Uniformly accelerating frame of reference

[sweet springs]

In an inertial frame of reference let numerous standard rockets that load synchronized standard clocks place on all the space lattice. Simultaneously the rockets start to move with same intrinsic acceleration to the same direction. In other words there exists a common instantaneous inertial frame of reference to all the rockets.
May I understand it as a description of uniformly acceｌeｒating frame of reference ? 　If so I worry Rindler coordinates cover only part of the frame of reference. Is it OK?

 

[Orodruin]

In other words there exists a common instantaneous inertial frame of reference to all the rockets.

No, there does not. The clocks will not generally keep being synchronised and the space ships will have different velocities in the instantaneous rest frame of any of them. I suggest looking up and trying to understand Bell's spaceship paradox.

 

[A.T.]

n other words there exists a common instantaneous inertial frame of reference to all the rockets.

No, there does not. The clocks will not generally keep being synchronised and the space ships will have different velocities in the instantaneous rest frame of any of them.

At any time point of the acceleration, you can construct an inertial frame, where all the ships are instantaneously at rest. But the clocks of the ships have offsets in such a frame.

 

[sweet springs]

Thanks.

At any time point of the acceleration, you can construct an inertial frame, where all the ships are instantaneously at rest. But the clocks of the ships have offsets in such a frame.

Very interesting. Do the corresponding clocks in an inertial frame and the ships differ with same amount or various values according to where they are ?

 

[Dale]

At any time point of the acceleration, you can construct an inertial frame, where all the ships are instantaneously at rest.

I don't think that this is correct. Given the problem description it seems that the rockets all have the same velocity in the original frame. So if you transform to any other frame the rockets will have different velocities. So they would not all be at rest wrt each other.

 

[jbriggs444]

Very interesting. Do the corresponding clocks in an inertial frame and the ships differ with same amount or various values according to where they are ?

When are you doing the comparison? Note that there is a can of worms in the word "when".

 

[Dale]

In other words there exists a common instantaneous inertial frame of reference to all the rockets.

As you have described the setup this is not true.

You should go through the math on this one and learn how to calculate this kind of thing.

 

[A.T.]

I don't think that this is correct. Given the problem description it seems that the rockets all have the same velocity in the original frame. So if you transform to any other frame the rockets will have different velocities. So they would not all be at rest wrt each other.

Indeed, Dale is right here.

 

[pervect]

In an inertial frame of reference let numerous standard rockets that load synchronized standard clocks place on all the space lattice. Simultaneously the rockets start to move with same intrinsic acceleration to the same direction. In other words there exists a common instantaneous inertial frame of reference to all the rockets.
May I understand it as a description of uniformly acceｌeｒating frame of reference ? 　If so I worry Rindler coordinates cover only part of the frame of reference. Is it OK?

Assuming that when you say " Simultaneously the rockets start to move with same intrinsic acceleration", you mean what I think you mean that the simultaneity is defined by the initial inertial frame that the rockets are initially in, and that "intrinsic acceleration" means "proper acceleration", then you have Bell's spaceship paradox. The rockets will NOTkeep a constant distance apart, the distance between any pair of rockets will decrease with time. If you assign labels to the rockets based on their initial position coordinates in their initial inertial frame and use those labels as spatial coordinates, you'll have a perfectly valid coordinate system, but it won't be the Rindler coordinate system. Because the distance between any pair of rockets changes with time, the value of $g_{00}$ in this coordinate system will be a function of time. If this coordinate system has a name of its own, I haven't seen it. It's definitely not what's usually meant by a "uniformly accelerating frame". What's usually meant by a "uniformly accelerating frame" in the literature is an arrangement of rockets that keep a constant distance from each other, adjusting their acceleration profile as needed. This ensemble of rockets keeping a constant distance from each other defines the Rindler frame.

 

[Orodruin]

The rockets will NOTkeep a constant distance apart, the distance between any pair of rockets will decrease with time

This depends on the frame you consider things in. In the instantaneous rest frame of one of the rockets, the distance between rockets will increase. In the original rest frame it will stay the same.

