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Given a Schwarzschild radius $r_s=2 G M/c^2$, the escape velocity (equal to speed if falling from infinity) will be $\sqrt{2 G M/r}=\sqrt{r_s c^2/r}$ where the radial distance "r" is the point at which the measured circumference is $2 \pi r$. Lets assume that we set its speed equal to the local escape velocity when throwing our probe into the black hole from a relatively safe distance of $100 r_s$.

As you get closer to the event horizon, space is expanded in the radial direction by $1/\sqrt{1−r_s/r}$ so even as its speed gets closer to $c$ relative to an outside stationary observer, it will slow down asymptotically relative to the Schwarzschild radial distance.

What I would like to know is if there exists an exact formula for r(t) (either for radial distance or total measured distance fallen in frame of observer with same velocity as black hole), or if you have to numerically integrate the formula for speed (divided by space expansion) as a function of radial distance.

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  • $\begingroup$ See en.wikipedia.org/wiki/… , the equations where it says "When $E=mc^2$ and $h=0$..." As you get closer to the event horizon, space is expanded in the radial direction Not true. Gravity in GR is a curvature of spacetime, not a compression or expansion of space. $\endgroup$
    – user4552
    Dec 18 '18 at 22:56
  • $\begingroup$ The quantity I am interested in in this case is the BH rest-frame measurement of radial distance traveled compared to what you would expect given the change in measured circumference. The radial distances balloon to infinity (when you use the rest frame) while the circumference stays at reasonable values so I am making the conceptual simplification of thinking of it as identical to the flat-space value. From this point of view, there appears to be radial space expansion near the event horizon (or contraction from the point of view of a distant observer). $\endgroup$ Dec 18 '18 at 23:33
  • $\begingroup$ The quantity I am interested in in this case is the BH rest-frame measurement of radial distance This is not a definition that makes sense. GR doesn't have global frames of reference, only local ones. Your statements about radial distances are simply wrong. $\endgroup$
    – user4552
    Dec 18 '18 at 23:58
  • $\begingroup$ The "rest frame" I am referring to is the velocity (presumably maintained by continuous firing of rockets) at which an object would be neither moving closer or further away from the black hole and is not moving around it (assuming the BH doesn't have angular momentum so no frame dragging). Do you know a technical term for this which fits better into the usual mathematical formulation of GR? $\endgroup$ Dec 19 '18 at 0:28
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When you solve for a radial timelike geodesic, outside the horizon of a Schwarzschild black hole, for a particle falling in from rest at infinity, you can't get a nice formula for $r(t)$, but you can get one for $t(r)$:

$$t(r)-t_0=\frac{\sqrt{2}}{3}M\left[\sqrt\frac{r_0}{M}\left(6+\frac{r_0}{M}\right)-\sqrt\frac{r}{M}\left(6+\frac{r}{M}\right)\right]-4M\left[\tanh^{-1}\sqrt\frac{2M}{r_0}-\tanh^{-1}\sqrt\frac{2M}{r}\right] \tag{1}$$

in units with $c=G=1$. The integration constants have been chosen to make $r=r_0$ at $t=t_0$. The inverse hyperbolic tangent makes $t\rightarrow\infty$ as $r\rightarrow 2M$.

Addendum: The OP clarified that he is more interested in $r(\tau)$ for radial in-fall from rest at infinity, where $r$ is the radial Schwarzschild coordinate and $\tau$ is the proper time along the geodesic (i.e., the time as measured by the infalling probe). This is given by the simpler formula

$$r(\tau)=\left[\frac{9M(\tau_0-\tau)^2}{2}\right]^{1/3} \tag{2}$$

where $\tau_0$ is the proper time at which the probe reaches the singularity at $r=0$.

This is the result of integrating

$$\frac{d^2r}{d\tau^2}+\frac{M}{r^2}=0 \tag{3}$$

or

$$\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2-\frac{M}{r}=0. \tag{4}$$

Remarkably, (3) and (4) are identical in form to the Newtonian versions, with the Euclidean radial coordinate replaced by the Schwarzschild radial coordinate, and the absolute time replaced by the proper time.

