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In special relativity, we make a big fuss about setting up inertial frames of reference, and then constructing coordinate systems using networks of clocks and rulers. This gives an unambiguous physical definition of the spacetime point $(x, t)$. In general relativity, you can take whatever coordinate system you want, but then I don't know what the coordinates mean.

For example, the event horizon of a Schwarzschild black hole is $r = 2GM$. Naively interpreting $r$ as a radial coordinate, this suggests that the event horizon is "$2GM$ away from the center of the black hole", but that statement doesn't make sense either mathematically (the distance $\int ds$ is not $2GM$ at all) or physically (you can't extend your ruler network inside the black hole). But the books I've seen seem to treat $r$ just like the radial coordinate, and talk about "the radius of a stable circular orbit" or stuff like that.

In the rest of physics, we relentlessly focus on how mathematical quantities can be measured, but I don't know how that works here, for the Schwarzschild coordinate $r$.

  • Can the statement "the event horizon is at $r = 2GM$" be phrased in a coordinate-independent way?
  • How can the coordinate $r$ be measured?
knzhou
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    What you can measure is the space time interval given by the formula for ds^2. This defines time intervals or spatial distances. The coordinates have no meaning, they are analogous to points on a map. E.g. on a map of the US, you can put arbitrary dots and give them labels to distinguish them. These labels don't (necessarily) have any meaning. What matters is the distance between the points. – Count Iblis Mar 12 '16 at 22:22
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    Sure, using this method you can measure differences in $r$ values. But then how do we measure absolute $r$ values? We can't just go to the origin and start measuring out from there because, well, it's inside the black hole. – knzhou Mar 12 '16 at 22:24
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    You can use the properties of the Schwarzschild metric to deduce the value of r using only local measurement up to a scaling factor. – Count Iblis Mar 12 '16 at 22:30
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    Oh, I see. Like, you can compute the curvature in your local area. I'm still not totally satisfied because this doesn't tell me how $r$ gets interpreted as a radial coordinate. – knzhou Mar 12 '16 at 22:31
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    If you look at the formula for ds^2, you see that dr^2 appears with a factor which depends on r, so r cannot be taken as the radial distance. If you integrate the square root of that term from r1 to r2, you obtain the actual radial distance from r1 to r2. – Count Iblis Mar 12 '16 at 22:38
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    There is no such thing as "absolute distance" in a geometric sense. Just as in special relativity the geometric distance between two points is observer (movement) dependent. You can pick an arbitrary observer (like a person dangling from infinity on a string) and ask how a (radial) distance might occur to them, but the answer will not be the same as that for someone who is in free fall or on in a (meta-) stable orbit. – CuriousOne Mar 12 '16 at 22:40
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    I dont think there is any "absolute" value in a coordinate (not to be confused with "absolute value"). As CuriosOne mentioned though, we can asses a distance between points according to the perspective of a particular observer. – IntuitivePhysics Mar 12 '16 at 22:59

4 Answers4

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Coordinates can be measured in GR, though all too often this fact is overlooked or even contradicted by people getting caught up in coordinate invariance.

As you well note, in Schwarzschild $r$ isn't really a radius in the "integrate at constant angle from the center and recover this value" sense. It is, however, radial in the sense of being orthogonal to the angular coordinates, Moreover, it matches Euclidean intuition with regards to circumferences and areas at fixed $r$.

How can the coordinate $r$ be measured?

One measurement procedure you can adopt is this: Sit in your rocket with a fixed amount of thrust pushing directly away from the black hole, so that you are hovering at constant $r$. Get all your friends to do the same around the black hole, everyone experiencing the same acceleration. Everyone can then lay down rulers in a circle passing through all the rockets, and the sum of the readings (assuming you've adjusted positions so as to maximize this value) is in fact $2\pi r$.

Can the statement "the event horizon is at $r =2GM$" be phrased in a coordinate-independent way?

Sort of, though perhaps not in as direct a way as you would want. Certainly the event horizon is simply the surface delineating what events can influence future null infinity -- no coordinates involved.

Using the discussion above, though, we could say that for any $r > 2GM$ that the surface of constant $r$ is the locus of points such that rockets with a prescribed radial acceleration hold stationary there, with the event horizon being the limit of such surfaces.

In general, what I'm pushing is the idea that coordinates can be measured as long as you can come up with some experiment where they appear in the formula. This is slightly broader than the notion of measurement of "integrate $\sqrt{g_{\mu\mu}}$ along a line where all coordinates except $x^\mu$ are constant" that suffices for simple spaces.

