# Gullstrand–Painlevé coordinates

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Posted (edited)

A worldline cannot be associated with two different reference frames, so if Lemaître coordinates define a reference frame, Painlevé coordinates do not.

The Painlevé coordinate system does not appear to constitute a relativistic reference frame. Indeed, in the (dt,dr) coordinate system of Schwarzschild, the coordinate system associated with Gullstrand–Painlevé (dT, dr + √(Rs/r)dT) does not form a reference frame because the light cone tilts. In relativity, the light cone must not tilt. To maintain isotropy in another reference frame, the simultaneity line must be redefined by a boost. Painlevé metric is ds² = dT² - (dr +√(Rs/r)dT)²

Edited by externo

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Posted (edited)

You seem to have a fundamental misunderstanding of the term reference frame. A worldline connects two reference frames. A reference frame can be inertial or non inertial.

ALice has one reference frame Bob has his own reference frame. The worldine is the transition between Alice and Bob's reference frames. The choice of coordinate systems does not change this detail due to invariance of coordinate choice. That is a fundamental principle of the Einstein field equations.

Ds^2 is the separation  distance between the two events Alice and Bob. Ds^2 is not a reference frame but the spacetime path.

Every event (observer, emitter ) is it's own reference frame. The coordinate choice doesn't alter that detail

Edited by Mordred
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Posted (edited)
1 hour ago, Mordred said:

You seem to have a fundamental misunderstanding of the term reference frame. A worldline connects two reference frames. A reference frame can be inertial or non inertial.

ALice has one reference frame Bob has his own reference frame. The worldine is the transition between Alice and Bob's reference frames. The choice of coordinate systems does not change this detail due to invariance of coordinate choice. That is a fundamental principle of the Einstein field equations.

Ds^2 is the separation  distance between the two events Alice and Bob. Ds^2 is not a reference frame but the spacetime path.

Every event (observer, emitter ) is it's own reference frame. The coordinate choice doesn't alter that detail

Do you agree with this definition of a reference frame:

"By definition a 1D time-like foliation of a 4D Riemannian spacetime is a reference frame"

Edited by externo
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Posted (edited)
2 hours ago, externo said:

Do you agree with this definition of a reference frame:

"By definition a 1D time-like foliation of a 4D Riemannian spacetime is a reference frame"

No, any more than the bijective map R maps to Rn is although it is an index since they have the same count of members.

A coordinate system has a richer structure to allow for the connections between the points in the form of functions, derivatives etc.

Edited by studiot
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Posted (edited)
2 hours ago, externo said:

Do you agree with this definition of a reference frame:

"By definition a 1D time-like foliation of a 4D Riemannian spacetime is a reference frame"

Along with what Studiot posted There are numerous reference frames under GR. With regards to the Painleve chart one can utilize the Lenaitre reference frame.

Here is a better article covering Lemaitre frames with regards to Gullstrand- Painleve coordinates

Edited by Mordred
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11 hours ago, Mordred said:

You seem to have a fundamental misunderstanding of the term reference frame. A worldline connects two reference frames. [...]

ALice has one reference frame Bob has his own reference frame. The worldine is the transition between Alice and Bob's reference frames.

Where are you getting your information? How is this consistent with anyone else's use of the term "world line"? Can you please provide a reference?

https://en.wikipedia.org/wiki/World_line "The world line (or worldline) of an object is the path that an object traces in 4-dimensional spacetime." How can that possibly be a transition between Alice and Bob's reference frames?

How many dozens of people have to come to this site and be told by an "expert" that they "have a fundamental misunderstanding", and then be given such utterly embarrassing nonsense from you? How many go away believing it, knowing only that an expert told them something beyond the possibility of understanding, and never get the truth?

You really should request that the site removes your "expert" status, for the sake of everyone. I would like to request that they do that. Is there an official way to do that?

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Posted (edited)
19 hours ago, externo said:

A worldline cannot be associated with two different reference frames, so if Lemaître coordinates define a reference frame, Painlevé coordinates do not.

The choice of coordinates is not necessarily the same as a choice of physical observers. Of course, particular coordinate choices may naturally correspond to the “point of view” of particular observers (in the sense that the coordinates correspond to what clocks and rulers in that frame physically read), making the maths simpler; for example, Schwarzschild coordinates naturally describe the point of view of a stationary observer very far from the central mass, whereas Gullstrand-Painlevé coordinates correspond to the point of view of an observer freely falling towards such a mass. But you’re not restricted to those choices - you are free to pick different coordinates to describe the same physical scenario, so long as they are related by valid diffeomorphisms. Thus you’ll have many reference frames coexisting in the same spacetime. So there’s no problem with describing the same world line in different coordinate charts - you’ll find that its geometric length is an invariant, it is always the same no matter what coordinates you choose. There are also valid coordinate charts that nevertheless don’t correspond to any physical observer - such as eg Kruskal-Szekeres.

The Schwarzschild chart and the Gullstrand-Painlevé chart are diffeomorphisms of one another, so they cover the exact same spacetime (though the latter covers a larger part of the manifold than the former does).

