# Are the space and time axes of Schwarzschild metric uncurved?

Schwarzschild metric is commonly considered as an expression of curved spacetime:

$$\mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 \left(\mathrm d\Theta^2 + \sin^2 \Theta ~\mathrm d\Phi^2\right)$$

However, looking at this equation, I find that the metric on the left side is curved (distorted) but not the coordinate axes dt and dr on the right (which are given in the equation).

dt and dr are the displacement coordinates. The Schwarzschild equation does not calculate them, but they are given. For calculating the warped metric on the left side, they are subject to distortion (multiplication/ division by a factor). But in no way, the space and time coordinate axes are warped. It is only the distance between events which deviates from the distance of Minkowski metric.

It seems to be the same situation as for Minkowski diagrams, where the metric ds is different from the Euclidean metric, but the coordinate axes for space and time (horizontal and vertical) are identic with those already Newton could have used.

## 1 Answer

The coordinates are just labels that we use as labels points in spacetime. They need not have any physical significance and indeed sometimes don't. So when you ask about whether coordinates like $t$ and $r$ are curved this is a somewhat meaningless question.

As it happens the Schwarzschild $t$ and $r$ coordinates do have a physical meaning, though for the $r$ coordinate it is a little subtle. The $t$ coordinate is simply the time as measured by a clock that is stationary at $r=\infty$, but the $r$ coordinate is defined in a less obvious way. It is the circumference of a circle centred on the mass divided by $2\pi$, so it not a radial distance in the sense of a distance measured from the mass.

Incidentally, you should note that polar coordinates are curved even in flat spacetime because the metric is:

$$ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2\sin^2\theta d\phi^2$$

not:

$$ds^2 = -dt^2 + dr^2 + d\theta^2 + d\phi^2$$

This is an example where the spacetime is flat but the coordinates are curved.

The point of all this that it doesn't really make sense to ask whether the coordinates are flat because coordinates don't necessarily have any physical meaning.