# The Painleve–Gullstrand chart

1. Jul 13, 2010

### Passionflower

The Painleve-Gullstrand chart, which describes the Schwarzschild solution, has a flat spatial section. All the curvature is in the time-time and time-space parts of the metric. From this it seems we could conclude that the r coordinate in the PG chart relates to the circumference as it does in flat space. This is obviously not the case for the Schwarzschild chart, here the spatial section is not flat and thus the r coordinate does not relate to the circumference as it does in flat space.

One can convert the Schwarzschild chart to the PG chart without integration while one can only calculate a physical radius using the Schwarzschild chart by integration because the spatial section is not flat.

Now is this consistent?

2. Jul 13, 2010

### Ich

Of course it does. That's how it is defined.

3. Jul 13, 2010

### Passionflower

I see how my description of the question causes confusion.

Let me restate it:

Because space in the Schwarzschild chart is not flat the physical radius is not equal to r, r is instead related to Gaussian curvature. However in a GP chart space is flat and so r seems to directly refer to the physical radius.

One can convert the Schwarzschild chart to the PG chart without integration while one can only calculate a physical radius using the Schwarzschild chart by integration because the spatial section is not flat.

Now is this consistent?

4. Jul 13, 2010

### Ich

If both calculations were supposed to give a single result, "physical radius", yes, that would be strange.
But there is no such thing as a single physical radius. You calculate a totally different thing in the PG chart than in the Schwarzschild chart. So it's not a miracle that the mathematical effort is different, too.

5. Jul 13, 2010

### Mentz114

An interesting difference between GP and Schwarzschild coords is that a Minkowski observer in Schwarzschild coords has zero expansion ( ie the ball of coffee grounds does not change volume) but in GP coords the expansion is negative, so it gets smaller with decreasing r. Neither oberver experiences any shear velocities.

6. Jul 13, 2010

### nutgeb

In Schwarzschild coordinates, an object plunging radially toward the massive origin at exactly escape velocity also measures the local space to be spatially flat. In effect, the SR radial length contraction and Schwarzschild positive spatial curvature exactly offset each other at this particular velocity. Taylor & Wheeler refer to this as the "rain frame" metric in their textbook "Exploring Black Holes."

Painleve-Gullstrand coordinates generalize this same effect to the coordinate chart as a whole. It has been referred to as the "river" model because it can be interpreted as if space itself is flowing toward the massive origin at exactly the escape velocity at each coordinate location.

Last edited: Jul 13, 2010
7. Jul 14, 2010

### Passionflower

Insightful!

But consider a rod A>-----<B free falling in such a way. I think that you are saying that both the A and the B end will 'see' the space as flat, right? If so, how do we explain the tidal effects, as without it the rod ought to behave as a Born-rigid rod.

8. Jul 14, 2010

### nutgeb

Let's make sure we are saying the same thing. I said that the freefalling observer at escape velocity will see local space as being 'spatially' flat. But that does not mean that the local 'spacetime' is flat. Those are two very different concepts.

In Schwarzschild curved spacetime, tidal effects follow the same metric for a freefaller at escape velocity as they do for a freefaller at any other velocity. Tidal effects (at least as the term is normally used) don't occur because of local spatial curvature; they occur because the gravitational field is non-uniform. I.e., all gravitational acceleration vectors aim toward a single point at the center of the Schwarzschild mass, and the strength of the acceleration is inversely proportional to the square of the radial distance due to Gauss' Law. Slightly separated freefalling particles therefore experience different acceleration vectors.

A rigid rod falling lengthwise toward the Schwarzschild mass will experience internal tidal stresses as the differential gravitational acceleration tries to stretch the rod lengthwise. If the rod's length is perpendicular to the radial path along which it falls, then the differential gravitational acceleration tries to tidally squeeze the rod lengthwise.

