Why Does Hawking Radiation Reduce The Mass Of A Black Hole?

The metric around a Schwarzschild black hole is:

$ds^2 = -(1-\frac{2M}{r})dt^2+(1-\frac{2M}{r})^{-1}dr^2+r^2(d\theta^2+sin^2(\theta)d\phi^2)$

As the metric is independent of co-ordinate time, due to noether's theorem there is a conserved quantity which can be represented by the killing vector $\xi^\mu = (1,0,0,0)$ and the contravariant form simply

$\xi_{\mu} = g_{\mu\nu}\xi^\mu = g_{\mu t} = -(1-\frac{2M}{r})(1,0,0,0)$

The quantity $\xi_{\mu}P^{\mu}$ is hence conserved.

Therefore when a particle and an antiparticle are created from vacuum, $\xi_{\mu}P^{\mu} + \xi_{\mu}\bar{P^{\mu}} = 0$ to conserve total $\xi_{\mu}P^{\mu}$ .

Taking the situation where the particle forms outside the horizon and antiparticle within

Outside the horizon where $r>2M$ , the metric can be seen to have signature $(-,+,+,+)$ therefore $\xi_{\mu}$ is timelike and hence $\xi_{\mu}P^{\mu}$ is an energy, therefore it must be positive. However within the horizon the signature can be seen to be purely spacelike $(+,+,+,+)$ , hence $\xi_{\mu}\bar{P^{\mu}}$ is not an energy and can be negative, allowing pair production while still conserving $\xi_{\mu}P^{\mu} + \xi_{\mu}\bar{P^{\mu}} = 0$

This argument is clearly the same in the case where the particle forms within the horizon and antiparticle outside, so an equal number of particles and antiparticles will be emitted.

Minus the mass squared of an object is the four momentum squared, therefore:

$-m^2 = (P_{BH}+\bar{P})^2$ $-m^2 = P_{BH}^2+\bar{P}^2+2P_{BH}\bar{P}$ $-m^2 = -m_{BH}^2-\bar{m}^2+2P_{BH}\bar{P}$ However as the metric is spacelike within the blackhole, $2P_{BH}\bar{P}$ is purely positive, decreasing the magnitude of $m^2$