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I Help with Newtonian Gravity as Limit case of General Relativ

  1. Nov 28, 2016 #1
    In Schutz says When we have weak gravitaional fields then the line element *ds* is
    $$
    ds^{2}=-(1+2\phi)dt^{2}+(1-2\phi)(dx^{2}+dy^{2}+dz^{2})
    $$
    so the metric is
    $$
    {g_{\alpha\beta}} =\eta_{\alpha\beta}+h_{\alpha\beta}= \left( \begin{array}{cccc}
    -(1+2\phi) & 0 & 0 & 0\\
    0 & (1-2\phi) & 0 & 0\\
    0 & 0 & (1-2\phi) & 0\\
    0 & 0 & 0 & (1-2\phi)\end{array} \right)
    $$
    where
    $$
    \phi=\frac{M}{r}
    $$
    so *h* is

    $$
    {h_{\alpha\beta}} = \left( \begin{array}{cccc}
    -2\phi & 0 & 0 & 0\\
    0 & -2\phi & 0 & 0\\
    0 & 0 & -2\phi & 0\\
    0 & 0 & 0 & -2\phi\end{array} \right)
    $$
    the element

    $$
    h_{00}= -2\phi
    $$

    and the elements out of the diagonal are zero because the condition weak gravitational fields it implies

    $$
    T_{i,j}=0
    $$

    but i dont get it how in the book do

    $$
    h_{xx}=h_{yy}=h_{zz}=-2\phi
    $$

    i believe they use de definition of *trace reverse*

    $$
    \bar h^{\alpha\beta}=h^{\alpha\beta}-\frac{1}{2}\eta^{\alpha\beta}h
    $$
    and the *trace* definition

    $$
    h = h^{\alpha}_{\alpha}
    $$
    but how they do? what im missing?

    Thanks
     
  2. jcsd
  3. Nov 28, 2016 #2
    We are just pulling each diagonal term out of h. h1 = hx. Repeated subscript is an element along the diagonal.
     
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