Inconsistent forms of the metric in a uniform field

[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