# What is the maximum time dilation factor when orbiting a rotating black hole?

Suppose one spaceship is stably orbiting a rotating black hole and another is far away from the black hole. What is the maximum time dilation factor between the two ships? Can it be made arbitrarily large, and if so does that require the black hole to be maximally rotating?

The innermost stable circular orbit (ISCO) for a maximally rotating Kerr black hole is for a prograde orbit at $r=m$. This is supposed to be the event horizon, which would make the time dilation factor infinite. However, the text around equation (22) in these lecture notes says that this "is an artifact of the coordinate system." Does this mean that the time dilation factor is not unbounded?

• I found the time dilation equation for an observer at a fixed coordinate position in a commonly-used coordinate system for Kerr black holes (Boyer-Lindquist coordinates), and in that case the time dilation factor can become arbitrarily large at the same radius as a stable circular orbit, but I couldn't find an equation for the ratio of proper time to coordinate time for the orbiting object itself. Anyone here know it, or would be able to derive it without too much trouble? – Hypnosifl Nov 11 '14 at 19:23
• (presumably the time dilation would be less extreme for an observer orbiting in the same direction as the rotation of the black hole than one at fixed position coordinate, since the fixed observer has to oppose the frame-dragging of space around the hole, and as his radius approaches r=2GM/c^2 in the equatorial plane his speed would have to approach the speed of light relative to local free-falling observers, since that's the boundary of the ergosphere) – Hypnosifl Nov 11 '14 at 19:26

You can get the time dilation factor by computing the redshift of a radial photon emitted by someone on a circular orbit, compared to the frequency measured by someone at rest at infinity. The derivation of this formula is a bit involved, but the answer is not too complicated: $$\frac{\omega_{emit}}{\omega_\infty}=\frac{r^{3/2}+a}{\left(r^2(r-3)+2ar^{3/2}\right)^{1/2}}$$ This is for a prograde orbit, and I'm using units where $G=c=M=1$. For an extremal black hole, $a=1$ and the ISCO is at $r=1$, so you can see this factor diverges and you can get arbitrarily high redshifts coming from orbits near the horizon.
It's also interesting to consider the nearly extremal black hole, where $a=1-\epsilon$. In that case the ISCO is located at (again, from a somewhat involved calculation): $$r_{ISCO} \approx 1+(4\epsilon)^{1/3}$$ Using these formulas, we can compute the time dilation factor coming from the ISCO to lowest order: $$\frac{\omega_{emit}}{\omega_\infty}\approx\left(\frac{2}\epsilon\right)^{1/3}$$
So it is diverging as $\epsilon\rightarrow 0$, but it is doing so rather slowly. For example, say for some reason you wanted 1 hour in the orbit to correspond to 7 years at infinity, which is a factor of about $60000$. This requires an $\epsilon \approx 10^{-14}$, so it requires the black hole to be very close to extremal. Also if you consider Kip Thorne's bound that says astrophysical black holes can only ever reach $\epsilon\approx .002$, the maximum time dilation factor you can achieve is around $10$.
• @Hypnosifl No I think the bound applies to all black holes, even supermassive ones. Here's a link to the paper if you are interested: adsabs.harvard.edu/abs/1974ApJ...191..507T It seems like he was considering how accreting plasma spins up the black hole, and you would think it could spin all the way up to $a=1$, but then radiative effects stop the spin up at $a=0.998$. – asperanz Nov 14 '14 at 4:59
• Having looked over Thorne's book "The Science of Interstellar", I see he actually mentions the calculated difference in rotation from an extremal black hole, he says it's "less than the maximum by about one part in 100 trillion", same as your answer of $\epsilon \approx 10^{-14}$. And from your formula, $r_{ISCO}=1.000034$ in these units ($r=1$ would be a radius of $GM/c^2$) Also in these units the event horizon should be at $r = 1 + \sqrt{1 - a^2}$ or $r=1.00000014$. – Hypnosifl Nov 15 '14 at 22:18