# What's the meaning when Kerr-Newman metric's mass is zero?

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?

• 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. Commented Oct 9, 2014 at 10:49
• 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.
– Void
Commented Oct 9, 2014 at 11:29
• 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. Commented Oct 9, 2014 at 14:49
• @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. Commented Oct 9, 2014 at 15:15
• 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 $a^2 + Q^2 < M^2$.
– user10851
Commented Oct 9, 2014 at 15:20

$$\rm M=\sqrt{\frac{16 M_{\rm irr}^4+8 M_{\rm irr}^2 \ Q^2+Q^4}{16 M_{\rm irr}^2-4a^2}}$$