Schwartzschild exterior and interior solutions

[PAllen]

1) This is your misunderstanding, all the time I've just pointed to an existing difference between 2 types of definition of AF

But both definitions define an intrinsic feature of geometry. Whether they allow exactly the same geometries, I don't know, but they qualify geometries, not coordinate systems. The coordinate definition can only be applied in certain types of coordinates, but if it is true, the geometric fact is true no matter what coordinate transforms are used.

2)this also misinterprets what I've been saying, changing coordinates can only change the geometry if the change introduces some further modification such as a conformal transformation, and I cite from a current text-book: General relativity from Hobson et al. page 51: "A conformal transformation is not a change of coordinates but an actual change in the geometry of a manifold" I've maintained that the introduction of the tortoise coordinate r to get Kruskal line element is equivalent to a conformal transformation of the Scwartzschild line element that is actually possible given the general covariance of the GR equations, but that doesn't fulfill coordinate-dependent AF. It is known that the Einstein field equations per se, without further conditions allow many different geometries, as the collection of cosmological solutions that historically have been derived from them since 1915 makes evident.

The transform to introduce Kruskal is not a conformal transform. It does not change geometry. You can choose to extend the geometry, or not, but the transform itself is just a coordinate transform. You have not provided any reference or argument that it is a conformal transform. References you have provided discuss applying a conformal tansform *after* the coordinate transform to produce e.g. Penrose-Carter diagrams.

4) coordinate singularities might or might not be significant, I thinks this is the common understanding.

6)I don't know what exactly you are calling r and R here.

R is the event horizon radius, r the Schwazschild coordinate for this geometry.

 

[JesseM]

When you say:"Boundary conditions are normally understood to have some physical meaning" , that perfectly summarizes my point in this thread.

Also, just to be clear, do you agree that one version of ingoing Eddington-Finkelstein coordinates does not match your definition of "coordinate-dependent AF" while the other does? If so, do you actually question whether the two are physically equivalent in some sense, or do you agree this is just a coordinate issue with no physical implications?

Yes, and I really don't know if it is just a coordinate issue or it has physical meaning.

It seems to me this may just be a semantic issue where the answer would just depend on how you chose to define "physical meaning". In non-quantum relativistic theories I think of the "physical truth" as just being frame-independent facts about each point in spacetime, like what two clocks read at the moment they pass each other, or what readings show on a particular measuring-instrument after it takes a reading, so two coordinate systems with line elements would be "physically equivalent" if they predicted exactly the same set of local facts of this type. Originally I thought your point was that you were suspicious that spacetime actually continued past the r=2GM boundary of exterior Schwarzschild coordinates (as suggested by the title you chose for the thread), but if you have exactly the same objections to viewing the two forms of ingoing Eddington-Finkelstein coordinates as "physically equivalent" then I really don't know what you mean by that term, since both forms of the coordinate system cover exactly the same region of spacetime, and they don't disagree about coordinate singularities.

If we lived in the spacetime described by the second form of EF coordinates at the bottom of the wiki article (the one whose line element approaches the Minkowski line element as r approaches infinity), it would be a trivial matter to just relabel each physical event with the coordinates of the first form of EF coordinates (the one whose line element has the extra 2dvdr in its line element so it doesn't approach the Minkowski line element), obviously a simply relabeling would not change the local facts about what occurs at any point in the spacetime. So in what sense are the two forms of EF not "physically equivalent", how can they have a different "physical meaning"? Can you provide clear definitions of what you mean by these terms? And if you can, do you think there is any compelling reason why physicists "should" adopt your definitions (as opposed to my definition of 'physically equivalent' above, for example), or would you agree that this is a basically arbitrary choice of which definitions we find more aesthetically appealing and so the whole thing boils down to semantics?

 

[TrickyDicky]

The transform to introduce Kruskal is not a conformal transform. It does not change geometry. You can choose to extend the geometry, or not, but the transform itself is just a coordinate transform.

