I wondered how time would be dilated for an object in the middle of 2 black holes, however https://en.wikipedia.org/wiki/Gravitational_time_dilation only provides a formula for one source of gravity

i tried approximating it by multiplying the time dilation factors, however when i used 1 single source with twice the mass the results differed from 2 objects with half the mass quite drastically

the examples:
mass $1,00E+036$kg distance $3000000000$m

mass $2,00E+036$kg distance $3000000000$m

approximation: $0,7105905766*0,7105905766=0,5049389676 $ which is off by a factor of 5 and would be even more off the closer the distance gets to the schwarzschild radius

so how can the time dilation of multiple objects be approximated ?


1 Answer 1


In the static weak-field approximation, the metric is approximated in terms of the the Newtonian gravitational potential $\Phi$ as: $$\mathrm{d}s^2 = -\left(1+2\Phi\right)\mathrm{d}t^2 + (1-2\Phi)\mathrm{d}S^2\text{,}$$ where $\mathrm{d}S^2$ is the metric for Euclidean $3$-space. A more general situation requires one to solve a multi-body problem in general relativity, although the parametrized post-Newtonian formalism provides a canonical approximation scheme. So the completely general answer to your question is "it's very difficult problem of numerical general relativity that doesn't have a neat analytic solution."

However, at large distances from slowly-moving objects (i.e., large compared to their Schwarzschild radii, we can approximate a multi-body time dilation relative to a stationary observer at infinity as just given by the sum of their gravitational potentials: $$\frac{\mathrm{d}\tau}{\mathrm{d}t} = 1 - \frac{1}{c^2}\sum_k\frac{GM_k}{r_k} \text{.}$$ Note that this is consistent with the Schwarzschild formula for large radial coordinates, since by the Taylor-MacLaurin expansion, $$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1-\frac{2GM}{rc^2}} = 1 - \frac{GM}{rc^2} + \mathcal{O}\left(\frac{R^2}{r^2}\right)\text{,}$$ where $R = 2GM/c^2$ is the Schwarzschild radius. This also means that simply multiplying the factors is a pretty good approximation as long as those factors are close to $1$.


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