# Compactification of Minkowski spacetime

I'm studying Ray D'Inverno's book "Introducing Einstein's relativity". I'm having trouble understanding Fig. 17.7 (pag. 236), which is an illustration of compactified Minkowski spacetime. Especially, I don't understand how $$r'$$ coordinate is measured on the cylinder: there are two vertical straight lines on the cylinder labelled $$r' = 0$$ and $$r' = \pi$$, but if I want to individuate, say, $$r' = \frac{\pi}{3}$$, how should I proceed? Thank you for your answers.

• I've included it just now.
– Al01
Sep 21, 2023 at 19:56
• Are you wondering which way around you should go? It wouldn’t matter as far as I can tell. Sep 21, 2023 at 20:02
• I believe $r'$ is considered as an angular coordinate. In the book, on the page preceding this figure, he states that constant $t$ coordinate slices have the topology of a three sphere, so in that case $r'$ would be treated as an angular coordinate. Sep 21, 2023 at 20:04
• So do the points on the cylinder with $r' = c$, where $c$ is a constant and $0 \leq c \leq \pi$, lie on the vertical straight line making an angle with $r'=0$ equal to $c$ when measured from the axis of the cylinder, no matter the direction (clockwise or counterclockwise)?
– Al01
Sep 21, 2023 at 20:10
• Yes, except the direction (clockwise or counterclockwise) does matter. Just pick one of these directions for angles between $0$ and $\pi$. Then the other direction is for angles between $\pi$ and $2\pi$. Sep 21, 2023 at 20:50

Although in the cylinderical diagram the time coordiante $$t'$$ has the range $$-\infty < t' < \infty$$

the relavent range is only from $$-\pi< t' < \pi$$

As such, the conformally compactified region of the Minkowski spacetime is only the diamond region of the cylinder with boundaries ($$\mathscr{I}^{+}, \mathscr{I}^{-}$$).

In addition we pick any one side of this diamond with $$0 \leq r' \leq\pi$$. As two dimensions are suppressed, on its own the compactified region would appear like shown in the image below.

In this image $$r' \in (0, \pi)$$ and $$r' = \pi/3$$ would lie at a distance 1/3 from the LHS along the lines $$t =$$ const.

$$r'$$ is just the angular position on the cylinder's surface. In the Penrose diagram (Fig. 17.9), it's the horizontal coordinate.

From comments, I think you're confused because there is a second copy of the Penrose diagram running from $$0$$ to $$-π$$ (or $$π$$ to $$2π$$). That's a peculiarity of the 1+1 dimensional case. In $$d+1$$ dimensions, the compactification lives on the surface of $$S^d\times \mathbb R$$, and the points at a fixed $$r'$$ ($$0) form a $$(d-1)$$-sphere of radius $$\sin r'$$. If $$d>1$$, that's a continuous surface, and you can cover it with the $$θ,φ$$ of the original Minkowski space. If $$d=1$$, it's two points, and it's easiest to allow $$r'$$ to take negative values—the same thing that happens if you try to put polar coordinates on 1D Euclidean space.

• When you say "the horizontal coordinate" do you mean simply that I have to draw on Fig. 17.9 (that one showing the Penrose diagram) a vertical straight line at a distance from the left vertical edge equal to the value of the coordinate? As if I had a cartesian (x,y) 2D coordinate system?
– Al01
Sep 22, 2023 at 6:57