How does this affect the length contraction paradox?

[Fantasist]

But what about a different scenario? Assume the observer is actually based outside the gravitational field of the Earth (e.g. at one of the Lagrange points) and observes the free-falling rod attracted by the earth. Clearly, both the observer and the rod have zero proper acceleration, so you should not observe any length contraction of the rod, despite the fact that its velocity with regard to the observer increases.

Sure. Rindler coordinates are for flat spacetime only. The spacetime around the Earth is not flat. You would have to use GR, not SR for this scenario.

Any reference for this? I would have thought that the length contraction due to GR just adds to that of SR (after all that is how it is e.g. for the time dilation of GPS- satellite clocks).

The gravitational length contraction

$$dr=ds\sqrt{1-\frac{2GM}{rc^2}}$$

does obviously not contain any velocity dependent terms at all, in fact not even any time dependent terms if you consider a stationary rod on the earth. Yet

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At)^2\right)}$

would predict a change of the length of the stationary rod with time as viewed from outside the gravitational field of the earth.

I do not see how the first effect could negate the second one.

 

[Dale]

Textbook definitions don't appear to be sufficient to unambiguously answer questions like whether an electron accelerated in an electric field occupies an inertial reference frame or not (assuming that the charge of the electron is distributed homogeneously within its volume (i.e. the electron does not 'feel' the Coulomb force as the latter acts on all parts of it and no internal stress forces are set up)).

Nonsense. The textbook definitions are unambiguous on topics like this. You appear to disagree with the textbooks, but your disagreement does not indicate any ambiguity whatsoever.

That only works because

1) the external references are programmed into the system in advance
2) it is assumed that the external references don't change

Sure, and the same is true of any navigation based on coordinate acceleration.

3) any change in proper acceleration is interpreted as a change in coordinate acceleration

This is not true. For example, in Schwarzschild coordinates you can have a change in proper acceleration without any coordinate acceleration for an observer whose r coordinate decreases at a constant rate.

Your stated preference for coordinate acceleration is unreasonable, all of your objections are either flat out wrong, or they apply equally to coordinate acceleration.

 

[Dale]

Any reference for this?

Yes. The Wikipedia entry I pointed you to on Rindler coordinates clearly shows the derivation from the flat spacetime metric.

 

[Fantasist]

Textbook definitions don't appear to be sufficient to unambiguously answer questions like whether an electron accelerated in an electric field occupies an inertial reference frame or not (assuming that the charge of the electron is distributed homogeneously within its volume (i.e. the electron does not 'feel' the Coulomb force as the latter acts on all parts of it and no internal stress forces are set up)).

Nonsense. The textbook definitions are unambiguous on topics like this. You appear to disagree with the textbooks, but your disagreement does not indicate any ambiguity whatsoever.

Really? Take the Wikipedia definition:

All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration.

This is not only ambiguous but even a contradiction in one sentence: if you have two accelerated rockets moving with a constant velocity relatively to each other, they would be inertial systems according to the first part of the sentence, but non-inertial systems according to the second part.

3) any change in proper acceleration is interpreted as a change in coordinate acceleration

This is not true. For example, in Schwarzschild coordinates you can have a change in proper acceleration without any coordinate acceleration for an observer whose r coordinate decreases at a constant rate.

Only that the cruise missile wouldn't have a clue about it. If it sits on the ground and suddenly the local gravitational field increases due to some restructuring in the Earth's crust, it will interprete this as an indication that it is is being lifted upwards with a constant acceleration. In this respect the measured proper acceleration is always ambiguous.

 

[Fantasist]

Yes. The Wikipedia entry I pointed you to on Rindler coordinates clearly shows the derivation from the flat spacetime metric.

I meant a reference that explicitly gives the equivalent of the flat spacetime length contraction formula

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At)^2\right)}$

in curved spacetime

 

[Mentz114]

All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration.

This is not only ambiguous but even a contradiction in one sentence: if you have two accelerated rockets moving with a constant velocity relatively to each other, they would be inertial systems according to the first part of the sentence, but non-inertial systems according to the second part.

