## Penrose Diagram January 31, 2009

Posted by keithkchan in Relativity.
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In a paper I encounter the Penrose diagram. Since I am waiting for my adviser to do something, I don’t have anything urgent to do these days, I decide to learn a little bit Penrose diagram. I learn it mainly from this article. Here I summarize what I learn about Penrose Diagram. For more details, see the article (Of course, I assume the reader of the Blackboard don’t bother to go to the lengthy article :))

First, Penrose Diagram is named after Roger Penrose, who first (or one of the first) used this type of diagram in research. Penrose diagram is a neat way to depict the causal structure of spacetime. The diagram is 2D, and so each point is in fact a 2-sphere. Each point of the diagram, the metric is proportional to the true metric, up to a scale factor. That is the metric in the diagram is a conformal transformation of the original metric: $g_{\mu \nu} \rightarrow \omega^2 g_{\mu \nu}$. Conformal transformation is widely used in gravity. Note that it is not a coordinate transformation, and the physics is changed! There is often debate in the literature about which frame is physical.

Let’s work through constructing the Penrose diagram of the flat Minkowski spacetime. The Minkowski metric is
$ds^2 = -dt^2 + dr^2 + r^2d\Omega^2$.
We can factorize the differential elements in t and r by defining new variables u=t+r, v=t-r (incidentally, these are called null coordinates since u=0 and v=0 are null geodesics): $d^2 = -du dv + \frac{( u-v)^2 }{ 4 } d\Omega^2$.
In these new coordinates, u and v are symmetric, both run from minus infinity to plus infinity. The aim is to come up with some coordinates with the scale factor factored out. The new coordinates in the Penrose diagram are finite. Let’s try u=tan(p), v=tan(q) (at least infinities are mapped to finite numbers in these transformations). Plugging in these transformations, we get $ds^2= \sec^2p \sec^2q (-dp dq + \frac{1}{4}\sin^2(p-q)d\Omega^2 )$. Very good! We can do it better by separating p and q by reversing null coordinates, i.e. let p+q=T , p-q=R, the line element becomes
$ds^2= \frac{1}{4} \sec^2(\frac{ T+R}{2} ) \sec^2(\frac{T-R}{2}) ( -dT^2 + dR^2 + \sin^2R d\Omega^2)$
Note that the prefactor is always positive, so it is a genuine conformal transformation. We now throw the prefactor to the hell and retain only
$ds^2= -dT^2 + dR^2 + \sin^2R d\Omega^2$.
This is the metric in the Penrose diagram. We summarize the transformations we have made
$T= \tan^{-1}(t+r) + \tan^{-1}(t-r) ; \newline R= \tan^{-1}(t+r) - \tan^{-1}(t-r) .$

OK. Let’s now draw the Penrose diagram. Recall that we only draw the R-T diagram, with the 2-sphere suppressed. We first work out how infinities are mapped to the points in the Penrose diagram. From the above transformations, it is easy to see $(t,r)=(+\infty,r) \rightarrow (\pi,0); (-\infty,r )\rightarrow (-\pi,0); (t,\infty)\rightarrow (0,\pi).$

These points correspond to the top, bottom and right corner of the Penrose diagram shown. (The figure is from Wikipedia.)  Note that $R + T = 2 \tan^{-1} \frac{t +r }{ 2 }$, so that when $r+t \rightarrow \infty$, $R+T = \pi$ and we should connect $(0,\pi )$ and $(\pi,0 )$ by a straight line (null geodesics). Similar analysis yields that we should also connect $(0,-\pi)$ with  $(\pi,0)$ by a null geodesic.

We can now draw the curves correspond t=const, r increases from 0 to infinity. R=0 when r=0. Obviously, for t=0, T=0 and R goes from 0 to $\pi$ when r goes from 0 to infinity. For $t /= 0$, one can figure it out whether it is convex or concave by looking at the transformations. For small r, T changes much slower than changes in R, so you may guess it is concave. Of course, you’d better draw it using computer or do more careful analysis.

Because of the symmetry between T and R, we can draw r=const, t goes from 0 to $\pm \infty$ by reflecting the about the 45 deg lines passing through the origin. Moreover, from the metric in T and R, one can immediately see that the null geodesics are still $45 \deg$. The final product is the following beautiful diagram.

I may talk about Penrose diagram in one more post. I think this post contains most equations among all the posts.