19 hours ago, externo said:

In relativity, the light cone must not tilt.

This isn’t true - in curved spacetimes, light cones will naturally “tilt” along geodesics.

Edited by Markus Hanke
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Posted (edited)
5 hours ago, md65536 said:

Where are you getting your information? How is this consistent with anyone else's use of the term "world line"? Can you please provide a reference?

https://en.wikipedia.org/wiki/World_line "The world line (or worldline) of an object is the path that an object traces in 4-dimensional spacetime." How can that possibly be a transition between Alice and Bob's reference frames?

The null geodesic aka worldline is a field treatment under GR the interaction is readily described by the perturbation  tensor h_{\mu\nu} which acts upon the metric.

I won't bother with thr rest of your commentary not worth my effort. Not to mention potentially hijacking someone else's thread

Edited by Mordred
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Posted (edited)
3 hours ago, Markus Hanke said:

The choice of coordinates is not necessarily the same as a choice of physical observers. Of course, particular coordinate choices may naturally correspond to the “point of view” of particular observers (in the sense that the coordinates correspond to what clocks and rulers in that frame physically read), making the maths simpler; for example, Schwarzschild coordinates naturally describe the point of view of a stationary observer very far from the central mass, whereas Gullstrand-Painlevé coordinates correspond to the point of view of an observer freely falling towards such a mass. But you’re not restricted to those choices - you are free to pick different coordinates to describe the same physical scenario, so long as they are related by valid diffeomorphisms. Thus you’ll have many reference frames coexisting in the same spacetime. So there’s no problem with describing the same world line in different coordinate charts - you’ll find that its geometric length is an invariant, it is always the same no matter what coordinates you choose. There are also valid coordinate charts that nevertheless don’t correspond to any physical observer - such as eg Kruskal-Szekeres.

The Schwarzschild chart and the Gullstrand-Painlevé chart are diffeomorphisms of one another, so they cover the exact same spacetime (though the latter covers a larger part of the manifold than the former does).

This isn’t true - in curved spacetimes, light cones will naturally “tilt” along geodesics.

In the Schwarzschild metric, the light cone does not tilt but closes. A tilted light cone would imply an accelerating speed during fall, contradicting the isotropic nature of light speed relative to stationary objects. Regardless of spacetime curvature, the light cone cannot tilt in relativity, as this would indicate a physically anisotropic light speed. Painlevé coordinates do not constitute a reference frame according to relativity because, in Schwarzschild coordinates (dt, dr), Painlevé coordinates (dT, dr + √(Rs/r)dT) are not Minkowski orthogonal. Relativity tells us that when an object is in motion the light cone remains straight and that this object has a new simultaneity represented by a line orthogonal to the world line according to Minkowski orthogonality. However, Painlevé coordinates deviate from this by tilting the light cone and defining a new simultaneity (dT, dr + √(Rs/r)dT) orthogonal to dT in Euclidean terms. This stems from the fact that the decomposition (dT, (dT, dr + √(Rs/r)dT)) does not form a relativistic reference frame in the (dt, dr) frame. Consequently, I am surprised by the equivalence attributed to Painlevé and Lemaître coordinates. Lemaître coordinates constitute a genuine reference frame as the new simultaneity is derived from boosts preserving the light cone.

This has critical implications. In Painlevé coordinates, the tilted light cone prevents clock desynchronization for a freely falling observer. Conversely, Lemaître's falling observer requires clock resynchronization to maintain isotropic light speed. These coordinate systems yield differing physics, with Lemaître coordinates contradicting both the physical world and the equivalence principle due to its clock desynchronization.

Edited by externo
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Posted (edited)
1 hour ago, externo said:

In the Schwarzschild metric, the light cone does not tilt but closes.

By “tilted” we mean a change of orientation of the “cone opening” when drawn onto a spacetime diagram.

Here’s such a spacetime diagram for Schwarzschild spacetime, with light cones drawn in:

As you can see, the opening of light cones at events with smaller r-values is both rotated in spacetime and narrower, relative to light cones at larger r-values; on the diagram they are thus facing more towards the horizon/singularity. Whether you wish to call this “tilting” or “narrowing” is really just a matter of semantics.

A light cone encodes causal structure - everything “inside” the cone are time-like world lines, whereas the “walls” are traced out by null geodesics. As you approach the horizon, there are fewer and fewer future-directed time-like paths available that would lead you back out and away from the black hole (necessitating more and more extreme acceleration profiles), whereas more and more of the physically available paths lead into the black hole - IOW, the future light cones in the above diagram are increasingly facing towards the horizon. Very close to the horizon, only some null geodesics - and only those - lead potentially back out. Below the horizon, the singularity is always in the future of any time-like or null world line, so there’s no escape.

1 hour ago, externo said:

A tilted light cone would imply an accelerating speed during fall, contradicting the isotropic nature of light speed relative to stationary objects.

The walls of the cone are always null geodesics by definition, so the locally measured value of c is of course not affected.

