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Equation for time dilation of body in orbit around Kerr black hole?

  1. Nov 13, 2014 #1

    JesseM

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    Inspired by the movie Interstellar which featured a planet orbiting a rotating supermassive black hole with an extremely high time dilation factor (slowed by a factor of 60,000 relative to observers far from the black hole), I was wondering if anyone knows of an equation for time dilation of orbiting bodies in Boyer-Lindquist coordinates (which seem to be most commonly used for describing Kerr black holes), or would be able to derive it without too much difficulty. For simplicity assume we're just talking about circular orbits in the equatorial plane, and with the orbit being in the same direction as the rotation (a 'prograde' rather than 'retrograde' orbit).

    I was able to find an expression for the ratio between proper time and coordinate time for an observer at fixed position coordinates in the Boyer-Lindquist system, see p. 27 of the thesis here--the expression is [itex]d\tau = \sqrt{1 - 2mr/\rho^2} dt[/itex], where m is a constant corresponding to half the Schwarzschild radius for a non-rotating black hole of the same mass M, given by [itex]m = GM/c^2[/itex], and [itex]\rho[/itex] is another constant defined by this equation:

    [itex]\rho^2 = r^2 + a^2 cos^2 \theta[/itex]

    Here [itex]\theta[/itex] is one of the angular coordinates of the observer at radius r, and a is yet another constant defined by a = J/Mc, where J is the black hole's angular momentum. As it turns out, in the equatorial plane where [itex]\theta = \pi/2[/itex], this reduces to the equation for time dilation near a non-rotating Schwarzschild black hole at a fixed position in Schwarzschild coordinates, [itex]d\tau = \sqrt{1 - 2GM/rc^2} dt[/itex]. This would indicate that the time dilation goes to infinity as you approach the Schwarzschild radius [itex]r = 2GM/c^2 = 2m[/itex]. But for a rotating black hole the event horizon is not actually at the Schwarzschild radius--p. 28 mentions that Boyer-Lindquist coordinates, the Kerr black hole's event horizon would be located at [itex]r = m + \sqrt{m^2 - a^2}[/itex], which reduces to the Schwarzschild radius r=2m only if a=0 (meaning the angular momentum J is 0), and is otherwise at a smaller radius than the Schwarzschild radius (for an "extremal" Kerr black hole with the maximum rotation possible before it becomes a naked singularity, [itex]a^2 = m^2[/itex], which means the event horizon is at r=m, half the Schwarzschild radius). I assume that the time dilation going to infinity has to do with the fact that to maintain a constant position you have to oppose the frame-dragging effect around the rotating black hole, and at r=2m you would actually have to move at the speed of light (as measured by local free-falling observers) to maintain a constant position, and if you were any closer than that it would be impossible to maintain a fixed position without moving faster than light (i.e. you would be inside the ergosphere, which is defined in exactly this way). P. 33 of the thesis seems to back this up, it says that the "stationary limit" which marks the boundary of the ergosphere is located at [itex]r = m + \sqrt{m^2 - a^2 cos^2 \theta}[/itex], and in the equatorial plane where [itex]\theta = \pi/2[/itex] this would reduce to r=2m.

    So, this would imply that even though the time dilation goes to infinity at r=2m for a stationary observer in the equatorial plane, it wouldn't need to do so for an orbiting observer who was orbiting in the same direction as the black hole's rotation (I also found an equation for the radius of the innermost stable orbit on http://relativity.livingreviews.org/Articles/lrr-2013-1/articlese2.html [Broken]). So again, what would the actual formula be for such an orbiting observer?
     
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  3. Nov 13, 2014 #2
    I was looking at this myself. I think what you're looking for is the reduction factor. The reduction factor for an object in free fall in Kerr metric is-

    [tex]\alpha=\sqrt{\Delta}\ \rho/\Sigma[/tex]

    where

    [tex]\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta[/tex]
    [tex]\Delta= r^{2}+a^{2}-2Mr[/tex]
    [tex]\rho^2=r^2+a^2 \cos^2\theta[/tex]

    You'll find that this goes to zero at the outer event horizon as apposed to zero at 2M.

