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Scaling Paradox.

  1. Oct 6, 2009 #1
    Differential geometry (which includes general relativity) often introduces the length differential, expressed as ds2=gabdxadxb, to introduce the covariant form of the metric tensor gab. However, this formulation scales ds2 incorrectly. The appearance of an index as a superscript, as in dxa, indicates that under a dilation, dxa scales as a contravariant tensor. That is, if we double the length of our measuring rod, then the value of dxa falls in half. For a covariant tensor, such as a gradient, the index appears as a subscript, as in dxa. If we double the length of our measuring rod, then the value of the of dxa doubles. Rotations and other lorentz transformations do NOT affect covariant and contravarient vectors differently because the projection of one unit vector on another is reflexive. The common representation of a lorentz transformation as [tex]\gamma[/tex]ab displays invariance of a lorentz transformation under dilation. If ds2=gabdxadxb, then ds2 is invarient under dilation. Then ds does NOT describe the length of a differential displacement, which should scale contravariantly, like dxa, because in another coordinate system, ds=dx'a. Writing ds2=dxadxa for Cartesian coordinates gives ds proper scaling for a differential displacement. But, to describe a Minkowski space, the sign on dr2 must differ from that on dt2. So one of them must have an explicit factor of i. This sign appears in the metric if we adopt David Hestenes' formulation gab=[tex]\gamma[/tex]a[tex]\gamma[/tex]b in terms of the basis vectors [tex]\gamma[/tex]a. This formulation leaves the metric in a familiar form. If the negative sign attaches to dt2, then [tex]\gamma[/tex]a[tex]\gamma[/tex]a=3-1=2 instead of 4. Then if [tex]\gamma[/tex]a[tex]\gamma[/tex]b=g(ab)+g[ab] operates on [tex]\gamma[/tex]b, then 2[tex]\gamma[/tex]a=[tex]\gamma[/tex]a+g[ab][tex]\gamma[/tex]b, or [tex]\gamma[/tex]a=g[ab][tex]\gamma[/tex]b, so that the antisymmetric part g[ab] transforms tensors without dilation beyond that in the basis vectors, like the symmetric part g(ab). This symmetry might say something about why we observe variation in only one time dimension rather than the expected three. If [tex]\gamma[/tex]a[tex]\gamma[/tex]a=1-3=-2, then -[tex]\gamma[/tex]a=g(ab)[tex]\gamma[/tex]b and -[tex]\gamma[/tex]a=g[ab][tex]\gamma[/tex]b, to preserve the absence of further dilation.
  2. jcsd
  3. Oct 6, 2009 #2
    There was a lot said there, and no question marks. So I'll just try to respond to some of it and hope it helps.

    [itex]ds^2[/itex] is the invariant line element, it is neither covariant or contravariant. You should expect it to not scale/change at all with a change in coordinates.

    Lengths are indeed contravarient (ie [itex]x^\mu[/itex] instead of [itex]x_\mu[/itex]). The displacement would be something like [itex]r = x^\mu e_\mu[/itex] where 'e' are the basis. As the basis are obviously covariant, written this way it is clear the 'lengths' should be contravariant.

    Since that (hopefully) helps clear up much of the beginning of your comments, does that help 'fix' whatever the main issue is?
    If not, can you try describing more succinctly what your question is for us?
  4. Oct 6, 2009 #3
    Thx Re: Scaling Paradox.

    Thank you, Justin Levy. I wrote a lot because I wanted my paradox to be understood by someone who is new to differential geometry. Jargon can speed or impede communication. Your reply confirmed a suspicion of mine which I have not seen explicitly stated in a text before. It certainly clears up the paradox for me. But, it leaves another: Does the invariant length suggest a preferred coordinate system, and therefore a preferred reference frame? It would be the coordinate system that gives the same value for a length as the value given by the invariant length.
  5. Oct 8, 2009 #4
    Re: Thx Re: Scaling Paradox.

    Even restricting ourselves to inertial coordinate systems, that would not specify a unique coordinate system. Since it is not unique, I would not consider it preferred in any sense.

    "Preferred frame" can just be the colloquial meaning that there is a frame that the physics looks simpler in (is easier to do the calculations in). In that sense, yes there are preferred frames all the time. But the term is usually reserved for theories in which one frame is physically unique, in that a background field or constant makes the physics different in that frame. Since we are only discussing relativity itself here (which is essentially just the requirement of a particular symmetry), no invariant can cause a preferred frame to appear in this sense here.
  6. Oct 8, 2009 #5
    Thank you. My comment about the preferred reference frame is a little embarrassing to me. There is nothing wrong with preferring a frame which matches geodesics closely. Furthermore, the distances which one measures in that frame are invariant with respect to metric, in that picking a different coordinate system will not affect that measurement.
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