I want to calculate the shifting of the time in a star that orbits a supermassive black hole, i already know that if the star is orbiting at 5000 km/s with respect to earth the time will be 1/(1-(5000^2/300000^2)^0,5) = 1.016 times faster than earth, but i am not taking into account how much the black hole's gravity changes the time.

  • $\begingroup$ "5000 km/s with respect to earth" should be "with respect to the black hole". With respect to earth the star's speed will fluctuate between +5000 and -5000, so there's no effective time difference. $\endgroup$
    – hdhondt
    Jul 7, 2019 at 23:06
  • $\begingroup$ i am comparing how fast the time passes on the star vs how fast the time passes on the black hole so in the special relativity analysis the only speed that matters is the 5000 km/s in any direction. I am supposing that the black hole is static with respect to earth $\endgroup$ Jul 8, 2019 at 0:00

1 Answer 1


The redshift factor for a circular (prograde) equatorial orbit around a spinning (Kerr) black hole with mass $M$ and spin $a$ is given by

$$ \frac{r^{3/2}+aM^{1/2}}{r^{3/4}\sqrt{(r-3M)r^{1/2}+2aM^{1/2}}},$$

using "geometrical units" such that $G=c=1$.

That is, for every second that passes for the object, this many seconds pass for a distant observer. (This includes both "gravitational" and "doppler" redshifts, which really cannot be disentangled)


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