# About tidal forces

snoopies622
Recently while searching through some old threads in this section of the forums, I came across a few statements that do not seem to me to be entirely self-consistent. I was wondering if someone could tell me which - if any - of these declarations is not correct:

-- in a uniformly accelerating frame of reference - like inside a rocket moving with constant proper acceleration - points "lower" in the induced gravitational field experience greater acceleration than ones at "higher" locations

-- such a frame of reference is described by the Rindler metric

-- tidal forces are equivalent to curved spacetime

-- the curvature tensor of the Rindler metric is zero

Thanks.

## Answers and Replies

Staff Emeritus
Gold Member
-- in a uniformly accelerating frame of reference - like inside a rocket moving with constant proper acceleration - points "lower" in the induced gravitational field experience greater acceleration than ones at "higher" locations
-- such a frame of reference is described by the Rindler metric
In these first two statements, you should keep in mind that a uniform gravitational field is not something that really exists in GR. By the equivalence principle, we expect that a uniform gravitational field is equivalent to a uniformly accelerating frame. Therefore there is an ambiguity in the notion of a uniformly accelerated frame in GR. These ambiguities are the subject of Bell's spaceship paradox http://en.wikipedia.org/wiki/Bell's_spaceship_paradox . There are other metrics that are candidates for playing the role of "the" (really "a") uniform field in GR, e.g., the Petrov metric http://arxiv.org/abs/0802.4082 .

-- tidal forces are equivalent to curved spacetime
Tidal forces are one type of curved spacetime, measured by one part of the Riemann tensor. There is another part of the Riemann tensor, expressed by the Ricci tensor, which measures non-tidal curvature. If we're looking for something that is going to play the role of a uniform field in GR, we probably want a vacuum solution, so in that special case it makes sense to focus on tidal forces.

-- the curvature tensor of the Rindler metric is zero
The gravitational field is not a tensor in GR. At a given point in space, an observer can make the gravitational field be anything he likes, including zero, by a choice of coordinates. Therefore we should not expect that a uniformly accelerating frame, or a frame with a uniform gravitational field, should correspond in any particular way to intrinsic properties of spacetime as expressed by tensors like the Riemann tensor. The curvature tensor of the Petrov metric is not zero, but I don't think that has anything in particular to do with whether one or the other (Petrov or Rindler) is a better candidate for a uniformly accelerating frame/uniform field.

Last edited by a moderator:
Mentz114
snoopies622 said:
-- tidal forces are equivalent to curved spacetime

I don't know where you found that but it begs the question - which curved spacetime ? As it happens there are no tidal forces in the Rindler spacetime because the Riemann ( and hence Weyl ) tensor is all zero.

The Rindler chart is a set of coordinates in Minkowski space-time used by an observer with constant proper acceleration. It is a trivial solution of the Einstein field equations but it seems not to to correspond to a ST created by a source.

snoopies622
I just found the thread where I got that idea, although I vaguely remember seeing it elsewhere, long ago:

Entry #62 quotes Kip Thorne, "tidal gravity is a manifestation of spacetime curvature," after which the PF member says,

"thus it is impossible to have tidal forces without spacetime curvature".

Looking at it again I notice that these two statements are actually not logically equivalent.

snoopies622
..there are no tidal forces in the Rindler spacetime..

Wait, does this mean that my first statement about the differences in acceleration at different "heights" in the spaceship is not correct?

Staff Emeritus
Gold Member
Wait, does this mean that my first statement about the differences in acceleration at different "heights" in the spaceship is not correct?

Your original statement was correct. This is analyzed in more detail here: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken] In the Rindler coordinates, the proper acceleration of a particle with an instantaneous coordinate velocity of zero varies with height. This is a good argument against the Rindler coordinates as GR's version of a uniform field. The Petrov metric is better by this criterion. In the Petrov spacetime, the proper acceleration of a particle with an instantaneous coordinate velocity of zero is independent of the "height" r.

Last edited by a moderator:
Mentz114
Of course there are tidal effects. The 'ball of coffee grounds' would stretch in the direction of the acceleration. I keep applying GR concepts to the flat Rindler space and coming up with no-no's. There must be a lesson here but I'm not sure what it is ... Ben, I can't find a Petrov metric. Do you have a reference ?

Staff Emeritus
Gold Member
Ben, I can't find a Petrov metric. Do you have a reference ?

The one I gave in #2 is the reference on the Petrov metric that I think is the easiest to understand and most accessible online.

Staff Emeritus
Gold Member
Here is a FAQ entry I wrote up on this topic.

FAQ: In general relativity, what does a uniform gravitational field look like? What about a uniformly accelerating frame of reference?

The equivalence principle tells us that (a) these two are really the same question, and (b) both questions are frame-dependent.

There is no metric and set of coordinates that gives a completely satisfactory uniform gravitational field. The two possibilities that are generally regarded as coming the closest are (1) flat spacetime in Rindler coordinates, and (2) the Petrov metric.

Rindler coordinates are the coordinates defined by an observer in flat space with a constant proper acceleration a. The line element is (1+ax)^2dt^2-dx^2. This metric describes a flat spacetime, since it is derived from the Minkowski metric by a change of coordinates. It is not a perfect embodiment of our concept of a uniform gravitational field, because a test particle released with zero coordinate velocity at some height x has a proper acceleration that depends on x; only if it is released at x=0 does it have acceleration a.

The Petrov metric was first found by Lewis in 1932, but is now named after Petrov, who rediscovered it in 1962. It is a vacuum solution of the Einstein field equations, and it is unique because of its high degree of translational symmetry; in technical terms, it is "[t]he only vacuum solution of Einstein’s equations admitting a simply-transitive four-dimensional maximal group of motions" (Gibbons 2008). This translational symmetry is what we want in a uniform field. Unlike the Rindler coordinates, the Petrov metric has the property that the proper acceleration of a test particle released with zero coordinate velocity is the same no matter where it is released. However, the Petrov metric has some strange physical properties, including closed timelike curves, which make it also not satisfactory as "the" uniform gravitational field in general relativity. The most accessible reference for information on this spacetime is Gibbons 2008.

T. Lewis, Proc. Roy. Soc. Lond. A136 (1932) 176

Petrov, in "Recent Developments in General Relativity," 1962, Pergamon, p. 371

Gibbons and Gielen, "The Petrov and Kaigorodov-Ozsváth Solutions: Spacetime as a Group Manifold," http://arxiv.org/abs/0802.4082

Last edited:
Mentz114
Ben,
thanks a lot. That's helpful. I gathered the paper in a sweep of Gibbons' papers in the arXiv earlier, but I hadn't seen the ref in #2. I doubt if I'll be able to follow it all but it looks interesting.

