Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 3 / a

Table of contents
- Solution a

Solution

Write the Kruskal–Szekeres plane with

\overset{r}{^}

horizontal and

\hat{t}

vertical. The relation

\hat{t}^{2} - \overset{r}{^}^{2} = - e^{r / (2 M)} (r - 2 M)

shows that constant-

r

curves are hyperbolae.

For

r > 2 M

, their right and left exterior branches satisfy

\overset{r}{^}^{2} - \hat{t}^{2} = e^{r / (2 M)} (r - 2 M) > 0.

For

0 < r < 2 M

, their future and past interior branches satisfy

\hat{t}^{2} - \overset{r}{^}^{2} = e^{r / (2 M)} (2 M - r) > 0.

The event horizons

r = 2 M

are the null diagonals

\hat{t} = \pm \overset{r}{^},

and the curvature singularities

r = 0

are the spacelike hyperbolae

\hat{t}^{2} - \overset{r}{^}^{2} = 2 M

in the normalization stated in the question.

Constant Schwarzschild-

t

curves are straight rays through the origin. In the exteriors their slopes obey

\frac{t ^}{r ^} = tanh \frac{t}{4 M},

while in the interiors

\frac{r ^}{t ^} = tanh \frac{t}{4 M} .

Thus the qualitative diagram is the usual four-region Kruskal diagram: two exterior wedges separated from black-hole and white-hole interiors by the two null horizons, with spacelike singularities bounding the interior wedges.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version