Rest length in general relativity

[jostpuur]

I would be interested to know how the author thinks that all observers agree on rest length in general relativity.

Is it the case, that in general relativity the rest length is not absolute?

How is the rest length defined?

Is it the problem, that it is not clear how one should choose the rest frame of some object, if the object is large?

 

[gendou2]

In general relativity, is the rest length absolute?

What do you mean "absolute"?
Assuming only intelligent observers, all observers in all reference frames will agree on what the rest length of an object is.
Is that what you were wondering?

How is the rest length defined?

The rest length of an object is how long it is when you measure it in it's reference frame.

How should one choose the rest frame of a large object?

No different from choosing the rest frame of a small object?
Take the Earth, for example.
Ignoring it's rotation, when you stand still on the surface, you're in it's rest frame.

 

[MeJennifer]

The rest length of an object is how long it is when you measure it in it's reference frame.

How do you define a reference frame in a non-stationary spacetime, which is the spacetime of our universe?

 

[gendou2]

How do you define a reference frame in a non-stationary spacetime, which is the spacetime of our universe?

This is how I define a reference frame:
http://www.google.com/search?q=define:+reference+frame

I don't know about stationary versus non-stationary spacetime.
Please explain how it follows that one cannot define reference frames within a non-stationary spacetime.

 

[Fredrik]

The problem isn't to define what a coordinate system is, it's to choose which one to use. It's definitely not a trivial question.

Proper length is the integral of \sqrt{dx^2+dy^2+dz^2-dt^2} along a space-like curve, but which one? If the object is a spinning rod for example, one of the endpoints of the curve is an event on the world line of one of the endpoints of the rod, but what's the other endpoint of the curve? It's an event on the world line of the other endpoint of the rod, but which one? And even if you have an answer to that, do you know which curve connecting the two events you should use?

 

[Mentz114]

The problem isn't to define what a coordinate system is, it's to choose which one to use. It's definitely not a trivial question.

Proper length is the integral of \sqrt{dx^2+dy^2+dz^2-dt^2} along a space-like curve, but which one? If the object is a spinning rod for example, one of the endpoints of the curve is an event on the world line of one of the endpoints of the rod, but what's the other endpoint of the curve? It's an event on the world line of the other endpoint of the rod, but which one? And even if you have an answer to that, do you know which curve connecting the two events you should use?

Observables in GR are coordinate independent because they are scalars obtained by contracting tensors. The rest length of a rod is such an observable, so it doesn't matter what coordinate system is chosen.

You have given the SR definition of proper length, incidentally. As you know it is different in GR.

 

[jostpuur]

Observables in GR are coordinate independent because they are scalars obtained by contracting tensors. The rest length of a rod is such an observable, so it doesn't matter what coordinate system is chosen.

Not precisely. The expression for length also involves an integral

<br /> \int \sqrt{g_{ij}\frac{dx^i}{d\alpha}\frac{dx^j}{d\alpha}} d\alpha<br />

which is supposed to be carried out on a given instant, so it depends on the chosen frame. The rest length is supposed to be invariant, which could be achieved if there existed a unique frame where the integration is supposed to be carried out. If I understood correctly what MeJennifer and Fredrik were saying, then this unique frame doesn't exist, although I don't yet understand why this is the case.

 

[JesseM]

This is how I define a reference frame:
http://www.google.com/search?q=define:+reference+frame

I don't know about stationary versus non-stationary spacetime.
Please explain how it follows that one cannot define reference frames within a non-stationary spacetime.

You can certainly define a coordinate system where an object is at rest in general relativity, the problem is that you can construct all different kinds of coordinate systems where this is true and which nevertheless disagree about the object's coordinate length, and all these coordinate systems are equally valid. This is different from SR, where only inertial coordinate systems are equally valid, and there is a set procedure for constructing inertial coordinate systems which guarantees that if two inertial coordinate systems agree an object is at rest, they will agree on its coordinate length.