 

[sweet springs]

Thanks all for advice. Let me summarize my understanding here.

In an inertial frame of reference(IFR) let numerous standard space ships that load synchronized standard clocks place on all the space lattice, labelled with the coordinates. Simultaneously the ships start to move with same proper acceleration to the same direction.

a. There always is an instataneous IFR where all the ships are at rest.
b. In the instataneous IFR the distances between the ships in that direscion increase.
c. In the instataneous IFR the clocks are not syncronized in that direction. Faster forward, delayed backward in proportion to distance.

 

[Dale]

a. There always is an instataneous IFR where all the ships are at rest.

No. This is not correct. See posts 5, 7, and 8.

 

[Orodruin]

There always is an instataneous IFR where all the ships are at rest.

No, this is wrong.

No. This is not correct. See posts 5, 7, and 8.

And 2 ...

 

[sweet springs]

Thanks. I try correction.

a(revised). There always is an instataneous IFR where ships on a traverse line are at rest and others have different speeds in that direction. Faster forward, backward faster backward in proportion to distance.

 

[Dale]

And 2 ...

Oops! Yes.

 

[sweet springs]

Give me one more clarification please.

d. Let the ships have come to the initial fire state by having kept accerelation of the same manner at past. In an instantaneous IFR, distance of the ships in that direction were longer. The ships were coming closer.

It seems that a kind of bounce out from contraction to inflation takes place at the initial time.

 

[stevendaryl]

In an inertial frame of reference let numerous standard rockets that load synchronized standard clocks place on all the space lattice. Simultaneously the rockets start to move with same intrinsic acceleration to the same direction. In other words there exists a common instantaneous inertial frame of reference to all the rockets.
May I understand it as a description of uniformly acceｌeｒating frame of reference ? 　If so I worry Rindler coordinates cover only part of the frame of reference. Is it OK?

Let me clarify the situation (hopefullly) by pointing out that there are two different notions of a line of rockets accelerating together, and they're not the same in Special Relativity (although they are the same in Newtonian physics):

1. The line of rockets start off at rest in some rest frame, and then they all accelerate with the same acceleration profile (that is, they all accelerate such that they all "feel" the same g-forces).
2. The line of rockets start off at rest in some frame, and then they all accelerate in such a way that the distance between the rockets remain constant--as viewed by the travelers on board the rockets.

These aren't the same thing. If the rockets all have the same acceleration profile, then

• According to observers in the original rest frame, the distance between the rockets remains constant.
• According to observers on board the rockets, the distance between the rockets steadily increases.

If the rockets keep the same distance apart (which is called "Born-rigid acceleration"), then

• According to observers in the original rest frame, the distance between the rockets shrinks with time.
• According to observers on board the rockets, the distance between rockets remains constant.

In the case of Born-rigid acceleration, the rockets do NOT have the same acceleration profile: the rockets in the rear have a larger acceleration (and feel greater g-forces) than the rockets in the front.

 

[sweet springs]

"Born-rigid acceleration" was new to me. Thanks.

"Uniformly accerelerating frame of reference" has uniformity of proper acceleration in time. I have misunderstood that Uniformity also means that all the ships share common proper acceleration.

In this context, proper accerelation values of not only one ship but also of another one (a higher or lower ship) toghether with the distance between are necesary to identify the "uniformly accerelerating frame of reference". Am I right?

 

[stevendaryl]

"Born-rigid acceleration" was new to me. Thanks.

The intuitive meaning is that an object undergoing Born-rigid (named after the physicist Born, not like "Born to run" or "Born to be wild") acceleration will feel no stretching or compression forces. In contrast, if you have a line of rockets accelerating with the same profile, then a string connecting the rockets will be stretched and will eventually break (that's Bell's spaceship paradox).

 

[sweet springs]

In this context, proper accerelation values of not only one ship but also of another one (a higher or lower ship) toghether with the distance between are necesary to identify the "uniformly accerelerating frame of reference". Am I right?

