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What happens to the form basis after making the metric time orthogonal

  1. Sep 16, 2013 #1
    Given a basis for spacetime ##\{e_0, \vec{e}_i\}## for which ##\vec{e}_0## is a timelike vector. Of these vectors one can make a new basis for which all vectors are orthogonal to ##\vec{e}_0##. I.e. the vectors $$\hat{\vec{e}}_i = \vec{e}_i - \frac{\vec{e}_i \cdot \vec{e}_0}{\vec{e}_0 \cdot \vec{e}_0} \vec{e}_0 =\vec{e}_i - \frac{g_{i 0}}{g_{00}} \vec{e}_0 $$

    Now at page 35 in the following lecture notes: http://www.uio.no/studier/emner/matnat/fys/FYS4160/v06/undervisningsmateriale/kompendium.pdf

    one use the basis ##\{\vec{e}_\hat{0}, \hat{e}_i\}## basis to define a "spatial metric" ##\gamma_{ij} = \hat{\vec{e}}_i \cdot \hat{\vec{e}}_j## for which it is trivial to show that

    $$\gamma_{ij} = g_{ij} - \frac{g_{i0}g_{j0}}{g_{00}}$$ such that in this basis the spacetime metric becomes

    $$ ds^2 = - d\hat{t}^2 + dl^2$$

    where ##dl^2 = \gamma_{ij} dx^i dx^j## and ##\hat{\vec{e}}_0## is normalized. Note that ##\gamma_{i0} = \gamma_{00} = 0##. Now what the author does next is to express ##d \hat{t}^2## in terms of the arbitrary basis by

    $$d \hat{t}^2 = dl^2 - ds^2 = (\gamma_{\mu \nu} - g_{\mu \nu})dx^\mu dx^\nu$$

    and by plugging in the result above for ##\gamma_{ij}## he arrives at

    $$d\hat{t}^2 = - g_{00}(dx^0 + \frac{g_{i0}}{g_{00}} dx^i)^2. $$

    Here is my problem with this derivation: when writing
    $$dl^2 - ds^2 = (\gamma_{\mu \nu} - g_{\mu \nu})dx^\mu dx^\nu$$
    one is using the same forms ##dx^\mu## to express ##dl^2## for the spatial metric as one uses to express ##ds^2## (the not time-orthogonal metric). Is there a rationale behind doing this? I.e. can one use the same basis of forms for expressing the spatial metric as one does with the spacetime metric when using components defined by ##\vec{e}_i \cdot \vec{e}_j## and ##\hat{\vec{e}}_i \cdot \hat{\vec{e}}_j## respectively?
     
  2. jcsd
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