1 hour ago, externo said:

Painlevé coordinates do not constitute a reference frame according to relativity because, in Schwarzschild coordinates (dt, dr), Painlevé coordinates (dT, dr + √(Rs/r)dT) are not Minkowski orthogonal.

I have no idea what you mean by this, especially since we’re not in Minkowski spacetime. Can you restate this mathematically?

1 hour ago, externo said:

Consequently, I am surprised by the equivalence attributed to Painlevé and Lemaître coordinates.

What exactly do you mean by “equivalence”? They are not the same coordinate chart, but they are related by a valid diffeomorphism (so equivalent perhaps in that sense), so they describe patches of the same spacetime. The physical meaning of the time coordinate in Lemaître coordinates is the same as in GP coordinates, but the physical meaning of the radial coordinate is different between the two.

Edited by Markus Hanke
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12 minutes ago, Markus Hanke said:

The walls of the cone are always null geodesics by definition, so the locally measured value of c is of course not affected.

I have no idea what you mean by this, especially since we’re not in Minkowski spacetime. Can you restate this mathematically?

Minkowksi spacetime is tangent, in everything I write I speak of instantaneous tangent spaces.

In the Schwarzschild metric, dt and dr are orthogonal: ds² = (1-Rs/r)dt² - dx²/(1-Rs/r) In relativity, if you change the reference frame, you must perform a boost. The new reference frame will also form an orthogonal frame and light will be isotropic in both reference frames. The worldline of light will be the bisector of the axes of the frame. This is what is achieved when switching to Lemaître coordinates. Lemaître coordinates are: ds² = dT² - (Rs/r)dρ² They are obtained by performing Lorentz transformations on (dt, dr) In these coordinates, the cone does not tilt, instead, the coordinate dρ is defined so that the worldline of light is the bisector of dT and dρ. The speed of light is thus isotropic for both (dt, dr) and (dT, dρ) But the Painlevé coordinate system is not an identical decomposition, the speed of light is not isotropic for both (dt, dr) and (dT, dR) = (dT, dr +√(Rs/r)dT)) For it to be isotropic with respect to (dT, dR) = (dT, dr +√(Rs/r)dT)), the cone must tilt, and suddenly the light is no longer isotropic with respect to (dr, dt)

We can see that the Painlevé coordinate system is not a reference frame according to Einstein's relativity. It is a reference frame in which the isotropy of light in (dT, dR) comes with the loss of isotropy in (dt, dr).

12 minutes ago, Markus Hanke said:

What exactly do you mean by “equivalence”? They are not the same coordinate chart, but they are related by a valid diffeomorphism (so equivalent perhaps in that sense), so they describe patches of the same spacetime. The physical meaning of the time coordinate in Lemaître coordinates is the same as in GP coordinates, but the physical meaning of the radial coordinate is different between the two.

As the radial coordinate is not the same, you can see that both cannot be reference frames according to relativity.

I am talking about physical equivalence. The explanation I gave on the desynchronization of clocks shows that they are not physically equivalent. Lemaître's coordinates predict that clocks accelerated by free fall desynchronize in the free fall reference frame while Painlevé's predict that they do not desynchronize.

12 minutes ago, Markus Hanke said:

The walls of the cone are always null geodesics by definition, so the locally measured value of c is of course not affected.

In order for the walls to remain null geodesics the simultaneity must be changed and dr is no longer the line of simultaneity for stationary objects.

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Posted (edited)

I thought I answered your actual question, given in your post #2, shortly after you posted it.

But I do not seem to have a reply.

Edited by studiot
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Posted (edited)
28 minutes ago, studiot said:

I thought I answered your actual question, given in your post #2, shortly after you posted it.

But I do not seem to have a reply.

This definition of a reference frame is not mine. It is that of someone in another forum. I just wanted to know what people thought about it here, but that wasn't the topic of the thread.

2 hours ago, Markus Hanke said:

By “tilted” we mean a change of orientation of the “cone opening” when drawn onto a spacetime diagram.

Here’s such a spacetime diagram for Schwarzschild spacetime, with light cones drawn in:

As you can see, the opening of light cones at events with smaller r-values is both rotated in spacetime and narrower, relative to light cones at larger r-values; on the diagram they are thus facing more towards the horizon/singularity. Whether you wish to call this “tilting” or “narrowing” is really just a matter of semantics.

This is not a Schwarzschild diagram, in a Schwarzschild diagram the cone doesn't tilt :

Edited by externo
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Posted (edited)

I assume your aware that PG coordinates uses a different time called PG time correct ? The Schwartzchild metric the relation between coordinate time and proper time is given as follows

$d\tau=\sqrt{1-\frac{2GM}r}dt$

in the above dt becomes infinite at the horizon. PG introduces a new time coordinate to address that issue for notation I will simply use capital T for PG time.

so you need to find T in terms of the Schwarzschild metric. Now due to symmetry all directions of space are symmetric so we can only depend on t and r.

$dT=\frac{\partial T}{\partial{t}}dt+\frac{\partial{T}}{\partial{r}}dr$

for an object dropped at rest at infinity in the Schwarzschild coordinates this gives the energy per unit mass as e=1 and the angular momentum $$\ell=0$$.

