Can Every Spherically Symmetric Spacetime Be Represented by a Penrose Diagram?

[mersecske]

Can you recommend a good book or pdf file to learn how to draw Penrose diagrams in general cases? It is possible to draw "precise" well-defined Penrose diagrams for every spacetimes?

 

[Rasalhague]

Can you recommend a good book or pdf file to learn how to draw Penrose diagrams in general cases? It is possible to draw "precise" well-defined Penrose diagrams for every spacetimes?

I bought The Road to Reality earlier this year, partly to find out more about these. It's taken me a while to understand some aspects of the notation, but I think that's mainly because I've also been struggling to learn the concepts that it expresses. But looking at it recently after a long break, I've found some things starting to fall into place. It gives precise details, as far as I know, but I'm very much a beginner, and not really qualified to judge. The basic notation is introduced quite briefly but recurs throughout the book, so you get to see plenty of examples. Further details, such circles around tensors for their covariant derivatives are explained later along with the concepts they represent, so its not really aimed at someone who knows the concepts and is just looking for a guide to the notation, but still it shouldn't be too hard to track stuff down with the index. On the other hand, from my position of ignorance, I don't know all the things you might want to use them for, so I can't really say how complete the information is.

On a less formal level, LukeD started an interesting thread here in June introducing his own version of Penrose diagrams [ https://www.physicsforums.com/showthread.php?t=407776 ].

 

[DrGreg]

I bought The Road to Reality earlier this year, partly to find out more about these. It's taken me a while to understand some aspects of the notation, but I think that's mainly because I've also been struggling to learn the concepts that it expresses. But looking at it recently after a long break, I've found some things starting to fall into place. It gives preceise details, as far as I know, but I'm very much a beginner, and not really qualified to judge.

On a less formal level, LukeD started an interesting thread here in June introducing his own version of Penrose diagrams [ https://www.physicsforums.com/showthread.php?t=407776 ].

I don't think that's what mersecske meant. I think he/she is referring to conformal diagrams (p.723 in Road to Reality). I can't answer the original question, but the Wikipedia article gives some references that might be worth trying.

 

[Rasalhague]

I don't think that's what mersecske meant. I think he/she is referring to conformal diagrams (p.723 in Road to Reality). I can't answer the original question, but the Wikipedia article gives some references that might be worth trying.

Oh, I see. Sorry Mersecske. I should paid more attention to "for every spacetime".

 

[mersecske]

Time-like linear lines on the Carter-Penrose diagram of the extended Schwarzschild space-time has some special meaning? They are not geodetic curves, are they?

 

[JesseM]

At least in the case of a Penrose diagram of the Schwarzschild metric, there is an exact transformation from a Kruskal-Szekeres coordinates to a coordinate system where the diagram looks like a Penrose diagram:

http://casa.colorado.edu/~ajsh/schwp.html#penrose

This pdf also discusses the above in more detail, along with an explanation of the coordinate transformation to get a Penrose diagram from Minkowski spacetime:

http://www.ift.uni.wroc.pl/~blaschke/master/Felinska.pdf

And here's one that discusses some other cases like an Einstein static (closed) universe, a de Sitter spacetime, and a Reissner-Nordstrom charged black hole:

http://www.hep.phys.soton.ac.uk/~g.j.weatherill/Articles/Penrose.pdf

Couldn't find anything on the coordinate transformation for a rotating Kerr black hole, although a google books search turned up a page of this book which says that Penrose diagrams for Kerr black holes are discussed more in section 6.6 (not available for preview).

 

[mersecske]

If this is an answer to the "Penrose diagrams in general cases"
yes i can found description on special cases like Schwarzschild, de Sitter, Reissner-Nordstrom, and so on, but I would like some general description. Yes I know these are the most common spherically symmetric spacetimes, but it is possible to construct other ones by matching these once together trough (for example) a timelike hypersurface.

And my other question is independent from the previous one:
Time-like linear lines on the Carter-Penrose diagram of the extended Schwarzschild space-time has some special meaning? They are not geodetic curves, are they?

 

[JesseM]

If this is an answer to the "Penrose diagrams in general cases"
yes i can found description on special cases like Schwarzschild, de Sitter, Reissner-Nordstrom, and so on, but I would like some general description.

The two pdf files I linked to both seem to say that a Penrose/conformal diagram should have two basic features--1) that the worldlines of light beams (null geodesics) are represented as straight lines, usually at 45 degrees, so that the causal structure is obvious, and 2) that these geodesics all have finite length in the diagram, even if a "typical" coordinate representation would have the light beams traveling for an infinite amount of coordinate time (note that although light worldlines have finite length, and any region of spacetime separated from others by event horizons has a finite area, you can have spacetimes where the entire Penrose diagram is infinite and there exist timelike worldlines of infinite length, see the Penrose diagrams for the maximally extended Reissner-Nordstrom and Kerr spacetimes here). It may be that these are the only two basic requirements, and that any coordinate transformation satisfying them qualifies as a valid Penrose diagram, although I'm not sure about this.

And my other question is independent from the previous one:
Time-like linear lines on the Carter-Penrose diagram of the extended Schwarzschild space-time has some special meaning? They are not geodetic curves, are they?

I don't know the answer to that, but since the papers I've read mention that null geodesics should be straight lines but don't say anything about timelike worldlines, and since lines of constant position in Minkowski coordinates (which are timelike geodesics) look curved in the Penrose diagram for Minkowski spacetime, I would guess that straight timelike worldlines in a Penrose diagram don't need to be geodesics. Maybe someone else can confirm this though...

 

[arkajad]

See also my new paper: http://arxiv.org/abs/1008.4703"

 

[mersecske]

Let us suppose that a timelike worldline goes thru the Schwarzschild horizon.
Its 4-velocity (u) is (\dot{T},\dot{R},0,0), where \dot{} means d/dtau.
This means u = \dot{T} * e_t + \dot{R} * e_r
\dot{T} diverges at the horizon while \dot{R} stays finite.
But e_t is ortogonal to the horizon on the Penrose diagram,
therefore u has to be ortogonal to the horizon also,
which means that it has to be null-like. Something is wrong whith this...

 

[JesseM]

Let us suppose that a timelike worldline goes thru the Schwarzschild horizon.
Its 4-velocity (u) is (\dot{T},\dot{R},0,0), where \dot{} means d/dtau.
This means u = \dot{T} * e_t + \dot{R} * e_r
\dot{T} diverges at the horizon while \dot{R} stays finite.
But e_t is ortogonal to the horizon on the Penrose diagram,
therefore u has to be ortogonal to the horizon also,
which means that it has to be null-like. Something is wrong whith this...

A timelike worldline can't be orthogonal to the horizon on the Penrose diagram, since the horizon is inclined at 45 degrees from the vertical so any line orthogonal to it would be too, but only null worldlines can have an angle of 45 degrees from the vertical, and timelike worldlines must always be closer to vertical than that.

(incidentally, if you want LaTeX code to display you need to put "tex" in square brackets in front, and "/tex" in square brackets at the end of the equation)

 

[mersecske]

Yes I know! Thats why I said "something is wrong"!
But what is wrong in my statements!

 

[JesseM]

Yes I know! Thats why I said "something is wrong"!
But what is wrong in my statements!

OK, but it looked like you just asserted "e_t is orthogonal to the horizon" without explaining why. In general your notation isn't very clear to me--are T and R supposed to be time and radial coordinates in Schwarzschild coordinates, or in the coordinate system used to draw the Penrose diagram? (defined here in terms of a transformation from Kruskal-Szkeres coordinates, which are themselves usually defined in terms of a transformation from Schwarzschild coordinates) Likewise, what coordinate system is the 4-velocity u defined relative to? And what do e_t and e_r represent, just unit vectors in the T and R directions in the same coordinate system? (that's the only way I can make sense of the claim that u equals the vector (dT/d\tau , dR/d\tau, 0, 0) but u also equals the sum dT/d\tau * e_t + dR/d\tau * e_r)

 

[mersecske]

T and R are the Schwarzschild coordinates.
e_t and e_r are the unit vectors respectively
(orthogonal unit vectors to the T=const. and R=const. coordinate curves)
Close to the horizon T=const curves tends to R=2M line,
e_t is orthogonal to r=2M (I think), and e_r also
But now, I figure out the answer:
\dot{T} is infinity at the horizon,
but the length of e_t on the Penrose diagram goes to zero
Am I right?

 

[JesseM]

T and R are the Schwarzschild coordinates.
e_t and e_r are the unit vectors respectively

e_t is a unit vector in the direction of the Schwarzschild time coordinate, right? In that case it should be parallel to the horizon, not orthogonal. And likewise, your 4-velocity u would be in Schwarzschild coordinates--so even though u approaches being parallel to the horizon as an object approaches the horizon in Schwarzschild coordinates, the 4-velocity u' in the coordinates used for the Penrose diagram would not behave the same way, it would be neither parallel to nor orthogonal to the horizon.

 

[mersecske]

No, e_t is a unit vector in the tangent space belongs to the t-coordinate, which means it is orthogonal to the t=const. curves. For example in Minkowski space e_t is parallel to the t-axis.

 

[JesseM]

No, e_t is a unit vector in the tangent space belongs to the t-coordinate, which means it is orthogonal to the t=const. curves. For example in Minkowski space e_t is parallel to the t-axis.

So if you're talking about the t-coordinate in Schwarzschild coordinates, then e_t should be orthogonal to t=constant curves and parallel to the t-axis in Schwarzschild coordinates, right? In Schwarzschild coordinates the horizon is parallel to the t-axis, so e_t should be parallel to the horizon.

 

[mersecske]

On the Penrose diagram t=const curves are parallel to the r=2M horizon, so e_t is orthogonal to it, isn't it?

 

[George Jones]

What are the integral curves for \partial_t?

 

[JesseM]

On the Penrose diagram t=const curves are parallel to the r=2M horizon

No, a t=const curve would have constant t but varying r, while the horizon has constant r at all values of t, so the horizon is orthogonal to t=const curves and parallel to the t-axis in Schwarzschild coordinates.

 

[George Jones]

At any event p, \partial_t is tangent to the r=constant curve (parametrized by t) that passes through p.

 

[George Jones]

No, a t=const curve would have constant t but varying r, while the horizon has constant r at all values of t, so the horizon is orthogonal to t=const curves and parallel to the t-axis in Schwarzschild coordinates.

Hi Jesse. The comments that I have made are with respect to a diagram for Kruskal-Szekeres coordinates. For such a diagram, what are the possible values for Schwarzschild coordinates "on" the horizon"? I have used scare quotes, because Shwarzschild coordinates don't actually cover the horizon.

 

[mersecske]

Both the r=const and t=const curves are parallel to the horizon 'asymptotically' (or near) at the horizon, I think.

 

[mersecske]

Are there any freely available gnuplot script for generating Penrose-Carter diagram?
Or Maple, Mathematica?

 

[George Jones]

Let us suppose that a timelike worldline goes thru the Schwarzschild horizon.
Its 4-velocity (u) is (\dot{T},\dot{R},0,0), where \dot{} means d/dtau.
This means u = \dot{T} * e_t + \dot{R} * e_r
\dot{T} diverges at the horizon while \dot{R} stays finite.
But e_t is ortogonal to the horizon on the Penrose diagram,
therefore u has to be ortogonal to the horizon also,
which means that it has to be null-like. Something is wrong whith this...

Schwarzschild coordinates do not cover the horizon, and thus e_t is not defined on the horizon. In order to talk about a particle's 4-velocity on the horizon, a coordinate system that does cover the horizon should be used, e.g, Kruskal-Szekeres, Eddington-Finkelstein, Painleve-Gullstrand.

e_t is a unit vector in the direction of the Schwarzschild time coordinate, right? In that case it should be parallel to the horizon, not orthogonal.

On the horizon, a vector orthogonal to the horizon also is parallel (tangent) to the horizon!

 

[JesseM]

On the horizon, a vector orthogonal to the horizon also is parallel (tangent) to the horizon!

Do you mean "orthogonal to the horizon" in some sense other than Schwarzschild coordinates, like a vector orthogonal to the horizon in a locally inertial frame at the horizon? Because it seems to me that if you consider the 4-velocity in Schwarzschild coordinates of a particle approaching the horizon radially as mersecske was doing (i.e. u = (dT/d\tau, dR/d\tau, 0, 0) with T and R being Schwarzschild coordinates), and consider the limit of points on the particle's worldline closer and closer to the horizon, then in this limit the vector u approaches being perfectly parallel to the horizon, not orthogonal to it. Is that incorrect?

 

[DrGreg]

A null vector is orthogonal to itself!

 

[JesseM]

A null vector is orthogonal to itself!

Gotcha, I was just thinking of the direction rather than the magnitude. OK, so suppose we consider the vector u' = (dT/d\tau, dR/d\tau, 0, 0) / | u |, where | u | is just the magnitude of the 4-velocity vector u = (dT/d\tau, dR/d\tau, 0, 0) (and again T and R are in Schwarzschild coordinates). In this case for any nonzero u, u' will always be a unit vector. So if we consider u' at various points along the worldline of a particle falling towards the event horizon, in this case it would be true that in the limit as the T approaches infinity and R approaches 2GM, u' approaches being parallel to the horizon, right?

