I am wondering if there exist closed form-expressions for the time dilation experienced by an observer in different orbits around a Schwarzschild black hole, outside the event horizon, relative to some distant observer sitting fixed relative to the black hole.

I am no expert in GR, but I know these quickly turn into some complicated integrals. Any point in the right direction, even toward an approximate answer, would be much appreciated.

  • $\begingroup$ This attracted great answers, and the user did accept one of them, which shows legit engagement with the community. I'm surprised by the net score of 0, and wouldn't be surprised if this had something to do with the asker having only made 4 posts in total in 8 years. If my questions were treated this way I'd probably do the same! $\endgroup$ Feb 1, 2022 at 7:14

2 Answers 2


Indeed there is such an expression, which is dependent on the value of the Schwarzschild metric at various values of the coordinate $r$ and does not actually involve complicated integrals at all! We need to be careful in how we talk about distances here, because the radial coordinate does not, strictly speaking, correspond to the distance from some "center point" of the black hole. Recall that the Schwarzschild metric, which describes the spacetime outside of any spherically symmetric mass distribution, is given by

$$g_{\mu \nu} = \text{diag}\left(-\left(1-\frac{r_s}{r}\right), \left(1-\frac{r_s}{r}\right)^{-1}, r^2, r^2\sin^2(\theta)\right)$$

Just to be clear, I am using a metric with signature (-+++), and $r_s$ is the Schwarzschild radius, $r_s \equiv \frac{2GM}{c^2}$ (with $G$ Newton's gravitational constant, $M$ the mass of the star/black hole/whathaveyou, $c$ the speed of light). Now we can consider the proper times experienced by a stationary observer at infinity and an observer orbiting the black hole. When we take the radial coordinate to infinity, this metric reduces to that for a flat (Minkowski) spacetime, and the observer here will have a proper time equivalent to the coordinate time. To find how an observer in orbit around the black hole, at some finite radius, will measure time, we need to consider that observer's 4-velocity.

The 4-velocity is given by $$u^{\mu} = \left(\gamma c, \gamma \vec{v}\right)$$ where $\gamma$ is the Lorentz coefficient defined by $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ and $\vec{v}$ is the familiar velocity vector from Newtonian mechanics. We can also rewrite this 4-velocity as a derivative of the particle's position with respect to a new variable which we call proper time, $\tau$. This proper time is the time experienced by the particle whose motion we are considering. In this form $u^{\mu}$ is simply $\frac{d}{d\tau} x^{\mu}$ where $x^{\mu} = (t, x, y, z)$ or $(t, r, \theta, \phi)$, etc. depending on your coordinates, is the position vector of the particle. It is this aforementioned $\gamma$ coefficient which will allow us to compare the passage of an observer's proper time to the coordinate time that a distant observer will measure, through the formula $dt = \gamma d\tau$. What is particularly nice about this 4-velocity is that we can form a relativistic invariant by taking the inner product between this vector and itself $$u_{\mu} u^{\mu} = g_{\mu \nu} u^{\mu} u^{\nu} \equiv -c^2 $$

Here I am using the Einstein summation convention, which allows us to omit the summation symbol with the implied understanding that whenever we find a pair of upper and lower indices, we sum over those values from 0 to 3. What we really want is to write this out in an explicit coordinate form so that we can solve for the time dilation factor.

$$-\gamma^2 c^2 \left(1-\frac{r_s}{r}\right) + \left(\frac{dr}{d\tau}\right)^2 \left(1 - \frac{r_s}{r}\right)^{-1} + r^2 \left(\frac{d\theta}{d\tau}\right)^2 + r^2 \sin^2(\theta)\left(\frac{d\phi}{d\tau}\right)^2 = -c^2 $$

Now this may look daunting, but we can simplify this expression. All particle orbits in the Schwarzschild metric occur in a plane, and we are free to choose the orientation of our coordinate axes so that the orbits we are considering occur in the $\theta = \pi/2$ plane. Moreover, this value of $\theta$ is constant and therefore independent of the proper time $\tau$, so we can throw out the term $r^2 \left(\frac{d\theta}{d\tau}\right)^2$. At the moment I am not willing to jump in and perform the necessary work for orbits where $r$ may vary (non-circular orbits). I may return to this later or perhaps someone with more experience can jump in here and provide the more general solution. As such, I will take $\frac{dr}{d\tau} = 0$. With this the equation reduces to

$$-\gamma^2 c^2 \left(1-\frac{r_s}{r}\right) + r^2\left(\frac{d\phi}{d\tau}\right)^2 = -c^2 $$

We still have an unknown term $\frac{d\phi}{d\tau}$, but we will see shortly how this can be dealt with.

There are two conserved quantities in the Schwarzschild metric which correspond to the fact that it is a spherically symmetric, time independent system, and are associated with angular momentum and energy in Newtonian physics. These are usually denoted $l \equiv r^2 \sin^2\theta \frac{d\phi}{d\tau}$ and $e \equiv \left(1 - \frac{r_s}{r}\right) \gamma c^2$. We can relate these quantities to the angular velocity of the orbiting observer as measured by the observer at infinity by the equation

$$\Omega \equiv \frac{d\phi}{dt} = \frac{d\phi/d\tau}{\gamma} = \frac{c^2}{r^2}\left(1-\frac{r_s}{r}\right)\left(\frac{l}{e}\right) $$

I will not prove this, but for circular orbits the ratio $l/e$ is equal to $\sqrt{GMr/c^4}\left(1- \frac{r_s}{r}\right)^{-1}$, which gives us

$$\Omega = \sqrt{\frac{GM}{r^3}}$$

Now, we can solve for $\frac{d\phi}{d\tau}$ and plug this back in to our invariant equation above, which gives us

$$-\gamma^2c^2\left(1-\frac{r_s}{r}\right) + r^2 \frac{GM\gamma^2}{r^3} = -c^2$$

And we are finally ready to solve for the time dilation factor.

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

To sum up, if you have an observer executing a circular orbit at radius $r$ around a star or black hole of mass $M$, that observer will experience time at a rate of ${\sqrt{1-\frac{3GM}{rc^2}}}$ times that experienced by a stationary observer far away from the star. I should reiterate that this only works for an object in a circular orbit, because there are some strong constraints imposed on the system by that requirement, like the ratio of $l$ and $e$ and the fact that $\frac{dr}{d\tau}$ can be set to zero. I ought also to acknowledge that this simplified problem as I have presented it is solved in texts on relativity and I drew heavily on the solution from Hartle's book.

  • $\begingroup$ Thanks very much; this was a very informative calculation. Unfortunately I realized today that the case where r is constant is not interesting for my problem (as the time dilation factor is also a constant). Do you know how much harder this is to extend to more general orbits (i.e. elliptical, parabolic, or hyperbolic)? $\endgroup$
    – Adrian
    Feb 4, 2014 at 18:22
  • $\begingroup$ @Adrian I don't know exactly how much more difficult it would be, but if you had an explicit form for the orbit you could in principle find the $r$ and $\phi$ components of the 4-velocity and use the same idea that the inner-product of the 4-velocity with itself is both an invariant and a constant of the motion to find the time dilation factor. $l$ and $e$ would still be conserved as well. $\endgroup$
    – Jordan
    Feb 5, 2014 at 19:36
  • $\begingroup$ Slightly puzzled that the time dilation can be constant for something in orbit. Why isn't the standard doppler effect, which depends on the phase of the orbit also a factor? i.e. Forget about GR for a moment, why isn't there a component of the time dilation just due to the fact that you have something moving along a circular path with non-zero speed? $\endgroup$
    – ProfRob
    Apr 2, 2016 at 14:21
  • $\begingroup$ @RobJeffries I'm not sure what you mean with the doppler effect, but the time dilation factor derived above includes the fact that the object has nonzero speed. If you took $d\phi/d\tau = 0$ in the above derivation you'd get something different. $\endgroup$
    – Jordan
    Oct 19, 2016 at 20:17
  • $\begingroup$ Jordan you are quite correct. I had a blind spot about this, which I even asked a question about physics.stackexchange.com/questions/246960/… $\endgroup$
    – ProfRob
    Oct 19, 2016 at 20:24

Since in Schwarzschild coordinates the metric components are independent of Schwarzschild time, $\partial_t$ must also be Killing vector field, and therefore every orbit with four-velocity $u$ has conserved specific energy $$\epsilon = -\partial_t\cdot u = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{.}$$ This you can read off directly from the metric components and $u^\alpha = \mathrm{d}x^\alpha/\mathrm{d}\tau$. In the case of a timelike orbit, the factor $\mathrm{d}t/\mathrm{d}\tau$ can be interpreted as time dilation comparing the Schwarzschild time $t$ of a stationary observer at infinity to the orbit's proper time $\tau$.

For some other stationary observer at Schwarzschild radial coordinate $R$, the Schwarzschild metric gives their time $\tau'$ as $\mathrm{d}\tau' = (1-2M/R)\,\mathrm{d}t$, so the above can be directly related to any other such observer, not necessarily at infinity.

The other obvious Killing vector field is $\partial_\phi$, since the metric components are likewise independent of $\phi$, which leads to conserved specific angular momentum for the orbits, $$l = \partial_\phi\cdot u = r^2\sin^2\theta\,\frac{\mathrm{d}\phi}{\mathrm{d}\tau}\text{,}$$ which combined with the timelike four-velocity condition $u\cdot u = -1$ can be derive an an effective potential for the orbits in the Schwarzschild spacetime.


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