Let I be in a Rindler space. I do not where I am or my coordinate z and acceleration parameter of a of the system.
I obeserve metric here and get $\sqrt{g_{00}(z)}=\alpha$.
I climb up h and observe metric and get $\sqrt{g_{00}(z+h)}=\beta$
Applying them to the formula of metric,
$1+\frac{az}{c^2}=\alpha$
$1+\frac{a(z+h)}{c^2}=\beta$
Thus I get
$\frac{a}{c^2}=\frac{\beta-\alpha}{h}$, $z=\frac{(\alpha-1)(\beta-\alpha)}{h}$

I have never seen such a way in textbooks. Do not we need such two point observation to identify Rindler space?

Or let us define time of the Rindler space be proper time where I am so that $\alpha=1$ thus z=0 here, a is acceleratin I am feeling. Does it work?

 

[Dale]

There always is an instataneous IFR where ships on a traverse line are at rest and others have different speeds in that direction. Faster forward, backward faster backward in proportion to distance.

Yes, although I don't know if it is in proportion to the distance or the square of the distance or something. I would have to work out the math.

 

[Dale]

It seems that a kind of bounce out from contraction to inflation takes place at the initial time.

Yes, that sounds right.

 

[Orodruin]

Yes, although I don't know if it is in proportion to the distance or the square of the distance or something. I would have to work out the math.

It is relatively easy to work out, but you need to decide which distance you are talking about. As a matter of fact, I just gave essentially this problem to my students for their homework problems.

 

[pervect]

This depends on the frame you consider things in. In the instantaneous rest frame of one of the rockets, the distance between rockets will increase. In the original rest frame it will stay the same.

I was regarding the "frame of reference" as being defined by the congruence of worldlines of the rockets, as described in the original post:

http://arxiv.org/abs/gr-qc/9908061][/PLAIN]
(c) Monad’s methods
. A frame of reference is identified with a non-null (non-isotropic) (time-like) contravariant vector field interpreted as the velocity of an observer (material point).

The point is that if the "material observers", i.e. the rockets all of whom share a constant proper acceleration, measure their distance to their nearest neighbors as changing with time. I.e. we consider some rocket A that's part of the congruence of worldlines, and some nearby rocket B. The rockets measure their distance to each other by exchanging radar signals - or, in Bell's spaceship paradox, by means of a string. In either case, the distance changes with time - in the Bell spaceship paradox, the string breaks.

The other point is that in the Rindler congruence, "the strings don't break", but the proper acceleration of the rockets varies with position. At the Rindler horizon, the required proper acceleration to maintain a constant distance to the other nearby members of the congruence becomes infinite.

 

[sweet springs]

Thanks for good discussions. Let me deepen my understanding.

a. There always is an instataneous IFR where all the ships are at rest.

The wrong answer a. seems to turn out to be right in another case that the ships are arranged to represent lattices of uniformly accelerating frame of reference. i.e.,

In an inertial frame of reference(IFR) let numerous standard space ships that load synchronized standard clocks place on all the space lattice, labelled with the coordinates.
Simultaneously the ships start to move to the same direction with constant proper acceleration. Proper accerelation of ships are arranged so that ships on a traverse line have same proper acceleration and nearby traverse lines keep constant distance.

a. There always is an instataneous IFR where all the ships are at rest.
b. In the instataneous IFR the distances between the ships in that direcion are constant.
c. In the instataneous IFR the clocks are not syncronized in that direction. Faster forward, delayed backward.
d. Let the ships have come to the initial fire state by having kept accerelation of the same manner at past. In the instantaneous IFR, distance of the ships in that direction were constant.

Rindler coordinate is interpreted as continuous get off and on the instantaneous IFRs.

 

[sweet springs]

Or let us define time of the Rindler space be proper time where I am so that $\alpha=1$ thus z=0 here, a is acceleratin I am feeling. Does it work?

Let any place be z=0 where z-axis is the direction of acceleration, proper time there be time of the system and proper acceleration there be parameter of the system. Now I know that it describes Rindler space.
Thanks. .