 

[mersecske]

Thanks your answers. I have another problem. What kind of coordinates are used for compactificated Penrose diagram (V,U)? I know that they are compactificated Kruskal coordinates (v,u). But what is the exact formula? I've tried V=tanh(v) and U=tanh(u), but this is not the same. r=0 is not a horizontal line (its curved), and r=+infinity is a vertical line instead of ">" shape. I also tried arctan, atanh.

 

[JesseM]

Thanks your answers. I have another problem. What kind of coordinates are used for compactificated Penrose diagram (V,U)? I know that they are compactificated Kruskal coordinates (v,u). But what is the exact formula? I've tried V=tanh(v) and U=tanh(u), but this is not the same. r=0 is not a horizontal line (its curved), and r=+infinity is a vertical line instead of ">" shape. I also tried arctan, atanh.

The links I gave in post #6 gave specific equations--for example, see this link.

 

[mersecske]

Thanks.

If Penrose-Carter diagram means:

1) compactificated space-time diagram and
2) null-like world lines are 45 degree lines

than, every spherically symmetric space-time has Carer-Penrose diagram?
If Yes, can you cite a paper about proving the existence?

Else:
on the 2D Carter-Penrose diagram only radial motion can be studied,
what about 3D=2+1 Penrose-Carter diagrams? Can you draw it?
The 2D diagram is simply rotated?

 

[JesseM]

Thanks.

If Penrose-Carter diagram means:

1) compactificated space-time diagram and
2) null-like world lines are 45 degree lines

than, every spherically symmetric space-time has Carer-Penrose diagram?

Yes, I found a reference confirming this--see p. 48 of 'Gravitational Collapse and Spacetime Singularities on google books, in the paragraph right before section 2.7.4 the author says:

The structure of infinity for any spherically symmetric spacetime can be depicted by a similar Penrose diagram.

Likewise, p. 20 of http://www.mittag-leffler.se/preprints/0809f/files/IML-0809f-14.pdf talks about how to define the black hole interior region as a complement of the exterior region described in terms of a Penrose diagram, and says:

For spherically symmetric spacetimes arising as solutions of the Cauchy problem for (2), one can show that there always exists a Penrose diagram, and thus, a definition can be formalised along precisely these lines (see [60]).

Here reference [60] is M. Dafermos http://iopscience.iop.org/0264-9381/22/11/019/

If Yes, can you cite a paper about proving the existence?

I think the Dafermos paper above probably proves it, although understanding it would probably require more knowledge of topological reasoning in GR than I have...

on the 2D Carter-Penrose diagram only radial motion can be studied,
what about 3D=2+1 Penrose-Carter diagrams? Can you draw it?
The 2D diagram is simply rotated?

No, it can't be a simple rotation, that would imply that in the maximally extended Schwarzschild solution one could travel around in a circle from the exterior region I on the right to the other exterior region III on the left of the diagram, which is supposed to represent a "different universe"! In fact I would guess (though I don't know this for a fact) that if Penrose diagrams are specific to spherically symmetric spacetimes, then they would always be two dimensional, with the radial dimension shown and the angular coordinates suppressed. It may help to think in terms of the fact that any if you take any spacelike surface through a Penrose diagram (or a Kruskal-Szkeres diagram, which looks basically identical except that it doesn't compress the spacetime to a finite size), which will just be any line closer to the horizontal than 45 degrees, there will be an http://www.bun.kyoto-u.ac.jp/~suchii/embed.diag.html showing the curvature of space in that spacelike surface. This embedding diagram will depict one of the two angular dimensions in addition to the radial dimension depicted on the Penrose/Kruskal-Szekeres diagram. For example, p. 528 of Gravitation by Misner/Thorne/Wheeler shows embedding diagrams for various spacelike slices through the Kruskal-Szekeres diagram:

Each point on a Penrose/Kruskal-Szekeres diagram actually represents a spherical region of space at constant radius (with both angular coordinates allowed to vary), so in the embedding diagrams each point on the spacelike surface becomes a circle (since the embedding diagram only shows 2 of the 3 spatial dimensions)--you can see that a cross-section of any of the above embedding diagrams would be a circle (several of these embedding diagrams illustrate a Schwarzschild wormhole or 'Einstein-Rosen bridge', see also the animations in the section of http://casa.colorado.edu/~ajsh/schww.html titled 'Instability of the Schwarzschild Wormhole'). Here, on the other hand, you can see on the right a Kruskal-Szekeres diagram for a more realistic black hole that forms from a collapsing star, with no white hole interior region or second exterior region on the left:

And here was my guess about what embedding diagrams for spacelike slices through this diagram would look like: