# Time dilation factor for the circular orbit at 3/2 Schwarzschild radius

What would the time dilation factor be if a massive (as in rest mass>0) point particle orbiting a Schwarzschild black hole in the photon sphere? If I understand correctly, this is the only possible orbit for photons, but it is also the closest possible orbit for a massive particle. So this is the same thing as asking what the maximum physically possible time dilation for a circular orbit is.

That point is 3/2 times the Schwarzschild radius. Any closer than that, and no free-fall path goes to infinity or completes a full orbit. It's also unstable. However, an exactly timed maneuver could put a particle into orbit there for a large number of orbits, so I don't think the instability affects the meaningfulness of the question. One could even realistically transition from r=infinity to the outer edge of this orbit, complete several orbits, and then escape back to r=infinity.

Reason for asking: a simplistic application of general relativity circular orbit time dilation tells me that the factor is infinity. In other words, time doesn't pass for a particle in this orbit.

I can't even begin to rationalize that. How could the universe be frozen still for such an observer? I don't think that makes any sense.

• Time dilation isn't normally a concept that is used in general relativity. How are you proposing to define it in this example?
– user4552
May 4, 2019 at 14:33

I believe that the answer above by Jerry is wrong and that the time dilation factor is really zero or infinite depending on how you define it at the photon radius. In order to stay in a circular orbit at the photon sphere radius you have to be moving at the speed of light and your time will move infinitely slow as compared to a distant observers time.

The time dilation for a circular orbit should be $$d\tau=dt\sqrt{1-\frac{3GM}{rc 2}}$$ or $$d\tau=dt\sqrt{1-\frac{3M}{r}}$$ depending on your units.

From the metric you get:

$$\frac{d\tau^2}{dt^2}=\left(1-\frac{2GM}{rc^2}\right)-\frac{r^2}{c^2}\frac{d\theta^2}{dt^2}$$

assuming we are moving in a plane with $$\phi=0$$. Interpreting $$rd\theta/dt=v$$ you can write $$d\tau=dt\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}$$. The velocity of an object in a circular orbit is the same in GR as classically (in $$t$$ but not in $$\tau$$) so you can write $$v=\sqrt{GM/r}$$ and get to the first expression above.

The time dilation factor is given by $\sqrt{1-\frac{2M}{r}}$, which is decidedly finite for $r=3M$. The last stable circular orbit around a nonspinning black hole is at $r=6M$. If the black hole is spinning, then stability depends on whether the object is co-rotating or counter-rotating, and the co-rotating orbits can get closer, but not as close as the unstable photon orbit.

• Isn't that the factor for stationary points? The factor for circular orbits is given as $\sqrt{1-\frac{3 M}{r}}$. But obviously I don't entirely buy the applicability of that either. Nonetheless, that formula at r=3M is what I'm talking about. Apr 26, 2014 at 0:52
• I think Jerry has overlooked the time dilation due to the orbital velocity. However he's quite correct that there is no stable orbit for $r < 3r_s$ for massive particles. The orbit at $3r_s$ is, as you say, unstable wrt any perturbations. Apr 26, 2014 at 10:34