, and so are the "forces" (there are no "internal" forces because the object has no internal structure).

OTOH, if by "forces" above you mean actual "external" forces that have a non-zero net resultant on the body as a whole, as I noted above, in standard GR a body in free fall has *no* such forces acting on it. In particular, "gravity" is *not* such a force; an (idealized, "point-like"--see below) object moving solely under the influence of "gravity" has zero force on it and feels zero acceleration. "Gravity" in standard GR is the same as "fictitious forces" like centrifugal force; it only appears if you adopt a certain coordinate system and insist on referring "forces" to that system instead of adopting a coordinate-free, physical definition of what a "force" is (a force is "something you can feel", something that causes a non-zero net proper acceleration of a body, which gravity does not do).

(I put "gravity" in quotes above because once you've removed the "Newtonian" part of gravity, which is not a force in GR, you still have tidal gravity left over, which in Newtonian terms is the difference in the Newtonian "acceleration due to gravity" as you move in space--in GR it's somewhat more complicated than that, but the Newtonian language will do for right now. Even tidal gravity, however, only appears as a "force" in an extended object which has non-gravitational internal forces between its parts; those internal forces can be non-zero if an object is large enough for the "natural" motions of its parts relative to one another to be affected by tidal gravity, so that the internal forces have to be non-zero to keep the parts at the same relative distances. Again, we're idealizing that away right now in this thread by considering the motion of "point-like" objects which are too small to "see" any tidal gravity and have no internal forces between parts because they have no parts.)

So it would be the increase in the amount of acceleration due to gravity, which is tidal force. Proper acceleration in free-fall is felt as tidal force. I don't think it can be idealised away. Even an impossible point-like object would feel that acceleration in free-fall, although technically it wouldn’t be tidal force.

 

[PeterDonis]

I seem to have offended you for some reason

No, I'm not offended. I was just wondering when you were going to actually answer the question. I do appreciate your reminding me of my previous post...

and I have answered...and so have you

...but that post was aiming at a different question, the question of what the "acceleration of the river relative to the river bed" would be; that acceleration does not correspond to an acceleration felt by anyone, in either standard GR or your model. However, your next comment...

So it would be the increase in the amount of acceleration due to gravity, which is tidal force.

...does propose an answer to the question I was asking about what acceleration objects should actually feel. However, if you're actually proposing that as an answer, then I can see at least two objections:

(1) You said in a previous post that the "river bed" wasn't fixed; and the whole point of using the "acceleration of the river relative to the river bed" and its rate of change was that the "river bed" in the standard "river model" *is* fixed. (That doesn't imply a "preferred frame" in the standard river model because the "river bed" is not a physical object; it's just a convenience for visualizing the model.) If the "river bed" is not fixed, I have no idea how to calculate the "acceleration of the river relative to the river bed", let alone its radial rate of change. (In particular, as you'll see below, the "acceleration of the river relative to the river bed" is *not*, as I thought it might be in that previous post, the same as the standard formula for "acceleration due to gravity".)

(2) The tidal force on an object, calculated in the standard way, depends on the size of the object (since it's the difference in "acceleration due to gravity" from one end of the object to the other). But we're talking about a proper acceleration that should be felt at the object's center of mass, even for an idealized "point-like" object with zero length. So the length of the object can't enter into the calculation. So again, I don't know how to calculate an analogue of the "tidal force" that would be usable as a proper acceleration at an idealized "point-like" object's center of mass.

As an addendum to the above, let me actually check what I said I would check in my previous post. In the standard "river model", the "velocity of the river relative to the river bed" (this assumes, btw, that the "river bed" is fixed--as I noted above, I have no idea how to calculate this if the "river bed" is not fixed) is given by the Newtonian "escape velocity" formula:

[tex]v = \sqrt{\frac{2 G M}{r}}[/tex]

Note that the variable "r" here, however, is *not* the physical radial distance, but the Schwarzschild radial coordinate. That means that when r = 2GM (at the horizon), v = c. The "standard" river model allows v to be greater than c (because the river bed isn't physical, that doesn't violate SR), but I'm not going to tackle that issue in this post.

What we need to check is the radial derivative of the velocity; it is

[tex]\frac{\partial v}{\partial r} = - \sqrt{\frac{G M}{2 r^{3}}}[/tex]

This is *not* the same as the "acceleration due to gravity", either the Newtonian version or the relativistic version. So the equality I postulated might hold in my previous post does *not* hold: the "acceleration of the river relative to the river bed" is *not* the same as the "acceleration due to gravity".

 

[A-wal]

The riverbed can't be fixed because the acceleration of the riverbed isn't felt, because you’re being pulled along with it. But the increase in acceleration relative to the riverbed would be felt, as if the riverbed were always the same speed as the river is at that exact moment. So really what's felt is the rivers acceleration relative to itself. Tidal force also matches that description. A longer object feels more tidal force because there's more difference between the acceleration at either end.

 

[PeterDonis]

So the equality I postulated might hold in my previous post does *not* hold: the "acceleration of the river relative to the river bed" is *not* the same as the "acceleration due to gravity".

I realized on re-reading that I had done this incorrectly; the "acceleration of the river relative to the river bed" is not dv/dr, but dv/dt, where "t" is some "natural" time coordinate for the river bed. Fortunately, to correct my derivation, we don't have to know exactly what "t" is; we just have to know that v = dr/dt, so we can use the chain rule:

[tex]a = \frac{dv}{dt} = \frac{dv}{dr} \frac{dr}{dt} = v \frac{dv}{dr}[/tex]

This means that my previous post was right after all: the "acceleration of the river relative to the river bed" *is* the same as the *Newtonian* "acceleration due to gravity":

[tex]a = \frac{G M}{r^{2}}[/tex]

(Note that I also got the sign of v wrong in my last post; v should be negative since it's radially inward, so the sign of a is positive, as just above.)

Since a is the *Newtonian* acceleration, though, *not* the relativistic acceleration, it does *not* diverge as the horizon is approached; it is finite at the horizon, and all the way down to r = 0. So the "tidal acceleration" t *is* the radial derivative of the "acceleration of the river relative to the river bed".

P.S.: Am I the only one who is seeing the raw TeX formulas instead of the nice-looking equations?

 

[PeterDonis]

The riverbed can't be fixed because the acceleration of the riverbed isn't felt, because you’re being pulled along with it. But the increase in acceleration relative to the riverbed would be felt, as if the riverbed were always the same speed as the river is at that exact moment. So really what's felt is the rivers acceleration relative to itself. Tidal force also matches that description. A longer object feels more tidal force because there's more difference between the acceleration at either end.

That's the standard description of tidal force, yes. But under that description, an idealized point-like object with zero length feels zero acceleration. You've claimed that even such an object feels a non-zero acceleration due to "tidal force". How does that work?

 

[A-wal]

That's the standard description of tidal force, yes. But under that description, an idealized point-like object with zero length feels zero acceleration. You've claimed that even such an object feels a non-zero acceleration due to "tidal force". How does that work?

I just assumed tidal force could be thought of as proper acceleration. I think you're right though. The hoverers are feeling all of the acceleration in this situation. They can't accelerate to c. You don't need to accelerate relative to a fixed riverbed (hovering observers) but it's still handy because you can't reach c relative to it, or relative to any observers. The hoverers could accelerate away at almost c relative to the black hole and a free-faller still wouldn't be moving at c relative to them. It doesn't change the fact that the speed of light is constant and can't be reached.

 

[PeterDonis]

I just assumed tidal force could be thought of as proper acceleration. I think you're right though. The hoverers are feeling all of the acceleration in this situation.

Does this mean you now agree that the idealized, "point-like" freely falling objects in both of my scenarios feel exactly zero acceleration? Just to recap, the two scenarios are:

(1) Two radially infalling objects, which are mutually at rest at slightly different heights at some instant;

(2) Two objects in circular orbits at slightly different altitudes, which pass each other and thus both lie on the same radial line (moving in the same direction but at slightly different speeds) at some instant.

