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If you were to try to find the time dilation in a region of space near a black hole you would use the equation

$$t_r=\sqrt{1-\frac{2GM}{rc^2}}$$

Would the time dilation from two black holes be this?

$$t_r=\sqrt{1-\frac{2G}{c^2}\left(\frac{M_1}{r_1}+\frac{M_2}{r_2}\right)}=\sqrt{1+\frac{2P_C}{c^2}}$$

Where $P_C$ is the classical gravitational potential.

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  • $\begingroup$ That equation makes no sense. The left side is a time and the right side is dimensionless. $\endgroup$
    – G. Smith
    Dec 25, 2018 at 23:44
  • $\begingroup$ The left side is the ratio of the rate at which time passes at a point to the rate time passes at infinity. Both sides of the equation are unitless. The top formula is the formula for gravitation time dilation around a black hole. $\endgroup$
    – Laff70
    Dec 26, 2018 at 0:40
  • $\begingroup$ OK, thank you for explaining what you meant. I think using the symbol $t_r$ to mean the ratio of two time intervals is confusing, and I have never seen this usage anywhere. $\endgroup$
    – G. Smith
    Dec 26, 2018 at 0:52
  • $\begingroup$ I think your formula is correct when $M_1/r_1<<1$ and $M_2/r_2<<1$. When this is not the case, I don’t think the formula is known, because there is no analytic solution for two black holes. The nonlinearity of the Einstein equations means that you can’t superpose them. $\endgroup$
    – G. Smith
    Dec 26, 2018 at 1:00
  • $\begingroup$ Well I just tested out the second equation for a distribution of mass that is a hollow spherical shell. It gave a constant gravitational time dilation inside the shell and the correct gravitation time dilation outside the shell. So it seems that my equation is correct at least for spherically symmetric distributions. $\endgroup$
    – Laff70
    Dec 26, 2018 at 1:25

2 Answers 2

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You need to distinguish two different regions. As long as you are not too close to either black hole their gravitational potentials will just add. This is the linear region. Close to either black hole this linearity breaks down and we enter the non-linear region.

In the linear region the time dilation (relative to an observer at infinity) is well approximated by the weak field expression:

$$ t_r = \sqrt{1 +\frac{2\Phi}{c^2}} \tag{1} $$

where $\Phi$ is the Newtonian gravitational potential. In fact for a single black hole the Newtonian potential is:

$$ \Phi = -\frac{GM}{r} $$

and substituting this in equation (1) gives us the relativistic expression that you cite:

$$ t_r = \sqrt{1 - \frac{2GM}{c^2r}} \tag{1} $$

However be cautious about assigning too much significance to this as it's just a coincidence. The meaning of the variable $r$ is different in the Newtonian and relativistic theories.

Anyhow, provided we are in the linear region to calculate the time dilation for two (or more) black holes we need only calculate the total gravitational potential, and this is just the sum of the two separate potentials:

$$ \Phi_t = -\frac{GM_1}{r_1} - \frac{GM_2}{r_2} $$

and substituting this in equation (1) is going to give the expression that you suggest in your question:

$$ t_r = \sqrt{1 +\frac{2\Phi_t}{c^2}} = \sqrt{1 -\frac{2G}{c^2}\left( \frac{M_1}{r_1} + \frac{M_2}{r_2} \right)} \tag{2} $$

This works for the linear region, but there is still the non-linear region to consider i.e. close to one or both of the black holes. The trouble is that there is no analytic solution for the metric that describes two orbiting black holes. All we can say is that equation (2) will become increasingly inaccurate the closer we approach. We would need to do a numeric calculation to get an accurate result.

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    $\begingroup$ Actually if we are in the linear (weak field) region square roots are useless. You may simply write $t_r = 1+\Phi_t/c^2$. $\endgroup$
    – Elio Fabri
    Dec 26, 2018 at 10:33
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    $\begingroup$ @Laff70 In hypothetical scenarios anything can happen. You can't imagine one contradicting basic physical laws and then demand to draw physical conclusions in it. $\endgroup$
    – Elio Fabri
    Dec 26, 2018 at 10:33
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    $\begingroup$ You have made a few sign errors in your answer. $\endgroup$
    – A.V.S.
    Dec 26, 2018 at 14:43
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    $\begingroup$ @Laff70 you could ask a new question about the numerical calculation, but it's formidably hard. Basically if you need to ask you can't do it. But then the linear approximation works surprisingly well as long as you're more than a handful of Schwarzschild radii away from both horizons. If your mass distribution isn't dense enough to contain a horizon then the linear approximation is effectively perfect. $\endgroup$
    – John Rennie
    Dec 26, 2018 at 16:32
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    $\begingroup$ @Laff70 someone has just asked about numerical computation here $\endgroup$
    – John Rennie
    Dec 26, 2018 at 16:37
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No, it would not. Time dilation is a relative effect from one frame of reference and location to another. First define those two points, then the total effect will be that due to the cumulative difference in gravity at those points (assuming they are stationary).

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  • $\begingroup$ When one is talking about gravitational time dilation, they usually have the reference frame be at infinity. This can be assumed for this question. Also the system is static. It would be helpful if you were to give the correct equation for the time dilation. $\endgroup$
    – Laff70
    Dec 25, 2018 at 23:26
  • $\begingroup$ @Laff70 If 2 black holes are close enough to each other for there to be a region between them with non-negligible combined gravitational time dilation, then the system won't be static. And binary black hole systems are notoriously nonlinear. $\endgroup$
    – PM 2Ring
    Dec 25, 2018 at 23:44
  • $\begingroup$ This is a hypothetical scenario for crying out loud!!! Imagine these are black holes(which don't evaporate in this scenario) with the masses of pebbles spaced meters apart from each other. Let's also say all contributions are non-negligible. Also, for good measure, lets say we have some super scifi tech that lets you anchor black holes in place. $\endgroup$
    – Laff70
    Dec 26, 2018 at 0:37

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