Inconsistent forms of the metric in a uniform field

[Altabeh]

Ben,

[aside]
Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction?

Not in any direction! Let's say along a time-like geodesic, for instance, because acceleration remains constant and the coffee never splashes!

So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart?

Correct if we believe in the equivalence principle and the locally flatness!

I suspect that the metric above has these properties but I'm too sleepy now to work it out.

Both can be found in the above metric because it doesn't admit a gravitational field and so it is flat!

 

[yuiop]

Also, Rindler coordinates are not a uniform gravitational field (at least that's what Rindler's text claims).

I think this is right because if I recall correctly, the acceleration in Rindler coordinates is proportional to 1/x.

... These effects are the ones that are involved in Bell's spaceship paradox...

I think Ben has touched on an important aspect here and I think the physical implications of the effects should be made clear. If we take an array of accelerating rockets in flat spacetime, all with equal proper acceleration in the positive x direction, such that there mutual spatial separation remains constant according to an inertial Minkowski observer, then the mutual spatial separation of the rockets is continually increasing in the x direction, according to the accelerating rocket observers. By the EP, this implies that observers at rest in a uniform gravitational field will see distances constantly increasing as measured by radar signals and any physical rulers that attempt to remain at rest with the uniform gravitational field will be torn apart. This makes defining distance or even defining what "at rest" means very difficult because distance is always changing over time. Certainly, we can not create a physical grid of rulers and clocks in such a space-time.

Also, could someone make it clear if by "uniform gravitational field" we mean a uniform non-zero proper acceleration in a one dimensional way, such as along the x axis, or if we mean uniform non-zero proper acceleration in all 3 directions? I guess the latter would be represented by the FLRW metric on the cosmological scale.

In an old thread we discussed the gravitational fields of some hypothetical objects and I think concluded that an infinitely long cylindrical massive object would have an external gravitational field that varied proportional to 1/r and that a thin planar gravitational mass extending infinitely in the y and z directions would have an external field that is independent of x.

As far as I can tell, the term "uniform gravitational field" is used in the context where we consider a small enough volume of space-time that deviations from uniformity are negligible and this is a local approximation.

[aside]
Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction ?

From the above considerations, the ball of coffee grounds will not change shape in a uniform field, according to an inertial observer that is co-free-falling with the coffee grounds and will be expanding according to an accelerating observer that is at rest with gravitational field.

So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart ?

Yes, (if the observer is released at the same time). Presumably, simultaneity is not an issue here, because clocks at different heights in a uniform gravitational field run at the same rate.

I suspect that the metric above has these properties but I'm too sleepy now to work it out.

All out of coffee-grounds?

 

[bcrowell]

In an old thread we discussed the gravitational fields of some hypothetical objects and I think concluded that an infinitely long cylindrical massive object would have an external gravitational field that varied proportional to 1/r and that a thin planar gravitational mass extending infinitely in the y and z directions would have an external field that is independent of x.

This is only true in the weak-field limit, if the source isn't rotating. Under those circumstances, you can just find the gravitational field with Gauss's law, and you don't need relativity.

The Petrov solution can be interpreted as the exterior field of an infinitely long cylinder of dust that rotates rigidly at \omega =c/R, where R is the radius of the cylinder. Because it's a strong field, you can't find the field using Gauss's law. The external region has a constant scalar curvature everywhere, so there definitely isn't a 1/r field.

Can we can define 'uniform' in a gravitational field as meaning a ball of coffee-grounds in free-fall will not change shape at all, in any direction ? So neighbouring geodesics stay parallel and two particles released simultaneously wrt to an observer half-way between them but from different 'heights' stay the same distance apart ?

I think that's equivalent to saying that there is no tidal curvature, i.e., a vanishing Weyl tensor. If we required that, then the Petrov metric would be disqualified, because it has a nonzero Weyl tensor.

If we want it to be a vacuum solution, then we need the Ricci curvature to vanish. If we demand that the Weyl tensor vanish as well, then we're requiring the space to be flat -- only a flat space has both a vanishing Weyl tensor and a vanishing Ricci tensor. If you want a flat space described in a uniformly accelerating frame, then Rindler coordinates do that, but as we've seen, they don't have all the properties you'd want in a uniform field.

If you change the ball of coffee grounds into two spaceships connected by a thread, then you have Bell's spaceship paradox.

The thing that I'm currently trying to understand about the Petrov metric is how you would define a proper acceleration in it. In e.g., Rindler coordinates this is fairly simple. You define a hovering observer as one whose coordinate velocity is zero, and then that observer releases a test particle initially at rest, and observes its acceleration relative to her. A couple of problems come up when you try to use this definition in the Petrov metric. One is that you have two different coordinates that can play the role of time, depending on r. The metric is stationary but not static, so there is a preferred time coordinate, but this preferred time coordinate is not \phi or t. The other problem is that I'm not even sure material particles can have a coordinate velocity of zero.

I want to play around with this stuff this weekend by doing numerical simulations of geodesics.

 

[bcrowell]

Wiki shows how in the Rindler chart, the Minkowski line element becomes
ds^2=-a^2x^2+dx^2+dy^2+dz^2.
Repeating the calculation I did previously with this I get \ddot{x}=-a.

Which is independent of x. The line element is a coordinate transformation from Minkowski space, but the original space and time have been mixed, so the x coordinate is not pure space. I suppose one could consider the line-element to be the result of some source.

For the line element, I assume you meant
ds^2=-a^2x^2dt^2+dx^2+dy^2+dz^2
(with a dt2 in the first term).

This confuses me, because a^2x^2 is the same as (1+ax)^2 except for adding a constant to the x coordinate. So I don't see how the proper acceleration can come out to be independent of x in the first case, but dependent on x in the second case.

 

[bcrowell]

OK, re Lut's #48, I calculated the Christoffel symbols for both metrics, using Maxima in an attempt to avoid computational errors.

load(ctensor);
dim:2;
ct_coords:[t,x];
lg:matrix([(1+a*x)^2,0],
          [0,-1]);
cmetric();
christof (lcs);

This gives \Gamma\indices{^x_{tt}}=-a(1+ax).

load(ctensor);
dim:2;
ct_coords:[t,x];
lg:matrix([(a*x)^2,0],
          [0,-1]);
cmetric();
christof (lcs);

This gives \Gamma\indices{^x_{tt}}=-a^2x.

In both cases, the Christoffel symbol can be interpreted as the instantaneous value of d^2x/d\tau^2, where x is the position of a particle that has an instantaneous coordinate velocity of zero, and \tau is the proper time of that particle.

Both of these results for the acceleration vary with x, and they're consistent with each other when you do the coordinate transformation x \rightarrow x \pm 1/a.

I think this is all consistent with the time-dilation interpretation I gave in #47.

 

[Altabeh]

I think Ben has touched on an important aspect here and I think the physical implications of the effects should be made clear. If we take an array of accelerating rockets in flat spacetime, all with equal proper acceleration in the positive x direction, such that there mutual spatial separation remains constant according to an inertial Minkowski observer, then the mutual spatial separation of the rockets is continually increasing in the x direction, according to the accelerating rocket observers. By the EP, this implies that observers at rest in a uniform gravitational field will see distances constantly increasing as measured by radar signals and any physical rulers that attempt to remain at rest with the uniform gravitational field will be torn apart. This makes defining distance or even defining what "at rest" means very difficult because distance is always changing over time. Certainly, we cannot create a physical grid of rulers and clocks in such a space-time.

Weiss's idea is flawed physically, as already was discussed, since it suffers lack of a globally compatible definition of constant proper acceleration along geodesics due to dependency on spatial coordinates so that making use of the idea of "instantaneously at rest" particles and instantaneous co-moving observers to overcome this problem for such a curved metric is like you are trying to patch a huge rupture with a highly unphysical tool (ideal clock) around a gravitational field which naturally isn't uniform!

Also, could someone make it clear if by "uniform gravitational field" we mean a uniform non-zero proper acceleration in a one dimensional way, such as along the x axis, or if we mean uniform non-zero proper acceleration in all 3 directions? I guess the latter would be represented by the FLRW metric on the cosmological scale.