 

[pervect]

Thanks for good discussions. Let me deepen my understanding.The wrong answer a. seems to turn out to be right in another case that the ships are arranged to represent lattices of uniformly accelerating frame of reference. i.e.,

In an inertial frame of reference(IFR) let numerous standard space ships that load synchronized standard clocks place on all the space lattice, labelled with the coordinates.
Simultaneously the ships start to move to the same direction with constant proper acceleration. Proper accerelation of ships are arranged so that ships on a traverse line have same proper acceleration and nearby traverse lines keep constant distance.

a. There always is an instataneous IFR where all the ships are at rest.

The "frame of reference" in which all the ships are "at rest" isn't an inertial frame, because it has pseudo-gravitational forces due to the ships acceleration. Inertial frames of reference, by defintion, don't have pseudo-gravitational forces, objects in them obey Newton's laws. Because "frames of reference" is rather ambiguous, I prefer to say there is a coordinate system (rather than a frame of reference) and that in this coordinate system all of the above ships have constant coordinates (rather than saying they are at rest).

This coordinate system has a space-time metric, the Rindler metric.

This coordinate system doesn't cover all of space-time. There are various ways of "explaining"
this, one of them is to point out that space-ships must have a finite proper acceleration, and that the required proper acceleration to hold station becomes infinite at what is known as the "Rindler horizon". Another way of looking at it is to look at the metric coefficients, and note that $g_{00}$ becomes equal to zero.

 

[sweet springs]

The "frame of reference" in which all the ships are "at rest" isn't an inertial frame, because it has pseudo-gravitational forces due to the ships acceleration.

Thanks. The "frame of reference" in which all the ships are "at rest" INSTANTANEOUSLY is an inertial frame. The ships with their proper accelerations do not stay at rest in this IFR. Continuous transfer to such instantaneous IFRs make Rindler coordinate with gravitation.

 

[pervect]

Second try:

If you pick one specific observer from the congruence of observers, at any instant of time that observer has an instantaneously co-moving inertial reference frame, henceforth abbreviated as ICMIRF specific to that observer at that instant of time. Establishing that all the other observers in the congruence share the same instantaneously co-moving frame is something that needs to be determined. In general it's not true. The only way I know of checking this involves a fair amount of fairly advanced math. You represent an inertial reference frame by its basis vectors, $\hat{e_x}, \hat{e_y}, \hat{e_z}$. To compare inertial reference frames, you need the concept of the connection coefficients and parallel transport. Given a metric, you can compute the connection coefficients and determine if it's true that the basis vectors in the second frame (which we will indicate by a prime symbol) are the same as the basis vectors in the original frame - i.e. that $\hat{e'_x}=\hat{e_x}, \hat{e'_y}=\hat{e_y}, \hat{e'_z}=\hat{e_z}$. Fortunately, in the context of special relativity, parallel transport is path independent, so the notion of comparison of the basis vectors is well defined and path independent.

Some counterexamples of frames where the ICMIRF's are not shared are rigid rotating frames, where every point in the rigid rotating frame has a different ICMIRF, and expanding frames, such as the "Bell Spaceship" frame .

 

[sweet springs]

Thanks perverct. So I try to correct. In uniformly accelerating system the following applies again,

a(revised). There always is an instantaneous IFR where ships on a traverse line are at rest and others have different speeds in that direction. Faster forward, backward faster backward in proportion to distance.

where not "in proportion to distance" but in a subtle way so that the Bell's threads are not broken.
It is interesting that the speed of the both ends of thread are different in the ICMIFRs but the threads are not torn apart. The speed of the bottom line of the ships reach light speed.

 

[pervect]

That's still not what I said :(.

There is always an IFR at any point on the worldline of an observer in arbitrary motion. I didn't say anything about the relative velocity of the space-ships in the Rindler congruence, the remarks a made were about their acceleration.

