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Suppose that we have a non-rotating black hole with mass $M$. We know that the time dilation $\Delta t'$ at a distance $r$ to the center of the black hole is given by $\Delta t' = \Delta t \sqrt{1 - \frac{2GM}{rc^2}}$.

Now, we want to consider a rotating black hole with mass $M$ and specific angular momentum $a = 1 - \epsilon$. How can we obtain the time dilation $\Delta t'$ at a distance $r$ to the center of the black hole?

In a previous question, we already had the time dilation for the special case of the innermost stable circular orbit (ISCO) for a maximally rotating black hole, but I have been unable to find a solution to the more general problem. I would like to be able to work out, for instance, the time dilation of a ship in orbit of the fictional Gargantua when $M = 10^{8}\ M_\odot = 2 \times 10^{38}\ kg,\ \epsilon = 10^{-14},\ r - R = 2\ AU = 3 \times 10^{11}\ m$.

In the comments, there are relevant answers to questions about another previous question: First, can $r$ be anything other than ISCO? Second, is there an explanation why mass is not involved at all in a rotating black hole, unlike a non-rotating one? Third, when a rotating black hole is said to have 100 million solar masses, does it typically refer to its irreducible mass?

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    $\begingroup$ This question was already answered at physics.stackexchange.com/questions/557123/… to compare with Schwarzschild simply set a=0 $\endgroup$
    – Yukterez
    Dec 20, 2023 at 19:11
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    $\begingroup$ Kenneth Kho asked: "First, can r be anything other than ISCO?" - Yes, r can be anything, not only the ISCO, but remember that the coordinate radius r is not the physical radius, and in case of the rotating Kerr BH not even the circumference divided by 2π, so make sure to know what you compare. - Kenneth Kho asked: "Second, is there an explanation why mass is not involved at all in a rotating black hole?" - It is, we just set G=M=c=1 for brevity so that lengths are in GM/c² and times in GM/c³, but if you want the metric including M see here $\endgroup$
    – Yukterez
    Dec 20, 2023 at 19:20
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    $\begingroup$ With rotating black holes there's also a difference between the total mass M=E/c² as measured at infinity (this is what we set to 1) and the irreducible mass ℳ which doesn't contain the mass equivalent of the rotational energy, see physics.stackexchange.com/a/779659/24093 so if you compare Schwarzschild with Kerr make sure you know if you compare equal rest mass or equal energy black holes. $\endgroup$
    – Yukterez
    Dec 20, 2023 at 19:28
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    $\begingroup$ No, normally they mean the M as it is measured from far away and gives the Newtonian g=GM/r² in the limit of very large r. In most cases they don't even know the exact spin, until now the measured or inferred spin always had large uncertainty bars. The irreducible mass can go as low as ℳ=M√0.5=0.707M (with a=1). $\endgroup$
    – Yukterez
    Dec 20, 2023 at 19:44

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The first answer in the question you linked includes the solution to the more general problem you are looking for right there before the special case for the ISCO, it is the first formula in asperanz's answer. It's just a matter of algebra. You just need to change your $r$ to be in terms of the minimal ISCO for your black hole in question.

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