# Why does the Kruskal diagram extend to all 4 quadrants?

Why is it that the Kruskal diagram is always seen extended to all 4 quadrants when the definitions of the $U,V$ coordinates don't seem to suggest that the coordinates are not defined in, say, the 3rd quadrant. I can see that combining the definitions give equations that seem to allow that, but shouldn't the individual definitions of $U,V$ matter...? Physically, does it make sense?

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## 1 Answer

You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then someone could have come along later and defined (r,t) in terms of (U,V). We would then say that a disadvantage of the (r,t) coordinates is that they don't cover all four quadrants.

In general, if you have a solution to the Einstein field equations, it makes sense to extend it as much as possible. Physically, if a test particle reaches a nonsingular point in a finite proper time, then its world-line can and should be extended beyond that point, not just terminated there. It's only at singular points that the laws of physics break down and geodesics can't be extended. In the case of the Schwarzschild spacetime, extending it in this way results in all four quadrants of the Kruskal diagram.

The maximally extended Schwarzschild spacetime is not a realistic model of a black hole, however. When a black hole forms by gravitational collapse, you don't get quadrants III and IV.

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Thank you very much, Ben, for you explanation! :D – Helen May 10 '13 at 12:08