I didn't make any statement about whether or not the ships were at rest. I believe we can probably say that the space-ships in the Rindler congruence are at rest, but it would involve a rather lenghtly discussion and I"m not sure we share the same notion of what "at rest" even means. (I'd be thinking of "at rest" in terms of parallel transport - is this something you're familiar with at all?).

What I was saying was not a statement about the ships being "at rest" or not, but rahter about the ships having different proper accelerations. Proper accelerations are not velocities.

I also said that the issue of determining if two observers shared the "same" IFR was rather tricky, and again I suspect we're not using the same conceptual underpinnings.

 

[sweet springs]

I am puzzled. As for the ships representing RIndler coordinates, in the instantaneous IFR of one ship to be at rest, that is GLOBAL NOT LOCAL, we cannot say anything about velocity of other ships, you say? I agree with you that all the ships have their own
proper accelerations. We can calculate accelerations of the ships in the instantaneous GLOBAL IFR by their proper accelerations and velocities.

 

[pervect]

I am puzzled. As for the ships representing RIndler coordinates, in the instantaneous IFR of one ship to be at rest, that is GLOBAL NOT LOCAL, we cannot say anything about velocity of other ships, you say? I agree with you that all the ships have their own
proper accelerations. We can calculate accelerations of the ships in the instantaneous GLOBAL IFR by their proper accelerations and velocities.

It rather depends on what you mean by "global". Any specific observer has an IFR, and also a specific notion of simultaneity - what events happen at the same time in the IFR of that observer. (I hope you are familiar with the second point and just forgetting about it, there are a number of threads on the topic, most of which drag out to great lengths).

However, in general different observers can and do have different IFR's and different notions of simultaneity. Because notions of simultaneity can vary, and because we do have acceleration, it's not at all clear if objects that are at rest in one observers IFR using that observers notion of simultaneity are "at rest" in another observer's IFR which in general may have a different notion of simultaneity

[add]This is all rather abstract, so I'll just briefly mention a specific example where I know this is an issue, the issue of rotating frames. There is no single inertial frame that covers all of a rotating frame.

The only way I'm aware of involves the mathematical machinery designed to compare inertial frames and "glue them together". This mathematical machinery consists of describing inertial frames of reference by their basis vectors, defining maps from one frame's basis vectors to another frame's basis vectors (the maps are called connection coefficients), and another set of concepts called "parallel transport" that describes how you transport basis vectors.

Some of the above concepts are not esseintial to learn about the Rindler congruence,so your focus on "at rest" is a bit of a diversion from the minimum necessary to understand the Rindler coordinates. A textbook reference that goes into some detail in derving the Rindler coordinates is MTW's textbook "Gravitation", which goes through all the math. You definitely need the concept of basis vectors to use MTW's approach. It would be very helpful to understand the concept of connection coefficients and transport laws, but you might be able to get a partial understanding by ignoring the fine details on that part of the text involving the transport laws, and simply assuming that the text give you a good recipie (which is called fermi walker transport) as to how to accomplish this task.

 

[sweet springs]

It rather depends on what you mean by "global". Any specific observer has an IFR, and also a specific notion of simultaneity - what events happen at the same time in the IFR of that observer. (I hope you are familiar with the second point and just forgetting about it, there are a number of threads on the topic, most of which drag out to great lengths)..

I mean GLOBAL IFR as familiar one in special relativity. In special relativity we care about IFR but do not care where in this IFR observers, who share the same simultaneity, are. They can be at elsewhere. They are not crews of space ships who feel gravity.

 

[pervect]

We seem to be going around in circles. Let's try this.

Assume we have some global IFR in special relativity, with coordinates (t,x,y,z). Then we can assign Rindler coordinates (T,X,Y,Z) in the following manner

$$T = (x+1/g) \sinh gt \quad X = (x + 1/g) \cosh gt \quad Y = y \quad Z=z$$

See MTW page 173, though I've adapted their notation slightly. If you want (t,x,y,z) = (0,0,0,0) to map to (T,X,Y,Z) = (0,0,0,0) you need to modify the MTW formulas slightly with a subtractive constant:

$$T = (x+1/g) \sinh gt \quad X = (x + 1/g) \cosh gt -1/g \quad Y = y \quad Z=z$$

(T,X,Y,Z) are not the coordinates in any "inertial frame". They are generalized coordinates in the Rindler frame.

