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

[PeterDonis]

In what way does it seem incoherent? The model is identical apart from the extra relative component of gravity, which to my mind tidies it up nicely. The standard black hole description is so messy. Normally the difference is marginal.

Your model makes definite predictions that are very different from the standard model. It predicts that the horizon can't be reached, and that the "rope experiment" will give different results. But I have seen no logical structure from you that leads to those predictions; all I see is your intuitive sense that they "make sense". You certainly haven't stated a coherent model that is "identical" to standard GR "apart from the extra relative component of gravity"; I'm not exactly sure what that means, but I do know that you can't arrive at a consistent model that makes the predictions you're making just by "adding in" some simple extra ingredient to standard GR. Standard GR is a very precise, specific logical structure, and adding anything to it like that would make it inconsistent. You may think you have stated a coherent model, but you haven't; all you've done is made some intuitive, hand-waving statements that don't form a coherent logical structure.

Can you clarify this please? I get the idea but are you saying there's a definite upper limit on the amount of energy that can be exerted onto an object? What happens if you use a thicker rope once this limit's been reached, or two ropes?

I'm saying there's a finite upper limit on the *breaking strength* of any material. There's no limit on the amount of force you can exert on the material (at least, not in principle), so any material will eventually break.

However, just saying "finite breaking strength" may not be the best way of stating the limit, because it suggests questions like you asked, about adding more ropes or making the rope thicker. Previously, I stated the limit as "the speed of sound in the material must be less than the speed of light", which should make it clear that in the case we've been discussing, adding more ropes or making the rope thicker won't help, because it won't change the fastest speed at which any force exerted on one end of the rope can propagate to the other end. That speed must be less than the speed of light, and as long as it is, the rope must break.

But even that way of stating it may still be misleading, because it makes it seem like it's the speed of *inward* propagation of the force through the rope that's critical. That's not quite right. The real problem is that, once the lower end of the rope drops below the horizon, *it* would have to move faster than light to keep up with the portion of the rope that's still above the horizon. It can't do that, so the rope has to break. The "critical time" I've been referring to is simply the last time when a force applied at the top end of the rope, by the "hovering" observer, can propagate down the rope, at the speed of light, and reach the free-falling observer at the lower end before he crosses the horizon--meaning, before he would have to move faster than light to keep up.

"The three infinite spatial dimensions are all encompassed in the radial coordinate r" seems contradictory to me.

Consider ordinarly 3-dimensional Euclidean space. It is infinite in all three spatial dimensions. If we describe it in Cartesian coordinates, x, y, z, then all three coordinates have infinite range, so it's "obvious" in these coordinates that all three spatial dimensions are infinite.

Now look at the same space in spherical coordinates, r, theta, phi. Only r has an infinite range; theta and phi are limited to a finite range (the normal convention is theta from 0 to pi, and phi from 0 to 2 pi). But it's the same space as before, so all three spatial dimensions are still infinite. It's just that the coordinates make it harder to see because the infinite range of all three spatial dimensions is "tied up" in the infinite range of the r coordinate alone. (A vector pointing in the direction of increasing r can point in *any* direction in the space, depending on the theta and phi coordinates.)

Describing spacetime around a black hole works the same way; you're just adding a t coordinate, so you have four infinite dimensions now instead of three, but only two coordinates (t and r) with infinite range. The r coordinate encompasses the infinity of three of the dimensions of the spacetime (all the spatial ones). Strictly speaking, this only refers to the exterior portion of the spacetime (outside the horizon); the portion inside works differently (see below).

Do you mean it's only infinite from the inside? <snip> Maybe that's not what you meant?

You're right, it's not what I meant. The spacetime as a whole is infinite in extent, but if you consider just the interior, the portion inside the horizon, I'm pretty sure that portion is finite. (I haven't actually done the calculation of its 4-volume to be certain.)

 

[A-wal]

Your model makes definite predictions that are very different from the standard model. It predicts that the horizon can't be reached, and that the "rope experiment" will give different results. But I have seen no logical structure from you that leads to those predictions; all I see is your intuitive sense that they "make sense". You certainly haven't stated a coherent model that is "identical" to standard GR "apart from the extra relative component of gravity"; I'm not exactly sure what that means, but I do know that you can't arrive at a consistent model that makes the predictions you're making just by "adding in" some simple extra ingredient to standard GR. Standard GR is a very precise, specific logical structure, and adding anything to it like that would make it inconsistent. You may think you have stated a coherent model, but you haven't; all you've done is made some intuitive, hand-waving statements that don't form a coherent logical structure.

I respectfully disagree. It’s based on the idea that the curvature created by matter is indistinguishable from the curvature created by energy accept that the curvature from energy is *c squared greater than the curvature from matter. That’s why gravity’s so weak.

I'm saying there's a finite upper limit on the *breaking strength* of any material. There's no limit on the amount of force you can exert on the material (at least, not in principle), so any material will eventually break.

However, just saying "finite breaking strength" may not be the best way of stating the limit, because it suggests questions like you asked, about adding more ropes or making the rope thicker. Previously, I stated the limit as "the speed of sound in the material must be less than the speed of light", which should make it clear that in the case we've been discussing, adding more ropes or making the rope thicker won't help, because it won't change the fastest speed at which any force exerted on one end of the rope can propagate to the other end. That speed must be less than the speed of light, and as long as it is, the rope must break.

But even that way of stating it may still be misleading, because it makes it seem like it's the speed of *inward* propagation of the force through the rope that's critical. That's not quite right. The real problem is that, once the lower end of the rope drops below the horizon, *it* would have to move faster than light to keep up with the portion of the rope that's still above the horizon. It can't do that, so the rope has to break. The "critical time" I've been referring to is simply the last time when a force applied at the top end of the rope, by the "hovering" observer, can propagate down the rope, at the speed of light, and reach the free-falling observer at the lower end before he crosses the horizon--meaning, before he would have to move faster than light to keep up.

Nice. Especially the third paragraph.

Consider ordinarly 3-dimensional Euclidean space. It is infinite in all three spatial dimensions. If we describe it in Cartesian coordinates, x, y, z, then all three coordinates have infinite range, so it's "obvious" in these coordinates that all three spatial dimensions are infinite.

Now look at the same space in spherical coordinates, r, theta, phi. Only r has an infinite range; theta and phi are limited to a finite range (the normal convention is theta from 0 to pi, and phi from 0 to 2 pi). But it's the same space as before, so all three spatial dimensions are still infinite. It's just that the coordinates make it harder to see because the infinite range of all three spatial dimensions is "tied up" in the infinite range of the r coordinate alone. (A vector pointing in the direction of increasing r can point in *any* direction in the space, depending on the theta and phi coordinates.)

Describing spacetime around a black hole works the same way; you're just adding a t coordinate, so you have four infinite dimensions now instead of three, but only two coordinates (t and r) with infinite range. The r coordinate encompasses the infinity of three of the dimensions of the spacetime (all the spatial ones). Strictly speaking, this only refers to the exterior portion of the spacetime (outside the horizon); the portion inside works differently (see below).

I think we’re getting further from the point that from the exterior it has a finite three dimensional spherical shape so I see absolutely no reason why it should be any different in the other one.

You're right, it's not what I meant. The spacetime as a whole is infinite in extent, but if you consider just the interior, the portion inside the horizon, I'm pretty sure that portion is finite. (I haven't actually done the calculation of its 4-volume to be certain.)

I would have thought it would have to be.

 

[PeterDonis]

I respectfully disagree. It’s based on the idea that the curvature created by matter is indistinguishable from the curvature created by energy accept that the curvature from energy is *c squared greater than the curvature from matter. That’s why gravity’s so weak.

I'm not sure what you're disagreeing with. In standard GR, curvature created by matter *is* indistinguishable from curvature generated by energy; in fact, "matter" and "energy" are really the same thing, just measured in different units, and the speed of light squared is just a conversion factor between the different units.

So if you mean that your own model is based on this idea as well, then your model should give the same predictions as standard GR, and it doesn't. So your model can't be "based on" this idea and still be consistent. If you mean something else, please clarify.

I think we’re getting further from the point that from the exterior it has a finite three dimensional spherical shape so I see absolutely no reason why it should be any different in the other one.

The *horizon* has a finite three dimensional spherical shape (more precisely, the intersection of the horizon with a slice of constant time has that shape). But the horizon is not the entire spacetime. When I was talking about dimensions being infinite in extent, I was talking about the entire spacetime (more precisely, the portion exterior to the horizon).

I would have thought it would have to be.

I think so too, but it's not entirely obvious because the portion of the spacetime inside the horizon still covers an infinite range of "time" coordinates. (This is true not only in the interior Schwarzschild coordinates, but in Painleve coordinates and in Kruskal-Szeres coordinates.) But I believe (though I haven't confirmed by explicit calculation--maybe one of the experts on these forums has) that the integral for the 4-volume of the interior portion of the spacetime converges fast enough as the "time" coordinate goes to infinity to make the total 4-volume finite, for reasons similar to the reasons that the integrals for proper time to fall to the horizon and proper distance to the horizon converge (as I showed much earlier in this thread).

(Also, I put "time" in scare-quotes because inside the horizon, the "time" coordinates of the first two coordinate systems I mentioned, Schwarzschild interior and Painleve, are not timelike! In Schwarzschild interior coordinates, the radial coordinate r is timelike inside the horizon. In Painleve coordinates, all four coordinates are spacelike inside the horizon! George Jones discusses that in this post from a thread where the subject came up:

https://www.physicsforums.com/showpost.php?p=3000266&postcount=121

Kruskal coordinates are more "normal" in this respect, there is one timelike coordinate and three spacelike--one "radial" and the two angular coordinates--and the timelike/spacelike nature of each coordinate remains the same throughout the full range.)

 

[A-wal]

I'm not sure what you're disagreeing with. In standard GR, curvature created by matter *is* indistinguishable from curvature generated by energy; in fact, "matter" and "energy" are really the same thing, just measured in different units, and the speed of light squared is just a conversion factor between the different units.

So if you mean that your own model is based on this idea as well, then your model should give the same predictions as standard GR, and it doesn't. So your model can't be "based on" this idea and still be consistent. If you mean something else, please clarify.

I know one of the principles of general relativity is that the curvature is the same, but it then goes on to describe a type of curvature that's different from the curvature in special relativity. It should be relative, just like velocity, which is why you can't feel the difference between being in gravitational fields of different strengths when in free-fall. In special relativity there is a limit to how fast you can go relative to anything else. You can keep on accelerating but you'll never reach c. I think it should work exactly the same for gravity. You shouldn't be able to curve it beyond 90 degrees whether the force is created by energy or matter. There should be an upper limit equivalent to c (event horizon) even though you can keep on accelerating just by free-falling (tidal force). One way of looking at it is that gravity can and does reach a value that exceeds c but that would be unreachable for any observer because the world-line between you and the horizon would be potentially infinite. Not actually infinite but a line that actually reaches the horizon would be infinite, so it's impossible. Its four-dimensional volume(?) would depend on its mass and the distance of the observer from the horizon. That's not new. In standard GR the size of the Earth in all four dimensions decreases the closer you get to it, but it's normally marginal. In the case of a black hole its volume in all four dimensions would always reach zero before anything could reach the horizon. The fact that the curvature is caused by matter rather than energy means the equivalent curvature from energy would be much greater. So the question of why gravity is so weak is really asking why E=mc squared, and we have special relativity to answer that.

The *horizon* has a finite three dimensional spherical shape (more precisely, the intersection of the horizon with a slice of constant time has that shape). But the horizon is not the entire spacetime. When I was talking about dimensions being infinite in extent, I was talking about the entire spacetime (more precisely, the portion exterior to the horizon).

