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I know that already exist a question on How to get the four-velocity components from a given metric tensor?; the conclusion of the answers were that 4-velocity and the metric are quite unrelated concepts, concerning calculations of 4-velocity components.

Nevertheless, I would like to know if there's tricks and tips to calculate these components. For instance, it seems that the normalisation constrain:

$$ u_{\mu}u^{\mu} = -1 \tag{1}$$

plays an desicive role on how calculate 4-velocity components in an arbitrary spacetime. Furthermore, even though the metric appears just as an tensor operation:

$$ g_{\mu\nu}u^{\mu}u^{\nu} := u_{\mu}u^{\mu} = -1 \tag{2}$$

we have, still, a relationship between these two tensors.

So, I would like to ask:

If someone give me a metric tensor and ask: "what are the components of 4-velocity?" how can I calculate the components?

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$[1]$ How to get the four-velocity components from a given metric tensor?

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If someone give me a metric tensor and ask: "what are the components of 4-velocity?" how can I calculate the components?

You can’t. A massive particle at any point in any metric can have any time-like four-velocity, and a massless particle can have any lightlike four-velocity.

All the metric lets you do is compute the covariant components of the four-velocity from the contravariant components, or vice versa.

One scalar equation $g_{\mu\nu}u^{\mu}u^{\nu}=-1$ can’t determine all four components of a four-vector. All it can do is impose a constraint so that only three of the components are independent. Usually one thinks of the three $u^i$ as independent and $u^0$ as determined by the constraint.

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  • $\begingroup$ Ok. But what about the information of $u_{b}$? in $u^{0} = g^{0b}u_{b}$. I not confusing tensor algebra here, rather I'm little confused about the physics. $\endgroup$
    – M.N.Raia
    Commented Apr 8, 2020 at 21:31
  • $\begingroup$ Why are you writing a spatial index $b$ instead of a spacetime index? I would describe $u^\mu$ and $u_\mu$ as having the same amount of “information”. I consider $u^\mu$ to be physically intuitive (since coordinates are contravariant) and $u_\mu$ to be formalism. $\endgroup$
    – G. Smith
    Commented Apr 8, 2020 at 21:37
  • $\begingroup$ Ok. But what about the information of $u_{\nu}$? in $u^{\mu} = g^{\mu\nu}u_{\nu}$. I not confusing tensor algebra here, rather I'm little confused about the physics. I'm inclined to think that you can always state that $u_{\mu} = (-1,0,0,0)$ then, calculate the $u^{\mu}= g^{\mu\nu}u_{\nu}$ and find some different compoenents $u^{\mu} = (a,b,c,d)$. Actually I'll transform this into a question. $\endgroup$
    – M.N.Raia
    Commented Apr 8, 2020 at 21:41
  • $\begingroup$ I think you are getting confused by considering a rest frame. In general, $u_\mu$ is not $(-1,0,0,0)$. $\endgroup$
    – G. Smith
    Commented Apr 8, 2020 at 21:46

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