A bit of algebra will give you the Rindler line element (metric), if you take the line element in your global inertial frame $-dt^2 + dx^2 + dy^2 + dz^2$, and substitute in the above formula, after a lot of algebra you get the line element in terms of the Rindler coordinates, $-(gX+1)dT^2 + dX^2 + dY^2 + dZ^2$.

You can also say that spatial part of the metric in coordinates (X,Y,Z) is Euclidean, so the subspace of T=constant represents a Euclidean spatial sub=manifold of space-time, thus (X,Y,Z) can be interpreted as having the usual physical signficance in the Rindler frame, similar to the signficance that (x,y,z) have in the global inertial frame. Note that t is not equal to T, so surfaces of constant t are not surfaces of constant T.

This much I can vouch for. I can't vouch for any remarks about "things being at rest" without some precise defintions of what is meant by "at rest", it's just too vague without a mathematical statement (of very rigorous non-mathematical one) as to exactly what you mean. The above equations give you an operational defintion of Rindler coordinates, so perhaps you can attempt to answer your own question. As for the derivation of the above transformations, I can refer you to MTW which derives the above, but you'll need to be familiar with (at a minimum) basis vectors. It's not terribly complicated stuff (though the notation can be confusing and needs careful explanation). It would be a lot of work to attempt to summarize the textbook, and I feel like we've been going around in circles on much simpler matters, so I'm reluctant to attempt it, though it's possible I could be convinced otherwise.

 

[sweet springs]

To me at rest is having zero velocity. At rest or not depends on which FR you take. I am afraid it is different from your "things being at rest"

Please take a look at a drawing to express the situation I take.
.

Dick keeps at rest in the uniformly acceleration system. Dick is at rest instantaneously in the IFR. Do they contradict?
Any teaching is appreciable. I still wonder Tom and Harry are also at rest instantaneously or not. I try math you suggested.
Thanks.

 

[sweet springs]

Thanks for good discussions. Let me deepen my understanding.The wrong answer a. seems to turn out to be right in another case that the ships are arranged to represent lattices of uniformly accelerating frame of reference. i.e.,

In an inertial frame of reference(IFR) let numerous standard space ships that load synchronized standard clocks place on all the space lattice, labelled with the coordinates.
Simultaneously the ships start to move to the same direction with constant proper acceleration. Proper accerelation of ships are arranged so that ships on a traverse line have same proper acceleration and nearby traverse lines keep constant distance.

a. There always is an instataneous IFR where all the ships are at rest.
b. In the instataneous IFR the distances between the ships in that direcion are constant.
c. In the instataneous IFR the clocks are not syncronized in that direction. Faster forward, delayed backward.
d. Let the ships have come to the initial fire state by having kept accerelation of the same manner at past. In the instantaneous IFR, distance of the ships in that direction were constant.

Rindler coordinate is interpreted as continuous get off and on the instantaneous IFRs.

Now I check math and I was right.

 

[sweet springs]

We seem to be going around in circles. Let's try this.
This much I can vouch for. I can't vouch for any remarks about "things being at rest" without some precise defintions of what is meant by "at rest", it's just too vague without a mathematical statement (of very rigorous non-mathematical one) as to exactly what you mean.

I will try to explain two possible meaning of "at rest".

In instataneous IFR, one ship has no velocity at the instant. Other ships would have velocities of non zero. Acceleration does not matter.

In proper frame of reference, or comoving frame of reference, all the ships keep zero velocity and zero acceleration. Metric of the frame would be function of time-space coordinates thus it could not be an IFR.

I take the former one. The latter might be of your concern.

 

[pervect]

I will try to explain two possible meaning of "at rest".

