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Four-Velocity and Schwartzchild Metric

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    What is the Schwartzchild metric. Calculus the 4-velocity of a stationary observer in this spacetime (u). Show that u2 = c2.

    2. Relevant equations
    Schwartzchild Metric
    [tex]d{s^2} = {c^2}\left( {1 - \frac{{2\mu }}{r}} \right)d{t^2} - {\left( {1 - \frac{{2\mu }}{r}} \right)^{ - 1}}d{r^2} - {r^2}d{\theta ^2} - {r^2}{\sin ^2}\theta d{\phi ^2}[/tex]

    3. The attempt at a solution
    I already turned this in... I just want to make sure I understand this properly. I'm pretty sure that [itex]\mathbf{u}\cdot\mathbf{u}[/itex] is always equal to c^2 in the rest frame of a particle... So I think the point of the problem is to figure out the components of the four-velocity based on this fact. So here is my attempt at a solution... how ever I actually arrived at the solution in going from the bottom to the top.

    So here's how I attempted to solve it:
    From the Schwartzchild metric:
    [tex]g_{00} = c^2\left(1 - \frac{2\mu }{r}\right)[/tex]

    The four-velocity u of a stationary observer is given by:
    [tex]
    \begin{gathered}
    \left[ {{u^\mu }} \right] = \left( {{{\left( {1 - \frac{{2\mu }}{r}} \right)}^{ - 1/2}},0,0,0} \right) \hfill \\
    {c^2}\left( {1 - \frac{{2\mu }}{r}} \right){\left[ {{{\left( {1 - \frac{{2\mu }}{r}} \right)}^{ - 1/2}}} \right]^2} = {c^2} \hfill \\
    \hfill \\
    \end{gathered} [/tex]

    It follows that:
    [tex]{\mathbf{u}} \cdot {\mathbf{u}} = {g_{ab}}{u^a}{u^b} = {g_{00}}{\left( {{u^0}} \right)^2} = {c^2}\left( {1 - \frac{{2\mu }}{r}} \right){\left[ {{{\left( {1 - \frac{{2\mu }}{r}} \right)}^{ - 1/2}}} \right]^2} = {c^2}[/tex]
     
  2. jcsd
  3. Oct 26, 2016 #2

    Orodruin

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    Looks fine to me.
     
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