Is Gravity Fictitious? A Look at Victor Stenger's The Fallacy of Fine Tuning

[bobsmith76]

This is a quote from Victor Stenger's the Fallacy of Fine Tuning, let me know if you think it is true. He claims that gravity is fictitious (typos are due to the fact that I scanned and OCR'ed the book:

One of the several great insights of Einstein was that gravity is indistinguishable from acceleration. Imagine a closed chamber far out in space accelerating at exactly 9.81 meters per second per second, the acceleration we experience from gravity on the surface of Earth. If you were to wake up inside that chamber,
which has no windows to the outside, with no memory of how /ou got there, you would think you were just sitting in a room :>n Earth.
Now, technically, if you had some accurate measuring equipment, you would be able to detect the fact that the paths of falling objects were not converging as they would on Earth, pointing to the center of Earth. So this equivalence is “local,” that is, limited to an infinitesimal volume.

Consider the law of motion F = ma for a freely falling body in terms of a coordinate system falling along with the body. Since the body does not change position in its own reference frame, both its velocity and its acceleration in that reference frame are zero. Zero acceleration means that a freely falling body experiences no external force. An observer in that refer-snce frame has no sense of gravity.
Next, let us consider a second coordinate system fixed to a second body such as Earth. This could be any coordinate system accelerating with respect to the first. An observer on Earth witnesses a body accelerating toward Earth and interprets it as the action of a “gravitational force.”

Ask yourself this: If the gravitational force can be transformed away by going to a different reference frame, how can it be “real”? It can’t.
We see that the gravitational force is an artifact, a “fictitious” force just like the centrifugal and Coriolis forces. These forces ire introduced in physics so that we can still write a law of motion like F = ma in an accelerating reference frame. We sit on a spinning Earth, which puts us in an accelerated reference frame. By introducing the centrifugal and Coriolis forces we >till can use all the machinery of Newtonian mechanics.
The point, though, is that if we want to make all our models point-of-view invariant, then we should not include non-invariant concepts in our basic models. We can still use the notion of a gravitational force mat has proved so powerful from

the time of Newton, as long as we keep in mind that it is fictitious.

 

[Diffeomorphic]

I've actually read Stenger's book and I'm not a big fan of the way he just makes philosophical assertions about the nature of knowledge and the philosophy of science without any real good argumentation. Ad hominems aside, first off "Imagine a closed chamber far out in space accelerating at exactly 9.81 m/s". What does acceleration mean to you? The reason you experience weight is because you have a mass and you are located in a relatively uniform gravitational field on Earth where the particular acceleration due to gravity is 9.81 m/s. Note, "Acceleration due to gravity". What causes this acceleration? From my understanding, it is the mass of the Earth that causes a "gravity well" or a change in the local configurations of the physical space in that area (known as a diffeomorphism).

Are you then positing that the entire Earth is moving "up" at 9.81 m/s^2 if gravity is fictitious and acceleration is fundamental?

From my opinion, gravity can be completely explained by a sophisticated account of how mass causes a diffeomorphism in a manifold and the specifics of what type of manifold we really live on (is there actually any curvature?).

 

[PeterDonis]

I'm fine with all the stuff about gravity being indistinguishable from acceleration, as far as it goes (since there's a caveat that it's only true in a sufficiently small local region of spacetime). However, just saying "gravity is a fictitious force" by itself doesn't address a couple of important things:

(1) If we consider a freely falling body near the Earth, and say that the "force of gravity" is causing it to fall, I would agree that this "force" we are calling gravity is indeed fictitious. However, that does not mean there is no force present; it's just that we are looking for a force in the wrong place. The right place to look for a force is the person standing at rest on the Earth's surface, watching the object fall. Why? Because he feels weight, whereas the falling object doesn't. (We're assuming, of course, that all this is taking place in vacuum, so air resistance is not present. Maybe the whole thing should be set up on the Moon.) If we adopt a pragmatic, physical definition of force, as "something that causes you to feel weight", then there is indeed a force present, but it's not "gravity"; it's the mundane force that prevents solid objects from collapsing. This definition of force is the one GR adopts; the degree to which it clears up confusion, if adopted consistently, is evidenced by Einstein's famous statement that the thought he had in 1907, that "if a person falls freely, he will not feel his own weight", was "the happiest thought of my life".