In all three dimensions! Here we have been talking about Rindler metric in 2 dimensions, so the second axis would be x or y or z along which the uniformity of gravitational field is considered. The FLRW metric is not static, but dependent on time if we take all coordinates to be fixed and pin the changes to time only which enters in the metric by scale factor, so particles moving on geodesics will either accelerate or decelerate according to which epoch the cosmos is getting through: the epoch at which the universe contracts, or at the one it expands as we see it right now! This is simply seen, for example, for radial motion that the popper acceleration depends on t, so it is not constant at all! Furthermore, FLRW metric is curved and it does not admit any uniform gravitational field everywhere but locally, as expected.

In an old thread we discussed the gravitational fields of some hypothetical objects and I think concluded that an infinitely long cylindrical massive object would have an external gravitational field that varied proportional to 1/r and that a thin planar gravitational mass extending infinitely in the y and z directions would have an external field that is independent of x.

In the case of a weak gravitational field, this can be read off the Poisson equation directly. But if the central or self-gravitating body has a very strong gravity, then this won't hold until one says that spacetime is not closed and at sufficiently large distances, the field starts getting closer to zero in proportionality with 1/r as in the asymptotically flat spacetimes.

As far as I can tell, the term "uniform gravitational field" is used in the context where we consider a small enough volume of space-time that deviations from uniformity are negligible and this is a local approximation.

Completely right! This is because a "uniform gravitational field" within a spacetime makes all points undergo the same changes geometrically and proper accelerations of particles moving along paths change point-by-point equivalently so this means that in order to have such a field, the (time-like) geodesics must be independent of spatial coordinates (in the case of static metrics) which cannot be seen in any curved spacetime throughout the whole manifold except in a very small region of it!

AB

 

[bcrowell]

Also, could someone make it clear if by "uniform gravitational field" we mean a uniform non-zero proper acceleration in a one dimensional way, such as along the x axis, or if we mean uniform non-zero proper acceleration in all 3 directions? I guess the latter would be represented by the FLRW metric on the cosmological scale.

What I have in mind personally is one that has uniformity in all three dimensions. I don't think an FLRW metric qualifies, because an object released with a coordinate velocity of zero (in the usual coordinates) maintains a coordinate velocity of zero indefinitely.

As far as I can tell, the term "uniform gravitational field" is used in the context where we consider a small enough volume of space-time that deviations from uniformity are negligible and this is a local approximation.

Any field is uniform in this weaker local sense. IMO the only thing that makes the problem interesting and hard is that it's so hard to find anything that looks *globally* like a uniform field in GR. The Petrov metric is globally uniform (although it has lots of other strange properties that make it not necessarily the global solution that we'd like to say is *the* uniform field in GR).

 

[yuiop]

This is only true in the weak-field limit, if the source isn't rotating. Under those circumstances, you can just find the gravitational field with Gauss's law, and you don't need relativity.

Your right. We did use Gauss's law and it probably does not apply here outside the weak field limit.

If you change the ball of coffee grounds into two spaceships connected by a thread, then you have Bell's spaceship paradox.

So do you agree that physical rulers would be torn apart in such a spacetime, like the thread in Bell's spaceship paradox?

In all three dimensions! Here we have been talking about Rindler metric in 2 dimensions, so the second axis would be x or y or z along which the uniformity of gravitational field is considered. The FLRW metric is not static, but dependent on time if we take all coordinates to be fixed and pin the changes to time only which enters in the metric by scale factor, so particles moving on geodesics will either accelerate or decelerate according to which epoch the cosmos is getting through: ..

What I have in mind personally is one that has uniformity in all three dimensions. I don't think an FLRW metric qualifies, because an object released with a coordinate velocity of zero (in the usual coordinates) maintains a coordinate velocity of zero indefinitely.

I see now that my casual guess that the FLRW metric might qualify was incorrect. Thanks guys.

OK, so we are talking about constant acceleration in all 3 dimensions. Now there is another problem. Since there is equal acceleration in all directions, how does a particle know which way to accelerate or does everyting simply expand? Do we have to assume a coordinate centre to such a spacetime and assume everything either accelerates directly towards (or away?) from that coordinate centre? It seems such a gravitational system could not be isotropic, in the sense that any arbitary location could be chosen as the centre.

 

[yuiop]

This is probably irrelevant, but there are also the x and z coordinates. I've seen a nice elementary argument by Born that because a measuring rod in a rotating frame of reference experiences a Lorentz contraction when oriented in the transverse direction, by the equivalence principle a rod's length varies with height in a gravitational field. This would presumably apply to x and z, not y.

The only trouble is that we know length contraction does not occur in the x and z directions in the Schwarzschild metric. (I assume you mean horizontal by x and z in the gravitational context)

Here are some comparisons of rotational and gravitational length measurements. All measurements in a small region to a first order aproximation. I am also using z as the radial coordinate as you have done in your later post.

Z COORDINATE:

Radial coordinate of a cylinder:
Parallel to the felt acceleration.
NOT length contracted according to radar or ruler measurements.(A rotating and non rotating observer will measure this to be the same.)
Increasing time dilation in the positive y direction.

Vertical coordinate in a gravitational field:
Parallel to the felt acceleration.
Length contracted according to radar or ruler measurements. (A stationary local observer will measure this distance to be longer than the distance calculated by an observer at infinity.)
Decreasing time dilation in the positive y direction.


As you can see there are obvious differences (and more than I mention here) between the rotating disc and the gravitational field.

On the face of it, Born's statement appears to be incorrect, or he he saying the transverse coordinates (orthogonal to the radius) of a disc are equivalent to the vertical radial coordinate of a gravitational field?

I found the passage in the Born book that I assume is the one I'd seen summarized elsewhere. It's at p. 320 in the 1962 Dover edition. It's part of a discussion of spacetime on a rotating disk, and it's actually very brief. He discusses the impossibility of global clock synchronization, talks about the interpretation in terms of the equivalence principle, and then says:

In a gravitational field a rod is longer or shorter or a clock goes more quickly or more slowly according to the position at which the measuring apparatus is situated.

This seems like somewhat of a leap to me, since he's generalizing from the rotating disk to gravitational fields in general. But it does seem to tie in correctly with the fact that generalizing the 1+1 metric to 3+1 by simply adding -d x^2 - d y^2 gives unphysical results.

In the 2+1 carousel setup, rulers oriented in the transverse direction are shorter when they're lower in the gravitational field (closer to the rim). This means that the xx part of the metric should decrease with z. Generalizing to 3+1, it's not obvious to me whether the contraction should apply to both x and y or only to x.

I am going to assume x, y and z in the context of a cylinder rotating around its long axis, as that is easier to visualise than a disc or sphere rotating around two axes at the same time, although that is technically possible. For the sake of argument let's say the x coordinate is circular and goes around the circular perimeter of the cylinder (equivalent to d\theta *r in the Schwarzschild metric) and the y coordinate is distances measured parallel to the long rotational axis of the cylinder.

X COORDINATE:

Circumference of the cylinder:
Orthogonal to the felt acceleration.
Length contracted according to radar and ruler measurements.(A rotating observer will measure the distance between two marks on the perimeter to be longer than the distance measured by the non rotating observer.)

Horizontal coordinate in a gravitational field:
Orthogonal to the felt acceleration.
NOT length contracted according to radar and ruler measurements. (A stationary local observer will measure the distance between two marks to be the same as the distance calculated by an observer at infinity.)

Y COORDINATE:

Parallel to the long rotational axis of the cylinder:
Orthogonal to the felt acceleration.
NOT length contracted according to radar or ruler measurements.(A rotating and non rotating observer will measure this to be the same.)

There is no equivalent of the cylindrical Y coordinate in the Schwarzschild metric. In general there is also no equivalence for length contraction between a rotating system and a gravitational system and that is not what the EP claims. It simply implies that measurements conspire to make it very difficult to determine if you are in a rotational or gravitational field if you restricted to measurements in an infinitesimally small volume of spacetime.

In the case of a rotating sphere, it is impossible to arrange a rotation scheme around any number of axes simultaneously, whereby an asymmetry would be undetectable to an observer at rest with the sphere, so it is difficult to imagine that there is an a rotational equivalent of a 3+1 uniform gravitational field.

If nothing else, I wanted to clear up what you mean by the x, y and z coordinates. In your first post you seam to be using y as your radial coordinate and the later post you seem to using z as the radial coordinate. That might to be due to different usages in the texts you are referring to. Maybe we should have our own coordinates for this thread, such as (p,q,r) for (x,y,z) as defined in this post?

