# Mass change of a black hole during the Penrose process

I have been reading about Kerr black holes, specifically via Hobson et. al, General Relativity: An Introduction for Physicists. When discussing the Penrose process, the book considers a particle A emitted from infinity, which decays into particles B and C in the ergoregion of the metric. (Throughout I am referring to Boyer-Lindquist coordinates - see here)

Within this region, the basis vector $$\textbf{e}_t$$ is spacelike, and so the component $$\textbf{e}_t \cdot \textbf{p} = p_t$$ of either particle's 4-momentum represents a spatial component of momentum. We then suppose that particle C escapes the ergoregion, and returns to infinity, and particle B enters the black holes event horizon. We have that $$E^{(C)} = E^{(A)} -p^{(B)}_t$$ by 4 momentum conservation at the decay event, and so the energy of C may exceed the energy of A. All of this I am fine with.

I am confused however, with what the book writes about the change in mass of the black hole given that particle B enters:

$$M \rightarrow M +p^{(B)}_t$$ If $$p^{(B)}_t$$ is a component of spatial momentum, then surely this does not make sense in terms of the change in mass of the black hole, and instead it should be the time-like component of $$\textbf{p}^{(B)}$$ that contributes?

I think I have solved this by realising that we can define the energy by the invariant $$E^{(B)} = \textbf{u}_{obs} \cdot \textbf{p}^{(B)}$$, where $$\textbf{u}_{obs}$$ is the observer who is measuring the energy. In this case, we have implicitly placed the observer fixed at infinity, such that $$[u_{obs}^{\mu}] = (1,0,0,0)$$, and so clearly $$E^{(B)}$$ must still be given by $$p_t^{(B)}$$, which is therefore also the mass taken from the black hole, as measured by the observer at infinity.