You have misinterpreted this ( it is rather inelegant). It means that the necessary and sufficient conditions are 1) constant relative velocity 2) both frames have zero accelerometer reading.

 

[Dale]

Really? Take the Wikipedia definition:

All inertial frames are in a state of constant, rectilinear motion with respect to one another; an accelerometer moving with any of them would detect zero acceleration.

This is not only ambiguous but even a contradiction in one sentence: if you have two accelerated rockets moving with a constant velocity relatively to each other, they would be inertial systems according to the first part of the sentence, but non-inertial systems according to the second part.

First, Wikipedia isn't a textbook. Second, that statement is not a contradiction. It says that if two frames are inertial then they are in a state of rectilinear motion wrt each other. It does not say that if two frames are in rectilinear motion to each other then they are inertial. Your rockets are therefore not a counterexample.

Only that the cruise missile wouldn't have a clue about it. If it sits on the ground and suddenly the local gravitational field increases due to some restructuring in the Earth's crust, it will interprete this as an indication that it is is being lifted upwards with a constant acceleration. In this respect the measured proper acceleration is always ambiguous.

So what? The same is true of a navigation system based on coordinates. If your map is incorrect or outdated you will have errors in either case. Garbage in, garbage out. It is true, but hardly a problem unique to proper acceleration.

 

[Dale]

I meant a reference that explicitly gives the equivalent of the flat spacetime length contraction formula

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At)^2\right)}$

in curved spacetime

Not that I know of. I don't even know of a general definition of length contraction in curved spacetime.

 

[Mentz114]

I meant a reference that explicitly gives the equivalent of the flat spacetime length contraction formula

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At)^2\right)}$

in curved spacetime

Do you mean a velocity dependent contraction or a grvitational length contraction ?

Not that I know of. I don't even know of a general definition of length contraction in curved spacetime.

The best we can do is consider two worldlines in curved spacetime that coincide at a point p, and calculate the coordinate transformation that connects the local frames at that point. It turns out that this is the Lorentz transformation . So the length contraction formula is x'₁-x'₂=γβ(x₁-x₂).

 

[Fantasist]

You have misinterpreted this ( it is rather inelegant). It means that the necessary and sufficient conditions are 1) constant relative velocity 2) both frames have zero accelerometer reading.

So you are saying an observer in free-fall in the gravitational field of the Earth and another in space (at rest relatively to the earth) are not inertial (as they don't move with constant velocity relatively to each other)?

 

[Fantasist]

Do you mean a velocity dependent contraction or a grvitational length contraction ?

The velocity dependent contraction of an object free-falling under gravity but observed from outside the gravitational field.

 

[WannabeNewton]

So you are saying an observer in free-fall in the gravitational field of the Earth and another in space (at rest relatively to the earth) are not inertial (as they don't move with constant velocity relatively to each other)?

The presence of a gravitational field changes what constitutes an inertial frame. A freely falling object is locally inertial in a gravitational field but a static object in a gravitational field is obviously not inertial, it is an accelerating frame of reference. A static frame accelerates relative to a local inertial frame i.e. it accelerates according to local accelerometer measurements.

 

[WannabeNewton]

The velocity dependent contraction of an object free-falling under gravity but observed from outside the gravitational field.

You can't insulate yourself from a gravitational field unless you're in an asymptotically flat space-time and relegated to spatial infinity. Even then, kinematical length contraction is local not global so the kinematical length contraction formula would only make sense between two locally separated observers in the gravitational field.

EDIT: don't confuse local Lorentz contraction i.e. kinematical length contraction (which is velocity dependent) with non-trivial proper lengths for ideal rods/rulers due to the presence of a gravitational field.

 

[Fantasist]

The presence of a gravitational field changes what constitutes an inertial frame. A freely falling object is locally inertial in a gravitational field but a static object in a gravitational field is obviously not inertial, it is an accelerating frame of reference. A static frame accelerates relative to a local inertial frame i.e. it accelerates according to local accelerometer measurements.

