# A Distance of two shells in Gullstrand-Painleve coordinates

1. Oct 13, 2016

### Sonderval

I am a bit confused by the fact that in GP-coordinates
https://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates
the spatial part is flat.
I try to imagine the following experiment:
First create two rigid shells at two coordinates r1 and r2 outside of the event horizon.
The observer on the outside shell has a long ribbon. The ribbon is slowly spooled down. After each millimeter or so of downspooling the ribbon, the outer shell observer sends at r2 a signal to the observer at r1 to ask whether the ribbon has reached r1 or not. When the signal comes back, the next millimeter is spooled from the ribbon, until the ribbon reaches r1. At this point, the totally spooled length is noted at r2 and defined as distance between r2 and r1.

Using the Schwarzschild metric, the distance ds will obviously be greater than (r2-r1) due to the "funnel shape".

The same result for ds should come out using the GP coordinates (same amount of ribbon spooled), but due to the exchange of light signals and the slow (could be arbitrarily slow) process of spooling of the ribbon, I do not see how the dr dT-term in the GP metric enters the calculation to yield the same result, since dT can basically be arbitrary. Somehow the difference in the time coordinate at the two shells must enter here, but I am not sure what the correct way to understand/calculate this is considering that after the ribbon reaches from the outer to the inner shell, everything is stationary.

Probably I'm making a very simple mistake here, but I'm not able to find it, and previous forum entries, although touching the subject, did not really clarify things.

2. Oct 14, 2016

### Ibix

Forgive me if I misunderstand, but you seem to be interested in the interval along a path that lies in a surface of constant GP time. But in that case it doesn't lie in a surface of constant Schwarzschild time, no? So the $dr^2$ term is the only non-zero term in the interval in GP coordinates, but not in Schwarzschild coordinates - so the fact that the $dr^2$ terms aren't identical isn't significant of anything.

Or am I missing something?

3. Oct 14, 2016

### Staff: Mentor

By which you mean, I assume, create two shells that are static at r1 and r2. In other words, their r coordinates do not change with time (their own proper time, but also Schwarzschild or GP coordinate time).

Notice that by making this specification, you have already imposed a particular constraint on the scenario: there will be a particular family of spacelike hypersurfaces that are orthogonal to the worldlines of all the points on both shells. These spacelike hypersurfaces will be surfaces of constant coordinate time in one of the two coordinate charts you are studying, but not the other. A very helpful exercise is to answer the question: which?

How is this measured? I.e., a millimeter according to what observer? That's a critical point in your scenario.

"Same amount" by what definition?

In cases like this, it is very helpful to actually write down the math, instead of trying to reason heuristically in words. Write down the actual coordinates for the events in question in Schwarzschild coordinates. Compute the invariants. Then transform to GP coordinates and compute the invariants again. You will find that the invariants haven't changed; but other things will.

4. Oct 15, 2016

### pervect

Staff Emeritus
Basically, coordinates don't have anything directly to do with distance, and I would guess this is the source of your confusion, thinking that they do, somehow. Defining distance in the most general case is a bit tricky in GR, but when there is a convention for which points in space-time occur "at the same time", and rotation is absence, it's more straightforwards to compute than in the general case. One takes a curve in space-time where all points occur "at the same time", and one considers the path that is "straight" (the formal name is a geodesic) in the resulting spatial slice, the length of this curve (measured by integrating the Lorentz interval of the curve) is the distance. There are still some technical issues that may need more study to be clear if you intend to actually carry out the calculation, the most basic is "what is the Lorentz interval", followed by "what do we mean by a straight line or geodesic", with a side order of explaining the difference between a geodesic in the 3-d spatial slice of constant time and the 4-d geodesic in space-time.

The resulting formula will be independent of the choice of coordinates, but it will involve the metric components for those coordinates that one chooses, rather than a simple subtraction. I'm not sure it'd be useful to work out all the details, without knowing what sort of background you have, I hope this rather general discussion will be sufficient.

The presence of rotation makes things a lot trickier, hopefully we can avoid discussing that.

5. Oct 15, 2016

### Sonderval

Yes. This is the construction explained in the book draft of "Exploring black holes". If I understand correctly, the r-coordinate of GP and Schwarzschild are identical, so I assume that this should work for both.
Yes, that is exactly the point that is confusing me. To make it more specific, let's count atoms along the ribbon. The number of atoms should be the same for both observers; so I assume that in going from SS to GP coordinates, somewhere a length contraction enters (so that they observe atoms of different size), and the drdT-Term in the GP coordinates is responsible for this.

Assume that the coordinate distance between the shells is 100 atom lengths (as measured on each of the shells in the tangential direction) - then the SS observer will, after laying down the ribbon, count a larger number of atoms between the shells, say 101. (This is the excess radius that is for example also explained in the Feynman lectures and that is responsible for the funnel shape of embedding spacetime diagrams).
Let the ribbon be static and stay in its position for an arbitrarily long time. Then the GP observer should also count 101 atoms between the shells (since the shell coordinates are the same). OTOH, if the GP-observer measures the distance between the end points at a fixed T, it will be equal to r2-r1 (100 atom lengths). Thus, the drdT-term somehow has to enter into this calculation to yield the same result of 101 atoms counted.