Of course, I may be completely off target here and you are using cartesian coordinates with one coordinate axis being the axis of rotation and the other two being radial axes that are orthogonal to each other and the rotation axis, in which case the two radial axes are completely equivalent as far as length contraction is concerned.

 

[yuiop]

This confuses me, because a^2x^2 is the same as (1+ax)^2 except for adding a constant to the x coordinate. So I don't see how the proper acceleration can come out to be independent of x in the first case, but dependent on x in the second case.

(1+ax)^2 expands to 1 + 2ax + a^2x^2 and the additional (1+2ax) is not a constant.

 

[Altabeh]

OK, so we are talking about constant acceleration in all 3 dimensions. Now there is another problem. Since there is equal acceleration in all directions, how does a particle know which way to accelerate or does everything simply expand? Do we have to assume a coordinate centre to such a spacetime and assume everything either accelerates directly towards (or away?) from that coordinate centre? It seems such a gravitational system could not be isotropic, in the sense that any arbitrary location could be chosen as the centre.

I think the problem is that the definition of constant proper acceleration in a 4D spacetime is not so much well established but it is rather baffling and thus it takes me a few minutes writing some lines to get you to know how particles can find which way to accelerate.

I assume you've taken a look at Semay's metric introduced http://arxiv.org/pdf/physics/0601179v1" (early posts in this thread can also help you to get acquainted with it):

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

You can simply see that the curvature tensor vanishes for this metric, giving rise to a flat spacetime wherein a uniform gravitational potential can be considered locally which chalks up to equivalence principle. To verify this, let's check out the geodesic equations:

\ddot{x}=-a(1+ax)\dot{t}^2,
\ddot{y}=\ddot{z}=0,
\ddot{t}=\frac{-2a\dot{x}\dot{t}}{1+ax},

where the dot being the derivative wrt proper time. The aim is to show that for this metric, freely falling particles which follow timelike geodesics, are all accelerated the same locally and the proper acceleration is \ddot{x}=-a. As the second equations demonstrate, along y and z axes the proper accelerations are all zero. This is why the freely falling particles are able to find the path on which they can accelerate and fall in almost every other metric (sometimes as in Schwarzschild metric it gets so much twisted to get such a result and that is because the g_{rr} is not considered to be constant in contrast to this case)!

Solving the last equation for \dot{t} gives

\dot{t}=\frac{1}{(1+ax)^2}.

Introducing this into the first equation yields

\ddot{x}=\frac{-a}{(1+ax)^3}.

Therefore, for ax<<c^2, one gets \ddot{x}=-a that confirms the particles following timelike geodesics have constant proper acceleration locally. Sounds like particles know how to accelerate in Semay's spacetime! I see people use hovering particles at a constant x=x_0 to get global proper accelerations along the geodesics throughout the spacetime . Roughly speaking, this makes the motion boil down to a 2D spatial one so neglecting the general 3D motions in space! I think now you are convinced that on planes parallel to the yz plane the motion gains a constant proper acceleration, but as long as the planes turn to have a slope, the constancy of proper acceleration is guaranteed if one uses equivalence principle.

Now suppose the geodesic equations in a special Rindler's metric,

ds^2=e^{2ay}dt^2-dx^2-dy^2-dz^2.

Following calculations similar to those done for Semay, we can get

\ddot{x}=-ae^{-2ax}.

This is again for ax<<c^2, the result shown by Semay and thus we have a locally uniform gravitational field. But note that this metric is not flat, so that would not be possible to make it coincide with Semay's metric up to all orders in ax and thus you have to be careful not to think both are flat since the Taylor expansion of e^{2ax} is
1+2ax up to order 1 in ax. Rest assured that in this case, freely falling particles will find the way they can accelerate!

AB

 

[bcrowell]

OK, so we are talking about constant acceleration in all 3 dimensions. Now there is another problem. Since there is equal acceleration in all directions, how does a particle know which way to accelerate or does everyting simply expand?

There are two different issues:

(1) Does the field vary with position in space?

(2) What direction does the field point?

The notion of a uniform field refers to #1, not #2.

 

[bcrowell]

Now suppose the geodesic equations in a special Rindler's metric,

ds^2=e^{2ay}dt^2-dx^2-dy^2-dz^2.

This is not a vacuum solution when you add the -dx^2-dy^2 terms. See Lut's #16 and my #29.

 

[bcrowell]

The only trouble is that we know length contraction does not occur in the x and z directions in the Schwarzschild metric.

The Schwarzschild isn't a uniform field, it's a spherically symmetric field. That clearly is going to have all kinds of nontrivial effects on the transverse stuff. If you want to check whether Born's statement holds in a specific GR example, I think it makes more sense to do it in a GR example that has translational symmetry in the transverse directions. The Petrov metric has that kind of symmetry, and Born's statement does hold for the Petrov metric. (The Petrov metric has four killing vectors. Three of them are translations of the time and transverse space coordinates, while the fourth is a kind of rescaling that connects the radial coordinate to the others.)

On the face of it, Born's statement appears to be incorrect, or he he saying the transverse coordinates (orthogonal to the radius) of a disc are equivalent to the vertical radial coordinate of a gravitational field?

I think he's making the analogy where (r,phi) on the disk corresponds to (vertical,horizontal) in a uniform field.

I don't claim that Born's argument is anything more than a heuristic. It just provides a nice motivation for trying out a metric of the Kasner form, whose results end up being very similar to the Petrov metric. One way to see that it definitely can't be any more than a heuristic is that he's basing it on the analogy with a rotating coordinate system, but the rotating coordinate system has a nonuniform spatial metric (the scalar curvature depends on r).

 

[bcrowell]

Yes, (if the observer is released at the same time). Presumably, simultaneity is not an issue here, because clocks at different heights in a uniform gravitational field run at the same rate.

No, there is gravitational time dilation. They run at different rates.

 

[Altabeh]

This is not a vacuum solution when you add the -dx^2-dy^2 terms. See Lut's #16 and my #29.

I think talking about vacuum solutions makes no sense when it comes to local inertia! I just was trying to show that particles move along timelike paths that determine the direction of acceleration in two examples so the problem of "which way to accelerate" is dependent on the geodesic equations and thus the metric itself!

There are two different issues:

(1) Does the field vary with position in space?

(2) What direction does the field point?

The notion of a uniform field refers to #1, not #2.

I think he meant something else! In all static curved spacetimes, the field varies with position and in all flat spacetimes there is no gravitational field unless one makes use of EP and says 'it's just a small region and the flatness is guaranteed within it' so there the spacetime can admit a uniform gravitational field! The notion of the direction of the proper acceleration refers to both (1) and (2)!

AB

 

[yuiop]

No, there is gravitational time dilation. They run at different rates.

Ah, of course. I realized that a while after I posted. I made the basic mistake of assuming equal acceleration equals equal time dilation which is of course wrong.

 

[yuiop]

Ok, let us assume we have a 3D grid of rulers and clocks in flat space. They all accelerate simultaneously according to a non accelerated observer S and they all have equal proper acceleration according to S and constant acceleration (a) as measured by the accelerated grid observers. (All grid points follow parallel paths in S). Now, assuming acceleration in multiple directions can resolved to a single direction and assuming the x-axis can always be chosen such that it aligns with the direction of acceleration, then as far as I can tell the line element is described by:

d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2

Where x and d\tau are measured in S and dt, dx, dy and dz are measured in the accelerating frame. (Could someone check that?)
This would seem to be in agreement with the time dilation factor (1+ax) given by the relativistic rocket FAQ http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html . In other words, when dx=dy=dz=0, dS/dt = (1+ax).

The situation is essentially the Rindler metric with a = constant, substituted for the usual a = 1/x. The situation is also essentially the Bell's rocket paradox so the observers on the accelerated grid will see themselves getting further apart in the x direction (which eventually tears the grid apart) while transverse distances (y and z) remain constant according to the accelerating grid observers (ruler or radar) and the y,z measurements are measured to be the same according to both the accelerated and non accelerated observers. (No transverse length contraction). This can be confirmed by looking at the Rindler metric http://en.wikipedia.org/wiki/Rindler_coordinates where the transformation from Minkowski to Rindler is Y=y and Z=z. The transverse measurements are simply the same as in linear motion in Special Relativity which makes sense in the constext of instantaneous co-moving observers or what is sometimes referred to as instantaneously "at rest" observers which is a bit more confusing.

