Rest length in general relativity

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In general relativity, the concept of rest length is not absolute and is dependent on the observer's chosen reference frame. The rest length is defined as the length of an object measured in its rest frame, but selecting this frame can be complex, especially for large or non-stationary objects. Observers can agree on the rest length if they are in the same frame, yet in non-stationary spacetimes, the rest length may vary over time due to spatial contraction. The proper length is determined through integrals that are frame-dependent, leading to ambiguity in defining a unique rest length. Ultimately, while the rest length can be invariant, the lack of a unique frame complicates its measurement in general relativity.
  • #31
Mentz114 said:
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 :smile:

I didn't know what MeJennifer was talking about in her post, so I decided to ask something.
 
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  • #32
pmb_phy said:
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
 
  • #33
Mentz114 said:
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.
 
  • #34
JesseM said:
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 dependent as mentioned above.

Pete
 
  • #35
pmb_phy said:
All that would mean is that the rest length is time dependent 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.
 
  • #36
JesseM said:
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 dependent 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
 
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  • #37
JesseM said:
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
 
  • #38
pmb_phy said:
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.
pmb_phy said:
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.
 
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  • #39
Mentz114 said:
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.
Mentz114 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 ?
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.
 
  • #40
JesseM said:
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
 
  • #41
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
 
  • #42
pmb_phy said:
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?
pmb_phy said:
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.
pmb_phy said:
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.
 
  • #43
Mentz114 said:
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.
 
  • #44
Mentz114 said:
You have given the SR definition of proper length, incidentally. As you know it is different in GR.
Oops. :redface: I should post less when I'm tired.
 
  • #45
Jesse,
I just saw this, sorry for the lateness of this response.

JesseM said:
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
 
  • #46
Mentz114 said:
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?
 
  • #47
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. :)
JesseM said:
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
 
  • #48
JesseM said:
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
 
  • #49
pmb_phy said:
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.
pmb_phy said:
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).
pmb_phy said:
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.
pmb_phy said:
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?
 
  • #50
pmb_phy said:
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.
 
  • #51
Mentz114 said:
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.
But how can you define how it "changes" without giving length as a function of t, which gives the problem of defining "instantaneous length" at a precise time t? How do you measure this instantaneous length, given that every light signal takes some finite time interval to cross?
JesseM said:
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?
Mentz114 said:
Not relevant, we only measure the time of flight of the light, not its frequency.
I wasn't talking about frequency. I meant that if you send a signal from the bottom of the tower to the top and then bounce it back to the bottom and look at the time elapsed on a clock at the bottom, this might end up being different from the time elapsed on a clock at the top when you send a signal from the top to the bottom and then bounce it back to the top, since clocks at top and bottom tick at different rates due to gravitational time dilation.
 
  • #52
gel said:
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.
This should tell you that you made a very serious error somewhere along the way. If an inherently invariant quantity like ds changes under a coordinate transformation then you can bet your horses that there was an error made somewhere.

Pete
 
  • #53
pmb_phy said:
If an inherently invariant quantity like ds changes under a coordinate transformation then you can bet your horses that there was an error made somewhere.

No error. You keep saying that ds is invariant, but it isn't.

The two frames do not agree on simultaneity of points in space time, so they imply a different curve to integrate along when calculating the length, and therefore get a different length. s would only be invariant if it was calculated along the same curve in spacetime in both cases, but defining such a curve in an invariant way is the problem
 
  • #54
JesseM said:
But how can you define how it "changes" without giving length as a function of t, which gives the problem of defining "instantaneous length" at a precise time t? How do you measure this instantaneous length, given that every light signal takes some finite time interval to cross?
There is no such thing as instantaneous measurement of length, and I did not introduce it. I just proposed an operational definition of length. Obviously the longer the object the longer it will take to measure it.

I wasn't talking about frequency. I meant that if you send a signal from the bottom of the tower to the top and then bounce it back to the bottom and look at the time elapsed on a clock at the bottom, this might end up being different from the time elapsed on a clock at the top when you send a signal from the top to the bottom and then bounce it back to the top, since clocks at top and bottom tick at different rates due to gravitational time dilation.
Point taken, I'll think about it.

M
 
  • #55
JesseM - As challenging and interesting as this is there is a point where one bows out. Therefore please do not take my lack of response to your comments as me avoiding the issue or me thinking that this is a waste of time or not a good use of my time because it is! Its a cool discussion. I'd love to return to this in the future to analyze where your misakes are but I am letting my other work fall behind.