$\frac{dr}{d\tau}=-\sqrt{2GM}{r}$

$\frac{dt}{d\tau}=(1-\frac{2GM}{r})^{-1}$

gives

$\frac{dr}{dt}=-\sqrt{\frac{2GM}{r}}(1-\frac{2GM}{r})$

however to work out $$\partial T/\partial r$$ we need to consider two events that occur at the same time in S time dt=0 but at slightly different radii however PG time is not the same as S time to fall the same distance. The total difference in PG time is

$dT=\sqrt{\frac{r}{2GM}}[1-(1-\frac{2GM}{r})^{-1}]dr$

$=\sqrt{\frac{r}{2GM}}\frac{2GM}{r-2GM}dr$

$=-\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dr$

with $$\partial T/\partial r$$ as the rate of change with respect to increasing r we get

$\frac{\partial T}{\partial r}=\sqrt{2GM}{r}\frac{1}{1-2GM/r}$

and the differentials are

$dT=dt+\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dt$

$dT=dt-\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dt$

which is symmetric for infalling and outfalling under change of sign that should highlight the distinction between the time components of the Schwarzschild metric as opposed to the PG coordinates which is a class of solution to remove the coordinate singularity of the EH (thought also not the only solution other coordinate systems do as well)

Edited by Mordred
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2 hours ago, externo said:

This definition of a reference frame is not mine. It is that of someone in another forum. I just wanted to know what people thought about it here, but that wasn't the topic of the thread.

Oh.

Marcus in particular and also Mordred like to deal in the abstract.

I like to ask the question "What does that posh statement mean, can i construct a simple down-to-earth example ?"

I offered you such but thank you for not wanting it.

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Posted (edited)
1 hour ago, studiot said:

Oh.

Marcus in particular and also Mordred like to deal in the abstract.

I like to ask the question "What does that posh statement mean, can i construct a simple down-to-earth example ?"

I offered you such but thank you for not wanting it.

lol sometimes I love the abstract other times not so much. When it comes to GR extremes such as the different metric treatments involved applied at a BH for example Markus is far more familiar with those metrics than I am. I rarely look into them as they aren't particularly applicable for Cosmology or particle physics. I tend to think in terms of Langrangian solutions as opposed to how GR handles spacetime. So I can sometimes give confusing statements such as my earlier post as a result. Simply put I think in terms of the Principle of least action when it comes to geodesics. (including Null geodesics). In the OPs case from what I see of the discussion between the OP and Markus knowing the distinction of how time is handled between the S metric and the PG metric is essential.

atm I'm seeing if Mathius Blau has a better coverage of PG coordinates

Edited by Mordred
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Posted (edited)
3 hours ago, Mordred said:

I assume your aware that PG coordinates uses a different time called PG time correct ?

The Painlevé coordinates (dT, dr + √(Rs/r) taken at time dt do not form a Minkowskian reference frame, so the river model is outside the scope of Einstein's theory.
It is exactly like instead of defining moving frames of reference in this way:
you defined them in this way:
that is to say with a Euclidean rotation instead of a hyperbolic one.

Edited by externo
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Posted (edited)

For proper acceleration you would have hyperbolic rotation

Do you for some reason feel there is no proper velocity with regards to PG coordinates ? Which if you look above I've shown for. Recall that the gradient term for gravity is under GR/SR a free fall state where the gravity term is the divergence (tidal force) so where is there an issue describing the freefall of particles under PG ? You have no additional force term being applied to cause acceleration. The line element is the freefall null geodesic hence why the setting for angular momentum terms is set at zero.( top of my post.)

World lines in Minkowsii space are straight lines where a null (lightlike) vector satisfies the following causal structure

four dimensional vector

$v-(ct,x,y,z)=(ct,r)$

$$c^2t^2>r^2$$ timelike

$$c^2t^2<r^)$$ spacelike

$$v=c^2t^2=r^2$$ null

the last condition must apply to be a worldline after all the speed of light is constant for all observers given by c.

I assume you already know that spherical coordinates and cylindrical coordinates are still Euclidean though not Cartesian (Cartesian is also Euclidean).

Edited by Mordred
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Posted (edited)
1 hour ago, Mordred said:

For proper acceleration you would have hyperbolic rotation

Do you for some reason feel there is no proper velocity with regards to PG coordinates ? Which if you look above I've shown for. Recall that the gradient term for gravity is under GR/SR a free fall state where the gravity term is the divergence (tidal force) so where is there an issue describing the freefall of particles under PG ? You have no additional force term being applied to cause acceleration. The line element is the freefall null geodesic hence why the setting for angular momentum terms is set at zero.( top of my post.)

World lines in Minkowsii space are straight lines where a null (lightlike) vector satisfies the following

c2t2=r2

The Painlevé metric can be written as: ds² = dT² - (dr + √(Rs/r)dT)²
which means that dT and dr + √(Rs/r)dT are orthogonal.
But it turns out that in the Schwarzschild frame (dt,dr), dT and dr + √(Rs/r)dT are not orthogonal according to the Minkowski metric, but they are orthogonal according to the Euclidean metric.
The Lemaître coordinates are written as: ds² = dT² - (Rs/r)dρ²
and in the frame (dt,dr), dT and dρ are orthogonal according to the Minkowski metric.

