# 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 '19 at 14:33

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. – Alan Rominger Apr 26 '14 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. – John Rennie Apr 26 '14 at 10:34
The time dilation for a circular orbit should be $$d\tau=\sqrt{1-\frac{3GM}{rc 2}}$$ or $$d\tau=\sqrt{1-3M/r}dt$$ depending on your units.
From the metric you get: $$\frac{d\tau^2}{dt^2}=(1-\frac{2GM}{rc^2})-\frac{r^2}{c^2}\frac{d\theta^2}{dt^2}$$, assuming we are moving in a plane with $$\phi=0$$. Interpreting $$rd\theta/dt=v$$ you can write $$d\tau=\sqrt{1-\frac{2GM}{rc^2}-v^2/c^2}dt$$. 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.