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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?

Qmechanic
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relston mendonsa
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1 Answers1

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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 $c$.

John Rennie
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