<p>Why does the Sun not rotate so fast? Because it has lost the vast majority of the specific angular momentum of the material it has formed from. It has to, otherwise it could not form at all.</p>
<p>The problem of how protostars lose their angular momentum is still not fully solved. Likely mechanisms include jets and winds from the accretion disc, that are coupled to the star by a very strong early stellar magnetic field. This may lock the star and disc into co-rotation, allowing the transfer of angular momentum outwards.</p>
<p>Once the disc cleared, loss of angular momentum continues via the stellar wind and magnetic field. The wind is held in corotation by the magnetic field until it &quot;drops off&quot; the field lines at the Alfven radius (well beyond the orbit of Mercury). The field is then dragged through the wind, applying a torque to the Sun and taking away angular momentum. The classic reference is <a href="https://ui.adsabs.harvard.edu/abs/1967ApJ...148..217W/abstract" rel="noreferrer">Weber &amp; Davis (1967)</a></p>
<p>We can be reasonably sure that the young Sun was rotating at 5-50 times faster when it reached the main sequence by observing such stars in young open clusters.</p>
<p>You could reasonably ask why such mechanisms don't operate for Jupiter. Well, the mass loss rate is negligible and it is also empirically observed that lower mass stars and brown dwarfs have <em>much</em> longer spin down timescales. Again, for not fully understood reasons.</p>

CC BY-SA 4.0
2024-10-13T08:34:03.497Z
2024-10-13T08:45:19.450Z
2024-10-13T08:45:19.450Z
ProfRob
<p>The sun is so big — almost a million miles wide — that it doesn't have to rotate as fast to contain a huge amount of angular momentum.</p>
<p>You are floating outside the space station with a pair of weightlifting barbells. One has 10 pound wights and one has 50 pound weights</p>
<p>If you apply the same amount of energy to make them each spin, the light one will rotate much faster than the heavy one.</p>
<p>Same with Jupiter.</p>

CC BY-SA 4.0
2024-10-13T18:42:10.807Z
It would be interesting to compare the rotational angular momentum of the Sun with the angular momenta of Jupiter, both rotational and orbital.
PM 2Ring
CC BY-SA 4.0
2024-10-14T16:45:35.773Z
It doesn't matter if there are a billion particles or just one.  Each individual particle carries its angular momentum with it as it's pulled towards the center.
Greg Miller
CC BY-SA 4.0
2024-10-15T13:20:32.277Z
@GregMiller Right, as long as the particle remains discrete. But when it merges with the whole, it gives up its individual momentum  and adds its vector to the collective, just like the borg.
Miss Understands
CC BY-SA 4.0
2024-10-13T18:23:29.890Z
2024-10-13T18:23:29.890Z
Miss Understands
<p>The Sun has two things Jupiter essentially doesn't have: solar wind and sunlight.</p>
<p>The escaping solar wind interacts with the Sun's magnetic field to apply a torque to the Sun. The escaping photons slow down the Sun's photosphere due to a Poynting-Robertson-like drag.</p>
<p>The sunlight emitted by the Sun acts as a parasitic drag on the Sun's rotation rate, carrying some of the Sun's angular momentum away from the Sun. The interaction between the solar wind and the magnetic field carries angular momentum away from the Sun, to a greater extent compared to photons.</p>
<p>While Jupiter does emit some photons due to its non-zero (kelvin) temperature, because this is a <span class="math-container">$T^4$</span> relationship, Jupiter emits far fewer photons that does the Sun, not just &quot;far fewer&quot; but orders and orders of magnitude fewer.</p>
<p>This is loss of angular velocity is not unique to the Sun. Younger stars tend to rotate much faster than do older stars due to loss of angular momentum to escaped particles interacting with the star's magnetic field, and also photons. The Sun is no longer a young star; it is about halfway between when it formed and when it will run out of hydrogen at its core.</p>