 

[A-wal]

Does this mean you now agree that the idealized, "point-like" freely falling objects in both of my scenarios feel exactly zero acceleration? Just to recap, the two scenarios are:

(1) Two radially infalling objects, which are mutually at rest at slightly different heights at some instant;

(2) Two objects in circular orbits at slightly different altitudes, which pass each other and thus both lie on the same radial line (moving in the same direction but at slightly different speeds) at some instant.

I think so yea, to both. It's still relative movement between free-fallers and hoverers though. How would you calculate velocity between the two if one's relative and one isn't?

 

[PeterDonis]

I think so yea, to both. It's still relative movement between free-fallers and hoverers though. How would you calculate velocity between the two if one's relative and one isn't?

I'm not sure what you mean by "one's relative and one isn't". Velocity is always relative; there's no such thing as "absolute" velocity. (At least, there isn't in standard GR or SR. If there is in your model, then it's up to you to figure out how to calculate it.) If you mean calculate the relative velocity between a free-faller and a hoverer at a certain radial coordinate r, that's straightforward as long as you know the height at which the free-faller was at rest. In the simplest case (mathematically speaking), where the free-faller is at rest "at infinity", the free-faller's inward velocity relative to a hoverer who is at a constant radius r is just the Newtonian "escape velocity", [itex]\sqrt{2GM/r}[/itex].

This formula only applies at values of r where a hoverer can exist--i.e., for values of r outside the horizon. At or inside the horizon, there are no hoverers, but one can still calculate the relative velocity of a free-faller and an accelerated observer, as long as you specify how the latter is accelerating (and, again, the height at which the free-faller was at rest). The easiest way to do that is usually to set up a local inertial frame around the free-faller and figure out the acceleration of the other observer in that frame.

Of course, all this is the standard GR picture, which you don't accept. But it's perfectly consistent. And since you now agree that free-fallers feel zero acceleration (at least in the idealized, "point-like" case), I'm not sure how you propose to construct an alternate model where spacetime somehow "stops" at the horizon, since an ingoing free-faller, you now agree, is in a state of motion that's indistinguishable from floating in empty space. It would seem to follow that he could float right through the horizon without expending any effort at all. True, the horizon is "moving at the speed of light", but it's moving *outward* at the speed of light, so an *outgoing* observer would have to move at c to catch up to it. But an *ingoing* observer just sees the horizon move *towards* him at c as he falls; it's not moving *away* from him (so that he would be required to "catch up" to it to cross it) until it's passed him and he's inside.

 

[A-wal]

I'm not sure what you mean by "one's relative and one isn't". Velocity is always relative; there's no such thing as "absolute" velocity. (At least, there isn't in standard GR or SR. If there is in your model, then it's up to you to figure out how to calculate it.) If you mean calculate the relative velocity between a free-faller and a hoverer at a certain radial coordinate r, that's straightforward as long as you know the height at which the free-faller was at rest. In the simplest case (mathematically speaking), where the free-faller is at rest "at infinity", the free-faller's inward velocity relative to a hoverer who is at a constant radius r is just the Newtonian "escape velocity", [itex]\sqrt{2GM/r}[/itex].

No, far from it. I don't think movement can ever be absolute. It's always relative to c, but in the standard model that doesn't hold because objects have to move faster than c to cross the horizon. If movement’s relative to c then moving faster than c has to be absolute movement.

This formula only applies at values of r where a hoverer can exist--i.e., for values of r outside the horizon. At or inside the horizon, there are no hoverers, but one can still calculate the relative velocity of a free-faller and an accelerated observer, as long as you specify how the latter is accelerating (and, again, the height at which the free-faller was at rest). The easiest way to do that is usually to set up a local inertial frame around the free-faller and figure out the acceleration of the other observer in that frame.

I bet it always holds. I don't see how you can't accelerate away if you're at rest. The acceleration's coming from the hoverer, but it still can't reach c.

Of course, all this is the standard GR picture, which you don't accept. But it's perfectly consistent. And since you now agree that free-fallers feel zero acceleration (at least in the idealized, "point-like" case), I'm not sure how you propose to construct an alternate model where spacetime somehow "stops" at the horizon, since an ingoing free-faller, you now agree, is in a state of motion that's indistinguishable from floating in empty space. It would seem to follow that he could float right through the horizon without expending any effort at all. True, the horizon is "moving at the speed of light", but it's moving *outward* at the speed of light, so an *outgoing* observer would have to move at c to catch up to it. But an *ingoing* observer just sees the horizon move *towards* him at c as he falls; it's not moving *away* from him (so that he would be required to "catch up" to it to cross it) until it's passed him and he's inside.

I don't think it would be accelerating outwards. It moves out at c, but that's no time at all if you were there. It would always be accelerating away at c. What you call a white hole.

 

[PeterDonis]

No, far from it. I don't think movement can ever be absolute. It's always relative to c

I have no idea what this means. By "relative velocity" I mean "the relative velocity of two objects"--two observable things. If by "relative to c" you mean "relative to a light ray", first, you have to specify *which* light ray (light rays are particular objects with particular worldlines, just like other objects), and second, light rays always move at c (locally--the concept of "relative velocity" only applies locally, where it can be directly observed) relative to timelike objects (objects with mass), so the "relative velocity" of a timelike object and a light ray is always c; it doesn't "approach" c as you accelerate further in a particular direction. Please clarify what you mean here.

I don't think it would be accelerating outwards. It moves out at c, but that's no time at all if you were there.

This is not correct. If a light ray is moving past you in ordinary empty space, far from any gravitating body, does your clock stop? Free-falling inward past the horizon works the same way; your clock keeps on ticking normally, you just see the horizon flash past you as an outgoing light ray. There is a (strained) sense in which "zero time elapses" along the light ray's worldline (because it's a null worldline, with a "length" of zero), but that doesn't affect you, since you're not traveling along with the light ray; it's just passing you.

 

[A-wal]

I have no idea what this means. By "relative velocity" I mean "the relative velocity of two objects"--two observable things. If by "relative to c" you mean "relative to a light ray", first, you have to specify *which* light ray (light rays are particular objects with particular worldlines, just like other objects), and second, light rays always move at c (locally--the concept of "relative velocity" only applies locally, where it can be directly observed) relative to timelike objects (objects with mass), so the "relative velocity" of a timelike object and a light ray is always c; it doesn't "approach" c as you accelerate further in a particular direction. Please clarify what you mean here.

I just mean you always move relative to a constant speed of light. It determines the distance between objects in the frame that you’re in.

This is not correct. If a light ray is moving past you in ordinary empty space, far from any gravitating body, does your clock stop? Free-falling inward past the horizon works the same way; your clock keeps on ticking normally, you just see the horizon flash past you as an outgoing light ray. There is a (strained) sense in which "zero time elapses" along the light ray's worldline (because it's a null worldline, with a "length" of zero), but that doesn't affect you, since you're not traveling along with the light ray; it's just passing you.

It’s the equivalent to accelerating in flat space-time. If the hoverer and the free-faller start off 1 light-year apart and you measure their speed when they cross then you would get the same answer as if there was the same amount of space separating them and the accelerator had the same starting speed as the free-faller (who is now inertial) had in flat space-time.

At no point can anything reach c. It doesn’t mean there will ever be a point when no amount of acceleration will move you away. Whatever proper time you work out for the free-faller to hit the horizon will be wrong because ‘distance shortening’ means distances don’t add together like they do in flat space-time if you try to reach the horizon, and the ‘distance shortening’ at the horizon would be infinite.

 

[PeterDonis]

I just mean you always move relative to a constant speed of light. It determines the distance between objects in the frame that you’re in.

Sorry, but even if it seems "obvious" to you what this means, it's not well-defined as it stands. You can't just wave your hands and say "a constant speed of light". You have to specify moving relative to some *particular* light ray--some particular null worldline, or family of null worldlines. And as soon as you try to specify that explicitly, you'll see the point I've been trying to make all along: in the spacetime surrounding a central gravitating mass, there is *no* family of such null worldlines that works the way a "constant speed of light" works in the flat spacetime of SR, because of the tilting of the light cones. But don't take my word for it; try it. Specify for me a particular family of null worldlines that I can use to define moving "relative to a constant speed of light".

