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Absolute Space and Time in Einstein's General Theory of Relativity

  1. Aug 18, 2003 #1
    The Special Theory of Relativity, we teach our students, did away with Absolute Space and Absolute Time, leaving us with no absolute motion or rest, and also no absolute time order. General Relativity is viewed as extending the "relativity of motion" applicable to curved spacetimes, and General Relativity's most probable models of our actual spacetimes (the big-bang models) appear to re-introduce a privileged "cosmic" time order, and a definite sense of absolute rest. In particular, some of the same kinds of effects whose *absence* led to rejection of Newtonian absolute space are present in these models of GTR.

    End of quote. Colloquium for 13-NOV-97 Abstract, UCR.
    See http://physics.ucr.edu/Active/Abs/abstract-13-NOV-97.html [Broken]

    I'm delighted that common sense is finally being recognized in the physics community. When do you think it will be realized that an absolute time order precludes the possibility of anything falling into a black hole?

    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Aug 18, 2003 #2


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    It does? That's news to me. Wanna point out how before you run wild with this?
  4. Aug 19, 2003 #3
    The colloquium started without me. I don't even know the academician who delivered the talk. The posted sketch seems eminently believable, based on the elementary global theorem of the second link. If you believe the analysis is false, how then do you resolve the paradox of the great illumination?
  5. Aug 19, 2003 #4
    Let's start with the basics. A frame is an absolute frame of reference if it is physically distinguished from all other frames and if it is totally independent of the distribution of matter/energy in the universe.

    The simplest and easiest to conceive universe that has an absolute frame of reference is the cylinder SxR. The physics of the SxR universe is dealt with at length in these links:

    http://cornell.mirror.aps.org/abstract/PRD/v8/i6/p1662_1 [Broken]

    The reader is required to comprehend trivial model universes or advanced mathematics. The space part of SxR is a circle S.

    Let an observer in the circle universe be at rest in her own "inertial frame of reference." Let her pick a positive and negative direction. Suppose she has a nice watch on her wrist to note the time. Other than just looking nice and being able to see how old she's getting, let her do experiments. Let t1 be the time she measures for a photon to circumnavigate the universe in the positive direction. Let t2 be the time she measures for a photon to circumnavigate the universe in the negative direction. Being only a one-dimensional creature she is still smart enough to realize the impossibility of all frames agreeing on a frame independent law of light propagation. Consequently, if t1 doesn't equal t2, then she is moving at some velocity v with respect to an absolute frame of reference. Isn't it obvious, based on the global theorem, that the velocity v is given by the equation: t1/t2 = (c+v)/(c-v) ?
    Last edited by a moderator: May 1, 2017
  6. Aug 19, 2003 #5
    There will never be any time difference to the observer. The only time difference that will be measured will be from the person sitting in your "absolute frame of reference". Everyone in other frames will get different time differences and they will all think they are in what you claim to be the "absolute frame of reference". No one will agree on anything.

    This is an argument I've seen raised many times and refuted each time. Also, from what I've just seen of "www.everythingimportant.com", it doesn't look like a reliable scientific reference.

  7. Aug 19, 2003 #6


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    Actually, that's incorrect.

    For example, assume the obvious structure on SxR, and suppose in the "absolute" frame of reference it takes light one year to circumnavigate the universe.

    Now, if an observer travelling 0.5c in the clockwise direction (WRT the "absolute" frame) emits a clockwise travelling photon, it will be 1.73 years before he sees it again, yet only 0.58 years for an emitted counterclockwise travelling photon.

    (as computed by actually drawing SxR and the worldlines, and directly computing the proper time observed by the observer)

    However, the everythingimportant website has its problems; it all starts to go downhill with the statement "It’s obvious from the notion of measure that the distance around the universe is the same in both directions"; it's not even obvious how to measure the distance around the universe, let alone obvious that it must be the same.

    One way to do this is to integrate spatial displacement dx along a circumnavigating worldline (or any circumnavigating path, for that matter) in a particular globally minowski frame. The "absolute" reference frame will indeed find, when measuring in this way, that the distance around the universe will be a constant no matter what path is taken. However, in every other reference frame, the measured distance around the universe will vary depending on the chosen path. In particular, for the previously mentioned observer, if he used the worldlines of the two photons to measure the distance around the univesre, he'd find it the clockwise-travelling photon underwent a spatial displacement of 1.73 light-years, and the counterclockwise-travelling photon underwent a spatial displacement of 0.58 light-years (results that are unsurprisingly consistent with the speed of light).

    I don't see any justification in calling the "absolute" reference frame absolute; it does have one particular property that the others do not, but that's it.

    Anyways, the section "A Global Theorem" depends crucially on there being a well-defined "distance around the universe" in any reference frame, but such a thing has not been demonstrated to exist.
  8. Aug 19, 2003 #7

    You got the ratio correct but the numbers wrong. t1 = 1.5 and t2 = .5

    Use the transformation equations:



    With these equations and the following rules you'll see that my system is free of contradictions.

    Let d be the distance around the universe in the absolute frame of reference. Let d' be the distance around the universe according to a "stationary" observer in a moving frame of reference. Then d'=Y(v)d

    To track the motion of light rays in a moving frame of reference, use the equation, distance =rate*time and take special note of the positive and negative directions:

    C(v) is the velocity of light in the positive direction.

    C(v) is the velocity of light in the negative direction.

    C(v) = (Y(v)^2)(c-v)

    C(v) = (Y(v)^2)(c+v)

    Be bold. Assume that distance in a moving frame makes sense. (I made that assumption). It then follows that "distance around the universe is the same in both directions."

    More importantly, can you prove that there are any contradictions in my interpretation of relativity on SxR?

    Eugene Shubert
    Last edited: Aug 20, 2003
  9. Aug 20, 2003 #8
    The observer was described as being in the motion frame. Just terminology against convention in the original stated problem.

  10. Aug 20, 2003 #9


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    I didn't use any transformation. I computed the proper time experienced by the moving observer using the formula for proper time:

    (c d&tau;)2 = (c dt)2 - dx2

    If you draw the worldlines (using the "absolute" coordinates), while waiting for the clockwise travelling photon to return, the moving observer undergoes a spatial displacement &Delta;x = 1 light-year, and undergoes a (coordinate) temporal displacement of &Delta;t = 2 years, so plugging into the formula for proper time you get an elapsed proper time &Delta;&tau; = [squ]3. For the counter-clockwise travelling photon, &Delta;x = 1/3, &Delta;t = 2/3, and &Delta;&tau; = 1/[squ]3

    Which brings us to your transformation equations; they do not preserve the form of the metric. (t', x') is not an inertial reference frame.

    Basically, you have thrown away the whole of relativity just so you could have a constant "distance around the universe" in every reference frame.
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