1. Mar 8, 2010

### 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.

2. Mar 8, 2010

### bcrowell

Staff Emeritus
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 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 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: Apr 24, 2017
3. Mar 8, 2010

### Mentz114

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.

4. Mar 8, 2010

### 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.

5. Mar 8, 2010

### snoopies622

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

6. Mar 8, 2010

### bcrowell

Staff Emeritus
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: May 4, 2017
7. Mar 8, 2010

### 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 ?

8. Mar 8, 2010

### bcrowell

Staff Emeritus
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.

9. Mar 8, 2010

### bcrowell

Staff Emeritus
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: Mar 8, 2010
10. Mar 8, 2010

### 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.

11. Mar 8, 2010

### bcrowell

Staff Emeritus
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.

12. Mar 8, 2010

### JesseM

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.

13. Mar 9, 2010

### Mentz114

I don't know. Can there be free-fall in this scenario ? The coffee-grounds are being individually rocket propelled.

14. Mar 9, 2010

### 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.

15. Mar 9, 2010

### JesseM

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?

16. Mar 9, 2010

### Mentz114

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.

17. Mar 9, 2010

### Altabeh

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

18. Mar 9, 2010

### JesseM

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.
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):

19. Mar 9, 2010

### JesseM

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.
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:
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: Mar 9, 2010
20. Mar 9, 2010

### Altabeh

This means that the observers have zero proper velocities momentarily.

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.

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

21. Mar 9, 2010

### JesseM

How are you defining "proper velocity"? I haven't seen that term.
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:
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?
But you'd agree that if we transform from Rindler coordinates into an inertial frame, the grounds are all moving inertially, right?

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22. Mar 9, 2010

### JesseM

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: Mar 9, 2010
23. Mar 10, 2010

### snoopies622

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: Mar 10, 2010
24. Mar 10, 2010

### 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.

25. Mar 10, 2010

### Altabeh

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.

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!

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?

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!

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.

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

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