Ok, let's heuristically investigate this:
For radially moving null-geodetic light ray spherical coordinates theta and phi are constant, so their differentials vanish, if we consider this in the Kruskal line element, we have a conformal scaling factor multiplying a 2 dimensional flat space, this doesn't happen in the original Schwartzschild metric, and actually leads to:considering the Kruskal spacetime in the vanishing angular components derivatives case as conformally flat spacetime for light null geodesics.
Would there be thefore in this particular case no Weyl curvature and no bending of light possible??no vacuum solution actually??

 

[PAllen]

Ok, let's heuristically investigate this:
For radially moving null-geodetic light ray spherical coordinates theta and phi are constant, so their differentials vanish, if we consider this in the Kruskal line element, we have a conformal scaling factor multiplying a 2 dimensional flat space, this doesn't happen in the original Schwartzschild metric, and actually leads to:considering the Kruskal spacetime in the vanishing angular components derivatives case as conformally flat spacetime for light null geodesics.
Would there be thefore in this particular case no Weyl curvature and no bending of light possible??no vacuum solution actually??

I don't follow much of what you suggest. What I see is that if I consider a purely radial slice, light paths have the coordinate form U=V or U=-V (plus displacements). These, of course, have zero interval along them. Further, if I compute ds along any radial timelike path, I get the same interval as using Schwarzschild coordinates. Similarly for spacelike paths. This can be said to prove they are the same geometry.

I don't follow your analysis of curvature. The U,V plane with angles constant is not flat because r is complicated function U and V. When computing metric differentials to define curvature, you get non-trivial terms from the derivatives of r by U and V.

To analyze light bending, there is no escaping using varying angles, so I don't see what you can discern without analyzing using the complete line element.

[EDIt: To clarify how wrong it is to assume that a the U,V plane metric is flat, consider 2-d Euclidean signature geometry. I define a line element:

ds^2 = f(x,y) ( dx^2 + dy^2)

Suppose f(x,y) is coordinate distance from a circle of radius 5. Then the circumference of coordinate circle of radius 5 is zero, while its radius is not zero. On the other hand, it can be seen that this metric is asymptotically Euclidean at infinity.
]

 

[TrickyDicky]

Take a look at this link (the first 3 pages,especially equation (9) and the next paragraph, and you'll understand better what I wrote in a hurry.
http://www.ast.cam.ac.uk/teaching/undergrad/partii/handouts/GR_14_09.pdf

 

[PAllen]

I found this page (don't know if anyone gave it earlier in this thread). It derives Kruskal directly from EFE, without starting from Schwarzschild. It also shows (if your read it all through) that the curvature of Kruskal does not vanish. But to see this, you must read all the way through because the first 2/3 end up deriving 'Kruskal like' coordinates for flat spacetime (e.g. they answer the question I raised in answer to Jessem, of what limiting metric would you use if you wanted to define a coordinatea criteria for AF directly in Kruskal coordinates; that is, you would ask if the Kruskal metric approaches the flat form as r(U,V)->infinity).

http://www.mathpages.com/home/kmath655/kmath655.htm

Note that this and most derivations of this geometry don't explicitly assert *any* boundary conditions at infinity. This is because spherical symmetry + vacuum solultion imply asymptotic flatness as noted here:

http://staff.science.uva.nl/~jpschaar/report/node5.html

 

[TrickyDicky]

Wow, that's sync, at the same minute!

 

[TrickyDicky]

I found this page

The link?

Note that this and most derivations of this geometry don't explicitly assert *any* boundary conditions at infinity. This is because spherical symmetry + vacuum solultion imply asymptotic flatness as noted here:

http://staff.science.uva.nl/~jpschaar/report/node5.html

Sure, if the two conditions are met, there is AF.

 

[PAllen]

Take a look at this link (the first 3 pages,especially equation (9) and the next paragraph, and you'll understand better what I wrote in a hurry.
http://www.ast.cam.ac.uk/teaching/undergrad/partii/handouts/GR_14_09.pdf

Ok, fine, so they define a concept of a 2 space that has curvature but also has a conformally flat metric by their definition. So what? There is not the slightest hint of a claim or argument that any of the transforms they have done change geometry. All they are saying is the metric takes this form in these coordinates. And...

Note, especially, they say things like:

"We can find lots of different coordinate transformations that preserve the conformal nature
of the 2-space defined by equation (5)."

