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

[stevebd1]

Using Autocad, I decided to draw the black hole and planet to scale, see attached drawings. In the first image, BH_1, the outer curves are the boundary of the ergosphere and the circle is the event horizon. The line at the equator represents the accretion disk. The small box at the junction between the accretion disk and the EH is 3.0e+7 by 2.0e+7 km and is represented in the second image BH_2. The smaller box in the centre of the image BH_2 is 1.5e+6 by 1.0e+6 km and is represented in BH_3. In BH_3, you can just make out the planet at the centre of the image (I've used a 10,000 kmm dia for the planet). This demonstrates just how close to the EH the planet needs to be to get that kind of time dilation.

 

[George Jones]

I want to bump this thread.

I didn't want to look at this until I knew story, and, last weekend, I read the novelization of movie. I also have a copy of Kip Thorne's book.

I want to attempt my own systematic time dilation calculation for an orbiting observer. Yesterday and today, I did the background reading, identified my line of attack, and started setting up the problem. Over the next few days (or weeks) I will try to gradually chip away at the problem. (Family responsibilities will prevent me from spending much time on this on any particular day.) It looks like my main reference will be be the treatment of rotating black holes in "General Relativity: An Introduction for Physicists" by Hobson, Efstathiou, and Lasenby.

Possibilities: 1) I will abandon my attempt; 2) I will get the wrong answer; 3) I will get right answer.

 

[George Jones]

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

I think that the time dilation factor for a circular orbit about a Schwarzschild black hole is

$$\frac{dt}{d \tau} = \frac{1}{\sqrt{1 - \frac{3M}{r}}}.$$

 

[WannabeNewton]

Possibilities: 1) I will abandon my attempt; 2) I will get the wrong answer; 3) I will get right answer.

The take home final for the first semester of GR at my university this past semester contained a problem that was verbatim the calculation you seek to do so if you post yours I can compare it with the calculation from the problem. I can't post the problem itself of course, for obvious reasons. Have fun!

 

[George Jones]

I got the result.

 

[George Jones]

if you post yours I can compare it with the calculation from the problem. I can't post the problem itself of course, for obvious reasons. Have fun!

I got the result.

I did have fun working on this for a few mornings at my local coffee shop. Your method probably is more elegant, as I just used brute force.

I didn't want to use numbers, I wanted to derive Thorne's expression for gravitational time dilation given at the bottom of page 292 of "The Science of Interstellar",

$$\delta = \frac{16 S^3}{3 \sqrt{3}}.$$

Here, gravitational time dilation is given by $S = \left( dt / d \tau \right)^{-1}$, where coordinate time $t$ is the proper time of a distant, stationary observer, and $\tau$ is the proper time for an observer in the innermost co-rotating stable circular orbit of a very rapidly spinning black hole. The limiting spin parameter for a rotating black hole is $a = M$, and

$$\delta = \frac{\mu - a}{\mu}$$

is the fractional difference between the black hole's spin parameter and the limiting value.

According to the beautiful classic 1972 paper by Bardeen, Press, and Teukolsky,

http://www.google.ca/url?sa=t&rct=j...Avu_sGBpcyN6_VFk--cifew&bvm=bv.82001339,d.cGU

the innermost co-rotating stable circular orbit is given by $r = \left( 1 + \left(4 \delta \right)^{1/3} \right) M$.

Define $x = \delta^{1/3}$, so that $a = \left(1 - x^3 \right) M$ and $r = \left( 1 + 4 ^{1/3} x \right) M$. These equations will be used repeatedly.

According to HEL (13.40), on a timelike geodesic in the equatorial plane,

$$\frac{dt}{d\tau} = \frac{1}{\Delta} \left[\left( r^2+ a^2 +\frac{2Ma^2}{r} \right) E- \frac{2Ma}{r} L \right],$$

and, in a co-rotating circular equatorial orbit,

$$E = \frac{1-\frac{2M }{r}+a\sqrt{\frac{M}{r^3}}}{\sqrt{1-\frac{3M }{r}+2a\sqrt{\frac{M}{r^3}}}}$$

$$L = \frac{\sqrt{M r}\left( 1+\frac{a^{2}}{r^2}-2a\sqrt{\frac{M }{r^3}}\right) }{\sqrt{1-\frac{3M }{r}+2a\sqrt{\frac{M }{r^3}}}} $$

Using the expressions for $a$ and $r$, and then taking first-order Taylor series approximations gives

$$E = \frac{1}{\sqrt{3}}\left( 1 + 4^{1/3} x\right) $$

$$L = \frac{2}{\sqrt{3}}\left( 1+2^{\frac{2}{3}}x\right) M. $$

Using the expressions for $a$ and $r$ in $\Delta = r^2-2M r+a^2$, and keeping only the lowest order term, gives $\Delta = 2^{4/3} M^2 x^2$.