I tried to follow your advice to actually calculate this - as expected, the calculation works out, but my intuition does not:
I assume that $r_2-r_1\ll r_2$. The distance $d \sigma$ is then (calculated at equal time in SS coordinates)
$$d\sigma = \frac{d r}{\left(1-\frac{2M}{r}\right)^{1/2}}$$
(eq. 3-16 from http://www.eftaylor.com/exploringblackholes/Curving160401v1.pdf [Broken]
To get $d\sigma=1.01dr$ as in my example above, $2M/r\approx 0.0197$.
If I choose M=1 (in natural units), I am at r=101.523.
So to be specific, let $r_2=102$ and $r_1=101$; the change in the factor in front of the dr is then small and I assume it to be constant. I then have $d\sigma=1.01$

In GP coordinates, the r coordinate is the same as in SS, the T-coordinate of GP is related to the t-coordinate of Schwarzschild by
$$T= t + 4M \sqrt(r/2M) - 2M \ln\frac{ 1+\sqrt(2M/r)}{1-\sqrt(2M/r)}$$
(Formula 25 from http://www.eftaylor.com/exploringblackholes/InsideBH160402v1.pdf [Broken]
If the SS t-coordinate is 0, the T-ccordinates are then $T(r_2)=28.0018$, $T(r_1)=27.8587$, resulting in $dT=0.143$

For the distance I thus get, using the "rain metric" (formula 32 from the same chapter), again a value of $d\sigma=1.01$ within my precision. (I calculated this using gnuplot.)
So if the SS observer has a (red) ribbon with atoms of size 0.01, he will see 101 of these atoms between the two points in space, and the GP observer will agree that there are indeed 101 atoms between the two points (which to her are at different times).

Now I do the reverse scenario: The GP observer uses another, identical ribbon (make it blue) and lays it down between r2 and r1. For her, at constant T, the distance in the GP (or rain) metric is $d\sigma=1$ (no calculation needed since dT is zero and the spatial part of the metric is flat).

The SS observer sees the two (simultaneous, from the GP observers POV) endpoints of the blue ribbon with a time difference of $dT=0.143$ (same as before), and will thus calculate in SS metric (formula 6 from the first link) the same distance of 1 as the GP observer.

So the calculation works out - so far, so good.

However, I'm still puzzled by what an observer at the two ribbons would actually see. If I am fixed at r=101.5 in the middle, I see the red and the blue ribbon. The red atoms are a bit length-contracted wrt to the blue ones according to the calculation.
But OTOH, if both observers let the ribbons hang unmovingly after spooling them down (let's say the ribbons are fixed at their ends in the relative frames), the atoms of both ribbons should be stationary at r=101,5, or would I somehow see them moving relative to each other? (and if so, how would the movement change over time)? Is there a way to understand/calculate this?

Last edited by a moderator: May 8, 2017
6. Oct 15, 2016

### Staff: Mentor

In the sense that every event in the spacetime has the same $r$ coordinate in both charts, yes. But you should note carefully that the basis vector $\partial / \partial r$ is not the same in the two charts.

Along the ribbon between what two points? I'll assume that you mean between the two shells; so far so good.

But now we get to the real issue: when we count atoms, what are we counting? You are implicitly assuming that we are counting points in a spacelike slice of constant coordinate time. But that means we are counting points in two different spacelike slices for the two charts. See below.

If you want to view "counting atoms" as counting points in a spacelike slice, then yes, this is one way of viewing what's going on. The spacelike slice of constant GP time is "tilted" in spacetime as compared to the spacelike slice of constant Schwarzschild time. That means the portion of the GP slice between the two shells, containing the same number of atoms, looks length contracted from the point of view of the Schwarzschild slice.

But there is a difference between this and the usual SR length contraction. The usual SR length contraction is symmetric: each frame looks length contracted to the other. That's not the case here; the Schwarzschild distance being longer than the GP distance between the two shells is an invariant, the same for both observers. Why is that? The answer is that we set up the shells to be stationary, i.e., to be at the same $r$ coordinate, and the spacelike slices of constant Schwarzschild time are the unique set of spacelike slices that are orthogonal to the worldlines of objects that keep the same $r$ coordinate. The spacelike slices of constant GP time are not orthogonal to those worldlines, so if we are defining "distance" as "coordinate distance between the shells", the Schwarzschild distance will be the maximum possible such distance. (The slices of constant GP time are orthogonal to the worldlines of observers free-falling inward from rest at infinity, which are a different set of worldlines.)

Another (better IMO) way to look at this is to define "counting atoms" as counting worldlines, not points. If we assume the atoms in the ribbon are also stationary (each one keeps the same $r$ coordinate), then the worldlines of all the atoms behave just like the worldlines of the shells--they are orthogonal to the slices of constant Schwarzschild time, not the slices of constant GP time. So the Schwarzschild slices "cut" the family of worldlines orthogonally, but the GP slices cut at an angle, so to speak, and thereby are shorter in terms of distance (spacelike interval between the shells). (Note that "cutting at an angle" corresponds to shorter spacelike distance, not longer, because the geometry of spacetime is locally Minkowskian, not Euclidean.)

It's not "responsible" for it, because coordinate-dependent quantities can't be "responsible" for anything; only invariants can. But the dr-dT cross term in GP coordinates is a sign of the non-orthogonality I described above: the slices of constant GP time are not orthogonal to worldlines of constant $r$.

I haven't looked in detail at your calculations, but hopefully the above will help to clarify what is going on.

7. Oct 16, 2016

### Sonderval

Of course!
Thanks for the explanation - I'll have to think about it a bit more to be sure that I really get it, but it seems you hit the nail on the head (as usual).
You're great!