If the hoverer and the free-faller start off 1 light-year apart and you measure their speed when they cross then you would get the same answer as if there was the same amount of space separating them and the accelerator had the same starting speed as the free-faller (who is now inertial) had in flat space-time.

Have you actually done the calculations to check this? Or are you just guessing? When I do the calculation, I get formulas for the two speeds that do not look at all the same.

Whatever proper time you work out for the free-faller to hit the horizon will be wrong because ‘distance shortening’ means distances don’t add together like they do in flat space-time if you try to reach the horizon, and the ‘distance shortening’ at the horizon would be infinite.

You keep on making this assertion even though I've proved it false, by posting the actual derivation that shows the distance is finite. It's true that "distances don't add together like they do in flat spacetime", in the sense that the integral determining the distances in question has an extra factor in it due to spacetime curvature; but that doesn't mean the integral just magically becomes infinite or zero. You're waving your hands again instead of actually checking your logic to see if it's correct.

 

[Dale]

You're waving your hands again instead of actually checking your logic to see if it's correct.

Hi Peter, you are forgetting that he doesn't want logic, he firmly believes that his ignorance and avoidance of math qualify him to fix some supposed errors in the theory. He thinks that logic gets in the way of understanding physics, and that using logic and study are counter-productive.

 

[A-wal]

Sorry, but even if it seems "obvious" to you what this means, it's not well-defined as it stands. You can't just wave your hands and say "a constant speed of light". You have to specify moving relative to some *particular* light ray--some particular null worldline, or family of null worldlines. And as soon as you try to specify that explicitly, you'll see the point I've been trying to make all along: in the spacetime surrounding a central gravitating mass, there is *no* family of such null worldlines that works the way a "constant speed of light" works in the flat spacetime of SR, because of the tilting of the light cones. But don't take my word for it; try it. Specify for me a particular family of null worldlines that I can use to define moving "relative to a constant speed of light".

You’re in free-fall and you turn your head lights on. What speed does the light move away from you? It starts at c, then slows down as it moves into a higher gravitational field than you. If you could reach the event horizon then those light rays wouldn’t be moving away from you and you would start to overtake them inside the horizon. That’s not how it works though because the speed of light’s constant. The light cones can’t tilt to 90 degrees.

Have you actually done the calculations to check this? Or are you just guessing? When I do the calculation, I get formulas for the two speeds that do not look at all the same.

The hoverer/accelerator has to maintain a constant acceleration to stay stationary relative to the black hole. A free-faller gradually speeds up at a faster and faster rate relative to the hoverer because the hoverer is accelerating towards them. The situation is no different to acceleration in flat space-time. You can accelerate as much as you like and you still can’t reach the event horizon/c. You think you can because you’ve broken this symmetry.

You keep on making this assertion even though I've proved it false, by posting the actual derivation that shows the distance is finite. It's true that "distances don't add together like they do in flat spacetime", in the sense that the integral determining the distances in question has an extra factor in it due to spacetime curvature; but that doesn't mean the integral just magically becomes infinite or zero. You're waving your hands again instead of actually checking your logic to see if it's correct.

It would be exactly the same formula adding speeds together in flat space-time to try to reach c, because the situation is equivalent.

Hi Peter, you are forgetting that he doesn't want logic, he firmly believes that his ignorance and avoidance of math qualify him to fix some supposed errors in the theory. He thinks that logic gets in the way of understanding physics, and that using logic and study are counter-productive.

Hi Dale.

 

[PeterDonis]

You’re in free-fall and you turn your head lights on. What speed does the light move away from you? It starts at c, then slows down as it moves into a higher gravitational field than you.

This is incorrect. The light does not "slow down". More precisely, if you are free-falling radially inward, and you shine light radially inward, the light will always seem to you to be moving away from you (towards r = 0) at c.

You may be confusing (once again) ingoing with outgoing light rays; there is a (strained) sense in which outgoing light rays are "slowed down" close to the horizon, because the outgoing sides of light cones are tilted almost vertical there, so it takes "longer" (in the sense of Schwarzschild coordinate time) for light close to the horizon to move outward by a certain increment of radius than it does far away from the hole. (Exactly at the horizon, the light cones are tilted to exactly vertical, so outgoing light at the horizon stays at the horizon forever.) But this only applies to outgoing light, not ingoing light. The asymmetry is because gravity pulls inward, not outward.

However, even in the case of outgoing light rays, if you shine light outward, it will seem to you to be moving away from you at c, no matter how hard you accelerate to try and catch up with it.

One last point about this: I'm curious why you think the light would slow down as gravity gets stronger, instead of speeding up. After all, a massive object falling inward speeds up, doesn't it?

If you could reach the event horizon then those light rays wouldn’t be moving away from you and you would start to overtake them inside the horizon.

Incorrect. You can never "close the gap" between yourself and a light beam you shine away from you. The light is always moving away from you at c.

That’s not how it works though because the speed of light’s constant. The light cones can’t tilt to 90 degrees.

Why not? I know it seems "obvious" to you that they shouldn't be able to, but what logical argument do you have that *proves* they can't? Unless you can give such an argument, starting from premises we both accept (e.g., you can't *assume* that spacetime stops at the horizon, or anything else that's logically equivalent to what you're trying to prove), then there's nothing here that I can respond to.

It would be exactly the same formula adding speeds together in flat space-time to try to reach c, because the situation is equivalent.

Instead of just asserting this without proof, you should actually check to see if it's true. You will find that it isn't; the situations are *not* equivalent. One hint of something that's different: in the case with gravity (the free-faller falls towards the black hole while the hoverer stays at constant radius above it), the "acceleration due to gravity" experienced by the free-faller changes in the course of his fall--it is *not* always the same (mathematically) as the acceleration of the hoverer. In the flat spacetime case (the free-faller stays at a constant x-coordinate, while the accelerator accelerates in the positive x-direction), the equivalent "acceleration due to gravity" that the free-faller sees does *not* change; it's always equal to the acceleration of the accelerator.

 

[PeterDonis]

in the case with gravity (the free-faller falls towards the black hole while the hoverer stays at constant radius above it), the "acceleration due to gravity" experienced by the free-faller changes in the course of his fall--it is *not* always the same (mathematically) as the acceleration of the hoverer. In the flat spacetime case (the free-faller stays at a constant x-coordinate, while the accelerator accelerates in the positive x-direction), the equivalent "acceleration due to gravity" that the free-faller sees does *not* change; it's always equal to the acceleration of the accelerator.

I realized on re-reading that I should go into this in more detail, both because I didn't quite state it right in the quote above, and because it may help to illustrate what I'm asking for when I ask for logical arguments.

Take the flat spacetime case first. Here we have a free-faller at a constant (large) x-coordinate, and an accelerator who starts at a (much smaller) x-coordinate and accelerates uniformly (meaning, the acceleration he feels, and which he measures with his accelerometer, is constant) in the positive x-direction.

In this case, the quantity which is the best analogue to "acceleration due to gravity" seen by the free-faller is the appropriate "hovering" acceleration at whatever distance above the Rindler horizon the free-faller is currently at. This will start out much *smaller* than the accelerator's proper acceleration, and will gradually get larger until, at the instant the accelerator passes the free-faller, it is equal to the accelerator's proper acceleration. It will then continue to increase, until it goes to "infinity" at the instant the free-faller crosses the Rindler horizon. (By "Rindler horizon" here I always mean the Rindler horizon of the accelerator, which is the analogue to the black hole horizon.)

I didn't quite state things right in my previous post because I said that the free-faller's "acceleration due to gravity" was constant; but the above isn't. Sorry about that. What I should have said is this: consider the case with gravity, where we have a free-faller starting at rest at a (large) radial coordinate, and falling towards (and eventually passing) a hoverer at a constant (much smaller, but still above the horizon) radial coordinate.

