 Solution

The stress-energy tensor is already first order, so its energy flux can be evaluated on the unperturbed Schwarzschild event horizon $r=2M_0$, with the infinity-normalized Killing vector field $\xi ={\partial }_v$. On this background $g^{rv}=1$, and henceThe conserved stress-energy current from a Killing vector is $J^a=-T^a{}_b{\xi }^b$. Its radial component is negative, so the positive inward energy flux is $-J^r$. Using the directed null-surface measure $r_H^{2}dvd\Omega$ givesThe initial and final event horizon areas are $A_0=16\pi M_0^2$ and $A_1=16\pi (M_0+\Delta M{)}^2$. Thus $\Delta A=32\pi M_0\Delta M+O((\Delta M{)}^2)$. With Schwarzschild surface gravity ${\kappa }_0=1/(4M_0)$, the Physical-process first law of black-hole mechanics follows:The flux is Killing energy, including gravitational redshift, rather than the energy measured by a sequence of static observers arbitrarily close to the event horizon.