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I was reading this question:

Are gravitational time dilation and the time dilation in special relativity independent?

And JohnRennie's answer: (sorry for the syntax in formulas, I don't know how to make them look good)

Now consider general relativity, and the effect of gravity. But first let me rewrite the special relativity equation for the line element in polar co-ordinates:

$$\mathrm ds^2 = -\mathrm dt^2 +\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2\theta~\mathrm d\phi^2) $$

and now I'll write the equation for the line element near a black hole, i.e. the Schwarzschild metric:

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

Question:

So what happens when the mass M and the distance are set so that 2M/r=1? If you are at r=2M distance from a M mass black hole then (1-2M/r)=0 and what will that mean?

This will be zero −(1−2Mr)dt2.

And this will be infinite dr2(1−2Mr).

What does that mean?

Community
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Árpád Szendrei
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2 Answers2

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What does that mean?

It means that for $r \gt 2M$, an infinitesimal displacement $\mathrm{d}r$ is space-like while for $r \lt 2M$, an infinitesimal displacement $\mathrm{d}r$ is time-like. The horizon is the boundary.

Put another way, inside the horizon, moving forward in time means decreasing in $r$, i.e., for the same reason that we move forward in time, one must move toward $r = 0$ once inside the horizon.

Alfred Centauri
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So what happens when the mass M and the distance are set so that $2M/r=1$? If you are at $r=2M$ distance from a $M$ mass black hole then $(1-2M/r)=0$ and what will that mean? This will be zero $−(1−2Mr)dt^2$.

The correspondence between the Schwarzchild metric and the flat scale metric implies that the the time coordinate used here $t$ is the same as the time coordinate used by a distant. observer.

And this will be infinite $dr^2(1−2M/r)$.

When $r$ equals $2m$, you get a singularity, the value of $r$ that causes this is the Schwarzchild radius, in the case of the Sun, this is deep inside the body of the Sun, about 3 Km from the core.

What you must do is ensure that this is not a singularity caused by your choice of coordinates, remember you can use any coordinate, but it would make sense to use the most convenient one.

The method used to check for coordinate singularites is is to calculate, $$R_{abcd}R^{abcd} = \frac {48m^2}{r^6} $$

At $r =0$, this is still a singularity, a scalar that is the same in all systems, so it's a true singularity.

What do you mean by $r=2M$ is a coordinate singularity (metric becomes singular) but not a true singularity (no invariants blow up)" What does that mean? what is a true singularity then?

From ACuriousMind's comment below:

What you get at $r=2M$ is a coordinate singularity (metric becomes singular) but not a true singularity (no invariants blow up). There are other coordinates in which the metric remains finite, the event horizon is not a true singularity, although it is a region of some significance. The only true singularity in the Schwarzschild case is located at the center of the black hole $r = 2M$ is precisely the location of the event horizon of the black hole.