 

[jostpuur]

I just realized the integral might have as well be written as

<br /> \int\sqrt{g_{\mu\nu}\frac{d x^{\mu}}{d\alpha}\frac{d x^{\nu}}{d\alpha}}d\alpha,<br />

and then you don't need to be in the rest frame.

So... hmhmh... even this integral is not coordinate independent? Even though the integration path in the physical space would be fixed? hmhmh... the integral looks invariant to me The rest length is not unique despite that this integral is invariant?

 

[pmb_phy]

This is how I define a reference frame:
http://www.google.com/search?q=define:+reference+frame

I don't know about stationary versus non-stationary spacetime.
Please explain how it follows that one cannot define reference frames within a non-stationary spacetime.

A stationary spacetime is a spacetime for which there exists a coordinate system in which the components of the metric tensor are not explicit functions of time.

..which is supposed to be carried out on a given instant, so it depends on the chosen frame.

The quantity that you just defined is an invariant. It is independant of the coordinate system used. The value of this quantity, known as the proper distance, depends on the worldline, not on the coordinate system. However for small spatial distances this quantity is unique.

Pete

 

[jostpuur]

The quantity that you just defined is an invariant. It is independant of the coordinate system used. The value of this quantity, known as the proper distance, depends on the worldline, not on the coordinate system. However for small spatial distances this quantity is unique.

I noticed I made a mistake in the post #7, and continued about it in the post #9, but I got new questions there, which still puzzle me.

 

[jostpuur]

Is it so that the integration path cannot be defined uniquely when determining the rest length of an object?

 

[pmb_phy]

Is it so that the integration path cannot be defined uniquely when determining the rest length of an object?

No. While the proper length of a path is path dependant the length is path dependant and its that path that you choose and that should be unique. I can't think of a counter example.

Pete

 

[jostpuur]

the matter clear?

In order to find the rest length of an object, we first choose a rest frame of the object, and set the length to be

<br /> \int \sqrt{g_{ij}\frac{dx^i}{d\alpha}\frac{dx^j}{d\alpha}} d\alpha,<br />

where the integration is carried out only in the spatial space, with fixed time. Since dx^0/d\alpha=0, the integral can as well be written as

<br /> \int\sqrt{g_{\mu\nu}\frac{d x^{\mu}}{d\alpha}\frac{d x^{\nu}}{d\alpha}}d\alpha,<br />

and then the integral does not depend on the chosen frame anymore, and is invariant.

The problem with uniqueness rises from the fact, that if the rest frame cannot be chosen uniquely in the beginning, then the integration path is not unique either. So this is the reason why there is not unique rest length in general relativity?

 

[jostpuur]

No. While the proper length of a path is path dependant the length is path dependant and its that path that you choose and that should be unique. I can't think of a counter example.

Pete

It could be I don't want to see the counter example. I'm not devoting my life for the general relativity

MeJennifer and JesseM seemed to be confident that the rest frame cannot be chosen uniquely, so I think I'm satisfied with it, assuming that they have learned this from some reliable source.

 

[Mentz114]

The rest length of an object is its length as measured and reported by an observer at rest wrt to the measured object. That's it. There is no ambiguity in this. It is not observer dependent because we chose the observer.

What are we all talking about ?

 

[jostpuur]

The rest length of an object is its length as measured and reported by an observer at rest wrt to the measured object. That's it. There is no ambiguity in this. It is not observer dependent because we chose the observer.

What are we all talking about ?

It is not possible to be intuitively capable of telling what is happening if there is a large object in space with non-trivial time-depending metric. The observer might have to travel along the object in order to measure its length. Or if the observer has installed measuring devices all over the object before the measurement, he will have to be able to synchronize them somehow, and then be able to interpret the data. It's not like the observer who is in rest with the object merely measures the distance, and that's it.

