How could I calculate the time it will take for light and mass to go towards a black hole and come back, to and from constant radial distances?

Question

If you have a "perfect mirror" and a "perfect trampoline" at some constant distance outside a black hole's event horizon:

a) How would a shell observer at some distance farther from the black hole calculate the time it takes for a dropped object to fall, bounce on the trampoline, and come back?

b) Same question as a) but for a light beam to reflect off the mirror and comeback.

I am not concerned with the proper time of the falling object, only the time measured by the shell observer. Bonus points if we can get the equation for the cases when the observer throws the object with different initial velocities. I am only looking for the cases of radial motion.

Answer 1

Since everything is time reversible in your scenario with a stationary mirror or trampoline where we assume ideal conditions, the way down takes the same time as the way up. To keep the equations simple this answer uses natural units of $\text{G=M=c=1}$ and assumes purely radial trajectories.

---

For light set $\rm ds=0$ and solve the line element, then you get

$$\rm \frac{dr}{dt}=\pm \frac{r-2}{r}$$

with the minus for ingoing and the plus for outgoing light rays. If you integrate that from $\rm r_1$ to $\rm r_2$ you get the coordinate time

$$\rm t=r_2-r_1-2 \ ln(r_1-2)+2 \ln(r_2-2)$$

To get from the coordinate time $\rm t$ to the shell time $\rm T$ at $\rm r_1$ multiply $\rm t$ with the $\rm \sqrt{g_{tt}}$ at $\rm r_1$.

---

If it isn't light but an object dropped from rest or with initial local velocity $\rm v$ you have to set $\rm ds=1$ and integrate twice:

$$\rm \frac{d^2 r}{d \tau^2}=-\frac{1}{r^2}$$

with the initial condition

$$\rm \frac{dr}{d\tau}= v_0 \ E , \ E=\sqrt{\frac{1-2/r}{1-v_0^2}}$$

which gives

$$\rm \frac{dr}{d\tau}\pm\sqrt{\frac{2 \ (r_2-r_1)}{r_1 \ r_2}+\left(v_0 \ E\right)^2}$$

and the transformation from proper time $\rm \tau$ to coordinate time $\rm t$

$$\rm \frac{d\tau}{dt}=\sqrt{(1-2/r)(1-v^2)}$$

---

If the objects travel with the escape velocity ($\rm E=1, \ v=\pm \sqrt{2/r}$) we have

$$\rm \frac{dr}{dt}=\pm \frac{\sqrt{2} \ (r-2)}{r^{3/2}}$$

and the coordinate time to get from $\rm r_1$ to $\rm r_2$ or vice versa is

$$\rm t=4 \ arctanh \left(\sqrt{\frac{2}{r_1}}\right)-4 \ arctanh \left(\sqrt{\frac{2}{r_2}}\right)-\sqrt{\frac{2}{9}}\left(6 \ \sqrt{r_1}+r_1^{3/2}-\sqrt{r_2} \ (6+r_2)\right)$$

Those are the worldlines of objects dropped from infinity (the ingoing ones have the negative escape velocity, and the outgoing ones the positive):

---

With initial velocity $\rm v_0=0$ the coordinate time $\rm t$ to fall from $\rm r_2$ to $\rm r_1$ is

$$\rm t=\frac{r_1 \ r_2 \ \sqrt{1-2/r_2} \ \sqrt{2/r_1-2/r_2}}{2}+$$

$$\rm +2 \ \sqrt{\frac{r_2}{2}-1} \ \left(\frac{r_2}{2}+2\right) \ arctan\left(\sqrt{\frac{r_2}{r_1}-1}\right)+$$

$$\rm +4 \ arctanh \left(\sqrt{\frac{2/r_1-2/r_2}{1-2/r_2}}\right)$$

The solution for the proper time $\tau$ is simpler:

$$\rm \tau=\frac{r_2^{3/2} \ arctan \left(\sqrt{\frac{r_2}{r_1}-1}\right)}{\sqrt{2}}+\frac{r_1 \ r_2 \ \sqrt{2/r_1-2/r_2}}{2}$$

For the detailed equations for all the other different kind of scenarios for free falling or accelerated objects see here under "equations" or here under "free fall time".

Answer 2

This is actually a rather simple calculation: you calculate the geodesic on which your particle will travel using the standard geodesic equation and your black hole metric of choice, and then choose some value $s_0$ for the curve parameter along the geodesic to place your reflector. You can then use the standard method of calculating arc length between the initial point and $s_0$. Assuming the reflector is truly perfect, meaning the particle goes back the exact path it came from, the total distance would simply be twice the distance you calculated.

EDIT: I seem to have overlooked the fact that you asked for the time it will take the particle. However, this is another standard calculation which you should be able to perform.