Okay but I don't know how anything could be infinite in extent. You say the horizon has a finite three-dimensional spherical shape so why wouldn't it have a four-dimensional spherical shape?

I think so too, but it's not entirely obvious because the portion of the spacetime inside the horizon still covers an infinite range of "time" coordinates. (This is true not only in the interior Schwarzschild coordinates, but in Painleve coordinates and in Kruskal-Szeres coordinates.) But I believe (though I haven't confirmed by explicit calculation--maybe one of the experts on these forums has) that the integral for the 4-volume of the interior portion of the spacetime converges fast enough as the "time" coordinate goes to infinity to make the total 4-volume finite, for reasons similar to the reasons that the integrals for proper time to fall to the horizon and proper distance to the horizon converge (as I showed much earlier in this thread).

(Also, I put "time" in scare-quotes because inside the horizon, the "time" coordinates of the first two coordinate systems I mentioned, Schwarzschild interior and Painleve, are not timelike! In Schwarzschild interior coordinates, the radial coordinate r is timelike inside the horizon. In Painleve coordinates, all four coordinates are spacelike inside the horizon! George Jones discusses that in this post from a thread where the subject came up:

https://www.physicsforums.com/showpost.php?p=3000266&postcount=121

Kruskal coordinates are more "normal" in this respect, there is one timelike coordinate and three spacelike--one "radial" and the two angular coordinates--and the timelike/spacelike nature of each coordinate remains the same throughout the full range.)

See, messy!

 

[PeterDonis]

I know one of the principles of general relativity is that the curvature is the same, but it then goes on to describe a type of curvature that's different from the curvature in special relativity.

Huh? There is no curvature in special relativity; in SR, spacetime is flat.

It should be relative, just like velocity, which is why you can't feel the difference between being in gravitational fields of different strengths when in free-fall.

Yes, this is called the equivalence principle, and is one of the cornerstones of general relativity. It is perfectly consistent with being able to reach and pass through a black hole's horizon.

In special relativity there is a limit to how fast you can go relative to anything else. You can keep on accelerating but you'll never reach c.

Fine so far, but...

I think it should work exactly the same for gravity. You shouldn't be able to curve it beyond 90 degrees whether the force is created by energy or matter.

What does this have to do with velocity? And what does "curve it beyond 90 degrees" mean? Are you talking about the tilting of the light cones (see my next comment)? If so, you're wrong that they can't tilt beyond "90 degrees" (meaning "vertical" in the sense I describe below). They can.

There should be an upper limit equivalent to c (event horizon) even though you can keep on accelerating just by free-falling (tidal force).

There is an upper limit to velocity in GR, just as in SR; but it's expressed in a more general form that works when spacetime is curved (the form "you can't go faster than c relative to anything else" only works when spacetime is flat). In GR, we say that worldlines of objects can't go outside the light cones. That still works the same in flat spacetime, because all the light cones are aligned with each other (the "sides" of the light cones tilt upward at a 45 degree angle, to the left and right if you're looking at a spacetime diagram showing one time and one space coordinate). But it carries over to curved spacetime, where light cones at different events can be tilted with respect to each other. At the horizon of a black hole, the light cones are tilted over so that the radially outgoing sides of the cones are vertical--they stay at r = 2M forever. The ingoing sides of the cones, however, still tilt inward, just as they do far away from the horizon.

One way of looking at it is that gravity can and does reach a value that exceeds c but that would be unreachable for any observer because the world-line between you and the horizon would be potentially infinite.

Except that it isn't, as has already been shown in this thread.

The fact that the curvature is caused by matter rather than energy means the equivalent curvature from energy would be much greater.

Huh? Energy and matter are just different forms of the same thing; a given amount of energy causes exactly the same curvature as the equivalent amount of matter (with the formula E = mc^2 defining what is "equivalent").

Okay but I don't know how anything could be infinite in extent. You say the horizon has a finite three-dimensional spherical shape so why wouldn't it have a four-dimensional spherical shape?

Because it doesn't. That's not what the mathematics of a black hole solution to the Einstein Field Equation describes. There are other solutions that (sort of) describe a four-dimensional sphere (for example, the kind of "Euclidean" models that Stephen Hawking talks about in his "no boundary" proposal for the beginning of the universe), but they don't describe black holes.

See, messy!

How so? Different coordinate systems are useful for different purposes, so they have different properties. This is not something that's particular to GR; it happens in ordinary geometry too. Look at maps of the Earth in different projections--Mercator, stereographic, etc.--and compare to an actual globe. The different projections distort the globe in different ways in order to represent particular properties on a flat map. You can't make a flat map of a curved surface and not have it distorted, so "messiness", if you insist on calling it that, is unavoidable unless you want to carry a globe around with you all the time. And in the case of spacetime, we don't have that option; there's *no* way to represent a 4-dimensional manifold in three or fewer dimensions without distortion, and we choose maps that have different distortions for different purposes. What's the problem?

 

[A-wal]

Huh? There is no curvature in special relativity; in SR, spacetime is flat.

Space-time is flat. Objects can follow curved paths through it using energy to accelerate. The curvature from acceleration can never get you to c, and the curvature from gravity can never get you to c. You said that the rope would always break because anything beyond the event horizon has broken the light barrier. If you can’t use energy to do it then what makes you think you can do it with the curvature from matter? The strength of the gravitational field makes no difference, just like velocity. The only difference is that you accelerate harder in free-fall when in higher gravity, but you’ll never be able to accelerate hard enough to reach c.

Yes, this is called the equivalence principle, and is one of the cornerstones of general relativity. It is perfectly consistent with being able to reach and pass through a black hole's horizon.

Only if you choose to ignore or have the cheek to disagree with what I just said.

What does this have to do with velocity? And what does "curve it beyond 90 degrees" mean? Are you talking about the tilting of the light cones (see my next comment)? If so, you're wrong that they can't tilt beyond "90 degrees" (meaning "vertical" in the sense I describe below). They can.

A velocity of c as a right-angle on a space-time diagram like the one I described before. Gravitational curvature shouldn’t be able to move you to 90 degrees any more than energy curvature can. You say it can, but why when we know that they’re equivalent?

There is an upper limit to velocity in GR, just as in SR; but it's expressed in a more general form that works when spacetime is curved (the form "you can't go faster than c relative to anything else" only works when spacetime is flat). In GR, we say that worldlines of objects can't go outside the light cones. That still works the same in flat spacetime, because all the light cones are aligned with each other (the "sides" of the light cones tilt upward at a 45 degree angle, to the left and right if you're looking at a spacetime diagram showing one time and one space coordinate). But it carries over to curved spacetime, where light cones at different events can be tilted with respect to each other. At the horizon of a black hole, the light cones are tilted over so that the radially outgoing sides of the cones are vertical--they stay at r = 2M forever. The ingoing sides of the cones, however, still tilt inward, just as they do far away from the horizon.

Why would using the two different causes of curvature together be any different from using more of one?

Except that it isn't, as has already been shown in this thread.

Except that it is, as has also already been shown in this thread, in a slightly less technical and more hand-wavy way.

Huh? Energy and matter are just different forms of the same thing; a given amount of energy causes exactly the same curvature as the equivalent amount of matter (with the formula E = mc^2 defining what is "equivalent").

Right, so why is it such a mystery that the strength of electro-magnetism is so much greater than gravity when electro-magnetism is curvature from energy and gravity is from matter?

Because it doesn't. That's not what the mathematics of a black hole solution to the Einstein Field Equation describes. There are other solutions that (sort of) describe a four-dimensional sphere (for example, the kind of "Euclidean" models that Stephen Hawking talks about in his "no boundary" proposal for the beginning of the universe), but they don't describe black holes.

Perhaps they do. There’s no complex mechanism to consider when talking about the life span of a black hole, just time-dilation and length contraction which are equivalent. So it should have same length in all four dimensions.

How so? Different coordinate systems are useful for different purposes, so they have different properties. This is not something that's particular to GR; it happens in ordinary geometry too. Look at maps of the Earth in different projections--Mercator, stereographic, etc.--and compare to an actual globe. The different projections distort the globe in different ways in order to represent particular properties on a flat map. You can't make a flat map of a curved surface and not have it distorted, so "messiness", if you insist on calling it that, is unavoidable unless you want to carry a globe around with you all the time. And in the case of spacetime, we don't have that option; there's *no* way to represent a 4-dimensional manifold in three or fewer dimensions without distortion, and we choose maps that have different distortions for different purposes. What's the problem?

It’s messy because from what I’ve seen, there’s so many different ways of measuring it in the standard version that you can manipulate it to show anything you want.

 

[PeterDonis]

Space-time is flat. Objects can follow curved paths through it using energy to accelerate. The curvature from acceleration can never get you to c, and the curvature from gravity can never get you to c.

You are using very confusing terminology, which may help to explain why your thinking is confused. Specifically, you are using the same word, "curvature", to refer to two distinct concepts that should not be conflated. Let me try to restate what I think you mean by the two uses of the word "curvature" to see if I'm understanding you correctly:

(1) What you are calling "curvature due to matter" or "curvature from gravity" is curvature of *spacetime* itself. As I've pointed out, spacetime curvature can actually be caused by matter *or* energy; it can also be caused by pressure and stress in a material. When the word "curvature" is used in standard GR without qualification, it is this kind of curvature that is meant; and every time I've used the word "curvature", including the expression "curvature due to energy", I've been using it to mean this.

(2) What you are calling "curvature due to energy" or "curvature from acceleration" is curvature of the *worldline* of an object *within* a spacetime (which may itself be either flat or curved). The word "curvature" is *never* (as far as I've seen) used in standard GR to mean this; the word "acceleration" (or the term "proper acceleration" if more precision is needed) is used. You are correct that, in order to follow such a path, an object must normally expend energy, for example by firing a rocket engine. However, this is not always required; we don't have to expend any energy to remain stationary on the Earth's surface, but our worldlines are still accelerated; we are not in free fall.

The reason it's important to make the distinction I've just made is that, as far as standard GR is concerned, there is *no* useful analogy between the two things I've just described, curvature of spacetime and having an accelerated worldline. They are simply different things, and have to be analyzed and thought about separately. Your mental model appears to be based on drawing a close analogy between these two phenomena that simply is not justified.

You said that the rope would always break because anything beyond the event horizon has broken the light barrier.

No, that's not what I said. I said that the lower end of the rope, once it drops below the horizon, would have to move faster than light to *keep up* with the portion of the rope that's above the horizon. That's no different than saying that, in flat spacetime, an object that is beyond the Rindler horizon of a uniformly accelerating observer would have to move faster than light to keep up with that observer. It's simply a consequence of the way the light cones are laid out in the spacetime. You simply look at where the two objects are, and ask, "Would object A have to move outside the light cone in order to keep up with object B?" If the answer is yes, then object A can't keep up. So the object below the horizon has not "broken the light barrier"; quite the opposite. It's precisely the fact that it *can't* break the light barrier that prevents it from keeping up with objects that are hovering above the horizon.

If you can’t use energy to do it then what makes you think you can do it with the curvature from matter? The strength of the gravitational field makes no difference, just like velocity. The only difference is that you accelerate harder in free-fall when in higher gravity, but you’ll never be able to accelerate hard enough to reach c.