This is just a nice metric form that can be achieved by coordinate transforms. Reading any of this to say these coordinate transforms change the geometry is a false reading of this material.

 

[PAllen]

The link?

Sorry, forgot the link, it's there now.

 

[PAllen]

As I understood the references you gave before about a type of conformal transform that changed geometry, the meaning was completely differfent: these were not just coordinate transforms, they introduced 'points at infinity' which certainly does change the geometry.

 

[TrickyDicky]

Ok, fine, so they define a concept of a 2 space that has curvature but also has a conformally flat metric by their definition. So what? There is not the slightest hint of a claim or argument that any of the transforms they have done change geometry. All they are saying is the metric takes this form in these coordinates. And...

Note, especially, they say things like:

"We can find lots of different coordinate transformations that preserve the conformal nature
of the 2-space defined by equation (5)."

This is just a nice metric form that can be achieved by coordinate transforms. Reading any of this to say these coordinate transforms change the geometry is a false reading of this material.

Let's not jump into conclusions yet (me included), do you agree that the line element in (9), when considering only null geodesics (light), in other words considering the angular components of the metric constant, is that of a conformally flat spacetime?

Conformally flat spacetimes have a vanishing Weyl curvature, any vacuum solution is incompatible with a vanishing Weyl curvature.

If so (and I insist considering a radiation only situation), how could this line element be a vacuum solution for light?

 

[PAllen]

Let's not jump into conclusions yet (me included), do you agree that the line element in (9), when considering only null geodesics (light), in other words considering the angular components of the metric constant, is that of a conformally flat spacetime?

Conformally flat spacetimes have a vanishing Weyl curvature, any vacuum solution is incompatible with a vanishing Weyl curvature.

If so (and I insist considering a radiation only situation), how could this line element be a vacuum solution for light?

I am not familiar with the formal definition and properties of conformal flatness. Your reference defines that this metric is conformally flat, noting carefully, that it still has curvature using the precisely the same argument about derivatives that I gave in my post #94.

I am not very familiar with the Weyl curvature tensor. A quick look up in Wikipedia suggests that in dimension >=4, vanishing Weyl tensor is necessary and sufficient for conformal flatness. However, we are talking about conformal flatness of a 2-d slice of a manifold. The same Wiki article (on Weyl tensor) observes that *any* 2-d manifold is conformally flat, so all your other conclusions are unfounded. That is, I conclude from all this that you can take any 2-d slice you want of any 4-d manifold, and you can find coordinates that express the metric in a conformally flat form on that slice.

Obviously, the conclusion that KS can't be a vacuum solution is wrong, as I gave you a reference to a complete derivation of it directly from the vacuum field equations.

 

[TrickyDicky]

I am not familiar with the formal definition and properties of conformal flatness.
I am not very familiar with the Weyl curvature tensor.

That says a lot.

Obviously, the conclusion that KS can't be a vacuum solution is wrong, as I gave you a reference to a complete derivation of it directly from the vacuum field equations.

Your reference doesn't even derive the conformal nature of the line element, is put by hand in the very first mathematical expression by saying the metric must have a diagonal form.

 

[George Jones]

Let's not jump into conclusions yet (me included), do you agree that the line element in (9), when considering only null geodesics (light), in other words considering the angular components of the metric constant, is that of a conformally flat spacetime?

No, (9) does not give a conformally flat spacetime. Why should the angular components be constant for null geodesics?

 

[PAllen]

That says a lot.

Yes, but so what? None of this is really relevant to questions of whether a coordinate transform can change geometry (in which case almost all books on GR are wrong), or what it means to apply the coordinae definition of AF.

Be that as it may, I looked information up and found, I think, a key point.

Your reference doesn't even derive the conformal nature of the line element, is put by hand in the very first mathematical expression by saying the metric must have a diagonal form.

So what? The point is that it verifies that the KS metric form is vacuum solution of the field equations, and can be directly derived from them. Putting the metric in a desired general form and then seeing if you can find a solution of that form is a standard GR technique. Further, given my finding on conformal flatness of any 2-d manifold, I think it follows that putting the metric in this form is not restrictive of possible geometries of the solutions.