To derive (3), one combines

$$\frac{d^2r}{d\tau^2}+\frac{M}{r^2}\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2-\frac{M}{r^2}\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2=0, \tag{5}$$

which is the $r$-component of the geodesic equation

$$\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=0 \tag{6}$$

in the case of a purely radial geodesic, with the equation

$$1=\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2-\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2, \tag{7}$$

which comes from the expression for the proper time as given by the metric,

$$d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2. \tag{8}$$

Eliminating $dt/d\tau$ between the (5) and (7) gives (3).

With the $G$'s and $c$'s restored, $r(\tau)$ is

$$r(\tau)=\left[\frac{9GM(\tau_0-\tau)^2}{2}\right]^{1/3}. \tag{9}$$

The derivation of (1) is only a bit more complicated. From (4) one has

$$\left(\frac{dr}{d\tau}\right)^2=\frac{2M}{r}. \tag{10}$$

Substituting this into (7) gives

$$\left(\frac{dt}{d\tau}\right)^2=\left(1-\frac{2M}{r}\right)^{-2} \tag{11}$$

Thus

$$\frac{dr}{dt}=\frac{\frac{dr}{d\tau}}{\frac{dt}{d\tau}}=-\left(\frac{2M}{r}\right)^{1/2}\left(1-\frac{2M}{r}\right). \tag{12}$$

Integrating this gives

$$t=-\int\left(\frac{2M}{r}\right)^{-1/2}\left(1-\frac{2M}{r}\right)^{-1}dr \tag{13}$$

or

$$\frac{t}{2M}=\int u^{-5/2}(1-u)^{-1}\;du \tag{14}$$

where $u=2M/r$. This integral gives

$$\frac{t}{2M}=-2u^{-1/2}-\frac{2}{3}u^{-3/2}+\log(1+u^{1/2})-\log(1-u^{1/2})+C \tag{15}$$

or

$$\frac{t}{2M}=-2\left(\frac{r}{2M}\right)^{1/2}-\frac{2}{3}\left(\frac{r}{2M}\right)^{3/2}+\log\frac{(\frac{r}{2M})^{1/2}+1}{(\frac{r}{2M})^{1/2}-1}+C. \tag{16}$$

This is just (1) in a different form, if one uses the identity

$$\tanh^{-1}z=\frac{1}{2}\log\frac{1+z}{1-z}. \tag{17}$$

In dimensionless coordinates $r^\prime=r/2M$ and $t^\prime=t/2M$ scaled by the Schwarzschild radius $2M$, we have, dropping the primes,

$$t=-2r^{1/2}-\frac{2}{3}r^{3/2}+\log\frac{r^{1/2}+1}{r^{1/2}-1}+C. \tag{18}$$

Unfortunately, this nice expression for $t(r)$ can't be inverted to get a nice expression for $r(t)$.

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  • $\begingroup$ How do you add the constants c and G back in so you can make a calculation? $\endgroup$ Dec 18 '18 at 23:44
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    $\begingroup$ Replace the $M$'s in front of the brackets with $GM/c^3$, because they must be a time. Replace the $M$'s inside the brackets with $GM/c^2$, because they must be a distance. $\endgroup$
    – G. Smith
    Dec 18 '18 at 23:55
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    $\begingroup$ No, that $t$ is the $t$ in the Schwarzschild metric, so it is time measured by an observer at infinity. I thought that’s what you wanted because you used the symbol $t$. There is a simpler formula for $r$ in terms of the proper time $\tau$ measured by the infalling probe, which I will post later. $\endgroup$
    – G. Smith
    Dec 19 '18 at 2:42
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    $\begingroup$ I've added an addendum to my answer, with the other formula that you want, and a derivation of both formulas. $\endgroup$
    – G. Smith
    Dec 19 '18 at 8:19
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    $\begingroup$ Yes, but that’s a question about coordinates, not units. I see you restored the $G$ and $c$ on the right side of the equation. The left side would be $dr/cdt$. $\endgroup$
    – G. Smith
    Dec 19 '18 at 18:18

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