As I said above, acceleration of a particular worldline (for example, by firing a rocket engine, or standing on the surface of a planet) is simply a different thing from curvature of spacetime itself due to the presence of matter (more precisely, to the presence of stress-energy, which includes all the things I listed above that can cause spacetime curvature). An object in free fall is *not* accelerating (this is another confusing terminology you keep insisting on using, to use the word "acceleration" to refer to an object that is in free fall--avoiding this sort of confusion is a big reason why the word "acceleration" is *not* used in standard GR to refer to what is called "acceleration due to gravity" in Newtonian physics). Since the object is in free fall, there's no question of being able to "accelerate hard enough to reach c"; the object isn't accelerating at all. It's just following the straightest possible path in the spacetime it's in. If spacetime is flat, that path (worldline) will be a "straight line". If the spacetime is curved, the worldline of the object will be a geodesic of that curved spacetime--the analogue of a "straight line" in a curved manifold (where there are no "straight lines" in the strict Euclidean sense). If following such a geodesic worldline takes the object into portions of the spacetime where the light cones are tilted differently enough, then it may "look like" the object *is* moving "faster than light" if you insist on using coordinates that don't take proper account of the different tilting of the light cones. But that's a problem with your coordinates, not with the physics. The physics is simply that the free-falling object goes wherever the geodesic worldline takes it, just as it does in flat spacetime; it requires no more "energy" or "effort" than it does for an object in flat spacetime to remain in the same inertial frame for all time.

Only if you choose to ignore or have the cheek to disagree with what I just said.

I'll take option 2.

A velocity of c as a right-angle on a space-time diagram like the one I described before. Gravitational curvature shouldn’t be able to move you to 90 degrees any more than energy curvature can. You say it can, but why when we know that they’re equivalent?

They're not equivalent, once we clarify the terminology as I did above. What you have been calling "curvature due to energy" is *not* the same thing as what *I* have been calling "curvature due to energy". What you have been calling "curvature due to energy" is *not* equivalent to what you have been calling "gravitational curvature". I wasn't able to make that clear before because I didn't understand your non-standard use of the expression "curvature due to energy".

Why would using the two different causes of curvature together be any different from using more of one?

Because they're causing different kinds of "curvature", as I explained above--one of which should not even be described by the term "curvature".

Except that it is, as has also already been shown in this thread, in a slightly less technical and more hand-wavy way.

Except that your "hand-wavy" way is based on a false analogy and confusing terminology that obfuscates crucial distinctions.

Right, so why is it such a mystery that the strength of electro-magnetism is so much greater than gravity when electro-magnetism is curvature from energy and gravity is from matter?

Now you're using the term "curvature from energy" in yet *another* different way, which I don't understand, but which doesn't appear to me to be compatible with *either* of the uses of the term I described above.

Perhaps they do.

No, they don't. There is no event horizon in any of the cosmological models I referred to, so they can't describe a black hole.

There’s no complex mechanism to consider when talking about the life span of a black hole, just time-dilation and length contraction which are equivalent. So it should have same length in all four dimensions.

The spatial size of the horizon (i.e., the intersection of the horizon with a slice of constant time) is not finite because of "length contraction". It's finite because it's at a finite radius, r = 2M. But the horizon extends through an infinite range of "time" coordinates, so its behavior in time is *not* "equivalent" to its behavior in space.

It’s messy because from what I’ve seen, there’s so many different ways of measuring it in the standard version that you can manipulate it to show anything you want.

I don't understand where you're getting this from. The underlying geometry--either of the Earth's surface, or of a black hole spacetime--is the same *regardless* of what coordinate system you use to map it. Different coordinate systems are useful for different purposes, but they all describe the same underlying geometry, and they all give the same answers for all physical predictions within the portion of the geometry that they cover. You certainly can't "manipulate" the answers just by changing coordinates; if you get a different answer for a physical prediction in one coordinate system vs. another, then you've done something wrong.

 

[A-wal]

You are using very confusing terminology, which may help to explain why your thinking is confused. Specifically, you are using the same word, "curvature", to refer to two distinct concepts that should not be conflated.

Yes, that's intentional. I'm trying to make a point.

Let me try to restate what I think you mean by the two uses of the word "curvature" to see if I'm understanding you correctly:

(1) What you are calling "curvature due to matter" or "curvature from gravity" is curvature of *spacetime* itself. As I've pointed out, spacetime curvature can actually be caused by matter *or* energy; it can also be caused by pressure and stress in a material. When the word "curvature" is used in standard GR without qualification, it is this kind of curvature that is meant; and every time I've used the word "curvature", including the expression "curvature due to energy", I've been using it to mean this.

(2) What you are calling "curvature due to energy" or "curvature from acceleration" is curvature of the *worldline* of an object *within* a spacetime (which may itself be either flat or curved). The word "curvature" is *never* (as far as I've seen) used in standard GR to mean this; the word "acceleration" (or the term "proper acceleration" if more precision is needed) is used. You are correct that, in order to follow such a path, an object must normally expend energy, for example by firing a rocket engine. However, this is not always required; we don't have to expend any energy to remain stationary on the Earth's surface, but our worldlines are still accelerated; we are not in free fall.

Yes we do have to expend energy to remain stationary on the surface. It's called our sodding weight!

The reason it's important to make the distinction I've just made is that, as far as standard GR is concerned, there is *no* useful analogy between the two things I've just described, curvature of spacetime and having an accelerated worldline. They are simply different things, and have to be analyzed and thought about separately. Your mental model appears to be based on drawing a close analogy between these two phenomena that simply is not justified.

They're the same. Same effect, different (but equivalent) causes.

No, that's not what I said. I said that the lower end of the rope, once it drops below the horizon, would have to move faster than light to *keep up* with the portion of the rope that's above the horizon. That's no different than saying that, in flat spacetime, an object that is beyond the Rindler horizon of a uniformly accelerating observer would have to move faster than light to keep up with that observer. It's simply a consequence of the way the light cones are laid out in the spacetime. You simply look at where the two objects are, and ask, "Would object A have to move outside the light cone in order to keep up with object B?" If the answer is yes, then object A can't keep up. So the object below the horizon has not "broken the light barrier"; quite the opposite. It's precisely the fact that it *can't* break the light barrier that prevents it from keeping up with objects that are hovering above the horizon.

I know what you meant. You explained it very well but that description has the problem of there being an infinite (well c anyway) amount of difference between something fractionally outside the horizon and something fractionally inside it. The Rindler horizon in flat space time is the equivalent to an object in free-fall that can't be caught by an object (as long as it stays in free-fall) if it's a certain distance away - the Rindler horizon. An object doesn't have to cross the event horizon for this to happen. The event horizon is always the value of c more curved at the horizon than it is away from it. That's why the energy needed to escape at the horizon suddenly goes from finite to infinite. Any transition would have to be smooth. I know that it represents moving at c, so you could argue that it's a smooth transition but the fact remains that if it was a fixed event horizon you would need an finite amount of energy to escape from just outside the horizon and it would suddenly go up to infinity one Planc length later. How can you be in a position because one force of a certain strength has put you there, yet a different force can't move you away no matter how strong it is? You would need to travel faster than c to escape, which means gravity has effectively accelerated you past c. You should always be able to escape with a velocity of less than c because c is unreachable. It represents a right angle, and I don't think you can use curvature due to energy or gravity to get to that point, or a combination of both.

As I said above, acceleration of a particular worldline (for example, by firing a rocket engine, or standing on the surface of a planet) is simply a different thing from curvature of spacetime itself due to the presence of matter (more precisely, to the presence of stress-energy, which includes all the things I listed above that can cause spacetime curvature). An object in free fall is *not* accelerating (this is another confusing terminology you keep insisting on using, to use the word "acceleration" to refer to an object that is in free fall--avoiding this sort of confusion is a big reason why the word "acceleration" is *not* used in standard GR to refer to what is called "acceleration due to gravity" in Newtonian physics). Since the object is in free fall, there's no question of being able to "accelerate hard enough to reach c"; the object isn't accelerating at all. It's just following the straightest possible path in the spacetime it's in. If spacetime is flat, that path (worldline) will be a "straight line". If the spacetime is curved, the worldline of the object will be a geodesic of that curved spacetime--the analogue of a "straight line" in a curved manifold (where there are no "straight lines" in the strict Euclidean sense). If following such a geodesic worldline takes the object into portions of the spacetime where the light cones are tilted differently enough, then it may "look like" the object *is* moving "faster than light" if you insist on using coordinates that don't take proper account of the different tilting of the light cones. But that's a problem with your coordinates, not with the physics. The physics is simply that the free-falling object goes wherever the geodesic worldline takes it, just as it does in flat spacetime; it requires no more "energy" or "effort" than it does for an object in flat spacetime to remain in the same inertial frame for all time.

My definition of acceleration due to gravity is tidal force, not free-fall. I agree that being in free-fall is the same as being at rest. I also think there's an upper relative limit that no object can reach and it's the equivalent to c, and it's called an event horizon. I don't think you can gravity to break the light barrier either, if that's what you meant.

I'll take option 2.

I thought you might. Don't worry, I was prepared for that eventuality.

They're not equivalent, once we clarify the terminology as I did above. What you have been calling "curvature due to energy" is *not* the same thing as what *I* have been calling "curvature due to energy". What you have been calling "curvature due to energy" is *not* equivalent to what you have been calling "gravitational curvature". I wasn't able to make that clear before because I didn't understand your non-standard use of the expression "curvature due to energy".

In what way are they not equivalent?

Because they're causing different kinds of "curvature", as I explained above--one of which should not even be described by the term "curvature".

Apart from the fact that one curves outward (pushes you away from the source) and the other curves inward (pulls you towards the source), and their strength, what's the difference?

Except that your "hand-wavy" way is based on a false analogy and confusing terminology that obfuscates crucial distinctions.

My point is that I don't see why those distinctions actually exist.

Now you're using the term "curvature from energy" in yet *another* different way, which I don't understand, but which doesn't appear to me to be compatible with *either* of the uses of the term I described above.

No, same way. Energy curves your path through space-time. Matter curves your path through space-time. You can be at rest or you can accelerate. Whether the acceleration is coming from energy or from the tidal force of free-falling into a higher gravitational field doesn't make any difference. The effect is the same. Can accelerate in the opposite direction to the pull of gravity to cancel it out as we do on Earth. You can always match the acceleration from one with the other. If there's no such thing as absolute velocity then there's no such thing as absolute gravity. The fact that we're pulled down towards the ground is because of the difference in the strength of gravity between our heads and our feet. That's why it's more comfortable to lay down. There's no absolute strength of gravity which is what would be needed to cross an event horizon.

No, they don't. There is no event horizon in any of the cosmological models I referred to, so they can't describe a black hole.

I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe. We're inside a big black hole from this way of looking at it. And guess what; you can't reach the horizon.

The spatial size of the horizon (i.e., the intersection of the horizon with a slice of constant time) is not finite because of "length contraction". It's finite because it's at a finite radius, r = 2M. But the horizon extends through an infinite range of "time" coordinates, so its behavior in time is *not* "equivalent" to its behavior in space.

Why? What makes it different? And what does r=2M actually mean? Two what? And why two? I didn't mean it's finite because of length contraction.

I don't understand where you're getting this from. The underlying geometry--either of the Earth's surface, or of a black hole spacetime--is the same *regardless* of what coordinate system you use to map it. Different coordinate systems are useful for different purposes, but they all describe the same underlying geometry, and they all give the same answers for all physical predictions within the portion of the geometry that they cover. You certainly can't "manipulate" the answers just by changing coordinates; if you get a different answer for a physical prediction in one coordinate system vs. another, then you've done something wrong.

So why is it okay to switch between Rindler coordinates where objects can cross the horizon and Schwarzschild coordinates where nothing ever can? Just use Schwarzschild coordinates. Rindler ones don't take into account the fact that gravity is relative. You're at rest when in free-fall so you'll never even be able to start approaching the point when no amount of energy would be enough to escape. And it's messy precisely because you have to use multiple coordinate systems to describe it properly, and even then it contradicts itself.