Staff Emeritus
Gold Member
Ben,
thanks a lot. That's helpful. I gathered the paper in a sweep of Gibbons' papers in the arXiv earlier, but I hadn't seen the ref in #2. I doubt if I'll be able to follow it all but it looks interesting.

Yeah, a lot of it is technical stuff I don't understand either, but for me it was relatively intelligible if I didn't obsess over the details.

Of course there are tidal effects. The 'ball of coffee grounds' would stretch in the direction of the acceleration. I keep applying GR concepts to the flat Rindler space and coming up with no-no's. There must be a lesson here but I'm not sure what it is ... Wait, are you saying a freefalling "ball of coffee grounds" would stretch? That doesn't seem right to me, since a freefalling ball in Rindler coordinates can be transformed into an inertially-moving ball in an inertial coordinate system on the exact same spacetime.

Mentz114
Wait, are you saying a freefalling "ball of coffee grounds" would stretch? That doesn't seem right to me, since a freefalling ball in Rindler coordinates can be transformed into an inertially-moving ball in an inertial coordinate system on the exact same spacetime.
I don't know. Can there be free-fall in this scenario ? The coffee-grounds are being individually rocket propelled.

snoopies622
1. Since the coffee grounds at the "lower" end of the sphere accelerate at a higher rate then those at the "upper" end, the sphere must stretch

2. Since the Wely tensor and the Ricci tensor of the Rindler metric are both zero, the sphere must maintain its shape and size

3. Since the observer's speed with respect to the sphere is increasing, from his/her perspective the sphere must squash

This sounds like one of those SR paradoxes that can only be resolved through a detailed mathematical analysis using spacetime diagrams.

I don't know. Can there be free-fall in this scenario ? The coffee-grounds are being individually rocket propelled.
But that's what I'm asking, what's the scenario? From the perspective of the accelerating observers at rest in Rindler coordinates, are the coffee grounds also supposed to be held at rest (and if so how? Individual rockets? But in that case wouldn't each individual particle be at rest too, so the shape wouldn't change at all?) or do the Rindler observers see the coffee grounds free-falling in the pseudo-gravitational field, so that they're moving towards the Rindler horizon?

Mentz114
JesseM said:
... or do the Rindler observers see the coffee grounds free-falling in the pseudo-gravitational field, so that they're moving towards the Rindler horizon?

In this picture the test bodies would be seen from the Rindler frame to be falling away.
In GR I would calculate the tidal forces in a frame by transforming the Riemann tensor to the frame, then projecting out with the 4-velocity of the observer, to give a second rank tensor of the tidal forces. So, because the Riemann tensor is zero, the tidal forces will also be zero.

But I'm almost certainly wrong as usual.

I don't know how the grounds would look to a Minkowski observer, assuming they are individually propelled.

Altabeh
But that's what I'm asking, what's the scenario? From the perspective of the accelerating observers at rest in Rindler coordinates, are the coffee grounds also supposed to be held at rest (and if so how? Individual rockets? But in that case wouldn't each individual particle be at rest too, so the shape wouldn't change at all?) or do the Rindler observers see the coffee grounds free-falling in the pseudo-gravitational field, so that they're moving towards the Rindler horizon?

Wait a minute! What do you mean by "accelerating observers at rest"? This is only correct if we take the frame of comoving observers to be instantaneously at rest which means their proper velocity is zero momentarily so in this case one leads to $$\ddot{t}=\frac{-2a\dot{x}\dot{t}}{1+ax}=0$$ for the Rindler metric

$$ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2$$

and from this point it is obvious that $$\dot{t}= const.=T$$ so that

$$\ddot{x}=-aT^2(1+ax).$$

But since the proper acceleration of particles is still dependent on the position, so the coffee grounds are not going to retain their initial shape while staying at rest by definition. This strongly suggests that the Rindler metric does not generally admit a uniform gravitational field unless assuming the local discussion and the EP.

AB

In this picture the test bodies would be seen from the Rindler frame to be falling away.
In GR I would calculate the tidal forces in a frame by transforming the Riemann tensor to the frame, then projecting out with the 4-velocity of the observer, to give a second rank tensor of the tidal forces. So, because the Riemann tensor is zero, the tidal forces will also be zero.

But I'm almost certainly wrong as usual.
Well, the "free-falling in the pseudo-gravitational field seen in Rindler coordinates" scenario is equivalent via a coordinate transformation to having the grounds all moving inertially in an inertial frame, so at least in the inertial frame there will be no change in shape and no tidal forces. It's possible that there is some coordinate-based change in shape seen in Rindler coordinates, although this wouldn't imply any actual physical stresses if the ball were semi-solid rather than just a cloud of noninteracting coffee grounds.
Mentz114 said:
I don't know how the grounds would look to a Minkowski observer, assuming they are individually propelled.
Again, when you talk about them being "individually propelled", do you mean they are propelled in such a way that each ground remains at rest in Rindler coordinates? In that case the ends of the cloud would get closer together just like the family of accelerating observers at rest in Rindler coordinates. This page shows a diagram of how the worldlines of these Rindler observers look in Minkoski coordinates (along with the surfaces of simultaneity from Rindler coordinates drawn in gray, and the Rindler horizon drawn as a dotted line): Wait a minute! What do you mean by "accelerating observers at rest"?
The full phrase was "accelerating observers at rest in Rindler coordinates"...but I was speaking sloppily, I meant that they are accelerating as seen in any inertial frame (and thus have nonzero proper acceleration), but they are observers "at rest in Rindler coordinates". See the diagram I posted above, which shows the worldlines of observers at rest in Rindler coordinates as seen in an inertial frame.
Altabeh said:
But since the proper acceleration of particles is still dependent on the position, so the coffee grounds are not going to retain their initial shape while staying at rest by definition. This strongly suggests that the Rindler metric does not generally admit a uniform gravitational field unless assuming the local discussion and the EP.
I think "uniform gravitational field" has a confusing meaning in relativity, it doesn't actually mean a field where the proper acceleration is the same everywhere--see this page from John Baez's twin paradox page, which says that the time dilation of clocks at different heights in a "uniform" field would actually be different:
Uniform "gravitational" time dilation: Say you have two identically constructed clocks. One is deep down in a uniform "gravitational" potential well (or "pseudo-potential", if you prefer); the other is higher up. If the two clocks compare rates by sending light signals back and forth, then both will agree that the lower clock runs slower than the higher clock. This can be rephrased as "Time runs slower as you descend into the potential well of a uniform pseudo-force field."
This implies that clocks at rest in this uniform field have different velocities as seen in an inertial frame (which is true for observers at rest in Rindler coordinates who are undergoing Born rigid acceleration, which is designed to ensure that the distance between them remains constant in their own instantaneous inertial rest frame at each moment...again, see the diagram I posted above for what this should look like).

