# Angular momentum of an oblate spheroid

1. Apr 2, 2008

### stevebd1

Can anyone shed some light on how to calculate the angular momentum of an oblate spheroid (i.e. a rapidly rotating neutron star). I'm aware the following applies for various shapes-

$$J=I\omega$$

where

$$\omega=\frac{v}{r}$$

and for a solid sphere-

$$I=\frac{2}{5}mr^{2}$$

for a solid disk or cylinder-

$$I=\frac{1}{2}mr^{2}$$

for a ring or loop-

$$I=mr^{2}$$

where J is the angular momentum, I is the inertia, v is the velocity of the outer edge, m is the mass and r is the radius to the outer edge of the object.

In some cases, I've seen the equation for the solid sphere apply to oblate spheroids but I've also seen ellipticity allowed for. In most cases, this is expressed as the following-

$$\epsilon=\frac{a-b}{a}$$

where epsilon is the factor of ellipticity and a is the long radius and b the short radius of the spheroid. I've seen it expressed as the following in equations for oblate spheroids-

$$I=\frac{2}{5}mr^{2}(1+\epsilon)$$

in some cases, a and b are expressed as inertias as if the inertia had been calculated for each radius as solid spheres and then plugged into the ellipticity equation in order to get a ratio of some kind.

On a slightly different note, if ellipticity is allowed for based on a and b being radii and the neutron star has a high rotation, the fraction comes out high and pushes the 2/5 (0.4) factor for a solid sphere up to ~0.7 which is higher than the disk or cylinder fraction (0.5) which seems at odds.

Also, would it be safe to assume that the interior is rotating slightly faster than the crust and that also due to different densities, it might be worthwhile calculating the seperate angular momentums for the crust and interior and adding them to get a more realistic figure for J?

regards
Steve

Last edited: Apr 2, 2008
2. Apr 3, 2008

### VelocideX

Do it the same way you do a sphere, although the answer is a bit messy.. You know the inertia of a disk is given by I = 1/2 mr^2.

So for a differential disk, dI = 1/2 dm r^2.

A little bit of thought using the property that an oblate spheroid is a rotated ellipse will give you r as a function of z. In particular, you have

x^2/p^2 + y^2/p^2 + z^2/q^1 = 1
gives
r^2 = x^2+y^2 = (1-z^2/q^2)p^2

Now use the fact that the oblate sphere has a circular cross section in the xy plane. So dm = rho dV (rho = density) = pi r^2 dz.

so dI = (pi rho)/2 r^4 dz

plug in r^2 from above, integrate from -z to z. You then need to get rid of the density and get the total mass back in. For this, you need the volume of an oblate spheroid (which is well known: http://mathworld.wolfram.com/Volume.html ).

I haven't done this to completion, but it should give you an answer. It may not be neat though. the r^4 term gives you a heap of terms from the integral, the highest power of which is z^9, the lowest of which is z. It will be correct, however.

Allowing for density as a function of distance from the origin is difficult, because the integral is done over z. Again, it's still possible if you can formulate it.

This is the "disk" approach. The other approach is the cylindrical approach, where you take the inertia of a cylinder, get a radial differential, then construct your oblate spheroid out of cylinders. I don't know if this is more tractable, but you can do the integral over r directly. On the downside, you're going to have to solve for the contact point between the cylinder boundary and the oblate spheroid, which makes the integration limits a little trickier.

Does this help?

3. Apr 3, 2008

### VelocideX

I should also point out that Wikipedia gives the moments of inertia for an ellpsoid, of which the oblate spheroid is a restricted case

http://en.wikipedia.org/wiki/Ellipsoid

4. Apr 3, 2008

### stevebd1

Thanks for the response Velocidex, all useful stuff. There was also another useful equation under 'Equatorial bulge' regarding flattening-

$$\textit{f}=\frac{a-b}{a}=1-b:a \approx\frac{3\pi}{2GT^{2}\rho}$$

where a is the long radius (equatorial), b is the short radius (polar) G is the gravitational constant, T is the rotation period (in seconds) and $$\rho$$ is density

Regarding the density, while it's accepted that neutron stars flatten as they rotate faster, it also seems that the density reduces as well (this might seem obvious but it's worth quantifying). One bit of info I found states that a 2 sol mass neutron star with a 12 km radius flattens at 1650 Hz to an equatorial radius of 22 km and a polar radius of 7 km, while this is just an increase of 7% in the cross-section area (area x 1.07), it's nearly a 100% increase in volume (volume x 1.96) which I imagine would reduce the density by approx. half*.

This makes the flattening equation tricky as you have a number of variables- the radii, rotation period and density.

*http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.2708v2.pdf page 7 figs. 5 & 6

regards
Steve

Last edited: Apr 3, 2008
5. Apr 7, 2008

### stevebd1

Regarding the inertia of oblate spheroids, I've put together profiles of neutron stars at various rates of rotation (a) at a critical mass of 3 sol superimposed over the Kerr black hole they would produce relevant to rotation.

As

$$a=\frac{cJ}{GM^{2}}$$

so

$$J=\frac{aGM^{2}}{c}$$

a is the angular momentum factor which ranges from 0 (no spin) to 1 (maximum spin or unity). In the case of a = 1, this would equal 7.926x10^42 Nms for a 3 sol mass. Using this as a maximum and after a couple of trial and errors using-

$$v=\frac{J}{MR(2/5)}$$

for rapidly rotating objects, R is the elongated equatorial radius

and the flattening equation-

$$\textit{f}=\frac{a-b}{a}\approx\frac{3\pi}{2GT^{2}\rho}$$

In the case of the flattening equation, a is the long radius (equatorial) and b is the short radius (polar).

For a 3 sol mass neutron star, I calculated a flattening ratio of approx. 0.455 at a = 1 based on a diameter of 9000 m at a = 0 (just outside the Schwarzschild radius); based on the fact that the cross-section area hardly changes, this is an equatorial radius of 12200 m and a polar radius of 6645 at a = 1. Interestingly, I also tried a radius of 11500 m at a = 0 and came up with a flattening ratio of 0.467 at a = 1 which implies that as the neutron star's radius reduces in collapse, it's density increases proportionally more and flattening is less (diagram attached shows the profiles for 9000 m radius at a = 0).

The red line is the outline of the neutron star on the brink of collapse and the very faint yellow line indicates the equatorial radius when the rotational velocity at the edge of the spheroid would equal c (based on v = J / MR 0.4). This occurs within the outer event horizon (R+) below a = 0.6 and outside the event horizon above a = 0.6 (rotational v = c occurs exactly on the gravitational radius at a = 0.4 and exactly on the Schwarzschild radius at a = 0.8). This seems to coincide to some degree with a paper I found which shows the outer part of a star with rapid rotation collapsing outside the apparent horizon (possibly due to the inability to attain a velocity of c), forming a temporary disk (which presumably eventually falls into the black hole due to its proximity).

http://people.sissa.it/~rezzolla/Whisky/WhiskyI/

It also seems there's a chance that a 3 sol mass neutron star above a = 0.6 wouldn't collapse as the high spin might keep the density below critical (though this is more likely with the radius of 11500 m at a = 0 rather than the 9000m at a = 0).

No surprise there's a strong resemblance between the profile of the neutron star and the static limit profile for each quantity of a.

Steve

EDIT: B+W image added, frequencies indicate rotation of neutron star close to moment of collapse based on elongated equatorial radius.

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Last edited: Apr 7, 2008