 

[Dmitry67]

So why is it okay to switch between Rindler coordinates where objects can cross the horizon and Schwarzschild coordinates where nothing ever can?

Here is a toy model for you (attached)
Red bug is moving conter-clockwise on a circle from the south pole to the north pole. It emits light every seconds.

I also have a second coordinate system - a projection from a center of a circle to the black line. There is a second, black bug moving at that line to the right. Somehow, their communication is posisble only via that projection line (it is a toy model).

For the black bug moving along the horisontal line, red bug becomes more and more 'dilated' in time: in gets the second splash at 1.2 seconds, thrird - at about 2 seconds, etc.

Now black bug claims that red bug would NEVER reaches the north pole, because for it (black bug) it takes forever to go to the end of the line, and still red bug will cross only a half of a distance.

 

[PeterDonis]

Yes, that's intentional. I'm trying to make a point.

If the point you are trying to make is that there somehow *is* a useful analogy between what you are calling "curvature due to acceleration" and what you are calling "curvature due to gravity", then your point is simply not valid. If you are trying to make some other point, please clarify what it is.

Yes we do have to expend energy to remain stationary on the surface. It's called our sodding weight!

Objects stationary on the surface of a planet, like the Earth, don't have to expend any energy just to remain stationary and have weight. Yes, the Earth exerts a force on the object, but the force is a static force (because the object is stationary) and no work is done, so no energy is expended. This isn't even particular to relativity; the analysis works the same way in simple Newtonian physics.

Perhaps you are referring to the fact that you and I, being bipedal living organisms, do have to "expend energy" to remain standing. But that's just because our bodies are unstable in a standing position. You can lie down and expend no energy to remain stationary. Or take the simplest possible example: a rock. It expends no energy--no chemical reactions take place in the rock, no nuclear reactions, no fuel being burned, no rocket engine being fired--yet it remains stationary on the Earth's surface just fine.

Or make it even simpler, since the Earth is an "active" planetary body with magnetic fields inside it, molten magma beneath the surface, tectonic plates moving about, etc. Consider a rock on the surface of the Moon. The Moon itself is just a big spherical rock, and is as static and non-reactive as the rock I described above. Yet a rock can sit on the Moon's surface, and have weight, when neither the rock or the Moon are expending any energy.

If it seems like I'm belaboring this point, it's because I am surprised that I need to make it at all. The fact that static forces do no work and expend no energy is very, very basic physics.

They're the same. Same effect, different (but equivalent) causes.

In standard GR, neither the causes nor the effects are equivalent. If you want to claim they are, you are going to have to give a detailed argument for why you think so. Just saying that it seems obvious to you won't do.

I know what you meant. You explained it very well but that description has the problem of there being an infinite (well c anyway) amount of difference between something fractionally outside the horizon and something fractionally inside it.

No, this is wrong. There's only a small amount of difference, because for an object that's just a little bit outside the horizon, the object has to move radially outward at a speed just a little bit less than c to avoid falling below the horizon. Any lesser speed, even if it's directly radially outward, will result in the object falling below the horizon.

(I've snipped the rest of your comments about the Rindler horizon because I address your misconception about that further below.)

My definition of acceleration due to gravity is tidal force, not free-fall. I agree that being in free-fall is the same as being at rest.

Fine so far.

I also think there's an upper relative limit that no object can reach and it's the equivalent to c, and it's called an event horizon.

If you mean by this that no objects can move outside the light cones, you are correct. But if you mean that there is a limit to how much light cones at a given event can be "tilted" relative to light cones somewhere else, that's wrong.

I don't think you can gravity to break the light barrier either, if that's what you meant.

It wasn't.

In what way are they not equivalent

...

Apart from the fact that one curves outward (pushes you away from the source) and the other curves inward (pulls you towards the source), and their strength, what's the difference?

...

My point is that I don't see why those distinctions actually exist.

Curvature of spacetime itself determines which paths in spacetime are geodesics (freely falling worldlines) and which are not. Being a geodesic or not a geodesic is an intrinsic property of a path in a particular spacetime geometry. The geometry itself depends on global specifications of the quantities that feed into the Einstein Field Equation: what (if any) matter or energy is present (globally), and what the boundary conditions are (for example, a black hole spacetime is asymptotically flat).

"Curvature due to acceleration" is just a way of saying the particular path you are following is not a geodesic. Specifying which path you are following is a property of the *object*; you figure it out by figuring out the object's physical situation and what it is doing (is it firing rockets, resting on the surface of a planet, etc. as opposed to floating freely in space).

These are two distinct concepts because you can "mix and match" them freely in any combination. You can have a geodesic path in flat spacetime, and an object moving on it (e.g., an inertial observer in Minkowski spacetime); you can have a non-geodesic path in a flat spacetime, and an object moving on it (e.g., a "Rindler observer" uniformly accelerating in Minkowski spacetime); you can have a geodesic path in curved spacetime, and an object moving on it (e.g., an observer freely falling into a black hole), and you can have a non-geodesic path in curved spacetime, and an object moving on it (e.g., an observer firing rockets to hover at a constant radius r > 2M above a black hole's horizon). So knowing just one thing doesn't impose any restrictions on the other thing.

No, same way. Energy curves your path through space-time. Matter curves your path through space-time. You can be at rest or you can accelerate. Whether the acceleration is coming from energy or from the tidal force of free-falling into a higher gravitational field doesn't make any difference. The effect is the same. Can accelerate in the opposite direction to the pull of gravity to cancel it out as we do on Earth. You can always match the acceleration from one with the other.

No, you can't. If you are below the horizon of a black hole, no amount of acceleration will prevent you from continuing to fall inward. It is true that *outside* the horizon, there is always some finite acceleration that will just "cancel" the inward "acceleration" due to the presence of the hole, but there is no requirement that that must be true throughout the spacetime. (And no, the fact that it seems "obvious" to you that you should always be able to "cancel gravity with acceleration" is *not* a proof that that *must* be true. If you think you can concoct an actual proof, please post one. But you can't just *assume* that it's true.)

The fact that we're pulled down towards the ground is because of the difference in the strength of gravity between our heads and our feet. That's why it's more comfortable to lay down.

Tidal gravity is *not* the same as the apparent "force of gravity" that pulls us towards the center of the Earth. Again, this isn't particular to relativity; the same is true in Newtonian physics. Tidal gravity is the *rate of change* of the apparent "force of gravity" (note that in relativity, the rate of change is in *spacetime*, not just space; there can be tidal gravity in the "time direction" as well as in the "space directions").

The case you give, of standing vs. lying down, actually illustrates the difference. Yes, the tidal gravity on your body is less when you're lying down, but you still weigh the same. The reason it's more comfortable to lie down is that your body is unstable standing up, and your muscles have to continually compensate to keep you upright, which tires you out.

I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe. We're inside a big black hole from this way of looking at it. And guess what; you can't reach the horizon.

You are misunderstanding the standard cosmological model of the Big Bang; it does *not* say we are inside a big black hole. (And the universe does not have an "edge"; it is either finite but unbounded, like the surface of the Earth--that's if the universe is closed--or it is infinite in extent and has no edge, if it is open. The current "best fit" model has the universe being infinite, but the error bars are large enough that there is still an outside chance we will end up judging that it is closed when we have more data.)

It is true that, if you pick different worldlines coming out of the Big Bang, and go back in time close enough to the Big Bang (how close depends on how far apart the worldlines are), you will come to a point where there had not been enough time since the Big Bang for light to travel between the two worldlines. The term "horizon" is sometimes used to refer to this phenomenon, but it's not the same as a black hole event horizon, because as time goes on, more and more worldlines "come into contact" with each other, meaning there has been enough time for light to travel between them. A black hole's event horizon remains the same through time; the region inside can *never* send light signals to the region outside, while the region outside can send signals to the region inside at any time.

what does r=2M actually mean? Two what? And why two?

r = 2M is just the standard expression for the radius of a black hole's horizon in terms of its mass, where the mass is expressed in "geometric units"--i.e., mass and length have the same units. In conventional units, the expression is [itex] r = 2 G M / c^{2} [/itex], where G is Newton's gravitational constant and c is the speed of light. M is then the mass of the hole in conventional units.

I didn't mean it's finite because of length contraction.

You were trying to argue (as I understood you) that the black hole's "size in time" should be the same as its "size in space" because time dilation and length contraction are equivalent. This argument only makes sense if the hole's finite "size in space" is due to length contraction. Since it isn't, the argument fails; time dilation and length contraction are irrelevant.

So why is it okay to switch between Rindler coordinates where objects can cross the horizon and Schwarzschild coordinates where nothing ever can? Just use Schwarzschild coordinates.

Objects can't cross the Rindler horizon in Rindler coordinates. Those coordinates only cover the region "above" the horizon (where all the accelerating observers are) in flat spacetime. To "see" objects crossing the horizon in flat spacetime, you have to use Minkowski coordinates, since those cover the entire spacetime. So, can objects cross the Rindler horizon in flat spacetime? Of course they can. It's just that Rindler coordinates aren't good coordinates to analyze that case.

Similarly, in a black hole spacetime, to "see" objects crossing the horizon, you have to use, for example, Kruskal coordinates, which cover the entire spacetime in a way similar to the way Minkowski coordinates cover the entire spacetime when spacetime is flat. So once again, can objects cross the horizon in a black hole spacetime? Of course they can. It's just that Schwarzschild coordinates (which are analogous to Rindler coordinates) aren't good coordinates to analyze that case.

 

[PeterDonis]

No, this is wrong. There's only a small amount of difference, because for an object that's just a little bit outside the horizon, the object has to move radially outward at a speed just a little bit less than c to avoid falling below the horizon. Any lesser speed, even if it's directly radially outward, will result in the object falling below the horizon.

I realized after re-reading this that I should clarify: the above statement applies to objects that are moving inertially (i.e., they may have a very large velocity, approaching c, but their velocity is constant; they are not accelerating, in the invariant, coordinate-independent sense of "accelerating"). Such an object, just outside the horizon, must have an outward radial velocity just short of c to avoid falling through the horizon.

An *accelerating* object just outside the horizon can be stationary (i.e., can "hover" at a constant radial coordinate), as long as its acceleration is large enough--the required acceleration increases without bound ("approaches infinity") as the horizon is approached. However, as I noted elsewhere in my previous post, there is no requirement that one must be able to "hover" stationary at a finite acceleration everywhere in a black hole spacetime. It just happens to be the case that you can do so anywhere outside a black hole's horizon.

 

[A-wal]

Here is a toy model for you (attached)
Red bug is moving conter-clockwise on a circle from the south pole to the north pole. It emits light every seconds.

I also have a second coordinate system - a projection from a center of a circle to the black line. There is a second, black bug moving at that line to the right. Somehow, their communication is posisble only via that projection line (it is a toy model).

For the black bug moving along the horisontal line, red bug becomes more and more 'dilated' in time: in gets the second splash at 1.2 seconds, thrird - at about 2 seconds, etc.

Now black bug claims that red bug would NEVER reaches the north pole, because for it (black bug) it takes forever to go to the end of the line, and still red bug will cross only a half of a distance.

Isn't that describing a Rindler horizon rather than an event horizon?

 

[A-wal]

If the point you are trying to make is that there somehow *is* a useful analogy between what you are calling "curvature due to acceleration" and what you are calling "curvature due to gravity", then your point is simply not valid. If you are trying to make some other point, please clarify what it is.

