Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

According to the General Theory of Relativity, the coordinate time distance per spacetime distance traveled by a particle freely falling into a black hole gets closer and closer to $0$ as the particle approaches the event horizon

$$\mathrm{d}s^2 ~=~ -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left( 1-\frac{2M}{r} \right)^{-1}\mathrm{d}r^2 + \dots.$$

Equivalently, an observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily long time. However, Hawking radiation implies that black holes don't last forever, but instead shrink and fade away? What happens to everything that is "currently" falling into the black hole during that time?

share|improve this question
Related: physics.stackexchange.com/q/95366/2451 and links therein. –  Qmechanic Mar 20 '14 at 0:32
I am unsure how this question was closed as a duplicate of a completely unrelated question. I am voting to re-open. –  Kyle Kanos Mar 21 '14 at 0:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.