Here the "acceleration due to gravity" seen by the free-faller is the appropriate "hovering" acceleration at whatever radial coordinate the free-faller is currently at. This seems like an exact analogue of the above, but it isn't, and that's what I was trying to get at in my previous post (but didn't quite state right). The reason it isn't is twofold: (i) the radial coordinate is *not* a direct measure of distance like the x coordinate is in the flat spacetime case; and (ii) the variation of the "acceleration due to gravity" with distance is not the same as it is in the flat spacetime case. In the flat spacetime case, the "hovering" acceleration goes as the inverse distance (it is c^2 / x, where x is the x coordinate); in the gravity case, it goes as the inverse *square* of the radial coordinate. (It varies even more sharply with physical distance, but that's due to the "length contraction" factor which I accounted for in item (i) just now.)

So unlike A-wal, I would *not* expect that the speeds at which the free-faller and the accelerator/hoverer would pass each other would be the same in the two cases, given the same initial conditions (meaning, the same proper acceleration for the accelerator, and the same physical distance between them when they start out at mutual rest). (I actually know they aren't the same, because I've done the math and derived both formulas; but the above is a way of seeing that you shouldn't *expect* them to be the same even before you've done the actual calculation.)

Postscript: I suppose I should also comment on that bit about the free-faller's acceleration "going to infinity" as he crosses the accelerator's Rindler horizon. One could also claim, on similar grounds, that the free-faller would have to "move at c" to cross the accelerator's Rindler horizon (because his speed relative to successive accelerators that he passes as he gets closer to the horizon approaches c as a limit). But of course the free-faller in this scenario is just sitting at rest in a global inertial frame in flat spacetime, and it's obvious that he *can* cross the Rindler horizon; he just has to sit there! (Another way of seeing this: the Rindler horizon is simply the 45-degree line t = x in the global inertial frame. Obviously the free-faller crosses this line, since he just sits at the same x coordinate forever, so he will cross it at a coordinate time t equal to his x-coordinate--or his x-coordinate times c in conventional units.) A-wal is simply unwilling or unable to see that a free-faller in the presence of gravity can cross the black hole horizon in exactly the same way. (But we went all through this a few hundred posts ago; I'm just reiterating because I mentioned that bit about acceleration "going to infinity" above.)

 

[A-wal]

This is incorrect. The light does not "slow down". More precisely, if you are free-falling radially inward, and you shine light radially inward, the light will always seem to you to be moving away from you (towards r = 0) at c.

You may be confusing (once again) ingoing with outgoing light rays; there is a (strained) sense in which outgoing light rays are "slowed down" close to the horizon, because the outgoing sides of light cones are tilted almost vertical there, so it takes "longer" (in the sense of Schwarzschild coordinate time) for light close to the horizon to move outward by a certain increment of radius than it does far away from the hole. (Exactly at the horizon, the light cones are tilted to exactly vertical, so outgoing light at the horizon stays at the horizon forever.) But this only applies to outgoing light, not ingoing light. The asymmetry is because gravity pulls inward, not outward.

However, even in the case of outgoing light rays, if you shine light outward, it will seem to you to be moving away from you at c, no matter how hard you accelerate to try and catch up with it.

It would seem to you to be moving away from you at c but slowing down is as it moves closer to the event horizon.

One last point about this: I'm curious why you think the light would slow down as gravity gets stronger, instead of speeding up. After all, a massive object falling inward speeds up, doesn't it?

Light isn’t a massive object. It doesn’t have acceleration due to gravity. Time goes slower if you’re in a higher gravitational field, so why wouldn’t it slow down?

Incorrect. You can never "close the gap" between yourself and a light beam you shine away from you. The light is always moving away from you at c.

Yes, that’s exactly my point. It doesn’t hold past the horizon. The light is always moving away from you at c, to start with. A distant observer would say the light’s moving away from the free-faller slower than c, slower and slower as it approaches the horizon. If objects could cross the horizon then it would have to overtake the light, which it can’t do because light always moves away from you at c, to start with. The light moves slower because time’s moving slower. Light can’t reach the horizon, so neither can anything else.

Why not? I know it seems "obvious" to you that they shouldn't be able to, but what logical argument do you have that *proves* they can't? Unless you can give such an argument, starting from premises we both accept (e.g., you can't *assume* that spacetime stops at the horizon, or anything else that's logically equivalent to what you're trying to prove), then there's nothing here that I can respond to.

Because you can’t reach c, whether you’re in flat space-time or not makes no difference.

Instead of just asserting this without proof, you should actually check to see if it's true. You will find that it isn't; the situations are *not* equivalent. One hint of something that's different: in the case with gravity (the free-faller falls towards the black hole while the hoverer stays at constant radius above it), the "acceleration due to gravity" experienced by the free-faller changes in the course of his fall--it is *not* always the same (mathematically) as the acceleration of the hoverer. In the flat spacetime case (the free-faller stays at a constant x-coordinate, while the accelerator accelerates in the positive x-direction), the equivalent "acceleration due to gravity" that the free-faller sees does *not* change; it's always equal to the acceleration of the accelerator.

The free-faller starts off far enough away that we can ignore the effects of gravity on them. To start with all the acceleration of the hoverer is being used to maintain a constant distance between them. Eventually gravity does have an effect on the free-faller and the gravitational difference between them decreases as the distance between them decreases and the hoverer starts to accelerate towards the free-faller. When they pass each other we can work out what speed they would have passed each other in flat space-time if the hoverer had used the same amount of acceleration and the difference will be the acceleration due to gravity.

I want to know how you add those two velocities together to reach c at the horizon. It should be no different to adding velocities in flat space-time and you could even work out what that would be very easily by keeping track of the acceleration that the hoverer would need to maintain a constant distance from the free-faller, which would decrease as they got closer. You would then know all the proper acceleration and acceleration due to gravity and you could work out what their relative velocities would be in flat space-time, and it would never reach c. You can’t reach an event horizon in curved space-time any more than you can reach c in flat space-time.

I didn't quite state things right in my previous post because I said that the free-faller's "acceleration due to gravity" was constant; but the above isn't. Sorry about that. What I should have said is this: consider the case with gravity, where we have a free-faller starting at rest at a (large) radial coordinate, and falling towards (and eventually passing) a hoverer at a constant (much smaller, but still above the horizon) radial coordinate.

Here the "acceleration due to gravity" seen by the free-faller is the appropriate "hovering" acceleration at whatever radial coordinate the free-faller is currently at. This seems like an exact analogue of the above, but it isn't, and that's what I was trying to get at in my previous post (but didn't quite state right). The reason it isn't is twofold: (i) the radial coordinate is *not* a direct measure of distance like the x coordinate is in the flat spacetime case; and (ii) the variation of the "acceleration due to gravity" with distance is not the same as it is in the flat spacetime case. In the flat spacetime case, the "hovering" acceleration goes as the inverse distance (it is c^2 / x, where x is the x coordinate); in the gravity case, it goes as the inverse *square* of the radial coordinate. (It varies even more sharply with physical distance, but that's due to the "length contraction" factor which I accounted for in item (i) just now.)

You mean it varies even more sharply because there’s length contraction from gravity as well as acceleration? Adding them together doesn’t allow you to reach c.

Take the flat spacetime case first. Here we have a free-faller at a constant (large) x-coordinate, and an accelerator who starts at a (much smaller) x-coordinate and accelerates uniformly (meaning, the acceleration he feels, and which he measures with his accelerometer, is constant) in the positive x-direction.

In this case, the quantity which is the best analogue to "acceleration due to gravity" seen by the free-faller is the appropriate "hovering" acceleration at whatever distance above the Rindler horizon the free-faller is currently at. This will start out much *smaller* than the accelerator's proper acceleration, and will gradually get larger until, at the instant the accelerator passes the free-faller, it is equal to the accelerator's proper acceleration. It will then continue to increase, until it goes to "infinity" at the instant the free-faller crosses the Rindler horizon. (By "Rindler horizon" here I always mean the Rindler horizon of the accelerator, which is the analogue to the black hole horizon.)

