What is the difference between proper time and the observer time?

Whilst thinking about Black holes, when we see the Schwarzschild metric

$$c^2\tau ^2 = \left ( 1 - \frac{r_{s}}{r} \right )c^2t^2 - \frac{r^2}{1-\frac{r_{s}}{r}} - r^2d\Omega ^2,$$

and compare it with Einstein's special relativity equation, $$c^2\tau ^2 = c^2t^2 - x^2,$$

we find that at the horizon of a black hole or at the schwarzschild radius for any infinitesimally small time spend by any object at the horizon the observers time tends to infinite

Why and how is this so? it doesnt make sense whilst trying to imagine it?

  • $\begingroup$ Why is it so? Because the Schwarzschild metric says it is so. If you believe in GR, you have to believe in the solutions. The difference between proper time and observer time is that the former is the invariant length of a path in spacetime, and the latter is the proper time of the observer. $\endgroup$ – ACuriousMind Jul 4 '14 at 21:38
  • $\begingroup$ i asked for a explanation in words as in ... why does the observer time tend to infinity .... is it because the light from the object doesn't reach the observer ? but then if so how can the observer see the object atall ? $\endgroup$ – relston mendonsa Jul 4 '14 at 21:41
  • 1
    $\begingroup$ That's the reason why black holes are black - no light emitted from their horizon or further inwards can ever leave the hole in a finite amount of time, and the closer you come to the horizon, the longer it takes for the light to escape. $\endgroup$ – ACuriousMind Jul 4 '14 at 21:50

This is essentially the same effect that you get in special relativity as the velocity approaches the speed of light.

If you take a clock and accelerate it towards the speed of light then it will run slowly. If you could get the clock to the speed of light (which you can't of course) then it would stop completely. To use your words for any infinitesimally small time spend by any object approaching the speed of light the observers time tends to infinity.

Indeed there is a sense in which any object falling into a black hole crosses the horizon at the speed of light. Suppose you are hovering just outside the event horizon at some radial coordinate $r$. We call this type of observer a shell observer. If you now watch an object falling freely from infinity then the speed the object passes you is:

$$ v = - \left( \frac{2M}{r} \right)^{1/2} $$

where I'm using geometric units so $c = 1$ and the event horizon is at $r_s = 2M$. As $r \rightarrow r_s$ the speed the falling object passes you approaches $c$.


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