# Spin of Black Hole Directly Measured?

1. Mar 7, 2014

### D English

Last edited by a moderator: May 6, 2017
2. Mar 9, 2014

### Staff: Mentor

My question is what is spinning when we talk about black hole spin?

3. Mar 10, 2014

### PhysicoRaj

I think it might indicate the spin of the star's core, if it were there at the moment.. conservation of angular momentum..?

4. Mar 10, 2014

### Chronos

A spinning black hole belongs to a class known as kerr black holes.

5. Mar 10, 2014

### PhysicoRaj

6. Mar 10, 2014

### Staff: Mentor

What exactly is spinning?

7. Mar 10, 2014

### PhysicoRaj

And what does OP mean by 'half the speed of light'? Aren't they measured in rotations per time?

8. Mar 10, 2014

### D English

If I read the article right, what is being measured is the inner edge of the accretion disk. They looked at the changes in X-Ray spectrum, and the number of fluctuations in that spectrum (the fluctuations being driven by the gravity of the black hole, or perhaps more accurately, the gravity at the edge of the EH).

The larger the fluctuations, the closer the accretion disk is to the black hole. In this case, they say the inner edge of the accretion disk is so close to the EH that the only way the disk can remain intact is if the black hole is spinning extremely rapidly.

Regarding spinning at half the speed of light, I think that it is exactly what it seems to mean: The Earth spins at ~ 1,000mph, the black hole spins at 93,000 miles per second.

So, wouldn't this mean that if the spin is created by conservation of angular momentum, that the back hole is incredibly small?

9. Mar 10, 2014

### Chronos

The term 'spin' may be confusing the issue here. I do not believe it is correct to view it like the spin of a planet rotating on its axis. A more appropriate term is angular momentum, which is technically one of the basic properties of a black hole. An analogous case would be the spin of an electron which, like a singularity, is a point particle. The initial singularity formed by a collapsing star would retain the angular momentum of its precursor star [at least the portion that was not ejected].

10. Mar 10, 2014

### Staff: Mentor

Is there any difference between the effect a neutron star's rotational has on spacetime, and a black hole's?

11. Mar 10, 2014

### Chronos

They both induce frame dragging in their immediate vicinity.

12. Mar 10, 2014

### Staff: Mentor

So no difference then?

13. Mar 11, 2014

### D H

Staff Emeritus
The OP got that directly from the popsci article about the Nature article. The popsci article is a bad science article in that regard.

The Nature article writes in terms of a black hole's angular momentum parameter $a$. In metric units, $a=\frac{Lc}{GM^2}$ where $L$ is the angular momentum. This is a unitless parameter; it's the same regardless of the system of units one uses. The black hole becomes "naked" (the event horizon becomes zero; i.e., vanishes) if $a=1$. If $a>1$ the event horizon becomes imaginary. There's a hypothesis, the cosmic censorship hypothesis, that the universe doesn't allow such nastinesses to happen. This hypothesis says that there's an upper limit on angular velocity, $a<1$. This is a hypothesis for now.

I didn't pay to read the Nature article. The abstract says nothing about $a$ being restricted to being between 0 and 1. It merely says that the angular momentum parameter of the black hole in question is at least 2/3 (five sigma value).

In other words, this is a rapidly spinning black hole, but not enough for Hawking to lose his updated bet on the cosmic censorship hypothesis.

14. Mar 11, 2014

### stevebd1

I think they are talking about tangential velocity of the frame-dragging at the event horizon as observed from infinity. This would be based on-
$$v_T=\omega R$$
where
$$\omega=\frac{2Mra}{\Sigma^2}$$
and
$$R=\frac{\Sigma}{\rho}\sin\theta$$
where $v_T$ is tangential velocity, $\omega$ is the angular velocity of the frame-dragging and $R$ is the reduced circumference, see link below for complete equations on page 379

If a/M=1 (max spin) then the tangential velocity at the event horizon $(r_+)$ is calculated as 1c, anything where a/M<1 would be observed as less than the speed of light. It's worth noting that this speed is not the actual speed locally. If you want the actual local speeds, then you need to divide by the reduction factor $(\alpha)$.

The spin parameter $(a)$ can be established by looking to see where the marginally stable orbit is (MSO), the inner edge of the accretion disk, which can range from 6M for a static BH to 1M for a maximal BH (a/M=1), see link below, equations on page 392

Source-
'Compact Objects in Astrophysics' by Max Camenzind

Last edited: Mar 11, 2014
15. Mar 12, 2014

### PhysicoRaj

I see a clearer picture now. And Steve's link is awesome! Thanks!