OK, but this means the condition "All inertial frames are in a state of constant, rectilinear motion with respect to one another" is in general incorrect. The only working definition seems to be that a comoving accelerometer measures zero acceleration. But this in turn would mean that one can never say by purely theoretical arguments whether a reference frame is inertial or not for a given scenario. It is always an actual experimental measurement necessary.

 

[WannabeNewton]

The theoretical definition of a (locally) inertial frame is simply a frame which has zero proper acceleration; proper acceleration can be defined mathematically and measured experimentally. If the "all inertial frames are in a state of constant, rectilinear motion with respect to one another" statement is unappealing and/or confusing to you then just ignore it. All you need to know is that if a frame has zero proper acceleration then it is (locally) inertial.

 

[PAllen]

OK, but this means the condition "ll inertial frames are in a state of constant, rectilinear motion with respect to one another" is in general incorrect. The only working definition seems to be that a comoving accelerometer measures zero acceleration. But this in turn would mean that one can never say by purely theoretical arguments whether a reference frame is inertial or not for a given scenario. It is always an actual experimental measurement necessary.

It's always experimental, even without gravity. If two rockets are both accelerating with profiles such that each measures the other to be at constant relative speed, that does not mean they are inertial. To pick out, in the real world, what is inertial motion, you have to make measurements.

 

[Fantasist]

You can't insulate yourself from a gravitational field unless you're in an asymptotically flat space-time and relegated to spatial infinity. Even then, kinematical length contraction is local not global so the kinematical length contraction formula would only make sense between two locally separated observers in the gravitational field.

EDIT: don't confuse local Lorentz contraction i.e. kinematical length contraction (which is velocity dependent) with non-trivial proper lengths for ideal rods/rulers due to the presence of a gravitational field.

I am not confusing it. The question is why the velocity dependent part should disappear for non-local problems (as other posters here have claimed).

 

[WannabeNewton]

It's not that Lorentz contraction disappears; it just doesn't make any sense mathematically to define Lorentz contraction non-locally in curved space-time.

 

[Fantasist]

I think it makes perfect sense to ask whether a co-moving non-local observer sees the same velocity dependence for the length contraction as the local observer or not.

 

[WannabeNewton]

No it doesn't. 3-velocity is a locally measured quantity.

 

[Nugatory]

I think it makes perfect sense to ask whether a co-moving non-local observer sees the same velocity dependence for the length contraction as the local observer or not.

In flat spacetime, yes.

In curved spacetime, not so much. You can't talk about "comoving observers" or the "velocity dependence" observed by these observers until you've defined the relative velocity, and that's not so easy for a remote observer to do in curved space.

 

[Dale]

I think it makes perfect sense to ask whether a co-moving non-local observer sees the same velocity dependence for the length contraction as the local observer or not.

In curved spacetime there is no way to uniquely identify co-moving non-local observers from non-co-moving non-local observers. The problem is that parallel transport is path dependent in curved spacetime.

EDIT: as Nugatory said

 

[Dale]

The question is why the velocity dependent part should disappear for non-local problems (as other posters here have claimed).

By the way, I never said that the velocity dependent part disappeared. I said that I don't know what the formula is in curved spacetime, and I said that to calculate it would require GR.

Perhaps you are talking about someone else, but I didn't see anyone else say that either.

 

[pervect]

If I may add something, the whole argument about proper acceleration sort of misses the point.

If we consider an object moving with x(t) = (1/2) a t^2, such a motion is possible for low t , and impossible for t >= c/a, because it requires superluminal velocity for t >= c/a (or alternatively it's subluminal only for at<c). So it's not a "good" relativistic motion in general, but it's OK for "small t".

For small t, the motion represents the motion of an object moving with a constant coordinate acceleration. While possible in principle for small t, you won't find a lot of discussion in textbooks. In the case where t << a/c, a taylor series expansion of the motion of constant proper acceleration for x(t) will show that it's nearly equivalent to constant coordinate acceleration, as one would expect. In the intermediate range where t < a/c but of the same order, the two motions differ, and when t>=a/c constant coordinate acceleration becomes imposibile because at> c, and noting can move as fast or faster than light.

The errors in the first post were in not applying relativity properly.

You start with a single x(t), representing the motion of an observer.

