Let's say a spinning star radiates mass-energy only from it's pole regions. How does the loss of mass-energy effect the angular momentum of the star?
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Energy has momentum, too. Lets say that the energy is emited in the form of photons. Then, the photons will have $p=\frac{h}{\lambda}=\frac{h}{\frac{hc}{E}}=\frac{E}{c}=mc$ where E is energy of the mass, and $\lambda$ is thte wavelength of light. Now, this applies to a particle at rest. Actually, the energy will be radiated in all directions, thus conserving net momentum, Give it a velocity, and the light will be emitted more in one direction. Similarly, one can calculate angular momentum of the photon. This will also be conserved.
If the energy is radiated radially from the poles, the photons will have no angular momentum at all, and thus the star must spin faster. But if they are radiated tangentially from any other point on the star, the star will keep its angular speed as the new photons will carry off enough angular momentum to balance the loss of mass.
Update: Just to clarify, when I say "more light is emitted in the forward direction", I mean "more light energy". The number of photons emitted in each direction is the same, just that the photons going forward are doppler-blueshifted (hence getting more energy-->more momentum). The ones going backwards are similarly redshifted, leading to an asymmetry in momentum distribution.
answered Feb 4, 2012 at 17:40
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I found this question rather challenging, and I begun by answering the question that it was not asking. I'll leave that answer because I think my writing offers some info I don't see elsewhere.
For a sun radiating from the full surface
Imagine viewing the sun from a distance, and you are stationary relative to the sun. Say its axis of rotation is up and down and it is rotating such that the right half is rotating toward you and the left half away from you.
Since the surface on the right half of the sun is moving toward you, the light will be blue shifted. Picture a color gradient that goes from blue on the right to red on the left. These photons impart momentum to you, but this momentum is not directly in the direction opposite of the sun's center of mass (CM). It will push you away from the CM and slightly to the left because the blue-shifted photons have more momentum.
If you imagine that all of the sun's light is absorbed by objects similar to yourself (no photons left flying in space), then all of the angular momentum that the sun lost since it began radiating energy will then exist in the objects that absorbed the photons. Say all of the sun's energy was released in the form of photons. Angular momentum is conserved.
For a sun radiating from the poles only
Obviously, this problem can reduce to the prior one if the emission was over a non-zero surface instead of a point. That would dodge the question, so I will attempt to answer this given that photons are emitted from 2 single points on either pole.
The mechanism of blue-red shift can not apply in this case. The photons are emitted from a rotating, but otherwise stationary, reference frame. So, given a population of photons emitted from a single point, can they carry away angular momentum? No, they can not. If you do the $\vec{r} \times \vec{p}$ integral centered around the point of emission, then the vectors $\vec{r}$ and $\vec{p}$ are in the same direction, giving zero.
Is there another mechanism for transferring angular momentum? Can a single photon have angular momentum? A photon is massless but still has momentum, can the same apply for angular momentum? Probably not. You can take the photon to have zero mass, but it's not clear (to me at least) if you can take it to have zero size. A body of zero mass with a finite size (say, with a diameter equal to the wavelength) could potentially rotate in a way that yields angular momentum; consider a photon orbiting a black hole. In spite of the fact that this may be possible, I don't think it's real. If you consider the angular speed of the planet, it is very very slow because the planet has a large moment of inertia. Even if a photon could carry away angular momentum, it is definitely not clear how this would happen. Even if it did, it wouldn't be enough to make a difference. The photons can "see" the angular speed, but not the angular momentum of the body. Again, the angular speed is low (from the photon's point of view) but the angular momentum is large, even if the sun is on the verge of ripping itself apart.
For a sun or a black hole that emitted no light other than at the poles, the angular momentum would not decrease even though the total mass does decrease. Provided the moment of inertia decreased (which it probably would), it would continue to rotate faster. Eventually, this will lead to some instability. I don't know what, but for a gaseous, gravitationally bound body like a star, it would break up into pieces. Either that, or there would be some pressure to release energy at some location other than the poles. In the case of a black hole, I don't know what kind of instability it encounters from rotating too fast, but there are several physical mechanisms by which it can get rid of angular momentum. Due to my arguments here, radiating massless energy at the poles is not one of these.
Ball dropped from an airplane
I don't think this is a particularly insightful physical case. The angular momentum will be decreased due to air friction as it falls. Some of the angular momentum will be converted to linear momentum when it hits the ground, as you're familiar with when you spin a ball and drop it.
That's pretty much it.
answered Feb 5, 2012 at 15:51
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$\begingroup$ Ah, example: Consider a water balloon hung from a string, with a hole opposite the string connection. It spins about the string-hole axis. As water exists the hole, no angular momentum is lost. The balloon will, in fact, spin faster. $\endgroup$ Commented Feb 5, 2012 at 20:40
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$\begingroup$ No no flywheel is dropped, but a flywheel based energy storage device gadget, consisting of bearings, enclosure and a flywheel. $\endgroup$– kartsaCommented Feb 6, 2012 at 5:47
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1$\begingroup$ I have a correction to make to the above post, although I don't think this affects the answer to the question: individual photons certainly can carry angular momentum. They are spin 1 particles, and so can carry angular momentum of $\pm \hbar$ $\endgroup$ Commented Feb 8, 2012 at 3:10
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It depends on the direction of the radiation. If mass-energy is radiated from the poles along the axis of the star, the angular momentum of the star with respect to the center of the star does not change.
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$\begingroup$ @kartsa: I don't quite see any reasons for that - the radiated mass carries no angular momentum. $\endgroup$ Commented Feb 5, 2012 at 7:10
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1$\begingroup$ @akhmeteli kartsa is saying the angular velocity will speed up because the angular momentum is not lost but mass/energy is. This is true in this scenario unless the loss of energy causes the star to expand enough to cancel this tiny effect out. $\endgroup$– N. VirgoCommented Feb 5, 2012 at 13:35
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$\begingroup$ @Nathaniel: I am afraid I am not convinced yet. Let us consider the following example: the star consists of just three equal point masses $A$, $B$, and $O$. Mass $O$ is on the rotation axis, and masses $A$ and $B$ are at a distance $r$ from the axis each, and $O$ is in the center of mass of the star (so $O$ is between $A$ and $B$). Let us assume that the star rotates with an angular velocity $\omega$. Let us imagine that mass $O$ disappears ("radiates"along the rotation axis). Why should the angular velocity increase? $\endgroup$ Commented Feb 5, 2012 at 17:50
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$\begingroup$ Ok, the star will speed up unless it expands, or unless it only loses mass from points along the axis of rotation. But in the case of a star the latter seems highly unlikely to me... $\endgroup$– N. VirgoCommented Feb 5, 2012 at 19:02
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I see, it's the friction resisting the Coriolis effect, that removes the angular momentum from the stuff leaving the star. Coriolis effect
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$\begingroup$ I don't think that makes sense. We can still draw the boundary around the star, and it's only vacuum with no contact and no emission above the surface (per the question). $\endgroup$ Commented Feb 7, 2012 at 22:33
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Since the star is losing its mass while not losing its angular momentum, it will spin faster and faster. At some point it will be impossible to be integer anymore.
answered Feb 7, 2012 at 21:44
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1$\begingroup$ Agree, but "At some point it will be impossible to be integer anymore." doesn't make sense to me. $\endgroup$ Commented Feb 7, 2012 at 22:34