Since the situation I have described does not have spherical or cylindrical symmetry, it does not need to be local and can be extended arbitrarily far. Interestingly, there are still event horizons in this spacetime, but the location of the event horizon depends upon where the accelerating observer is.

A ball of free-falling coffee grounds in this spacetime will obviously remain unchanged from the point of view of a co-free-falling observer, but to the accelerated observers on the grid the ball of coffee grounds will appear to be shrinking in the x direction.

If the physical description I have given does not match what you have in mind in this thread, please could you give a physical description of what you do have in mind. Also, if it could be made clear which frame the measurements are made in it might help clear up why there appears to be inconsistent forms of the metric.

 

[Altabeh]

Ok, let us assume we have a 3D grid of rulers and clocks in flat space. They all accelerate simultaneously according to a non accelerated observer S...

Yeah, but not in every four dimensional flat spacetime, e.g. non-Minkowski, it would be true for a a non accelerated observer S to observe all particles accelerate simultaneously because the proper acceleration probably depends on the position of particles! This of course does not occur for flat spacetimes for which all Christoffel symbols vanish! I assume that your flat space is like these!

Now, assuming acceleration in multiple directions can resolved to a single direction and assuming the x-axis can always be chosen such that it aligns with the direction of acceleration, then as far as I can tell the line element is described by

d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2

Where x and d\tau are measured in S and dt, dx, dy and dz are measured in the accelerating frame. (Could someone check that?) This would seem to be in agreement with the time dilation factor (1+ax) given by the relativistic rocket FAQ http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html . In other words, when dx=dy=dz=0, dS/dt = (1+ax).

Except for \tau, dt and dx^i for i=1,2,3 are all measured by an at-rest observer and here d\tau is measured by the accelerating observers. This is reminiscence of time dilation (assuming dx^i=0) where

c^2d\tau^2=c^2(1+ax/c^2)^2 dt^2,

and thus

d\tau=(1+ax/c^2) dt, with (1+ax/c^2)^{-1}=\gamma. Nevertheless, all observers moving relative to each other in arbitrary ways agree on the quantity d\tau^2.

The situation is essentially the Rindler metric with a = constant, substituted for the usual a = 1/x.

I don't see this. Please explain it a little bit!

The situation is also essentially the Bell's rocket paradox so the observers on the accelerated grid will see themselves getting further apart in the x direction (which eventually tears the grid apart) while transverse distances (y and z) remain constant according to the accelerating grid observers (ruler or radar) and the y,z measurements are measured to be the same according to both the accelerated and non accelerated observers.

This may be incorrect because we don't know about the proper velocities along y and z axes relative to an observer moving inertially in the spacetime! They can be anything constant equally since from the two geodesic equations we have \ddot{z}=\ddot{y}=0 so along a geodesic particles can uniformly move parallel to y and z axes relative to the observers. To a non accelerated (inertial) observer moving in the direction of x axis, you are right.

(No transverse length contraction). This can be confirmed by looking at the Rindler metric http://en.wikipedia.org/wiki/Rindler_coordinates where the transformation from Minkowski to Rindler is Y=y and Z=z. The transverse measurements are simply the same as in linear motion in Special Relativity which makes sense in the constext of instantaneous co-moving observers or what is sometimes referred to as instantaneously "at rest" observers which is a bit more confusing.

The transverse measurements of velocity, as I said, are not the same unless they are zero according to all observers or to the observers that are all moving inertially parallel to the x axis! But instantaneous co-moving observers are also able to overcome the problem as they are moving parallel to the geodesics with the same acceleration and velocity.

Since the situation I have described does not have spherical or cylindrical symmetry, it does not need to be local and can be extended arbitrarily far.

In a flat spacetime, for example the Semay's spacetime, that the geodesic equations are dependent on the position at least the measurement of proper acceleration is always local from every observer's perspective.

AB

 

[bcrowell]

If the physical description I have given does not match what you have in mind in this thread, please could you give a physical description of what you do have in mind.

I don't think it totally works as a uniform field, because an observer hovering at a fixed location relative to the lattice you described can release a test particle from rest, and the acceleration of the test particle is -a(1+ax). This isn't constant, so the field isn't uniform.

I don't think there is any metric in GR that can be called a uniform field and that satisfies every criterion you could put on a wish list. They all have certain undesirable properties. But anyway this is the undesirable property of this particular metric.

Also, if it could be made clear which frame the measurements are made in it might help clear up why there appears to be inconsistent forms of the metric.

The whole notion of a gravitational field is frame-dependent. If you're looking for a metric that in some sense embodies a uniform field with a particular value of g, then all you can hope for is a metric that accomplishes that in one particular set of coordinates. Those are the coordinates you use to write down the line element.

That doesn't mean that there are no frame-independent properties that you should hope for. In particular, I think it's desirable for the scalar curvature to be constant, and for the metric to be vacuum solution. Both the Rindler coordinates and the Petrov spacetime satisfy these criteria.

 

[bcrowell]

This of course does not occur for flat spacetimes for which all Christoffel symbols vanish!

This is incorrect. For example, the Rindler coordinates, with ds^2=(1+ax)^2dt^2-dx^2, give \Gamma\indices{^x_{tt}}=-a(1+ax) and \Gamma\indices{^t_{xt}}=a(1+ax), but the spacetime is flat.

 

[yuiop]

I don't think it totally works as a uniform field, because an observer hovering at a fixed location relative to the lattice you described can release a test particle from rest, and the acceleration of the test particle is -a(1+ax). This isn't constant, so the field isn't uniform.

Hi Ben. I think the accelerating grid in #68 is uniform from the point of view that you can pick any point on the grid at random at any random time and the local acceleration would be a. Because you can pick any random time this means that simultaneity is less of an issue as you can pick any plane of simultaneity you like and the accelerometers will all be reading the same. On the other hand, any individual observer on the lattice will consider the other observers at rest with the lattice to be accelerating away from them in the x direction so if we have a notion of distance on the lattice that keeps all the observers at rest with some artificial coordinate system then the apparent acceleration of free falling test particles might come out different from the calculation you give.

I don't think there is any metric in GR that can be called a uniform field and that satisfies every criterion you could put on a wish list. They all have certain undesirable properties. But anyway this is the undesirable property of this particular metric.

I would agree there is no real gravitational field that is equivalent to the accelerating latice and maybe the reason GR can not produce a metric that can be called a uniform field is that GR concerns itself with real gravitational fields generated by masses. Not sure about that though.

Anyway, do you agree about my conclusions for the accelerated lattice in #68 about the coffee grounds and specifically about who measures what?

d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2

Where x and d\tau are measured in S and dt, dx, dy and dz are measured in the accelerating frame.

Also, I said:

while transverse distances (y and z) remain constant according to the accelerating grid observers (ruler or radar) and the y,z measurements are measured to be the same according to both the accelerated and non accelerated observers. (No transverse length contraction).

I think it would be fair to add that the radar transverse distance starts to differ from the transverse ruler distance over larger distances and this disagreement of radar distance from ruler distance appears to be a characteristic of any accelerating field.

The whole notion of a gravitational field is frame-dependent. If you're looking for a metric that in some sense embodies a uniform field with a particular value of g, then all you can hope for is a metric that accomplishes that in one particular set of coordinates. Those are the coordinates you use to write down the line element.

I agree that would be ideal, but it would appear the text are using "mixed coordinates". For example the FAQ on the relativistic rocket gives the the time dilation factor as (1+ ad) and makes it clear that a is measured in the accelerating frame (proper acceleration and distance d is measured in the non accelerating frame, so two different frames in one equation. It is also obvious by inspection that dx, dy and dz are not measured in the non accelerating frame because if they were, there would be no acceleration or time dilation when those values are set to zero.

 

[bcrowell]

I just noticed that earlier in the week someone (who wants to remain anonymous) sent me some very detailed and helpful information on this topic via PM. I thought it might be helpful to pass on some of the general ideas. Any mistakes in summarizing his info is purely my fault, not his.

