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Given a 4-velocity $u^0$, how do you find $u_0$? Do you use $u_{\alpha}u^{\alpha} = -1$?

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    $\begingroup$ Do you know what the metric is? $\endgroup$ – Kyle Kanos May 20 '15 at 11:10
  • $\begingroup$ Yes, it is the Schwarzschild metric. $\endgroup$ – user145974 May 20 '15 at 11:39
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    $\begingroup$ I had meant in general, do you know what the metric is, not specifically in this case. The metric (and it's inverse) allows you to change the position of the indices. This is a really trivial issue that is mentioned in every GR text, AFAIK. $\endgroup$ – Kyle Kanos May 20 '15 at 11:46
  • $\begingroup$ @user145974: I think Kyle's point is that this seems an awfully large hole in your knowledge of differential geometry, and you should fill it asap. Having said that, the co and contravariant distinction and the whole business of raising and lower indices confused the hell out of me at first :-) $\endgroup$ – John Rennie May 20 '15 at 15:45
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Use:

$$ u_{\alpha}= g_{\alpha\beta}u^{\beta} $$

where $g_{\alpha\beta}$ is the metric tensor.

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  • $\begingroup$ Don't forget summing over beta. $\endgroup$ – Horus May 21 '15 at 12:19
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You can always transform to the coordinates of the traveler which experiences no motion in space but only in time (with proper time!). That way $ d\tau=dt $ and $ dx=dy=dz=0 $ . $ |u|^2 = \eta_{\alpha \beta} u^{\alpha} u^{\beta}=-1 $ - which is a scalar. Scalars are invariant under any coordinate transformation, so even after you return to your original coordinates - the norm of the 4-velocity vector stays the same.

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