By the way, I had an idea about the ball of coffee grounds. Although it's true that if they are free-falling in Rindler coordinates they must be moving inertially in an inertial frame, individual grounds at different ends of the ball might still have different velocities in the inertial frame so the ball as a whole could change shape in the inertial frame. And if you require that the grounds were all initially at rest in Rindler coordinates before beginning to fall (held up by individual rockets which are all turned off simultaneously in Rindler coordinates, for example), wouldn't this in fact imply different velocities in the inertial frame? (edit: actually looking at the gray lines of simultaneity in the diagram above, it looks like q=0 in Rindler coordinates would give a line of simultaneity parallel to the t=0 plane (i.e. the x-axis) in the inertial Minkowski frame, and if the worldlines of the grounds at that moment are tangent to the worldlines of the accelerating observers at t=0, then all the worldlines of the grounds should just be vertical in the inertial frame, meaning they are all moving at the same velocity, so that idea of mine above probably isn't the right answer.)

Last edited:
Altabeh
The full phrase was "accelerating observers at rest in Rindler coordinates"...but I was speaking sloppily, I meant that they are accelerating as seen in any inertial frame (and thus have nonzero proper acceleration), but they are observers "at rest in Rindler coordinates". See the diagram I posted above, which shows the worldlines of observers at rest in Rindler coordinates as seen in an inertial frame.

This means that the observers have zero proper velocities momentarily.

I think "uniform gravitational field" has a confusing meaning in relativity, it doesn't actually mean a field where the proper acceleration is the same everywhere--see this page from John Baez's twin paradox page, which says that the time dilation of clocks at different heights in a "uniform" field would actually be different:

I don't see anything in contrast to my understanding of a uniform gravitational field: There it is also said that "...by "uniform" we mean that the force felt by each object is independent of its position." Or, equivalently, the proper acceleration is independent of position. This is a funadamental property of spacetimes if the EP and thus local inertia are taken into account.

And if you require that the grounds were all initially at rest in Rindler coordinates before beginning to fall (held up by individual rockets which are all turned off simultaneously in Rindler coordinates, for example), wouldn't this in fact imply different velocities in the inertial frame?

If the grounds were all initially at rest, they all would stay at the same level locally as the instantaneous observer being at rest does only keep being at rest just in a very short time dt and there everything will look like the Minkowski metric. But as long as time elapses much, the grounds start having their own levels as they experience different proper accelerations. If a bus hits a bump in the street, due to the rapid change of acceleration, your body may jump upwards and your handbag would then be in a different shape for moments of being in jump.

AB

This means that the observers have zero proper velocities momentarily.
How are you defining "proper velocity"? I haven't seen that term.
Altabeh said:
I don't see anything in contrast to my understanding of a uniform gravitational field: There it is also said that "...by "uniform" we mean that the force felt by each object is independent of its position." Or, equivalently, the proper acceleration is independent of position.
You're right, they do say that, but I'm confused about how that fits with what they said about time dilation...after all if you have a family of observers with the same proper acceleration, then in an inertial frame where they all begin to accelerate simultaneously, they will all have the same velocity as a function of time and thus their clocks won't get out of sync. So if they're talking about this type of acceleration rather than Born rigid acceleration, why then do they say this:
Say you have two identically constructed clocks. One is deep down in a uniform "gravitational" potential well (or "pseudo-potential", if you prefer); the other is higher up. If the two clocks compare rates by sending light signals back and forth, then both will agree that the lower clock runs slower than the higher clock.
Is it because although they don't get out of sync in the inertial frame, we're considering a non-inertial frame whose definition of simultaneity at each moment matches up with the definition of simultaneity in one of the observer's instantaneous inertial rest frame at that moment, so in the non-inertial frame the clocks are getting progressively further out-of-sync?

Also, if they all had the same proper acceleration, then in each observer's instantaneous inertial rest frame, the instantaneous velocity of the other observers would be nonzero, so if there were measuring-rods connecting the observers they would be physically stretching (experiencing changing internal stresses, as in the [URL [Broken] spaceship paradox[/url])...only Born rigid acceleration guarantees that each observer sees the other observers' instantaneous velocity as zero in their own instantaneous inertial rest frame (and thus the distance between neighboring observers doesn't change from one moment to the next if you consider their instantaneous inertial rest frame at each moment), correct? So if you want to draw an equivalence between a small accelerating lab and a small laboratory in a gravitational field where you can put up a ruler to measure the distance between the ceiling and floor and the ruler can have stable structure (its internal stresses aren't increasing until it breaks), don't you need to assume the accelerating lab is experiencing Born rigid acceleration rather than uniform proper acceleration at both the ceiling and floor? Or does it not really matter because you're only considering what would be measured in each lab during an infinitesimally brief span of time?
Altabeh said:
If the grounds were all initially at rest, they all would stay at the same level locally as the instantaneous observer being at rest does only keep being at rest just in a very short time dt and there everything will look like the Minkowski metric. But as long as time elapses much, the grounds start having their own levels as they experience different proper accelerations. If a bus hits a bump in the street, due to the rapid change of acceleration, your body may jump upwards and your handbag would then be in a different shape for moments of being in jump.
But you'd agree that if we transform from Rindler coordinates into an inertial frame, the grounds are all moving inertially, right?

Last edited by a moderator:
(edit: actually looking at the gray lines of simultaneity in the diagram above, it looks like q=0 in Rindler coordinates would give a line of simultaneity parallel to the t=0 plane (i.e. the x-axis) in the inertial Minkowski frame, and if the worldlines of the grounds at that moment are tangent to the worldlines of the accelerating observers at t=0, then all the worldlines of the grounds should just be vertical in the inertial frame, meaning they are all moving at the same velocity, so that idea of mine above probably isn't the right answer.)
Following the idea I suggested here, I drew in some vertical worldlines whose velocities were tangent to the worldines of some accelerating Rindler observers at t=0 in the inertial frame (which would also correspond to the q=0 simultaneity plane in Rindler coordinates, so the worldlines started out with 0 velocity at a single instant in Rindler coordinates): Here the blue lines represent worldlines of individual grounds at the "top" and "bottom" of the ball, the red lines connect events on these worldlines which are deemed simultaneous in Rindler coordinates. This diagram seems to show the distance in Rindler coordinates between the top and bottom actually shrinks rather than expands as the ball "falls" in the pseudo-gravitational field, since the first red line at q=0 extends from a middle Rindler observer to that observer's neighbors on either side, but the second red line at q=1 does not reach the simultaneous position of either observer to the side of the middle observer's worldine that the red line crosses through.