So while Lemaître's coordinate system does indeed produce a set of instantaneous reference frames in the sense of relativity, Painlevé's does not produce such frames.

Now, something else: in SR if two stationary clocks accelerate at the same time and in the same way they become desynchronized for an observer moving with them but they remain synchronized for a stationary observer. The same phenomenon occurs in the Schwarzschild coordinates for an object in free fall, because the world line of the faller changes orientation but the cone does not change. This means that the observer in free fall sees the clocks become desynchronized in his Lemaître reference frame. But this does not happen in reality.
On the other hand, in the Painlevé "reference frame", which is orthogonal according to Euclidean orthogonality, the clocks will not become desynchronized from the point of view of the faller, because the light cone tilts as the clocks's world lines change orientation.
So I see a contradiction between the Lemaître coordinates and Painlevé's. In the real world clocks do not go out of sync, so it seems that Lemaître's frame of reference does not correctly describe physical reality.

5 hours ago, studiot said:

Oh.

Marcus in particular and also Mordred like to deal in the abstract.

I like to ask the question "What does that posh statement mean, can i construct a simple down-to-earth example ?"

I offered you such but thank you for not wanting it.

I don't agree with this definition either, but it seems that this definition exists.

Edited by externo
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Posted (edited)

I think your missing an important detail the four velocity follows the curve Walds General Relativity book has an excellent section detailing that. Section 4.2 if you have a copy. ( if not I can post the relevant relations for you here)

One also has to be careful of which synchronization convention is being applied. You have proper velocity so have Lorentz invariance.

see section 9.2.1 onward

PS latex for this site is fairly easy just use \( prior with the same \ and close bracket at the end for inline.