 

[TrickyDicky]

No, (9) does not give a conformally flat spacetime. Why should the angular components be constant for null geodesics?

I'm considering a specific case, might look unnatural but is the case used by GR textbooks to introduce the concept of conformal flatness and it is the line element structure that appears when the EF coordinates and the tortoise r coordinate are introduced in the Schwartzschild metric:
From Hobson GRtext: "The form of the line element (9)has an important consequence for studying the paths of radially moving photons (for which dphi=dtheta=0). Since the conformal factor that multiplies (dv^2−du^2) is just a scaling, it does not change the lightcone structure and so the latter should just look like that in Minkowski space."
Further every diagram to understand the Kruskal space obviates the angular components so that every point in the diagram represents a 2-sphere.
To derive the EF coordinates p and q (which are just the integration constants of the null geodesics in Schwartzschild metric)from which later the tortoise coordinate make use to derive the Kruskal line element, only null geodesics are used, the advanced or ingoing photon and the retarded or outgoing photon.

So in this context, considering only radially moving light, I guess it is fine to make this simplification in which the (9) is conformally flat for radiation.

 

[TrickyDicky]

Not sure if absence of rebuttal to my last post means compliance or lack of interest in answering, which would be kind of odd in a forum that is all about debating and learning.
To be clear, I'm not saying the Kruskal line element is conformally flat per se, only that when the simplification made by textbooks (see also Ryder General relativity chapter7 on Kruskal coordinates) to introduce the change of coordinates of the Scwartzschild line element, namely that for radially ingoing and outgoing null geodesics we can neglect the angular components.
If that simplification is valid to introduce the new coordinates, it must be valid to say that using that same working assumption the Kruskal line element is conformally flat with respect to radial radiation, and since the solution should be spherically symmetric wrt any radiation.
But then how can we deal with light bending effects,etc?

 

[TrickyDicky]

An alternative way to view this conformal flatness is to remember that in the Kruskal spacetime describing the eternal symmeric black hole, hypersurfaces of constant time t are straight lines through the origin.
To better understand this t=constant hypersurfaces we make use of the isotropic transformation of coordinates from the Schwartzschild metric where it is easy to see that the metric in isotropic coordinates is conformally flat for the t= constant hypersurfaces.
And being the Kruskal an static spacetime and therefore invarian for t, wouldn't be valid to consider the manifold conformally flat?
Can someone help with this?

 

[PAllen]

Not sure if absence of rebuttal to my last post means compliance or lack of interest in answering, which would be kind of odd in a forum that is all about debating and learning.
To be clear, I'm not saying the Kruskal line element is conformally flat per se, only that when the simplification made by textbooks (see also Ryder General relativity chapter7 on Kruskal coordinates) to introduce the change of coordinates of the Scwartzschild line element, namely that for radially ingoing and outgoing null geodesics we can neglect the angular components.
If that simplification is valid to introduce the new coordinates, it must be valid to say that using that same working assumption the Kruskal line element is conformally flat with respect to radial radiation, and since the solution should be spherically symmetric wrt any radiation.
But then how can we deal with light bending effects,etc?

No, it means I think I have already answered all your objections and there was nothing more to say since you haven't provided any new substantive responses. The observations about the ability to make any chosen 2-d slice have conformally flat metric form with no loss of generality is fundamental, and all other points have been answered previously.

As to your new point above, your difficulty seems hard to understand. Radial light rays don't bend. Off radial light rays bend. The former can be treated ignoring theta and phi, the latter cannot. What's the issue here?

Maybe you are thinking that conformal flatness of metric form in two coordinates of a spherically symmetric solution means the whole metric is conformally flat? This is simply wrong from what I read about conformal flatness. Even intuitively, all you can conclude is that radial rays from any direction go straight through to the singularity; this says nothing about off radial rays, that start out with a direction involving changing theta/phi.

 

[PAllen]

An alternative way to view this conformal flatness is to remember that in the Kruskal spacetime describing the eternal symmeric black hole, hypersurfaces of constant time t are straight lines through the origin.
To better understand this t=constant hypersurfaces we make use of the isotropic transformation of coordinates from the Schwartzschild metric where it is easy to see that the metric in isotropic coordinates is conformally flat for the t= constant hypersurfaces.
And being the Kruskal an static spacetime and therefore invarian for t, wouldn't be valid to consider the manifold conformally flat?
Can someone help with this?

