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Could a GR expert look at this?

  1. Jan 14, 2005 #1

    selfAdjoint

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    I posted one quantum theory about this paper: http://www.arxiv.org/abs/quant-ph/0501034

    Chen uses a double Kalusza-Klein mod to GR with the two extra dimensions being timelike. He claims this introduces QM into the modified GR, so he has interacting QM and gravity, then he claims to support Dirac spinors in his formalism.

    These are bold claims and he shows his work. I wonder if that work is correct and don't myself have the chops to say yea or nay. Would some of you gurus just take a look? Thanks.
     
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  3. Jan 15, 2005 #2

    Garth

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    It seems to work, although as Chen's English is a little quaint the paper is difficult to understand in places. Basically you can add further forces that obey certain symmetries to gravitation and explain them geometrodynamically by using more dimensions, the more complicated the forces the greater the number of dimensions needed, hence ending up with 10, 11 or even 26 dimensions. He claims to use only six of which three are 'time' dimensions, which is elegant and efficient to say the least!

    However I have three questions:

    1. Chen's metric is:
    ds2 = dx20 − dx21 - dx22 − dx23 + e−(2i/h) (paxa − m0x5)dx24 - dx25 (Chen's equation 18)

    and he says
    Yet mathematically he treats the sixth dimension x5 as a space dimension in that his metric has a signature [+,-,-,-,+,-]. This is important as he uses both time dimensions to explain quantum non-locality by suggesting (in my 'hand waving' explanation) that particles can 'nip round the back' in the other time dimensions to reappear so that they seem to be at two places at once. (At least I think that was what he was saying) Therefore in what way is x5 actually a time dimension?

    2. Chen’s fifth dimension x4 is complex and that makes the whole metric complex. What does this mean in practice? Solving a quadratic equation to obtain two complex roots normally means that a solution does not exist in the real world; does the same apply to Chen’s metric? He seems to leave the question open for future resolution as he says:
    3. Although Chen claims his theory incorporates some essential features of quantum theory he does not address whether, as a gravitational theory, it is concordant with the standard tests of GR and with the present cosmological constraints. So does it really work?

    In conclusion, with these caveats above, I would only be convinced by it if the theory made some testable prediction that was then verified by experiment.

    Garth
     
  4. Jan 15, 2005 #3

    selfAdjoint

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    Garth, thanks for digging into it. I hadn't spotted that complex dimension, and I wonder if the signature on the other one is just a typo or not. I agree that his English is difficult to follow.
     
  5. Jan 15, 2005 #4

    pervect

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    I'm still trying to get some sort of feeling for what "two time dimensons" would mean. If we write dx^2+dy^2+dz^2 = dt1^2 + dt2^2, we can see that light travels in a sphere only when dt2 = 0. We will get apparent superluminal effects when d2t is not equal to zero - dx^2 + dy^2 + dz^2 > dt1^2

    It's not quite clear to me what the physical interpretation of the second time dimension in this paper is. Is the author "rolling it up into a ball", or is he doing something else? He seems to argue that the second time dimension doesn't have to be small - but if it's not small, how do we explain that light appears to travel at "c" on a macroscopic scale? I can only start to make sense of the theory if I assume that the second time dimension is very small, or otherwise constrained to be almost unobservable on a macroscopic scale.
     
  6. Jan 15, 2005 #5

    selfAdjoint

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    I originally thought that both the extra time dimensions were real, and compacted on small circles; that would make the two together a torus, which would have its own effects. But this complex dimension has thrown me. The Riemann sphere cross a circle?
     
  7. Jan 16, 2005 #6

    Aether

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    It is well known (e.g. zitterbewegung) that one component of the instantaneous speed of all particles is equal to the speed of light, but the direction of that motion cycles very quickly.

    Using spherical polar coordinates, take dt1 as the magnitude (e.g. dR) of the time vector [tex]\bold{\alpha}[/tex] rather than assuming proper time to be a scalar (let t and dt always refer exclusively to cosmological time which is a scalar). Then the other two "dimensions" of time simply encode the instantaneous direction of the time vector, and [tex](\bold{\alpha},\bold{\beta})}[/tex] replaces [tex](1,\bold{\beta}})=\frac{(c,\bold{v})}{c}[/tex]. There is more symmetry to this than there is with 4-vectors, and a framework for nonlocal dynamics emerges that is absent with 4-vectors.

    The no preferred frame doctrine is enforced by taking proper time to be a scalar (e.g., proper time is automatically isotropic in all frames). Taking proper time to be a vector provides a framework for the identification of a preferred frame.
     
    Last edited: Jan 16, 2005
  8. Jan 16, 2005 #7

    pervect

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    If we consider a group of observers at a single point in space, moving with different velociities, all of them will have different time vectors (they will all point in different directions in the space-time manifold). However, according to standard theory, no observer will have more than one time vector. Everyone will have an orthonormal basis of vectors which consists of three space vectors and one time vector. Chen's theory has 6 (or maybe 7) vectors in it's orthornormal base (ONB) of vectors - one of one time, four of space (judging by the metric signature rather than Chen's unclear English), and one which Chen calls time but is actually a complex number (!) - I would guess that there is some messy decomposition of Chen's theory into a 7 dimensional space as the term "dimension" is usually definied.

    It's not very clear what you mean by "cosmological time", I would assume given no further information that it would be intended to be the time vector of an observer who sees the cosmic microwave background radiation as being isotropic.

