# Modulus of four-speed in Schwarzschild metric

( 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:

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

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 ?

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