Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 37E / b / ii

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- Solution ii

Solution

A radial null geodesic satisfies

0 = - (1 - \frac{2 M}{r}) d t^{2} + (1 + \frac{2 M}{r}) d r^{2},

so for the outgoing ray

\frac{d t}{d r} = 1 + \frac{2 M}{r} + O (M^{2} / r^{2}) .

The coordinate-time interval is therefore

Δ t = (r_{2} - r_{1}) + 2 M lo g \frac{r _{2}}{r _{1}} .

The stationary observer at

r_{2}

measures proper time

d T = 1 - 2 M / r_{2} d t

. Hence

T = (r_{2} - r_{1}) + 2 M lo g \frac{r _{2}}{r _{1}} - \frac{M}{r _{2}} (r_{2} - r_{1}) .

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