I'll try one more time. Since any 2-d manifold is conformally flat (see the Wikipedia article on the Weyl tensor), and this can be made manifest by change of coordinates, the it follows that a coordinate change (with no loss of generality) can give a conformally flat metric form to the family of 2-d slices obtained by holding two coordinates of a 4-d metric constant. The ability to do this says nothing about the conformal flatness of the 4-d manifold.

This is similar to the fact that lines have no intrinsic curvature, so all lines embedded in a 2-manifold have no intrinsic curvature. Yet that says nothing about the curvature of the 2-manifold.

Note also that the reference you gave for conformal transforms that changed geometry were talking about something completely different: a mapping *plus* adding points at infinity. It is specifically adding points at inifinity that change geometry. Schwarzschild to Kruskal is just a coordinte change; further, you can arrive at Kruskal directly, as in the reference I gave. As to your objections to that derivation, I already answered them.

 

[TrickyDicky]

I'll try one more time. Since any 2-d manifold is conformally flat

My last example is of 3-d hypersurface.

 

[TrickyDicky]

Radial light rays don't bend. Off radial light rays bend. The former can be treated ignoring theta and phi, the latter cannot.

Correct, all I'm saying is the change of cordinates is done under the assumption that we can ignore theta and phi, so it only refers to radial light rays, it can't be applied to off radial light rays.

 

[TrickyDicky]

further, you can arrive at Kruskal directly, as in the reference I gave.

That derivation is anything but a rigorous mathematical proof, the quantity of unwarranted assumptions thru the long and winding path it takes to the KS line element can be used to demonstrate just about anything.

 

[PAllen]

My last example is of 3-d hypersurface.

Then I am confused. The V=constant hypersurfaces of KS coords are certainly not conformally flat in metric form (and obviously in geometry). If you are claiming that a surface of constant Schwarzschild t is conformally flat in KS cordinates, I don't see this either. Constant t is U/V constant, so expressing one in terms of the other to get a 3-metric form, the result is not at all conformally flat.

Ok, I think you are referring to istotropic Schwarzschild coordinates, where t=constant hypersurface is conformally flat. That is interesting. However, I don't think it is true that the existence of conformally flat embedded 3-surfaces implies anything at all about the geometry of the 4-manifold. I can fill Euclidean 3-space with conscentric embedded 2-spheres. Does that make the 3-space spherical in geometry? Obviously not. I think what is going on here is equivalent. In general, you can't conclude very much about the geometry of a manifold from the geometry of embedded lower dimensional manifolds.

 

[TrickyDicky]

Ok, I think you are referring to istotropic Schwarzschild coordinates, where t=constant hypersurface is conformally flat. That is interesting. However, I don't think it is true that the existence of conformally flat embedded 3-surfaces implies anything at all about the geometry of the 4-manifold. I can fill Euclidean 3-space with conscentric embedded 2-spheres. Does that make the 3-space spherical in geometry? Obviously not. I think what is going on here is equivalent. In general, you can't conclude very much about the geometry of a manifold from the geometry of embedded lower dimensional manifolds.

Only that is not what I'm saying as I specifically stressed in my last posts, I'm not claiming the KS is conformally flat, only that if we maintain the assumptions used to introduce the new coordinates, that is that we can neglect angular components, but then if you consider spherical symmetry, all radial UV planes are included and the assumption extends to all radial light.

In the case of the Schwarzschild metric in isotropic coordinates, the fact that the solution must be static implies that all 3d hypersurfaces are t=constant and thus the whole manifold must be conformally flat.

 

[PAllen]

Only that is not what I'm saying as I specifically stressed in my last posts, I'm not claiming the KS is conformally flat, only that if we maintain the assumptions used to introduce the new coordinates, that is that we can neglect angular components, but then if you consider spherical symmetry, all radial UV planes are included and the assumption extends to all radial light.

I don't understand what you are getting at. What is it, exactly, you are saying about radial light? And what are you claiming that might imply about a claim that KS coordinates are geometrically different fomr Schwarzschild when restricted to the regions covered by the latter (say, regions I and II)?