for the line latex (autofills the entire line) its $same rule on the close bracket ] One thing I do agree on however is that the Minkoswkii metric is pseudo-Euclidean and not Euclidean. Edited by Mordred ##### Link to comment ##### Share on other sites Posted (edited) 18 hours ago, externo said: Minkowksi spacetime is tangent, in everything I write I speak of instantaneous tangent spaces. Ok, if you’re looking at a purely local (ie small and short enough) region, then you’ll have Minkowski spacetime, and only Minkowski spacetime - it makes no sense to refer to GP coordinates at all there. 18 hours ago, externo said: In relativity, if you change the reference frame, you must perform a boost. While curved spacetimes are everywhere locally Minkowskian, globally these frames at different events are related in non-trivial ways, so you need to account for this. 18 hours ago, externo said: Lemaître coordinates are: ds² = dT² - (Rs/r)dρ² They are obtained by performing Lorentz transformations on (dt, dr) The transformation between the Lemaître chart and the Schwarzschild chart isn’t a Lorentz transformation. 18 hours ago, externo said: The speed of light is thus isotropic for both (dt, dr) and (dT, dρ) No, it’s not. In Schwarzschild coordinates, the coordinate (!) speed of light goes to zero as the horizon is approached. In curved spacetimes you have to be very careful with coordinate and proper quantities, and their physical meanings. 18 hours ago, externo said: We can see that the Painlevé coordinate system is not a reference frame according to Einstein's relativity. GP coordinates naturally correspond to the frame of a test particle freely falling from far away towards a central mass in Schwarzschild spacetime; this is very much a valid frame, irrespective of which coordinates you choose to describe it in. Note also that orthogonality of coordinate axis is not a necessary condition for the existence of reference frames in GR; many useful metrics have off-diagonal components, another obvious example being the Kerr metric. 18 hours ago, externo said: As the radial coordinate is not the same, you can see that both cannot be reference frames according to relativity. That statement makes no sense. For example, if you’re standing still on the surface of the Earth, you can describe this in spherical coordinates, or you can choose to describe it in Cartesian coordinates centered on the Sun - all coordinate expressions are different here, but they still describe the same physical situation. 18 hours ago, externo said: Lemaître's coordinates predict that clocks accelerated by free fall desynchronize in the free fall reference frame while Painlevé's predict that they do not desynchronize. Both coordinate charts share the same time coordinate, with proper time between events just being the geometric length of the free fall world line. This is obviously different from the coordinate time of a clock left stationary far away - unsurprisingly so, since there’s no global notion of simultaneity in curved spacetimes. 18 hours ago, externo said: In order for the walls to remain null geodesics the simultaneity must be changed and dr is no longer the line of simultaneity for stationary objects. Of course, that’s the whole point of being in a curved spacetime - there’s no global concept of simultaneity. Time is a purely local concept here. 17 hours ago, externo said: This is not a Schwarzschild diagram Yes it is - it’s from visualrelativity.com, and originally taken from a Scientific American article by Roger Penrose. It’s important to remember though that the shape and orientation of the cones on the various diagrams you find in textbooks depend on your chosen coordinate system, since light cones are always drawn in a particular set of coordinates. It’s always possible to find coordinates where the cones don’t get rotated. PS. It seems to me that you’re forgetting that we’re in a curved spacetime here. Locally things are Minkowski, but the relationships between those local patches are non-trivial - you can’t just Lorentz-transform between distant frames, there’s no global notion of simultaneity, and the difference between coordinate and proper quantities is even more crucial than in SR. Like studiot has pointed out to you, a “reference frame” - and the relationships between them - are richer than just having a isolated set of coordinate axis. You might also wish to take a look at the tetrad formalism of GR, which builds on this fact. Edited by Markus Hanke ##### Link to comment ##### Share on other sites Posted (edited) 8 hours ago, Markus Hanke said: The transformation between the Lemaître chart and the Schwarzschild chart isn’t a Lorentz transformation. A local Lorentz boost is performed. We switch from the instantaneous basis (dt,dr) to (dT, dρ) by a hyperbolic rotation. 8 hours ago, Markus Hanke said: No, it’s not. In Schwarzschild coordinates, the coordinate (!) speed of light goes to zero as the horizon is approached. In curved spacetimes you have to be very careful with coordinate and proper quantities, and their physical meanings. The radial velocity of light in Schwarzschild and Lemaître coordinates is isotropic, meaning it descends as fast as it ascends. On the other hand, in Painlevé coordinates, it descends faster than it ascends. 8 hours ago, Markus Hanke said: Both coordinate charts share the same time coordinate, with proper time between events just being the geometric length of the free fall world line. This is obviously different from the coordinate time of a clock left stationary far away - unsurprisingly so, since there’s no global notion of simultaneity in curved spacetimes. You misunderstood. Lemaître's coordinates predict that free-falling observers will see free-falling clocks desynchronize while Painlevé's observers predict that free-falling clocks will not desynchronize. So they do not predict the same physics. See below. 8 hours ago, Markus Hanke said: Yes it is - it’s from visualrelativity.com, and originally taken from a Scientific American article by Roger Penrose. A Schwarzschild diagram is in Schwarzschild coordinates otherwise it is not a Schwarzschild diagram. 8 hours ago, Markus Hanke said: It’s important to remember though that the shape and orientation of the cones on the various diagrams you find in textbooks depend on your chosen coordinate system, since light cones are always drawn in a particular set of coordinates. It’s always possible to find coordinates where the cones don’t get rotated. PS. It seems to me that you’re forgetting that we’re in a curved spacetime here. Locally things are Minkowski, but the relationships between those local patches are non-trivial - you can’t just Lorentz-transform between distant frames, there’s no global notion of simultaneity, and the difference between coordinate and proper quantities is even more crucial than in SR. Like studiot has pointed out to you, a “reference frame” - and the relationships between them - are richer than just having a isolated set of coordinate axis. You might also wish to take a look at the tetrad formalism of GR, which builds on this fact. There's something important: in SR, the light cone never tilts, and when an object accelerates, the speed of light changes relative to it, implying that accelerated clocks will desynchronize for the accelerating observer comoving with them. The synchronization procedure must be reinitiated. However, from the moment the cone tilts simultaneously with the accelerating object, its clocks will not desynchronize in their accelerated reference frame. According to Schwarzschild coordinates, the cone does not tilt, and so the falling object undergoes kinematic time dilation, which implies that the falling clocks will desynchronize from the viewpoint of the faller. However, this does not happen in practice for a falling object. Schwarzschild coordinates therefore contradict the physical world. Edited by externo ##### Link to comment ##### Share on other sites Posted (edited) 16 hours ago, externo said: A local Lorentz boost is performed. We switch from the instantaneous basis (dt,dr) to (dT, dρ) by a hyperbolic rotation. The transformation between the charts is explicitly given here, under the “Metric” section. This is not a Lorentz transformation. 16 hours ago, externo said: The radial velocity of light in Schwarzschild and Lemaître coordinates is isotropic, meaning it descends as fast as it ascends. This is not the meaning of “isotropy”. 16 hours ago, externo said: Lemaître's coordinates predict that free-falling observers will see free-falling clocks desynchronize while Painlevé's observers predict that free-falling clocks will not desynchronize. You talk about synchronisation as if there was some meaningful notion of global simultaneity here. But there isn’t, because we’re in a curved spacetime. A clock stationary far away will never be synchronous with a clock in free fall towards the horizon, irrespective of coordinate choices. 16 hours ago, externo said: A Schwarzschild diagram is in Schwarzschild coordinates otherwise it is not a Schwarzschild diagram. A Schwarzschild spacetime diagram is a diagram of Schwarzschild spacetime - unsurprisingly. You are free to choose your coordinates as you wish, but it still remains Schwarzschild spacetime. If you draw the diagram in GP coordinates, the cones both rotate and distort; if you draw it in SS coordinates, the cones just become narrower, but don’t rotate. That’s a consequence of how these coordinate charts work, but you’re always in the same spacetime. 16 hours ago, externo said: when an object accelerates, the speed of light changes relative to it The locally measured speed of light is always c. It’s only the coordinate speed that will differ in (eg) Rindler coordinates - which is why I pointed out earlier that one must carefully distinguish between these. 16 hours ago, externo said: Schwarzschild coordinates therefore contradict the physical world. In curved spacetime, notions of space and time are purely local. Schwarzschild coordinates represent an observer who remains stationary far away from the central mass, and this coordinate system describes well the local physics associated with this observer. But the point is that they are only locally physical - if you try to use Schwarzschild coordinates to draw physical conclusions about distant frames (like eg a test particle in free fall), you’ll quickly run into problems. So I wouldn’t say they are unphysical, you just need to be very careful how you apply them in practice. In curved spacetimes, the difference between local and global is crucial. In particular, you can’t use Schwarzschild coordinate time to draw conclusions about what distant clocks record in their own frames; there’s simply no global notion of simultaneity here that can form a basis for this. —- Let’s return to your original claim that GP coordinates can’t be associated with a physically valid reference frame. I think we agree that the GP metric is a mathematically valid solution to the Einstein equations; if you disagree, it’s up to yourself to provide mathematical proof that it’s not. The question then is first and foremost what “reference frame” even means, mathematically speaking. The precise definition is given in (eg) Wu/Sachs, General Relativity for Mathematicians (1977), which is the one I’m using below: Suppose we are given a spacetime, being a semi-Riemannian manifold endowed with a metric and the Levi-Civita connection. An observer in that spacetime is then defined to be a future-oriented time-like curve that is everywhere smooth and differentiable. Finally, a reference frame is a vector field in that spacetime whose integral curves are observers. Straight away we notice that a reference frame isn’t the same as a coordinate chart. So what is a Gullstrand-Painlevé observer? It’s a free-fall geodesic of our spacetime (not necessarily purely radial) that connects an event far away to another event spatially closer to the central mass in a time-like manner, with the express boundary condition that at t=0 the observer be at rest. This geodesic gives our future-oriented time-like curve. So what is the vector field? It’s simply the 4-velocity field given by those very geodesics in spacetime. Recall that there’s no proper acceleration in free fall, thus (we’re in a curved spacetime, so covariant derivatives must be used): \[\frac{D^2x^{\mu}(\tau)}{d\tau^2}=0$