 

[pmb_phy]

It could be I don't want to see the counter example. I'm not devoting my life for the general relativity

MeJennifer and JesseM seemed to be confident that the rest frame cannot be chosen uniquely, so I think I'm satisfied with it, assuming that they have learned this from some reliable source.

MeJennifer was referring to non-stationary spacetimes. A body in such a spacetime would, in general, have its rest length a function of time because the spatial contraction would be a function of time. In that sense it is not unique, but (I think) its still invariant.

Hurkyl pointed out something else that might be bothering MeJennifer. The fact that the proper length of a body would depend on the bodies orientation in a gravitational field

However the FAQ that started this inquiry was about SR although the author didn't say that, unfortunately.

Pete

 

[Hurkyl]

The rest length of an object is its length as measured and reported by an observer at rest wrt to the measured object. That's it. There is no ambiguity in this. It is not observer dependent because we chose the observer.

What are we all talking about ?

Objects have spatial extent.

In special relativity, there's no reason different parts of the object must be at rest with each other, and therefore there's no reason to expect an observer can be at rest with all parts of the object.

In general relativity, we cannot even define what it means for things at two different locations in space-time to be at rest with each other, so "observer at rest with the mesaured object" is, strictly speaking, nonsense.


AFAIK, the best we can do is assume some sort of smallness condition that ensures all reasonable methods all give approximately the same answer.

 

[jostpuur]

MeJennifer was referring to non-stationary spacetimes. A body in such a spacetime would, in general, have its rest length a function of time because the spatial contraction would be a function of time. In that sense it is not unique, but (I think) its still invariant.

Hurkyl pointed out something else that might be bothering MeJennifer. The fact that the proper length of a body would depend on the bodies orientation in a gravitational field

However the FAQ that started this inquiry was about SR although the author didn't say that, unfortunately.

Pete

MeJennifer's comment caught my attention, but I thought continuing it in the FAQ thread would have been a little bit off topic. But with new thread, it doesn't matter anymore how off topic this is!

"Rest length of a spacelike curve" is not ambiguous. "Rest length of an object" is ambiguous, because there is no canonical way to choose which spacelike curve expresses the rest length of the object.

You get the same problem in special relativity too -- there are only a few special cases which determine 'obvious' sorts of spacelike curve to measure. e.g. an inertially traveling object, or one undergoing uniform constant acceleration.

Well this is a new twist in the story. Is this because the velocity of the object is not unique even in some fixed frame, necessarily? Some parts of the object are moving with different velocities, making the rest frame difficult concept?

So this thing never had much to do with the general relativity?

 

[jostpuur]

Yeah. I wasn't seeing the Hurkyl's post #19 when writing my previous post.

 

[pmb_phy]

In general relativity, we cannot even define what it means for things at two different locations in space-time to be at rest with each other, so "observer at rest with the mesaured object" is, strictly speaking, nonsense.

What justification do you have for this assertion?

Pete

 

[Mentz114]

Hurkyl,

In general relativity, we cannot even define what it means for things at two different locations in space-time to be at rest with each other, so "observer at rest with the mesaured object" is, strictly speaking, nonsense.

Exactly. So the question is nonsense too. Or at least ill-defined. The use of 'rest' in GR does not have the meaning it does in SR, and can only be defined over a small region. But, my definition could be said to be valid because it's operational.

M

 

[jostpuur]

What justification do you have for this assertion?

I suppose it's the usual: You need to do parallel transporting to compare velocities at different space time points. The relative velocity depends on the chosen path of parallel transport.

Exactly. So the question is nonsense too. The use of 'rest' in GR does not have the meaning it does in SR. But, my definition is valid because it's operational.

Your definition attempt about rest frame in SR is not valid, because some parts of the object are not necessarily in rest with respect to other parts of the object.

 

[pmb_phy]

I suppose it's the usual: You need to do parallel transporting to compare velocities at different space time points. The relative velocity depends on the chosen path of parallel transport.

This is altogether different. Parallel transport has nothing to do with proper distance.