Finally, plugging everything back into the expression for $dt/d\tau$, and taking a first-order Taylor of the stuff that multiplies $1/\Delta$ gives

$$\frac{dt}{d\tau} = \frac{1}{S} = \frac{2^{4/3}}{\sqrt{3}} \frac{1}{x}, $$

which is the desired result.

I suspect that all of this can be streamlined substantially.

 

[Matterwave]

I think that the time dilation factor for a circular orbit about a Schwarzschild black hole is

$$\frac{dt}{d \tau} = \frac{1}{\sqrt{1 - \frac{3M}{r}}}.$$

Ah, I realized I was only considering stationary observers...

 

[George Jones]

Taylor series

I should add that I used a computer algebra package to do the Taylor series stuff.

 

[stevebd1]

Here's a few miscellaneous points relating to this thread-

Another form of the Kerr metric is-
ds^2=\alpha dt^2-\varpi^2(d\phi-\omega dt)^2-\frac{\rho^2}{\Delta}dr^2-\rho^2 d\theta^2
which shows some key components of the Kerr metric clearly, such as \alpha as the redshift/time dilation/reduction factor, \omega which is the frame dragging rate and \varpi as the reduced circumference (sometimes written as R) where \varpi=\Sigma\sin \theta/\rho (incidently, regardless of spin and mass, when r=r₊ the reduced circumference equals 2M at the equator and 2M_ir at the poles where M_ir is the irreducible mass).The equation featured in the link from JesseM's post #8 (see below) is exactly equivalent to 1/A (when using M=1 for r) (presumably \omega in this case means wavelength)
\frac{\omega_{emit}}{\omega_\infty}=\frac{r^{3/2}+a}{\left(r^2(r-3)+2ar^{3/2}\right)^{1/2}}On a side note, \Omega_s is related to v_s by the following equation-
v_s=(\Omega_s-\omega)\frac{R}{\alpha}
v_s=(\Omega_s-\omega)\frac{\Sigma^2\sin\theta}{\rho^2\sqrt{\Delta}}
http://www.icra.it/MG/mg12/talks/apt1_slany.pdf page (there's also a variation of the equation for A on page 5)It's also worth pointing out that coordinate radius of the inner event (Cauchy) horizon is 147,684,962 km where a weak singularity is supposed to occur, this is a coordinate difference of 42 km which means Cooper would have hit this in about 0.00014 seconds once passing the outer event horizon.

 

[JesseM]

One point where his numbers don't seem to agree with mine: in this section he also says that "Miller's planet moves at 55 percent of the speed of light, 0.55c", whereas I got that it orbits at 0.500014c. But maybe he was assuming that 1 hour for every seven years on Earth translates to something different than the ratio of 61362 that I assumed.

Looking over the book again, I think the approximation he used to get 55% rather than 50% may have been a very simple one: in chapter 17 he writes that "Einstein's laws dictate that, as seen from afar, for example, from Mann's planet, Miller's planet travels around Gargantua's billion-kilometer-circumference orbit once each 1.7 hours". If you divide a billion kilometers by 1.7 hours = 6120 seconds, you get a velocity of 163399 km/s, which rounded off to the nearest percentage point is indeed 55% the speed of light. While 1.7 hours agrees with the figure I got for the orbital period in Boyer-Lindquist coordinates of 1.72 hours, a circumference of a billion kilometers is a rough approximation of the circumference at the gravitational radius (and the circumference at the innermost stable circular orbit would only be larger than the circumference at the gravitational radius by 1.000037636343 according to the numbers I got in post #8), or 2*pi*492.7 light-seconds = 3096 light-seconds = 928.2 million kilometers. If you divide 0.928 billion kilometers by 1.72 hours, in that case you do get the 50% light speed figure that I got in post #8. So I think the issue is probably just that he used more accurate numbers when calculating other quantities like the rotation rate and the time dilation factor at the innermost stable circular orbit, but then rounded off the circumference to the nearest hundred billion kilometers when calculating speed.