Some last comments
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.
The original inertial frame? Your coodinate transformation is merely a mathematical one, i.e. all you have done is to relable your clock readings. Sort of set some clocks ahead etc. It doesn't appear as if you have synchronized clocks according to the appropriate clock synchronization procedure (which is not coordinate dependent, only frame dependent - there's an imporant by subtle difference there). It does not represent a new inertial frame. This should be obvious since the body has not changed position in this new inertial frame.
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.
But you've changed the very meaning of "synchronized." Its as if you think clocks are always synchronized when they have a reading that a math equation says to have. That is not the case.

I'm hoping someone else can help explain this better than I can. I'm getting a headache (AC is making my eyes dry and that is giving me a headache).

Best wishes and good luck

Pete
 
  • #56
Mentz114 said:
There is no such thing as instantaneous measurement of length, and I did not introduce it. I just proposed an operational definition of length. Obviously the longer the object the longer it will take to measure it.
But if the length is not constant, how do you define the way it's changing, if you can't measure length as a function of time?
 
  • #57
JesseM said:
But if the length is not constant, how do you define the way it's changing, if you can't measure length as a function of time?
I don't care if/how it is changing. I'm looking for something all observers can agree on - even if it is changing. Anyhow, I need time to think about this and it's getting late. I'll be back.

M
 
  • #58
pmb_phy said:
The original inertial frame? Your coodinate transformation is merely a mathematical one, i.e. all you have done is to relable your clock readings. Sort of set some clocks ahead etc. It doesn't appear as if you have synchronized clocks according to the appropriate clock synchronization procedure (which is not coordinate dependent, only frame dependent - there's an imporant by subtle difference there). It does not represent a new inertial frame. This should be obvious since the body has not changed position in this new inertial frame.
I never said the new coordinate system was a new "inertial frame", in fact I specifically said it wasn't in post #38 where I suggested I could provide a flat spacetime example:
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.
Surely now that you understand what I mean when I talk about creating arbitrary coordinate systems in SR, you can imagine that one could do the same thing in GR, yes? If I have some set of clocks around a spherical planet whose times match those of Schwarzschild coordinates, I could find some coordinate transformation equation which would allow me to reset them in a way that produced a new smooth coordinate system for this spacetime. In GR unlike in SR, there are no physically preferred coordinate systems, agreed?
pmb_phy said:
But you've changed the very meaning of "synchronized." Its as if you think clocks are always synchronized when they have a reading that a math equation says to have. That is not the case.
The math equation was only used in the process of setting the clocks, once they are set you don't need to use any equations, you just look at their readings. (Simple math equations are also used in setting clocks according to the Einstein synchronization convention--for example, if a ray of light leaves clock A when it reads tA, is reflected by clock B when clock B reads tB, and returns to clock A when it reads t'A, then you want to set clock B such that tB = (tA + t'A)/2.) And are you suggesting that in GR where there is no preferred set of coordinate systems like inertial ones in SR, there are some physical procedures for setting clocks which make them "synchronized" and other procedures for setting clocks under which they are not truly synchronized, but just have a "reading that a math equation says to have"? This would seem fundamentally at odds with the principle of diffeomorphism invariance which says all GR coordinate systems are on equal footing physically.

On the other hand, if your point about some forms of synchronization being more physical than others was only intended to apply to SR, please understand that my point was about GR and whether the notion of "proper length" still makes sense in curved spacetime given the fact that you can produce infinitely many different coordinate systems in which an object is at rest that all have different definitions of simultaneity...as I said earlier I only used a flat spacetime example because I'm not sufficiently familiar with the math of GR.
 
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  • #59
I haven't read this in any detail... but it seems relevant to this discussion:
http://arxiv.org/abs/gr-qc/9512006 (Gen.Rel.Grav. 28 (1996) 899-903)
"How to measure spatial distances?" H-J Schmidt.
 
  • #60
robphy said:
I haven't read this in any detail... but it seems relevant to this discussion:
http://arxiv.org/abs/gr-qc/9512006 (Gen.Rel.Grav. 28 (1996) 899-903)
"How to measure spatial distances?" H-J Schmidt.

I think that when they say that the velocity vector (of the dust - or body in this thread) is orthogonal to the t=const surfaces, this gives a reasonable definition of length.

If you do this to measure the length of a rod, then it just means that the space-like curve along which you calculate the proper distance is orthogonal (locally) to the time-like velocity of the rod. Although the length may change over time (which is unavoidable), at least different observers would agree what the length is, but may not agree on the time at which that length occurs.
 

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