Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 54 / 2 / e / Solution

Let $\Delta M=M_1-M_0$ and retain only terms first order in $\mu /M_0$ and $\rho /M_0$. The outgoing event horizon equation expands toHere ${\kappa }_0$ is the initial Schwarzschild surface gravity. The integrating factor $e^{-{\kappa }_{0}v}$ gives $d(e^{-{\kappa }_{0}v}\rho )/dv=-e^{-{\kappa }_{0}v}\mu /(2M_0)$. Future stationarity imposes $\rho =2\Delta M$ for $v>v_0$ and removes the growing homogeneous solution. Integrating from $v$ to infinity yields the teleological response of a Vaidya event horizon:The integral is finite because $\mu$ is bounded. Integrating by parts also givesFor $v>v_0$, this reduces to $2\Delta M$. For $v<0$, the first form becomesConsequently $r_H\to 2M_0$ as $v\to -\infty$. The positive constant $C$ shows that this event horizon is already expanding in the initially vacuum region. For monotone accretion, $\rho -2\mu \ge 0$, so it lies outside the instantaneous apparent horizon to first order. All displayed radius corrections are understood up to $O((\Delta M{)}^2/M_0)$.