# B Stationary frames of reference

1. Jan 4, 2017

### name123

What determines whether a frame of reference can be considered stationary? I assume it is not allowed that the earth be considered stationary and the universe is moving around it for example, as I would have thought that would lead to observation of faster than light movement.

In https://en.wikipedia.org/wiki/Fictitious_force it states:

A fictitious force, also called a pseudo force,[1] d'Alembert force[2][3] or inertial force,[4][5] is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference frame.

Does the detection or absence of the detection of a "fictitious force" determine whether the mass need needs to have its motion described using a non-intertial frame of reference, such that it cannot be considered stationary?

2. Jan 4, 2017

### phinds

I assume you mean inertial. A frame of reference is inertial if it is not accelerating. So, for example, when taken locally, the surface of the Earth can be used as an inertial frame of reference for things like Einstein's train thought experiment, but globally it's a rotating (non-inertial) frame.

3. Jan 4, 2017

### Staff: Mentor

In principle, any object can be considered "stationary" with an appropriate choice of coordinates. Choosing non-inertial coordinates in which an object with nonzero proper acceleration is at rest does not preclude you from treating that object as "stationary". It just means the coordinates are non-inertial.

4. Jan 4, 2017

### name123

But what establishes that it cannot be considered inertial and that the universe is orbiting around it?

5. Jan 4, 2017

### Staff: Mentor

The fact that it has nonzero proper acceleration. That can be measured directly by an accelerometer attached to the object.

6. Jan 4, 2017

### name123

So it could be considered that the Earth was stationary and that the universe was orbiting around it?

7. Jan 4, 2017

### Staff: Mentor

No, it wouldn't. The coordinate speed of objects far enough away would be greater than the number $c$, but the coordinate speed of light itself at those distances would also be greater than $c$ (and by a larger margin). So no object would move faster than light beams at the same location, which is the correct way of formulating the requirement that nothing moves faster than light.

8. Jan 4, 2017

### name123

But if the coordinate speed of objects far enough away were observed at a number greater than $c$ as well as the light at those distances, then are not those objects and the light at those distances being observed as travelling faster than $c$?

You seem to be stating that moving faster than $c$ is not the same as moving faster than light, but the observation of light moving faster than $c$ would at least indicate that light can be observed to move faster than $c$ would it not?

9. Jan 4, 2017

### Staff: Mentor

Sure, but "traveling faster than $c$" in this sense is not forbidden by the laws of physics.

Same answer as above.

10. Jan 4, 2017

### Staff: Mentor

Yes, if you adopt appropriate coordinates. We use such coordinates in, for example, navigation all the time.

11. Jan 4, 2017

### name123

So could you please give an example what happens in the Special Relativity equations, as would they not start involving imaginary numbers? The calculation of Gamma for example would involve the square root of a negative number would it not as v > c ?

12. Jan 4, 2017

### Staff: Mentor

The equations you refer to assume an inertial frame. If you have a non-inertial frame they don't apply.

13. Jan 4, 2017

### phinds

That equation applies within the same inertial frame of reference and for proper motion. Far distant objects that are receding from us at greater than c they are not IN the same inertial frame and so the equation does not apply.

EDIT: I see Peter beat me to it.

Also, I should have added, recession velocity is not proper motion.

14. Jan 4, 2017

### Staff: Mentor

Careful. If we are assuming flat spacetime, which the OP seems to be, all objects are "in" every inertial frame, since inertial frames are global. For inertial frames to be confined to a certain small patch of spacetime, spacetime has to be curved, and it doesn't seem like we're getting into that here.

Again, this concept applies to our expanding universe, which is not flat spacetime, so I'm not sure it's within the scope of this discussion.

15. Jan 4, 2017

### phinds

Now I'm confused. If I understand the thread correctly, you are saying that the Lorentz transformation doesn't apply to far distance receding galaxies BECAUSE they are receding at an accelerating rate, not because they are not in the same inertial frame as us and not because their motion is not proper motion. Have I got that right?

16. Jan 4, 2017

### pervect

Staff Emeritus
I'm not sure what sense the word "stationary" is being used in this thread. If it's the same as in the wiki article on stationary space times, then all you need is an (asymptotically) time-like Killing vector. Or to put it another way, a set of metric coefficients must exist describing the space time that are independent of time.

Static space times are more restrictive, basically they're not allowed to rotate. A Schwarzschild space-time would be static and stationary, a Kerr space-time would not be static, but would be stationary.

If some other sense of the word "stationary" is being used, all bets are off.

17. Jan 4, 2017

### Staff: Mentor

No. I wasn't talking about receding galaxies at all, since that involves curved spacetime. I was talking about a hypothetical non-inertial coordinate chart set up in flat (or flat to a good enough approximation) spacetime, in which the rotating Earth was at rest and the "rest of the universe" (meaning here distant objects placed in flat spacetime) was rotating around it. The "faster than $c$" motion would be the tangential coordinate velocity of a sufficiently distant object in this non-inertial chart. That is based on my understanding of what the OP was asking about; yes, it's a highly idealized scenario, but it's the one I took the OP to be describing.

18. Jan 4, 2017

### Staff: Mentor

My understanding is that the OP just used that term to mean "at rest in some chosen coordinate chart". The term does have the technical meaning you describe, but I don't think the OP intended that usage.

19. Jan 4, 2017

### phinds

Ah, I missed that and obviously had the wrong idea about what you were saying. Thanks.

20. Jan 5, 2017

### Staff: Mentor

@name123 , you appear to have missed this previous reply by Peter regarding inertial vs non-inertial frames:
I'll say it in a way that may speak to you better: velocities are relative, but rotations are absolute. That's how we know it is the earth rotating and not the universe (even if for some purposes we consider the Earth to not be rotating).