Stationary frames of reference

[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?

 

[phinds]

What determines whether a frame of reference can be considered stationary?

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.

 

[PeterDonis]

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?

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.

 

[name123]

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.

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

 

[PeterDonis]

what establishes that it cannot be considered inertial

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

 

[name123]

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.

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

 

[PeterDonis]

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.

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.

 

[name123]

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.

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 traveling 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?

 

[PeterDonis]

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 traveling faster than $c$?

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

the observation of light moving faster than ccc would at least indicate that light can be observed to move faster than ccc would it not?

Same answer as above.

 

[PeterDonis]

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

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

 

[name123]

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

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 ?

 

[PeterDonis]

could you please give an example what happens in the Special Relativity equations

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

 

[phinds]

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 ?

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.

 

[PeterDonis]

Far distant objects that are receding from us at greater than c they are not IN the same inertial frame

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.

recession velocity is not proper motion.

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.

 

[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?

 

[pervect]

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.

 

[PeterDonis]

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

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.

 

[PeterDonis]

I'm not sure what sense the word "stationary" is being used in this thread.

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.

 

[phinds]

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.

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

 

[russ_watters]

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

@name123 , you appear to have missed this previous reply by Peter regarding inertial vs non-inertial frames:

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

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).

 

[Battlemage!]

Okay here's question regarding acceleration. The proper acceleration, according to wikipedia (and according to a derivation that gives you the relativistic energy equation) is:

$$α=γ^3(u) \frac{du}{dt}$$

Now, wikipedia says this acceleration is "the acceleration of the object in its rest frame." But how is it at rest if it's accelerating?

Is it true that this type of acceleration is the smallest absolute value of acceleration an object can measure? (like how proper time is the shortest time interval). If so, that would give a different way to distinguish it. I'm just trying to see some way of distinguishing it in a way that goes beyond "this is what you actually feel."

 

[Dale]

Now, wikipedia says this acceleration is "the acceleration of the object in its rest frame." But how is it at rest if it's accelerating?

You are right. That should be "momentarily comoving inertial frame" instead of "rest frame"

Is it true that this type of acceleration is the smallest absolute value of acceleration an object can measure? (like how proper time is the shortest time interval).

This is not true for either proper acceleration or proper time.

 

[PeterDonis]

how is it at rest if it's accelerating?

The same way you, standing on the surface of the Earth, are at rest yet accelerating. "Accelerating" here means "feeling acceleration", i.e., feeling weight. You can feel weight and still be at rest.

The concept of "being at rest" is frame-dependent, and you can always choose a frame in which any chosen object is at rest, regardless of whether it is feeling weight or not.

I'm just trying to see some way of distinguishing it in a way that goes beyond "this is what you actually feel."

There isn't one. "What you actually feel" is the direct observable; that's the best way of distinguishing it that you're going to get.

 

[name123]

@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).

Thanks for that, and that is also to the others that were explaining.

 

[name123]

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

Thanks, and a couple more questions if you don't mind.

1) What equations would apply to an accelerating object? One reason I ask is that I had learned in mathematics that a curve can be approximated by a series of straight lines, and I assume you are suggesting that with relativity this is not the case (as it seems to me each straight line would involve an imaginary number in the Earth at rest, universe orbiting scenario).

2) If gravity curves space, does acceleration occur if an object follows the space curvature at a constant speed?

 

[PeterDonis]

What equations would apply to an accelerating object?

You can describe an accelerating object using an inertial frame; it will just have a worldline whose slope is not constant.

If you want to construct a non-inertial frame, you can, but it won't work the same. For one thing, it might have coordinate singularities that are not present in an inertial frame. See below.

