# The Reissner-Nordstrom metric

1. Sep 26, 2009

### Orion1

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:
$$Q(r) = \int_0^r 4 \pi j^0 e^{\frac{(\nu + \lambda)}{2}} dr$$

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

What Einstein-Maxwell charge effects with respect to the metric components $$\nu$$ and $$\lambda$$ 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 $$r^2$$ dimensions?

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

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. Sep 30, 2009

### Orion1

total mass of charged neutron star...

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

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
3. Oct 2, 2009

### Orion1

Last edited: Oct 2, 2009
4. Oct 4, 2009

### Orion1

nu metric component...

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

The quantity $$J^{\kappa}$$ 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 $$J^1$$, which implies that the only non-vanishing component of $$F^{\kappa \mu}$$ is $$F^{01}$$:

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

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

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

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

Therefore:
$$\boxed{j^0 = \rho_{ch} r^2}$$

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

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

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]