No, I'm trying to make the point that you consider invalid.

Objects stationary on the surface of a planet, like the Earth, don't have to expend any energy just to remain stationary and have weight. Yes, the Earth exerts a force on the object, but the force is a static force (because the object is stationary) and no work is done, so no energy is expended. This isn't even particular to relativity; the analysis works the same way in simple Newtonian physics.

If the Earth exerts a force on the object then obviously the object exerts a force on the Earth. The energy we need to remain stationary is felt as our weight. Two opposing forces cancelling each other out doesn't mean there's no force. We only feel the acceleration, not the gravity pulling us the other way because gravity is relative, just like velocity. Is there a test that you can perform, even in principle to show that you're in a gravitational field? Is there a test you can perform to show that there's a moving time-line? Is there a test to see if you're moving at all? It's all the same thing.

Perhaps you are referring to the fact that you and I, being bipedal living organisms, do have to "expend energy" to remain standing. But that's just because our bodies are unstable in a standing position. You can lie down and expend no energy to remain stationary. Or take the simplest possible example: a rock. It expends no energy--no chemical reactions take place in the rock, no nuclear reactions, no fuel being burned, no rocket engine being fired--yet it remains stationary on the Earth's surface just fine.

You weigh the same whether you’re standing or not. The "expended energy" needed stays the same. I don't know what you mean by a static force, other than that we remain stationary relative to the Earth, and I really don't see how that's relevant. The only reason I can see for it to be considered what you call a static force is because gravity and electromagnetism are balanced. Or to put it another way the outward curvature from your acceleration matches the inward curvature from gravity.

Or make it even simpler, since the Earth is an "active" planetary body with magnetic fields inside it, molten magma beneath the surface, tectonic plates moving about, etc. Consider a rock on the surface of the Moon. The Moon itself is just a big spherical rock, and is as static and non-reactive as the rock I described above. Yet a rock can sit on the Moon's surface, and have weight, when neither the rock or the Moon are expending any energy.

Yes they are, through their weight, in the same way that our bodies and everything else on the planet needs to expend energy to stop it from being crushed.

If it seems like I'm belaboring this point, it's because I am surprised that I need to make it at all. The fact that static forces do no work and expend no energy is very, very basic physics.

Ah, that explains it. I know next to nothing about very, very basic physics. I've never even heard of a static force before, other than static electricity.

In standard GR, neither the causes nor the effects are equivalent. If you want to claim they are, you are going to have to give a detailed argument for why you think so. Just saying that it seems obvious to you won't do.

What the hell do you think I'm trying to do? Space-time can be curved in one of two ways. Energy can accelerate an object away from its source which causes time dilation and length contraction because everything has to move through space-time at c and c has to remain constant. You can use tidal force to accelerate you towards an object, but it's a much weaker force because it comes from matter rather than energy and E=mc^2. To me this is very, very basic physics. If you want to claim there's some fundamental difference between these two processes to make one behave differently from the other one then I think the onus should be on you to show that there is.

No, this is wrong. There's only a small amount of difference, because for an object that's just a little bit outside the horizon, the object has to move radially outward at a speed just a little bit less than c to avoid falling below the horizon. Any lesser speed, even if it's directly radially outward, will result in the object falling below the horizon.

What did I say just after that? The energy required to escape from the black hole jumps from finite to infinite in an instant. Surely that doesn't make sense to you?

I realized after re-reading this that I should clarify: the above statement applies to objects that are moving inertially (i.e., they may have a very large velocity, approaching c, but their velocity is constant; they are not accelerating, in the invariant, coordinate-independent sense of "accelerating"). Such an object, just outside the horizon, must have an outward radial velocity just short of c to avoid falling through the horizon.

An *accelerating* object just outside the horizon can be stationary (i.e., can "hover" at a constant radial coordinate), as long as its acceleration is large enough--the required acceleration increases without bound ("approaches infinity") as the horizon is approached. However, as I noted elsewhere in my previous post, there is no requirement that one must be able to "hover" stationary at a finite acceleration everywhere in a black hole spacetime. It just happens to be the case that you can do so anywhere outside a black hole's horizon.

Nothing just happens. You can always hover at a finite acceleration because nothing can produce infinite force. If you need infinite energy to escape from within the horizon then the black hole has infinite gravitational force past the horizon.

(I've snipped the rest of your comments about the Rindler horizon because I address your misconception about that further below.)

What misconception? If I'm understanding the concept correctly (I don't entirely trust my interpretation of the Rindler horizon because I had to look it up rather than figure it out for myself) then a free-falling object has crossed the Rindler horizon when it can't ever be caught by a specific more distant accelerating object, whereas a free-falling object has crossed the event horizon when it can't ever be caught by any more distant accelerating object. It would mean gravity accelerates the closer object faster than any amount of energy could accelerate the further object, no matter how close it is. That's very clever. It's so clever that I don't think it's exaggerating to call it magical.

Fine so far.

So you agree that tidal force is equivalent to acceleration in flat space-time? So why can tidal force accelerate you to the equivalent of c and beyond when energy can only accelerate you to other relative velocities?

If you mean by this that no objects can move outside the light cones, you are correct. But if you mean that there is a limit to how much light cones at a given event can be "tilted" relative to light cones somewhere else, that's wrong.

Is it? Any particular reason or is it just wrong? Why is it okay for gravity to tilt the light cone past 45 degrees but not energy? What makes gravity so special? If they can tilt past 45 degrees then would they be moving backwards through time?

It wasn't.

Okay, but that's basically what you said whether you meant it or not. You said that in order to escape from the event horizon you would have to move faster than c, which means that gravity has effectively moved you past c to get there in the first place.

Curvature of spacetime itself determines which paths in spacetime are geodesics (freely falling worldlines) and which are not. Being a geodesic or not a geodesic is an intrinsic property of a path in a particular spacetime geometry. The geometry itself depends on global specifications of the quantities that feed into the Einstein Field Equation: what (if any) matter or energy is present (globally), and what the boundary conditions are (for example, a black hole spacetime is asymptotically flat).

The only practical difference I can see other than their direction and strength is that energy doesn't tend to effect an object for long. It's there and then it's gone. Matter lasts long enough for us to map its effect.

"Curvature due to acceleration" is just a way of saying the particular path you are following is not a geodesic. Specifying which path you are following is a property of the *object*; you figure it out by figuring out the object's physical situation and what it is doing (is it firing rockets, resting on the surface of a planet, etc. as opposed to floating freely in space).

And gravity is also just a way of saying the particular path you are following is not a geodesic from a distance.

These are two distinct concepts because you can "mix and match" them freely in any combination. You can have a geodesic path in flat spacetime, and an object moving on it (e.g., an inertial observer in Minkowski spacetime); you can have a non-geodesic path in a flat spacetime, and an object moving on it (e.g., a "Rindler observer" uniformly accelerating in Minkowski spacetime); you can have a geodesic path in curved spacetime, and an object moving on it (e.g., an observer freely falling into a black hole), and you can have a non-geodesic path in curved spacetime, and an object moving on it (e.g., an observer firing rockets to hover at a constant radius r > 2M above a black hole's horizon). So knowing just one thing doesn't impose any restrictions on the other thing.

The fact that you can “mix and match” them doesn't prove they are distinct concepts. If anything the opposite is true.

No, you can't. If you are below the horizon of a black hole, no amount of acceleration will prevent you from continuing to fall inward. It is true that *outside* the horizon, there is always some finite acceleration that will just "cancel" the inward "acceleration" due to the presence of the hole, but there is no requirement that that must be true throughout the spacetime.

I think it is a requirement for the universe to make sense.

(And no, the fact that it seems "obvious" to you that you should always be able to "cancel gravity with acceleration" is *not* a proof that that *must* be true. If you think you can concoct an actual proof, please post one. But you can't just *assume* that it's true.)

Bugger!

Tidal gravity is *not* the same as the apparent "force of gravity" that pulls us towards the center of the Earth. Again, this isn't particular to relativity; the same is true in Newtonian physics. Tidal gravity is the *rate of change* of the apparent "force of gravity" (note that in relativity, the rate of change is in *spacetime*, not just space; there can be tidal gravity in the "time direction" as well as in the "space directions").

Tidal gravity as the "rate of change of the apparent force of gravity" is exactly what I've been saying forever.

The case you give, of standing vs. lying down, actually illustrates the difference. Yes, the tidal gravity on your body is less when you're lying down, but you still weigh the same. The reason it's more comfortable to lie down is that your body is unstable standing up, and your muscles have to continually compensate to keep you upright, which tires you out.

1). It was a joke. 2). There is a difference in tidal force as you said, but it's so marginal that you could never feel it. That was the joke. 3). I know you weigh the same standing up as laying down, and the reason it's more comfortable is because your weight's spread out. The question is why laying down spreads out your weight. There's nothing to determine the direction that gravity pulls you other than the direction in which it's strongest. In other words the difference between the gravity at your head and your feet. In other other words, tidal force. The reason your more comfortable laying down is because all the weight goes to your feet when you’re at a lengthwase angle to the direction of the tidal force. So in fact you are more comfortable when laying down because the gap between the top and bottom parts of your body are at their lowest.

You are misunderstanding the standard cosmological model of the Big Bang; it does *not* say we are inside a big black hole. (And the universe does not have an "edge"; it is either finite but unbounded, like the surface of the Earth--that's if the universe is closed--or it is infinite in extent and has no edge, if it is open. The current "best fit" model has the universe being infinite, but the error bars are large enough that there is still an outside chance we will end up judging that it is closed when we have more data.)

Why do you always assume I've misunderstood the principles before you assume that you misunderstood what I meant? I know it doesn't have that kind of an edge. How would that even work? I didn't actually say we were inside a black hole, although you could definitely look at it in that way. If it's closed then I don't think it's wrong as far as the standard model goes, just the same thing from the inside, accept you can’t even start to approach the horizon. And I'd say there's more than an outside chance that it's closed.

It is true that, if you pick different worldlines coming out of the Big Bang, and go back in time close enough to the Big Bang (how close depends on how far apart the worldlines are), you will come to a point where there had not been enough time since the Big Bang for light to travel between the two worldlines. The term "horizon" is sometimes used to refer to this phenomenon, but it's not the same as a black hole event horizon, because as time goes on, more and more worldlines "come into contact" with each other, meaning there has been enough time for light to travel between them. A black hole's event horizon remains the same through time; the region inside can *never* send light signals to the region outside, while the region outside can send signals to the region inside at any time.

That one-way road situation is one of the many reasons why it doesn't make sense to have a fixed crossable horizon. And if there wasn't enough time for light to travel between the two world-lines then how could there have been enough time for the two world lines to get that far apart?

r = 2M is just the standard expression for the radius of a black hole's horizon in terms of its mass, where the mass is expressed in "geometric units"--i.e., mass and length have the same units. In conventional units, the expression is [itex] r = 2 G M / c^{2} [/itex], where G is Newton's gravitational constant and c is the speed of light. M is then the mass of the hole in conventional units.

But not its length in time? What makes time so special?

You were trying to argue (as I understood you) that the black hole's "size in time" should be the same as its "size in space" because time dilation and length contraction are equivalent. This argument only makes sense if the hole's finite "size in space" is due to length contraction. Since it isn't, the argument fails; time dilation and length contraction are irrelevant.

I don't think its finite volume is due to length contraction. I think the whole thing is length contraction and time dilation, caused by the singularity. If you can't reach the black hole then what actually is it? It's just a four-dimensional bubble of nothingness that nothing can ever reach, because there's always too much space and time between it and you, just like the edge of the universe (it still has an edge, just not a fixed one).

