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Time synchronization in arbitrary reference frames.

  1. May 23, 2013 #1
    I'm having a bit of trouble with digesting an argument for which the conclusion is that it's not possible, in general, to Einstein synchronize clocks around closed curves. The argument goes as follows;

    (Beginning)
    Consider a four-velocity field ##e_{\hat{0}}## of the reference particles in a reference frame R.
    In an arbitrary basis ##\{e_\mu\}## where we choose ##e_0## parallel to ##e_{\hat{0}}## the vectors ##e_i## need not be parallel to ##e_0##. We thus introduce

    $$ e_{\perp i} = e_i - \frac{e_i \cdot e_0}{e_0 \cdot e_0} e_0 $$

    and define

    $$ \gamma_{ij} = e_{\perp i}\cdot e_{\perp j} = g_{ij} - \frac{g_{i0}g_{j0}}{g_00}$$

    where ##g_{\mu \nu}## is the components of the metric tensor corresponding to the original general basis. Written in terms of the time orthogonal basis ##\{e_{\hat 0}, e_{\perp,i}\}## then we get

    $$ds^2 = - d\hat t^2 + \gamma_{ij}dx^i dx^j.$$

    We can thus write

    $$ d\hat t^2 = \gamma_{ij}dx^i dx^j - ds^2 = \gamma_{ij}dx^i dx^j - g_{\mu \nu} dx^\mu dx^\nu = [(-g_{00})^{1/2}(dx^0 + \frac{g_{i0}}{g_{00}} dx^i)]^2.$$

    Therefore ##d\hat t=0## corresponds to

    $$ dx^0 = - \frac{g_{i0}}{g_{00}} dx^i$$

    which is not a perfect differential. Thus the line integral of the coordinate time around a closed curve is in general not zero. The author then concludes that since ##d\hat t = 0## corresponds to simultaneity on Einstein synchronized clocks it is not in general (##g_{i0} \neq 0##) possible to Einstein synchronize clocks around closed curves.
    (End)

    Questions:
    (1) Why does ##d\hat t = 0## correspond to Einstein synchronization of clocks?
    (2) Is so bad that the line integral of the coordinate time is different from zero? Does this generally implies something other than a bad choice of time coordinate?
     
  2. jcsd
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