On the other hand, SR length contraction does not introduce internal stresses in a rigid rod falling lengthwise. However, I'm not sure whether positive spatial curvature (e.g., freefalling at less than escape velocity) introduces its own (non-tidal) internal stresses (radial stretching/transverse squeezing) in a rigid rod falling from a great distance into a Schwarzschild mass. I tend to think that it does, but that seems counterintuitive if the SR and spatial curvature contributions are inconsistent regarding whether they introduce (potentially offsetting) internal stresses.

Last edited: Jul 14, 2010
9. Jul 14, 2010

### Passionflower

Of course.

Of course they follow the same metric, but are the effects similar? Gravitational strength depends on relativistic mass, thus, it seems, that an object traveling faster or slower than escape velocity should encounter a different kind of tidal effect. No?

It seems to me that that is true for Newtonian physics, in the Schwarzschild metric the gravitational strength also increases when the approaching test particle reaches relativistic speeds right?

I think you are right, in the Schwarzschild solution when the test particle approaches the center of gravity and reaches relativistic speeds then in addition to the Newtonian increase of gravitational strength there is also an increase in gravitational strength due to an increasing relativistic mass.

Last edited: Jul 14, 2010
10. Jul 15, 2010

### nutgeb

As you suggest, a relativisitic freefall velocity would increase the magnitude of the gravitational acceleration on the rigid rod and therefore would increase the internal tidal stresses, all other factors remaining the same. The rod also will experience internal stresses arising from Borne rigidity effects.

However, those effects have nothing to do with spatial curvature per se, which is not normally included in the term "tidal effect". I still am not sure of the answer to my specific question, which is whether the increasing Schwarzschild positive spatial curvature, as the rod approaches the central Schwarzschild mass, introduces its own internal stresses in a rigid rod freefalling radially at less than escape velocity.

To the extent that the rod's relativistic freefall velocity exceeds escape velocity (e.g., having initially been boosted toward a black hole event horizon), the effect of SR length contraction exceeds the effect of positive spatial curvature. I'm not sure whether it is accurate to extrapolate from this that the local space at the rod appears to have negative spatial curvature. In any event, a second rigid rod located radially immediately above or below the first and falling at almost the same relativistic velocity definitely would not appear to the first rod to be experiencing internal stresses due specifically to the excess of both rods' Lorentz contraction over the Schwarzschild positive spatial curvature. So I'm inclined to think that the converse must also be true -- that a a rigid rod freefalling at less than escape velocity, which observes local space to have net positive curvature, will not experience internal stresses specifically caused by the increasing spatial curvature.

11. Jul 18, 2010

### nutgeb

C'mon, somebody here must have some insight about whether a rigid rod would experience internal stresses due solely to the positive local spatial curvature, when freefalling at less than escape velocity toward a Schwarzschild mass, in addition to the normal tidal stresses and stresses introduced by borne rigidity effects.

I see two possible kinds of internal stress due to spatial curvature.

The first is a tidal stress due to differences in the amount of spatial curvature between the two ends of the rod. I'm less interested in that, and its existence really depends on the second kind of stress.

The second is a stress to the rod as a whole, as the increasing spatial curvature during freefall seeks to change the ratio between the radial dimension and the transverse dimension within the rod (as compared to the ratio between those dimensions when the rod started falling from infinity, where the spatial curvature was zero. I'm interested in whether that second kind exists, and I'm tending to think it does not.

Last edited: Jul 18, 2010
12. Jul 19, 2010

### Mentz114

Too much handwaving. Why don't you do some calculations to try and answer your question ?

13. Jul 19, 2010

### nutgeb

Hmmm, sorry about the handwaving, but I don't know how else to approach this question. I think this is a matter of heuristics, not something that is straightforward to answer through calculations. Or at least I haven't figured it out.

It's analagous to the question whether Lorentz contraction introduces internal stresses into a rigid rod which is in inertial relativistic motion relative to an observer. In that case the answer is definitely not. But how would one go about demonstrating that mathematically? One tends to answer the question by shifting one's perspective to the rod's inertial rest frame, in which the rod is not contracted, and therefore it can be deduced that it feels no stress.