I had learned in mathematics that a curve can be approximated by a series of straight lines, and I assume you are suggesting that with relativity this is not the case

No. Calculus still works in a non-inertial frame. See below.

it seems to me each straight line would involve an imaginary number in the Earth at rest, universe orbiting scenario

No. You have to be careful not to confuse coordinate quantities with actual physical quantities. For example, if we assume that spacetime is flat (which means we are ignoring the Earth's gravity), we can write the metric for a non-inertial frame in which the rotating Earth is at rest as follows (in cylindrical coordinates):

$$
ds^2 = - \left( 1 - \omega^2 r^2 \right) dt^2 + 2 \omega r dt d\phi + dz^2 + dr^2 + r^2 d\phi^2
$$

where $\omega$ is the angular velocity of rotation of the Earth. An object at rest on the rotating Earth will be at constant spatial coordinates $z$, $r$, $\phi$ in this chart. There will be some value of $r$ (much larger than the radius of the Earth), at which the quantity $1 - \omega^2 r^2$ vanishes. This is often described as "the universe is rotating at the speed of light" at this radius. But that is not really correct. What is actually happening is that an object which is at rest in this rotating coordinate chart (i.e., one "rotating with the Earth", not one "rotating with the universe") would have to "move at the speed of light" at this radius--more precisely, a curve of constant $z$, $r$, $\phi$ at this value of $r$ becomes null instead of timelike at this radius (and becomes spacelike at a larger radius). But the worldline of an object which is "at rest relative to the universe", i.e., which is "rotating" relative to the Earth, remains timelike at this radius (and indeed at any radius whatsoever); this is easy to check by using the fact that such a worldline satisfies the equation $\phi + \omega t = \Phi_0$, where $\Phi_0$ is a constant.

So what all this is really telling us is that the "rotating" object, like the Earth, must be limited in spatial extent; it can't reach or exceed a radius large enough that its surface would have to move at or faster than the speed of light (more precisely, follow a null or spacelike worldline). Which of course makes perfect sense physically. But there is no such restriction on objects at rest "relative to the universe"; even though they appear to be "rotating" in this chart, and even though at large enough radius their coordinate "speed of rotation" appears to be faster than light, they are still moving on timelike worldlines, and approximating those worldlines, which look curved in this chart, by straight lines locally does not involve any imaginary numbers or other mathematical peculiarities. It just requires paying careful attention to the actual meaning of the various quantities involved.

If gravity curves space

Gravity curves spacetime, not just space.

does acceleration occur if an object follows the space curvature at a constant speed?

I'm not sure what you mean by this.

 

[PeroK]

Thanks, and a couple more questions if you don't mind.

1) What equations would apply to an accelerating object? One reason I ask is that I had learned in mathematics that a curve can be approximated by a series of straight lines, and I assume you are suggesting that with relativity this is not the case (as it seems to me each straight line would involve an imaginary number in the Earth at rest, universe orbiting scenario).

2) If gravity curves space, does acceleration occur if an object follows the space curvature at a constant speed?

One of your difficulties is that you are considering more advanced topics. Question 2) asks about General Relativity, but in such as way that you show that you have no clear idea about the theory.

Whereas, all this about inertial and accelerating reference frames is there at the heart of classical physics: what equations does someone in an accelerating car use? Or, someone on a merry-go-round? And, you have the Corilolis force caused by the Earth's rotating (accelerating) reference frame.

As I see it, your core concern is that by using an accelerating reference frame you disrupt the laws of physics. That's true insofar as any equations derived specifically for an inertial reference frame cannot be used. But, that in itself, is of little consequence.

 

[Nugatory]

If gravity curves space, does acceleration occur if an object follows the space curvature at a constant speed?

Proper acceleration, the kind that an accelerometer measures and that is the same no matter what coordinate system I choose, happens or not according to whether the object is in free fall or not. Freefall paths through flat spacetime appear in space like the straight lines of Euclidean geometry; freefall paths through curved spacetime do not. eiether way they are freefall paths and there is no proper acceleration.

Coordinate acceleration happens or not according to the coordinates you choose. I can choose coordinates in which an object is accelerating or not accelerating - no matter what the trajectory of the object, no matter whether it's moving through curved spacetime (note - spacetime not space!) or flat, no matter whether a force is acting on it or not. Thus, coordinate acceleration has no physical significance in its own right.

