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The 4-velocity in contravariant form is given by $$V^\mu=\frac{dx^\mu}{d\tau}$$ for some general co-ordinates $x^\mu$ and proper time $\tau$.

Is the 4-velocity in covariant form given by

$$V_\nu=V^\mu g_{\mu\nu}=\frac{dx^\mu}{d\tau}g_{\mu\nu}=\frac{dx_\nu}{d\tau}?$$

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  • $\begingroup$ Yes, that is correct. In the context of general relativity these are better understood when you consider their transformations! $\endgroup$
    – twisted manifold
    Commented Jan 22, 2020 at 17:26
  • $\begingroup$ So $V_\nu=dx_\nu/d\tau$ is the correct notation? $\endgroup$
    – John Eastmond
    Commented Jan 22, 2020 at 17:28
  • $\begingroup$ I wouldn't say "correct", since it's notation. It is what we usually use though. $\endgroup$
    – twisted manifold
    Commented Jan 22, 2020 at 17:29
  • $\begingroup$ Ok - great. I was worried that I was somehow assuming that $g_{\mu\nu}$ could go inside the $\tau$ differential $\endgroup$
    – John Eastmond
    Commented Jan 22, 2020 at 17:30
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    $\begingroup$ I would say it is incorrect as explained in comments here. It is at best confusing/misleading notation. $\endgroup$
    – Qmechanic ♦
    Commented Jan 22, 2020 at 19:41

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Yes. However, the last term is a bit 'icky' because it looks like $$ \frac{dx_\nu}{d\tau} = \frac{d}{d\tau} x_\nu $$ while what is actually meant is $$ \frac{dx_\nu}{d\tau} = \left(\frac{dx}{d\tau}\right)_\nu $$ It doesn't really make sense to 'lower' the index of a coordinate of the base manifold. However, while I wouldn't recommend it, I think I've seen other people use that notation as well.

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