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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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  • Is the star close enough to the black hole for the black hole spin to matter? – TimRias Oct 06 '22 at 09:25
  • @TimRias let's avoid that complication for now - Schwarzschild spacetime. – ProfRob Oct 06 '22 at 09:39
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    I'm pretty sure the approximation in this answer isn't what you want, but it may be helpful to future readers ending up here. – uhoh Oct 07 '22 at 07:57
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    On a related note, please see The JPL Planetary and Lunar Ephemerides DE440 and DE441, Ryan S. Park et al (2021) AJ 161 105. Section 2.3 gives the details of the conversion from TAI, which is mean atomic time on the geoid to TDB, which is time in a frame comoving with the solar system barycentre, but in flat spacetime. This calculation is rather complicated (compared to the "simple" Schwarzschild solution) because it includes the contributions of the gravitational potential of a large number of bodies. – PM 2Ring Oct 07 '22 at 15:44

2 Answers2

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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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  • How does an observer far away measure the radial velocity? – TimRias Oct 06 '22 at 09:22
  • I was expecting an answer that depended on the orbital parameters. As Tim says you can't in general measure either $v_r$ or $\omega$ but can infer them from the orbital parameters I suppose. – ProfRob Oct 06 '22 at 09:24
  • The Schwarzschild radius $r$ is measured by measuring the circumference of a circle centred on the black hole and passing through your object, then dividing by $2\pi$. That is it is simply given by $C = 2\pi r$. To find the radial velocity repeat the measurement a short time $dt$ later and divide $dr$ by $dt$. – John Rennie Oct 06 '22 at 09:24
  • @ProfRob in principle the radial and angular velocities can be measured by the Schwarzschild observer, though it would be impractical to actually do. But as you say, in the weak field limit just take the usual orbital parameters, calculate the velocities from them and that will give you an excellent approximation. The weak field limit would be an excellent approximation for the stars orbiting Sag A*, which it what I would guess motivated the question. – John Rennie Oct 06 '22 at 09:29
  • You can measure a line of sight velocity (from the Doppler effect) and estimate a plane-of-sky velocity (both of which are subject to time dilation effects). What I want is to be able to estimate what the time dilation effects should be at any point in the orbit of a star with a given semi-major axis and eccentricity. I upvote in any case, because this certainly is an answer to the question. – ProfRob Oct 06 '22 at 09:32
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    Offhand I don't know what the Newtonian expressions for $v_r$ and $\omega$ are in terms of $a$ and $e$ but I assume they exist. Plugging them in would give an approximation that was effectively perfect for all orbits outside a few tens of Schwarzschild radii. Do do better you'd need to solve the geodesic equation for the orbit but I don't think this has a simple closed form. – John Rennie Oct 06 '22 at 09:35