edit: Though from the discussion on the page I am not actually sure if the hyperbolas are meant to represent observers who are equally-spaced in Rindler coordinates themselves, it's possible that pairs of neighboring hyperbolas closer to the Rindler horizon have a greater coordinate distance in Rindler coordinates than pairs further from the Rindler horizon...the author doesn't specify if the s=1, s=2 etc. in the diagram actually represent Rindler position coordinates or something else, though this page does seem to indicate that lines of constant time-coordinate in Rindler coordinate look like straight lines that go through the origin as with the lines of constant q in the diagram. Also, Rindler coordinates might be similar to Schwarzschild coordinates in that nearby pairs of observers hovering at constant position coordinate, and with equal coordinate spacing between them, would actually measure a greater ruler distance if they were closer to the horizon than farther away...I know that in terms of ruler distances, it's actually possible to have an infinite series of observers hovering above the black hole horizon with equal distances to their neighbors, though the Schwarzschild coordinate distance between neighbors becomes smaller and smaller as you approach the horizon.

edit 2: scratch that last part, I just realized that since all the accelerating observers are initially at rest in the inertial frame, then if they are equally-spaced in the inertial frame, they'll also be equally-spaced at that moment according to ruler distance in their instantaneous rest frame, and observers at rest in Rindler coordinates have a ruler distance to one another that doesn't change over time.

Last edited:
snoopies622
...the distance in Rindler coordinates between the top and bottom actually shrinks rather than expands as the ball "falls" in the pseudo-gravitational field...

If I am understanding it correctly, the diagram on the right hand side of entry #13 in this thread

is consistent with this conclusion. "..the solid black lines are the worldlines of the inertial Minkowski observers", says Kev. The free-falling objects begin equally spaced from one another, then form hyperbolic secant curves, each asymptotic to the same line - the accelerated observer's event horizon.

So then, what are the fallacies of arguments 1 and 2 I made back in entry #14?

Last edited:
snoopies622
Regarding argument 1, it just occurred to me that

the proper (upward) acceleration experienced by a particle that is being held stationary in a Rindler coordinate system

and

the instantaneous (downward) acceleration observed of the same particle once it is released, that is - observed by someone who is also being held stationary in the Rindler coordinate system but may be at a different height -

are not necessarily the same thing.

Altabeh
How are you defining "proper velocity"? I haven't seen that term.

There is a couple of definitions for the velocity of an object in relativity: 1- coordinate velocity which is measured by an observer at rest and thus equals $$v=dx^{\alpha}/dt$$. 2- proper velocity relative to an observer divides observer-measured distance by the time elapsed on the clocks of the traveling object and thus equals $$v=dx^{\alpha}/d\tau.$$ When an instantaneously at rest observer is assumed, then the proper velocity is zero. This automatically suggests from $$dx^{\alpha}/d\tau \gamma^{-1}$$ that the coordinate velocity vanishes as well.

You're right, they do say that, but I'm confused about how that fits with what they said about time dilation...after all if you have a family of observers with the same proper acceleration, then in an inertial frame where they all begin to accelerate simultaneously, they will all have the same velocity as a function of time and thus their clocks won't get out of sync. So if they're talking about this type of acceleration rather than Born rigid acceleration,

This is correct! Since in an inertial frame all observers at rest move with a uniform velocity, then their proper accelerations are the same, thus the clocks will remain in sync!

Is it because although they don't get out of sync in the inertial frame, we're considering a non-inertial frame whose definition of simultaneity at each moment matches up with the definition of simultaneity in one of the observer's instantaneous inertial rest frame at that moment, so in the non-inertial frame the clocks are getting progressively further out-of-sync?

Two scenarios can be imagined:

1-It is just because they want to call, for example, Rindler coordinates fallaciously 'uniform' (or similarly inertial) and justify why such position-dependent field would make the time dilation find its meaning through the height difference!

2- They define some sort of non-inertial frame wherein they map all information at each moment into an observer's instantaneous inertial rest frame to only have the definition of simultaneity met in a possible way. But since the non-inertia of the first frame does not keep the travelling clock in sync with the observer's clock, then the time dilation occurs!

Both scenarios are flawed. The First one has this problem that a uniform field makes the proper acceleration stay at a constant everywhere so all clocks will experience the same time dilation at any point! This only works if the spacetime is Minkowski! The second one is contradictory: if the clocks are simultaneously set with an inertial observer's clock, then how does the property of "being in sync" from frame to frame stay in agreement with the fact that the non-inertial clocks are getting out of sync?

Also, if they all had the same proper acceleration, then in each observer's instantaneous inertial rest frame, the instantaneous velocity of the other observers would be nonzero, so if there were measuring-rods connecting the observers they would be physically stretching (experiencing changing internal stresses, as in the [URL [Broken] spaceship paradox[/url])...only Born rigid acceleration guarantees that each observer sees the other observers' instantaneous velocity as zero in their own instantaneous inertial rest frame (and thus the distance between neighboring observers doesn't change from one moment to the next if you consider their instantaneous inertial rest frame at each moment), correct?

Correct! But as bcrowell said, we don't have a spacetime that admits "uniform field" globally in GR. This can only be discussed along with a consideration borrowed from the equivalence principle that in a small region of spacetime the field can be uniform thus the local inertia and Born rigidity get resurrected again!

So if you want to draw an equivalence between a small accelerating lab and a small laboratory in a gravitational field where you can put up a ruler to measure the distance between the ceiling and floor and the ruler can have stable structure (its internal stresses aren't increasing until it breaks), don't you need to assume the accelerating lab is experiencing Born rigid acceleration rather than uniform proper acceleration at both the ceiling and floor? Or does it not really matter because you're only considering what would be measured in each lab during an infinitesimally brief span of time?

Of course it does not matter because as I earlier said, both the implications of "uniform acceleration" and "Born rigid acceleration" are the same in a small region of spacetime. But I recall that if you have a spacetime admitting a uniform gravitational field, then it is neccessary for proper accelerations to be position-independent and thus both the above accelerations will be the same! The only difference between the two pictures taken into account by Born and by the uniform gravitational field is that considering an instantaneously co-moving inertial rest frame in Born's picture is mandatory, but in the latter we already have the constant distances between particles everywhere.

But you'd agree that if we transform from Rindler coordinates into an inertial frame, the grounds are all moving inertially, right?

Of course! But remember that such a coordinate transformation changes the nature of spacetime, if considered globally true, and thus is completely artificial. In the equivalence principle, we do a metric transformation to reach the Minkowski metric in a small region, but this does not destroy the nature of spacetime because we have already intuited this in the real world that in small regions the flatness is approximately guaranteed.