Pete

 

[MeJennifer]

proper distance.

Note that term 'proper distance' in cosmology has a particular meaning for FRW spacetimes.

 

[Mentz114]

jost,

Your definition attempt about rest frame in SR is not valid, because some parts of the object are not necessarily in rest with respect to other parts of the object.

This is getting circular - the object I'm measuring in my rest frame has ALL its parts at rest by definition ! If not, it is not in my rest frame. For heavens sake.

The concept you use of 'rest length' in your question suffers the same flaw. What do you mean by it ? Are you saying that some of your 'rest' frame is not at rest ? That is contradictory.

I just think you are using the word 'rest' incorrectly. Are you asking about the possibility that the measured length changes if the orientation to the gravitation changes ?

M

PS it's good to see you here in the relativity section - I enjoyed your many contributions in QM.

 

[pmb_phy]

Note that term 'proper distance' in cosmology has a particular meaning for FRW spacetimes.

This is news to be. Can you be more specific?

Pete

 

[MeJennifer]

See for instance:

Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe
Authors: Tamara M. Davis, Charles H. Lineweaver
http://arxiv.org/abs/astro-ph/0310808

 

[robphy]

This is altogether different. Parallel transport has nothing to do with proper distance.

Pete

So, given two 4-velocities at different events in a general spacetime,
how do you compare those unit-vectors to see if they are parallel
(which is presumably one way to characterize that those particles are at rest with respect to each other)?

 

[jostpuur]

The concept you use of 'rest length' in your question suffers the same flaw. What do you mean by it ?

I'm not sure. Actually I asked in the opening post, that what are we supposed to mean by it

I didn't know what MeJennifer was talking about in her post, so I decided to ask something.

 

[pmb_phy]

MeJennifer was referring to non-stationary spacetimes. A body in such a spacetime would, in general, have its rest length a function of time because the spatial contraction would be a function of time.

Ummmm ... at least this is what I think she was referring to. Is this correct MJ?

Pete

 

[JesseM]

This is getting circular - the object I'm measuring in my rest frame has ALL its parts at rest by definition ! If not, it is not in my rest frame. For heavens sake.

What is "your rest frame" in general relativity? One can use arbitrary coordinate systems, so you could find one coordinate system where you were at rest but the object (or one part of the object) was not, and another where you were both at rest. There isn't a specific procedure for constructing your frame as there is in special relativity where you use inertial rulers and clocks and synchronize clocks using the Einstein synchronization convention.

 

[pmb_phy]

What is "your rest frame" in general relativity? One can use arbitrary coordinate systems, so you could find one coordinate system where you were at rest but the object (or one part of the object) was not, ...

All that would mean is that the rest length is time dependant as mentioned above.

Pete

 

[JesseM]

All that would mean is that the rest length is time dependant as mentioned above.

Does "time dependent" mean it depends on a particular coordinate system's definition of simultaneity? Of course in GR surfaces of simultaneity can bend and wave pretty much arbitrarily (as long as the surface is spacelike) from one end of the object to another depending on what coordinate system you choose, they aren't neat planes as in SR, so if this is true it seems like you could make the "rest length" of an object pretty much anything you wanted depending on your choice of coordinate system.

 

[pmb_phy]

Does "time dependent" mean it depends on a particular coordinate system's definition of simultaneity? Of course in GR surfaces of simultaneity can bend and wave pretty much arbitrarily (as long as the surface is spacelike) from one end of the object to another depending on what coordinate system you choose, they aren't neat planes as in SR, so if this is true it seems like you could make the "rest length" of an object pretty much anything you wanted depending on your choice of coordinate system.

We are talking about proper distance. There is no ambiguity in the definition of proper length that I;'m aware of or that I've seen so far in this thread. By definition, the proper length of an object is the lenth as measured in the objects rest frame. In this frame it is the distance between the end points as measured at the same time in the rest frame. Granted, this may not be a constant but it is an invariant. However I could of course be wrong. As such Please given a solid example of what you're talking about so that we can work with a specific example rather in the abstract.