Objects can't cross the Rindler horizon in Rindler coordinates. Those coordinates only cover the region "above" the horizon (where all the accelerating observers are) in flat spacetime. To "see" objects crossing the horizon in flat spacetime, you have to use Minkowski coordinates, since those cover the entire spacetime. So, can objects cross the Rindler horizon in flat spacetime? Of course they can. It's just that Rindler coordinates aren't good coordinates to analyze that case.

I thought objects could cross the event horizon using Rindler coordinates. It doesn't matter. The point was that you need to use two different points of reference to describe the same thing fully because one shows something the other doesn't, which in this case, is just another way of saying they contradict each other. Using that logic, whether or not you fall into a black hole is dependent on what coordinate system you choose to measure your decent. Why is it so acceptable to switch between two entirely views when one shows something that's automatically assumed wrong? It's very easy to say "of course they can" but to my mind it's obvious that they can't.

Similarly, in a black hole spacetime, to "see" objects crossing the horizon, you have to use, for example, Kruskal coordinates, which cover the entire spacetime in a way similar to the way Minkowski coordinates cover the entire spacetime when spacetime is flat. So once again, can objects cross the horizon in a black hole spacetime? Of course they can. It's just that Schwarzschild coordinates (which are analogous to Rindler coordinates) aren't good coordinates to analyze that case.

Schwarzschild coordinates cover the entire space-time external to the black hole. It's just that time and space are compressed to c/infinity at the horizon. It's different than the way it happens on a standard graph showing acceleration up to c. What that doesn't show is that it would look different if you were the one accelerating, just like it would look different if you were the one falling and accelerating through tidal force. Length contraction means you have to travel further and further, and time dilation means you have less and less time to do it the closer you get to the horizon. You think that you can use a different type of acceleration to outrun this process and reach the horizon but they're equivalent. You could use standard acceleration through energy to move towards it faster but than your left with exactly the same problem. If you accelerate in the opposite direction you can cancel it out or overcome it. There'll never be a point when one take somewhere the other can't take you away from. Gravity's just as relative as velocity, but you're trying to use it to break the light barrier. Not a chance in hell son.

 

[PeterDonis]

If the Earth exerts a force on the object then obviously the object exerts a force on the Earth.

Yes.

The energy we need to remain stationary is felt as our weight.

No, weight is a force, not energy. You feel the force of the Earth pushing on you. No energy is expended (see below).

Two opposing forces cancelling each other out doesn't mean there's no force.

True.

We only feel the acceleration, not the gravity pulling us the other way...

Not quite. We feel the *force* of the Earth pushing on us. The Earth, similarly, "feels" the force of us pushing on it. You're correct that there is no need to describe either of these forces as "gravity". They're not. However...

...because gravity is relative, just like velocity.

This is not quite right either, at least not in the way you use the term "gravity" (because you say tidal gravity and "acceleration due to gravity" are the same thing, which they're not). There is a sense in which "gravity is relative", meaning the "acceleration due to gravity" is relative, because you can always find a freely falling reference frame in which, locally, that acceleration disappears. But *tidal* gravity is *not* relative; it is a frame-invariant, genuine physical quantity.

Is there a test that you can perform, even in principle to show that you're in a gravitational field?

There is no *local* test you can perform to show that you're "being accelerated by gravity" if you are in free fall (i.e., not feeling any force). (Obviously you can use non-local information, such as the fact that you are above a large planetary body like the Earth, to infer that that body's mass is causing you to fall towards it. But that's not a local test; it requires information about distant objects.) But there *are* tests you can perform to show whether *tidal* gravity is present or absent between you and neighboring objects, even if you and all those other objects are all in free fall.

You weigh the same whether you’re standing or not. The "expended energy" needed stays the same.

Only in the sense that the "expended energy" is zero in both cases.

I don't know what you mean by a static force, other than that we remain stationary relative to the Earth, and I really don't see how that's relevant.

You have the correct definition of static force: it's a force which does not result in any relative motion of either body. It's relevant because at least one body has to move for any work to be done and any energy to be expended. For example, if you stand up from a prone position, you *do* expend energy, because you have to do work to raise your body's mass from a lower to a higher position (i.e., you move your body further away from the Earth--more precisely, your body's center of mass is further from the Earth's center of mass). Or if you are far out in space in a rocket and you fire the rocket's engine, the rocket does work by expelling exhaust out the back--the exhaust and the rocket are moving relative to each other, so work is done and energy is expended. There has to be relative motion.

The only reason I can see for it to be considered what you call a static force is because gravity and electromagnetism are balanced. Or to put it another way the outward curvature from your acceleration matches the inward curvature from gravity.

There is no "force of gravity" anywhere in the problem. The electromagnetic force of the atoms in your body pushes against the electromagnetic force of the atoms in the ground, and the ground pushes back.

Yes they are, through their weight, in the same way that our bodies and everything else on the planet needs to expend energy to stop it from being crushed.

The Moon and the rock on the Moon are not expending any energy. How can they? Where is it going to come from? There are no chemical reactions going on, no nuclear reactions going on, no fuel being burned. Where does the energy come from? The answer is that it doesn't have to come from anywhere because no energy is being expended. The Moon and the rock are in a state of static equilibrium, with no work being done, and they will stay that way indefinitely, with no work being done and no energy expended. By your logic, the rock and the Moon would eventually "run out" of energy and collapse somehow. They don't.

What the hell do you think I'm trying to do? Space-time can be curved in one of two ways.

No, spacetime can be curved in only one way, by the presence of stress-energy (matter, energy, pressure, and internal stresses of materials--the stuff that goes into the stress-energy tensor).

An object's *path* through spacetime can also be curved, in the frame-invariant sense of "curved", in one way: by an object having a force (again, in the frame-invariant sense of "force") exerted on it, causing it to accelerate (in the frame-invariant sense of "accelerate"). You can tell whether an object's path is curved in this sense by seeing whether it feels weight.

A path can "look" curved without being curved (in the frame-invariant sense), or be curved (in the frame-invariant sense) without "looking" curved, if you adopt a coordinate system that obscures the frame-invariant curvature. For example, in Schwarzschild exterior coordinates, the paths of objects "hovering" at constant radius above a black hole's horizon look straight, but they are curved in the frame-invariant sense. (The same is true of your path through spacetime when you're standing motionless on the Earth's surface.)

If you want to claim there's some fundamental difference between these two processes to make one behave differently from the other one then I think the onus should be on you to show that there is.

I've done this. I've explained how:

(a) Curvature of *spacetime* is different and distinct from curvature of an object's *path* through spacetime.

(b) Tidal gravity is different and distinct from "acceleration due to gravity".

What did I say just after that? The energy required to escape from the black hole jumps from finite to infinite in an instant. Surely that doesn't make sense to you?

There is no jump. The outward velocity required to escape (for an object that's in free fall, not accelerated) increases smoothly to c as the horizon is approached and reached, and the energy required to escape (again, for an object that's in free fall, not accelerated) increases smoothly to infinity. This works the same as an object's energy going to infinity as its speed approaches c (in a given reference frame) in flat spacetime. There's no jump.

Edit: I should clarify that by "works the same" I mean only that, mathematically, the limiting process works the same. I do *not* mean that the horizon can't be reached and passed. Let me know if I need to elaborate on this.

If you need infinite energy to escape from within the horizon then the black hole has infinite gravitational force past the horizon.

There is no "force" anywhere. It's just geometry. See below.

a free-falling object has crossed the Rindler horizon when it can't ever be caught by a specific more distant accelerating object, whereas a free-falling object has crossed the event horizon when it can't ever be caught by any more distant accelerating object.

No. You keep on getting this backwards. The Rindler horizon is the path of a light ray that just fails to catch up to a family of uniformly accelerating observers outside (or "above") it. Similarly, a black hole's horizon is the path of a light ray that just fails to catch up to a family of "hovering" observers outside it. Neither horizon has anything to do with whether an object outside the horizon can catch up with an object that free-falls *into* the horizon. That's a separate question.

So you agree that tidal force is equivalent to acceleration in flat space-time?

No. I apologize for a misstatement there; I should have snipped the quote from your previous post more carefully. The only statement of yours I meant to agree with was the statement that being in free fall is equivalent to being at rest. I did not mean to imply agreement with your statement about tidal force.

Why is it okay for gravity to tilt the light cone past 45 degrees but not energy? What makes gravity so special?

"Energy" in the sense you are using the term (meaning energy expended by an object to accelerate itself so it feels weight) has no effect on the light cones. It can't tilt them at all. Also, "gravity" is not a separate "thing" that somehow tilts the light-cones. Gravity *is* the tilting of the light cones. They're the same thing (spacetime curvature).

If they can tilt past 45 degrees then would they be moving backwards through time?

No. The tilt is not "past 45 degrees". The *ingoing* sides of the light cones remain pointed at 45 degrees inward. The *outgoing* sides gradually point more and more *upward* (i.e., *less* than 45 degrees) until they become vertical (0 degrees) at the horizon. Inside the horizon, they tilt over the other way, which simply means objects can no longer avoid moving inward, even if they move at the speed of light. It does not require going backward in time.

Okay, but that's basically what you said whether you meant it or not. You said that in order to escape from the event horizon you would have to move faster than c, which means that gravity has effectively moved you past c to get there in the first place.

No, that doesn't follow, because in order to escape, you have to move *outward* faster than light. You do not have to move *inward* faster than light to get inside the horizon. You can free-fall there, without "moving" at all (you yourself said being in free fall is the same as being at rest, so all you have to do to get inside the hole is to stay at rest).

And gravity is also just a way of saying the particular path you are following is not a geodesic from a distance.

No, "gravity" is a way of saying that *spacetime* is curved. It does not say anything about particular *paths* in that spacetime. You can follow a geodesic path (freely falling path, weightless path) in a curved spacetime, where gravity is present. They are separate, distinct concepts.

I think it is a requirement for the universe to make sense.

To make sense, yes. To make sense in a way that satisfies all your intuitions, no.

Tidal gravity as the "rate of change of the apparent force of gravity" is exactly what I've been saying forever.

And yet you keep trying to claim that tidal force is the *same* as "acceleration due to gravity" or "apparent force of gravity". How can something be the same as its rate of change?

I know you weigh the same standing up as laying down, and the reason it's more comfortable is because your weight's spread out. The question is why laying down spreads out your weight.

Um, because a larger area of you is in contact with the ground?

There's nothing to determine the direction that gravity pulls you other than the direction in which it's strongest. In other words the difference between the gravity at your head and your feet. In other other words, tidal force.

But there are tidal forces in multiple directions, not just one. There can even be tidal force in the time direction. You can't extract a unique direction for gravity to pull you from tidal force.

The reason your more comfortable laying down is because all the weight goes to your feet when you’re at a lengthwase angle to the direction of the tidal force. So in fact you are more comfortable when laying down because the gap between the top and bottom parts of your body are at their lowest.

No, it's because your weight is spread more evenly over a larger area of your body. See above.

Why do you always assume I've misunderstood the principles before you assume that you misunderstood what I meant?

I wasn't assuming anything. I was responding to the actual words you used. You said: "I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe." You didn't elaborate on what you meant by "edge". How was I supposed to interpret it except in the obvious way?

That one-way road situation is one of the many reasons why it doesn't make sense to have a fixed crossable horizon.

Why not? I know it doesn't make sense in the mental model you have in your head. But can you give a logical argument, starting from premises we all accept, that shows that it *can't* make sense, in *any* consistent model? I don't think you can, but if you have such an argument, let's see it. Just claiming that it doesn't make sense won't do, because I have a consistent model where it does makes sense.