Postscript: I suppose I should also comment on that bit about the free-faller's acceleration "going to infinity" as he crosses the accelerator's Rindler horizon. One could also claim, on similar grounds, that the free-faller would have to "move at c" to cross the accelerator's Rindler horizon (because his speed relative to successive accelerators that he passes as he gets closer to the horizon approaches c as a limit). But of course the free-faller in this scenario is just sitting at rest in a global inertial frame in flat spacetime, and it's obvious that he *can* cross the Rindler horizon; he just has to sit there! (Another way of seeing this: the Rindler horizon is simply the 45-degree line t = x in the global inertial frame. Obviously the free-faller crosses this line, since he just sits at the same x coordinate forever, so he will cross it at a coordinate time t equal to his x-coordinate--or his x-coordinate times c in conventional units.) A-wal is simply unwilling or unable to see that a free-faller in the presence of gravity can cross the black hole horizon in exactly the same way. (But we went all through this a few hundred posts ago; I'm just reiterating because I mentioned that bit about acceleration "going to infinity" above.)

How will it "go to infinity" and how do you cross a Rindler horizon? Nothing crosses the accelerators Rindler horizon until the accelerator stops accelerating. You can move the Rindler horizon by increasing or decreasing the acceleration but how can it ever be crossed?

 

[PeterDonis]

Light isn’t a massive object. It doesn’t have acceleration due to gravity.

Incorrect. It does. Light bends when it goes by the Sun, for example.

Because you can’t reach c, whether you’re in flat space-time or not makes no difference.

What does "can't reach c" mean? Relative to what? That's what I keep on asking and you keep on not answering. You can't just say "can't reach c" without specifying *what* you can't reach c relative to. At first I thought perhaps it was the "river bed", but now you're saying the river bed isn't fixed, and you haven't explained how I can tell "how fast the river bed is moving", so I don't know what "can't reach c relative to the river bed" would mean.

This is a crucial point, which is why I keep harping on it. You seem to think it is somehow obvious what "can't reach c" means, but it isn't. You have to add more information to make it meaningful.

The free-faller starts off far enough away that we can ignore the effects of gravity on them. To start with all the acceleration of the hoverer is being used to maintain a constant distance between them. Eventually gravity does have an effect on the free-faller and the gravitational difference between them decreases as the distance between them decreases and the hoverer starts to accelerate towards the free-faller. When they pass each other we can work out what speed they would have passed each other in flat space-time if the hoverer had used the same amount of acceleration and the difference will be the acceleration due to gravity.

But you said that the situations were the same. Now you appear to be saying they're not (because you're implying that there *is* a difference in the observed speeds when the accelerator and the free-faller pass each other in the two cases). Which is it?

I want to know how you add those two velocities together to reach c at the horizon.

You don't, because there is no "hoverer" at the horizon. The horizon is an outgoing light ray (actually a lightlike surface when you include the angular directions). Relative to that light ray, of course the speed of the ingoing free-faller is c, since the speed of a light ray relative to any timelike object is c.

How will it "go to infinity" and how do you cross a Rindler horizon? Nothing crosses the accelerators Rindler horizon until the accelerator stops accelerating.

Wrong. I've explained this *many* times and you still don't get it. Read my description in the previous post again. The Rindler horizon is just the path of a light ray going in the positive x-direction in the global inertial frame (the one in which the free-faller is at rest). Obviously a light ray moving in the positive x-direction will pass any object at rest at a positive x-coordinate; or, equivalently, any object at rest at a positive x-coordinate will pass the light ray (cross the horizon).

 

[A-wal]

Incorrect. It does. Light bends when it goes by the Sun, for example.

Yes, that's what I'm saying. Its speed isn’t fixed at c if we’re talking about it moving into a higher gravitational field. That’s why it bends.

What does "can't reach c" mean? Relative to what? That's what I keep on asking and you keep on not answering. You can't just say "can't reach c" without specifying *what* you can't reach c relative to. At first I thought perhaps it was the "river bed", but now you're saying the river bed isn't fixed, and you haven't explained how I can tell "how fast the river bed is moving", so I don't know what "can't reach c relative to the river bed" would mean.

This is a crucial point, which is why I keep harping on it. You seem to think it is somehow obvious what "can't reach c" means, but it isn't. You have to add more information to make it meaningful.

Relative to anything, including a hypothetical riverbed made up of hovering observers. What would happen if there was a hover just outside the horizon and they accelerated away very sharply when a free-faller was passing them at almost c? Would they then be moving faster than c relative to the free-faller? Of course they wouldn’t. The same thing would happen if you calculate a hover even closer to the horizon. You can get as close as you like, just as you can get as close as you want to c. You won’t reach it.

But you said that the situations were the same. Now you appear to be saying they're not (because you're implying that there *is* a difference in the observed speeds when the accelerator and the free-faller pass each other in the two cases). Which is it?

It's different because the hoverer is in a higher gravitational field than the free-faller. You could use the proper time elapsed for the hoverer and match that in flat space-time to get the same answer when the cross. Work it out from the horizon again. Your way will give you c as the answer. In flat space-time you can see that it's not c.

You don't, because there is no "hoverer" at the horizon. The horizon is an outgoing light ray (actually a lightlike surface when you include the angular directions). Relative to that light ray, of course the speed of the ingoing free-faller is c, since the speed of a light ray relative to any timelike object is c.

You were allowed a point-like object. If you work out the velocity at the horizon you'll see that it can't reach c, so it isn't the horizon. The horizon moves back whenever you try to work out anything at the horizon. It moves back because length contraction and time dilation stop you from reaching it in exactly the same way they stop you from reaching c in flat space-time. You can say there's not enough 'distance shortening' to do that but there must be.

Wrong. I've explained this *many* times and you still don't get it. Read my description in the previous post again. The Rindler horizon is just the path of a light ray going in the positive x-direction in the global inertial frame (the one in which the free-faller is at rest). Obviously a light ray moving in the positive x-direction will pass any object at rest at a positive x-coordinate; or, equivalently, any object at rest at a positive x-coordinate will pass the light ray (cross the horizon).

I thought that if an object accelerates then it creates a Rindler horizon for an inertial observer because a light ray sent from them won't reach the accelerator until they stop accelerating, which is the equivalent to no object being able to the event horizon until the black hole's gone.

 

[PeterDonis]

Yes, that's what I'm saying.

You said light isn't subject to acceleration due to gravity. The fact that it bends shows that it is. So the fact that it bends does *not* support what you were saying. Do you read what you post?

Relative to anything, including a hypothetical riverbed made up of hovering observers.

Which then has to cover the entire spacetime in order to support your claims. So for you to be correct, you must *assume* that such a riverbed exists and covers the entire spacetime, which is equivalent to *assuming* that nothing can reach the horizon. So you are assuming what you claim to be proving.

What would happen if there was a hover just outside the horizon and they accelerated away very sharply when a free-faller was passing them at almost c? Would they then be moving faster than c relative to the free-faller? Of course they wouldn't.

Agreed; they would see the free-faller passing them at a speed a little *closer* to c than if they had just stayed hovering (I assume by "accelerated away very sharply" you mean "accelerated upward even more sharply than they would have to to maintain hovering just outside the horizon").

The same thing would happen if you calculate a hover even closer to the horizon. You can get as close as you like, just as you can get as close as you want to c. You won't reach it.

With reference to the hoverers, you are correct; they will never see a free-faller passing them at c (or faster). But this does not prove that free-fallers can't pass the horizon; as I noted above, what the hoverers can and can't see is only relevant if the hoverers can cover the entire spacetime, and assuming that is equivalent to *assuming* that free-fallers can't reach the horizon. But you can't *assume* that; you're supposed to be *proving* it.

You could use the proper time elapsed for the hoverer and match that in flat space-time to get the same answer when they cross.

