Kerr-Newman metric represents the spacetime of a charged and rotating black hole. If the mass parameter is zero, this metric is still not the Minkowski spacetime. What's the meaning of a charged and rotating but massless object's spacetime?
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3$\begingroup$ For the non-charged metric see problem 6 here. The $\mu \rightarrow 0$ limit is just Minkowski spacetime in some rather eccentric coordinates. I would guess that the Kerr-Newman metric will just give the $\mu \rightarrow 0$ limit of the Reissner-Nordstrom metric. $\endgroup$– John RennieCommented Oct 9, 2014 at 10:49
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$\begingroup$ I am very sure it will represent the electrovacuum space-time of a coulombic potential, but I don't recall the right transform from Boyer-Lindquist in the $M \to 0$ limit to recover the $M \to 0$ Reissner-Nordström. I guess $r^2 + a^2$ or $r^2 + a^2 + Q^2 $ spheroidal coordinate transform should work. $\endgroup$– VoidCommented Oct 9, 2014 at 11:29
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$\begingroup$ That's an interesting question. Is there a guarantee, that one can recover a trivial spacetime, at all? A point charge carries an infinite self-energy, which should give rise to some sort of non-trivial solution, even though I am afraid that it may be an unphysical one not just once, bit twice. $\endgroup$– CuriousOneCommented Oct 9, 2014 at 14:49
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$\begingroup$ @CuriousOne: it's easy enough to study the zero mass Nordstrom spacetime. I have no idea about how the central singularity behaves in that case. $\endgroup$– Zo the RelativistCommented Oct 9, 2014 at 15:15
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$\begingroup$ Are you trying to force charge and/or spin to be nonzero even as the mass vanishes? Because you're really only supposed to have $a^2 + Q^2 < M^2$. $\endgroup$– user10851Commented Oct 9, 2014 at 15:20
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1 Answer
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If you set M (the total mass equivalent including the irreducible rest mass, the rotational energy and the electrostatic field energy) to zero you would have to assign a negative sign to at least to one of the components. The relation in natural units of G=M=c=kc=1 is
$$\rm M=\sqrt{\frac{16 M_{\rm irr}^4+8 M_{\rm irr}^2 \ Q^2+Q^4}{16 M_{\rm irr}^2-4a^2}}$$
so if you set the total mass equivalent M and the irreduzible mass Mirr to 0, the electric charge Q also has to be 0. If Q≠0, Mirr would have a complex value. If you set everything but a to 0, there is no solution for a so setting M to zero without setting Q and a to 0 also does not give a physical solution.