In the case of the Schwarzschild metric in isotropic coordinates, the fact that the solution must be static implies that all 3d hypersurfaces are t=constant and thus the whole manifold must be conformally flat.

I don't see this at all. I see that all t=constant 3-surfaces in these coordinates are conformally flat. That clearly does not say the 4-manifold is conformally flat, because if you compute the Weyl tensor in these coordinates it does not vanish (obviously, it is actually sufficient to compute the Weyl tensor in any coordinates, because its vanishing / non-vanishing character is invariant).

 

[TrickyDicky]

I don't see this at all. I see that all t=constant 3-surfaces in these coordinates are conformally flat. That clearly does not say the 4-manifold is conformally flat

You seem a bright chap so I'm sure you'll understand this last reflection:
In isotropic coordinates, the Schwarzschild metric time 3d-slices (t constant) are conformal to 3d-Euclidean space, right so far?
The Schwarzschild 4d-manifold in isotropic coordinates must also be a static spacetime to be a vacuum solution so it must be time-invariant, right so far?
If a 4d manifold that is time invariant has time slices that are conformally flat, does that tell us anything about the conformally flat nature of the 4d-manifold?
That is my question.

 

[PAllen]

You seem a bright chap so I'm sure you'll understand this last reflection:
In isotropic coordinates, the Schwarzschild metric time 3d-slices (t constant) are conformal to 3d-Euclidean space, right so far?
The Schwarzschild 4d-manifold in isotropic coordinates must also be a static spacetime to be a vacuum solution so it must be time-invariant, right so far?
If a 4d manifold that is time invariant has time slices that are conformally flat, does that tell us anything about the conformally flat nature of the 4d-manifold?
That is my question.

I I've answered it several times, to my satisfaction but apperently not yours.

1) compute Weyl tensor in these coordinates; it does not vanish; therefore, by established theorems, the conclusion does not follow. That is really all that needs to be said. You have an intuition about what 'ought to be true'; it is wrong. I've been there countless times in my life. I respond by educating my intuition when it conflicts with proofs, or measurements. (In fact, as I freely admitted, I was not familiar with conformal flatness theory and the properties of the Weyl tensor before you raised them; so I educated myself and read about them).

2) Intuitive explanation: filling a manifold with embedded lower dimensional manifolds does not allow you to claim the geometry of lower dimensional manifolds carries over the the higher dimensional one. My example with concentric 2-spheres in flat euclidean 3-space makes the conclusion in (1) seem *not* counter-intuitive.

 

[TrickyDicky]

I I've answered it several times, to my satisfaction but apperently not yours.

1) compute Weyl tensor in these coordinates; it does not vanish; therefore, by established theorems, the conclusion does not follow. That is really all that needs to be said. You have an intuition about what 'ought to be true'; it is wrong. I've been there countless times in my life. I respond by educating my intuition when it conflicts with proofs, or measurements. (In fact, as I freely admitted, I was not familiar with conformal flatness theory and the properties of the Weyl tensor before you raised them; so I educated myself and read about them).

Fine, would you be so kind to show me the non-zero components of the weyl tensor in terms of the isotropic coordinates? I assume you've done or have access to that computation. It'd help me a lot.

2) Intuitive explanation: filling a manifold with embedded lower dimensional manifolds does not allow you to claim the geometry of lower dimensional manifolds carries over the the higher dimensional one. My example with concentric 2-spheres in flat euclidean 3-space makes the conclusion in (1) seem *not* counter-intuitive.

Your example of the concentric spheres in a 3-space is silly in this context, and has nothing to do with what I'm saying about the Schwarzschild manifold in isotropic coordinates.
First of all Euclidean 3-space is not invariant for any of their 3 components. While Schwarzschild spacetime is static and therefore invarian with respect to one of its 4 components.
The fact that a flat 3-space allows curved surfaces to exist,and is not itself curved is trivial but in fact you cannot have an x constant slice of 3-d euclidean that defines a curved surface because an x constant slice is a 2d-flat plane. But in my example you can have a time constant 3d-slice of the 4d-manifold that is conformally flat.