# For a giving metric in GR, how do we learn which observer the metric refer to?

For example, I have been told the Schwarzschild observer is far away from blackhole and events,(namely, I think, the observer is static at infinity of the coordinate.)

And the second example,the Gullstrand–Painlevé coordinates, its obeserver is a comoving observer of free-falling massive particle from infinity.

(Maybe that's, in fact,one example,because we could use transformation from Schwarzchild metric to Gullstrand–Painlevé metric to learn how does it go in Gullstrand–Painlevé coordinates. And that might be true, but that is not the answer I wish for.)

I have no idea about that for a giving metric,how to calculate where is its observers?

A metric need not refer to an observer at all.

When we talk about the Schwarzschild observer we mean the observer for whom the variables in the metric have a simple meaning - typically that they have the same meaning as in flat spacetime. So if you compare the Schwarzschild metric:

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

with the flat space metric:

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

then obviously they match when $r_s/r \rightarrow 0$ i.e. at $r \rightarrow \infty$.

The GP metric uses the Schwarzschild $r$, $\theta$ and $\phi$, but the time coordinate is the time measured by an observer falling freely from infinity:

$$ds^2=-\left(1-\frac{2M}{r}\right)dt_r^2+2\sqrt\frac{2M}{r}dt_rdr+dr^2 + d\Omega^2$$

So there is no observer who observes the metric variables directly, though since once again if we take the $r \rightarrow \infty$ limit it becomes the flat space metric you could argue that the GP observer is at infinity.

However take the Kruskal-Szekeres metric, which uses the somewhat abstract coordinates $u$ and $v$. Though $u$ is spacelike and $v$ is timelike there is no observer form whom $du$ matches what they would measure with a ruler or $dv$ what they would measure with a clock.

It's an important principle of GR that we can use any coordinate system, so we're able to choose coordinates that make our calculations simple and we don't need to worry whether they correspond to any physical measurements an observer could make.

• This answer beats around the bush and obfuscates the issues. I don't think the second paragraph is right. We could have many different coordinate systems that would asymptotically match up with flat-space spherical coordinates. IMO this could be improved simply by cutting the whole middle part out and leaving in only the first and last paragraphs. – user4552 May 2 '18 at 5:38

There is no corresponding observer, in general. In the context of accelerating observers or curved spacetime, we should only think about local observers in general, so picture an entire swarm of them filling spacetime. But still, there are no corresponding observers (plural) in general. Coordinates are arbitrary, and need not have "a simple metrical significance" as Einstein realised early on, meaning not (directly) the measurement of observers.

Having said that, we can argue on a case-by-case basis. For example in deriving their "radar metric" to measure spatial distance, Landau & Lifshitz assume observers who are comoving with a given coordinate system, meaning at constant $x^1$, $x^2$, and $x^3$-coordinates, if this is allowed. For your example of Gullstrand-Painleve coordinates, they are indeed nicely suited to observers who freefall "from rest at infinity" so-to-speak, because these observers are comoving in $\theta$ and $\phi$, the time coordinate is their proper time, and the $r$-coordinate is the spatial distance measured in the radial direction. (My last point is the least known, but see Taylor & Wheeler's Exploring Black Holes.)

Given a metric $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, the observer is, by construction, on the origin of the coordinate system $\{x^\mu\}$.

• That would put the Schwarzschild observer at the singularity. Since the singularity isn't part of the manifold that's the one place the observer absolutely can't be. – John Rennie Dec 18 '15 at 9:58
• One thing is the manifold, the metric is some additional structure on it. A singularity just means that the metric isn't defined at that point and it tells nothing about the topological manifold itself. This point isn't a singularity of the manifold. So besides the odd metric's behavior at its singularity, the manifold is completely fine at the metric's singularity and the observer could be there without problems. As always in physics, the observer is located at origin of some coordinate frame. That's all. – Mr. K Dec 18 '15 at 15:37
• No, the singularity is not considered part of the manifold. It isn't just that the metric is undefined there, the manifold is defined to exclude the singularity. This is general practice. For example the Big Bang singularity is also not part of the manifold. – John Rennie Dec 18 '15 at 17:18
• The given answer only applies to very limited cases, such as an inertial observer in Minkowski (flat) spacetime, for which we might typically choose coordinates with the observer is at the origin. But for black holes, one reference supporting John Rennie's excellent description is the Stanford Encyclopedia of Philosophy: plato.stanford.edu/entries/spacetime-singularities/#BouCon – Colin MacLaurin May 2 '18 at 2:05