(2) You can make the "force" of gravity disappear in a small local region, but you can't make gravity as a whole disappear. Tidal gravity, aka spacetime curvature, will still be present. In fact, GR basically *defines* gravity as tidal gravity, aka spacetime curvature. In the situation of falling bodies near a large body like the Earth, tidal gravity/spacetime curvature manifests itself as nearby freely falling bodies that start out at rest relative to each other not staying that way: they either converge or diverge with time. For example, you mentioned that the paths of falling bodies converge on the center of the Earth; that's one manifestation of tidal gravity. Another is that, if two bodies are released from rest at different heights, one exactly above the other, they will separate; the one closer to the Earth will fall faster.

Both of the above items, feeling weight and tidal gravity, are invariants; they don't depend on point of view, so they meet Stenger's criterion for being "real", not "fictitious". So even if gravitational "force" is fictitious, gravity as a whole is not.

 

[harrylin]

This is a quote from Victor Stenger's the Fallacy of Fine Tuning, let me know if you think it is true. He claims that gravity is fictitious (typos are due to the fact that I scanned and OCR'ed the book: [..]
Ask yourself this: If the gravitational force can be transformed away by going to a different reference frame, how can it be “real”? It can’t.
We see that the gravitational force is an artifact, a “fictitious” force just like the centrifugal and Coriolis forces. These forces ire introduced in physics so that we can still write a law of motion like F = ma in an accelerating reference frame. [..] We can still use the notion of a gravitational force mat has proved so powerful from the time of Newton, as long as we keep in mind that it is fictitious.

First of all, gravity doesn't necessarily mean "force"; it relates to an effect from matter that everyone observes and there can be no doubt about the fact that matter really has that effect on other matter.

Anyway, that book specifically discusses "gravitational force". It uses an argument which boils down to suggesting that if you can hide something from observation, it cannot be real. I don't buy that. And the illustration of fictitious forces such as the Coriolis force may be misleading: those are fudge factors to compensate for not accounting for acceleration. So, if "gravitational force" should be considered "real" depends on one's exact definitions of "force" and "acceleration". There can be no doubt that when you sit on your chair, you feel a real force. The force that you feel is due to the tendency of your body to accelerate towards the earth, and which is restrained by your chair. Some people might define such a tendency as "force", and since the tendency is real, then that force would be called "real" as well.

I now notice that PeterDonis made similar comments, but formulated differently.

 

[A.T.]

The naming "real force" vs. "fictitious force" is one of those terrible naming choices, that leads into pointless philosophical debates. I prefer the terms "interaction force" & "inertial force". Both are just mathematical models invented by humans, and thus none is more real than the other. Both can be used to model gravity (the phenomenon of mass attraction).

There can be no doubt that when you sit on your chair, you feel a real force. The force that you feel is due to the tendency of your body to accelerate towards the earth, and which is restrained by your chair.

The real force that you feel is the electromagnetic repulsion between you and chair. You could just as well say it is due to the tendency of your body to move inertially (have zero proper acceleration), and the frame of the Earth's surface being accelerated.

Some people might define such a tendency as "force", and since the tendency is real, then that force would be called "real" as well.

The term "real force" is just a certain model that obeys Newtons 3rd Law. If you use that model for gravity, then gravity is "real force". If you model gravity as an "fictitious force" then gravity is a "fictitious force".

 

[A.T.]

You can make the "force" of gravity disappear in a small local region, but you can't make gravity as a whole disappear. Tidal gravity, aka spacetime curvature, will still be present.

You can make the region of uniform gravity arbitrarily large. One way is a uniform sphere with a non-concentric spherical cavity. Inside the cavity the field is uniform.

In fact, GR basically *defines* gravity as tidal gravity, aka spacetime curvature.

Not really. GR predicts mass attraction in uniform gravitational fields, without any curvature of space time. The metric of space time is not Euclidean but it has no intrinsic curvature. The very popular statement "gravity in GR is curvature of space time" is quite misleading. In fact, the term "curvature" (for intrinsic curvature) is unfortunate itself, which makes the statement even more misleading.

 

[harrylin]

The naming "real force" vs. "fictitious force" is one of those terrible naming choices, that leads into pointless philosophical debates. I prefer the terms "interaction force" & "inertial force". Both are just mathematical models invented by humans [..]

"Force" is not a mathematical model, it's a reasonably well defined human concept. But as we already elaborated, for the question here the answer subtly depends on one's exact definition of "force". Let's see what www.dictionary.com says:

1.physical power or strength possessed by a living being: He used all his force in opening the window.
[..]
Physics .
a.an influence on a body or system, producing or tending to produce a change in movement or in shape or other effects.
b.the intensity of such an influence. Symbol: F, f
[..]