"Constant speed" means "zero coordinate acceleration" so is every bit as coordinate-dependent a notion as coordinate acceleration so you have to be a bit cautious about using it. Often you hear it used when what is really meant is "moving in such a way that there exists an inertial frame in which its position does not change"; that's enough of a mouthful that people can be forgiven for this sloppiness.

 

[name123]

Thanks, and a couple more questions if you don't mind.

1) What equations would apply to an accelerating object? One reason I ask is that I had learned in mathematics that a curve can be approximated by a series of straight lines, and I assume you are suggesting that with relativity this is not the case (as it seems to me each straight line would involve an imaginary number in the Earth at rest, universe orbiting scenario).

2) If gravity curves space, does acceleration occur if an object follows the space curvature at a constant speed?

No. You have to be careful not to confuse coordinate quantities with actual physical quantities. For example, if we assume that spacetime is flat (which means we are ignoring the Earth's gravity), we can write the metric for a non-inertial frame in which the rotating Earth is at rest as follows (in cylindrical coordinates):

$$
ds^2 = - \left( 1 - \omega^2 r^2 \right) dt^2 + 2 \omega r dt d\phi + dz^2 + dr^2 + r^2 d\phi^2
$$

where $\omega$ is the angular velocity of rotation of the Earth. An object at rest on the rotating Earth will be at constant spatial coordinates $z$, $r$, $\phi$ in this chart. There will be some value of $r$ (much larger than the radius of the Earth), at which the quantity $1 - \omega^2 r^2$ vanishes. This is often described as "the universe is rotating at the speed of light" at this radius. But that is not really correct. What is actually happening is that an object which is at rest in this rotating coordinate chart (i.e., one "rotating with the Earth", not one "rotating with the universe") would have to "move at the speed of light" at this radius--more precisely, a curve of constant $z$, $r$, $\phi$ at this value of $r$ becomes null instead of timelike at this radius (and becomes spacelike at a larger radius). But the worldline of an object which is "at rest relative to the universe", i.e., which is "rotating" relative to the Earth, remains timelike at this radius (and indeed at any radius whatsoever); this is easy to check by using the fact that such a worldline satisfies the equation $\phi + \omega t = \Phi_0$, where $\Phi_0$ is a constant.

So what all this is really telling us is that the "rotating" object, like the Earth, must be limited in spatial extent; it can't reach or exceed a radius large enough that its surface would have to move at or faster than the speed of light (more precisely, follow a null or spacelike worldline). Which of course makes perfect sense physically. But there is no such restriction on objects at rest "relative to the universe"; even though they appear to be "rotating" in this chart, and even though at large enough radius their coordinate "speed of rotation" appears to be faster than light, they are still moving on timelike worldlines, and approximating those worldlines, which look curved in this chart, by straight lines locally does not involve any imaginary numbers or other mathematical peculiarities. It just requires paying careful attention to the actual meaning of the various quantities involved.

Sorry I did not quite follow that, I was not totally familiar with cylindrical coordinates, and I did not know what s referred to, I assume t was time, but I did not recognise what was the gamma element. Could you use a non-rotating cartesian coordinate system and then use straight line approximation to the curves using coordinate speed as an approximation for the velocity for the observer and the observed? Because if you could then would not the gamma for the observed involve an imaginary number?

 

[PeterDonis]

I did not know what s referred to

$ds^2$ is the square of the physical "distance" between neighboring events. If $ds^2$ is negative, the events are timelike separated and the "distance" is a time--i.e., there will be some observer whose worldline could pass through both events and whose clock would show elapsed time of $ds$ between them; if $ds^2$ is positive, the events are spacelike separated and the "distance" is an ordinary distance. If $ds^2$ is zero, the events are null separated and the events will be on the worldline of some light ray.

I did not recognise what was the gamma element.

$\gamma$ isn't a metric coefficient; it's a property of a particular object's worldline. And it really only applies in an inertial frame, not in non-inertial coordinates.

Could you use a non-rotating cartesian coordinate system

Sure, that's just an inertial frame. But an observer who is on the rotating Earth will not be at rest in such a frame, and the universe will not be "rotating" in such a frame.

and then use straight line approximation to the curves using coordinate speed as an approximation for the velocity for the observer and the observed?

