The Schwarzchild geometry is defined as

$$ds^2=-\left(1-\frac{2GM}{r} \right)dt^2+\left(1-\frac{2GM}{r} \right)^{-1}dr^2+r^2(d\theta^2+\sin^2(\theta) d\phi^2)$$

Lets examine what happens close to and far away from a black hole.

For a stationary observer at $r=\infty$, we get

$$d\tau^2=-ds^2=\left(1-\frac{2GM}{\infty} \right)dt^2=dt^2 $$

so the time measured is the proper time. For an observer orbiting a black hole (assume circular where $\theta=\pi/2$) a distance $r=r_0$ away from the black hole, we get

$$d\tau^2=\left(1-\frac{2GM}{r_0} \right)dt^2-{r_0}^2d\phi^2$$

For a circular orbit, it can be shown that $r_0^2 d\phi^2=\frac{GM}{r_0}dt^2$ and hence

$$d\tau^2=\left(1-\frac{3GM}{r_0} \right)dt^2$$

Thus $d\tau^2$ (the time measured by an observer infinitely far away from a black hole) is less than $dt^2$ (the time measured by an observer orbiting a black hole), which appears to suggest that time moves faster close to black holes.

Would someone be able to point out the flaw in my logic here?


3 Answers 3


To expand on Javier's answer, the symbol $\tau$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $t$, which is the coordinate time. In this case, we've defined the coordinates so that $t$ is the time as measured by the observer at infinity.

To avoid confusion, let's use $\tau_\infty$ for proper time in the frame of the stationary observer and $\tau_{orbit}$ for proper time in the frame of the orbiting observer. Then as you correctly showed,

\begin{align} d\tau_\infty^2 &= dt^2 \\ d\tau_{orbit}^2 &= \left( 1-\frac{3GM}{r_0} \right) dt^2 \end{align}

It follows that

$$d\tau_{orbit}^2 = \left( 1-\frac{3GM}{r_0} \right) d\tau_\infty^2$$

which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).

  • 1
    $\begingroup$ Thank you for this answer, I think I understand what’s going on now. One little last point to make: in special relativity, $dt$ is not constant between reference frames, right? This appears to make time different in special relativity than in general relativity. For example, time between a stationary and moving reference frame in SR is $-dt’^2 = -dt^2 +dx^2$. It seems that in the GR metric, however, $dt$ is chosen to be constant in all reference frames. $\endgroup$ Dec 17, 2018 at 1:49
  • $\begingroup$ So it’s different than special relativity is what you’re implying? I think my major problem was that I was trying to draw similarities between SR and GR regarding time between reference frames. $\endgroup$ Dec 17, 2018 at 1:58
  • 2
    $\begingroup$ @LukePolson In GR we tend to speak of coordinate systems instead of reference frames. This is because the connection between observers and coordinates is more complicated. Here we have a single coordinate system, Schwarszschild coordinates, and two (or many) observers. We don't need to use an observer's reference frame to describe its motion, we can do it from any system we want. (...) $\endgroup$
    – Javier
    Dec 17, 2018 at 2:33
  • 2
    $\begingroup$ (...) In the Minkowski metric $t$ is coordinate time, but we don't allow just any time to be used as a coordinate: only proper time as measured by some observer. $t$ will then be proper time as measured by an observer at rest, but it won't be proper time as measured by a moving observer. $\endgroup$
    – Javier
    Dec 17, 2018 at 2:34
  • 2
    $\begingroup$ @LukePolson yes, of course, path length is different for different trajectories, just like it is in regular space. Don't confuse this with the statement that a given (single) path length is independent of the coordinates, which is what is usually meant by $ds^2 = ds'^2$. $\endgroup$
    – Javier
    Dec 17, 2018 at 11:54

You are mixing the two times up. Proper time $\tau$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $t$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.

  • $\begingroup$ If proper time is always measured by the observer you’re considering, how does $dS^2 =-d\tau^2$ work? If the observer is only a finite distance away, then your answer makes it seem like there are two different types of proper time. $\endgroup$ Dec 17, 2018 at 1:20
  • $\begingroup$ @LukePolson Well, there are infinitely many proper times, one for each possible world line (i.e. observer). Usually if it's not clear from context we specify which one we're talking about. Not sure if this is what you are asking about, though. $\endgroup$
    – Javier
    Dec 17, 2018 at 1:26
  • $\begingroup$ @Luke Polson this answer is correct $d\tau$ is the proper time for the observer at $r_0$ when you are using $r_0$. Just like it was proper time for the observer at infinity when you used infinity $\endgroup$
    – Dale
    Dec 17, 2018 at 1:42

As far as I can see, in all this discussion a point is lacking (unless I missed it). Nobody has clearly stated that in order to compare times an operational way of doing the comparison is required. Usually, when we're comparing times of clocks occupying different space locations, this is accomplished via light signals.

Only when the comparison procedure is specified it becomes meaningful to say "time ... is less than time ...". Or else "The quantity that does not change between frames is $t$". In latter case the obvious question is: "how do you know of coordinate time?" (Not on paper, but in the lab.)

In some cases there are obvious procedures which experts tend to let understood, but this should be avoided when less experts are involved. It's well known that here the most frequent cause of errors is lurking.

In present problem OP did simply interchange the meanings of $t$ and $\tau$. His final formula is right but its interpretation is upside down.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.