[Physics] Does time speed up or slow down near a black hole

black-holesdifferential-geometrygeneral-relativitymetric-tensortime

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?

Best Answer

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).

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