There isn't, at this point, any x1(t), or x2(t). There is only x(t), the motion of "the observer".

You then need to define "the coordinates" of an accelerated observer" SR has a prescription for this, based on using the momentarily co-moving inertial frame. The notion does not extend gracefully to GR. The notion has a well-known weakness even in SR regarding the uniquness of the coordinates that I'll mention but gloss over, because it would be too confusing to explain at this point to the OP and not really relevant to the point.

Given the worldlines x1(t) = constant and x2(t) = constant, the wordlines in the momentarily comoving inertial frame can be defined using the Lorentz transform.

You wind up with different coordinates X1(T) and X2(T), where T represents the transformed t coordinate, and X1 and X2 represent the transformed x1 and x2 coordinates.

The error in the original post (#1) in my opinion was in not applying the Lorentz transform, but using the Galilean transform

i.e. it used

X = x - vt
T = t

rather than

X = x - vt
T = t- vx/c^2

and it skipped a few important steps by assuming that the Gallilean transform was correct.

The lorentz transform is one of the basics of special relativity. http://en.wikipedia.org/wiki/Lorentz_transformation

So post #1 basically did a non-relativistic treatment of the problem. I'm afraid I don't really have the time to do a full treatment of the problem correctly, and I think trying to do this in a post for someone not already familiar with relativity is a bad idea. If they are motivated enough, they can find a textbook. One can hope that this discussion will so motivate them, it's probalby not practical to learn SR from reading posts to a forum.

I'd also like to encourage the OP to avoid trying to hammer everything into the "observer" framework if this is at all possible. I'm not sure how successful I'll be at that. It sounds like they have not realized that "observers" aren't really required or all that useful, and they would most likely proceed to defend and cling to the notion :( rather than think about alternatives to observers - which , to put it succinctly are generalized coordinates.

 

[Fantasist]

In curved spacetime there is no way to uniquely identify co-moving non-local observers from non-co-moving non-local observers. The problem is that parallel transport is path dependent in curved spacetime.

There isn't really any transport involved for determining the velocity:

if you consider a triangle on a sphere, the cosine rule gives you

cos(c) = cos(a)*cos(b) + sin(a)*sin(b)*cos(C)

Now if you choose c infinitesimally small (c=dx) and the other sides equal (b=a), this becomes

cos(dx) = cos²(a) + sin²(a)*cos(dC)

and if you expand this in a Taylor series

1-dx²/2 = 1-sin²(a) + sin²(a)*(1-dC²/2)

or

dx² = sin²(a)*dC²

i.e.

dx = dC*sin(a)

or with dx=v*dt

v = dC*sin(a)/dt (sphere)

where dC is the observed angular change of position during time dt.


Now in flat geometry, the cosine rule is

c² = a² + b² -2ab*cos(C),

so with the same definitions as above we get

dx² = 2a² - 2a²*(1-dC²/2)

i.e.

dx = a*dC

or

v = dC*a/dt. (flat)

If you compare the two expressions for the curved and flat geometry, it is evident that the derived velocity dependence is the same apart from a constant factor (sin(a) instead of a) associated with the different metric.
So a co-moving observer (v=0) in flat space is also a co-moving observer in curved space.

 

[Fantasist]

If I may add something, the whole argument about proper acceleration sort of misses the point.

If we consider an object moving with x(t) = (1/2) a t^2, such a motion is possible for low t , and impossible for t >= c/a, because it requires superluminal velocity for t >= c/a (or alternatively it's subluminal only for at<c). So it's not a "good" relativistic motion in general, but it's OK for "small t".

For small t, the motion represents the motion of an object moving with a constant coordinate acceleration. While possible in principle for small t, you won't find a lot of discussion in textbooks. In the case where t << a/c, a taylor series expansion of the motion of constant proper acceleration for x(t) will show that it's nearly equivalent to constant coordinate acceleration, as one would expect. In the intermediate range where t < a/c but of the same order, the two motions differ, and when t>=a/c constant coordinate acceleration becomes imposibile because at> c, and noting can move as fast or faster than light.

The errors in the first post were in not applying relativity properly.