This system of equations, along with our boundary conditions, determines both the geodesic of our observer, and the associated 4-velocity field. But here’s the thing - we know that geodesics parallel-transport their own tangent vectors, and, since 4-velocity precisely is the tangent vector at every point of our motion, we are by the above equation already guaranteed that the geodesic is in fact an integral curve of its own 4-velocity field. This is hardly surprising!

Thus, the GP observer (more precisely - his 4-velocity field) does indeed constitute a valid reference frame.

If you still don’t agree, you need to show us explicitly and mathematically how a free-fall geodesic (which is what a GP observer is) is not in fact an integral curve of its own 4-velocity field.

Edited by Markus Hanke
Fixed mistake
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Posted (edited)

Nice post couldn't agree more +1

34 minutes ago, Markus Hanke said:

The question then is first and foremost what “reference frame” even means, mathematically speaking. The precise definition is given in (eg) Wu/Sachs, General Relativity for Mathematicians (1977), which is the one I’m using below:

haven't read that one I may pick it up

One detail to add though in PG coordinates we are using a new time coordinate defined by

tr=t+a(r)

related metrics here

Edit the equation Markus posted still applies. Even under that new time coordinate definition.

Edited by Mordred
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Posted (edited)
5 hours ago, Markus Hanke said:

The transformation between the charts is explicitly given here, under the “Metric” section. This is not a Lorentz transformation.

Here is the detail of Gemini 1.5 pro :

Quote

The Lemaître transformation can be locally approximated by a Lorentz transformation.

Here's a more detailed explanation :

Let's consider a point P in spacetime with Schwarzschild coordinates (t₀, r₀). We will approximate the Lemaître transformation in the vicinity of this point.

* Taylor Expansion: We expand the transformation equations in a Taylor series around r₀, keeping only the linear terms in (r - r₀):

* τ ≈ t + 2√(2GM/r₀) + (2GM/c²) ln | (√(r₀/2GM) + 1) / (√(r₀/2GM) - 1) |  + (r - r₀) / √(2GM r₀)
* ρ ≈ t + √(2GM r₀) + (r - r₀) √(GM/2r₀)

* Change of Variables: Introduce new coordinates (T, R) centered at P:

*  T = τ - [t₀ + 2√(2GM/r₀) + (2GM/c²) ln | (√(r₀/2GM) + 1) / (√(r₀/2GM) - 1) |]
*  R = (r - r₀) / √(2GM r₀)

* Linear Approximation: In these new coordinates, the transformation equations become:

* T ≈ t - t₀ + R
* ρ - [t₀ + √(2GM r₀)] ≈ t - t₀ + R/2

These equations resemble a Lorentz transformation for a boost in the radial direction with a velocity v = c/2. A Lorentz transformation for a boost in the x-direction is:

* t' = γ(t - vx/c²)
* x' = γ(x - vt)

where γ = 1/√(1-v²/c²) is the Lorentz factor.