Obviously there is such an influence of gravity between material bodies, and its existence is observed independently of coordinate system or perspective.

Does that help?

 

[A.T.]

"Force" is not a mathematical model, it's a reasonably well defined human concept.

As long as we agree that it is a human invented concept, I'm fine. That means that it is pointless to argue if "real forces" are more real than "fictitious froces".

But as we already elaborated, for the question here the answer subtly depends on one's exact definition of "force". Let's see what www.dictionary.com says:

Exact definition of physics-terms from a dictionary?

...tending to produce a change in movement ...

"Change of movement" depends on the reference frame.

 

[Hootenanny]

Let's see what www.dictionary.com says:

One must be careful about using popular, non-scientific definitions in scientific discussions, particularly when (as in this case) the term in question is undefined.

 

[Dale]

Not really. GR predicts mass attraction in uniform gravitational fields, without any curvature of space time. The metric of space time is not Euclidean but it has no intrinsic curvature.

This is interesting. What exactly are you thinking of here?

 

[harrylin]

One must be careful about using popular, non-scientific definitions in scientific discussions, particularly when (as in this case) the term in question is undefined.

Indeed - that's why I cited the common physics definition.

 

[harrylin]

As long as we agree that it is a human invented concept, I'm fine. That means that it is pointless to argue if "real forces" are more real than "fictitious froces".

The one doesn't follow from the other except if you hold that the chair in which you are sitting (a chair is also a human concept) can be called "fictitious".

Exact definition of physics-terms from a dictionary?

That is the very purpose of dictionaries, similar to encyclopedias; PeterDonis already mentioned (without reference) another definition that is used in the context of general relativity.
On what do you base your definitions? A single textbook isn't good enough, as they all define it slightly differently. You should take the most popular ones and try to make a merge that reflects the most generic opinion, or if that is not possible, you must give several of those. And that is what you find in a good dictionary. The one that I cited misses the alternative definition that Peter already gave, but it reflects rather well one of the most common physics definitions.

"Change of movement" depends on the reference frame.

I mentioned another pertinent effect that does not depend on reference frame.

PS see also post #3 here:
https://www.physicsforums.com/showthread.php?t=537033

Anyway I'm not interested in arguing or debates. Are our opinions clear enough bobsmith?

 

[D H]

There can be no doubt that when you sit on your chair, you feel a real force. The force that you feel is due to the tendency of your body to accelerate towards the earth, and which is restrained by your chair.

The force you feel is the chair pushing up on you. You do not feel gravitation. You cannot measure the gravitational force by means of any local experiment. In a very real sense, the gravitational force is not "real". It is a fictitious force in general relativity.

Imagine yourself on the Vomit Comet. During the climb that precedes a parabolic arc you feel that force as almost twice as strong as normal. Gravity hasn't doubled; it is in fact decreasing by a tiny amount (about 0.003 m/s² for every kilometer of increased altitude) during the climb. The director announces "over the top" and you feel that force disappear. The gravitational force did not change by two g during that brief interval, yet the force that you feel did.

During that climb, there is no local experiment you could conduct that would tell you that the 1.8 g you are feeling is a result of sitting still on a planet whose surface gravity is 1.8 times that of the Earth versus being on a spaceship in deep space accelerating at 1.8 g relative to some inertial frame versus being on a planet accelerating upward at 0.8 g relative to the surface of the Earth. Similarly, during the 25 or so seconds of zero g flight, there is no local experiment that you could conduct that could attribute that feeling of weightlessness to the plane flying a zero g parabolic arc versus a non-accelerating spaceship in deep space.

 

[PeterDonis]

You can make the region of uniform gravity arbitrarily large. One way is a uniform sphere with a non-concentric spherical cavity. Inside the cavity the field is uniform.

Not really. GR predicts mass attraction in uniform gravitational fields, without any curvature of space time. The metric of space time is not Euclidean but it has no intrinsic curvature. The very popular statement "gravity in GR is curvature of space time" is quite misleading. In fact, the term "curvature" (for intrinsic curvature) is unfortunate itself, which makes the statement even more misleading.

I see DaleSpam already asked this question: what are you thinking of here? Do you have a reference for the solution of the EFE you are talking about?

 

[harrylin]

The force you feel is the chair pushing up on you. You do not feel gravitation. [..] Imagine yourself on the Vomit Comet. [..]

