4 ms·

Time from that matter's perspective. From the outside observer's perspective, it zips right in. Same way that if you travel near the speed of light your percep

by skj 12y ago

Time from that matter's perspective. From the outside observer's perspective, it zips right in. Same way that if you travel near the speed of light your perception of time slows until you are no longer traveling that fast. But to an outside observer you are zipping right along.

guard-of-terra 12y ago
Nope, the other way around. From matter's perspective, its time ticks as ever. But for outside observer, it just halted there.
- espadrine 12y ago
  At some point, the matter goes past the event horizon. Then the notion of outside observer stops having any meaning. Before that, known clocks in that matter slowed down for an outsider, and the outsider's clock went faster from that matter's perspective.
- 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 ago
    If 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.
- sebular 12y ago
  Nope, skj had it right and you've got it backwards. Otherwise we wouldn't perceive black holes to be black holes... If we observed matter falling into a black hole to be "halting there", well, we'd see it halted there on the spherical surface of the event horizon instead of what we actually see, which is that matter and light get completely sucked in.
  - guard-of-terra 12y ago
    "instead of what we actually see, which is that matter and light get completely sucked in" We actually never see that "Due to this effect, known as gravitational time dilation, an object falling into a black hole appears to slow down as it approaches the event horizon, taking an infinite time to reach it. At the same time, all processes on this object slow down, for a fixed outside observer, causing emitted light to appear redder and dimmer, an effect known as gravitational redshift. Eventually, at a point just before it reaches the event horizon, the falling object becomes so dim that it can no longer be seen." http://en.wikipedia.org/wiki/Black_hole http://en.wikipedia.org/wiki/Black_hole
    - skj 12y ago
      This is completely consistent with my statement :) Our time perception of the object remains the same, but the processes experienced by the object itself are much slower, relative to how we perceive them. Specifically, if the object falling in is emitting blue light, and let's say that the frequency of blue light is one per second for the sake of doing some arithmetic, then that object keeps emitting 1/second and looks blue to itself. But in our frame of reference, those 1/second actually become .1/second (its time is moving slower then our time), and appears red.

by skj 12y ago

guard-of-terra 12y ago

Nope, the other way around. From matter's perspective, its time ticks as ever. But for outside observer, it just halted there.

espadrine 12y ago
At some point, the matter goes past the event horizon. Then the notion of outside observer stops having any meaning. Before that, known clocks in that matter slowed down for an outsider, and the outsider's clock went faster from that matter's perspective.
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 ago
  If 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.
sebular 12y ago
Nope, skj had it right and you've got it backwards. Otherwise we wouldn't perceive black holes to be black holes... If we observed matter falling into a black hole to be "halting there", well, we'd see it halted there on the spherical surface of the event horizon instead of what we actually see, which is that matter and light get completely sucked in.
- guard-of-terra 12y ago
  "instead of what we actually see, which is that matter and light get completely sucked in" We actually never see that "Due to this effect, known as gravitational time dilation, an object falling into a black hole appears to slow down as it approaches the event horizon, taking an infinite time to reach it. At the same time, all processes on this object slow down, for a fixed outside observer, causing emitted light to appear redder and dimmer, an effect known as gravitational redshift. Eventually, at a point just before it reaches the event horizon, the falling object becomes so dim that it can no longer be seen." http://en.wikipedia.org/wiki/Black_hole http://en.wikipedia.org/wiki/Black_hole
  - skj 12y ago
    This is completely consistent with my statement :) Our time perception of the object remains the same, but the processes experienced by the object itself are much slower, relative to how we perceive them. Specifically, if the object falling in is emitting blue light, and let's say that the frequency of blue light is one per second for the sake of doing some arithmetic, then that object keeps emitting 1/second and looks blue to itself. But in our frame of reference, those 1/second actually become .1/second (its time is moving slower then our time), and appears red.