# Coordinate Singularity in Metric

Suppose I have some metric

$$ds^2=g(t)dt^2+\frac{1}{r}dr^2$$

which has a singularity at $r=0$.

However, if I make the coordinate transformation $u=\frac{1}{r}$, then I get:

$$ds^2=g(t)dt^2+r^3 du^2$$

which no longer has a singularity at $r=0$.

Does this mean that the singularity at $r=0$ is merely an artifact of the coordinate system choice? Or does one have to calculate the curvature and see if there is a singularity in the curvature?

I find it confusing when people say that the singularity of the Schwarzschild metric at the event horizon is not a singularity, because clearly something is happening at the horizon. When you go to Kruskel coordinates (T,X) the metric is non-singular at the horizon, but so what? If you calculate the $ds^2$ of someone who crosses the horizon then you should get infinity, in either (t,r) or (T,X) coordinates, so that the ''singularity'' at the horizon is correct and there seems to be nothing defective about the metric being infinity in the Schwarzschild coordinates: the Schwarzschild metric is giving the correct physics. With Kruskel coordinates you seem to be shifting the infinity at the horizon into dX and dT which are infinity at the horizon when $dr\neq 0$.

It seems if you have a singularity in some coordinate system, then physically there is some trajectory that will give an infinite proper time, which indicates that the singularity is physical and can be felt by someone on that trajectory.

• The difference is between a 'coordinate' and a 'physical' singularity. This might be helpful. – DilithiumMatrix Dec 11 '15 at 16:47
• The proper time when crossing the horizon is finite; you can check this in Schwarzschild coordinates. – Javier Dec 11 '15 at 17:03
• BTW, in the $(u,t)$ coordinates that you have there is still a coordinate singularity at $r = 0$. Coordinate singularity means both $g$ and $g^{-1}$ are well-defined. – Prahar Jan 21 '16 at 11:06
• The important physical question to ask, I think, is: does anything bad happen to an inertial observer at the horizon? Clearly the answer is no (in pure GR). So any 'singularity' at the horizon must be some artifact of our description, not of the physics. And that turns out to be true: the 'singularity' is because we are using rotten coordinates at the horizon. – tfb May 10 '16 at 12:13

If you calculate the ds2 of someone who crosses the horizon then you should get infinity, in either (t,r) or (T,X) coordinates

In the context of the Schwarzschild solution, with the exception of a single event, there are no Schwarzschild coordinates for the horizon; that is to say, for any finite $t$, the coordinates (t, 2M) map to one event on the horizon.

But any world lines through that event are otherwise within the past and future horizons so Schwarzschild coordinates are inadequate to describe "someone who crosses the horizon".

The Kruskal–Szekeres coordinates are non-singular at the horizon and so can be used there. For purely radial motion and at the horizon,

$$d\tau^2 = 16M^2(dT^2 - dX^2)$$

However, if I make the coordinate transformation u=1/r

which isn't defined for $r=0$.

The Schwarzschild metric is $ds^2=-(1-\frac{2m}{r})dt^2+(1-\frac{2m}{r})^{-1}dr^2$. Presumably if you write $dt=\frac{dt}{dr}dr$, where $\frac{dt}{dr}$ comes from solving the geodesic equation, you get that $ds^2=f(r)dr^2$, where $\int ds=\int^r_{r_0} \sqrt{f(r)}dr$ has no singularities at $r=2m$ and in fact has no singularities down to $r=0$.

Viewed like this, the Schwarzschild coordinate system is good even at $r=2m$, but to get sensible physics you have to add the geodesic equation.

But if I make up a trajectory that doesn't satisfy the equations of motion (nongeodesic), I can get an infinite proper time. For example, $dt=0$, $r_0=2m+2m$, and take that trajectory all the way down from $r_0=2m+2m$ to $r=2m+\epsilon$, where $\epsilon$ is infinitesimal. That trajectory blows up, and it doesn't matter whether you use Kruskal coordinates or not, as $ds^2$ is the same in all coordinate systems. In Kruskal coordinates, the metric is fine, but $dX$ and $dT$, which are linear combinations of $dt$ and $dr$ (and $dt=0$ in my example so actually just a function of $dr$) with coefficients that depend on $(r,t)$: these coefficients blow up as $r \rightarrow 2m$.

So it seems both Kruskal coordinates and Schwarzschild coordinates will blow up at the horizon unless you use the geodesic equation.