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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 ). So again, what would the actual formula be for such an orbiting observer?
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 ). So again, what would the actual formula be for such an orbiting observer?
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