Uniformly accelerating frame of reference

[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 definition 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).