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In the imdb.com goofs page of the movie Interstellar, I found this statement:

enter image description here

How does time dilation affect $g$-force? If this could be explained in Laymen's terms it would be much appreciated.

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  • $\begingroup$ I've always heard it stated the other way around (as "how does gravitation dilate time?"), so maybe you'd be looking for a "symmetry" that doesn't exist. Noether's Theorem, at en.wikipedia.org/wiki/Noether%27s_theorem, relates symmetries to conservation laws, of which the best-known may be "conservation of momentum". $\endgroup$
    – Edouard
    Jun 4, 2020 at 16:05
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    $\begingroup$ I think people at IMDB mistook g-force for tidal forces. $\endgroup$
    – ksousa
    Jun 4, 2020 at 22:37
  • $\begingroup$ The g-force is defined by how much the time dilation changes over the radial distance. $\endgroup$
    – safesphere
    Jun 27, 2020 at 5:26

1 Answer 1

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The radial coordinate for the innermost stable prograde orbit at which the planet could orbit the black hole would in natural units of $G=M=c=1$ be at

$$r = 3 + Z_2 - \sqrt{(3-Z_1)(3+Z_1+2Z_2)}$$

with the terms

$$Z_1 = 1 + \sqrt[3]{1-a^2} \left( \sqrt[3]{1+a} + \sqrt[3]{1-a} \right) \ , \ \ Z_2 = \sqrt{3a^2 + Z_1^2}$$

For a black hole with a spin parameter of $a=0.99999999999999$ which they used according to this source that would be

$$r = 1.000034191427736$$

where the gravitational component of the time dilation is

$$\sqrt{g^{tt}} = \surd\left({\frac{4 \left(\left(a^2+r^2\right)^2-a^2 \left(a^2+(r-2) r\right) \sin ^2 \theta \right) \left(a^2 \cos ^2 \theta +r^2\right)}{\left(a^2+(r-2) r\right) \left(a^2 \cos (2 \theta )+a^2+2 r^2\right)^2}}\right)$$

which for our spin parameter and in the equatorial plane where $\theta=\pi/2$ gives

$$\sqrt{g^{tt}} = 58494.69347667821$$

The prograde orbital velocity relative to a ZAMO is

$$v = \frac{a^2-2 a \sqrt{r}+r^2}{\sqrt{a^2+(r-2) r} \left(a+r^{3/2}\right)}$$

which is in units of $c$

$$v=0.500012818545588$$

so the total time dilation would be

$$\dot{t}=\sqrt{\frac{g^{tt}}{1-v^2}} = 67544.43127396276$$

that means you get $7.71$ years per hour, as it is claimed here. In the older version of the answer I wrote before I found out which spin parameter they used in they movie I went under the assumption that the black hole spun at the Thorne limit of $a=0.998$, but in the movie they used a much larger spin parameter which can only be achieved by artificially spinning up the black hole, but not naturally by accretion. The Imdb argument about the g-force is gish gallop though, since in orbit you do not feel any g-force.

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  • $\begingroup$ Nice. I imagine by g-force they were really talking about tidal force. But to discuss that you need to put a mass on your black hole. $\endgroup$
    – ProfRob
    Jun 6, 2020 at 20:00
  • $\begingroup$ For future reference things get a bit weirder, in order to get the time dilation required by Nolan, Kipp Thorne used special relativity; the planet was supposedly spinning at 99.9% light speed $\endgroup$
    – yolo
    Oct 29, 2020 at 13:23
  • $\begingroup$ A planet can't spin that high, but the black hole's spin was even larger, according to the fandom wiki it was a/M=0.99999999999999, which corresponds to a horizon rotation speed of v/c=0.99999985857865 $\endgroup$
    – Yukterez
    Feb 1 at 17:45

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