If the definition of proper length as coordinate dependant then that would mean that a change in coordinates in the integral which defines proper length will give a change in its value. If this is so then proper length wouldn't be an invariant and we know it is by the very geometric nature of its definition. So in your example please show that the proper lengthe integral changes under a coordinate transformation.

Thanks

Pete

 

[Mentz114]

What is "your rest frame" in general relativity? One can use arbitrary coordinate systems, so you could find one coordinate system where you were at rest but the object (or one part of the object) was not, and another where you were both at rest. There isn't a specific procedure for constructing your frame as there is in special relativity where you use inertial rulers and clocks and synchronize clocks using the Einstein synchronization convention.

I'm using a reducio ad absurdum argument to show that the idea of a 'rest frame' in GR is not viable, except in a small region, where the conditions can be met. Apart from that I agree with everything you've said.

Suppose we measure a length by sending some light from one end of our rod, bouncing it back and recording the round trip time. In a freely-falling frame in GR, would the observer be aware of a change in the length ?

M

 

[JesseM]

We are talking about proper distance. There is no ambiguity in the definition of proper length that I;'m aware of or that I've seen so far in this thread. By definition, the proper length of an object is the lenth as measured in the objects rest frame.

But in GR there are an infinite number of possible coordinate systems where every point on an object is at rest at all times, and yet which define simultaneity completely differently from one another, no? It isn't like SR where there's a set procedure for constructing an inertial coordinate system where a non-accelerating rigid object at rest, so that saying that you're using the object's "rest frame" uniquely determines what its coordinate length must be, whether two events at different ends of the object are simultaneous, etc.

As such Please given a solid example of what you're talking about so that we can work with a specific example rather in the abstract. Thanks.

I don't have the familiarity with GR needed to give an example in curved spacetime, but I could probably give some example of a coordinate system in SR which doesn't match the definition of an inertial coordinate system, but where every point on some non-accelerating rigid object has a fixed position coordinate, if that would help.

 

[JesseM]

I'm using a reducio ad absurdum argument to show that the idea of a 'rest frame' in GR is not viable, except in a small region, where the conditions can be met. Apart from that I agree with everything you've said.

Ah, I didn't catch that you were making a reductio ad absurdum type argument, my mistake.

Suppose we measure a length by sending some light from one end of our rod, bouncing it back and recording the round trip time. In a freely-falling frame in GR, would the observer be aware of a change in the length ?

Well, if we remove your caveat "except in a small region" from above, in curved spacetime I'd guess tidal forces would stretch or compress the object at least a little in ways that could be measured using round-trip travel time for light.

 

[pmb_phy]

But in GR there are an infinite number of possible coordinate systems where every point on an object is at rest at all times, and yet which define simultaneity completely differently from one another, no?

Not that I'm aware of. If that was the case then the defining integral wouldn't represent a geometric quantity, which it is. Please given an example, even if not with math.

I don't have the familiarity with GR needed to give an example in curved spacetime, but I could probably give some example of a coordinate system in SR which doesn't match the definition of an inertial coordinate system, but where every point on some non-accelerating rigid object has a fixed position coordinate, if that would help.

Sure.

Question: Where did you get the idea that proper length was not a well defined quantity? Did you read this somewhere? Did you come up with yourself? Did someone convince you of this?

Pete

 

[Mentz114]

It isn't like SR where there's a set procedure for constructing an inertial coordinate system where a non-accelerating rigid object at rest, so that saying that you're using the object's "rest frame" uniquely determines what its coordinate length must be, whether two events at different ends of the object are simultaneous, etc.

There is special set of frames, the freely-falling. In these the observers experience no forces, but are still being accelerated. Their size depends on the local curvature. I see them as mathematical constructs we hope and expect to agree with experience.

