Choice of scaling function for Penrose diagrams

[bcrowell]

The standard definition of coordinates on Penrose diagrams seems to be something like $\tan(u\pm v)=x\pm t$. This is what Wikipedia gives, and Hawking and Ellis also give a transformation involving a tangent function, although I haven't checked whether the factors of 2, etc. agree. Neither source comments on why a tangent function is used. It's clear that if the transformation is going to be of the form $f(u\pm v)=x\pm t$, then f has to be a homeomorphism from a finite, open interval of the reals onto the whole real line. But it seems to me that we could just as well have used $f=\tan^3$, or $f(x)=-x/[(x-1)(x+1)]$. Is there anything about the tangent function that makes it especially desirable? Any transformation of the form $f(u\pm v)=x\pm t$ will preserve the shape of light-cones, since it sends curves x-t=const to curves u-v=const, and similarly for x+t and u+v. I believe that f=tan makes particles at rest have world-lines that look like hyperbolas, but is there some special reason that a hyperbola is a desirable result? World-lines of moving particles are funky S-shapes.

 

[arkajad]

Tangent function comes in naturally from the cylindrical embedding. See e.g. Section 4 of http://arxiv.org/abs/1008.4703" .

 

[bcrowell]

Tangent function comes in naturally from the cylindrical embedding. See e.g. Section 4 of http://arxiv.org/abs/1008.4703" .

Ah, thanks for the reference. I think I see the idea. You can make the static Einstein universe by tiling a cylinder with Penrose diagrams, so then it becomes natural to use the angle on the cylinder as a coordinate. So if I'm understanding correctly, the use of f=tan is visually compelling based on that visualized embedding, but is otherwise completely arbitrary. Does that sound right?

 

[arkajad]

Not quite. Once you parametrize $$v^2+w^2=1$$ by sin and cos functions, which is rather natural, then tan comes automatically by projecting from 5 to 4 dimensions.

P.S. Of course for the graphical diagram purpose only - you can take whatever suits you.

 

[bcrowell]

Not quite. Once you parametrize $$v^2+w^2=1$$ by sin and cos functions, which is rather natural, then tan comes automatically by projecting from 5 to 4 dimensions.

P.S. Of course for the graphical diagram purpose only - you can take whatever suits you.

Hmm...but isn't the embedding in 5 dimensions purely extrinsic and without observable physical significance?

 

[atyy]

There's probably some freedom of choice. Maybe try 3.2.1 and 3.2.2 of Winitzki's notes?

http://sites.google.com/site/winitzki/index/topics-in-general-relativity

 

[arkajad]

Hmm...but isn't the embedding in 5 dimensions purely extrinsic and without observable physical significance?

You can say that complex numbers in quantum mechanics are also purely extrinsic and without observable physical significance. After all you can work only with real numbers! But, working with complex numbers helps you to discover a lot of relations which do have physical significance. The same may be the case with embeddings in 5 or 6 dimensions.

 

[bcrowell]

There's probably some freedom of choice. Maybe try 3.2.1 and 3.2.2 of Winitzki's notes?

http://sites.google.com/site/winitzki/index/topics-in-general-relativity

Aha! That's exactly what I needed! On p. 77, he explains why f is basically arbitrary, but does need to satisfy the condition that the derivative of f^-1 is proportional to x^-2 for large x. (What I'm referring to as f^-1 would be his f.)

Thanks, atyy!