    The reduction factor of an object in orbit is-

    [tex]A=\alpha\cdot\sqrt{1-v_{\pm}^2}[/tex]

    where A is the total reduction factor and [itex]v_{\pm}[/itex] is the tangential velocity of the orbiting object (the [itex]\pm[/itex] represents prograde and retrograde orbits)

    For a stable orbit in the equatorial plane-

    [tex]v_{s\pm}=\frac{r^2+a^2\mp 2a\sqrt{Mr}}{\sqrt{\Delta} \left[a\pm r\sqrt{r/M}\right]}[/tex]

    which is equivalent to-

    [tex]A=\sqrt{g_{tt} + 2\Omega g_{\phi t}+\Omega^2 g_{\phi \phi}}[/tex]

    where

    [tex]g_{tt}=1+2Mr/\rho^2[/tex]
    [tex]g_{t\phi}=2Mra\sin^2\theta/\rho^2[/tex]
    [tex]g_{\phi\phi}=-(r^2+a^2+[2Mra^2\sin^2\theta]/\rho^2)\sin^2\theta[/tex]

    and for a stable orbit in the equatorial plane-

    [tex]\Omega_{s\pm}=\frac{\pm\sqrt{M}}{r^{3/2}\pm a\sqrt{M}}[/tex]

    source- http://arxiv.org/abs/gr-qc/0407004
     
    Last edited: Nov 13, 2014
  4. Nov 13, 2014 #3

    JesseM

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    Thanks. By "total reduction factor" do you mean the reduction in the rate the orbiting clock is ticking relative to coordinate time, so [itex]A[/itex] is just equal to [itex]\frac{d\tau}{dt}[/itex]? If not, how would [itex]\frac{d\tau}{dt}[/itex] be found in terms of the quantities you gave?
     
  5. Nov 13, 2014 #4
    From my understanding, A and [itex]\alpha[/itex] are equal to [itex]\frac{d\tau}{dt}[/itex] though I've not seen it written definitively as this. When a=0 (i.e. no spin), the equations reduce to the solutions for time dilation and redshift for a Schwarzschild black hole.
     
    Last edited: Nov 13, 2014
  6. Nov 13, 2014 #5

    JesseM

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    Since [itex]A[/itex] and [itex]\alpha[/itex] have different values when the tangential velocity is nonzero, when you said that [itex]\alpha[/itex] was the reduction factor for an object in free fall, did you mean an object falling along a purely radial path?
     
  7. Nov 14, 2014 #6
    [itex]\alpha[/itex] is the reduction factor for a ZAMO (zero angular momentum observer), an object that falls into a black hole without any angular momentum though will still spiral into the Kerr black hole due to the frame dragging effect. [itex]A>\alpha[/itex] is when the object has angular momentum and therefore addition tangential velocity.

    It's worth noting that with the second equation [itex]A=\sqrt(g_{tt} + 2\Omega g_{\phi t}+\Omega^2 g_{\phi \phi})[/itex], in the case of a ZAMO, [itex]\Omega[/itex] would equal [itex]\omega[/itex] which is the frame dragging rate where-

    [tex]\omega=\frac{2Mra}{\Sigma^2}[/tex]

    In this case, [itex]A=\alpha[/itex]
     
  8. Nov 15, 2014 #7
    A few pages from Kip Thorne's new book 'The Science of Interstellar' are available as preview on google books which shed some light on the properties of the black hole from the movie-

    Source- The Science of Interstellar (there's no page numbers but the quote is a little over half way down)

    This means the BH has a spin parameter of a/M=1-0.00000000000001. He does go on to say-

    He also goes on to say that a more reasonable mass for the BH would be 200 million sol but wanted to keep the numbers simple so chose 100 million sol.
     
  9. Nov 16, 2014 #8

    JesseM

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    Thanks again for your help. I used your equations, combined with the equation for the radius of the ISCO given in section 2.5 http://relativity.livingreviews.org/Articles/lrr-2013-1/articlese2.html [Broken], and found that the black hole would need to have a rotation rate of about (1 - (1.33266 * 10^(-14))) times the rotation rate of an extremal black hole, in that case the ISCO would be located at a radius of 1.000037636343 times [itex]GM/c^2[/itex], the velocity would be 0.500014c making the period 1.72 hours (relative to the Boyer-Lindquist time coordinate, not the proper time for someone on the planet), and this would give 1/A = 61362.023, which is very close to the number of hours in seven years (since it's specified that the time dilation factor is such that one hour on the planet is equal to 7 years on Earth), or (24*365.25*7) = 61362.