In instataneous IFR, one ship has no velocity at the instant. Other ships would have velocities of non zero. Acceleration does not matter.

In proper frame of reference, or comoving frame of reference, all the ships keep zero velocity and zero acceleration. Metric of the frame would be function of time-space coordinates thus it could not be an IFR.

I take the former one. The latter might be of your concern.

I think I agree that in the first sense, the velocities are zero, but I took some shortcuts in the analysis.Using only 2 dimensions in Rindler coordinates, t,x, with $ds^2 = -(1+gx)^2 dt^2 + dx^2$ the 4-velocity of a ship with a constant Rindler coordinate x is $\partial t/\partial \tau = 1/(gx+1)$, $\partial x / \partial \tau = 0$

In Minkowskii coordinates of the true globallyi inertial frame, the 4-velocity the 4-velocity using the transforms from post #35 is
$\partial T / \partial \tau = (dT/dt) (dt/d\tau) = [(gx+1) \cosh gt] [1/(gx+1)] = \cosh gt$, $\partial X / \partial \tau = (dX/dt) (dt/d\tau) = \sinh gt$

So the 3-velocity $dX/dT$ is $tanh gt$. Using the equations from post #35 (the second set with the origin offset) this is just $dX/dT = T/(X+1/g)$.

So it's obvious that at T=0, all the velocities are zero. It's not particularly obvious what happens when we do a Lorentz boost to a co-moving frame at some time T that is nonzero, but here is where I take a shortcut. There isn't anything particularly special about T=0, so the velocities should be zero in the comoving frame, though the calculations would be rather tedious.

 

[PeterDonis]

There isn't anything particularly special about T=0

There isn't for the Rindler congruence (the set of worldlines in which the proper acceleration varies as 1 / x), but there is for the Bell congruence (the set of worldlines in which the proper acceleration is the same on every worldline); in the latter case, $T = 0$ in the global inertial frame is the only spacelike hypersurface in which all of the worldlines in the congruence are simultaneously at rest.

It looks like both congruences are being discussed in this thread; it might help to be very clear about which one is being talked about.

 

[sweet springs]

There isn't for the Rindler congruence (the set of worldlines in which the proper acceleration varies as 1 / x), but there is for the Bell congruence (the set of worldlines in which the proper acceleration is the same on every worldline); in the latter case, $T = 0$ in the global inertial frame is the only spacelike hypersurface in which all of the worldlines in the congruence are simultaneously at rest.

It looks like both congruences are being discussed in this thread; it might help to be very clear about which one is being talked about.

Thanks. I thought about the Rindlere congruence with rockets are tighten with threads. In the Rindler congruence threads are tight but not torn apart that is usulally reffered in Bell's congruence. It might have been a misconfusing saying.

 

[pervect]

There isn't for the Rindler congruence (the set of worldlines in which the proper acceleration varies as 1 / x), but there is for the Bell congruence (the set of worldlines in which the proper acceleration is the same on every worldline); in the latter case, $T = 0$ in the global inertial frame is the only spacelike hypersurface in which all of the worldlines in the congruence are simultaneously at rest.

It looks like both congruences are being discussed in this thread; it might help to be very clear about which one is being talked about.

Yes - my remarks and calculations were specific to the Rindler congruence. It would be better (but more work) not to take shortcuts (as I did) which might not apply in other circumstances. It's probably worth noting that the Bell congruence, the sliding block congruence and a rotating congruence would all likely NOT satisfy the "at rest" criterion from the viewpoint of an ICMIRF centered around one specific reference observer.

 

[PeterDonis]

It's probably worth noting that the Bell congruence, the sliding block congruence and a rotating congruence would all likely NOT satisfy the "at rest" criterion from the viewpoint of an ICMIRF centered around one specific reference observer.

Yes, I think this is correct. But it's also worth noting that the reason for this is different in the case of the Bell congruence, compared to the other two. The Bell congruence has nonzero expansion (but zero shear and vorticity); the other two have nonzero vorticity (but zero expansion and shear).