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Did Einstein have a Physical Interpretation of Frames?

  1. May 25, 2014 #1
    By a frame I mean four orthogonal vectors for which one is timelike and the others are spacelike.

    In a bypassing some time ago I read something about the basis vectors in a frame representing sticks and and a clock. However, the author also noted that this interpretation was problematic. I would like very much like to read a more elaborate discussion on the issue.

    Did Einstein have an interpretation, can you explain it, or can you refer me to some good literature on it?
  2. jcsd
  3. May 25, 2014 #2


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  4. May 25, 2014 #3


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    What is problematic about it? A reference would be useful.
  5. May 26, 2014 #4

    I think it is related to the problem of rigid motion in relativity. I read it in "principles of quantum general relativity" which was on Google books, but is no more.
  6. May 26, 2014 #5


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    Born rigidity would only be a problem when speaking of frame fields, not frames themselves. Note that when I say "frame" or "frame field" I mean specifically local Lorentz frames or fields thereof.

    More precisely, the issue only arises if the frame field ##\{e_{\alpha}\}##, with ##e_0 = u## the 4-velocity field of some time-like congruence, is such that ##\mathcal{L}_{u} e_{i} \neq 0## for if ##\mathcal{L}_{u} e_{i} =0## then the always valid interpretation of each local Lorentz frame as three orthogonal meter sticks (or gyroscopes) and an ideal clock can also be carried over to the frame field itself as ##\mathcal{L}_{u} e_{i} =0## guarantees that if a given observer initially locks his meter sticks to those of an infinitesimally neighboring observer then they will remain locked for all (proper) time. Using these axes one can then obviously construct a rigid coordinate chart consisting of a lattice of rigid meter sticks and clocks.

    Now if such a frame field exists for this time-like congruence then note that ##h(e_0, e_{\alpha}) = 0, h(e_i,e_j) = \delta_{ij}## since ##g(e_{\alpha},e_{\beta}) = \eta_{\alpha\beta}##, where ##h## is the spatial metric, and so we have simply that [tex]\mathcal{L}_u h(e_{\alpha},e_{\beta}) = 0 \\ \Rightarrow (\mathcal{L}_u h)(e_{\alpha},e_{\beta}) + h(e_{\alpha}, \mathcal{L}_u e_{\beta}) + h(\mathcal{L}_u e_{\alpha}, e_{\beta}) = (\mathcal{L}_u h)(e_{\alpha},e_{\beta}) = 0 \\ \Rightarrow \mathcal{L}_u h = 0[/tex] which is, by definition, Born rigidity.

    In other words, in order to construct a rigid coordinate chart i.e. a rigid lattice of meter sticks and clocks, which as stated above requires the existence of a frame field that is Lie transported by the 4-velocity field of the time-like congruence of interest, one requires that said congruence be Born rigid. This means that for non-Born rigid fields of observers, one can still attach to each observer a local Lorentz frame and interpret each as a set of three meter sticks and a clock, but the so obtained frame field will not correspond to a rigid lattice of such meter sticks adapted to the entire field of observers. This basically means that if we want to construct a rigid coordinate chart adapted to some family of observers, it would be necessary for them to be undergoing Born rigid motion e.g. they could be following orbits of a time-like Killing field.

    So the lack of Born rigidity does not pose a threat to the interpretation of frames themselves as three orthogonal meter sticks and an ideal clock.
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