I was reading Prof.Tong's notes where everything went smooth until when he just stated that for Kruskal coordinates $$U=-\exp(-u/4GM), V=\exp(v/4GM),$$ the horizon $r=2GM$ of Schwarzschild metric becomes $U=0$ or $V=0$. For reference, $u=t-r_\star,~v=t+r_\star$, where $$r_\star=r+2GM\ln|\frac{r}{2GM}-1|.$$

He proceeded to state that $U=0$ is the horizon for blackhole whereas $V=0$ is the horizon for whitehole. I tried to understand this statement and analysed on my own but got the opposite conclusion. Could someone please explain the correct approach to understand this concept?

To further illustrate the point, in his drawing Figure 48 on page 247 of the notes, Professor Tong correctly points out in later paragraphs that the singularity UV=1's positive component is the singularity of a blackhole. Now if we assume $U=0$ is the horizon for blackhole, given that our original spacetime outside of blackhole horizon lives in $U<0~, V>0$ part, we would not need to even cross the blackhole horizon to reach the blackhole singularity.

Another evidence I could think of, is that for blackhole horizon the original ingoing radial null geodesic should stay untouched as we move towards the horizon, i.e. there is no need to extend our coordinate system. So as we approach the horizon small $u$ goes to $\infty$ corresponding to the outgoing radial null geodesic in the ingoing Eddington Finkelstein coordinate system approaching the horizon from the outside. So this horizon should be a blackhole horizon.

Is this a typo? The future horizon of the blackhole should be $V=0$ and vice versa for whitehole?


1 Answer 1


You confused yourself.

In Fig.48, the line $U=0$ is the straight line from bottom-left to top-right, i.e., the coordinate line of $V$. On the top-right, it separates the black hole from outside.

The line $V=0$ is the straight line from bottom-right to top-left, i.e., the coordinate line of $U$. On the bottom-right, it separates the white hole from outside.

So Prof. Tong is right.


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