You cited a sentence that referred to the cause of that force, as was inferred by logical reasoning - not just based on sensations. For example if one pulls a car with a rope, that car feels the force from that rope. Do you think that in such cases the rope or the chair are the cause of the force that is felt? If so, why do a loose piece of rope, or a chair in space, not push?

 

[PAllen]

If someone has the motivation, it should be much easier to solve the Newtonian potential and gradient for a uniform sphere of matter with a spherical, off center, cavity, than the GR case. This should at least provide a good weak field approximation to the GR case.

At various times I have seen references in the literature to matter configurations in GR that achieve quasi-local flatness with finite 'potential' i.e. quasi-local approximation of a uniform field to some order. However, a few times in recent years I have looked for and not been able to track down any primary sources on this.

 

[A.T.]

This is interesting. What exactly are you thinking of here?

A uniform gravitational field. No tidal forces. 2nd metric derivates zero, thus no space time curvature. But: 1st metric derivates not zero, thus gravitational acceleration.

 

[A.T.]

Do you have a reference for the solution of the EFE you are talking about?

Not for the spherical off-center cavity. But it must be uniform in the weak field limit, beacuse that's what Newton predicts.

 

[Dale]

A uniform gravitational field. No tidal forces. 2nd metric derivates zero, thus no space time curvature. But: 1st metric derivates not zero, thus gravitational acceleration.

Do you mean just in some local region or globally? I don't know of any solutions to the EFE that would fit the bill globally.

 

[A.T.]

Do you mean just in some local region or globally? I don't know of any solutions to the EFE that would fit the bill globally.

The uniform field is an idealized example to show that space time curvature is not key to gravitational acceleration in GR. To model such a field in GR the spacetime metric would have no curvature. This is about how spacetime geometry and gravitational acceleration are connected.

Whether we know an analytical solution to the EFEs that connects this idealized spacetime geometry to some mass distribution is a different question.

 

[PeterDonis]

The uniform field is an idealized example to show that space time curvature is not key to gravitational acceleration in GR. To model such a field in GR the spacetime metric would have no curvature. This is about how spacetime geometry and gravitational acceleration are connected.

Whether we know an analytical solution to the EFEs that connects this idealized spacetime geometry to some mass distribution is a different question.

But you can't "model such a field in GR" if the field is not a valid solution of the EFE.

Also, even leaving out the EFE issue, I'm not sure you can construct a valid "spacetime geometry" that has a zero curvature but a non-zero, everywhere constant connection; my intuitive guess is that trying to do this will run afoul of constraints imposed by the general rules governing any spacetime geometry. (I'm assuming that, even without the EFE constraint, you would agree that anything that can properly be called a spacetime geometry has to at least have local Lorentz invariance, a manifold structure, etc.) I admit, though, that my differential geometry-fu is not strong enough to back this intuition up with actual theorems. I agree with your underlying point that spacetime curvature and "gravitational acceleration" are distinct (since the latter has to do with the connection, not the curvature).

 

[A.T.]

But you can't "model such a field in GR" if the field is not a valid solution of the EFE.

It seems that it is one:
http://www.mathpages.com/home/kmath530/kmath530.htm

But regardless if there is a known EFE solution that yields this field, you can still figure out what metric you would need to generate uniform acceleration. I merely want show which properties of space-time geometry are responsible for which observed effects. This part of the model it independent of the EFEs.

As I put it in simple (maybe inaccurate) terms:
- gravitational acceleration depends on the 1st derivates of the metric
- tidal forces and curvature depend on the 2st derivates of the metric

Therefore you can have:
- gravitational acceleration without curvature (uniform field, at infinite wall or inside the off-center spherical cavity in a uniform density sphere)
- curvature without gravitational acceleration (at the center of a uniform density sphere)

 

[pervect]

A uniform gravitational field. No tidal forces. 2nd metric derivates zero, thus no space time curvature. But: 1st metric derivates not zero, thus gravitational acceleration.

I think you're talking about the Rindler metriic?

(!+gz)^2 dt^2 - dx^2 - dy^2 - dz^2, for example

It's just a re-labelling of Minkowski space-time. It's got non-zero Christophel symbols and a zero Riemann.

 

[A.T.]

I think you're talking about the Rindler metriic?

(!+gz)^2 dt^2 - dx^2 - dy^2 - dz^2, for example

It's just a re-labelling of Minkowski space-time. It's got non-zero Christophel symbols and a zero Riemann.

Yes, due to the Equivalence Principle the metric in an uniformly accelerated frame is the same as in an uniform gravitational field.

