- #1

Mihai_B

- 10

- 1

_{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