    Aside from more significant figures, several of these numbers end up being identical to the ones Thorne gives in his book--in the "Some Technical Notes" at the end, in the section about ch. 6 he says that "Gargantua's actual spin is less than its maximum possible spin" by 1.3 * 10^(-14), and in ch. 17 he says the planet orbits once every 1.7 hours. He doesn't actually specify the radius of the planet's orbit, but he does say it's at the innermost stable point, and the fact that his other numbers agree with what I got suggest he was assuming the same radius.

    If anyone wants to play around with the numbers, I used the technique of combining several equations into long expressions that could be entered directly into the online calculator here, and then varying different parameters to see what effect it had on the output. So, I'll post the long expressions here, that way anyone who's curious can vary some of the parameters and see how it changes the answers. I wanted units where G=c=1, so I used seconds for time and light-seconds for distance, and I found that to get G=1, in this case I had to use a mass unit (call it an 'm.u.') such that 1 m.u. = 4.037256 * 10^35 kg, which meant that a black hole with a mass of 100 million Suns would have M=492.7 m.u. (and the gravitational radius [itex]GM/c^2[/itex] that appears in these equations would be 492.7 light-seconds in these units). In this case, the expression for the innermost stable orbit for this black hole, given the assumption that it's rotating at (1 - (1.33266 * 10^(-14))) times the maximum possible rate, was the gravitational radius of 492.7 light-seconds times this factor:

    (3 + (sqrt(3*((1 - (1.33266*10^(-14))))^2 + (1 + ((1 - ((1 - (1.33266*10^(-14))))^2)^(1/3))*((1 + ((1 - (1.33266*10^(-14)))))^(1/3) + (1 - ((1 - (1.33266*10^(-14)))))^(1/3)))^2)) - sqrt((3 - (1 + ((1 - ((1 - (1.33266*10^(-14))))^2)^(1/3))*((1 + ((1 - (1.33266*10^(-14)))))^(1/3) + (1 - ((1 - (1.33266*10^(-14)))))^(1/3))))*(3 + (1 + ((1 - ((1 - (1.33266*10^(-14))))^2)^(1/3))*((1 + ((1 - (1.33266*10^(-14)))))^(1/3) + (1 - ((1 - (1.33266*10^(-14)))))^(1/3))) + 2*(sqrt(3*((1 - (1.33266*10^(-14))))^2 + (1 + ((1 - ((1 - (1.33266*10^(-14))))^2)^(1/3))*((1 + ((1 - (1.33266*10^(-14)))))^(1/3) + (1 - ((1 - (1.33266*10^(-14)))))^(1/3)))^2)))))

    This gives an innermost stable orbit of 1.000037636343 times the gravitational radius, but anyone can do a find-and-replace to change 1.33266*10^(-14) into some other number to find how this changes the orbital radius.

    Likewise, here's the expression for 1/A, or how many hours a distant observer would see go by for every hour spent at an orbit of 1.000037636343 times the gravitational radius, given the rotation rate of (1 - (1.33266 * 10^(-14))) times the maximum:

    1/(sqrt((1.000037636343 * 492.7)^2 + ((1 - (1.33266*10^(-14)))*492.7)^2 - 2*492.7*(1.000037636343 * 492.7)) * (1.000037636343 * 492.7) / sqrt(((1.000037636343 * 492.7)^2 + ((1 - (1.33266*10^(-14)))*492.7)^2)^2 - 2*((1 - (1.33266*10^(-14)))*492.7)*sqrt(492.7*(1.000037636343 * 492.7)))*sqrt(1 - (((1.000037636343 * 492.7)^2 + ((1 - (1.33266*10^(-14)))*492.7)^2 - 2*((1 - (1.33266*10^(-14)))*492.7)*sqrt(492.7*1.000037636343 * 492.7))/(sqrt((1.000037636343 * 492.7)^2 + ((1 - (1.33266*10^(-14)))*492.7)^2 - 2*492.7*(1.000037636343 * 492.7))*(((1 - (1.33266*10^(-14)))*492.7) + 1.000037636343*492.7*sqrt(1.000037636343 * 492.7/492.7))))^2))

    For example, if you substitute 1.5 for 1.000037636343 to see how time would be passing if a ship were orbiting at 1.5 times the gravitational radius while explorers were on a planet at the innermost stable orbit, you'll find that the ship only experiences 1 hour for every 5.25 hours of Earth time, a much smaller difference.