What curves are you talking about? The worldlines of distant objects in the universe will be straight lines in non-rotating Cartesian coordinates, since those are just the coordinates of an inertial frame. See above.

 

[name123]

What curves are you talking about? The worldlines of distant objects in the universe will be straight lines in non-rotating Cartesian coordinates, since those are just the coordinates of an inertial frame. See above.

Apologies, you are right of course. What about if the Cartesian coordinate system was rotating with the observer? The observed objects could then be observed to be traveling at a speed greater > $c$ within that coordinate system, and so using that coordinate system in the equations, would not gamma (for the observed distant stars) involve imaginary numbers?

 

[name123]

Proper acceleration, the kind that an accelerometer measures and that is the same no matter what coordinate system I choose, happens or not according to whether the object is in free fall or not. Freefall paths through flat spacetime appear in space like the straight lines of Euclidean geometry; freefall paths through curved spacetime do not. eiether way they are freefall paths and there is no proper acceleration.

Thanks.

Sorry one more thing. Am I correct in thinking that in a spacecraft if one were to change direction using a thruster, the initial velocity would have no effect on the how much the thruster changed the direction, or to the readings of an onboard accelerometer?

 

[name123]

Apologies, you are right of course. What about if the Cartesian coordinate system was rotating with the observer? The observed objects could then be observed to be traveling at a speed greater > $c$ within that coordinate system, and so using that coordinate system in the equations, would not gamma (for the observed distant stars) involve imaginary numbers?

Sorry I think I overlooked that if the frame was not inertial, then you would have to use equations which took the angular velocity of the frame into account.

 

[PeterDonis]

What about if the Cartesian coordinate system was rotating with the observer?

Then it would not be an inertial coordinate chart and things would work differently. Plus the metric would look a lot messier than the one I wrote down; I used cylindrical coordinates precisely because they make things look much simpler in a rotating chart.

The observed objects could then be observed to be traveling at a speed greater > $c$ within that coordinate system

At a coordinate speed greater than $c$, but coordinate speed is not physical speed. That is the point I was making in my previous long post. The objects would be traveling on timelike worldlines. See below.

would not gamma (for the observed distant stars) involve imaginary numbers?

No, because the analogue of $\gamma$ in non-inertial coordinates is not based on coordinate speed, it's based on the worldline being timelike.

Am I correct in thinking that in a spacecraft if one were to change direction using a thruster, the initial velocity would have no effect on the how much the thruster changed the direction

How would you measure how much the thruster changed the direction?

or to the readings of an onboard accelerometer?

This would be independent of initial velocity (which is coordinate dependent anyway), yes.

I think I overlooked that if the frame was not inertial, then you would have to use equations which took the angular velocity of the frame into account.

Yes, and that's what I wrote down in my previous long post. Note that there is an extra "cross term" involving $dt d\phi$ in the metric. That is a reflection of "the angular velocity of the frame". So the analogue of $\gamma$ for a given worldline now has to take into account, not just $dt$, but $d\phi$ as well. That's one way of describing how the worldline of an object "at rest relative to the universe" (and therefore "moving" in this rotating frame) can still be timelike even though $1 - \omega^2 r^2$ is zero or negative.

 

[PeroK]

Sorry I think I overlooked that if the frame was not inertial, then you would have to use equations which took the angular velocity of the frame into account.

A simple example of why you must remain in a single inertial reference frame is as follows, and I think this also answers your question about what happens if you accelerate in discrete steps with periods of constant velocity.

If you measure the distance to an object at rest, then you accelerate towards it, stop accelerating and take a new measurement, then the distance may have reduced greatly owing to length contraction.

If you take these measurements and divide the difference by the elapsed time on your watch, then you have a velocity of sorts, which may be far in excess of the speed of light. But, because the initial and final measurements were taken in different inertial reference frames, this is not a valid velocity in terms of SR equations. You cannot take that "velocity" and calculate a gamma factor with it.

In physics as well as mathematics, context is vital. In SR in particular you have to be careful that the assumptions under which the equations were derived hold good for your experiment.