  • Of course your friends can't thrust hard enough to stay at constant $r$ if $r=2MG$. So they need to thrust to stay a larger $r$ then synchronize their clocks, then stop thrusting at an agreed upon time, then measure how far apart their friends are at an agreed upon time then compare he percentage wise change of that compared to the circumference they measured before and then assume their friends got the same percentage wise adjust result and then associate that global length of a spacelike curve with their local measurements. Not so straightforward. – Timaeus Mar 12 '16 at 23:12
  • And there isn't a local way to measure a classical event horizon. – Timaeus Mar 12 '16 at 23:13
  • That's how ONE of an infinity of special observers will see the world. It's as meaningless as declaring an absolute coordinate system in Newtonian mechanics by using Newton's grave as the reference point for coordinate zero. – CuriousOne Mar 13 '16 at 00:45
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    @CuriousOne It's no more or less meaningful than describing how to measure one's latitude and longitude. Yes, one can sail around the world without lat/long, but that doesn't make them worthless or ill-defined concepts. –  Mar 13 '16 at 01:10
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    @Timaeus Nor is there a local way to measure the size of the Earth in Newtonian mechanics. At some level you either need to travel to elsewhere in space(time), or have a model for how to interpret information coming from elsewhere (e.g. the light of distant stars). This has been true in physics since long before Newton. –  Mar 13 '16 at 01:14
  • Latitude and longitude coordinates are not physically relevant coordinates but they are simply affine point markers on a manifold. What you are pretending here is that the physics of a point doesn't depend on the motion status of the observer. Of course it does. It does so already in Newtonian mechanics. The observer resting on the surface of Earth feels a force, the falling observer doesn't. Not the same physics by any measure. The free falling observer is in an inertial system, the resting observer is not. We teach this in high school... – CuriousOne Mar 13 '16 at 01:17
  • @ChrisWhite When you are outside you could wait to hear everyone's results and add them up, but totaling up results over a small time interval (before you die) is a different issue. And I think it's important to describe local frames and local measurements and note that they are limited in time as well as space. The physics is all about the predictions and the observations. An inference -- like that other observers we aren't getting data from right now see similar results is important too, but it is distinct. It was a bit off putting that your first response to measuring r failed for $r=2MG$ – Timaeus Mar 13 '16 at 01:23
  • This is a fantastic answer. – Zo the Relativist Mar 15 '16 at 23:02
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Firstly, a coordinate chart doesn't have to cover the entire spacetime and the Schwarzschild coordinate system fails to cover as much spacetime as other coordinate charts cover.

In particular, the event horizon is not part of the spacetime covered by the Schwarzschild coordinate chart.

But the books I've seen seem to treat $r$ just like the radial coordinate, and talk about "the radius of a stable circular orbit" or stuff like that.

You can get better books. The $r$ coordinate of the Schwarzschild chart is an areal coordinate, not a radial distance. And that's a pretty silly idea anyway. When a shell of matter falls towards a star/planet the distance between the shell and the distant stars increases by more than the distance between the shell and the star/planet decreases. That's life. Don't define yourself by how far you are from something, it'll bite you.

In the rest of physics, we relentlessly focus on how mathematical quantities can be measured, but I don't know how that works here, for the Schwarzschild coordinate $r$.

You can't measure $\theta$ or $\phi$ in any spherically symmetric coordinate system. So I'm not sure why this seems like a deal breaker. And you can measure the Schwarzschild areal coordinate $r$, unlike $\theta$ or $\phi$ which are unmeasurable.

In Minkowksi spacetime with coordinates $(t,x,y,z)$ you can't find the origin or any of the coordinates.

  • Can the statement "the event horizon is at $r = 2GM$" be phrased in a coordinate-independent way?

It doesn't even really make sense. The coordinate chart only covers $r>2MG$ you need a different coordinate chart at the events on the horizon.

  • How can the coordinate $r$ be measured?

You could measure tidal forces over a small region of time and space and compare those to the tidal forces expected for regions with different values of $r.$

But because of the equivalence principle, if you fix the precision and tolerance of your measurements and considered a small enough region, then you wouldn't be able to tell. Over a small region the tidal forces are hard to detect.

This is a fundamental principle. It's how we can actually make predictions. We state that for a small region of space and time, it's just like Minkowski spacetime (where you can't tell any coordinates, even though some directions are still clearly timelike and others clearly spacelike). You do your physics in that region, and then before you get to another region you switch to their coordinates.

The whole point of writing the metric for a practical coordinate system is to allow you to use one coordinate system in a region larger than the local freely falling inertial frame would allow.

But you shouldn't locally be able to tell. And if you look at larger regions it is essential how you bring all the information together.

But really science is about using a theory to make models and extracting predictions from the model on the one hand, and making observations on the other hand, so that in the gripping hand you can test the predictions against the observations.

Measuring a coordinate could be part of that process, but it's not what it's all about. We never even have to use that coordinate system. And we should not use that coordinate system at the event horizon.

