4 ms·
No - think about the experience from the POV of a photon. To a photon, it arrives in the same frame of time as it left, and it doesn't experience time at all.
by timcederman 12y ago
No - think about the experience from the POV of a photon. To a photon, it arrives in the same frame of time as it left, and it doesn't experience time at all. Meanwhile, the an outside observer, they see light move from source to target.
- baghira 12y agoIf I'm correct you are suggesting that the external observer "sees" the object falling inside whilst the object itself perceives time as frozen. This is wrong, as far as I know. The proper time in the free-falling frame of reference ticks in the usual manner, and it is quite easy to compute the time it takes the object to fall inside a black hole (the textbook computation being usually done for an object which has zero velocity at infinity). The external observer on the other hand will see the falling object to slow down and the wavelength of the object's emitted photons to be redshifted by an infinite factor as it approaches the event horizon, so the object will disappear before crossing it. This is true both for massive and massless particles falling in. Also, to nitpick: thinking about massless particles i.e. objects moving at c has having frozen time is often misleading: the "usual" definition of proper time, i.e. the square root of time squared minus space squared is not a good parametrization for light-like geodesics.