From Wikipedia I read that the formula for calculating the time dilation caused by a mass is $t_0 = t_f{\sqrt{1-\frac{2GM}{rc^2}}}$


  • $t_0$ is the proper time between events A and B for a slow-ticking observer within the gravitational field
  • $t_f$ is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object
  • $G$ is the gravitational constant
  • $M$ is the mass of the object creating the gravitational field
  • $r$ is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate),
  • $c$ is the speed of light

The question is, what if $\frac{2GM}{rc^2}$ is greater than 1? This can happen if the mass is too big or if the radius is too small.

  • 3
    $\begingroup$ If the term equals one, you get a blackhole. $\endgroup$
    – jinawee
    Commented Dec 16, 2013 at 18:39

2 Answers 2


The value $r = \frac{2GM}{c^{2}}$ defines a special surface in the Schwarzschild spacetime called the event horizon. Observers inside this radius cannot be stationary with respect to points very distant from the black hole, and they cannot communicate with any observer outside this radius, so the notion of time dilation doesn't make sense for them.

  • $\begingroup$ Wonderful!!! :) $\endgroup$
    – Massimo
    Commented Dec 16, 2013 at 18:44

The problem is that when you write $r$ and $t$ in the equation for the time dilation you are using the Schwarzschild radial and time coordinates, which are part of a system of coordinates that works well far away from the black hole but fails at and within the event horizon.

To make things a bit clearer we'll rewrite your equation as:

$$ t_0 = t_f{\sqrt{1-\frac{r_s}{r}}} $$

where $r_s$ is the radius (in the Schwarzschild coordinates) of the event horizon. If you calculate the ratio $t_0/t_f$ as you move from far away towards the event horizon you'll find that when $r = r_s$ the time dilation becomes infinite at the event horizon. In other works it is impossible to cross or even reach the event horizon because it would take infinite time.

So, as long as you stick to the Schwarzschild coordinates your question has no answer. This probably seems very strange, but it's quite common in general relativity that a system of coordinates does not cover all of spacetime but rather just a patch (not a great link but I couldn't find a better one) of it.

If you wish to explore the physics at or within the event horizon you need to use a different coordinate system, and the one usually used to describe the Schwarzschild metric is the Kruskal-Szekeres coordinate system. The only problem is that this is unintuitive since the coordinates don't directly relate to anything an observer might experience.

If you search this site for Kruskal-Szekeres you'll find lots of informative questions and answers, of which of course my favourite is my own. There are also lots of questions relating to the infinite time taken to reach the event horizon.

  • $\begingroup$ Since your definition of 'patch' does not require connectedness, the Schwarzschild coordinates are defined on a disconnected subset of the Schwarzschild geometry, and there's no problem in using them either outside or inside the black hole. Rather, it's the disconnect at the horizon that's the problem in any situation which requires you to cross it. But it's probably more convenient to require connectedness, so there'd really be two distinct 'Schwarzschild coordinate patches' with no transition functions between them. $\endgroup$
    – Stan Liou
    Commented Dec 16, 2013 at 19:40

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