Conclusion:

By performing a Taylor expansion around a point and judiciously choosing new coordinates, we have shown that the Lemaître transformation can be locally approximated by a Lorentz transformation. This confirms the equivalence principle and the possibility of defining a locally inertial frame at each point in spacetime, even near a black hole.

This approximation is valid only locally, in an infinitesimal neighborhood of point P. The boost velocity (c/2 in our example) depends on the chosen point P, and the approximation becomes less accurate as we move away from P and the curvature of spacetime becomes more significant.

5 hours ago, Markus Hanke said:

You talk about synchronisation as if there was some meaningful notion of global simultaneity here. But there isn’t, because we’re in a curved spacetime. A clock stationary far away will never be synchronous with a clock in free fall towards the horizon, irrespective of coordinate choices.

The clocks of Lemaître's observers share the same global simultaneity and do not become desynchronized with each other.

5 hours ago, Markus Hanke said:

A Schwarzschild spacetime diagram is a diagram of Schwarzschild spacetime - unsurprisingly. You are free to choose your coordinates as you wish, but it still remains Schwarzschild spacetime.

If you draw the diagram in GP coordinates, the cones both rotate and distort; if you draw it in SS coordinates, the cones just become narrower, but don’t rotate. That’s a consequence of how these coordinate charts work, but you’re always in the same spacetime.

Same spacetime, but not same physics.

5 hours ago, Markus Hanke said:

The locally measured speed of light is always c. It’s only the coordinate speed that will differ in (eg) Rindler coordinates - which is why I pointed out earlier that one must carefully distinguish between these.

When you speed up, your clocks get out of sync, the only physical explanation for this is a change in the speed of light. You can also call it a change in simultaneity, it doesn't change the fact that the speed of light in one direction is no longer the same as the other until you restart the synchronization procedure.

5 hours ago, Markus Hanke said:

Let’s return to your original claim that GP coordinates can’t be associated with a physically valid reference frame. I think we agree that the GP metric is a mathematically valid solution to the Einstein equations; if you disagree, it’s up to yourself to provide mathematical proof that it’s not.

The question then is first and foremost what “reference frame” even means, mathematically speaking. The precise definition is given in (eg) Wu/Sachs, General Relativity for Mathematicians (1977), which is the one I’m using below:

Suppose we are given a spacetime, being a semi-Riemannian manifold endowed with a metric and the Levi-Civita connection. An observer in that spacetime is then defined to be a future-oriented time-like curve that is everywhere smooth and differentiable. Finally, a reference frame is a vector field in that spacetime whose integral curves are observers.

Straight away we notice that a reference frame isn’t the same as a coordinate chart. So what is a Gullstrand-Painlevé observer? It’s a free-fall geodesic of our spacetime (not necessarily purely radial) that connects an event far away to another event spatially closer to the central mass in a time-like manner, with the express boundary condition that at t=0 the observer be at rest. This geodesic gives our future-oriented time-like curve.
So what is the vector field? It’s simply the 4-velocity field given by those very geodesics in spacetime. Recall that there’s no proper acceleration in free fall, thus (we’re in a curved spacetime, so covariant derivatives must be used):

D2xμ(τ)dτ2=0

This system of equations, along with our boundary conditions, determines both the geodesic of our observer, and the associated 4-velocity field. But here’s the thing - we know that geodesics parallel-transport their own tangent vectors, and, since 4-velocity precisely is the tangent vector at every point of our motion, we are by the above equation already guaranteed that the geodesic is in fact an integral curve of its own 4-velocity field. This is hardly surprising!

Thus, the GP observer (more precisely - his 4-velocity field) does indeed constitute a valid reference frame.

If you still don’t agree, you need to show us explicitly and mathematically how a free-fall geodesic (which is what a GP observer is) is not in fact an integral curve of its own 4-velocity field.

You give here for definition of a reference frame the same thing I gave above. 'Finally, a reference frame is a vector field in that spacetime whose integral curves are observers.' This definition is wrong. A reference frame requires a convention of simultaneity. If you do not give to the GP coordinates a simultaneity they do not form a reference frame. When one uses the GP coordinates the simultaneity used is that corresponding to dr + √(Rs/r)dT). The reference frame is thus formed by the vector fields dT and dr + √(Rs/r)dT

Now it turns out that this change of coordinate from (dt,dr) does not constitute a boost, unlike the (dT,dρ) of Lemaître. This is why I said that the Painlevé coordinates were not a reference frame according to relativity.

Now, here is: In the Schwarzschild coordinates the cone does not tilt thus the faller is not in inertia. There is a default. That contradicts the equivalence principle. The simultaneity of these coordinates is thus not correct. In the Painlevé coordinates, the cone tilts with the faller which is thus in inertia. Only the GP coordinates are conform to the equivalence principle. These coordinates are thus the physical reference frame. You say that the GP coordinates are better adapted, it is simply a euphemism for saying that they are physical."

Edited by externo

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