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The Reissner-Nordstrom metric

  1. Sep 26, 2009 #1

    Two citations referenced by Wikipedia with respect to General Relativity models for spinless charged neutron stars, reference 2 - eq. 24, states that the Reissner-Nordstrom metric relativistic Einstein-Maxwell Gauss law for the electric system charge is:
    [tex]Q(r) = \int_0^r 4 \pi j^0 e^{\frac{(\nu + \lambda)}{2}} dr[/tex]

    However, according to reference 3 - eq. 5, the electric system charge is:
    [tex]Q(r) = \frac{1}{r^2} \int_0^r 4 \pi r^2 \rho_{ch} e^{\frac{\lambda}{2}} dr[/tex]

    What Einstein-Maxwell charge effects with respect to the metric components [tex]\nu[/tex] and [tex]\lambda[/tex] are these equations describing?

    Why does the reference 2 equation have two metric components and the reference 3 equation has only one metric component?

    Why is the reference 2 equation missing the [tex]r^2[/tex] dimensions?

    And what are the International System of Units (S.I.) for [tex]j^0[/tex] and [tex]\rho_{ch}[/tex]?

    Please post links to this thread for subject equations cross-reference.

    Reference:
    http://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric" [Broken]
    http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.2708v2.pdf" [Broken]
    http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97332004000200038" [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 30, 2009 #2
    total mass of charged neutron star...


    How was this equation generated?

    Total mass of spinless charged neutron star at a radial distance r:
    [tex]\frac{dm(r)}{dr} = \frac{4 \pi r^2 \epsilon(r)}{c^2} + \frac{Q(r)}{c^2 r} \frac{dQ(r)}{dr}[/tex]

    How exactly does charge contribute to mass this way?

    Reference:
    http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.2708v2.pdf" [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Oct 2, 2009 #3

    Please post links to this thread for subject equations cross-reference.
     
    Last edited: Oct 2, 2009
  5. Oct 4, 2009 #4
    nu metric component...



    Given that [tex]F^{\mu \kappa}[/tex] satifies the covariant Maxwell field equation:
    [tex]\left[\sqrt{-g} F^{\kappa \mu} \right]_{, \mu} = 4 \pi J^{\kappa} \sqrt{-g}[/tex]

    The quantity [tex]J^{\kappa}[/tex] denotes the four-current which represents the electromagnetic sources in the star. For a static spherically symmetric system, the only non-zero component of the four-current is [tex]J^1[/tex], which implies that the only non-vanishing component of [tex]F^{\kappa \mu}[/tex] is [tex]F^{01}[/tex]:

    In this case I derived this formula for the relativistic electric field:
    [tex]F^{01}(r) = E(r) = \frac{Q(r)}{r^2} e^{- \frac{ \left( \nu + \lambda \right)}{2}}[/tex]

    The formula suggested by the remaining reference papers:
    [tex]F^{01}(r) = E(r) = \frac{Q(r)}{r^2}[/tex]

    Electric system charge:
    [tex]Q(r) = \int_0^r 4 \pi j^0 e^{\frac{ \left( \nu + \lambda \right)}{2}} dr[/tex]

    The formula suggested by the remaining reference papers:
    [tex]Q(r) = \int_0^r 4 \pi r^2 \rho_{ch} e^{\frac{\lambda}{2}} dr[/tex]

    Therefore:
    [tex]\boxed{j^0 = \rho_{ch} r^2}[/tex]

    Integration by substitution:
    [tex]F^{01}(r) = E(r) = \frac{e^{- \frac{ \left( \nu + \lambda \right)}{2}}}{r^2} \int_0^r 4 \pi j^0 e^{\frac{ \left( \nu + \lambda \right)}{2}} dr[/tex]

    Why does this solution have a [tex]\nu[/tex] metric component and two metric components relativistic electric_field modifier [tex]e^{- \frac{ \left( \nu + \lambda \right)}{2}}[/tex]?

    Reference:
    http://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric" [Broken]
    http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.2708v2.pdf" [Broken]
    http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-97332004000200038" [Broken]
    http://www.google.com/url?sa=t&sour...o.pdf&usg=AFQjCNG7gtMPDuwpTrqYwoZ2NuS1lk1nRw"
     
    Last edited by a moderator: May 4, 2017
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