Physics:Innermost stable circular orbit

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The innermost stable circular orbit (often called the ISCO) is the smallest marginally stable circular orbit in which a test particle can stably orbit a massive object in general relativity.[1] The location of the ISCO, the ISCO-radius ([math]\displaystyle{ r_{\mathrm{isco}} }[/math]), depends on the angular momentum (spin) of the central object.

The ISCO plays an important role in black hole accretion disks since it marks the inner edge of the disk.

Non-rotating black holes

For a non-spinning massive object, where the gravitational field can be expressed with the Schwarzschild metric, the ISCO is located at

[math]\displaystyle{ r_{\mathrm{ms}} = 6 \frac{GM}{c^2} = 3 R_S, }[/math]

where [math]\displaystyle{ R_S }[/math] is the Schwarzschild radius of the massive object with mass [math]\displaystyle{ M }[/math]. Thus, even for a non-spinning object, the ISCO radius is only three times the Schwarzschild radius, [math]\displaystyle{ R_S }[/math], suggesting that only black holes and neutron stars have innermost stable circular orbits outside of their surfaces. As the angular momentum of the central object increases, [math]\displaystyle{ r_{\mathrm{isco}} }[/math] decreases.

Circular orbits are still possible between the ISCO and the so called marginally bound orbit, which has a radius of

[math]\displaystyle{ r_{\mathrm{mb}} = 4 \frac{GM}{c^2} = {2 R_S}, }[/math]

but they are unstable.

For a massless test particle like a photon, the only possible but unstable circular orbit is exactly at the photon sphere.[2] Inside the photon sphere, no circular orbits exist. Its radius is

[math]\displaystyle{ r_{\mathrm{ph}} = 3 \frac{GM}{c^2} = {1.5 R_S}. }[/math]

The lack of stability inside the ISCO is explained by the fact that lowering the orbit does not free enough potential energy for the orbital speed necessary: the acceleration gained is too little. This is usually shown by a graph of the orbital effective potential which is lowest at the ISCO.

Rotating black holes

The case for rotating black holes is somewhat more complicated. The equatorial ISCO in the Kerr metric depends on whether the orbit is prograde (negative sign below) or retrograde (positive sign):

[math]\displaystyle{ r_{\mathrm{ms}} = \frac{GM}{c^2} \left( 3 + Z_2 \pm \sqrt{(3-Z_1)(3+Z_1+2Z_2)} \right) \le 9 \frac{GM}{c^2} = 4.5 R_S }[/math]

where

[math]\displaystyle{ Z_1 = 1 + \sqrt[3]{1-\chi^2} \left( \sqrt[3]{1+\chi} + \sqrt[3]{1-\chi} \right) }[/math]
[math]\displaystyle{ Z_2 = \sqrt{3\chi^2 + Z_1^2} }[/math]

with the rotation parameter [math]\displaystyle{ \chi=2a/R_S=cJ/(M^2G) }[/math].[3] As the rotation rate of the black hole increases [math]\displaystyle{ \chi \to 1 }[/math], the retrograde ISCO increases towards [math]\displaystyle{ r_{\mathrm{ms}} \to 9GM/c^2 }[/math] (4.5 times the a=0 horizon radius) while the prograde ISCO decreases. [4]

The marginally bound radius and photon sphere radius in the equatorial plane are respectively

[math]\displaystyle{ r_{\mathrm{mb}} = \frac{GM}{c^2}(1+\sqrt{1 \pm \chi})^2 \le \frac{3+\sqrt 8}{2}R_S \approx 5.828427 \frac{GM}{c^2} \approx 2.9142 R_S }[/math],
[math]\displaystyle{ r_{\mathrm{ph}} = 2\frac{GM}{c^2}(1+\cos(\tfrac{2}{3}\cos^{-1}(\pm \chi))) \le 4 \frac{GM}{c^2} = 2 R_S }[/math].

All these three radii drop to [math]\displaystyle{ r_{\mathrm{ms}} \to r_{\mathrm{mb}} \to r_{\mathrm{ph}} \to R_S/2 }[/math] for prograde rotation at an extremal black hole with [math]\displaystyle{ -\chi \to -1 }[/math], still logically and locally distinguishable though.

If the particle is also spinning there is a further split in ISCO radius depending on whether the spin is aligned with or against the black hole rotation.[5]

References

External links

  • Leo C. Stein, Kerr calculator V2 [1]