And if there wasn't enough time for light to travel between the two world-lines then how could there have been enough time for the two world lines to get that far apart?

Because the worldlines are expanding in different directions. Ned Wright's Cosmology FAQ, at the link below, briefly explains the kind of thing that's going on.

http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN

But not its length in time? What makes time so special?

I'm not sure what you're asking here. I was just explaining what r = 2M means, since you asked that; that radius is the radius of the horizon. Radius is not a quantity in the time dimension.

If you can't reach the black hole then what actually is it?

Since you can reach the black hole, this question is meaningless.

I thought objects could cross the event horizon using Rindler coordinates.

They can't.

It doesn't matter.

Oh, yes, it does. It's crucial, because it grounds the analogy between Rindler coordinates and Schwarzschild coordinates (as compared with Minkowski coordinates and Kruskal coordinates). See below.

Using that logic, whether or not you fall into a black hole is dependent on what coordinate system you choose to measure your decent.

I have never said that, and in fact I've explicitly said several times that whether or not you fall into the hole does *not* depend on the coordinates you use. What *does* depend on the coordinates is whether particular events on your worldline--such as the real, actual events of you crossing the hole's horizon and moving inward beyond that point--are "visible" in those coordinates. It's no different than saying that something that happens on the surface of the Earth at a point beyond your horizon--where you can't see--isn't visible to you.

In Rindler coordinates, the portion of a free-falling object's worldline that lies at or below the Rindler horizon is not visible, but it *is* visible in Minkowski coordinates. That's all there is to it; one coordinate system can see events that aren't visible in the other. I really don't see why this is so hard to grasp.

Similarly, in Schwarzschild exterior coordinates, the portion of a free-falling object's worldline that lies at or below the black hole horizon is not visible, but it *is* visible in Kruskal coordinates. Again, one coordinate system can see events that aren't visible in the other. What's impossible about that?

Why is it so acceptable to switch between two entirely views when one shows something that's automatically assumed wrong?

The only one who's automatically assuming it's wrong is you. I make no such assumption.

Schwarzschild coordinates cover the entire space-time external to the black hole.

Yes.

It's just that time and space are compressed to c/infinity at the horizon.

No, they're not. Nor do you have to "break the light barrier" to fall into the hole, because you're moving inward, not outward. The fact that you can't comprehend the consistent model that explains all this does not mean that model is false or inconsistent; it just means you can't understand it.

 

[Dmitry67]

Isn't that describing a Rindler horizon rather than an event horizon?

It is a toy model

It just shows the idea that infinities in one coordinate system do not always have any physical meaning. Like infinities in S. coordinates are a problem of that coordinate system, while infalling objects can cross the horizon without any problems.

 

[A-wal]

No, weight is a force, not energy. You feel the force of the Earth pushing on you. No energy is expended (see below).

Not quite. We feel the *force* of the Earth pushing on us. The Earth, similarly, "feels" the force of us pushing on it. You're correct that there is no need to describe either of these forces as "gravity". They're not. However...

Only in the sense that the "expended energy" is zero in both cases.

You have the correct definition of static force: it's a force which does not result in any relative motion of either body. It's relevant because at least one body has to move for any work to be done and any energy to be expended. For example, if you stand up from a prone position, you *do* expend energy, because you have to do work to raise your body's mass from a lower to a higher position (i.e., you move your body further away from the Earth--more precisely, your body's center of mass is further from the Earth's center of mass). Or if you are far out in space in a rocket and you fire the rocket's engine, the rocket does work by expelling exhaust out the back--the exhaust and the rocket are moving relative to each other, so work is done and energy is expended. There has to be relative motion.

Okay then, I should have said force instead of energy.

This is not quite right either, at least not in the way you use the term "gravity" (because you say tidal gravity and "acceleration due to gravity" are the same thing, which they're not). There is a sense in which "gravity is relative", meaning the "acceleration due to gravity" is relative, because you can always find a freely falling reference frame in which, locally, that acceleration disappears. But *tidal* gravity is *not* relative; it is a frame-invariant, genuine physical quantity.

Okay, if "you can always find a freely falling reference frame in which, locally, that acceleration disappears" what about at the event horizon? I didn't say tidal gravity is relative. I said it's the equivalent to acceleration in flat space-time. If I was using relative velocities as a measurement of the strength of gravity then it wouldn't be relative and the same area of space-time would change its properties depending on which observer makes the measurements. One of the reasons I have a problem with standard GR is exactly that. The properties of the same area of space-time shouldn't change depending on which direction you travel through it.

There is no *local* test you can perform to show that you're "being accelerated by gravity" if you are in free fall (i.e., not feeling any force). (Obviously you can use non-local information, such as the fact that you are above a large planetary body like the Earth, to infer that that body's mass is causing you to fall towards it. But that's not a local test; it requires information about distant objects.) But there *are* tests you can perform to show whether *tidal* gravity is present or absent between you and neighboring objects, even if you and all those other objects are all in free fall.

I wonder why. Do you think maybe it's equivalent to acceleration? Tidal force is the only way, and there's always tidal force because it has infinite range. That's acceleration though because it's the difference, or the rate of change as you put it, which is also the definition of acceleration.

There is no "force of gravity" anywhere in the problem. The electromagnetic force of the atoms in your body pushes against the electromagnetic force of the atoms in the ground, and the ground pushes back.

What? But why do they push against each other? Both together are the reaction to the force of gravity. The pull of gravity is being countered by acceleration by the combination of the Earth and our bodies both being solid objects.

The Moon and the rock on the Moon are not expending any energy. How can they? Where is it going to come from? There are no chemical reactions going on, no nuclear reactions going on, no fuel being burned. Where does the energy come from? The answer is that it doesn't have to come from anywhere because no energy is being expended. The Moon and the rock are in a state of static equilibrium, with no work being done, and they will stay that way indefinitely, with no work being done and no energy expended. By your logic, the rock and the Moon would eventually "run out" of energy and collapse somehow. They don't.

If you were to wait long enough then the rock would be crushed under its own weight. Admittedly, I'd have to invent a whole new set of epically large numbers to describe how long that would take, but I'm sure it would happen, eventually. If there's too much gravity for the object to take then it will collapse. If the object is close to the limit then it will collapse after a time. If it's not close then it will collapse after a long time. This is the same principle as gradually reaching infinity. I don't think anything can be strong enough to resist anything forever.

No, spacetime can be curved in only one way, by the presence of stress-energy (matter, energy, pressure, and internal stresses of materials--the stuff that goes into the stress-energy tensor).

You've just listed four. Space-time can be curved BY one of two main causes.

An object's *path* through spacetime can also be curved, in the frame-invariant sense of "curved", in one way: by an object having a force (again, in the frame-invariant sense of "force") exerted on it, causing it to accelerate (in the frame-invariant sense of "accelerate"). You can tell whether an object's path is curved in this sense by seeing whether it feels weight.

You would feel your weight if you were in free-fall in a strong enough gravitational field. Remember, gravity's very weak. It looks stronger than it really is because of the change it causes in relative velocity (yes I do mean it like that this time), but that doesn't count because it's mostly cause by the acceleration of object resisting gravity. Only acceleration into a higher gravitational field counts, but the object's being accelerated from a constantly increasing relative speed, making it look stronger.

A path can "look" curved without being curved (in the frame-invariant sense), or be curved (in the frame-invariant sense) without "looking" curved, if you adopt a coordinate system that obscures the frame-invariant curvature. For example, in Schwarzschild exterior coordinates, the paths of objects "hovering" at constant radius above a black hole's horizon look straight, but they are curved in the frame-invariant sense. (The same is true of your path through spacetime when you're standing motionless on the Earth's surface.)

It's curved any time you feel acceleration. It's the same thing.

I've done this. I've explained how:

(a) Curvature of *spacetime* is different and distinct from curvature of an object's *path* through spacetime.

(b) Tidal gravity is different and distinct from "acceleration due to gravity".

(a) Not really.

(b) It's the “acceleration of gravity”.

There is no jump. The outward velocity required to escape (for an object that's in free fall, not accelerated) increases smoothly to c as the horizon is approached and reached, and the energy required to escape (again, for an object that's in free fall, not accelerated) increases smoothly to infinity. This works the same as an object's energy going to infinity as its speed approaches c (in a given reference frame) in flat spacetime. There's no jump.

Which proves my point. You can't accelerate smoothly up to c in flat space-time can you? So why do you insist on being able to do the exact same thing using gravity instead? Infinity can't be reached gradually, so it can't be reached.

There is no "force" anywhere. It's just geometry. See below.

If you're going to look at it like that then acceleration is also just geometry. You can't claim that they're different types of curvature because one affects matter and the other affects things in it. That doesn't make sense. You can't affect space-time on its own because space-time is nothing on its own. It's just a measurement of the relative distances between matter. If what you're saying about it being dependant on the direction you're moving through it and all those other stupid things them maybe it would seem as though it could have properties in its own right. This is what happens when you let equations direct your intuition rather than the other way round. Of course there's a force.

No. You keep on getting this backwards. The Rindler horizon is the path of a light ray that just fails to catch up to a family of uniformly accelerating observers outside (or "above") it. Similarly, a black hole's horizon is the path of a light ray that just fails to catch up to a family of "hovering" observers outside it. Neither horizon has anything to do with whether an object outside the horizon can catch up with an object that free-falls *into* the horizon. That's a separate question.

I KNOW they're different questions! Are you doing this on purpose? My point before was exactly that. I was asked why I thought an object couldn't cross an event horizon when it could obviously cross a Rindler horizon. Not having the slightest clue what a Rindler horizon was, I looked it up. It's the only time I've done that so it's not surprising that I got it slightly wrong. It's not how I'd normally do it. My point here was that the event horizon marks the point when nothing from the outside can catch the free-faller because they've broken the light barrier. Silly isn't it?

No. I apologize for a misstatement there; I should have snipped the quote from your previous post more carefully. The only statement of yours I meant to agree with was the statement that being in free fall is equivalent to being at rest. I did not mean to imply agreement with your statement about tidal force.

Well it's a start. How is tidal force in a gravitational field any different from acceleration in flat space-time?

"Energy" in the sense you are using the term (meaning energy expended by an object to accelerate itself so it feels weight) has no effect on the light cones. It can't tilt them at all. Also, "gravity" is not a separate "thing" that somehow tilts the light-cones. Gravity *is* the tilting of the light cones. They're the same thing (spacetime curvature).

That's not what you said before. Didn't you say that acceleration in flat space-time was equivalent, so you could tilt light cones but not all the to 45 degrees from this view?

No. The tilt is not "past 45 degrees". The *ingoing* sides of the light cones remain pointed at 45 degrees inward. The *outgoing* sides gradually point more and more *upward* (i.e., *less* than 45 degrees) until they become vertical (0 degrees) at the horizon. Inside the horizon, they tilt over the other way, which simply means objects can no longer avoid moving inward, even if they move at the speed of light. It does not require going backward in time.

The angle depends on which side you are? Right.

No, that doesn't follow, because in order to escape, you have to move *outward* faster than light. You do not have to move *inward* faster than light to get inside the horizon. You can free-fall there, without "moving" at all (you yourself said being in free fall is the same as being at rest, so all you have to do to get inside the hole is to stay at rest).

I think you may have misunderstood what I meant. You don't have to move to get in but you have to move to get out? Free-fall is the equivalent to being at rest in the sense that you don't feel any acceleration and you can't reach reach if you're always at rest.

No, "gravity" is a way of saying that *spacetime* is curved. It does not say anything about particular *paths* in that spacetime. You can follow a geodesic path (freely falling path, weightless path) in a curved spacetime, where gravity is present. They are separate, distinct concepts.