I'm not sure I understand what you're suggesting here. It sounds like you're changing the conditions of the problem: you want to match the hoverer's/accelerator's proper time elapsed from "start", the time when the hoverer/accelerator and free-faller are mutually at rest, to the time when they cross, but that's not the same as matching the physical *distance* apart they are when they start out mutually at rest, which was the initial condition you specified before. Which is it?

You were allowed a point-like object. If you work out the velocity at the horizon you'll see that it can't reach c, so it isn't the horizon.

I don't know what you mean by this. What "velocity at the horizon"? The velocity of the free-falling object relative to the outgoing light ray that's just at the horizon. That *is* c.

The horizon moves back whenever you try to work out anything at the horizon.

I don't know what you mean by this either. The horizon is a fixed null surface in spacetime. It doesn't move.

It moves back because length contraction and time dilation stop you from reaching it in exactly the same way they stop you from reaching c in flat space-time. You can say there's not enough 'distance shortening' to do that but there must be.

Why "must" there be? Just because you say so? Do you have any actual physical reason? Any actual argument (that doesn't rely on assuming what you're supposed to be proving)?

I thought that if an object accelerates then it creates a Rindler horizon for an inertial observer because a light ray sent from them won't reach the accelerator until they stop accelerating, which is the equivalent to no object being able to the event horizon until the black hole's gone.

In other words, you still don't understand how a Rindler horizon works, despite claiming (many, many posts ago) that you were quite familiar with it and didn't need to have it explained to you; and you keep on making the same misstatement about it despite repeated corrections on my part.

I'll try once more, starting from scratch and spelling everything out. I'm sure a lot of this will seem "obvious" to you, but after hundreds of posts I can't assume anything about how you understand words relative to how I understand words. So I've got to dot all the i's and cross all the t's.

Consider a flat spacetime and a set of global inertial coordinates x, t on this spacetime (I'm using units in which c = 1, so x and t are in the same units). There are two families of observers in this spacetime. One family is of free-falling observers; each such observer is at rest (in the global inertial coordinate system) at some positive x-coordinate. The second family is of accelerating observers; each such observer starts out, at time t = 0, sitting at rest right next to one of the free-falling observers; but each accelerating observer accelerates with a proper acceleration (i.e., what's measured by his accelerometer) of c^2 / x, where x is the x-coordinate he is at at t = 0. Finally, there is a light ray that is emitted in the positive x-direction from x = 0 at time t = 0 (i.e., from the spacetime origin of the global inertial coordinates).

The following are all true:

(1) The light ray coincides with the Rindler horizon for *all* of the accelerating observers. This means that the light ray will never catch up to any of those observers; all of them will always be ahead of it (as long as they continue to accelerate at the given proper acceleration).

(2) Each of the free-falling observers passes the light ray (and therefore crosses the Rindler horizon) at a positive time t equal to his x-coordinate. Therefore, there are two regions in this spacetime: the region "outside" the horizon, which contains the starting points of all the observers at time t = 0, and the region "inside" the horizon, which is reached only by the free-falling observers, the accelerating observers never reach it (as long as they continue to accelerate). Therefore, the family of accelerating observers does *not* cover the same portion of the spacetime as the family of free-falling observers does; there is a portion (the region "inside" the horizon) that only the free-fallers cover.

(3) Once a free-falling observer crosses the Rindler horizon, there is no way for that observer to send any signal to the region of spacetime on the other side of the horizon (since such a signal would have to travel faster than light). That means no free-falling observer can send a signal to any of the accelerating observers after the free-falling observer has crossed the horizon. Which also means, of course, that no accelerating observer can "see" any free-falling observer after the free-falling observer has crossed the horizon. (All this applies as long as the accelerating observers continue to accelerate.)

(4) As a given free-falling observer passes each member of the family of accelerating observers, the velocity of the free-faller relative to the accelerating observers approaches c. However, it never reaches c, because there is no accelerating observer *at* the horizon; such an observer would have to move at c (since the horizon coincides with the path of a light ray), and no timelike observer can move at c. So no accelerating observer ever sees a freely falling observer pass him at c (or faster). Instead, from the point where each free-falling observer crosses the horizon, he simply stops passing accelerating observers (because there are no more for him to pass) and enters the region of spacetime that the accelerating observers don't cover.

The above is entirely in flat spacetime, it's entirely consistent with SR, and it clearly demonstrates several things: that free-falling observers *can* cross a Rindler horizon, that there *can* be an entire region of spacetime that accelerated observers can't "see" and which they don't cover, and that a free-falling observer not being able to send a light signal that can catch up with an accelerated observer is *not* equivalent to a free-falling observer not being able to cross the horizon.

I can already anticipate how you will reply to this post. You will say, "of course, that's how I've been saying a Rindler horizon works all along", and then proceed to claim that at least one of the three things I've demonstrated, which I listed in the previous paragraph, is false. If that's your inclination on reading the above, please think very carefully before posting a response. Don't even think about how the above translates into the spacetime surrounding a black hole, or what it implies or doesn't imply about a black hole horizon being reachable. Do you honestly believe, looking *just* at the flat spacetime scenario I've just described, as it stands, that anyone of the three things I listed in the previous paragraph is false? Because if so, I'm going to be hard pressed not to conclude that you and I can't have a sane discussion about this. It's one thing to be arguing about GR and how it applies to a black hole horizon. But if we can't even have a common understanding about how a simple scenario in SR works, I'm not sure what we *can* have a common understanding about.

 

[PeterDonis]

I've attached a graph illustrating what I described in my previous post. The horizontal axis is space (the x coordinate) and the vertical axis is time (the t coordinate). The dark magenta vertical lines are the free-fallers' worldlines. The blue lines are the accelerators' worldlines. The red line is the Rindler horizon.

 

[Dale]

I used to do three clubs at the same time and train every night for a while. Both times when I've had a space of years without doing it, I was much, much better at it after a little bit of practice.

I missed this earlier, and it seems completely reasonable to me. So going with the analogy, after you have done GR math every night for a while then you should stop for a space of years without doing it. When you return, then with a little bit of practice you should be much, much better.

Or do you really think that you would have magically achieved your "much, much better" skill level if you had never done the training in the first place? I have definitely experienced the effect that you are describing, but always in the way you experienced: having training and working hard at it first, followed by a break of some length, then returning to practice. I have never found myself suddenly good at something without the initial effort, which is what you are suggesting is possible with GR.

 

[Dale]

The horizon is a fixed null surface in spacetime. It doesn't move.

This is confusing in English, and I don't know how to clarify it w/o math. It is a null surface so it obviously moves at c relative to any local inertial frame, but it is a time-shift invariant surface in a static spacetime and it has a constant coordinate position in Schwarzschild coordinates so in those senses it doesn't move.

Btw, your Rindler horizon description is very good.

 

[PeterDonis]

The horizon is a fixed null surface in spacetime. It doesn't move.

This is confusing in English, and I don't know how to clarify it w/o math. It is a null surface so it obviously moves at c relative to any local inertial frame, but it is a time-shift invariant surface in a static spacetime and it has a constant coordinate position in Schwarzschild coordinates so in those senses it doesn't move.

Good point. I should at least attempt to clarify what I was getting at. I was responding to this statement of A-wal's:

The horizon moves back whenever you try to work out anything at the horizon.

Most of the time, A-wal appears to have a mental picture that more or less treats Schwarzschild coordinates as "fixed", so anything that's fixed in Schwarzschild coordinates doesn't move. That is the sense I meant when I said the horizon doesn't move; clearly, if you're working in Schwarzschild coordinates, the horizon is always at the same place, and you can do calculations without worrying that it's going to "move back" on you. (It's also worth noting, by the way, that this viewpoint of the horizon being "fixed" is the natural one to adopt for "hovering" observers, who accelerate so that they always stay at the same radial coordinate. It is true that the relationship between the radial coordinate and actual physical distance changes as you get closer to the horizon; but that doesn't change the fact that the horizon is fixed for "hovering" observers, and "hovering" observers closer and closer to the horizon will see their proper distance to the horizon decreasing in just the way you would expect if the horizon was fixed at r = 2GM/c^2.)