Similarly in the case of spatial curvature, for analysis purposes one needs to identify a reference frame in which local space at the rod is spatially flat. That indeed occurs if the rod is freefalling toward a Schwarzschild mass at exactly escape velocity. But if the rod is freefalling at less than escape velocity, local space at in the rod's own rest frame isn't spatially flat.

Alternatively one can consider the rest frame of an observer falling at escape velocity and passing alongside the falling rod (the rod is falling toward the Schwarzschild mass at less than escape velocity). That observer's frame is spatially flat, and we can assume that the nearby falling rod momentarily passes through the observer's flat local space. This observer sees the rod to be moving past itself at some radial velocity, and if that velocity is relativistic, he sees the rod to be Lorentz contracted. Therefore this observer can conclude that the Lorentz contraction arising from difference between the rod's freefall velocity and the observer's own velocity (toward the Schwarzschild mass) has in itself introduced no internal stress into the rod.

So I think through this analogy I've answered my own question. Spatial curvature itself does not introduce internal stresses into the freefalling rigid rod.

14. Jul 19, 2010

### Mentz114

I'm not sure if the effects of 'only' spatial curvature can be extracted. You've mentioned 'the usual tidal effects', which I take to mean the tidal tensor

$$T_{\hat{a}\hat{c}}=R_{\hat{a}\hat{b}\hat{c}\hat{d}}U^{\hat{b}}U^{\hat{d}}$$

where the hats indicate that the Riemann tensor is calculated in the observer frame and U is the time-coframe basis vector $\vec{\sigma}^0$. For the observer falling in with the 'river' in GP coords the spatial parts are $T_{rr}=-2m/r^3$, $T_{\theta\theta}=T_{\phi\phi}=m/r^3$. This gives the tidal strains for this observer. Now I suppose one could calculate this for different observers to see what the effect of different infall velocity is. I'm sure you've noticed that the 3-trace of T is zero, indicating that there will be no change in volume of a cloud of dust which is infalling.

However, I'm not sure if it's correct to extrapolate from dust to a solid body.

The problem with an extended body is that only some of it will be on geodesics, and the acceleration of different parts wrt to the rest have to found by calculating the geodesic deviations. Not a trivial problem.

Last edited: Jul 19, 2010
15. Jul 19, 2010

### nutgeb

Thanks. Tensor math is beyond me so I'm afraid I can only respond with more handwaving.

I'm trying to exclude all stresses caused by tidal effects. I want to find out if there are any non-tidal internal stresses that could be attributed to spatial curvature.

Clearly if spatial curvature doesn't cause internal stresses at all, then it won't cause tidal stresses in an extended body either. If spatial curvature caused internal stresses, most fundamentally they would result simply from the change in the object's radial dimension relative to its transverse direction. I suspect there is no specific tensor that can shed light on whether that change induces stresses, just like there is no non-trivial calculation (at least that I'm aware of) that shows whether a Lorentz-contracted rod experiences internal stresses in SR. However, if spatial curvature did cause non-tidal internal stresses (which I think it does not) then it would also cause additional tidal stresses in an extended object falling into a non-uniform gravitational field, which might be captured in the tidal tensor.

I think the analogy I gave in my last post is a reasonable demonstration that falling at a velocity below escape velocity does not result in internal stresses caused by spatial curvature.

The point you make about a cloud of dust not changing shape as it falls in GP coordinates is interesting. Perhaps this happens because, in effect, the GP spatial coordinates become radially/transversely distorted (compared to the Schwarzschild counterparts) as the radius decreases -- to a degree that exactly equals (and therefore offsets) the tidal stretching and squeezing that occurs in Schwarzschild coordinates. But that's just a handwaving suggestion.

16. Jul 19, 2010

### Mentz114

Caution, the signs in the tidal tensor in my previous post are reversed. I just checked the script.

It could be that every Minkowski observer in the Schwarzschild spacetime sees that tidal tensor, I'm not sure. The dust cloud will stretch along the r direction and shrink in the orthogonal directions while preserving its volume.

The tidal tensor for the Minkowski local frame in Schwarzschild coords has the same form as the one above. But it is not the same 'r', of course.

Last edited: Jul 19, 2010