Since this set includes just about everything in the cosmos, it doesn't help a lot. But one can calculate how two such observers will see each others frames. Adding non-geodesic motion doesn't complicate things a lot.

M

 

[JesseM]

Not that I'm aware of. If that was the case then the defining integral wouldn't represent a geometric quantity, which it is. Please given an example, even if not with math.

Well, look at the paragraph on this page which discusses the principle of "diffeomorphism invariance" from GR, immediately above the third animated diagram in which the same objects are assigned coordinates in a variety of arbitrary-looking coordinate systems with wavy axes. They write:

Closely related to background independence is another basic ingredient of general relativity, known by the imposing name diffeomorphism invariance. It concerns the coordinates physicists use to describe space and time. The principle of diffeomorphism invariance implies that, unlike in theories prior to general relativity, there are no additional structures in physics that allow us to distinguish preferred coordinate systems. As far as the laws of physics are concerned, no coordinate system is better than another, and one is free to choose. In terms of the simplified illustration above, there are infinitely many ways to choose a lattice - a few examples are shown here:

Do you agree with this paragraph?

Sure.

OK, say we have a standard inertial coordinate system in SR with coordinates x and t, and an object 10 light-seconds long at rest in these coordinates, with its back end at x=0 ls and its front end at x=10 ls. Now construct an x',t' coordinate system with the following transformation:

x' = x
and for t', if x < 5 ls, let t' = t + x*0.4/c; if x >= 5 ls, let t' = t + 2 s. According to the GR principle of diffeomorphism invariance, isn't this coordinate system just as valid as any other? And since x' = x, we know if every point on the object was at rest in the original inertial coordinate system, this will still be true in the new coordinate system. But the new coordinate system obviously defines simultaneity differently--the events (x=0 ls, t=2 s) and (x=10 ls, t=0 s) are not simultaneous in the inertial coordinate system, but in the new coordinate system they are assigned coordinates (x'=0 ls, t'=2 s) and (x'=10 ls, t'=2 s) so they are simultaneous.

Question: Where did you get the idea that proper length was not a well defined quantity? Did you read this somewhere? Did you come up with yourself? Did someone convince you of this?

No, just my own thought based on my understanding that diffeomorphism invariance allows you to consider all smooth coordinate systems equally valid in GR.

 

[JesseM]

There is special set of frames, the freely-falling. In these the observers experience no forces, but are still being accelerated. Their size depends on the local curvature.

But in curved spacetime, there will be some amount of tidal forces in any non-infinitesimal region, no? They may be too small to be worth worrying about, but the laws of physics in a freely-falling coordinate system in the region only become precisely like those of inertial frames in SR in the limit as the size of the region approaches zero.

 

[Fredrik]

You have given the SR definition of proper length, incidentally. As you know it is different in GR.

Oops. I should post less when I'm tired.

 

[Mentz114]

Jesse,
I just saw this, sorry for the lateness of this response.

Ah, I didn't catch that you were making a reductio ad absurdum type argument, my mistake.

Well, that was a fast shuffle from me. I've been arguing semantics and missing the real point.

Well, if we remove your caveat "except in a small region" from above, in curved spacetime I'd guess tidal forces would stretch or compress the object at least a little in ways that could be measured using round-trip travel time for light.

OK. If we define the length of a rod length as the time it takes light to travel from one end to the other, we may have something that all observers will agree on, regardless of the gravitational environment. I want to do the detailed calculations but I'm slow these days so maybe someone else can beat me to it.

M

 

[JesseM]

OK. If we define the length of a rod length as the time it takes light to travel from one end to the other, we may have something that all observers will agree on, regardless of the gravitational environment.

But isn't this problematic if the length is time-dependent? If the light takes a certain amount of time to cross from one end to another, and the object is being stretched or compressed as the light travels across it, is there any physical way to define a notion of the "instantaneous" length?