Timaeus
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  • If I recall correctly, the "locally Minkowski" of GR is at a point, not on an open set, so I'm not sure "locally you shouldn't be able to tell" is actually correct because measurements have some extent in time or space, hence do not happen at a point. – ACuriousMind Mar 16 '16 at 10:11
  • @ACuriousMind A limit isn't a thing that happens a point, it happens in a finite region whose size gets smaller and smaller. There are families of neighborhoods parameterized be $h$ such that the metric is Minkwoski and the first derivatives of the metric are zero, up to an order of h. That's literally how the limit exists to get that pointwise limit you refer to. – Timaeus Mar 16 '16 at 17:39
  • @ACuriousMind So when you fix your precision, and use that family of neighborhoods then eventually you reach a neighborhood where you can't distinguish the metric from the Minkowski metric. – Timaeus Mar 16 '16 at 17:41
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You can relate $r$ to the tidal forces an observer experiences, or use the reduced circumference or areal radius, etc. for observers with no angular momentum (see other answers). Many books do argue against calling $r$ a "radius", the word "radial" seems more popular and general, but to be extra cautious you can just say "Schwarzschild $r$-coordinate". I'm flexible personally; so long as it's understood you can't hang a static ruler from $r=0$ to $r=2M$, and that in relativity distance is relative to the observer.

However you can certainly use a moving ruler inside the horizon. Take an observer who fell freely from large $r$. Then the proper distance $\int ds$ along their spatial radial direction is indeed $dr$, and laying a bunch of such falling rulers end-to-end does indeed give a total length of $2M$. See Taylor & Wheeler, Exploring Black Holes (2000, $\S B.3$).

Colin MacLaurin
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In general relativity, you can take whatever coordinate system you want

That's what people tend say, but it isn't true. Imagine your coordinate system is blue. You can adopt a new coordinate system if you wish, but that's blue too. You can have any coordinate sysem you like, so long as it's blue. OK, now look at this black hole depiction:

enter image description here

It's black in the middle. That's where there is no coordinate system. People tend to say the event horion is a mere artefact, and you can adopt a new coordinate system that spans this central region, but you can't. Because gravitational time dilation system goes infinite.

For example, the event horizon of a Schwarzschild black hole is $r = 2GM$. Naively interpreting $r$ as a radial coordinate, this suggests that the event horizon is "$2GM$ away from the center of the black hole", but that statement doesn't make sense either mathematically (the distance $\int ds$ is not $2GM$ at all) or physically (you can't extend your ruler network inside the black hole). But the books I've seen seem to treat $r$ just like the radial coordinate, and talk about "the radius of a stable circular orbit" or stuff like that.

There's a definite distance involved. For simplicity think of space without a star, then the same space with a star in the middle. There's a distance from you to the surface of that star, and the to centre of the star. When you replace the star with a black hole, r is neither.

In the rest of physics, we relentlessly focus on how mathematical quantities can be measured, but I don't know how that works here, for the Schwarzschild coordinate $r$.

It doesn't quite work here, not the way people suggest. Have a look at the Wikipedia Schwarzschild radius article. Note the expression $\frac{t_r}{t} = \sqrt{1 - \frac{r_\mathrm{s}}{r}}$ and think about what happens when r = rs. Also note that you can measure time dilation using clocks at different elevations. OK, now unwind a very long cable out of your gedanken spaceship with light-clocks every ten metres. Let it dangle towards the black hole. Don't unwind it quite as far as the event horizon. This might take a few trial attempts, and some spare light-clock cables. Once you have unwound it, leave it there for a year so that the unwind and wind times are insignificant. Then plot all the clock readings. The clock readings lower down are lower than those higher up, and your plot is curved, much like the depiction above. You can extrapolate, whereupon you see the moot point: if you had been able to unreel your cable such that the lowest clock was at the event horizon, and if you had been able to properly cater for the unwinding and winding time, that lowest clock reading would come up zero. Because gravitational time dilation goes infinite. Then the truth ought to dawn: in truth you haven't made any measurement with that clock, because the event horizon is where your measurements stop, because that's where clocks stop.

A stopped clock can't go any slower than stopped, and since the force of gravity relates to the gradient in potential which in turn relates to those clock rates, you should be able to work out that at this place there is no gravity. You should also be able to appreciate that you cannot adopt a new coordinate system to make a stopped clock tick. Kruskals-Szekeres coordinates allegedly do this, but they effectively place a stopped observer in front of the stopped clock and claim he sees it ticking normally "in his frame". It's nonsense. The clock is stopped and so is he. He sees nothing. Eddington-Finkelstein coordinates are similar. Make sure you read the opening paragraph of the Wikipedia article which says they were dreamt up by Penrose and promoted by Misner/Thorne/Wheeler. Kip Thorne nowadays promotes time travel.

Can the statement "the event horizon is at $r = 2GM$" be phrased in a coordinate-independent way?

Yes. That's the place where all coordinate systems stop. You cannot get past this place. There's no falling past it either. See the second paragraph here and note the frozen star in Kevin Brown's Formation and Growth of Black Holes.

How can the coordinate $r$ be measured?

You can't practically measure it. But you can understand why not.

John Duffield
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