Of course they're not. What you call a geodesic path is one that goes with the curvature caused by gravity rather than trying to fight it. You could use an ongoing source of energy to replicate the exact same thing with outgoing rather than ingoing curvature.

To make sense, yes. To make sense in a way that satisfies all your intuitions, no.

We'll see.

And yet you keep trying to claim that tidal force is the *same* as "acceleration due to gravity" or "apparent force of gravity". How can something be the same as its rate of change?

Because being in free-fall is equivalent to being at rest, so you have to ignore the difference in relative velocities because that's all it is. The only thing that determines the direction of gravity is the direction it's strongest in, in other words the difference in gravity between the two end points of the object. The sharper the increase, the stronger the gravity. I'm not talking about the increase in velocity of an object free-falling in a gravitational field relative to something not free-falling.

Um, because a larger area of you is in contact with the ground.

But there are tidal forces in multiple directions, not just one. There can even be tidal force in the time direction. You can't extract a unique direction for gravity to pull you from tidal force.

No, it's because your weight is spread more evenly over a larger area of your body. See above.

If gravity is relative then the only thing that determines which way you are pulled is tidal force because that represent acceleration and it's that and only that that can really be classed as the force of gravity. That's all I was getting at. Stop wandering off.

I wasn't assuming anything. I was responding to the actual words you used. You said: "I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe." You didn't elaborate on what you meant by "edge". How was I supposed to interpret it except in the obvious way?

My point is that's not the obvious way, especially not in a relativity forum.

Why not? I know it doesn't make sense in the mental model you have in your head. But can you give a logical argument, starting from premises we all accept, that shows that it *can't* make sense, in *any* consistent model? I don't think you can, but if you have such an argument, let's see it. Just claiming that it doesn't make sense won't do, because I have a consistent model where it does makes sense.

No it doesn't. To describe one thing it needs multiple coordinate systems that completely contradict each other, it effectively accelerates you past c, it breaks the arrow of time and it has areas of space-time that have different values for gravity depending on which direction you travel through them. Apart from that it might be a consistent model that makes sense.

Because the worldlines are expanding in different directions. Ned Wright's Cosmology FAQ, at the link below, briefly explains the kind of thing that's going on.

http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN

Nothing can travel faster than light... accept when a physicist needs it to.

I'm not sure what you're asking here. I was just explaining what r = 2M means, since you asked that; that radius is the radius of the horizon. Radius is not a quantity in the time dimension.

Debatable. Length is though. A singularity has zero volume in all four dimensions, so it covers no space and no time. Why would the effect of the singularity be anything other than a four-dimensional sphere? Length contracts and time dilates, same thing.

Since you can reach the black hole, this question is meaningless.

Only by cheating and bending or ignoring altogether the rules that it contradicts. I don't think you can actually do it in real life. Doesn't it seem in the least bit silly to you?

They can't.

It still doesn't matter.

Oh, yes, it does. It's crucial, because it grounds the analogy between Rindler coordinates and Schwarzschild coordinates (as compared with Minkowski coordinates and Kruskal coordinates). See below.

I have never said that, and in fact I've explicitly said several times that whether or not you fall into the hole does *not* depend on the coordinates you use. What *does* depend on the coordinates is whether particular events on your worldline--such as the real, actual events of you crossing the hole's horizon and moving inward beyond that point--are "visible" in those coordinates. It's no different than saying that something that happens on the surface of the Earth at a point beyond your horizon--where you can't see--isn't visible to you.

Yes it is different. I can bring it into view by moving towards it. Nothing can witness an object crossing into black hole unless it crosses itself. But that object itself can't cross as far as any external observers can see, so no object crosses the event horizon from the outside

In Rindler coordinates, the portion of a free-falling object's worldline that lies at or below the Rindler horizon is not visible, but it *is* visible in Minkowski coordinates. That's all there is to it; one coordinate system can see events that aren't visible in the other. I really don't see why this is so hard to grasp.

It wouldn’t be a problem if that’s all there were to it. It stops working when you use infinity in one of them, because you can’t translate that into the other one without the two contradicting each other. I really don't see why that is so hard to grasp.

Similarly, in Schwarzschild exterior coordinates, the portion of a free-falling object's worldline that lies at or below the black hole horizon is not visible, but it *is* visible in Kruskal coordinates. Again, one coordinate system can see events that aren't visible in the other. What's impossible about that?

Nothing. Of course it's not impossible, provided they don't contradict each other. It's not okay two use two contradictory points of view and claim both are correct.

The only one who's automatically assuming it's wrong is you. I make no such assumption.

You serious?

Yes.

Schwarzschild coordinates cover the entire space-time external to the black hole and show that time dilates and length contracts to infinity at the horizon. At no point on any of that entire space-time external to the black hole does any object ever reach the horizon so I’d love to know why you think they can cross it.

No, they're not. Nor do you have to "break the light barrier" to fall into the hole, because you're moving inward, not outward. The fact that you can't comprehend the consistent model that explains all this does not mean that model is false or inconsistent; it just means you can't understand it.

Or maybe you just don't understand why it's not consistent?

 

[A-wal]

Use the river model, but add a river bed to act as a kind of ether. Not a real ether, just an imaginary one to compare our starting position with the horizon. We start far enough away for the gravitational pull of the black hole to be negligible enough to ignore. Our river bed provides us with an uncurvable surface that’s at rest relative to our starting position and the black hole. We wait a while and eventually notice we’re moving towards it as the river slowly moves us along relative to our starting position. We don't feel as though we're accelerating because we're still at rest relative to the river. That river moves relative to the river bed in exactly the same way any object using energy to accelerate would. It can’t reach c. The river changes in relation to the river bed as it moves faster, and the black hole changes with it. Time dilation and length contraction relative to the river bed mean that the its life span and size decrease exponentially as we approach the horizon. We would be traveling at the speed of light at the horizon if it made sense for anything to move that fast. You could try to use energy to accelerate you through the river to get you there faster, but that would just have the effect of adding very high velocities.

It is a toy model

It just shows the idea that infinities in one coordinate system do not always have any physical meaning. Like infinities in S. coordinates are a problem of that coordinate system, while infalling objects can cross the horizon without any problems.

I think it's a problem with the coordinate system you're using for the in-faller. They can't both be right.

 

[Dmitry67]

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

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

It can’t reach c.

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

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

 

[PeterDonis]

Okay, if "you can always find a freely falling reference frame in which, locally, that acceleration disappears" what about at the event horizon?

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

I didn't say tidal gravity is relative. I said it's the equivalent to acceleration in flat space-time.

Which it isn't.

If I was using relative velocities as a measurement of the strength of gravity then it wouldn't be relative and the same area of space-time would change its properties depending on which observer makes the measurements. One of the reasons I have a problem with standard GR is exactly that. The properties of the same area of space-time shouldn't change depending on which direction you travel through it.

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

I wonder why. Do you think maybe it's equivalent to acceleration? Tidal force is the only way, and there's always tidal force because it has infinite range. That's acceleration though because it's the difference, or the rate of change as you put it, which is also the definition of acceleration.

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

What? But why do they push against each other? Both together are the reaction to the force of gravity.

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

The pull of gravity is being countered by acceleration by the combination of the Earth and our bodies both being solid objects.

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

If you were to wait long enough then the rock would be crushed under its own weight. Admittedly, I'd have to invent a whole new set of epically large numbers to describe how long that would take, but I'm sure it would happen, eventually.

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

If there's too much gravity for the object to take then it will collapse. If the object is close to the limit then it will collapse after a time. If it's not close then it will collapse after a long time. This is the same principle as gradually reaching infinity. I don't think anything can be strong enough to resist anything forever.

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

You've just listed four. Space-time can be curved BY one of two main causes.

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

You would feel your weight if you were in free-fall in a strong enough gravitational field.

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

It's curved any time you feel acceleration. It's the same thing.

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

Which proves my point. You can't accelerate smoothly up to c in flat space-time can you? So why do you insist on being able to do the exact same thing using gravity instead?

I'm not insisting on any such thing.

If you're going to look at it like that then acceleration is also just geometry.

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

You can't affect space-time on its own because space-time is nothing on its own. It's just a measurement of the relative distances between matter.

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

My point here was that the event horizon marks the point when nothing from the outside can catch the free-faller because they've broken the light barrier.

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

Well it's a start. How is tidal force in a gravitational field any different from acceleration in flat space-time?

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

That's not what you said before. Didn't you say that acceleration in flat space-time was equivalent, so you could tilt light cones but not all the to 45 degrees from this view?

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

The angle depends on which side you are? Right.

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

I think you may have misunderstood what I meant. You don't have to move to get in but you have to move to get out?

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

Of course they're not. What you call a geodesic path is one that goes with the curvature caused by gravity rather than trying to fight it. You could use an ongoing source of energy to replicate the exact same thing with outgoing rather than ingoing curvature.

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

The only thing that determines the direction of gravity is the direction it's strongest in, in other words the difference in gravity between the two end points of the object.

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

My point is that's not the obvious way, especially not in a relativity forum.

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

To describe one thing it needs multiple coordinate systems that completely contradict each other, it effectively accelerates you past c, it breaks the arrow of time and it has areas of space-time that have different values for gravity depending on which direction you travel through them.

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

Nothing can travel faster than light... accept when a physicist needs it to.

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

Debatable. Length is though.

Length is a quantity in the time dimension?

A singularity has zero volume in all four dimensions, so it covers no space and no time.

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

Doesn't it seem in the least bit silly to you?

No.

It wouldn't be a problem if that's all there were to it. It stops working when you use infinity in one of them, because you can't translate that into the other one without the two contradicting each other. I really don't see why that is so hard to grasp.

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

Schwarzschild coordinates cover the entire space-time external to the black hole and show that time dilates and length contracts to infinity at the horizon. At no point on any of that entire space-time external to the black hole does any object ever reach the horizon so I'd love to know why you think they can cross it.

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

Or maybe you just don't understand why it's not consistent?

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

 

[PeterDonis]

Use the river model, but add a river bed to act as a kind of ether. Not a real ether, just an imaginary one to compare our starting position with the horizon.

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

 

[A-wal]

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

I've noticed. What does that tell you?

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

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

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

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

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

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

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

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

Which it isn't.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I'm not insisting on any such thing.

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

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

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

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

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

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

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

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

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

 

[A-wal]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Length is a quantity in the time dimension?

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

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

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

No.

Oh!

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

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Meaning what, exactly?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[DrGreg]

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

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

[tex]T = \frac{t}{v-x}[/tex]
[tex]X = x[/tex]
[tex]Y = y[/tex]
[tex]Z = z[/tex]​

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

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

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

 

[A-wal]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Meaning what, exactly?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

[tex]T = \frac{t}{v-x}[/tex]
[tex]X = x[/tex]
[tex]Y = y[/tex]
[tex]Z = z[/tex]​

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

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

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

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

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

 

[A-wal]

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

 

[PeterDonis]

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

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

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

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

Behave differently how?

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

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

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

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

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

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

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

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

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

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

Right so far.

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

Still ok.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

 

[A-wal]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Tidal force is movement into a stronger gravitational field!

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

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

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

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

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

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

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

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

Don't be silly.

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

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

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

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

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

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

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

Change rather than move then.

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

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

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

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

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

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

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

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

I'm not sure what else I can do.

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

 

[A-wal]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

 

[PeterDonis]

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

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

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

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

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

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

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

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

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

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

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

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

(2) The bending of light by the Sun?

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

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

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

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

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

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

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

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

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

 

[PeterDonis]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[Buckleymanor]

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

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