Also, if you don't want to worry about clock synchronization issues, you would probably want to define length as half the time it takes for light to go from one end to the other and back, as measured by a clock at the end it starts from and returns to. But even in a situation with no time-dependence, it might be that you'd get different answers depending on which end you chose...imagine a tower upright in a gravitational field with clocks at the top and bottom, doesn't gravitational time dilation indicate the clock at the bottom is ticking slower than the clock at the top?

 

[pmb_phy]

First off I'd like to thank you JesseM for providing me with a concrete example. I asked for one and you provided it. I wanted to let you know that I appreciate this. :)

Well, look at the paragraph on this page which discusses the principle of "diffeomorphism invariance" from GR, immediately above the third animated diagram in which the same objects are assigned coordinates in a variety of arbitrary-looking coordinate systems with wavy axes. They write:

Do you agree with this paragraph?

Of course. That's just the principle of general covariance. However I'd like to caution you on something before we go on. Just because all coordinate systems leave tensor equations covariant it doesn't mean that all coordinate systems are physically meaningful. Take as an example the Lorentz transformation as compared to the Galilean transformation. Each is a coordinate transformation but it is only the former that leaves Maxwell's equations covariant. The former has a physical reality while the later ... not so much.

OK, say we have a standard inertial coordinate system in SR with coordinates x and t, and an object 10 light-seconds long at rest in these coordinates, with its back end at x=0 ls and its front end at x=10 ls. Now construct an x',t' coordinate system with the following transformation:

x' = x
and for t', if x < 5 ls, let t' = t + x*0.4/c; if x >= 5 ls, let t' = t + 2 s.

Hold on there Dutch! What is the "s" in "t' = t + 2 s"? Does this mean that s is a variable or does it mean that "s" is a unit of time, i.e. add 2 seconds to t? I assume that its the later. I just want to make sure.

Note that this is not a continuos coordinate transformation and therefore the quantities which define the integral which then defines proper distance becomes problematic, i.e. dx' loses meaning.

According to the GR principle of diffeomorphism invariance, isn't this coordinate system just as valid as any other?

Valid as in what? Leaves a tensor equation covariant or has a physical meaning?

And since x' = x, we know if every point on the object was at rest in the original inertial coordinate system, this will still be true in the new coordinate system. But the new coordinate system obviously defines simultaneity differently--the events (x=0 ls, t=2 s) and (x=10 ls, t=0 s) are not simultaneous in the inertial coordinate system, but in the new coordinate system they are assigned coordinates (x'=0 ls, t'=2 s) and (x'=10 ls, t'=2 s) so they are simultaneous.

I question the meaning of simultaneous that you're using. Simultaneous is something that depends on how clocks are synchronized. Not on how you label the readings on clocks.

No, just my own thought based on my understanding that diffeomorphism invariance allows you to consider all smooth coordinate systems equally valid in GR.

Your example is not a smooth coordinate system.

I recommend that you do a coordinate transformation and see if the value of proper distance changs. Good luck.

Pete

 

[Mentz114]

But isn't this problematic if the length is time-dependent? If the light takes a certain amount of time to cross from one end to another, and the object is being stretched or compressed as the light travels across it, is there any physical way to define a notion of the "instantaneous" length?

I'm after something that all observers will agree on, it doesn't matter if it changes. For instance in SR all inertial observers agree on proper intervals along worldlines.

Also, if you don't want to worry about clock synchronization issues, you would probably want to define length as half the time it takes for light to go from one end to the other and back, as measured by a clock at the end it starts from and returns to. But even in a situation with no time-dependence, it might be that you'd get different answers depending on which end you chose...imagine a tower upright in a gravitational field with clocks at the top and bottom, doesn't gravitational time dilation indicate the clock at the bottom is ticking slower than the clock at the top?

Not relevant, we only measure the time of flight of the light, not its frequency. Of course, we have to assume that the speed of the light was c always.

I've been doing some rough calculation and it is not trivial to get the flight time ( for me ).