    And a formula for the coordinate velocity (in light-seconds per second) at the innermost stable orbit in this example:

    ((1.00003763634314 * 492.7)^2 + ((1 - (1.33266*10^(-14)))*492.7)^2 - 2*((1 - (1.33266*10^(-14)))*492.7)*sqrt(492.7*1.00003763634314 * 492.7))/(sqrt((1.00003763634314 * 492.7)^2 + ((1 - (1.33266*10^(-14)))*492.7)^2 - 2*492.7*(1.00003763634314 * 492.7))*(((1 - (1.33266*10^(-14)))*492.7) + 1.00003763634314*492.7*sqrt(1.00003763634314 * 492.7/492.7)))

    Lastly, note that someone in this physics stackexchange thread gave some simplified approximate formulas--if [itex]\epsilon[/itex] is the fraction by which the black hole's rotation rate differs from the maximum rate, i.e. the 1.33266*10^(-14) in my equations, then an approximate formula for the innermost stable orbit as a multiple of the gravitational radius would be [itex]1 + (4\epsilon )^{1/3}[/itex], and an approximate formula for 1/A at the innermost stable orbit would be [itex](2/\epsilon)^{1/3}[/itex]. For the first this would give 1.0000376345, close to the more exact answer of 1.000037636343, and for the second it would give 53143, not too far off from the more accurate answer of 61362.
     
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  10. Nov 30, 2014 #9
    Based on a/M=1-1e-14 and 1hr=7yrs, I get the following quantities (rounded up to the nearest km)-

    MSO=147,690,032 km

    A=0.0000162967

    vt=0.5000129c

    T=1.7196 hrs

    Based on the quantity for A, the orbit radius would be at 147,690,541 km (if I recall correctly, Kip Thorne mentioned that the planet was 'just outside the MSO'). The outer event horizon of the BH would be at 147,685,004 km which means the centre of the planet would be 5537 km from the EH! Even if the planet only had a radius of 5000 km, one side would be no less than just 537 km from the EH. It would make for an interesting sky. The MSO only exists at the equator and normally denotes the inner edge of the accretion disk so the planet would be in the disk (as apposed to outside it as shown in the movie). It's also deep withing the ergoregion so while the orbital velocity would be correct in respect of passage through spacetime, the local velocity relative to infinity would be much greater.
     
  11. Nov 30, 2014 #10

    Matterwave

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    I didn't read all the posts in this thread, because I have not had any experience working with the Kerr solution, so I figure much of the discussion is beyond me. The main questions I would have with this scenario in the movie though is 1) can a planet find a stable orbit close enough to Gargantua such that this 60,000:1 time dilation is possible? In a Schwarzschild scenario, this is impossible as there are no stable orbits within the photon orbit radius at r<3M and the time dilation at that radius is not anywhere near 60,000:1. 2) How far away would the orbiter have to be so that the orbiter ages roughly the same as Earth, but the shuttle ages 60,000 times slower? After all, the physicist who remained on the orbiter aged 23 years in the time Cooper aged 3 hours. It seemed highly implausible to me that a 60,000:1 time dilation factor could be achieved for the surface of a planet vs an orbiter which is quite close to the planet.
     
  12. Dec 1, 2014 #11
    The answers are given from my understanding of the thread:

    Yes. In the limit that the block hole is maximally rotating, the innermost stable orbit is at the event horizon unlike the Schwarzschild case where it is at three times the Schwarzschild radius. In order to get a 60000:1 time dilation one needs a rotation of (1 - (1.33266 * 10^(-14))) of the maximum value.

    Well, when one distances himself a few times the gravitation radius [itex]GM/c^{2}[/itex] the time dilation becomes of order unity.

    Using the formula given in the thread,
    For a distance of 1 [itex]GM/c^{2}[/itex] (147,685,004 km, about the Earth-Sun distance) from the planet the time dilation is 2.5:1, for a distance of 2 [itex]GM/c^{2}[/itex] the time dilation is 1.6:1, for a distance of 3 [itex]GM/c^{2}[/itex] the time dilation is 1.4:1, etc....


    I have a question about the tidal forces on the planet. Using a naive Newtonian estimation the tidal acceleration is [itex]~GMr_{p}/r_{g}^{3}[/itex]

    which is still about 10 times the tidal acceleration on the face of Sun. Can the planet stay intact? How much Does general relativity alter the strength of the tidal forces?

    Thanks
     
  13. Dec 1, 2014 #12

    Matterwave

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    These all seem like pretty ridiculous distances for an "orbiter" to orbit the planet...That's like saying I'm going to "orbit" the Earth by staying where Mars is...o.o

    I was expecting the orbiter to be what a few thousand km from the surface...
     