Another issue is that with the numbers I got, which as I mentioned before seem to agree with Thorne's numbers in all cases except for the one above, would put the innermost stable circular orbit awfully close to that of the event horizon. The event horizon radius for a Kerr black hole is given on p. 28 of this pdf by the equation $r = m + \sqrt{m^2 - a^2}$, where $m$ is the gravitational radius $GM/c^2$ = 492.7 light-seconds for a 100-billion-solar-mass-black hole, and $a$ is the rotation parameter $J / Mc$ where $J$ is the angular momentum; $a$ can equivalently be expressed as $m$ times the rotation rate as a fraction of that of an extremal black hole (in which case $a^2 = m^2$ as mentioned on p. 26 of the pdf), and in post #8 I got this fraction as (1 - (1.33266 * 10^(-14))), so plugging all this into the calculator here I get an event horizon radius of 492.700080437255 light-seconds. So if the innermost stable circular orbit is at 492.7*1.000037636343 = 492.718543426196 light-seconds, the difference between the two is only 0.018462988941 light-seconds, or about 5535 km, smaller than Earth's radius of 6371 km (and Miller's planet was supposed to have 1.3 times Earth's gravity). Of course you could imagine it was only the innermost surface of the planet that was at the radius of the ISCO rather than the center, but then the innermost surface wouldn't be orbiting as fast as a point particle at the ISCO and that would mess up the time dilation calculation.

In "some technical notes" at the end of the book, Thorne actually mentions that he fudged the mass of the black hole when calculating the tidal forces on Miller's planet--he finds an equation for the limiting case where the tidal force across the planet would be equal to the planet's own gravitational acceleration it its surface, showing that in this case the black hole's mass is $\sqrt{3 c^3} / \sqrt{2 \pi G^3 \rho }$, and assuming a density $\rho$ for Miller's planet of 10,000 kilograms/meter^3, he gets a limiting case of "3.4 * 10^38 kilograms for Gargantua's mass, which is about the same as 200 million suns--which in turn I approximate as 100 million suns". In terms of the units where c=1 (with seconds for time and light-seconds for distance) and G=1 that I've been using, a unit mass is equal to 4.037256 * 10^35 kg, so 3.4 * 10^38 kg is 842.2 in these units. But substituting this in for 492.7 in the equations I found in post #8 I find it makes no significant difference to the rotation rate needed for a time dilation factor of 61362 (1 hour for every 7 years experienced by distant observers) at the innermost stable circular orbit--the equation for innermost stable circular orbit radius as a multiple of the gravitational radius $GM/c^2$ only depends on the rotation rate as a fraction of the extremal rate, not on the actual value of the gravitational radius, so if we stick with a rotation fraction of (1 - (1.33266 * 10^(-14))) and an innermost stable circular orbit of 1.000037636343 times the gravitational radius, then the formula for the time dilation factor 1/A becomes:

1/(sqrt((1.000037636343 * 842.2)^2 + ((1 - (1.33266*10^(-14)))*842.2)^2 - 2*842.2*(1.000037636343 * 842.2)) * (1.000037636343 * 842.2) / sqrt(((1.000037636343 * 842.2)^2 + ((1 - (1.33266*10^(-14)))*842.2)^2)^2 - 2*((1 - (1.33266*10^(-14)))*842.2)*sqrt(842.2*(1.000037636343 * 842.2)))*sqrt(1 - (((1.000037636343 * 842.2)^2 + ((1 - (1.33266*10^(-14)))*842.2)^2 - 2*((1 - (1.33266*10^(-14)))*842.2)*sqrt(842.2*1.000037636343 * 842.2))/(sqrt((1.000037636343 * 842.2)^2 + ((1 - (1.33266*10^(-14)))*842.2)^2 - 2*842.2*(1.000037636343 * 842.2))*(((1 - (1.33266*10^(-14)))*842.2) + 1.000037636343*842.2*sqrt(1.000037636343 * 842.2/842.2))))^2))

...which, when plugged into the calculator here, gives the value of 1/A as 61362.065, still very close to the desired rate. And now the event horizon radius will be at 842.200137495953 while the innermost stable circular orbit radius will be at 842.231697328075, a difference of 0.031559832122 light-seconds or 9461.4 km, leaving room for an Earth-sized planet with its center at the radius of the innermost stable circular orbit.

 

[stevebd1]

I realized while posting in another thread that \alpha is not the redshift/time dilation for a ZAMO that has fallen from rest at infinity (as stated in post #26 & #28), it's for an object that is hovering at a specific radius, reducing to the Schwarzschild solution d\tau=dt\sqrt(1-2M/r) when a=0. This doesn't affect any of the equations, just the actual definition of what \alpha is.

The time dilation for a ZAMO that has fallen from rest at infinity would be-

A_{\text{ff}}=\alpha\cdot\sqrt{1-v_{\text{ff}}^2}

where

v_{\text{ff}}=\left(\frac{2M}{r}\right)^{1/2}\ \ \frac{r\ (r^2+a^2)^{1/2}}{\Sigma}

This would reduce to the Schwarzschild solution for an object falling from rest at infinity when a=0-

\frac{d\tau}{dt}=\left(1-\frac{2M}{r}\right)

where d\tau/dt=\sqrt(1-2M/r)\cdot\sqrt(1-v_{\text{ff}}^2) and v_{\text{ff}}=\sqrt(2M/r)