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Fundamental identity of SR valid in GR?

  1. May 28, 2007 #1
    From Relativity (Rindler) I learn that the fundamental identity

    [tex]c^2\text{d}{t'}^2 -\text{d}{x'}^2 -\text{d}{y'}^2 -\text{d}{z'}^2
    =
    c^2\text{d}{t}^2 -\text{d}{x}^2 -\text{d}{y}^2 -\text{d}{z}^2[/tex]

    relates co-ordinates of one inertial frame S' to another one called S. By putting the point in the origin of S' and re-arranging we get

    [tex]c^2(\frac{\text{d}{t'}}{\text{d}{t}})^2
    +(\frac{\text{d}{x}}{\text{d}{t}})^2
    +(\frac{\text{d}{y}}{\text{d}{t}})^2
    +(\frac{\text{d}{z}}{\text{d}{t}})^2
    = c^2[/tex]

    or, in short, [itex]|v|=c[/itex].

    This nicely shows time dilation, since the term [itex]\text{d}t'/\text{d}t[/itex] represents the speed of clocks in S', and [itex]\text{d}t' [/itex] is called, I think, proper time of S'. We see how [itex]\text{d}t'/\text{d}t[/itex] goes to zero, i.e. time stops, as spatial velocity approaches c. On the other extreme, if spatial velocity goes to zero, the term goes to 1, i.e. [itex]\text{d}t'=\text{d}t[/itex] showing that time in S' now develops as fast as time in S.

    In the latter situation, i.e. relative spatial velocity zero, put S and S' into an area with gravity such that S' suffers stronger gravity. Then afaik time runs slower in S', but this would mean that [itex]|v|=c[/itex] does not hold anymore. Consequently (?) the fundamental identity does not hold anymore. Not a surprise since the fundamental identity stems from SRT which does not take gravity into account.

    The question is: Is there a similarly easy to understand fundamental identity in GR and how does it look like?

    Thanks,
    Harald.
     
    Last edited: May 28, 2007
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  3. May 28, 2007 #2

    robphy

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    Look a little further in the Rindler book for the "metric" (or the related "line element", "interval").
     
    Last edited: May 28, 2007
  4. May 28, 2007 #3
    If anything, we can state that in a curved spacetime any two local Lorentz frames are generally not equivalent, thus:

    [tex]c^2\text{d}{t'}^2 -\text{d}{x'}^2 -\text{d}{y'}^2 -\text{d}{z'}^2
    \neq
    c^2\text{d}{t}^2 -\text{d}{x}^2 -\text{d}{y}^2 -\text{d}{z}^2[/tex]

    In curved spacetime Lorentz invariance is broken except locally. There is no such thing as global Lorentz invariance for curved spacetime.
     
    Last edited: May 28, 2007
  5. May 28, 2007 #4
    There would be no such thing if you thought GR were correct. GR believes in no absolute background but believes in absolute definition of time. If GR is wrong then the universe provides the absolute flat spacetime background and presents variant times defined by variant freely-falling frames. The new thought on gravity in the thread, therefore, is possible!
     
  6. May 28, 2007 #5

    cristo

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    Are you saying GR is incorrect?

    There is no "new thought" in this thread-- it was simply a question from one who is studying a GR text.
     
  7. May 28, 2007 #6

    marcus

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    I agree with Jennifer
    I disagree with realcreation
    In the context of cosmology, assuming an homogeneous isotropic universe, one can have simple solutions of GR where there is a lot of unrealistic symmetry and it is possible to define a single time. But when you get down to particular idealized solutions then you DO have a background!

    I don't think your statement makes sense. I never heard of an "absolute definition of time" in the context of the General Theory. Maybe I just don't understand you.
     
  8. May 28, 2007 #7
    GR is not proved by experiments. To prove GR means to prove that spacetime is curved.
    1. We can run experiments only in solar neighborhood where gravity is weak. Because it is weak, people always use the formulas of flat spacetime to test GR.
    2. Beyong solar system we have astronomic observation where, however, GR encounter deadly dificulties (see crisis in Cosmology topic of the forums).
    3. Even in solar neighborhood, we see the glimpses of failure of GR (see `alternative theories be tested by GPB experiment ..` in Astrophysics topic of the forums.

    The new thought is that SRT idea can expain gravity without the assumption of curved spacetime.
     
  9. May 28, 2007 #8
    Sorry. I dont like to use dogmatic terminology. I like to use the terminologies based on my understanding.

    "absolute definition of time" means the dogmatic assumption of GR that physics everywhere and whenever is THE SAME.
     
  10. May 28, 2007 #9

    cristo

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    Ok.. I'm not gonna bother with this discussion, since you clearly just want an argument!!
     
  11. May 28, 2007 #10

    Garth

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    Nonsense.
    There are always questions to be asked and resolved.
    There are published alternative theories being tested by the Gravity Probe B experiment, however at the moment the results so far have validated the GR prediction to within 1%.

    What is being directly measured by the geodetic precession is the curvature of space-time. Different theories predict different amounts of curvature, which is why this experiment was worth doing.
    Have you a published paper? Reference?

