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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?

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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.

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  • $\begingroup$ +1. This is good that you have worked out the answer to your question, and included it for others to see. $\endgroup$
    – joseph h
    Jul 11 at 1:35

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