But one could also take the other viewpoint you suggest, in which the horizon is a null surface and so it is moving at c relative to any timelike observer. However, that doesn't make A-wal's statement above correct, because he said the horizon moves "back", implying that it is an *ingoing* null surface--it would have to be to "move back" in such a way that a free-falling observer, who must be falling inward because of gravity, could never "catch up" to it. But of course the horizon is *not* an ingoing null surface; it's an *outgoing* null surface, and it therefore moves in the *opposite* direction to an ingoing free-falling observer. If we were to adopt this viewpoint at all, we would have to say that the horizon "moves toward" the free-faller, and eventually passes him.

Btw, your Rindler horizon description is very good.

Thanks!

 

[A-wal]

You said light isn't subject to acceleration due to gravity. The fact that it bends shows that it is. So the fact that it bends does *not* support what you were saying. Do you read what you post?

A lot of the time, yes. When I said that light isn't subject to acceleration due to gravity I meant that it doesn’t speed up like massive objects because it’s already at the maximum speed that time will allow under any circumstances, c! Because it always moves at the full speed of time, which is relative, which makes c constant.

Which then has to cover the entire spacetime in order to support your claims. So for you to be correct, you must *assume* that such a riverbed exists and covers the entire spacetime, which is equivalent to *assuming* that nothing can reach the horizon. So you are assuming what you claim to be proving.

You could have a hovering observer at every point up until the horizon but not at or inside the horizon. That we agree on. The speed that the in-faller would have to accelerate to relative to the hoverers would have to reach c in order to reach the horizon. Not possible!

Agreed; they would see the free-faller passing them at a speed a little *closer* to c than if they had just stayed hovering (I assume by "accelerated away very sharply" you mean "accelerated upward even more sharply than they would have to to maintain hovering just outside the horizon").

That’s exactly what I meant. A hoverer could accelerate away from the black hole as fast as they like and a free-faller would still always pass them at less than c. The same thing must happen with velocity at the horizon. When you try to approach c in flat space-time it slows down as you accelerate but you still can’t ever reach it because the rate that it slows down relative to your acceleration decreases the harder you accelerate, and it gets faster overall so that once you’ve stopped accelerating it’s increased its speed relative to your previous frame by the same amount as you have. When you try to approach an event horizon you can close the gap but you can’t ever reach it because the rate that you close the gap decreases if you’re free-falling, as you move into less ‘distance shortened’ space-time relative to where you were. If you were able to reach it then it would be moving at c, which is why you can’t reach it.

With reference to the hoverers, you are correct; they will never see a free-faller passing them at c (or faster). But this does not prove that free-fallers can't pass the horizon; as I noted above, what the hoverers can and can't see is only relevant if the hoverers can cover the entire spacetime, and assuming that is equivalent to *assuming* that free-fallers can't reach the horizon. But you can't *assume* that; you're supposed to be *proving* it.

What do you think seems more likely? The acceleration of gravity is able to do something that no amount of energy could do and pull you to a relative velocity greater than c so that no amount of acceleration in the opposite direction will actually move you in that direction, creating different rules depending on whether you're moving towards or away from it, creating a manifold that needs multiple and contradictory coordinate systems to describe it fully, breaking the law of conservation of energy and the arrow of time needing objects that can't form and existed for an infinite amount of time in the past to try to fix it. Or the event horizon is the closest light or anything else has got since to the singularity since it formed? You’re the one who needs to prove it. You’re making some outrageous claims but you feel safe doing it because it’s the consensus view.

I'm not sure I understand what you're suggesting here. It sounds like you're changing the conditions of the problem: you want to match the hoverer's/accelerator's proper time elapsed from "start", the time when the hoverer/accelerator and free-faller are mutually at rest, to the time when they cross, but that's not the same as matching the physical *distance* apart they are when they start out mutually at rest, which was the initial condition you specified before. Which is it?

Both. Why do I have to pick one? The answer stays the same no matter how you look at it. In flat space-time you could just match the proper time that a free-faller experiences before they reach the horizon and are traveling at c. Then you’d have an equivalent without using curvature. But there is no equivalent of reaching an event horizon because it would take infinite acceleration and infinite proper time to reach c in flat space-time. It’s no different in curved space-time. You can’t reach c because it recedes if you try so that it stays at a constant speed if you’re not accelerating.

I don't know what you mean by this. What "velocity at the horizon"? The velocity of the free-falling object relative to the outgoing light ray that's just at the horizon. That *is* c.

Yes, the velocity of an outgoing light ray is always c locally. It slows down as it approaches an event horizon. It slows down quicker the closer you are to the horizon so the light would build up in front of you as you fell as it takes a shorter and shorter amount of space/time to slow down. You’re saying you can break the light barrier and overtake the light that’s too ‘distance shortened’ to reach the horizon. Presumably this would create an optic boom as you smashed through your own light?

I don't know what you mean by this either. The horizon is a fixed null surface in spacetime. It doesn't move.

I mean that if you measure the horizon to be a mile in front of you and then move a mile (as measured at this distance) forwards towards the horizon you’d find that you haven’t reached the horizon because it’s moved back. This is because of the ‘distance shortening’ that you’d expect when accelerating or moving into a higher gravitational field. It’s exactly the same in flat space-time as saying “well this amount of energy got us to half c so the same again will allow us to reach c”. No it won’t. The event horizon moves away if you approach it in exactly the same way as the speed of light moves away as you try to reach it.

Why "must" there be? Just because you say so? Do you have any actual physical reason? Any actual argument (that doesn't rely on assuming what you're supposed to be proving)?

Because you need to reach c to reach the event horizon. That’s why no amount of energy of acceleration will allow you to pass from one side to the other. You have to stay outside of the horizon for the same reason you have to stay below c.

In other words, you still don't understand how a Rindler horizon works, despite claiming (many, many posts ago) that you were quite familiar with it and didn't need to have it explained to you; and you keep on making the same misstatement about it despite repeated corrections on my part.

I'll try once more, starting from scratch and spelling everything out. I'm sure a lot of this will seem "obvious" to you, but after hundreds of posts I can't assume anything about how you understand words relative to how I understand words. So I've got to dot all the i's and cross all the t's.

Consider a flat spacetime and a set of global inertial coordinates x, t on this spacetime (I'm using units in which c = 1, so x and t are in the same units). There are two families of observers in this spacetime. One family is of free-falling observers; each such observer is at rest (in the global inertial coordinate system) at some positive x-coordinate. The second family is of accelerating observers; each such observer starts out, at time t = 0, sitting at rest right next to one of the free-falling observers; but each accelerating observer accelerates with a proper acceleration (i.e., what's measured by his accelerometer) of c^2 / x, where x is the x-coordinate he is at at t = 0. Finally, there is a light ray that is emitted in the positive x-direction from x = 0 at time t = 0 (i.e., from the spacetime origin of the global inertial coordinates).

The following are all true:

(1) The light ray coincides with the Rindler horizon for *all* of the accelerating observers. This means that the light ray will never catch up to any of those observers; all of them will always be ahead of it (as long as they continue to accelerate at the given proper acceleration).

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

(2) Each of the free-falling observers passes the light ray (and therefore crosses the Rindler horizon) at a positive time t equal to his x-coordinate. Therefore, there are two regions in this spacetime: the region "outside" the horizon, which contains the starting points of all the observers at time t = 0, and the region "inside" the horizon, which is reached only by the free-falling observers, the accelerating observers never reach it (as long as they continue to accelerate). Therefore, the family of accelerating observers does *not* cover the same portion of the spacetime as the family of free-falling observers does; there is a portion (the region "inside" the horizon) that only the free-fallers cover.

The equivalent to a distant observer being able to receive a signal from an in faller but not the other way round. No one crosses a Rindler horizon. There is no Rindler horizon for the free-fallers. It’s created by the accelerating observers.