AB

Last edited by a moderator:
JesseM said:
But you'd agree that if we transform from Rindler coordinates into an inertial frame, the grounds are all moving inertially, right?
Of course! But remember that such a coordinate transformation changes the nature of spacetime, if considered globally true, and thus is completely artificial.
Quick question, what do you mean by "changes the nature of spacetime"? The spacetime geometry is exactly the same for Rindler coordinates and Minkowski coordinates, both being defined on the same globally flat spacetime, agreed?

Altabeh
Quick question, what do you mean by "changes the nature of spacetime"? The spacetime geometry is exactly the same for Rindler coordinates and Minkowski coordinates, both being defined on the same globally flat spacetime, agreed?

Yes, but I have this challenge that the geodesic equations of Rindler metric depend on position, while those of Minkowski don't! So it sounds like if you do a coordinate transformation, you're changing spacetime in such a way that would affect the motion of particles! Right?

Altabeh
-- in a uniformly accelerating frame of reference - like inside a rocket moving with constant proper acceleration - points "lower" in the induced gravitational field experience greater acceleration than ones at "higher" locations

If the uniform acceleration means the field is position-independent as claimed in http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html" [Broken], then this is erroneous!

-- such a frame of reference is described by the Rindler metric

If we have an agreement on the answer given above, then no, because Rindler coordinates is position-dependent.

-- tidal forces are equivalent to curved spacetime

We have spacetimes wherein Weyl tensor vanishes (they are said to be of type O), but they are curved. Such spacetimes are said to be "conformally flat" which means by doing a conformal transformation the curved spacetime can be transformed into Minkowski.

-- the curvature tensor of the Rindler metric is zero

Correct! This is because $$g_{00}$$ is the only non-constant metric component and its second derivative vanishes.

AB

Last edited by a moderator:
Yes, but I have this challenge that the geodesic equations of Rindler metric depend on position, while those of Minkowski don't! So it sounds like if you do a coordinate transformation, you're changing spacetime in such a way that would affect the motion of particles! Right?
I would say that's not really changing spacetime, just changing how you label points in spacetime. For me the spacetime "in itself" is just the spacetime geometry defined by the proper time along any possible path (which you get from the equations of the metric as expressed in your coordinate system). And for any possible path the Rindler metric and the Minkowski metric agree on its proper time (assuming you are using the transformation between Rindler coordinates and Minkowski coordinates to map physically identical points in each system, so you can decide what the 'same path' looks like in each system), so normally I think physicists would say they are the "same metric" just expressed in different coordinate systems. In contrast if you pick something like the Schwarzschild metric expressed in Schwarzschild coordinates, there's no possible coordinate transformation that would map points in Schwarzschild coordinates to points in Minkowski coordinates such that the Schwarzschild metric and the Minkowski metric would agree on the proper time of any "same path" defined by the mapping, so this is why the Schwarzschild metric and the Minkowski metric are viewed as physically different metrics describing different spacetime geometries.

Last edited:
Altabeh
I would say that's not really changing spacetime, just changing how you label points in spacetime. For me the spacetime "in itself" is just the spacetime geometry defined by the proper time along any possible path (which you get from the equations of the metric as expressed in your coordinate system). And for any possible path the Rindler metric and the Minkowski metric agree on its proper time (assuming you are using the transformation between Rindler coordinates and Minkowski coordinates to map physically identical points in each system, so you can decide what the 'same path' looks like in each system), so normally I think physicists would say they are the "same metric" just expressed in different coordinate systems. In contrast if you pick something like the Schwarzschild metric expressed in Schwarzschild coordinates, there's no possible coordinate transformation that would map points in Schwarzschild coordinates to points in Minkowski coordinates such that the Schwarzschild metric and the Minkowski metric would agree on the proper time of any "same path" defined by the mapping, so this is why the Schwarzschild metric and the Minkowski metric are viewed as physically different metrics describing different spacetime geometries.

Two points come to my mind here:

1-You have to discuss the Rindler metric apart from any other metric e.g. Minkowski. We can make any spacetime inertial locally and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally. But the result is non-sense in case the initial metric has non-constant components!!

2- A coordinate transformation is not allowed to be carried out on a given metric if it makes the gravitational and inertial accelerations go away in the geodesic equations unless it is just applied in a small region where EP holds! For example, I can make the Schwartschild metric be flat at some point and to the highest allowed degree i.e. in a small region this is approximately true! But doing this thing globally kills the second terms in the geodesic equations, leaving us with Minkowski spacetime.

Even if two metrics are of the same nature, but since the geodesic equations aren't the same you can't apply a global transformation in such a way that it allows you to get rid of gravitational fields present in the initial metric! If the transformation was between two flat spacetimes with only constant metric components, then you would do this!

AB

snoopies622
This diagram seems to show the distance in Rindler coordinates between the top and bottom actually shrinks rather than expands as the ball "falls" in the pseudo-gravitational field..

2. Since the Wely tensor and the Ricci tensor of the Rindler metric are both zero, the sphere must maintain its shape and size.

So what is the resolution to this apparent contradiction?

