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Schwarzschild and Reissner–Nordström metrics

  1. Nov 22, 2009 #1

    A non-rotating [itex]J = 0[/itex] and charge neutral [itex]Q = 0[/itex] spherically symmetric metric is defined by the Schwarzschild metric:
    [tex]c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \frac{dr^2}{1 - \frac{r_s}{r}} - r^2 d\theta^2 - r^2 \sin^2 \theta \, d\phi^2 \right)[/tex]

    The next metric form for a non-rotating [itex]J = 0[/itex] and charged [itex]Q \neq 0[/itex] spherically symmetric metric is defined as:
    [tex]c^2 {d \tau}^{2} = \left( 1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}} \right) c^2 dt^2 - \frac{dr^2}{1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}}} - r^2 d\theta^2 - r^2 \sin^2 \theta \, d\phi^2 \right)[/tex]

    Which reduces directly to the Schwarzschild metric for [itex]Q = 0[/itex].
    However, the formal definition for a non-rotating [itex]J = 0[/itex] and charged [itex]Q \neq 0[/itex] spherically symmetric metric is the Reissner–Nordström metric:
    [tex]c^2 {d \tau}^{2} = \left( 1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}} \right) c^{2} dt^{2} - \frac{dr^{2}}{1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}}} - r^{2} d\Omega^{2}[/tex]

    Where the solid angle is defined as:
    [tex]d \Omega^2 = d \theta^2 + \sin^2 \theta d \phi^2[/tex]

    The Reissner–Nordström metric:
    [tex]\boxed{c^2 {d \tau}^{2} = \left( 1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}} \right) c^2 dt^2 - \frac{dr^2}{1 - \frac{r_{s}}{r} + \frac{r_{Q}^{2}}{r^{2}}} - r^2 d\theta^2 - r^2 \sin^2 \theta \, d\phi^2 \right)}[/tex]

    Reference:
    http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken]
    http://en.wikipedia.org/wiki/Reissner-Nordström_black_hole" [Broken]
    http://en.wikipedia.org/wiki/Solid_angle" [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 23, 2009 #2

    George Jones

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    [tex]d \Omega^2 = d \theta^2 + \sin^2 \theta d \phi^2[/tex]
     
  4. Nov 24, 2009 #3

    cristo

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    It might just be me but... what's the point of this thread?
     
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