# Uniformly accelerating frame of reference

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1. Nov 13, 2015

### 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?

2. Nov 13, 2015

### Orodruin

Staff Emeritus
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 space ship paradox.

3. Nov 13, 2015

### A.T.

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.

4. Nov 13, 2015

### sweet springs

Thanks.

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 ?

5. Nov 13, 2015

### Staff: Mentor

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.

6. Nov 13, 2015

### jbriggs444

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

7. Nov 13, 2015

### Staff: Mentor

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.

8. Nov 13, 2015

### A.T.

Indeed, Dale is right here.

9. Nov 13, 2015

### pervect

Staff Emeritus
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.

10. Nov 13, 2015

### Orodruin

Staff Emeritus
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.

11. Nov 13, 2015

### 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.

Last edited: Nov 13, 2015
12. Nov 13, 2015

### Staff: Mentor

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

13. Nov 13, 2015

### Orodruin

Staff Emeritus
No, this is wrong.

And 2 ...

14. Nov 13, 2015

### 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.

Last edited: Nov 13, 2015
15. Nov 13, 2015

### Staff: Mentor

Oops! Yes.

16. Nov 13, 2015

### 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.

Last edited: Nov 13, 2015
17. Nov 13, 2015

### stevendaryl

Staff Emeritus
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.

18. Nov 13, 2015

### 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?

Last edited: Nov 13, 2015
19. Nov 13, 2015

### stevendaryl

Staff Emeritus
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).

20. Nov 14, 2015

### sweet springs

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?

Last edited: Nov 14, 2015