<p>Momentum is conserved. But it is the momentum of the system as a whole. The energy released by nuclear fusion escapes as a result of neutrinos and radiation. If these emissions are isotropic (the same in all directions), or at least with no hemispheric asymmetry, then NO net momentum is removed from the star, it is not accelerated and therefore its velocity with respect to any reference frame remains unchanged. It's momentum in that rest frame <em>is</em> then reduced by the percentage of mass that it has lost. The lost momentum exists in the escaping radiation and neutrinos which will appear anisotropic in any frame of reference other than the rest frame of the star.</p>

<p>In practice most radiation loss is fairly symmetric, if not spherically, then at least not in the sense that more radiation does not emerge from one hemisphere. I am not sure about neutrinos, but I see no reason for asymmetric emission and these are a fairly minor component in any case.</p>

CC BY-SA 3.0
2017-11-06T11:28:53.623Z
Agree & understand theory & analysis. Nicely explained. The question then is whether data agrees with theory. Clearly we observe Doppler, & when the math is done it will probably account for the lost 'p'. Has anyone done the math? I have never heard of Doppler shifts of neutrinos. That may be a measurement limitation. Still, we also observe stars accelerating, & DE is invoked. If the 'p' balance of a star, with actual measured flux and Doppler of photons & neutrinos, is done & comes us short, perhaps DE is unnecessary. Just wondering if anyone has actually done the math and compared to theory.
Tom Sullivan
CC BY-SA 3.0
2017-11-06T11:41:15.167Z
It may not even take actual measured values. A rough estimate might be doable with an understanding of the fusion reactions and necessary radiation/neutrino fluxes to maintain a star at a constant temperature.
Tom Sullivan
CC BY-SA 3.0
2017-11-06T12:26:51.623Z
@TomSullivan If you want to ask what the experimental verification of conservation of momentum is, then do that on Physics SE. I have no idea where you're going with questioning the need for DE. Galaxy redshifts do not come from individual stars and I emphasised, mass/radiation loss does not change stellar velocity.
ProfRob
CC BY-SA 3.0
2017-11-06T02:14:47.937Z
2017-11-06T02:14:47.937Z
ProfRob
<p>Stars move relative to many frames of reference.  We can use the CBR for this question, I suppose, but it may not matter.  Their momentum p=mass x velocity.</p>

<p>Stars lose mass as it is converted to energy. Percentage-wise it isn't much, but it is tons per second, so it isn't negligible.
What happens to the momentum of the star as it loses mass? Presumably it is conserved, so does the velocity of the star increase?  Or does the overall system carry off some momentum with the energy, leaving the stellar velocity constant?</p>

CC BY-SA 3.0
2017-11-06T02:00:41.750Z
This contains far too many questions. I'll answer the main one.
ProfRob
CC BY-SA 3.0
2017-11-06T03:08:19.750Z
This would apply to any hot object that radiates heat in all directions, not just stars.   It's an interesting application to conservation of energy that I'd never considered.   Ultimately I think the problem can be simplified by taking a moving object and imagining two photons flying off it in opposite directions.   From any frame of reference the photon traveling in the direction of the object would carry with it more energy and more momentum.  The speed should remain unchanged given equal photon emissions in opposite directions, but doing the math would be an interesting exercise.
userLTK
<star><astrophysics>
What happens to the momentum of a star as its mass is converted to energy?
Tom Sullivan
# What happens to the momentum of a star as its mass is converted to energy?

(Asked by: username_0 on 2017-11-06)

Stars move relative to many frames of reference. We can use the CBR for this question, I suppose, but it may not matter. Their momentum p=mass x velocity.

Stars lose mass as it is converted to energy. Percentage-wise it isn't much, but it is tons per second, so it isn't negligible.
What happens to the momentum of the star as it loses mass? Presumably it is conserved, so does the velocity of the star increase? Or does the overall system carry off some momentum with the energy, leaving the stellar velocity constant?

**Tags:** star, astrophysics
> **Comments:**
> **username_1**: This contains far too many questions. I'll answer the main one.  
> **username_2**: This would apply to any hot object that radiates heat in all directions, not just stars.   It's an interesting application to conservation of energy that I'd never considered.   Ultimately I think the problem can be simplified by taking a moving object and imagining two photons flying off it in opposite directions.   From any frame of reference the photon traveling in the direction of the object would carry with it more energy and more momentum.  The speed should remain unchanged given equal photon emissions in opposite directions, but doing the math would be an interesting exercise.  

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## 1 answers

### Answer by username_1 on 2017-11-06. Score: 2

Momentum is conserved. But it is the momentum of the system as a whole. The energy released by nuclear fusion escapes as a result of neutrinos and radiation. If these emissions are isotropic (the same in all directions), or at least with no hemispheric asymmetry, then NO net momentum is removed from the star, it is not accelerated and therefore its velocity with respect to any reference frame remains unchanged. It's momentum in that rest frame *is* then reduced by the percentage of mass that it has lost. The lost momentum exists in the escaping radiation and neutrinos which will appear anisotropic in any frame of reference other than the rest frame of the star.

In practice most radiation loss is fairly symmetric, if not spherically, then at least not in the sense that more radiation does not emerge from one hemisphere. I am not sure about neutrinos, but I see no reason for asymmetric emission and these are a fairly minor component in any case.

> **Comments:**
> **username_0**: Agree & understand theory & analysis. Nicely explained. The question then is whether data agrees with theory. Clearly we observe Doppler, & when the math is done it will probably account for the lost 'p'. Has anyone done the math? I have never heard of Doppler shifts of neutrinos. That may be a measurement limitation. Still, we also observe stars accelerating, & DE is invoked. If the 'p' balance of a star, with actual measured flux and Doppler of photons & neutrinos, is done & comes us short, perhaps DE is unnecessary. Just wondering if anyone has actually done the math and compared to theory.  
> **username_0**: It may not even take actual measured values. A rough estimate might be doable with an understanding of the fusion reactions and necessary radiation/neutrino fluxes to maintain a star at a constant temperature.  
> **username_1**: @TomSullivan If you want to ask what the experimental verification of conservation of momentum is, then do that on Physics SE. I have no idea where you're going with questioning the need for DE. Galaxy redshifts do not come from individual stars and I emphasised, mass/radiation loss does not change stellar velocity.  

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