I mentioned already earlier (post #27) that the assumption of a constant acceleration was only done for convenience. It is not material for the argument. One might as well have a time dependent acceleration for which the coordinates would then change according to

x1(t) = x1(0) - $\Delta X(t)$
x2(t) = x2(0) - $\Delta X(t)$

where

$$\Delta X(t) = \int_0^t dt' \int_0^{t'} dt'' a(t'')$$

You start with a single x(t), representing the motion of an observer.

There isn't, at this point, any x1(t), or x2(t). There is only x(t), the motion of "the observer".

Sorry, I don't get your argument: if you want to measure the length or distance of something you have to measure by definition the coordinates of two points x1 and x2. The length/distance is the difference of the two coordinates i.e. L=x2-x1. If you have e.g. a ruler free-falling past a very tall building with two markings on it, then the ruler coordinates of these markings give you the distance between them in the reference frame of the ruler (L(t)=x2(t)-x1(t)).

Given the worldlines x1(t) = constant and x2(t) = constant, the wordlines in the momentarily comoving inertial frame can be defined using the Lorentz transform.

You wind up with different coordinates X1(T) and X2(T), where T represents the transformed t coordinate, and X1 and X2 represent the transformed x1 and x2 coordinates.

The error in the original post (#1) in my opinion was in not applying the Lorentz transform, but using the Galilean transform

i.e. it used

X = x - vt
T = t

rather than

X = x - vt
T = t- vx/c^2

and it skipped a few important steps by assuming that the Gallilean transform was correct.

A length measurement implies by definition that the two coordinates are determined simultaneously. If you Lorentz-transform the coordinates of such a length measurement to a different reference frame, then this does not constitute a length measurement in the latter frame anymore. You need an independent measurement where the two coordinates are determined simultaneously in this frame as well.

 

[WannabeNewton]

Since you're so adamant about this without giving any textbook references regarding the well-definition of non-local 3-velocity measurements that you so stubbornly cling to, tell me mathematically and physically how a static observer at some $R >> 2M$ in Schwarzschild space-time would make an unambiguous measurement of the 3-velocity of an observer falling in from infinity radially when the infalling observer is at some $r = 2M(1 + \epsilon)$ where $\epsilon << 1$. I assure you that if you don't know the basics of differential geometry, you will find this futile.

Just to give you some background, if an observer has 4-velocity $\xi^{\mu}$ and another observer has 4-velocity $\eta^{\mu}$ then $\xi^{\mu}\eta_{\mu}$ is only defined when the two observer's worldlines intersect (i.e. they meet at an event coincident on both their worldlines) because in curved space-time you can only take the "inner product" of two 4-vectors when they both lie in the same tangent space to the space-time manifold (that is to say $\xi^{\mu}\eta_{\mu}$ is only defined if $\xi^{\mu}$ and $\eta^{\mu}$ are in the same tangent space to the space-time manifold at a given event). The relevance of this is that a 3-velocity measurement of $\eta^{\mu}$ by $\xi^{\mu}$ at a coincident event on both their worldlines is mathematically given by $\xi^{\mu}\eta_{\mu} = -\frac{1}{\sqrt{1 - v^2}}$.

EDIT: And seriously, start reading textbooks on SR and GR. Forum threads can only do so much for you.

 

[Dale]

There isn't really any transport involved for determining the velocity

Yes, there is. See chapter 3 here:

http://arxiv.org/abs/gr-qc/9712019

e.g. "two particles at different points on a curved manifold do not have any well-defined notion of relative velocity — the concept simply makes no sense" (p. 64)

 

[Dale]

One might as well have a time dependent acceleration for which the coordinates would then change according to

x1(t) = x1(0) - $\Delta X(t)$
x2(t) = x2(0) - $\Delta X(t)$

where

$$\Delta X(t) = \int_0^t dt' \int_0^{t'} dt'' a(t'')$$

If you have a non-uniform acceleration then there is no standard definition of the non inertial observers coordinate chart. There are several different options. My favorite definition is the Radar Coordinates used by Dolby and Gull:

http://arxiv.org/abs/gr-qc/0104077

 

[Fantasist]

