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Rest length in general relativity

  1. Jun 25, 2008 #1
    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?
     
  2. jcsd
  3. Jun 25, 2008 #2
    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?

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

    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.
     
  4. Jun 25, 2008 #3
    How do you define a reference frame in a non-stationary spacetime, which is the spacetime of our universe?
     
  5. Jun 25, 2008 #4
    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.
     
  6. Jun 25, 2008 #5

    Fredrik

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    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 [itex]\sqrt{dx^2+dy^2+dz^2-dt^2}[/itex] 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?
     
  7. Jun 25, 2008 #6

    Mentz114

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    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.
     
  8. Jun 25, 2008 #7
    Not precisely. The expression for length also involves an integral

    [tex]
    \int \sqrt{g_{ij}\frac{dx^i}{d\alpha}\frac{dx^j}{d\alpha}} d\alpha
    [/tex]

    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.
     
  9. Jun 25, 2008 #8

    JesseM

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    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.
     
  10. Jun 25, 2008 #9
    I just realized the integral might have as well be written as

    [tex]
    \int\sqrt{g_{\mu\nu}\frac{d x^{\mu}}{d\alpha}\frac{d x^{\nu}}{d\alpha}}d\alpha,
    [/tex]

    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 :confused: The rest length is not unique despite that this integral is invariant?
     
    Last edited: Jun 25, 2008
  11. Jun 25, 2008 #10
    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.
    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
     
  12. Jun 25, 2008 #11
    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.
     
  13. Jun 25, 2008 #12
    Is it so that the integration path cannot be defined uniquely when determining the rest length of an object?
     
  14. Jun 25, 2008 #13
    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
     
  15. Jun 25, 2008 #14
    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

    [tex]
    \int \sqrt{g_{ij}\frac{dx^i}{d\alpha}\frac{dx^j}{d\alpha}} d\alpha,
    [/tex]

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

    [tex]
    \int\sqrt{g_{\mu\nu}\frac{d x^{\mu}}{d\alpha}\frac{d x^{\nu}}{d\alpha}}d\alpha,
    [/tex]

    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?
     
    Last edited: Jun 25, 2008
  16. Jun 25, 2008 #15
    It could be I don't want to see the counter example. I'm not devoting my life for the general relativity :biggrin:

    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.
     
  17. Jun 25, 2008 #16

    Mentz114

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    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 ?
     
  18. Jun 25, 2008 #17
    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.
     
  19. Jun 25, 2008 #18
    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
     
  20. Jun 25, 2008 #19

    Hurkyl

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    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.
     
  21. Jun 25, 2008 #20
    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!

    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?
     
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