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.


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.

  • $\begingroup$ See Why can't I do this to get infinite energy? $\endgroup$
    – mmesser314
    Commented Feb 15, 2021 at 5:10
  • $\begingroup$ @mmesser314, I am interested in knowing the specific ratio as a function of the mass and radius. A number of formulas were given in the answer you linked to however it is not clear to me how to derive the answer to my question from them. If you could elaborate that would be great. Thanks for the help. $\endgroup$
    – Mathew
    Commented Feb 15, 2021 at 5:53
  • $\begingroup$ @TrixieWolf I think you're misreading that answer by ProfRob that you linked to. He explicitly says to multiply the "General Relativistic" and "Special Relativistic" time dilations together. $\endgroup$
    – PM 2Ring
    Commented Feb 15, 2021 at 19:14
  • $\begingroup$ @TrixieWolf, I have edited my question to address your comment. $\endgroup$
    – Mathew
    Commented Feb 15, 2021 at 19:15
  • $\begingroup$ @PM2Ring, if you think you know how to get the right formula out of ProfRob's answer, that would be great. I don't quite understand his answer well enough to get the right formula. $\endgroup$
    – Mathew
    Commented Feb 15, 2021 at 19:16

2 Answers 2


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.

  • $\begingroup$ I would like to accept your answer, however it differs from John Hunters, and I am unable to determine which is correct. Thanks to both of you for taking the time to look at my question. $\endgroup$
    – Mathew
    Commented May 3, 2021 at 18:26
  • $\begingroup$ The answers are the same to the first order in $r_s/r$. They would be completely identical if John Hunters did not use a Taylor series approximation at some steps. $\endgroup$
    – Pavlo. B.
    Commented May 3, 2021 at 18:35
  • $\begingroup$ Can you explain equation (4)? $\endgroup$
    – ProfRob
    Commented Apr 29, 2023 at 7:12
  • $\begingroup$ @ProfRob Sorry for the late response. Just treat the integrand in (3) as a Lagrangian, and use calculus of variations: en.wikipedia.org/wiki/Calculus_of_variations $\endgroup$
    – Pavlo. B.
    Commented Jun 8, 2023 at 15:08

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

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

  1. 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.

  • $\begingroup$ I would like to accept your answer, however it differs from Pavlo's, and I am unable to determine which is correct. Thanks to both of you for taking the time to look at my question. $\endgroup$
    – Mathew
    Commented May 3, 2021 at 18:26
  • $\begingroup$ The answers are the same to about $1 \times 10^{-15}$. If you put $r_1$ and $r_2$ as Neptune's and Mercury's radii in both formulae it's unlikely that there would be any noticeable difference. Glad the answers have answered your question. $\endgroup$ Commented May 3, 2021 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.