Numerical Relativity by Tomas Baumagrte and Stuart Shapiro page 10.

By adapting $h=c=1$, so in schwarzchild solution the areal radius $r=2M$ is the event horizon, and $r=3M$ is the photon orbit. But what happened between $2M<r<3M$ then? is that the distance where mass must falls in?

Also $E/\mu=\frac{(r-2M)^2}{r(r-3M)}$ where $\mu$ is the rest mass for particle.(Also one conservation for the angular momentum.) How could any matter pass through the photon orbit then?


1 Answer 1


$r=3M$ is where there is an (unstable) lightlike orbit around the black hole. for $2M < r < 3M$, the light ray will spiral into the black hole. In this range, nonaccelerating timelike paths can fall in, but they still can escape (i.e., by "firing their rockets).

This is different than paths that pass into the region $r \leq 2M$, where there are no timelike or null paths that escape back to infinity.

  • $\begingroup$ Thank you, but what happened to the energy conservation then? $r=3M$ both angular momentum and energy itself is infinity. $\endgroup$ Jan 30, 2019 at 18:26
  • $\begingroup$ @user9976437: it's the energy a timelike traveller has to consume accelerating to get back to an escape orbit. $\endgroup$ Jan 30, 2019 at 18:28
  • $\begingroup$ Still for free falling object it can never pass through $r=3M$ right? $\endgroup$ Jan 30, 2019 at 18:57
  • $\begingroup$ @user9976437: correct. see physics.stackexchange.com/questions/52315/… . I have also edited my answer to be more correct after thinking more. $\endgroup$ Jan 30, 2019 at 19:43
  • $\begingroup$ @user9976437 It depends on what you mean by falling. An object (or light) is considered to be in free-fall as long as gravity is the only thing influencing its motion, even if it is moving away from the black hole. Light starting at any $r>2M$ can escape by moving radially away from the black hole, even though it's "falling" in the sense that gravity is the only influence. For example, if an infalling object emits a omnidirectional flash of light just before it crosses the event horizon, some of that light will escape; in particular, the part travelling in the $+r$ direction will escape. $\endgroup$ Jan 31, 2019 at 1:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.