    I would not take dr/dt to be a "time vector", but a component of a velocity vector.

    I don't know much about Zitterbewegung, but a google search shows that it appears to be some sort of property of the solution of the Dirac equation. In such a case, it had better disappear for macroscopic objects.

    It's not possible to have a time that is isotropic in all frames - it's very well known that different observers have different notions of isotropy as the term is usually defined. To be very specific, a frame is isotropic in the sense I mean if two objects with an identical mass m and a velocity of +v and -v have a total momentum of zero. It's the necessary result of any "fair" velocity measurement which doesn't have a "preferred direction".
     
  9. Jan 16, 2005 #8

    Aether

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    Cosmological time is scalar Euclidean time, and it is a part of standard cosmology based on GR. The CMB monopole is locally Lorentz invariant but it evolves over time, and there are other such quantities all of which can be used to build a cosmological clock. Estimates of the age of the universe (e.g., 13.4Gyr from the WMAP team for example) are estimates of elapsed cosmological time, and in principle all that is required is a greater precision for such estimates to form a truly viable cosmological clock for use in SR-type experiments.

    dR/dt=c, and dr/dt=v.

    You mean, just like every other measurable aspect of QM and SR? I throw SR in there, tentatively, because all interferometer experiments (e.g., Michelson-Morley for example) depend on the assumed QM superposition of photons (e.g., each photon must travel along both orthogonal paths in the interferometer simultaneously!) to function. I wonder if that really is just as valid as comparing the time-of-flight of two different photons traveling along the two different paths within the interferometer.

    I suppose that proper time is automatically isotropic in all frames when it is represented by a scalar [tex]\tau[/tex]. Nevertheless, Cosmological time really is isotropic (and locally invariant over any Lorentz transform between frames) in all frames. The CMB monopole is isotropic (and locally invariant over any Lorentz transform between frames) in all frames, and can be (has already been) used to construct a cosmological clock.

    What the...? So, was Bob revealed to be a) a card counter, b) an aether theorist, or c) both?
     
    Last edited: Jan 16, 2005
  10. Jan 16, 2005 #9

    pervect

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    OK, this answers my question about what you mean by "cosmological time", but your main point still escapes me.

    You mean, just like every other measurable aspect of QM and SR?

    No, just like every other measuarable aspect of QM. In other words, QM must reduce to the classical limit.

    It's very easy to calculate interference fringes from the properties of waves - quantum mechanics is just not needed. One only needs the wave aspect of light to deal with interference, therfore a classical treatment of radiation with Maxwell's equations is perfectly adaquate to analyze the operation of interferometers.

    Because quantum mechanics must reduce to the classical limit for large number of particles, the results of a quantum analysis of an interferometer must reduce to the classical results.


    You seem to be slipping off into your own private theories here (possibly aggravated by non-standard usage of standard terms). Proper time is simply the time measured by an actual clock. With only one clock to deal with, synchronization issues do not arise, therfore the isotropy issue does not arise. So it doesn't make any sense to say that "proper time" is automatically isotropic. Isotropy is irrelvant to proper time.

    Cosmological time isn't defined in any particular frame of reference - and synchoronizing clocks in a specific inertial frame via cosmological time will not necessarily give one an isotropic clock syncronization.
     
  11. Jan 16, 2005 #10

    Aether

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    You are saying that "isotropic clock synchronization" makes sense, but that "isotropy of proper time" does not? Does it make any difference whether there is one time coordinate or three? Owning a set of cosmological clocks is what provides the isotropic clock synchronization. These may then be used to probe the isotropy of the proper time clock synchronizaton.
     
  12. Jan 17, 2005 #11

    pervect

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    Cosmological clocks really have nothing whatsoever to do with isotropy. Clocks are synchronized isotropically in an inertial frame when the speed of light is constant in all directions. (Note that it takes two clocks to measure the speed of light, one clock at the starting position and one at the ending position). If the clocks are not synchronized isotropically, the speed of light going in one direciton will be different than the speed of light going in the other direction.

    This clock synchronizatoin that makes the speed of light isotropic (the same in all directions) is the same clock synchronization that makes momentum isotropic - i.e. it makes the momentum of an object a function only of the magnitude of its velocity, and not the direction of travel.

    This notion of isotropic clock synchronization has nothing to do with cosmological clocks, or propre time. Proper time is the time measured on a single clock, so the issue of syncrhonization doesn't even arise. One can think of "proper time" as the time on an observer's watch, a clock that the observer carries wherever he goes.
     
  13. Jan 17, 2005 #12

    Aether

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    Cosmological clocks are absolute clocks measuring absolute Euclidean time regardless of location or inertial frame. This makes for an absolute definition of simultaneity (e.g., two events are simultaneous if they both occur at the same cosmological time). Now, either the speed of light is anisotropic, or proper time is anisotropic. They can not both be isotropic, in more than one preferred frame, in a closed universe where cosmological time is isotropic. So, if the speed of light is to be presumed isotropic, then three time coordinates will be needed to reconcile cosmological time with anisotropic proper time.

    Linear momentum has three components, one for each direction in space. Assuming that the speed of light is isotropic, then so must proper time have three components, one for each direction in space.

    Why not think of cosmological time as the time on an astute observer's watch from now on? Why not rely on an astute observer rather than a feral one? Single-coordinate cosmological clocks are OK, but single-coordinate proper time clocks will have to be replaced by three-coordinate proper time clocks.
     
    Last edited: Jan 17, 2005
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