  14. Dec 2, 2014 #13

    JesseM

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    They didn't actually say the orbiter was orbiting the planet, rather the diagram they drew showed it in a higher orbit around Gargantua.
     
  15. Dec 2, 2014 #14

    Matterwave

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    If their shuttles could travel AU distances in a matter of hours (as it seemed to me in watching the film, perhaps it took much longer), then why have the orbiter at all? And why did the orbiter take 2 years to reach Saturn from Earth?

    My only guess is they used the rotation of Gargantua to assist the shuttle in its entry and exit (indeed, they must have entered the ergosphere and then exited the ergosphere), but is it really possible to set up an optimal path that makes this so easy?
     
  16. Dec 2, 2014 #15

    JesseM

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    Kip Thorne explained in ch. 7 of The Science of Interstellar that in order to explain how they could get from one orbit to another despite the major differences in velocities, he was imagining that there were plenty of "intermediate mass black holes" orbiting Gargantua which were large enough (around 10,000 times the mass of our Sun) so that one could fly in close and get a large gravitational assist without the tidal forces becoming deadly. Apparently Christopher Nolan wanted to preserve the specialness of Gargantua for the audience so these other black holes weren't mentioned, but Thorne said that in his "science interpretation" of the movie that's how they did it even if they weren't mentioned onscreen (they did include a fudged version of this in a line about a gravitational assist from a neutron star, although it wouldn't really be massive enough to give the required velocity shift on its own).
     
  17. Dec 2, 2014 #16

    Matterwave

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    I should buy this book...but at this point it seems that the conditions are being set up just perfectly so that the things in the movie could transpire. The masses of these intermediate mass black holes would have to be just right, they would have to be spaced and distributed just right...I can't imagine how this system could have possibly formed in such a way, but ok, I guess that's...one way to get things done...
     
  18. Dec 2, 2014 #17

    JesseM

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    I think Kip Thorne would agree with you, it seems like his self-imposed rule was that the movie not slip into total fantasy by depicting things that explicitly violate basic physical laws (like true faster-than-light travel), but that for dramatic purposes it could be unrealistic in other ways (I suppose you could always imagine the system itself was engineered by the beings that created the wormhole). In another section where he talks about how Gargantua's spin could differ by only 1 part in 100 trillion from the maximum spin rate for an extremal rotating black hole, he notes that there's a physical argument (which he himself came up with in 1975) for thinking realistic rotating black holes would be unlikely to ever spin much faster than about 0.998 the rate of an extremal black hole, but goes on to say:
     
  19. Dec 2, 2014 #18

    Matterwave

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    My next concern would then be... can black holes exist in the ergosphere of another black hole? This is what one would require right? Intermediate mass black holes to be near the planet which is quite obviously within the ergo sphere? I believe the two body problem in GR has not been solved, but are there any indications that such a configuration would at all be possible?
     
  20. Dec 2, 2014 #19

    JesseM

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    Technically, if they haven't solved the two-body problem I suppose that means they can't even know that planetary orbits can be stable (around black holes or even around stars like our Sun), only those of test particles with infinitesimal mass. But I would think the test particle approximation would be better when one body is much more massive than the other (as is true with Gargantua vs. a planet, or Gargantua vs. an intermediate mass black hole), worse when their masses are more comparable. Do you know of any effects that would make a black hole orbiting within the ergosphere of a much larger rotating black hole any less plausible than a planet in the same orbit around the rotating black hole?

    A more likely problem, IMO, is that having multiple black holes with masses of around 10,000 suns orbiting within a few AU of one another would likely lead to their disrupting the orbits of much less massive bodies like planets. Maybe you could solve that by imagining the planets were really moons of these intermediate-mass black holes, and this just wasn't mentioned (that would also have the advantage that the travelers would not need to travel too large a distance from the planet in order to perform a gravitational slingshot, and thus fuel use could be minimized).
     
  21. Dec 2, 2014 #20

    Matterwave

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    I'm more concerned with the geometry of the configuration. I mean, when you have to use GR for both the central (Gargantua) and the orbiting (IM BH) bodies, I have a hard time believing something funky won't happen haha. For example, what if two black holes merge, how do the light cones within the area of overlap between the two event horizons behave? I have no idea. Already light cones behave weirdly when inside an ergo sphere, all future directed causal curves will move along with the rotation of the central body. What happens when you superimpose an event horizon somewhere in here? Can I move against the rotation of the central body then? It just seems some really weird physics is going to be involved...
     
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