I’m a little bit confused about how to get the four-velocity components from a given metric tensor (or line element). For instance, which are the components of the four-velocity in the Schwarzschild metric? Can anybody help me?
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3$\begingroup$ This is because the metric does not contain any information about four-velocity! The only "connection" that I am aware of, is using the derivatives of the metric (Christoffel symbols), each multiplied by two four-vectors, to calculate geodesic motion. en.wikipedia.org/wiki/Geodesics_in_general_relativity $\endgroup$– m4r35n357Commented Nov 11, 2018 at 10:51
2 Answers
As kalle has said, the metric and the four-velocity are not intrinsically related, you can have an object with velocity in any given direction and with any magnitude even if you fix the metric.
The thing you can do with the metric, however, is to normalize the velocity: in the $-+++$ metric convention the four-velocity of a massive object always has a norm of $u^2 = g_{\mu\nu}u^\mu u^\nu = -1$, which constrains the value of a component of the velocity as long as you know the values of the other three. For example, you can choose the three spatial components of the velocity to be whatever you like, and find out using the metric what the time component should be.
For a massless object the reasoning is the same, only with $u^2 = 0$.
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$\begingroup$ Why four-velocity is normalized to be u.u = -1 ? Any specific reason for that ? $\endgroup$– apkCommented Dec 23, 2023 at 10:09
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1$\begingroup$ Just convenience, it's an easy number - any other parameterization would be valid but potentially more cumbersome $\endgroup$ Commented Dec 24, 2023 at 21:33
The metric and the four velocity are not connected in such a way. For instance, you could consider a plasma around a black hole. You want to use the Schwarzschild metric. In the rest frame of the fluid you can write $u_j=\left(-\sqrt{B},0,0,0\right)$. As the plasma there is only governed by the magnetic field (in this very specific case!). You could as well use another frame and the 4-velocity would look different.
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$\begingroup$ But the desplacement vector is related to the metric $\endgroup$ Commented Nov 11, 2018 at 17:06