I am looking for some strange stars that is gravitationally blue shifted. A star surface redshift determines mass-radius relation for chargeless star. Can be possible that a charged neutron star has a negative gravitational surface redshift?


1 Answer 1


The simple answer is no, though the complicated answer is, well, complicated!

The spacetime outside a charged object is described by the Reissner-Nordstrom metric:

$$ ds^2=-\bigl(1-\frac{r_s}{r}+\frac{r_q^2}{r^2}\bigr)dt^2+\bigl(1-\frac{r_s}{r}+\frac{r_q^2}{r^2}\bigr)^{-1}dr^2+r^2d\Omega^2 $$


$$ r_s = \frac{2GM}{c^2} $$

$$ r_q^2 = \frac{Q^2G}{4 \pi \epsilon_0 c^4} $$

From this we can extract the expression for the time dilation at a radius $r$:

$$ \frac{d\tau}{dt} = \sqrt{1-\frac{r_s}{r}+\frac{r_q^2}{r^2}} $$

For an uncharged neutron star $r_q = 0$ and the right side is always less than one i.e. time always runs slower. However once $r_q^2 \gt 0$ this changes the time dilation and in principle it can become greater than one. This would mean that light would be blue shifted.

The problem is that the charge required is so high the neutron star could never be stable. The charge would cause it to disintegrate. So, sadly, there is no way for a neutron star to have a blue shift in the way you describe.

  • $\begingroup$ +1 but would be vastly improved if the final assertion was backed up with a reference. $\endgroup$
    – ProfRob
    Dec 22, 2016 at 13:44
  • $\begingroup$ Did you mean r_q^2 \gt r^2 $\endgroup$
    – user55944
    Dec 23, 2016 at 13:57
  • $\begingroup$ @user55944 Specifically I meant $\frac{r_q^2}{r^2}-\frac{r_s}{r} \gt 0$ $\endgroup$ Dec 23, 2016 at 15:08
  • $\begingroup$ all right, is it possible that matter mass of a star more than total gravitational mass? $\endgroup$
    – user55944
    Dec 24, 2016 at 18:56
  • $\begingroup$ John, What is the meaning of the zero gravitational redshift $z=0$ for a star. Is it possible? $\endgroup$
    – user55944
    Jan 9, 2017 at 0:26

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