(3) Once a free-falling observer crosses the Rindler horizon, there is no way for that observer to send any signal to the region of spacetime on the other side of the horizon (since such a signal would have to travel faster than light). That means no free-falling observer can send a signal to any of the accelerating observers after the free-falling observer has crossed the horizon. Which also means, of course, that no accelerating observer can "see" any free-falling observer after the free-falling observer has crossed the horizon. (All this applies as long as the accelerating observers continue to accelerate.)

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

(4) As a given free-falling observer passes each member of the family of accelerating observers, the velocity of the free-faller relative to the accelerating observers approaches c. However, it never reaches c, because there is no accelerating observer *at* the horizon; such an observer would have to move at c (since the horizon coincides with the path of a light ray), and no timelike observer can move at c. So no accelerating observer ever sees a freely falling observer pass him at c (or faster). Instead, from the point where each free-falling observer crosses the horizon, he simply stops passing accelerating observers (because there are no more for him to pass) and enters the region of spacetime that the accelerating observers don't cover.

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

The above is entirely in flat spacetime, it's entirely consistent with SR, and it clearly demonstrates several things: that free-falling observers *can* cross a Rindler horizon, that there *can* be an entire region of spacetime that accelerated observers can't "see" and which they don't cover, and that a free-falling observer not being able to send a light signal that can catch up with an accelerated observer is *not* equivalent to a free-falling observer not being able to cross the horizon.

No, a free-faller not being able to send a light signal that can catch up with a Rindler horizon is equivalent to an free-faller not being able to send a light signal that can catch up with an event horizon. Nothing reaches either horizon. They’re the same.

I can already anticipate how you will reply to this post. You will say, "of course, that's how I've been saying a Rindler horizon works all along", and then proceed to claim that at least one of the three things I've demonstrated, which I listed in the previous paragraph, is false. If that's your inclination on reading the above, please think very carefully before posting a response. Don't even think about how the above translates into the spacetime surrounding a black hole, or what it implies or doesn't imply about a black hole horizon being reachable. Do you honestly believe, looking *just* at the flat spacetime scenario I've just described, as it stands, that anyone of the three things I listed in the previous paragraph is false? Because if so, I'm going to be hard pressed not to conclude that you and I can't have a sane discussion about this. It's one thing to be arguing about GR and how it applies to a black hole horizon. But if we can't even have a common understanding about how a simple scenario in SR works, I'm not sure what we *can* have a common understanding about.

Give me a chance. You’re talking as if I’d already disagreed with you and as far as I can remember I haven’t done that once with flat space-time examples. Thanks for the description though, although I’m not sure what I was getting wrong. The quote that prompted this explanation of the Rindler horizon was correct. I don’t disagree with anything in the four examples in flat space-time, but I don’t agree with the implied interpretations of how they apply to curved space-time or with the conclusion afterwards. I’m sorry but if you’re going to claim that your way of looking at it is right and mine is wrong then I’m going to argue the point because I genuinely believe that the whole space-time is covered using just Schwarzschild coordinates/the light cones can’t tilt to 90 degrees/you can’t cross an event horizon because it’s moving inwards/you can’t reach c.

I've attached a graph illustrating what I described in my previous post. The horizontal axis is space (the x coordinate) and the vertical axis is time (the t coordinate). The dark magenta vertical lines are the free-fallers' worldlines. The blue lines are the accelerators' worldlines. The red line is the Rindler horizon.

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

I missed this earlier, and it seems completely reasonable to me. So going with the analogy, after you have done GR math every night for a while then you should stop for a space of years without doing it. When you return, then with a little bit of practice you should be much, much better.

Or do you really think that you would have magically achieved your "much, much better" skill level if you had never done the training in the first place? I have definitely experienced the effect that you are describing, but always in the way you experienced: having training and working hard at it first, followed by a break of some length, then returning to practice. I have never found myself suddenly good at something without the initial effort, which is what you are suggesting is possible with GR.

No I’m not. I said you need to empty your cup. I haven’t got as much crap in my head from learning it formally and I still needed a six month break. If I was interested in learning maths then yes, that probably would be the best way of doing it.

This is confusing in English, and I don't know how to clarify it w/o math. It is a null surface so it obviously moves at c relative to any local inertial frame, but it is a time-shift invariant surface in a static spacetime and it has a constant coordinate position in Schwarzschild coordinates so in those senses it doesn't move.

How can the horizon have a constant position in Schwarzschild coordinates when ‘distance shortening’ means the amount of space you need to cover and the time it would take to cover the same distance would of course lengthen as you get closer to it?

Most of the time, A-wal appears to have a mental picture that more or less treats Schwarzschild coordinates as "fixed", so anything that's fixed in Schwarzschild coordinates doesn't move. That is the sense I meant when I said the horizon doesn't move; clearly, if you're working in Schwarzschild coordinates, the horizon is always at the same place, and you can do calculations without worrying that it's going to "move back" on you. (It's also worth noting, by the way, that this viewpoint of the horizon being "fixed" is the natural one to adopt for "hovering" observers, who accelerate so that they always stay at the same radial coordinate. It is true that the relationship between the radial coordinate and actual physical distance changes as you get closer to the horizon; but that doesn't change the fact that the horizon is fixed for "hovering" observers, and "hovering" observers closer and closer to the horizon will see their proper distance to the horizon decreasing in just the way you would expect if the horizon was fixed at r = 2GM/c^2.)

It’s fixed for hovering observers but I don’t think they would see their proper distance to the horizon decreasing in the same way the proper distance would decrease in flat space-time. If you were to hover closer then the distance would increase. Obviously it would decrease overall because you’ve moved closer to it, but it wouldn’t decrease by as much as it would in flat space-time. That difference is what stops an object from reaching the horizon, because it gets more pronounced the closer you get to it. It’s exactly like adding velocities trying to reach c in flat space-time.

But one could also take the other viewpoint you suggest, in which the horizon is a null surface and so it is moving at c relative to any timelike observer. However, that doesn't make A-wal's statement above correct, because he said the horizon moves "back", implying that it is an *ingoing* null surface--it would have to be to "move back" in such a way that a free-falling observer, who must be falling inward because of gravity, could never "catch up" to it. But of course the horizon is *not* an ingoing null surface; it's an *outgoing* null surface, and it therefore moves in the *opposite* direction to an ingoing free-falling observer. If we were to adopt this viewpoint at all, we would have to say that the horizon "moves toward" the free-faller, and eventually passes him.

Of course it’s an ingoing null surface moving back in such a way that a free-falling observer could never catch up to it. It’s moving back at c because c sets the limit to how far you can get in a certain amount of time, so the event horizon represents how close you can get to the singularity at the time that you’re seeing it. When you accelerate in flat space-time c gets faster, not slower, relative to your previous frame which prevents you from reaching c. The event horizon would move away to prevent from reaching c. The event horizon is the point at which no amount of acceleration could slow you down, as if you were trapped above the speed of light rather than below. It’s not a problem because you can’t reach either, because they’re basically the same thing.

 

[A-wal]

How can gravity accelerate you to c when energy can’t? How can one force move something in a way that no amount of any other force can? I still honestly have no idea how you could possibly think that. The speed of light is the fastest anything can move and light can’t reach the horizon. If an object could cross then a distant observer would be able to see it, although there would be a delay. If light can’t cross from a distance then it can’t cross, and if light can’t cross then nothing can. The fact that you would need an infinite amount of force to escape from inside an event horizon shows that you would need an infinite amount of force to get inside one. If you can get in then you can get out and if you can’t get out...

 

[Dale]

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

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

How can the horizon have a constant position in Schwarzschild coordinates

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

How are you measuring the distance that you move through?

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

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

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

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

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

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

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

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

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

Incorrect. I already addressed this above.

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

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

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

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

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

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

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

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

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

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

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

Relative to what? Time according to what observer?

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

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

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

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

 

[PeterDonis]

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

 

[Passionflower]

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

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

 

[PeterDonis]

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

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

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

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

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