M

 

[JesseM]

Hold on there Dutch! What is the "s" in "t' = t + 2 s"? Does this mean that s is a variable or does it mean that "s" is a unit of time, i.e. add 2 seconds to t? I assume that its the later. I just want to make sure.

Yes, it was meant to be seconds, I thought t + 2 would be ambiguous.

Note that this is not a continuos coordinate transformation and therefore the quantities which define the integral which then defines proper distance becomes problematic, i.e. dx' loses meaning.

Well, it'd lose meaning at a single point--I thought in GR one could use coordinate systems with coordinate singularities, like Schwarzschild coordinates in a black hole spacetime. In any case, we could imagine other coordinate transformations which involve functions that don't involve coordinate singularities. For example, I think this would work:

x' = x
t' = t + (2 seconds)*sin[(pi/4)*(x/10 light-seconds)]

Here it would again be true that the events (x=0 l.s., t=2 s) and (x=10 l.s., t=0 s) are non-simultaneous in the original inertial frame, but in the new system they become (x' = 0 l.s., t'=2 s) and (x'=10 l.s., t'=2 s) which do have the same time coordinate. And I'm pretty sure surfaces of constant t' would still be spacelike...if not one could always choose a smaller multiplier for the sine function than 2 seconds (obviously in the limit as the multiplier goes to zero, the new coordinate system becomes the same as the original inertial one, where surfaces of constant t were definitely spacelike).

Valid as in what? Leaves a tensor equation covariant or has a physical meaning?

I question the meaning of simultaneous that you're using. Simultaneous is something that depends on how clocks are synchronized. Not on how you label the readings on clocks.

Sure, but if you define a new coordinate system in terms of a mathematical transformation on an existing coordinate system that's based on some well-defined physical recipe (whether an inertial coordinate system in SR or something like Schwarzschild coordinates in GR), it's trivial to just reset the clocks so that their readings now match those of the new coordinate system. This is a perfectly physical "synchronization" procedure, even if it's a lot more ungainly and inelegant than the original procedure that was used to synchronize clocks in the first coordinate system.

Your example is not a smooth coordinate system.

I recommend that you do a coordinate transformation and see if the value of proper distance changs. Good luck.

Well, see the new example above...anyway, the issue is not any specific coordinate system, the point is that one can come up with an infinite variety of smooth coordinate systems where an object is still at rest but they all define simultaneity (i.e. surfaces of constant t') differently. In SR we can say that the class of inertial coordinate systems is physically preferred, but when dealing with coordinate systems in curved spacetime there's really no basis for picking out any smooth coordinate systems as "special" in a physical sense, even if some are more elegant and convenient to use. The definition of proper distance in arbitrary coordinate systems in curved spacetime given by jostpuur looked like some kind of tensor equation so I don't know how to use it to calculate the proper distance in my flat spacetime coordinate system above, perhaps someone could give it a try and see if it still comes out to 10 light seconds?

 

[gel]

I recommend that you do a coordinate transformation and see if the value of proper distance changs. Good luck.

ok, let me have a go instead.
Suppose we have flat space and (t,x,y,z) are standard inertial coordinates.
Suppose that there is a long and thin rod at rest with respect to these coords, at y=z=0 and 0 <= x <= 1, so it has length 1.
Change to coords (t',x',y',z') with x'=x, y'=y, z'=z and t' = t + 0.6 x/c.
Note, the rod is still at rest in the new coord frame, and occupies the same region of space. However, proper length (s) along a curve with y=z=0 is,
<br /> ds^2 = dx^2-c^2dt^2 = (dx&#039;)^2 - (c dt&#039;-0.6 dx&#039;)^2<br />
Along a slice of constant t', use dt'=0 to get
<br /> ds^2 = (dx&#039;)^2 - 0.36 (dx&#039;)^2=(0.8 dx&#039;)^2<br />
So, the length of the rod measured using the (t',x',y',z') coord system is 0.8, not 1.