 

[PeterDonis]

It seems that it is one:
http://www.mathpages.com/home/kmath530/kmath530.htm

Yes, due to the Equivalence Principle the metric in an uniformly accelerated frame is the same as in an uniform gravitational field.

First, a clarification: on the mathpages page, are you referring to the "Newtonian" solution near the top of the page (which gives an acceleration independent of distance from the wall, but is not a valid GR solution); to the Rindler metric (which gives an acceleration that varies inversely with distance from the wall); or to the metric at the bottom of the page, describing a family of observers who all have the same constant proper acceleration (i.e., it doesn't vary with distance) in an empty spacetime? I assume the second (the Rindler metric), but confirmation would be nice.

Second, as just noted, the Rindler metric, describes a family of observers whose acceleration *changes* with distance from the "pivot point" at X = 0, so it isn't really "uniform". Also, that metric does not cover the entire spacetime; it only covers the regions with |t| < |x| (where x, t are the usual Minkowski coordinates).

Third, the last metric on the mathpages page, where the acceleration does not vary with position, would seem to be a better one to describe "uniform" acceleration, at least in the sense that Einstein was talking about in his thought experiments on the subject. However, as you can see, the metric in this form is both time-dependent and non-diagonal; after reading that page, I think that feature is what was behind my intuitive guess in an earlier post about "uniformly accelerating coordinates" raising issues.

But regardless if there is a known EFE solution that yields this field, you can still figure out what metric you would need to generate uniform acceleration. I merely want show which properties of space-time geometry are responsible for which observed effects. This part of the model it independent of the EFEs.

Another clarification: by "metric" here you mean, I take it, "metric as expressed in a particular coordinate system". The geometry itself doesn't change, and I would not say it's accurate to say that the geometry "generates uniform acceleration". For example, if the curvature is zero everywhere, which it is in both the "Rindler metric" case and the "constant proper acceleration" metric (the last one on the mathpages page), then the geometry is flat Minkowski spacetime. (You are correct, by the way, that this fact is independent of the EFE.) The term "uniform acceleration" really describes a family of worldlines within this spacetime, not the spacetime itself. Which particular family depends on the definition of "uniform acceleration" (Rindler or "constant"). Expressing the metric in coordinates that naturally correspond to a particular family of worldlines is a way to help understand the physical experience of observers traveling on those worldlines.

As I put it in simple (maybe inaccurate) terms:
- gravitational acceleration depends on the 1st derivates of the metric
- tidal forces and curvature depend on the 2st derivates of the metric

Therefore you can have:
- gravitational acceleration without curvature (uniform field, at infinite wall or inside the off-center spherical cavity in a uniform density sphere)
- curvature without gravitational acceleration (at the center of a uniform density sphere)

No argument about this; the only thing I would add here is that the first derivatives of the metric are not generally covariant, since they don't transform properly under changes of the coordinate system. The second derivatives of the metric do. So if you want the laws of physics to be generally covariant, they need to be written in terms of things that transform properly, like curvatures, not things that don't. For example, I observed above that if curvature is zero everywhere, the spacetime must be Minkowski spacetime. We would expect the laws of physics as applied in this spacetime to "comprehend" that. But we would not expect the laws to look different for the "uniformly accelerating" observers (by either definition) than for the inertial, freely falling ones. Their particular physical experiences will be different (the accelerating ones feel weight, while the inertial ones don't), but they should arrive at the same *laws* based on those experiences.

(A final note: some of your examples don't have curvature zero everywhere, just in a particular region; but the geometry is still independent of which particular coordinate system is used to describe it. So the general point that the laws of physics should be written in terms of things like curvature, not things like "gravitational acceleration", still holds.)

 

[A.T.]

Third, the last metric on the mathpages page, where the acceleration does not vary with position, would seem to be a better one to describe "uniform" acceleration, at least in the sense that Einstein was talking about in his thought experiments on the subject. However, as you can see, the metric in this form is both time-dependent and non-diagonal; after reading that page, I think that feature is what was behind my intuitive guess in an earlier post about "uniformly accelerating coordinates" raising issues.

Yes, that is the one I meant. I get your objection now.

For example, if the curvature is zero everywhere, which it is in both the "Rindler metric" case and the "constant proper acceleration" metric (the last one on the mathpages page), then the geometry is flat Minkowski spacetime.

Yes. It kind of follows from the EP that that a "globally uniform gravitational acceleration" is equivalent to an uniformly accelerating frame in flat space-time.

