Investigating The Photon Sphere

[.Scott]

TL;DR

Looking for 2 functions of the radius of a Schwarzschild Black Hole: the "vertical" escape velocity and the maximum angle for a photon to escape the photon sphere.

I've spent well over two hours searching the web for two functions of the radius of a Schwarzschild BH. One would give me the escape velocity of the BH assuming a perfectly vertical trajectory (so it isn't a normal escape velocity). The second relates to the trajectory of a photon that is outbound from inside the photon sphere. How far from vertical can its trajectory be and still escape?

For the first one, if I used the regular escape velocity formula - which happens to work at the event horizon, I get 0.8165 c ($(\sqrt{2/3})c$). But I have no faith in that result.

For the second one, clearly at the event horizon, only a vertical trajectory will allow the photon to reach the photon sphere - and even then, with no remaining energy ("fully" red-shifted). So if the radius is measured in Schwarzschild radii, then at the EH $f(1)=0$. At the photon sphere $f(1.5)=\pi/2$. But I would not hazard a guess as to what that function is. ... Or maybe I will: $f(r)=arctan((2r-2)/(3-2r))$ ?? Not likely.

BTW: I hope you don't mind layers because the reason I am trying to get these equations is to prepare another PF post.

 

[George Jones]

Summary:: Looking for 2 functions of the radius of a Schwarzschild Black Hole: the "vertical" escape velocity and the maximum angle for a photon to escape the photon sphere.

Velocity and angle with respect to what?

Consider an observer who uses a rocket to hover at constant $r$, $\theta$, $\phi$. With respect to this hovering observer, the escape velocity is

$$v = \sqrt{\frac{r_S}{r}}$$
and the maximum angle from vertical for escape by light is

$$\sin \theta = \frac{r_S}{2r} \sqrt{27 \left( 1 - \frac{r_S}{r} \right)}.$$
Here, $r_S$ is the Schwarzschild radius, and $r$ is the $r$ coordinate of the hovering observer.

This is in the first edition of "Exploring Black Holes" by Taylor and Wheeler. A lecture based on this book is
https://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect11_2004.pdf

 

[.Scott]

Thanks @George Jones :
I have downloaded that page and will read it in full (it's only 13 ages).

The "vertical escape velocity" I was asking about is the minimum delta V, proper to a hovering vehicle, that would launch the vehicle to infinity. Alternatively, it is the delta V relative to a vehicle that has been dropped towards the BH from infinity that would be required to stop that vehicle.

The equation that you provided is the one that I used to compute my guess at this vertical escape velocity at the photon sphere ($r=1.5\times r_s$). So I'll score myself a lucky guess.

I notice that the paper you cited for the second equation provides a full derivation for that equation. ... and I should be able to follow it

Thanks again.
@.Scott

 

[.Scott]

and the maximum angle from vertical for escape by light is

$$\sin \theta = \frac{r_S}{2r} \sqrt{27 \left( 1 - \frac{r_S}{r} \right)}.$$
Here, $r_S$ is the Schwarzschild radius, and $r$ is the $r$ coordinate of the hovering observer.

This is in the first edition of "Exploring Black Holes" by Taylor and Wheeler. A lecture based on this book is
https://web.stanford.edu/~oas/SI/SRGR/notes/SRGRLect11_2004.pdf

@George Jones

You and (I presume) the book are right.

But from the bottom of page 8 of the lecture notes you provided:
$$\sin \theta_{shell\space critical} = \frac{\sqrt{27}}{2}\frac{R_S}{r} \left( 1 - \frac{R_S}{r} \right)$$.

This is not the same as your equation - and it does not produce correct results. Also, those lecture notes match up 2.16GM/c^2 with a horizon view of 147 degrees. But to get 147, you need to start with about 2.18.

And thanks again - because even using your equation as a starting point in a Google search, I still can't find this online.