1-You have to discuss the Rindler metric apart from any other metric e.g. Minkowski.
Again, I think it would be common for physicists to say that they are the "same metric", just expressed in different coordinate systems. For example, googling "same metric" + relativity turns up this Google Books result from A First Course in General Relativity by Bernard Schutz, in which he writes:
For instance, suppose we have a stationary gravitational field. Then a coordinate system can be found in which the metric components are time independent ... Notice that coordinates can also be found in which the same metric has time-dependent components: any time-dependent coordinate transformation from the 'nice' system will do this.
In a footnote at the bottom of the page he also says "It is easy to see that there is generally no coordinate system which makes a given metric time independent", so again he is using 'given metric' to refer to a geometric entity which can be expressed using different equations in different coordinate systems.
Altabeh said:
We can make any spacetime inertial locally
What do you mean by that exactly? As I understand it the equivalence principle really only applies exactly in an infinitesimal region of curved spacetime, and even there the curvature tensor will not actually be zero as I understand it...there was some discussion of the subtleties of defining the equivalence principle on this thread and this one.
Altabeh said:
and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally.
Why does that make it "not logical"? Do you get any incorrect predictions about physical results? I would say that every set of metric equations describing a globally flat spacetime is the same metric describing the same spacetime, just expressed in different coordinate systems.
Altabeh said:
But the result is non-sense in case the initial metric has non-constant components!!
Why is it nonsense that a single spacetime geometry (representing a single set of predictions about all coordinate-independent physical measurements like proper time readings at particular events) can be expressed in terms of different metric equations in different coordinate systems, one having constant components and the other having non-constant ones? That's exactly the situation that was described above by Bernard Schutz, where he talked about the "same metric" having time independent components in one coordinate system and time-dependent components in a different one.
Altabeh said:
2- A coordinate transformation is not allowed to be carried out on a given metric if it makes the gravitational and inertial accelerations go away in the geodesic equations unless it is just applied in a small region where EP holds! For example, I can make the Schwartschild metric be flat at some point and to the highest allowed degree i.e. in a small region this is approximately true! But doing this thing globally kills the second terms in the geodesic equations, leaving us with Minkowski spacetime.
I'm certain you can't find a coordinate transformation between Schwarzschild and Minkowski coordinates such that, when you transform the Schwarzschild metric (expressed in Schwarzschild coordinates) using this coordinate transformation, you get back the Minkowski metric (expressed in Minkowski coordinates). Are you claiming this is possible? If you agree it's not possible, and if you also agree that a mere coordinate transformation isn't going to change your physical predictions about any coordinate-independent facts like proper time along worldlines, then I don't understand what you mean when you say certain coordinate transformations are "not allowed"...do you have a reference to a paper or textbook which states whatever rule you're suggesting here?
Altabeh said:
Even if two metrics are of the same nature, but since the geodesic equations aren't the same you can't apply a global transformation in such a way that it allows you to get rid of gravitational fields present in the initial metric!
Depends if you define "gravitational field" in terms of actual spacetime curvature (which seems to be the common practice among physicists nowadays) or if you include what this section of the twin paradox page calls "pseudo-gravitational fields". Under the first definition, no gravitational field is present anywhere in the Rindler metric.
Altabeh said:
If the transformation was between two flat spacetimes with only constant metric components, then you would do this!
The Rindler metric does describe a flat spacetime, even if the coordinate system it's expressed in includes non-constant components.

Altabeh
Again, I think it would be common for physicists to say that they are the "same metric", just expressed in different coordinate systems. For example, googling "same metric" + relativity turns up this Google Books result from A First Course in General Relativity by Bernard Schutz, in which he writes:

In a footnote at the bottom of the page he also says "It is easy to see that there is generally no coordinate system which makes a given metric time independent", so again he is using 'given metric' to refer to a geometric entity which can be expressed using different equations in different coordinate systems.

Did you find that footnote different from my point here? He is trying to simply say, for instance, the time-dependent scale factor in the FRW metric can't disappear via any coordinate transformation. This is right and I agree to it! But let me give you an example of what point I'm trying to make here.

First of all, I define the conditions that describe the qualities of two, say, flat metrics with certain identical signatures and belonging to the same family in my view:

1- Both metrics must have vanishing Riemann tensors,
2- In both metrics the particles moving on geodesics must have the same dynamical quiddity. I.e. if in one of them the proper acceleration is position-dependent, then so is in the other metric!

Consider the metric

$$ds^2 = dt^2 - dx^2 = dT^2 + 2X^2dTdX- (1-X^4)dX^2.$$

For this metric all Christoffel symbols vanish and thus it is an inertial frame for which and the Minkowski metric both conditions meet and thus they belong to the same family (so far it is obvious that they are actually the same by the use of the change of coordinates $$T = t - x^3/3, X = x$$.)

But for the Rindler metric the Christoffel symbols do not vanish thus this metric gets out of the family of the Minkowski metric. It may be weird that while these two are basically the same, but they act dynamically differently and this gives rise to why I insist giving up on your idea as a solution to getting constant proper acceleration.

What do you mean by that exactly?

See https://www.physicsforums.com/showpost.php?p=2599439&postcount=63" to know my viewpoint on the equivalence principle.

Do you get any incorrect predictions about physical results? I would say that every set of metric equations describing a globally flat spacetime is the same metric describing the same spacetime, just expressed in
different coordinate systems.

Not exactly the same spacetime in some cases. Spacetime by itself is nothing but a set of events; so I assume you also take into account the dynamics of particles in the "spacetime" as a part of it to give a physical meaning to these events which undergo effects of gravitational fields and curvature.

Why does that make it "not logical"?

Here I mean the statement is vacuously true. As long as you are ignoring the dynamics of particles in the entire spacetime, the result from a mathematical standpoint is correct though it suffers from the "change" of dynamics globally. It looks like you're applying the EP throughout the spacetime.

Why is it nonsense that a single spacetime geometry (representing a single set of predictions about all coordinate-independent physical measurements like proper time readings at particular events) can be expressed in terms of different metric equations in different coordinate systems, one having constant components and the other having non-constant ones?

Can you imagine the EP to occur in a very large region even in a flat spacetime e.g. Rindler spacetime? The geodesic equations are strongly coordinate-dependent in Rindler metric and of course in the cases of gravitational field making the geodesic equations coordinate-dependent, time dilation in turn would depend on position.

Still continues...

AB

Last edited by a moderator:
Did you find that footnote different from my point here?
Yes, I think it's different from your statement "You have to discuss the Rindler metric apart from any other metric e.g. Minkowski". Schutz would presumably say that the Rindler metric and the Minkowski metric are the "same metric" since either one can be obtained by a coordinate transformation on the other.
Altabeh said:
He is trying to simply say, for instance, the time-dependent scale factor in the FRW metric can't disappear via any coordinate transformation. This is right and I agree to it!
I agree as well, since the FRW metric is not a stationary one. But he is also saying that for any metric whose components are time-independent in one coordinate system, you can always transform into another coordinate system where the components are time-dependent, yet it is still the "same metric" in both cases--do you agree?
Altabeh said:
First of all, I define the conditions that describe the qualities of two, say, flat metrics with certain identical signatures and belonging to the same family in my view
It seems like you're using "same family" to discuss something different than the ordinary notion of spacetime geometry, though. Consider differential geometry of ordinary 2-dimensional spatial surfaces, like a sphere and a plane. On a given surface, you can place different coordinate systems and then the equations of the metric describing the curvature of the surface will be different. Still, geometrically a sphere is a sphere and a plane is a plane--you can define the difference between the two in intrinsic terms by talking about things like geodesic length-minimizing paths and the sum of the angles of a triangle formed by three geodesics crossing at three points. So, there is an equivalence class of metric equations which all describe a spherical geometry and thus would all count as the "same metric" in Schutz's terms, just expressed in different coordinate systems on a sphere...likewise there is an equivalence class of metric equations which all describe a planar Euclidean geometry.

