Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 309 / 2 / iv

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- Solution iv

Solution

Rearranging the outgoing radial null geodesics of the Vaidya metric equation along

r = r_{0} (v)

gives

M (v) = \frac{r _{0} ( v )}{2} (1 - 2 \frac{d r _{0}}{d v}) .

For the prescribed parametrization by

w

, the chain rule yields

\frac{d v}{d w} = 1 + 2 β n w^{n - 1} > 0, \frac{d r _{0}}{d v} = \frac{β n w ^{n - 1}}{1 + 2 β n w ^{n - 1}}, M (v (w)) = \frac{β w ^{n}}{2 ( 1 + 2 β n w ^{n - 1} )} .

Since

n > 1

,

v / w = 1 + 2 β w^{n - 1} \to 1

w \to 0^{+}

. The asymptotic equivalence

w \sim v

therefore implies

M (v) \sim \frac{β}{2} v^{n}, c_{1} = \frac{β}{2} .

More precisely

w = v + O (v^{n})

and

M (v) = (β /2) v^{n} [1 + O (v^{n - 1})]

, using Big O notation. The positivity of

d v / d w

justifies treating

w

as a single-valued function of

v

locally.

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