B Neutron Stars and Angular Momentum

lavinia

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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?
 
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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.
 
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Vanadium 50

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What happens during collapse if the angular momentum 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?
That would take infinite angular momentum, so we don't worry about it.
 

lavinia

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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.
 
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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.
 
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PeroK

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

lavinia

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lavinia

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

lavinia

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Where does the third power come from? In my opinion the correct equation (ignoring any relativistic effects) is just
##rv = r_0v_0##.
How do you show this?
 
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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.
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.
 
How do you show this?
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.
$$L = L_0$$
$${2 \over 5} mrv = {2 \over 5} mr_0v_0$$
$$rv = r_0v_0.$$
 
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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.
You're welcome. However, doesn't the collapse increase the surface speed? How can the initial super-luminal become post-collapse sub-luminal then?
 
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:
That would take infinite angular momentum, so we don't worry about it.
I don't know if the strong gravitation influences things (by e.g. gravitational time dilation).
 

Vanadium 50

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

[tex] \frac{v}{c} = \frac{\sqrt{J^2 - 1}}{J} ; J = \frac{L}{m_0r} [/tex]

Now, for what values of J is v/c > 1? What does that say about L?
 
Petr, take your equation and solve for v/c:

[tex] \frac{v}{c} = \frac{\sqrt{J^2 - 1}}{J} ; J = \frac{L}{m_0r} [/tex]

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

lavinia

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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.
$$L = L_0$$
$${2 \over 5} mrv = {2 \over 5} mr_0v_0$$
$$rv = r_0v_0.$$
right
 
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.
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.
 
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.
The rotation period is proportional to the radius squared (ignoring relativistic effects). If it were proportional to just the first power of radius (as you have written), the surface speed would be unchanged by the collapse (it was 5.75 times the speed of light before the collapse as well, see post #10) and the angular momentum would not be conserved.

Also note, as Vanadium 50 said, that as the surface speed approaches the speed of light, the angular momentum approaches infinity (see post #16). In other words, regardless of the angular momentum and radius, the surface speed never exceeds the speed of light.
 
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Chronos

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Let us assume the sun were to collapse to form a neutron star with a radius of 1.2x103m. It's precollapse radius is 7x108 m, precollapse with a spin rate of 1 per 26 days which works out to 4.45 x 10-7 s-1. Plugging this into the expression Vf= Vi(Ri/Rf)2yield Vf=4.45x10-7x(7x108/1.2x103)2, or
1.5x105s-1. Of course this exceeds the maximum permissible centripetal acceleration at the surface of the sun, not to mention faster than any known neutron star. Obviously, the sun must find a way to shed mass and angular momentum as it collapses - which commonly occurs during a supernova event. The progenitor star is typically at least 8 solar masses and the neutron star fragment ends up about 1 solar mass.
 
Let us assume the sun were to collapse to form a neutron star with a radius of 1.2x103m. It's precollapse radius is 7x108 m, precollapse with a spin rate of 1 per 26 days which works out to 4.45 x 10-7 s-1. Plugging this into the expression Vf= Vi(Ri/Rf)2yield Vf=4.45x10-7x(7x108/1.2x103)2, or
1.5x105s-1. Of course this exceeds the maximum permissible centripetal acceleration at the surface of the sun, not to mention faster than any known neutron star. Obviously, the sun must find a way to shed mass and angular momentum as it collapses - which commonly occurs during a supernova event. The progenitor star is typically at least 8 solar masses and the neutron star fragment ends up about 1 solar mass.
A static black hole's radius is rs = 3M/MSun km, so if the mass of the Sun shrunk to the radius of 1.2 km, as you have written, it would be a black hole. Neutron star radii are about 10 times larger, let us say 12 km. That would result in a spin rate f = 1500 s-1, still about twice as fast as the fastest neutron star known. Other properties of the resulting neutron star:
  • Mass M = MSun = 2 × 1030 kg
  • Surface speed v = 2πfr = 1.1 × 108 m/s = 0.38c
  • Surface gravity g = GM/r2 = 0.93 × 1012 m/s2
  • Surface centripetal acceleration a = v2/r = 1.1 × 1012 m/s2 (still exceeding surface gravity)
 
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Chronos

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Agreed, it was late and ... Not my first order of magnitude error, and most likely not my last. Probably explains why differential equations inflicted sleepless nights on me back in the day. A little dyslexia goes a long way in math.
 
Aside from that, your post explains things a lot, so thanks for it, Chronos.
 

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