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( Question will be: why modulus of obtained four-velocity is not $c^2$ ? )

In Schwarzschild metric, the trajectory of an object falling directly ($h=0$) into a planet from infinite, with no initial speed ($E=mc^2$) is, according to this wikipedia page:

$ \tau = \text{constant}\pm\frac{2}{3}\frac{r_{\rm s}}c\left(\frac r{r_{\rm s}}\right)^\frac{3}{2} $

where $r$ is the distance to planet center and $r_s$ Schwarzschild radius.

If we define $R_0=r(\tau=0)$, it can be expressed as:

$ r = \left[ R_0^{\frac{3}{2}}-\frac{3}{2}c\sqrt{r_s}\tau \right]^\frac{2}{3} $

Thus, the four-position in these coordinates is:

$ \left( c\tau, \left[ R_0^{\frac{3}{2}}-\frac{3}{2}c\sqrt{r_s}\tau \right]^\frac{2}{3}, 0 , 0 \right) $

and the four-velocity when $\tau=0$ is:

$ U(\tau=0) = \left( c, -\frac{4}{9}c\sqrt{\frac{r_s}{R_0}}, 0 , 0 \right) $

According to wikipedia page about four-velocity:

The value of the magnitude of an object's four-velocity, i.e. the quantity obtained by applying the metric tensor g to the four-velocity U is always equal to $±c^2$.

But it in this, taken into account the metric, the module results in:

$ c^2\left(1-\frac{R_0}{r_s} - \frac{16}{81}\frac{r_s}{R_0}\frac{1}{1-\frac{R_0}{r_s}} \right) $

different of the expected $c^2$.

Could someone say if I made a conceptual or calculus error ?

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Could someone say if I made a conceptual or calculus error ?

Conceptual. The first component of the four-position should be $c$ times the coordinate time $t$, not $c$ times the proper time $\tau$.

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