There is a family of metrics called the Weyl vacuums, which are of the form
ds^2 = -\exp(2u) \; dt^2 + \exp(2\,(v-u)) \; (dz^2+r^2) + \exp(-2u) \; r^2 \; d\phi^2 \qquad .
According to him, the metrics most commonly referred to as candidates for the GR version of a uniform field are: two special cases of the Weyl vacuum; the Rindler coordinates, which can be expressed as a special case of the Weyl vacuum; and the Petrov metric. The Weyl vacua, unlike the Petrov metric, are static. Although the Einstein field equations are nonlinear, there is a certain way of superimposing Weyl vacua in a nonlinear way that provides new solutions. In some cases you get strange artifacts called "struts," which are things that act like rigid rods but that don't have any gravitational mass. Although some of the Weyl vacua can be described in simple geometric terms, e.g., as the field of a line of mass, that is actually kind of misleading, because the interpretation of the spatial coordinates isn't what you'd think based on the description.

Re atyy's #2, he says that the Weyl vacuum for u=az is a vacuum solution, but it's not the same vacuum solution as the one atyy proposed.

 

[Altabeh]

This is incorrect. For example, the Rindler coordinates, with ds^2=(1+ax)^2dt^2-dx^2, give \Gamma\indices{^x_{tt}}=-a(1+ax) and \Gamma\indices{^t_{xt}}=a(1+ax), but the spacetime is flat.

1- What can be inferred about this: "This of course does not occur for flat spacetimes for which all Christoffel symbols vanish!"?

I said that if all Christoffel symbols vanish, then for that spacetime all particles accelerate simultaneously from every observer's point of view without knowing where they are at when observing particles! This sentence doesn't imply that, for example, for other four dimensional flat spacetimes whose Christoffel symbols do not vanish, there is no such thing as simultaneity of accelerations!

2- That spacetime you are in love with, , of course cannot have simultaneity for all particles, except for those that are instantaneously at rest according to an observer being at rest or a co-moving one!

3- I think you didn't read my post at all! Only skimmed some parts of it! But I enjoy reading all posts first then if there was something that can hit the spot I'd be glad to share it with you!

AB

 

[Altabeh]

I don't think it totally works as a uniform field, because an observer hovering at a fixed location relative to the lattice you described can release a test particle from rest, and the acceleration of the test particle is -a(1+ax). This isn't constant, so the field isn't uniform.

I don't think there is any metric in GR that can be called a uniform field and that satisfies every criterion you could put on a wish list. They all have certain undesirable properties. But anyway this is the undesirable property of this particular metric.

Oh God! This time bcrowell and I are on the same page!

The whole notion of a gravitational field is frame-dependent. If you're looking for a metric that in some sense embodies a uniform field with a particular value of g, then all you can hope for is a metric that accomplishes that in one particular set of coordinates. Those are the coordinates you use to write down the line element.

I guess you don't mean that line element is coordinate-dependent, do you?

That doesn't mean that there are no frame-independent properties that you should hope for. In particular, I think it's desirable for the scalar curvature to be constant, and for the metric to be vacuum solution. Both the Rindler coordinates and the Petrov spacetime satisfy these criteria.

As far as I can tell, every vacuum solution of Einstein field equations is a Ricci-flat spacetime! So every vacuum solution implies the vanishing of curvature scalar!

AB

 

[Altabeh]

Hi Ben. I think the accelerating grid in #68 is uniform from the point of view that you can pick any point on the grid at random at any random time and the local acceleration would be a. Because you can pick any random time this means that simultaneity is less of an issue as you can pick any plane of simultaneity you like and the accelerometers will all be reading the same. On the other hand, any individual observer on the lattice will consider the other observers at rest with the lattice to be accelerating away from them in the x direction so if we have a notion of distance on the lattice that keeps all the observers at rest with some artificial coordinate system then the apparent acceleration of free falling test particles might come out different from the calculation you give.

I'm sorry, but Ben is right! At different points of your grid, the position is a matter of importance to any kind of observer except for the instantaneous co-moving ones whose observations don't affect the spacetime to admit a uniform gravitational field! Nevertheless, the uniform gravitational field does only locally exist and GR has no such thing to stand for as a general solution or something!

I would agree there is no real gravitational field that is equivalent to the accelerating latice and maybe the reason GR can not produce a metric that can be called a uniform field is that GR concerns itself with real gravitational fields generated by masses. Not sure about that though.

True.

Anyway, do you agree about my conclusions for the accelerated lattice in #68 about the coffee grounds and specifically about who measures what?

It unfortunately is not correct! If it is, it will destroy your last conclusion concerning the non-existence of uniform gravitational fields globally!

I agree that would be ideal

So do I!

AB

 

[bcrowell]

As far as I can tell, every vacuum solution of Einstein field equations is a Ricci-flat spacetime! So every vacuum solution implies the vanishing of curvature scalar!

Ah, thanks for the correction. It would be more interesting to talk about having other curvature invariants that were constant, e.g., the Kretschmann scalar. I played around with a few of these in Maxima, and found that the Kretschmann scalar does vanish identically for the Petrov metric. That implies that the scalar invariant C^{abcd}C_{abcd} constructed from the Weyl tensor also vanishes identically.

 

[Altabeh]

Ah, thanks for the correction. It would be more interesting to talk about having other curvature invariants that were constant, e.g., the Kretschmann scalar. I played around with a few of these in Maxima, and found that the Kretschmann scalar does vanish identically for the Petrov metric. That implies that the scalar invariant C^{abcd}C_{abcd} constructed from the Weyl tensor also vanishes identically.

Hi

For a conformally flat spacetime, if the Ricci tensor vanishes, then the spacetime is flat. So the KS is also zero iff the Ricci tensor vanishes for conformally flat spacetimes! But, in general, the Ricci-flatness does not imply KS is zero!

Could you provide me with some available links that I can understand what exactly this Petrov metric is?

AB

 

[bcrowell]

I played around a little with the geodesic equations in the Petrov metric. Suppose you restrict yourself to a fixed z, and for convenience take an observer at an r such that \cos\sqrt{3}r=1, which means that the timelike coordinate is t. For a particle released at rest, the geodesic equations become
<br /> \ddot{t}=0<br />
<br /> \ddot{\phi}=0<br />
<br /> \ddot{r}=-\frac{1}{2}e^r \dot{t}^2<br />
where dots mean differentiation with respect to the affine parameter. If you take the affine parameter to be the particle's proper time, then \dot{t}=e^{-r/2} initially, and the particle accelerates toward lower values of r with a proper acceleration that is independent of r. If you go to a place where \cos\sqrt{3}r=-1, the timelike coordinate is again t, but I think the direction of time is reversed here as compared with at the place where \cos\sqrt{3}r=+1. Here the particle accelerates toward higher values of r. This seems to match up with what the Gibbons and Gielen paper describes about the geodesics: they oscillate back and forth in r.

So in this sense the Petrov metric seems to be a better realization of the uniform gravitational field than the Rindler coordinates, because the proper acceleration has the same magnitude everywhere. On the other hand, it does have strange properties like CTCs, and that's undesirable.

[EDIT] I think the constancy of the proper acceleration follows from the fact that in addition to the three obvious Killing fields, there is a fourth one involving r. If I've got that right, then the restriction of my calculation above to certain values of r can be relaxed, and the result still holds.

 

[bcrowell]

For a conformally flat spacetime, if the Ricci tensor vanishes, then the spacetime is flat. So the KS is also zero iff the Ricci tensor vanishes for conformally flat spacetimes! But, in general, the Ricci-flatness does not imply KS is zero!

I explicitly calculated the Kretschmann scalar and found that it was zero. For the assertion that this implies that C^{abcd}C_{abcd} vanishes, see http://en.wikipedia.org/wiki/Kretschmann_scalar . I didn't say that C^{abcd} vanished, just that that particular scalar constructed from it vanished.

[EDIT] In any case, I think it's more conclusive to look at the Killing vectors than to go through a list of curvature scalars and try to show that they're all constant.

Could you provide me with some available links that I can understand what exactly this Petrov metric is?

See #4.

 

[yuiop]

The situation is essentially the Rindler metric with a = constant, substituted for the usual a = 1/x.

I don't see this. Please explain it a little bit!

