# Radius of Ring Singularity in Kerr Black Hole

1. Oct 14, 2011

### edgepflow

I researched this some, but could not find a method to calculate the radius of the ring singularity in a Kerr black hole.

I would think it is a function only of black hole mass and angular velocity.

Please let me know if there is some reports or papers on this.

2. Oct 14, 2011

### PAllen

See the following, which gives the formula for the ring singulariy. Surprisingly, I could not find this anywhere in MTW (but it is possible I just missed it ...)

http://www.physics.mcmaster.ca/phys3a03/The Kerr Metric.ppt

3. Oct 15, 2011

### George Jones

Staff Emeritus

Consider the two-dimensional surface that results when $t$ is constant and $r=0$ in Kerr spacetime in standard Boyer-Lindquist coordinates. The metric for this two-dimensional surface can be found by setting $dt=dr=r=0$ in the Kerr metric. What is this metric?

4. Oct 15, 2011

### PAllen

So is there something wrong with slide 15 in the power point I linked to above?

5. Oct 15, 2011

### George Jones

Staff Emeritus
Oops, didn't see the link, just saw the reference to MTW. Now, that that you mention it, there is a mistake (simple typo) in the first equation of slide 15 .

This are a number of textbooks, including, I think, Carroll's that give this result. I'm not sure sure if there is an easy physics interpretation of this coordinate radius.

6. Oct 15, 2011

### PAllen

Should be a^2 not a ?

7. Oct 15, 2011

### George Jones

Staff Emeritus
Yes.

8. Oct 15, 2011

### edgepflow

Thank you for the link PAllen and input from George Jones.

I will have to work on the slides in this link and deal with the abstraction.

I saw the formula: x^2 + y^2 = a^2 which is the equation of a circle of radius a. I am not sure what the range of x and y will be.

At the end of the day, I would like to develop a formula I can enter values of black hole mass and angular velocity, and figure the radius of the ring singularity. Similar idea to the nice, tidy equation for Schwarzschild radius for a non-rotating, uncharged black hole. But maybe this idea does not make physical sense for a Kerr black hole.

9. Oct 15, 2011

### George Jones

Staff Emeritus
The parameter $a$ is given by $a = J/M$, where $J$ and $M$ are the angular momentum and mass of the black hole.
But this gives an expression for the event horizon, not the singularity.

10. Oct 15, 2011

### edgepflow

Thank you sir. Any theories about such a formula for the ring singularity?

11. Oct 15, 2011

### George Jones

Staff Emeritus
I'm not quite what you you're after.

Schwarzschild black hole: event horizon given by $r = 2M$; singularity given by $r=0$ (with some subtleties). Kerr: event horizon given by $r = M + \sqrt{M^2 + \left( J/M \right)^2 }$; singularity given by $r = 0 \Leftrightarrow x^2 + y^2 = \left( J/M \right)^2$. Note the correspondence when the angular momentum $J = 0$.

12. Oct 15, 2011

### edgepflow

Thanks again. I can work with this.

13. Oct 15, 2011

### PAllen

In other words, the radius is just J/M. What could be simpler?

14. Oct 15, 2011

### edgepflow

Yes sir, I noticed that.

15. Oct 17, 2011

### stevebd1

You might also be interested in http://en.wikipedia.org/wiki/Elliptic_coordinate_system" [Broken] which establishes a as r=0 and the outer edge of the ring singularity as suggested in these links-

http://pisces.as.utexas.edu/GenRel/KerrMetric.pdf [Broken]

Last edited by a moderator: May 5, 2017
16. Oct 18, 2011

### edgepflow

Thanks. Equations 7.20 to 7.22 of your 2nd link help clarify Post #11.

Last edited by a moderator: May 5, 2017
17. Oct 19, 2011

### edgepflow

Referring again to Equations 7.20 to 7.22 of stevebd1's 2nd link:

r+ = m + ($m^{2}$ + $a^{2}$)$^{1/2}$

a = 0 => $r_{+}$ = 2m and $r_{-}$ = 0 (Schwarzchild recovery)
a = m => $r_{+}$ = $r_{-}$ = m (largest ring = 1/2 Schwarzchild radius)

And if I did my math right, try:

a = m / 2 => $r_{+}$ = 1.866 m and $r_{-}$ = 0.134 m

Which shows the inner horizon is smaller than the ring singularity. But the outer horizon still hides the singularity and thus avoids a naked singularity. The solution to the quadratic yields two horizons which seems to me to be essestial to avoid naked singularities. Remarkable.

Let me know if this is correct.

18. Oct 20, 2011

### stevebd1

This is one reason why I favour elliptical coordinates. The idea that the ring singularity resides between r+ and r- doesn't seem right as this is spacelike and technically no stable r is possible. Using the following simple equations establishes elliptical coordinates and always puts the ring singularity within r-.

$$x=\sqrt{r^2+a^2}\sin \theta$$
$$y=r \cos \theta$$

Attached are two images for a black hole with a spin parameter of a/M=0.95. The top image uses regular spherical coordinates while the bottom image uses elliptical coordinates and defines the ring singularity with a coordinate radius of $a$ while the singularity still resides at r=0 and $\theta=\pi/2$. This also implies there is a new region of space (or extended Kerr geometry) at r<0.

I think I've seen one image where the ring singularity is between r+ and r- but I think the general census is that if it were to exist, it is within (or at maybe) r- unless there is some mathematical trick that allows it to hover at a specific radius within spacelike space which would be synonymous to something hovering at say 3.00pm in timelike space.

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