# B Neutron Stars and Angular Momentum

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1. Nov 11, 2016

### lavinia

Some observed neutron stars rotate hundreds of times per second. Speeds at the surface of these stars are as much as 15% the speed of light. These huge speeds are generated because angular momentum is conserved when a large rotating pre-super nova star collapses into a neutron star.

The question is: What happens during collapse if the angular momentum of the pre-collapse star is high enough to generate speeds as fast or faster than the speed of light at the surface of the neutron star? Does the collapsing star just rip apart?

Last edited: Dec 7, 2016
2. Nov 11, 2016

### Petr Matas

My guess is that the star collapses into an ellipsoid, a disc or even a ring with a radius large enough to accomodate for the angular momentum. The centrifugal force should prevent further collapse.

Last edited: Nov 11, 2016
3. Nov 11, 2016

Staff Emeritus
That would take infinite angular momentum, so we don't worry about it.

4. Nov 11, 2016

### lavinia

I was thinking naively that Conservation of Angular Momentum yields the equation

$r^{3}v = r^{3}_{0}v_{0}$

where $r_{0}$ and $v_{0}$ are the radius and surface velocity of the pre-collapse star. Physicists says that $r$, the radius of the neutron star, is about 10 kilometers so its variation doesn't seem large.

Stars that become neutron stars have a limited mass range so it would seem that unless something else happens their rate of rotation must be limited.

5. Nov 11, 2016

### |Glitch|

Just for fun I took the largest star we know, VY Canis Majoris (with a radius of 987,894,000 km), gave it an initial rotation of just one hour, then reduced the star (keeping its mass) to 10 km. It came out have a final rotational speed of 0.000187519 light-inches, or 0.00001537 nanoseconds, which is still slower than the speed of light.

Last edited: Nov 11, 2016
6. Nov 11, 2016

### PeroK

You would have to consider relativistic, not classical, angular momentum.

7. Nov 11, 2016

### lavinia

Right but how would that work?

8. Nov 11, 2016

### PeroK

9. Nov 11, 2016

### lavinia

I just listened to a youtube lecture presented by Victoria Kaspi. She said that the bulge in the star would at some point rip the star apart and this is what limits the speed of rotation of a neutron star. Still don't understand the detail but that seems to be a start.

10. Nov 11, 2016

### Petr Matas

Your values r = 987,894,000 km and T = 1 h yield v = 2πr/T = 1,724,200 km/s = 5,75c, i.e. the surface velocity is super-luminal even before the collapse.

11. Nov 11, 2016

### Petr Matas

Where does the third power come from? In my opinion the correct equation (ignoring any relativistic effects) is just
$rv = r_0v_0$.

12. Nov 11, 2016

### lavinia

How do you show this?

13. Nov 11, 2016

### |Glitch|

I did not consider that, thanks for pointing that out. I just used Tf = Ti(rf2 / ri2), and while that is a very small amount, it is still sub-luminal.

14. Nov 12, 2016

### Petr Matas

The moment of inertia of a solid homogeneous sphere is
$$I = {2 \over 5} mr^2.$$
Its angular momentum is
$$L = Iω = I{v \over r} = {2 \over 5} mrv,$$
which has to be conserved, i.e.
$$L = L_0$$
$${2 \over 5} mrv = {2 \over 5} mr_0v_0$$
$$rv = r_0v_0.$$

Last edited: Nov 12, 2016
15. Nov 12, 2016

### Petr Matas

You're welcome. However, doesn't the collapse increase the surface speed? How can the initial super-luminal become post-collapse sub-luminal then?

16. Nov 12, 2016

### Petr Matas

Consideration of special relativity requires $m$ to be replaced with $γm_0$, where $m_0$ is the rest mass, and the Lorentz factor
$$γ = {1 \over \sqrt{1 - {v^2 \over c^2}}}.$$
Unfortunately this factor depends on speed and therefore on position inside the body. A simplifying assumption that all mass is concentrated at equator (in a thin ring) yields angular momentum
$$L = mrv = γ m_0 r v = m_0 {r v \over \sqrt{1 - {v^2 \over c^2}}},$$
which again has to be conserved. Compare it to:
I don't know if the strong gravitation influences things (by e.g. gravitational time dilation).

17. Nov 12, 2016

Staff Emeritus
Petr, take your equation and solve for v/c:

$$\frac{v}{c} = \frac{\sqrt{J^2 - 1}}{J} ; J = \frac{L}{m_0r}$$

Now, for what values of J is v/c > 1? What does that say about L?

18. Nov 12, 2016

### Petr Matas

I have obtained a different result:
$${v \over c} = {1 \over \sqrt{{c^2 \over J^2} + 1}}$$
Anyway, v/c < 1 for any value of J or L or r.

19. Nov 13, 2016

### lavinia

right

20. Nov 13, 2016

### nikkkom

Something is off. Reducing radius from 987,894,000 to 10 should shorten the rotation period by 98789400. From 1 hour to 36 microseconds. In 36 us, light travels 10.8 km, but 10 km radius sphere's equator is longer: 63 km. Thus, in this example the star can't be compressed to 10 km radius without shedding some angular momentum.