Yes, there is. See chapter 3 here:

http://arxiv.org/abs/gr-qc/9712019

e.g. "two particles at different points on a curved manifold do not have any well-defined notion of relative velocity — the concept simply makes no sense" (p. 64)

That should be only the case if you leave the path undefined. However, any physical measurement will automatically define an unambiguous and unique path. In this case, if you observe the motion and length of the rod (located in the gravitational field of the earth) through the eyepiece of a telescope from outside the gravitational field, the path is clearly defined by the corresponding light path connecting the rod and the observer. I have shown above (post #60) what the latter would derive for the velocity in case of a spherical metric. It may be a bit more difficult in case of other metrics, but should still be straightforward as long as you know the curvature along the light path. In any case, for sufficiently small displacements of the rod (so that locality is preserved) and a stationary gravitational field, the observed velocity dependence of the length contraction should be independent of the metric.

 

[PAllen]

That should be only the case if you leave the path undefined. However, any physical measurement will automatically define an unambiguous and unique path. In this case, if you observe the motion and length of the rod (located in the gravitational field of the earth) through the eyepiece of a telescope from outside the gravitational field, the path is clearly defined by the corresponding light path connecting the rod and the observer. I have shown above (post #60) what the latter would derive for the velocity in case of a spherical metric. It may be a bit more difficult in case of other metrics, but should still be straightforward as long as you know the curvature along the light path. In any case, for sufficiently small displacements of the rod (so that locality is preserved) and a stationary gravitational field, the observed velocity dependence of the length contraction should be independent of the metric.

The fundamental difference is that with no gravity, there is no path dependence. You can say there is a comparison between distant vectors with no further qualification. In the presence of gravity, you have to specify a procedure - and different procedures yield different results.

What happens if between you and a ruler is a compact gravitating body of great mass. Then, your image of a ruler through a telescope will be a ring with struts coming out of it.

 

[Dale]

That should be only the case if you leave the path undefined. However, any physical measurement will automatically define an unambiguous and unique path.

Yes. If you specify a path you can have a well defined comparison between different vectors, and a physical measurement would definitely have a unique result.

In this case, if you observe the motion and length of the rod (located in the gravitational field of the earth) through the eyepiece of a telescope from outside the gravitational field, the path is clearly defined by the corresponding light path connecting the rod and the observer.

Yes, that is certainly a measurement which could be analyzed. However, such a measurement is unrelated to length contraction. I.e. Such a measurement would not result in length contraction in flat spacetime.

I have shown above (post #60) what the latter would derive for the velocity in case of a spherical metric.

No you didn't. I don't know why you would think you had.

In any case, for sufficiently small displacements of the rod (so that locality is preserved) and a stationary gravitational field, the observed velocity dependence of the length contraction should be independent of the metric.

Locally I agree. Non-locally you would have to do an in depth analysis using GR and a detailed experimental measure.

 

[pervect]

I mentioned already earlier (post #27) that the assumption of a constant acceleration was only done for convenience. It is not material for the argument.

Then let's not argue about it. The point that I think is relevant is that "acceleration" is ambiguous, you need to specify whether and when you are talking about coordinate acceleration and when you are talking about proper acceleration.

You haven't really disambiguated your text, so I'm not sure if the point has gotten through or not, or if you actually understand the distinction :-(.

The point is peripheral, but it could help you to understand the main issue. Since it is peripheral I won't go into more detail.

Sorry, I don't get your argument: if you want to measure the length or distance of something you have to measure by definition the coordinates of two points x1 and x2.

I agree, and the two points in question are X1 and X2, so there isn't any disagreement with how you calculate distance.

The difference between X1 and x1 is a matter of notation. I had reasons for using my notation, it's just a matter of consistently applying the Lorentz transform.

What is really important, and not just a matter of notation, is that x1 is a function of t, while X1 is a function of T.

The difference between t and T is *not* just a mater of semantics. What is represented by the symbol t is not the same thing that is represented by the symbol T.

I suspect that you are entrenched in the notion of absolute time, which is why you use only one symbol for time, "t", for all observers, rather than distinguishing between t and T.