Einstein (+Rosen) came to the conclusion that they have to change the sign for the energy tensor T(adsbygoogle = window.adsbygoogle || []).push({}); _{ik}:

"if we had taken the usual sign for T_{ik}, the solution would involve +ε^{2}instead of -ε^{2}. It would then not be possible, by making a coordinate transformation, to obtain a solution free from singularities."

And they came up with the solution :

c^{2}ds^{2}= 1 / ( 1 - r_{s}/r - r_{q}^{2}/r^{2}) dr^{2}+ r^{2}(dθ^{2}+ sin^{2}θ dΦ^{2}) - (1 - r_{s}/r - r_{q}^{2}/r^{2}) c^{2}dt^{2}

Now Kerr-Newman solution for a rotating charged mass involves the +r_{q}^{2}/r^{2}. But since Einstein came with the correction specified above I was wondering if "this" (see below) is how it would look like if it would have been applied to Kerr-Newman metric :

c^{2}ds^{2}= - (dr^{2}/Δ + dθ^{2})ρ^{2}+ (c dt - a sin^{2}θ dΦ)^{2}Δ/ρ^{2}- ((r^{2}- a^{2}) dΦ - a c dt)^{2}sin^{2}θ/ρ^{2}

r_{s}= 2mG/(rc^{2})

r_{q}^{2}= q^{2}G/(4πεc^{4})

a = J/(mc)

ρ^{2}= r^{2}+ a^{2}cos^{2}θ

Δ = 1 - r_{s}/r + a^{2}/r^{2}- r_{q}^{2}/r^{2}

And the correction is inside Δ where instead of + r_{q}^{2}we use - r_{q}^{2}in order to make the metric consistent with Einstein-Rosen "derivation".

My question : is it mathematically ok if one would just change the sign of r_{q}^{2}in Δ ? Will other signs change too ? I couldn't find anything else to change in the metric.

Thanks anyway!

References:

https://en.wikipedia.org/wiki/Kerr–Newman_metric

http://journals.aps.org/pr/abstract/10.1103/PhysRev.48.73

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# A Kerr Newman metric correction from Einstein-Rosen m

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