Do you agree that this notion of an equivalence class of metrics which which describe the same underlying geometry is meaningful in GR as well, and that this is presumably what Schutz meant when he talked about the "same metric" for a stationary gravity field having either time-dependent or time-independent components depending on what coordinate system it was written down in? And if so, do you agree that in this geometrical sense the Rindler metric is the same metric as the Minkowski metric? Finally, if you are using "same family" to describe a different equivalence class than the geometrical one I'm talking about, is this one that you've come up with yourself or one that you've seen discussed in textbooks or papers? What utility would this alternate equivalence class have?
JesseM said:
Do you get any incorrect predictions about physical results? I would say that every set of metric equations describing a globally flat spacetime is the same metric describing the same spacetime, just expressed in
different coordinate systems.
Altabeh said:
Not exactly the same spacetime in some cases. Spacetime by itself is nothing but a set of events; so I assume you also take into account the dynamics of particles in the "spacetime" as a part of it to give a physical meaning to these events which undergo effects of gravitational fields and curvature.
By "same spacetime" I was referring to this notion of the same spacetime geometry expressed in different coordinate systems (along with the same physical events in both coordinate systems of course). Hopefully you agree this is a meaningful notion?
JesseM said:
Why does that make it "not logical"?
Altabeh said:
Here I mean the statement is vacuously true.
What statement is vacuously true? The original "not logical" statement of yours was "We can make any spacetime inertial locally and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally."
Altabeh said:
As long as you are ignoring the dynamics of particles in the entire spacetime, the result from a mathematical standpoint is correct though it suffers from the "change" of dynamics globally. It looks like you're applying the EP throughout the spacetime.
What result from a mathematical standpoint is correct? And how am I "applying the EP"? I'm just saying that the Rindler metric equation and the Minkowski metric equation describe the same spacetime geometry (and would be the 'same metric' in different coordinate systems according to the way most physicists like Schutz seem to talk about things), and as such if you use either metric to make predictions about local coordinate-independent physical facts like what readings two clocks show when they pass next to one another, you'll get identical predictions. On the other hand if you used a metric equation that did not fall into the same equivalence class of describing a flat spacetime geometry, like the Schwarzschild metric, then there would be no possible coordinate transformation such that you'd get all the same physical predictions about coordinate-independent local facts if you used the Schwarzschild metric in Schwarzschild coordinates vs. the Minkowski metric in Minkowski coordinates. Thus, this notion of an equivalence class of metric equations describing the "same geometry" is a very useful one physically, since it distinguishes between cases where using different metric equations makes no difference at all in terms of any of your coordinate-independent physical predictions, and cases where it does make a difference.

Can you please tell me if you disagree with any of the above statements, and if so quote the specific ones you disagree with and explain why?
Altabeh said:
Can you imagine the EP to occur in a very large region even in a flat spacetime e.g. Rindler spacetime?
Depends how you define the EP I suppose, but certainly it would be possible to construct a "very large" system of rulers and clocks which are free-falling in the pseudo-gravitational field in Rindler spacetime, such that if you use this system to define a new coordinate system, then the equations of laws of physics as defined this coordinate system will be exactly the same as those seen in an SR inertial frame (in fact this free-falling frame is just an SR inertial frame as viewed in Rindler coordinates). Again, please tell me if you disagree with this.

Last edited:
Altabeh
Yes, I think it's different from your statement "You have to discuss the Rindler metric apart from any other metric e.g. Minkowski". Schutz would presumably say that the Rindler metric and the Minkowski metric are the "same metric" since either one can be obtained by a coordinate transformation on the other.

You seem to not even pay a little attention towards me. These two spacetimes are NOT the same everywhere. The Rindler line-element can be used as a line-element for a flat spacetime but we have circumferences under which Rindler does not reduce to Minkowski and this arises from

1- the existence of a horizon in Rindler metric at $$x=0$$,
2- the so-called "uniform" motion of different observers relative to each other.

The second statement says that the Rindler metric is in essence the Minkowski spacetime in "expanding coordinates" but one cannot account for it being the same as the Minkowski spacetime at any arbitrary point. Besides this, since the dynamical behavior, or the way geoemetry (gravitational field) allows particles to move, in the two metrics are different, you should be careful when applying an inverse transformation to get Minkowski from Rindler because, for instance, the points $$(t,0,y,z)$$ are not contained in the Rindler metric and for this reason I can say Rindler spacetime is a compactification of the Minkowski spacetime assuming x to be chosen from $$\textbf{R}-{0}$$. Thus such coordinate system is not equivalent to Minkowski everywhere. Agreed?

I agree as well, since the FRW metric is not a stationary one. But he is also saying that for any metric whose components are time-independent in one coordinate system, you can always transform into another coordinate system where the components are time-dependent, yet it is still the "same metric" in both cases--do you agree?

...is the same metric if no event horizon or a point appears where the metric acts irregularly.

It seems like you're using "same family" to discuss something different than the ordinary notion of spacetime geometry, though. Consider differential geometry of ordinary 2-dimensional spatial surfaces, like a sphere and a plane. On a given surface, you can place different coordinate systems

Haven't you ever herad they use the words "bad" and "good" to call coordinate systems? A bad coordinate system is one whose determinant vanishes at some point and the good one has a non-vanishing determinant everywhere. As for a bad coordinate system, I can name the spherical coordinates for which $$r=0$$ and $$\theta = \pi$$ generate singularity. For the Schwartschild metric, yet such singularities are present along with another coordinate singularity on the hyperplane $$2m=r$$. Due to this reason, we can't introduce an orthogonal metric basis that make the metric flat everywhere but only at points with properties mentioned above.

and then the equations of the metric describing the curvature of the surface will be different.

If a flat spacetime is described by another coordinate system, yet the curvature tensor must vanish. This is correct and I don't seem to lay my finger on it, but rather on the statement that in all these coordinates the metric remains ALWAYS the same and I think I gave a vivid example above.

Still, geometrically a sphere is a sphere and a plane is a plane--you can define the difference between the two in intrinsic terms by talking about things like geodesic length-minimizing paths and the sum of the angles of a triangle formed by three geodesics crossing at three points. So, there is an equivalence class of metric equations which all describe a spherical geometry and thus would all count as the "same metric" in Schutz's terms, just expressed in different coordinate systems on a sphere...likewise there is an equivalence class of metric equations which all describe a planar Euclidean geometry.

Yet this does not make the Rindler spacetime be the same as Minkowski everywhere!! Why don't you want to agree that the Rindler spacetime does not cover the whole of the Minkowski spacetime and it of course has a horizon and different dynamics? If you exclude this last thing, I would sort of agree with you! But This is physics and we have to take into account everything related to it!

Do you agree that this notion of an equivalence class of metrics which which describe the same underlying geometry is meaningful in GR as well, and that this is presumably what Schutz meant when he talked about the "same metric" for a stationary gravity field having either time-dependent or time-independent components depending on what coordinate system it was written down in?

