Velocity of object falling into black hole measured in Schwarzschild co-ordinates, depending on co-ordinate time

Question

This is a followup question to the answer given to a similar question where the velocity is given as

$$v(r) = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1}$$

where $r$ is the radius from the Schwarzschild co-ordinates and $r_s$ is the Schwarzschild radius.

What would be the velocity depending on $t$, the co-ordinate time, i.e.

$$ \frac{dr(t)}{dt} = \;?$$

Looking at the Wikipedia link given in the referenced answer, section "Local and delayed velocities" I would guess it is

\begin{align} \frac{dr(t)}{dt} &= \frac{dr}{d\tau}\frac{d\tau}{dt}\\ &= v_{\parallel}\gamma\sqrt{1-r_s/r}\cdot\frac{1}{\gamma} \sqrt{1-r_s/r}\\ &= v_{\parallel}(1-r_s/r) \end{align}

but I can't find what $v_{\parallel}$ is.

Answer 1

What would be the velocity depending on $t$, the co-ordinate time[?]

A free fall from infinity takes an infinite time. To define the velocity dependence on time, you can drop an object from a finite distance $R$ and start your clock $t$ at that moment.

A free fall to a Schwarzschild black hole from rest at a distance $R$ is given by the radial geodesic in geometrized units ($c=1$, $G=1$):

$$ t(r)=\sqrt{\dfrac{R}{2M}-1}\cdot\left(\left(\dfrac{R}{2}+2M\right)\cdot\arccos\left(\dfrac{2r}{R}-1\right)+\dfrac{R}{2}\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)+ $$

$$ +\, 2M\ln\left(\left|\dfrac{\sqrt{\dfrac{R}{2M}-1}+\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}{\sqrt{\dfrac{R}{2M}-1}-\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}\right|\right) $$

You can figure the coordinate velocity outside the horizon by differentiating this equation by $r$:

$$ v=-\dfrac{dr}{dt}=-\dfrac{1}{\dfrac{dt}{dr}} $$

Then for any $r$ you would know both the velocity $v$ and coordinate time $t$ to define the (non-explicit) dependence of velocity on time.

EDIT:

The following chart shows the speed $V$ of a free falling object as measured by hypothetical stationary observers hovering at the radius $r$ (where $\dfrac{dr}{dt}$ is from above and the expression in parenthesis is due to the hovering coordinates):

$$ V=-\dfrac{dr}{dt}\,(1-\dfrac{2M}{r})^{-1} $$

You can see that the speed $V$ of any free falling object always approaches the speed of light ($c=1$) at the horizon $r=2M$ regardless of the initial speed or at what distance the fall starts:

Answer 2

The Schwarzschild metric in Schwarzschild coordinates $(t, r, \theta, \phi)$ shows
$ds^2 = -(1 - 2M/r) dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2)$
where:
$c = G = 1$ natural units
$M$ black hole mass
$r_s = 2M$ Schwarzschild radius (event horizon)

If we drop an object at rest from infinity, we have a radial free fall $(d\theta = d\phi = 0)$ where
$dt = (1 - 2M/r)^{-1/2} d\tau_{stat}$ time dilation measured at infinity (far away from the horizon) vs. the proper time of a stationary observer at Schwarzschild radial coordinate $r = constant$
$dt = \gamma (1 - 2M/r)^{-1/2} d\tau_{free fall} = (1 - 2M/r)^{-1} d\tau_{free fall}$ time dilation measured at infinity (far away from the horizon) vs. the proper time of a radially free falling object
$\gamma = (1 - v^2)^{-1/2} = (1 - 2M/r)^{-1/2}$ Lorentz factor of a radially free falling object vs. a stationary observer at Schwarzschild radial coordinate $r = constant$
$v_{stat} = (2M/r)^{1/2}$ velocity relative to a stationary observer at Schwarzschild radial coordinate $r = constant$
$v_{Schw} = (1 - 2M/r) (2M/r)^{1/2}$ velocity relative to Schwarzschild coordinates

Coming to your first question, $dr/dt$ is $v_{Schw}$, i.e. the velocity measured in Schwarzschild coordinates. In the post it is equation (1).
As for your second question, $v_{\parallel}$ is $v_{stat}$, i.e. the velocity relative to a stationary observer.