A final note: some of your examples don't have curvature zero everywhere, just in a particular region; but the geometry is still independent of which particular coordinate system is used to describe it.

Yes, but the point of this example (off center spherical cavity) was merely to show that in an extended region of zero curvature , you still can have "gravitational acceleration" or "mass attraction".

So the general point that the laws of physics should be written in terms of things like curvature, not things like "gravitational acceleration", still holds.)

Yes, which I guess answers the OPs question: Gravitational acceleration is "fictitious" in GR.

 

[Dale]

It seems that it is one:
http://www.mathpages.com/home/kmath530/kmath530.htm

Am I reading that page correctly that the solution for an infinite plane of mass is the Rindler metric? If so, that's pretty cool!

 

[PeterDonis]

Yes. It kind of follows from the EP that that a "globally uniform gravitational acceleration" is equivalent to an uniformly accelerating frame in flat space-time.

I think that's what Einstein was getting at, yes, but as the mathpages page shows, there really isn't a way to actually set this up globally, only locally. Every example on that page fails to meet at least one condition that would seem necessary to really have a "globally uniform acceleration". As I noted before, I would say the last one comes closest, but even if we leave out the time-dependence and non-diagonal metric, observers "at rest" in these coordinates do not remain at a fixed proper distance from each other.

Yes, but the point of this example (off center spherical cavity) was merely to show that in an extended region of zero curvature , you still can have "gravitational acceleration" or "mass attraction".

Agreed.

 

[PeterDonis]

Am I reading that page correctly that the solution for an infinite plane of mass is the Rindler metric? If so, that's pretty cool!

I read it that way too, but with a couple of qualifications. First, it's only one possible solution, for the special case where all the metric coefficients except g_tt are constant. (The rest of the page covers the other possibilities pretty thoroughly.)

Second, for the "usual" case of the Rindler metric as just an alternate set of coordinates on Minkowski spacetime, that metric does not cover the entire spacetime, only "region I", where x > 0 and |t| < x (in Minkowski coordinates). In this spacetime, with an actual physical wall, unless I'm missing something, the metric is supposed to cover the entire spacetime, or at least all of it on one side of the wall (but by symmetry, the solution would be the same on the other side except for a parity transformation, so that the "acceleration vectors" point in the opposite direction in x). So this is not just flat Minkowski spacetime in disguise.

Another way to put this is: in the usual Rindler metric, the "Rindler horizon" at X = 0 is a null surface, but as I'm reading this, that is not the case for the wall; the infinite wall is a normal timelike surface, for which X = 0 and T, Y, Z can all range from minus infinity to plus infinity.

Still another way to put it: the solution assumes that the Ricci tensor vanishes, but strictly speaking, that only applies for X > 0 (and, by symmetry, for X < 0 with a parity flip). But if the wall is actually physically there, then the Ricci tensor should not vanish on the wall, at X = 0. That means the spacetime is not vacuum everywhere.

All of which makes me wonder if I'm missing something, because the metric does look, formally, exactly like the Rindler metric, which makes me think that if I computed its Weyl tensor, it would vanish (since the Weyl tensor certainly vanishes in Minkowski spacetime). But if the Weyl tensor vanishes, it's hard for me to see how the infinite wall at X = 0 can be a timelike surface instead of a null surface as the normal Rindler horizon is.

 

[PeterDonis]

it's hard for me to see how the infinite wall at X = 0 can be a timelike surface instead of a null surface as the normal Rindler horizon is.

Actually, after thinking this over, it's even more of an issue than I thought. If the Weyl tensor does vanish for X > 0, then the entire region X > 0 in these coordinates must be isometric to "region I" of Minkowski spacetime. But that means that the apparent 3-surface X = 0 in these coordinates is actually only a 2-surface (i.e., if we leave out the Y and Z coordinates, it is a point instead of a line--the point x = 0, t = 0 in standard Minkowski coordinates). So where is the wall? That is, where is the non-zero Ricci tensor region that is the "source" of the gravitational field? The Ricci tensor can't just be nonzero at that single point in spacetime; that would violate conservation of energy. It can't be nonzero along the entire null "boundary" of region I, because that would cause the Weyl tensor to be nonzero in region I itself (because of curvature "propagating" from the "past horizon" null surface, x = -t). But if the Ricci tensor is zero everywhere, then there is no wall at all; the derivation basically amounts to a proof that an infinite "wall" can only exist if its stress-energy tensor is identically zero. Was that actually the intent, or am I missing something?