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What would be the approach for finding the degree of time dilation for a star at any point in a general elliptical orbit around a supermassive black hole? Ideally I would be looking to work this out using the orbital parameters (semi-major axis, eccentricity and time since periapsis or equivalent).

Presumably, the maximum time dilation would be when the star was at periapsis?

Edit: Assume it's the Schwarzschild metric for simplicity.

ProfRob
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2 Answers2

12

For a Schwarzschild geodesic

$$ \frac{dt}{d\tau} = \frac{r}{r-r_s}E,$$

where $r$ is the radial position, $r_s$ is the Schwarzschild radius, and $E$ the specific energy of the orbit. The latter can be given in terms of the semi-major axis $a$ and $e$ eccentricity,

$$ E=\sqrt{\frac{(a (e-1)+r_s) (a(e+1)-r_s)}{a \left(a \left(e^2-1\right)+(e^2+3)r_s/2\right)}}.$$

analytic solutions for $r(\tau)$ exist, but are not very nice.

The semi-major axis $a$ and eccentricity $e$ are defined in terms of the periapsis and apoapsis distance $r_{\mathrm{min}}$ and $r_{\mathrm{max}}$ (Following Charles Darwin),

$$ a= \frac{r_{\mathrm{max}}+r_{\mathrm{min}}}{2},$$ $$ e = \frac{r_{\mathrm{max}}-r_{\mathrm{min}}}{r_{\mathrm{max}}+r_{\mathrm{min}}}. $$

Note that since these are expressed in terms of the Schwarzschild radial coordinate, they are generally coordinate dependent. Giving a coordinate independent characterization of the size and eccentricity of an orbit is a hard problem (see 2209.03390 for one possible solution based on the produced gravitational waves).

TimRias
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We can take the orbit to be in the equatorial plane, $\theta = \pi/2$, in which case for a Schwarzschild black hole we can write:

$$ d\tau^2 = (1 - r_s/r)dt^2 - \frac{dr^2}{c^2 (1 - r_s/r)} - \frac{r^2}{c^2} d\phi^2 $$

Then just write $dr = v_r dt$ and $d\phi = \omega dt$ and substitute to get:

$$ d\tau^2 = (1 - r_s/r)dt^2 - \frac{v_r^2 dt^2}{c^2(1 - r_s/r)} - \frac{r^2\omega^2}{c^2}dt^2 $$

Giving us the admittedly slightly messy expression for the time dilation:

$$ \frac{d\tau}{dt} = \sqrt{(1 - r_s/r) - \frac{v_r^2}{c^2(1 - r_s/r)} - \frac{r^2\omega^2}{c^2}} $$

where the radial velocity $v_r$ and the angular velocity $\omega$ are the values measured by the observer far from the black hole. For the Schwarzschild geometry we can define simple expressions for $v_r$ and $\omega$ in terms of the specific energy $E$ and specific angular momentum $L$:

$$ v_r = \frac{dr}{dt} = \frac{dr}{d\tau}\frac{d\tau}{dt} = \frac{dr}{d\tau} \sqrt{1 - r_s/r} $$

where:

$$ \frac{dr}{d\tau} = \sqrt{E^2 - \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{r^2}\right)} $$

and $\omega = L/r^2$.

TimRias
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John Rennie
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