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A Kerr Newman metric correction from Einstein-Rosen m

  1. Nov 26, 2016 #1
    Einstein (+Rosen) came to the conclusion that they have to change the sign for the energy tensor Tik :
    "if we had taken the usual sign for Tik, the solution would involve +ε2 instead of -ε2. 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 :
    c2 ds2 = 1 / ( 1 - rs/r - rq2/r2) dr2 + r2 (dθ2 + sin2θ dΦ2) - (1 - rs/r - rq2/r2) c2 dt2

    Now Kerr-Newman solution for a rotating charged mass involves the +rq2/r2 . 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 :

    c2 ds2 = - (dr2/Δ + dθ22 + (c dt - a sin2θ dΦ)2Δ/ρ2 - ((r2 - a2) dΦ - a c dt)2 sin2θ/ρ2

    rs = 2mG/(rc2)
    rq2 = q2G/(4πεc4)

    a = J/(mc)
    ρ2 = r2 + a2cos2θ

    Δ = 1 - rs/r + a2/r2 - rq2/r2

    And the correction is inside Δ where instead of + rq2 we use - rq2 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 rq2 in Δ ? Will other signs change too ? I couldn't find anything else to change in the metric.

    Thanks anyway!

  2. jcsd
  3. Dec 1, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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