Ah, are you interested in asking long question?! Yes, I agree that he is just talking about "equivalence class" of metrics via introducing coordinate systems that do transform the metric into other possible forms. But as I talked about earlier, in GR the curved Riemannian spacetimes have crucial bahaviour when it comes to the transformation! Almost every curved spacetime is only able to be transformed into Minkowski spacetime through an appropriate metric transformation (not coordinate transformation) which is only locally applied as in the case of Schwarzschild metric with Jaccobi coefficients

$$e_0^0=\frac{1}{c\sqrt{1-2m/r}},$$
$$(e_1^1)^2=-(1-2m/r),$$
$$(e_2^2)^2=-\frac{1}{r},$$
$$(e_3^3)^2=-\frac{1}{r\sin(\theta)}.$$

Such metric transformation occurs just locally at some given point P.

And if so, do you agree that in this geometrical sense the Rindler metric is the same metric as the Minkowski metric?

I agree due to reason I'm giving below!

Finally, if you are using "same family" to describe a different equivalence class than the geometrical one I'm talking about, is this one that you've come up with yourself or one that you've seen discussed in textbooks or papers? What utility would this alternate equivalence class have?

The most known classification of spacetimes is the Petrov classification and of course other much general classifications (for instance, some classify the spacetimes into categories with different signatures -Lorentzian, Pseudo Riemannian and etc.- by which the metric itself is mostly targeted) exist in the advanced textbooks and articles (just google them!). In the Petrov classification, both Rindler and Minkowski spacetimes have vanishing Riemann tensor so they are of type O thus in the same class! But such classification does not zoom into the dynamics of spacetime as they just look at the Weyl tensor and if it was zero (if the Ricci tensor vanishes so does the Weyl tensor), then the spacetime is of type O. And I think I told you that such classification is accepted at least in my own view.

By "same spacetime" I was referring to this notion of the same spacetime geometry expressed in different coordinate systems (along with the same physical events in both coordinate systems of course). Hopefully you agree this is a meaningful notion?

Yes, of course

What statement is vacuously true? The original "not logical" statement of yours was "We can make any spacetime inertial locally and of course it is not logical to apply a transformation that is going to map all points in the flat spacetime into the Minkowski metric because we can do this to every flat spacetime globally."

This actually refers to the fact that every coordinate transformation has its own "range of working" as in the Schwarzschild metric, you can't apply any coordinate transformation at essential singularities. Transforming coordinates blindly throughout the spacetime is not logical because we may be doing this at a point where the initial metric has a singularity there as in the above case!

However, I just found out that the singularity of Rindler metric is not essential and can be removed by defining the coordinates $$(u,v)$$

$$u=t-\ln(x),$$
$$v=t+\ln(x).$$

Now we get the differential of each coordinate:

$$du=dt-dx/x,$$
$$dv=dt+dx/x.$$

Multiplying these two equations by each other and rearranging terms give

$$x^2dt^2-dx^2=x^2dudv=e^{v-u}dudv,$$

where $$- \infty<u,v< \infty$$ and we used the fact that $$x=e^{v-t}$$ and $$x=e^{t-u}$$.

Finally introducing $$U=e^{-u}$$ and $$V=e^v$$ yields

$$ds^2=dUdV,$$

for the Rindler metric with $$0<U,V< \infty$$. But as we can see, there is no longer any singularity at $$U=0$$ or $$V=0$$. This suggests that we may now extend the spacetime by allowing $$V$$ and $$U$$ to be unrestricted and by applying a transformation of the form $$X=(V-U)/2$$, $$T=(V+U)/2$$ in this last line-element we get the Minkowski metric which is basically the extended spacetime not the Rindler metric itself! You got it?
What result from a mathematical standpoint is correct? And how am I "applying the EP"?

If you're applying a transformation over the entire manifold to make it flat, this is may be correct from a mathematical viewpoint, but since the dynamics of spacetime may also change, it is not true from a physical point of view because it resembles the situation where we apply EP throughout the spacetime! There is a long discussion in this https://www.physicsforums.com/showthread.php?t=377254" that is about answering a fundamental question: Do in a Rindler metric particles travelling along timelike geodesics experience a constant acceleration? The answer as given by bcrowell is yes only if we take instantaneously at rest particles with a comoving instantaneous observer with an ideal clock whose hands are not affected by gravitational field. Such assumption is made after battling with hardships that one needs to go through if he seeks out this property in the Rindler metric, not in the Minkowski metric where you arrive at through a simple coordinate transformation!!

I'm just saying that the Rindler metric equation and the Minkowski metric equation describe the same spacetime geometry (and would be the 'same metric' in different coordinate systems according to the way most physicists like Schutz seem to talk about things), and as such if you use either metric to make predictions about local coordinate-independent physical facts like what readings two clocks show when they pass next to one another, you'll get identical predictions.

I think it is now clear that the Minkowski spacetime is the extended Rindler metric! Agreed?

On the other hand if you used a metric equation that did not fall into the same equivalence class of describing a flat spacetime geometry, like the Schwarzschild metric, then there would be no possible coordinate transformation such that you'd get all the same physical predictions about coordinate-independent local facts if you used the Schwarzschild metric in Schwarzschild coordinates vs. the Minkowski metric in Minkowski coordinates.

Are you speaking locally? Or globally? You should specify this first! Any spacetime, as EP says, can be made flat approximately in a small region and can be made flat exactly at some given point! All the flat metrics are of type O, and fall into the same family but unfortunately there are other metrics within this class that are not flat, but conformally flat.

Thus, this notion of an equivalence class of metric equations describing the "same geometry" is a very useful one physically, since it distinguishes between cases where using different metric equations makes no difference at all in terms of any of your coordinate-independent physical predictions, and cases where it does make a difference.

It is sort of useful. But it does not tell everything about the metrics and thus it can't hold a candle to the Petrov classification.

Depends how you define the EP I suppose, but certainly it would be possible to construct a "very large" system of rulers and clocks which are free-falling in the pseudo-gravitational field in Rindler spacetime, such that if you use this system to define a new coordinate system, then the equations of laws of physics as defined this coordinate system will be exactly the same as those seen in an SR inertial frame (in fact this free-falling frame is just an SR inertial frame as viewed in Rindler coordinates). Again, please tell me if you disagree with this.

I disagree! In the Rindler metric the situation is the same as any other metric in which the EP is defined only within its common ranges, a small region where we can approximately make the spacetime flat and dreaming of a "very large" region as for the EP to hold in is just impossible because the geodesic equations are position-dependent!!

AB

Last edited by a moderator: