mersecske said: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 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 said: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 ].
DrGreg said: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.
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.mersecske said: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.
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...merseckske said: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?
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.mersecske said: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...
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 [tex](dT/d\tau , dR/d\tau, 0, 0)[/tex] but u also equals the sum [tex]dT/d\tau * e_t + dR/d\tau * e_r[/tex])mersecske said:Yes I know! Thats why I said "something is wrong"!
But what is wrong in my statements!
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 said:T and R are the Schwarzschild coordinates.
e_t and e_r are the unit vectors respectively
So if you're talking about the t-coordinate in Schwarzschild coordinates, then [tex]e_t[/tex] 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 [tex]e_t[/tex] should be parallel to the horizon.mersecske said:No, [tex]e_t[/tex] 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 [tex]e_t[/tex] is parallel to the t-axis.
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.mersecske said:On the Penrose diagram t=const curves are parallel to the r=2M horizon
JesseM said: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.
mersecske said: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 said: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.
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. [tex]u = (dT/d\tau, dR/d\tau, 0, 0)[/tex] 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?George Jones said:On the horizon, a vector orthogonal to the horizon also is parallel (tangent) to the horizon!
Gotcha, I was just thinking of the direction rather than the magnitude. OK, so suppose we consider the vector [tex]u' = (dT/d\tau, dR/d\tau, 0, 0) / | u |[/tex], where | u | is just the magnitude of the 4-velocity vector [tex]u = (dT/d\tau, dR/d\tau, 0, 0)[/tex] (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?DrGreg said:A null vector is orthogonal to itself!
The links I gave in post #6 gave specific equations--for example, see this link.mersecske said: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.