    Or do you want to submit your new theory to the IR Forum?

    Garth
     
    Last edited: May 28, 2007
  12. May 28, 2007 #11
    OK, The geodetic or frame-dragging formula is:
    DS/dt = O x S
    where O is derived on a generalized SR version of Laplace equation. Did not you see? You measured that time is curved???

    Schwarzchild metric is tested in the same way. People found hundrads of ways to derive the metric form in flat spacetime!!!
     
  13. May 28, 2007 #12
    I understand that there is no way to "fix" the equation, say, with an additional termto account for gravity? :frown:

    Thanks for the info,
    Harald.
     
    Last edited: May 28, 2007
  14. May 28, 2007 #13
    It is possible with static and stationary spacetimes but not in the general case.

    For example the Gullstrand-Painlevé coordinate chart of the Schwarzschild metric or the Doran chart of the Kerr metric both have a gravitational term. If you look at the relationship between the basis vectors and the tetrad frames of comoving observers in those charts you can see the gravitational term show up in the temporal part as a linear component of the 3-space.

    But in non-stationary spacetimes this is not the case as the gravitational terms from separate gravitational sources would "cross over" and create higher order non-linear terms. In fact it is impossible to algebraically reduce such solutions.

    It's not new :smile:
    Many, including Einstein, have actually tried that and failed. Einstein was not particularly eager to introduce curvature to his theory, but the conclusions were obvious, spacetime must be curved due to gravitation.
     
    Last edited: May 28, 2007
  15. May 28, 2007 #14

    Chris Hillman

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    Some caveats

    Hi Harald (not the User:Harald who edits the physics pages at Wikipedia, I take it?),

    Nothing in relativity theory (special or general) says "time runs slower here than there"! That wouldn't even make sense. Rather, when you somehow compare the times kept by observers in different states of motion (you need to say precisely how you do this!), you will generally find discrepancies. In particular, if two observers measure the time between two events (two flashbulbs going off, say), they may disagree on how much time elapsed between the events. That's all.

    Your question about gtr appears to relate to "coordinate speeds" in curved spacetimes. Two remarks may help:

    1. In working with curved spacetimes, it is essential to focus attention on physically/geometrically significant or "coordinate-free" notions, rather than "coordinate-dependent" notions. In particular, coordinate speeds are generally not operationally significant.

    2. There are operationally significant notions of "speed in the large". But you need to know that in curved spacetimes there are multiple operationally signficant methods of measuring distance "in the large", and thus "speed in the large". Thus to ask a question with an answer you need to specify one of these notions. This may require geometrical and physical insight.

    In particular, so-called "gravitational time dilation" really just means that outgoing null geodesics climbing away from the gravitational field of some isolated massive object will in general diverge due to the curvature of spacetime. This implies that distant observers may observe a redshifting of signals emitted by observers closer to the object.
     
  16. May 28, 2007 #15

    Garth

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    realcreation, you don't know what you are talking about.

    Try reading some basic texts on GR and then for this specific question read A General Treatment of Orbiting Gyroscope Precession by Ronald J.Adler and Alexander S. Silbergleit.

    Garth
     
    Last edited: May 28, 2007
  17. May 28, 2007 #16
    I know what I am talking about. I got master of science in modern differential geometry and PhD in astrophysics.

    Let us get down to physics: How to test physical theory (e.g., general relativity)? The answer is not math assumption. The answer is measuring equipment. Therefore, your space and time are given by physical equipment!!! If you study the most accurate equipments then you find out that they are generally electromagnetic waves!!! You know that waves are affected by gravity: then how do you say that waves measured curved spacetime. ALL honest astronomy theory or experiments including Gravity Probe B have the assumption that Local galaxy cluster gives a better inertial reference frame; Milky Way frame is worse; Solar frame more worse; Earth frame is the worst!!! It is very possible that The Universe is the absolute inertial reference frame, which, however, is denied by the requirement of curved spacetime assumption.

    Special Relativity provides one piece of physical truth: the perception of time and distance depends on the measuring reference frame, which, however, is rejected by GR. SR is apparently true because gravitational effect in waves is very weak. If SR is approximately true then GR is definitely wrong: not all reference frames are in equal footing; Curved spacetime is not neccesary.

    Have we measured curvature? No. The 4 classical tests of GR verified Schwarzschild Lagrangian form. Only Einstein equation involves curvature. However, its cosmologic solution suffers deadly difficulties. Gravitomagnetic effect will be given in December this year by GPB team.

    The above is the so-called GR (curved spacetime). Others like wormhole, black holes are simply imaginations based on curved spacetime. We should not be gooses following the leading one!

    I fully understand gr-qc/9909054 (Adler etc.).
     
  18. May 29, 2007 #17

    pervect

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    It's probably obvious to longer-term readers of Physics Forums why this thread has been locked, tagged, and bagged, but for newer readers benefit I will point out the PF forums guidelines:

    A sign of terminal crackpottery in my experience occurs when a poster makes claims that are completely contradictory with accepted science while simultaneously maintaining that he or she understands it fully.
     
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