# Penrose diagrams in general

1. Aug 16, 2010

### 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?

2. Aug 16, 2010

### Rasalhague

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 ].

Last edited: Aug 16, 2010
3. Aug 16, 2010

### DrGreg

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.

Last edited by a moderator: Apr 25, 2017
4. Aug 16, 2010

### Rasalhague

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

5. Aug 28, 2010

### 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?

6. Aug 28, 2010

### 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:

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 [Broken]

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).

Last edited by a moderator: May 4, 2017
7. Aug 28, 2010

### 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?

8. Aug 28, 2010

### JesseM

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.
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...

Last edited: Aug 28, 2010
9. Aug 30, 2010

Last edited by a moderator: May 4, 2017
10. Aug 30, 2010

### 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...

11. Aug 30, 2010

### JesseM

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)

12. Aug 30, 2010

### mersecske

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

13. Aug 30, 2010

### JesseM

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 $$dT/d\tau * e_t + dR/d\tau * e_r$$)

14. Aug 31, 2010

### 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?

15. Aug 31, 2010

### JesseM

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.

16. Sep 1, 2010

### 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 paralell to the t-axis.

17. Sep 1, 2010

### JesseM

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.

18. Sep 1, 2010

### 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?

19. Sep 1, 2010

### George Jones

Staff Emeritus
What are the integral curves for $\partial_t$?

20. Sep 1, 2010

### JesseM

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.