Time dilation due to gravitation of observers in circular orbit of a point mass

Question

Consider a point mass $M$ without angular momentum, and two observers Alice and Bob, each orbiting the mass in a circular orbit in the same plane and rotational direction at radii of $r_1$ and $r_2$ respectively. What is the relative time dilation between them? To be more specific, say each had a clock which they set to zero, then they sat in each of their orbits for some large amount of time, then compared how much time passed on each of their clocks. What would the ratio between their times be?

To be even more specific, say Alice and Bob start in orbits of radius $r_1$ and $r_2$ respectively. Alice plans to emit pulses of light which will be used to communicate when Bob should turn his clock on and off. Prior to doing this experiment, they agree upon a specific time $T$ that Alice will wait between emitting her pulses. Alice sets her clock to $0$, turns it on, and immediately emits a pulse of light which then is later received by Bob. Upon receiving the pulse, Bob sets his clock to $0$ and turns it on. Alice waits $T$ seconds after turning on her clock, at which point she emits another pulse. When this second pulse is received by Bob, he stops his clock with $T'$ seconds on it, and calculates the ratio $T'/T$. I would like to know $Q = \lim_{T\rightarrow\infty}(T'/T)$ as a function of $r_1,r_2$ and $M$.

Thanks for taking the time to read my question. If you find any of it unclear let me know and I will do my best to clarify. Also I would like to add that for an answer, I am hoping you can provide an equation relating the mass $M$ and radii $r_1,r_2$ with the ratio $Q$ between the clock times. For example, something like this: $Q = f(r_1,r_2,M)$ for some function $f$.

Also, if you would like to consider the more general case in which $M$ has angular momentum, please do. I suspect the angular momentum may change the answer.

@TrixieWolf, I have attempted to understand the question you linked to and I think the answer I am looking for is this:

$$ Q = \left(1 - \frac{3r_s}{2r}\right)^{1/2}\ .$$

Where $r_s=2GM/c^2$ is the Schwarzschild radius, $r$ is the radius of one of the orbits, and the other orbit is at infinity. I assume that if I wanted to compare to orbits not at infinity, I could compare each with infinity, then take the ratio of these. If you'd like to make this into an answer, please do and I will accept it.

Motivation

I thought I might add some extra context to further motivate the question. Say I live on Mercury and you live on Neptune. I want to know how much faster (or slower, I'm not sure which way it goes) my time goes by than your time. Ideally, an answer to this question would give my a formula where I can simply plug in the orbital radius of each planet and the mass of the sun. I'd image the fact that the orbits are actually elliptical would complicate this further, however analyzing this seems overly complicated to me, so I'm not asking for that for an answer. However, if you would like to try to analyze this, go ahead.

Answer 1

Your equation for time dilation is correct. Here is a derivation "from scratch".

We start with the Schwarzschild metric $$ d\tau^2=f(r)dt^2-\frac{1}{f(r)}dr^2-r^2d\theta^2,\tag{1} $$ where $f(r)=1-\frac{r_s}{r}$, where $r_s=2GM/c^2$. From the metric we obtain $$ \frac{d\tau}{dt}=\sqrt{f(r)-\frac{1}{f(r)}v_r^2-v_{\theta}^2 r^2},\tag{2} $$ where $\tau$ is the proper time of the observer that moves with the velocity $(v_r, v_\theta)$ in Schwarzschild's coordinates. Equations of motion, which are obtained by maximizing $$ \int_{t_1}^{t_2} d\tau = \int_{t_1}^{t_2}\frac{d\tau}{dt} dt=\int_{t_1}^{t_2}\sqrt{f(r)-\frac{1}{f(r)}v_r^2-v_{\theta}^2 r^2}dt,\tag{3} $$ yield for circular motion ($v_r=0$) $$ v_\theta^2 = \frac{f'(r)}{2r}.\tag{4} $$ After substituting this result into Eq. (2) and taking $v_r=0$, we obtain $$ \frac{d\tau}{dt}=\sqrt{f(r)-\frac{r f'(r)}{2}}=\sqrt{1-\frac{3r_s}{2r}}.\tag{5} $$ Considering two observers at distances $r_1$ and $r_2$ the last equation yeilds the ratio of their proper times: $$ \frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{3r_s}{2r_1}}{1-\frac{3r_s}{2r_2}}}.\tag{6} $$ Notice that the equations (5) and (6) are exact, and did not rely on any approximations.

Answer 2

There are two time dilations to take into account

1. due to the speed of the person in orbit, $v$, at radius $r$

2. due to the gravitational time dilation caused by mass $M$ at radius $r$.

We can do each separately, then multiply them together.

1. For a circular orbit

$\frac{mv^2}{r} = \frac{GMm}{r^2} $

$v^2 = \frac{GM}{r}$

the time dilation due to movement depends on the factor $\sqrt{{1- \frac{v^2}{c^2}}}$ so it's

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

2. Due to the gravitational field.

it's $$\sqrt{{1- \frac{2GM}{rc^2}}}$$

The time is slowed more (in both cases) for smaller $r$, multiplying gives

$\sqrt{{(1- \frac{GM}{rc^2}})({1- \frac{2GM}{rc^2}})}$ = $\sqrt{{1- \frac{3GM}{rc^2}}}$

(if $\frac{GM}{rc^2}$ is small, that's the case for the planets near the sun)

A binomial expansion gives the combined factor $${1- \frac{3GM}{2rc^2}}$$

For Alice and Bob the ratio is

$$\frac{1- \frac{3GM}{2r_1c^2}}{1- \frac{3GM}{2r_2c^2}}$$

with another binomial expansion it's

$$Q = 1- \frac{3GM}{2c^2}(\frac{1}{r_1}-\frac{1}{r_2})$$

With the time slowed more for the inner observer.