I am saying if you take the the equations for the relativistic rocket with constant proper acceleration and make the acceleration equal to 1/r where r is the displacement along the x-axis from the origin of the initial inertial refenence frame at t=0 then you have the Rindler metric, which has the desirable property that the accelerating observers consider their mutual spatial separation to be constant. If you make the proper acceleration a constant (independent of location) for a series of rockets then you get the get the accelerating grid with a form of uniform acceleration that I described earlier, but you lose the property of constant spatial separation of the observers.

It seems you and Ben have a wish list for a uniform gravitational field that is too long and impossible to meet in any realistic situation and I think my accelerating grid is as close as you are going to get. It might be helpful to define that wish list

I am not sure that the equation I gave:

d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2

precisely describes the accelerating grid.

I think we have yet to pin down why the other metric:

ds^2=e^{2gy}dt^2-dy^2

differs from the first.

I have also found other metrics that claim to describe the Rindler metric such as:

ds^2 = (1+2ax)dt^2 - \frac{dx^2}{(1+2ax)} - dy^2 - dz^2

from http://nedwww.ipac.caltech.edu/level5/Sept02/Padmanabhan/Pad10_1.html which seems to agree with this paper http://arxiv.org/PS_cache/physics/pdf/0601/0601179v1.pdf that derives an equation for a uniform field by taking a first order approximation of the Schwarzcshild metric in the weak limit and locally.

Wikipedia http://en.wikipedia.org/wiki/Rindler_coordinates gives the Rindler metric as:

ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2

which describes the Minkowski line element converted to the Rindler chart. I think this particular metric describes a free falling particle from the point of view of the accelerating Rindler observers and that is why there is no acceleration term in that metric.

Anyway, I think we should try and clear up the differences by determining exactly who measures what and if the various authors have different physical models in mind, or if it is just due to different points of view from observers in different states of motion?

 

[atyy]

How about

http://arxiv.org/abs/0904.4877
Lorentzian spacetimes with constant curvature invariants in four dimensions
Alan Coley, Sigbjorn Hervik, Nicos Pelavas

 

[yuiop]

This metric:

d\tau^2 = (1+ax)^2 dt^2 - dx^2 - dy^2 -dz^2

has been bothering me for while so I decided to have closer look at it.

From the equations for the relativistic rocket, it easy to work out the instantaneous velocity of rocket with constant proper acceleration (a) that has traveled a time \Delta t measured in the initial inertial reference frame (iirf) that the rocket was originally at rest in, from:

v= \frac{a\Delta t}{\sqrt{1+(a\Delta t/c)^2}}

From this we can work out the instantaneous time dilation factor from:

\gamma = \frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^2/c^2}} = \sqrt{1+(a\Delta t/c)^2}

Note that we do not need \Delta (x,y,z) because this is already implicit in a*t. In other words if we know the acceleration and the time it has been accelerating for, then all the information to calculate how far the rocket has traveled is already there.

So

d\tau = \frac {dt}{\sqrt{(1+(a\Delta t/c)^2)}}

Now the relativistic rocket equations give the relationship of distance traveled \Delta x and elapsed time \Delta t as:

\Delta t = \sqrt{( (\Delta x/c)^2 + 2\Delta x /a)}

and substituting this value into the equation above it gives:

d\tau = \frac {dt}{\sqrt{(1+(a\Delta x/c^2)^2 + 2 a \Delta x / c^2)}} = \frac{dt}{(1 + a\Delta x/c^2)}

This differs the normal presentation in a number of ways. The time dilation factor is inverted and it should be clear that this is correct because the proper time of the accelerating rocket is always less than the time of the inertial clock. Secondly there is no separate dx factor because this is already present in the \Delta x term. Up to now I have been assuming (and I think others have too) that the x in (1+ax) term of the usual metric is a location, rather than a distance traveled and the inclusion of \Delta x rather than just x makes it clear that the metric is independent of location.

This new metric can be expanded to 4 dimensions as:

d\tau = \frac{dt}{(1 + a\sqrt{(\Delta x+\Delta y + \Delta z)}/c^2)}

This last equation treats the proper acceleration as a scalar and does not really care which direction it is directed in. (The (\Delta x+\Delta y + \Delta z) term takes care of the direction.)

I hope I got that all right. Maybe someone can check it.

[EDIT] Just noticed that the above equations aare only valid when the intervals \Delta x or \Delta t are taken from t=0 in the iirf. It will need a bit more work to genralise the equations to any t.

 

[DrGreg]

The equations

\begin{array}{rclc}<br /> d\tau^2 &amp; = &amp; (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2 &amp; (1)\\<br /> ds^2 &amp; = &amp; a^2x^2 dt^2 - dx^2 -dy^2 -dz^2 &amp; (2)\\<br /> ds^2 &amp; = &amp; -x^2 dt^2 + dx^2 + dy^2 + dz^2 &amp; (3)<br /> \end{array} <br />​

are all variations on what is usually described as the "Rindler metric" (or, to be pedantic, should really be the "Minkowski metric expressed in Rindler coordinates").

They describe lattice points undergoing Born-rigid acceleration -- every pair of lattice points is a constant distance apart as measured by a comoving inertial observers at either point, and two events have the same t coordinate if they are simultaneous according to comoving inertial observers at either lattice point.

In (1), the nominal observer is at x=0 undergoing proper acceleration a.
In (2), the nominal observer is at x=1/a undergoing proper acceleration a.
In (3), the nominal observer is at x=1 undergoing proper acceleration 1.

(In all three cases, c=1, of course.) x is distance measured by comoving inertial observers, and t coincides with the proper time of the nominal observer.

The equation

<br /> ds^2 = (1+2ax)dt^2 - \frac{dx^2}{(1+2ax)} - dy^2 - dz^2<br />​

is not usually described as "Rindler coordinates" (in my limited experience), but also satisfies all the conditions above except that x no longer measures proper distance, but is a distorted distance. This particular metric resembles the Schwarzschild metric in many ways, and is an appropriate one to study if you are are applying the equivalence principle to Schwarzschild coordinates.

None of the above describe a "Bell's spaceship" lattice. In such a lattice, the lattice points are moving apart ("the string breaks"), so you would expect the coefficient of dx² to be time-dependent. I did spend half an hour with the back of an envelope trying to think what the metric would look like, but I gave up. Part of the problem is that there's no obvious definition of coordinate-simultaneity in this case. As two lattice point observers are moving apart, their comoving inertial observers disagree with each other over what is Einstein-simultaneous.

 

[bcrowell]

How about

http://arxiv.org/abs/0904.4877
Lorentzian spacetimes with constant curvature invariants in four dimensions
Alan Coley, Sigbjorn Hervik, Nicos Pelavas

That's very helpful -- thanks, atyy!

One nice thing about this paper is that it clearly characterizes the idea about curvature invariants that I was bumbling around trying to characterize earlier. They define a spacetime as CSI (constant scalar invariant?) if everything you can compute from the Riemann tensor by contraction and covariant differentiation is constant.

There is also a notion called "locally homogeneous," which they never seem to define explicitly. It is closely related to being CSI, but is not equivalent in all cases.

They prove that if a 3+1 spacetime is CSI, then it's either locally homogeneous or a certain type of Kundt spacetime.

They also define VSI, which means CSI with all the scalar invariants equaling zero. I don't know if the Petrov spacetime is VSI. It does have a vanishing Ricci scalar (because it's a vacuum solution) and a vanishing Kretschmann invariant.

They define a metric as I-degenerate if there are smooth transformations of the metric that leave the CIs the same while changing the spacetime intrinsically. If the Petrov metric is VSI, then maybe it's I-degenerate, because the strength of the gravitational field is a tunable parameter. (But I don't know if changing g actually makes the new spacetime inequivalent to the old one under a diffeomorphism, so maybe this isn't the case.)

This paper http://arxiv.org/abs/0901.0791 by the same authors has a flow-chart for determining I-degeneracy from curvature invariants (p. 20). On p. 27 they discuss Petrov type I (which is the type of the Petrov metric); they make some general statements, but say that there are exceptions for symmetric spacetimes, and these exceptions would presumably apply to the highly symmetric Petrov metric.

It seems you and Ben have a wish list for a uniform gravitational field that is too long and impossible to meet in any realistic situation and I think my accelerating grid is as close as you are going to get. It might be helpful to define that wish list

IMO there should be two main criteria:

(1) It should admit a set of global coordinates such that particles released anywhere with a coordinate velocity of zero have an initial coordinate acceleration that's equal in magnitude to some constant.

(2) In addition, it should satisfy some frame-independent criterion, like being CSI, having four simply transitive killing vectors, or being "locally homogeneous," whatever that means.

The Petrov metric satisfies 1, and satisfies 2 in the sense of Killing vectors, but has strange properties like CTCs. The Rindler coordinates don't satisfy 1.

At this point, I would be interested in figuring out whether the Petrov metric is VSI, and whether it's I-degenerate.

 

[Altabeh]

I am saying if you take the the equations for the relativistic rocket with constant proper acceleration and make the acceleration equal to 1/r where r is the displacement along the x-axis from the origin of the initial inertial refenence frame at t=0 then you have the Rindler metric, which has the desirable property that the accelerating observers consider their mutual spatial separation to be constant.

This is in your wish list, because we can't make such a desire happen to exist in Rindler or Semay metric. Putting a=1/r makes the acceleration position-dependent and the worst thing is that looking at the geodesic equations that describe the motion of lattice points along geodesics shows us that with your assumtion proper acceleration is still position-dependent unless we take r=x-x_0=x to be a constant (or consider DrGreg's representation of how the proper acceleration undergoes a constant proper acceleration a by putting x=0), which means the points on your lattice does remain motionless in the direction of x-axis while the acceleration is itself supposed to be along x-axis.

If you make the proper acceleration a constant (independent of location) for a series of rockets then you get the get the accelerating grid with a form of uniform acceleration that I described earlier, but you lose the property of constant spatial separation of the observers.

How can one make the proper acceleration a constant without assuming again no motion in the direction of x-axis? If you could have a position-independent proper acceleration, I would welcome the idea of "uniform acceleration" in your lattice-based version of the aforementioned metrics.

It seems you and Ben have a wish list for a uniform gravitational field that is too long and impossible to meet in any realistic situation and I think my accelerating grid is as close as you are going to get. It might be helpful to define that wish list

I myself only think of a uniform gravitational field in any spacetime locally! The reason is that until the proper acceleration depends on the position (or time), I can't agree to having a uniform gravitational field in a flat spacetime. (I think no such thing as gravity exist in a flat spacetime, either!) If we take the spacetime to be Minkowski, then all proper accelerations (in the direction of spatial coordinates) are equally zero; so we can say the gravitational field in this metric is uniform. But where is the gravity in such strictly flat spacetime!? The moral that I sign my wish list with it is: Equivalent principle leads to having a weakened form of a uniform field and vice versa.

Your lattice in my eyes seems to be working well locally as the above says. Though I think when it comes to local discussion of the whole stuff, your lattice gets as less valuable than all efforts done to this date in this zone as it can!

I am not sure that the equation I gave:

d\tau^2 = (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2

precisely describes the accelerating grid.

I think we have yet to pin down why the other metric:

ds^2=e^{2gy}dt^2-dy^2

differs from the first.

The two have a big difference: One is flat the other isn't!

AB

 

[yuiop]

This is in your wish list, because we can't make such a desire happen to exist in Rindler or Semay metric. Putting a=1/r makes the acceleration position-dependent and the worst thing is that looking at the geodesic equations that describe the motion of lattice points along geodesics shows us that with your assumption proper acceleration is still position-dependent

Indeed. I never claimed otherwise. The classic Rindler metric (with Born rigid acceleration) does not have uniform proper acceleration (as measured by local accelerometers on the rockets). That is why the classic Rindler metric has to be modified somehow to achieve a form of uniform acceleration.

How can one make the proper acceleration a constant without assuming again no motion in the direction of x-axis? If you could have a position-independent proper acceleration, I would welcome the idea of "uniform acceleration" in your lattice-based version of the aforementioned metrics.

Why do we need to assume no motion along the x axis?

I myself only think of a uniform gravitational field in any spacetime locally! The reason is that until the proper acceleration depends on the position (or time), I can't agree to having a uniform gravitational field in a flat spacetime. (I think no such thing as gravity exist in a flat spacetime, either!) If we take the spacetime to be Minkowski, then all proper accelerations (in the direction of spatial coordinates) are equally zero; so we can say the gravitational field in this metric is uniform. But where is the gravity in such strictly flat spacetime!? The moral that I sign my wish list with it is: Equivalent principle leads to having a weakened form of a uniform field and vice versa.

I am not sure what you are getting at here. I am sure we both agree that we can not have a gravitational field in flat spacetime, because any real gravitational field curves spacetime. I am sure you would also agree that we can create a simulation of a gravitational field by having a system of accelerating rockets, such that an observer confined to making local measurements inside a rocket artificially accelerating in flat spacetime would not be able to determine if he was in a gravitational field or not. That is the essence of the EP principle.

The two have a big difference: One is flat the other isn't!

This link would agree with you: http://www.dlugosz.com/files/PhysFAQ-edit/Relativity/SR/spaceship_puzzle.html

It illustrates the d\tau^2 = e^{2z}dt^2 - dx^2 - dy^2 - dz^2<br /> metric with this drawing:

and states that this metric represents a "uniform gravitational field" and can not be transformed to flat spacetime by a coordinate transformation. Presumably the diagram is the POV of a free falling observer. I am not sure why the wordlines cross over each other or how this would be consistent with their claim that in this spacetime the accelerating observers measure their mutual spatial separation to be constant. Hmmm.

The accelerating grid I had in mind would have worldlines more like this (parallel hyperbola):

This is the point of view of a free falling observer.

The equations

\begin{array}{rclc}<br /> d\tau^2 &amp; = &amp; (1+ax)^2 dt^2 - dx^2 -dy^2 -dz^2 &amp; (1)\\<br /> ds^2 &amp; = &amp; a^2x^2 dt^2 - dx^2 -dy^2 -dz^2 &amp; (2)\\<br /> ds^2 &amp; = &amp; -x^2 dt^2 + dx^2 + dy^2 + dz^2 &amp; (3)<br /> \end{array} <br />​

are all variations on what is usually described as the "Rindler metric" (or, to be pedantic, should really be the "Minkowski metric expressed in Rindler coordinates").

They describe lattice points undergoing Born-rigid acceleration -- every pair of lattice points is a constant distance apart as measured by a comoving inertial observers at either point, and two events have the same t coordinate if they are simultaneous according to comoving inertial observers at either lattice point.

In (1), the nominal observer is at x=0 undergoing proper acceleration a.
In (2), the nominal observer is at x=1/a undergoing proper acceleration a.
In (3), the nominal observer is at x=1 undergoing proper acceleration 1.

I did not see what you were getting at when I made this post last night, but I see now that given the special locations of the nominal observers the time dilation ratio dtau/dt does resolve to unity for dx=dy=dz=0 and I have now edited this post. Thanks for taking the time to clarify DrGreg. Cheers.

 

[Altabeh]

Why do we need to assume no motion along the x axis?

It is because the geodesic equations tell us that the proper acceleration along the x-axis is dependent on the coordinate x, so to assume a uniform acceleration we only have a choice ahead: take x=x_0 and consider the motion happens on the planes parallel to yz-plane! One would then be able to dig from this assumption along with the two other geodesic equations that say both proper accelerations along y and z axes are zero, that the proper acceleration is uniform which in turn again resurrects the local discussion!

I am not sure what you are getting at here. I am sure we both agree that we can not have a gravitational field in flat spacetime, because any real gravitational field curves spacetime. I am sure you would also agree that we can create a simulation of a gravitational field by having a system of accelerating rockets, such that an observer confined to making local measurements inside a rocket artificially accelerating in flat spacetime would not be able to determine if he was in a gravitational field or not. That is the essence of the EP principle.

Exactly! But, as you can see, the trouble is we can very barely find those coordinates that meet the demands mentioned in our wish list.

EP is equivalence principle so no need to be accompanied by a further principle!

and states that this metric represents a "uniform gravitational field" and can not be transformed to flat spacetime by a coordinate transformation. Presumably the diagram is the POV of a free falling observer. I am not sure why the wordlines cross over each other or how this would be consistent with their claim that in this spacetime the accelerating observers measure their mutual spatial separation to be constant. Hmmm.

I don't think their claim is true unless discussing the whole stuff just locally by having EP hold in a curved spacetime!

AB

 

[yuiop]

Exactly! But, as you can see, the trouble is we can very barely find those coordinates that meet the demands mentioned in our wish list.

It occurs to me that we could construct a system of concentric rotating rings, such that each ring has equal centripetal acceleration. This means the angular velocity (or instantaneous tangential velocity) of each ring would be different, with the smaller inner rings rotating with greater angular velocity w=sqrt(g/r) but lesser tangential velocity v=sqrt(gr) with g=constant (assuming the Newtonian relationship for angular acceleration still holds here.. not sure). This would give a form of constant acceleration which would be equal for any given r. Might be interesting to analyse and might be something like the gravitational field of the infinitely long, rotating cylinder of dust that Ben is fond of.

An interesting property of this structure is that it has constant vertical spatial separtion, even from the point of view of the accelerating observers, but the horizontal distances are a bit messy. Judging by the very complicated metrics for the cylinder of dust, I imagine the mutual horizontal distances are a bit messy there too.

I am coming around to your point of view that it impossible to have a gravitational field that is uniform everywhere and has constant spatial distances from the POV of the accelerated observers, except as a very local aproximation.

EP is equivalence principle so no need to be accompanied by a further principle!

I know!.. I know!.. kicks self

 

[Altabeh]

It occurs to me that we could construct a system of concentric rotating rings, such that each ring has equal centripetal acceleration. This means the angular velocity of instantaneous tangential velocity of each ring would be different, with the smaller inner rings rotating faster so v=sqrt(gr) or w=sqrt(g/r) with g=constant (assuming the Newtonian relationship for angular acceleration still holds here.. not sure). This would give a form of constant acceleration which would be equal for any given r. Might be interesting to analyse and might be something like the gravitational field of the infinitely long, rotating cylinder of dust that Ben is fond of.

This seems to be a good thing and Ben is currently working on it through Petrov metric (I guess)! I'm really busy and can really barely find a little free time to spend on this! But if I could catch the mainstream of the paper given at post #4 regarding the Petrov metric (PM) which is claimed to be the only rotating vacuum solution to Einstein field equations, then I would throw around my own conclusion of the gravitational uniform field and how it can be made globally existent! I would be interested in grappling with the geodesic equations of the PM and checking if they can offer us a coordinates by which the uniform acceleration is guaranteed everywhere! I do not have any hope for this!

Ben is missing now for 1 day and I think he'll have a good "start" with the new idea!

AB

 

[bcrowell]

I would be interested in grappling with the geodesic equations of the PM and checking if they can offer us a coordinates by which the uniform acceleration is guaranteed everywhere!

The coordinates in which the metric is normally written are coordinates in which the proper acceleration of a particle released at rest is the same everywhere. See #79. Of course if you want that acceleration to *remain* constant forever (i.e., to be constant regardless of the particle's current velocity) then that's impossible.

But if I could catch the mainstream of the paper given at post #4 regarding the Petrov metric (PM) which is claimed to be the only rotating vacuum solution to Einstein field equations,

No, they don't claim that. There are many rotating vacuum solutions to the EFE. What is unique about the Petrov metric is its symmetry. It is "The only vacuum solution of Einstein’s equations admitting a simply-transitive four-dimensional maximal group of motions..."

 

[Altabeh]

I played around a little with the geodesic equations in the Petrov metric. Suppose you restrict yourself to a fixed z, and for convenience take an observer at an r such that \cos\sqrt{3}r=1, which means that the timelike coordinate is t. For a particle released at rest, the geodesic equations become
<br /> \ddot{t}=0<br />
<br /> \ddot{\phi}=0<br />
<br /> \ddot{r}=-\frac{1}{2}e^r \dot{t}^2<br />
where dots mean differentiation with respect to the affine parameter. If you take the affine parameter to be the particle's proper time, then \dot{t}=e^{-r/2} initially, and the particle accelerates toward lower values of r with a proper acceleration that is independent of r. If you go to a place where \cos\sqrt{3}r=-1, the timelike coordinate is again t, but I think the direction of time is reversed here as compared with at the place where \cos\sqrt{3}r=+1. Here the particle accelerates toward higher values of r. This seems to match up with what the Gibbons and Gielen paper describes about the geodesics: they oscillate back and forth in r.

So in this sense the Petrov metric seems to be a better realization of the uniform gravitational field than the Rindler coordinates, because the proper acceleration has the same magnitude everywhere. On the other hand, it does have strange properties like CTCs, and that's undesirable.

[EDIT] I think the constancy of the proper acceleration follows from the fact that in addition to the three obvious Killing fields, there is a fourth one involving r. If I've got that right, then the restriction of my calculation above to certain values of r can be relaxed, and the result still holds.

If your calculations are all correct, then the assumption

\dot{t}=e^{-r/2}

is only in accordance with the first geodesic equation if one of the two conditions below holds:

1) r is taken to be a constant, so the motion is much restricted,
2) the proper velocity of particles along r is zero which is the case when an instantaneous comoving observer measures stuff regarding the motion of particles.

Of course the first condition would not be regarded as a tool of providing a constant proper acceleration globally. So talking about the other condition, and yet highly unphysical one, I don't see any better situation than when we were dealing with the Rindler metric because this is again dependent on position which is cured by setting the initial time of the observer's clock at the beginnig of the motion. And about the motion, there is no such thing along r though for an instantaneously at rest comoving observer along r at a constant z but with a varying \phi, one can have the proper acceleration \ddot{r} observed instantaneously and uniformly! But remember that this approach, I think, is so insubstantial in the context of global constancy of proper acceleration: we are actually binding time up into space through \dot{t}=e^{-r/2} in a way to have a uniform acceleration observed by some observer being instantaneously at rest while this is not essentially happening in the spacetime. I mean if you have z also involved as a freely varying coordinate, the situation gets much complicated and requires much more things that we need to mean in essence by 'global discussion' of the whole scenario than now!

Keeping in mind the global discussion, how one would show me all four elements can be adjusted somehow so as to make thee proper acceleration uniform in Petrov metric?

AB

 

[bcrowell]

If your calculations are all correct, then the assumption

\dot{t}=e^{-r/2}

is only in accordance with the first geodesic equation if one of the two conditions below holds:

1) r is taken to be a constant, so the motion is much restricted,
2) the proper velocity of particles along r is zero which is the case when an instantaneous comoving observer measures stuff regarding the motion of particles.

As stated explicitly in the text that you quoted, the result only holds for a particle that is instantaneously at rest. Please see my #37, in which I pointed out to you that I had already explicitly stated this twice, in #19 and #26.

Keeping in mind the global discussion, how one would show me all four elements can be adjusted somehow so as to make thee proper acceleration uniform in Petrov metric?

No adjustment is necessary. This was the result of the calculation in #79.

Folks, this has been a very enjoyable thread, but I think I'm going to stop following it now. Thanks, everyone, for sharing your valuable and insightful comments!

 

[Altabeh]

Let me give you a vivid picture of the globally uniform gravitational field and how it can be realized:

1- The metric must be Ricci-flat or a vacuum solution to the Einstein field equations;
2- In Cartesian coordiantes (t,x,y,z), the geodesic equations have the form

\ddot{t}, \ddot{x}, \ddot{y} and \ddot{z} must all be constant.
Or the Christoffel symbols be all constant.

It is not impossible to look for such a metric and I have some plan to do find one! I'll give it a go someday I find a little free time!

AB

 

[Altabeh]

As stated explicitly in the text that you quoted, the result only holds for a particle that is instantaneously at rest. Please see my #37, in which I pointed out to you that I had already explicitly stated this twice, in #19 and #26.

My bad, I didn't see 'em!

No adjustment is necessary. This was the result of the calculation in #79.

If not take instantaneously at rest as an "adjustment"!

Folks, this has been a very enjoyable thread, but I think I'm going to stop following it now. Thanks, everyone, for sharing your valuable and insightful comments!

Why? Anyways it's been too enjoyable for me, too and hope you are not leaving us forever!

AB

 

[Mentz114]

I'm signing off too with thanks to all posters. I have learned some things I should have known. I conjecture that -

1. The Rindler coords have nothing to do with gravity.
2. They are not global but apply locally to an accelerating observer in Minkowski space-time
3. The parameter 'a' is related to the observer's proper acceleration.
4. So far no-one has come up with a uniform gravitational field, either in vacuum or matter.