CC BY-SA 4.0
2024-10-12T19:13:31.437Z
I don't see how losing a particle reduces the rotation period.  Yes, it lost angular momentum because it lost mass.  But for conservation of momentum to slow the rotation period it would have to retain the mass and lose speed.  And AFAIK, most of the light, heat, activity, etc is powered from the nuclear fusion, so energy lost would be regenerated.
Greg Miller
CC BY-SA 4.0
2024-10-12T20:03:33.370Z
@Greg In the frame co-rotating with the Sun, the (angular and linear) momentum carried away by the emitted light is spherically symmetric. But that rotating frame is not inertial. In an inertial frame, the loss of momentum is *not* spherically symmetric.
PM 2Ring
CC BY-SA 4.0
2024-10-13T08:38:12.273Z
The post-accretion angular momentum loss from the Sun has nothing to do with radiation from the surface. Greg Miller's comment is also correct. A torque has to be applied to lose angular momentum.
ProfRob
CC BY-SA 4.0
2024-10-13T09:02:03.933Z
I think the most effective angular momentum loss mechanism is the interaction between the solar magnetic field and the solar wind. This is how the solar wind carries away disproportionately more angular momentum.
fraxinus
CC BY-SA 4.0
2024-10-13T12:15:57.917Z
@ProfRob So is my comment wrong? Or just irrelevant?
PM 2Ring
CC BY-SA 4.0
2024-10-13T12:54:00.153Z
@PM2Ring negligible. We assume the radiation from the Sun is isotropic so I do not see how it can apply a torque to the surface. There will be some tiny second order ($v^2,/c^2$) relativistic aberration effect.
ProfRob
CC BY-SA 4.0
2024-10-13T12:56:07.843Z
@It's relevant. A star loses angular momentum both by magnetic braking and by photonic braking, although the latter apparently is small for the Sun. (It might be more significant for brighter stars.) It's a relativistic effect, akin to Poynting-Robertson drag. See [Poynting-Robertson-like Drag at the Sun’s Surface](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.118.051102). I'll update my answer to reflect this.
David Hammen
CC BY-SA 4.0
2024-10-13T12:56:48.210Z
@ProfRob See the comment above.
David Hammen
CC BY-SA 4.0
2024-10-13T13:11:50.577Z
I would read the paper. It says that this tiny relativistic effect is orders of magnitude smaller than magnetic wind braking. And that is now, let alone when the Sun's field was orders of magnitude stronger. "Photon braking" is not why the Sun rotates slowly. Luminous stars rotate very fast. So it's not very effective there either.
ProfRob
CC BY-SA 4.0
2024-10-12T16:01:01.330Z
2024-10-13T13:06:56.117Z
2024-10-13T13:06:56.117Z
David Hammen
<p>The mainstream interpretation is that when Jupiter was being formed from a giant disk of gas and dust, the conservation of angular momentum dictated that it spun faster as it grew smaller.</p>
<p>At the same time the Sun rotates at a &quot;moderately&quot; 1~2 miles per second which is nowhere close to Jupiter's rotational speed. Obviously if you were to form a star your &quot;initial disk&quot; just had to be much, much larger than what you would a gas giant.</p>
<p>So, what happened to the sun? Was it because of its interaction with all the planets?</p>

<the-sun><jupiter><rotation><star-formation><planetary-formation>
Why does Jupiter spin so fast but not the Sun?
cream_pi
# Why does Jupiter spin so fast but not the Sun?

(Asked by: username_0 on 2024-10-12)

The mainstream interpretation is that when Jupiter was being formed from a giant disk of gas and dust, the conservation of angular momentum dictated that it spun faster as it grew smaller.

At the same time the Sun rotates at a "moderately" 1~2 miles per second which is nowhere close to Jupiter's rotational speed. Obviously if you were to form a star your "initial disk" just had to be much, much larger than what you would a gas giant.

So, what happened to the sun? Was it because of its interaction with all the planets?

**Tags:** the-sun, jupiter, rotation, star-formation, planetary-formation


---

## 3 answers

### Answer by username_5 on 2024-10-12. Score: 28 | Accepted answer

The Sun has two things Jupiter essentially doesn't have: solar wind and sunlight.

The escaping solar wind interacts with the Sun's magnetic field to apply a torque to the Sun. The escaping photons slow down the Sun's photosphere due to a Poynting-Robertson-like drag.

The sunlight emitted by the Sun acts as a parasitic drag on the Sun's rotation rate, carrying some of the Sun's angular momentum away from the Sun. The interaction between the solar wind and the magnetic field carries angular momentum away from the Sun, to a greater extent compared to photons.

While Jupiter does emit some photons due to its non-zero (kelvin) temperature, because this is a $T^4$ relationship, Jupiter emits far fewer photons that does the Sun, not just "far fewer" but orders and orders of magnitude fewer.

This is loss of angular velocity is not unique to the Sun. Younger stars tend to rotate much faster than do older stars due to loss of angular momentum to escaped particles interacting with the star's magnetic field, and also photons. The Sun is no longer a young star; it is about halfway between when it formed and when it will run out of hydrogen at its core.

> **Comments:**
> **username_4**: I don't see how losing a particle reduces the rotation period.  Yes, it lost angular momentum because it lost mass.  But for conservation of momentum to slow the rotation period it would have to retain the mass and lose speed.  And AFAIK, most of the light, heat, activity, etc is powered from the nuclear fusion, so energy lost would be regenerated.  
> **username_3**: @username_4 In the frame co-rotating with the Sun, the (angular and linear) momentum carried away by the emitted light is spherically symmetric. But that rotating frame is not inertial. In an inertial frame, the loss of momentum is *not* spherically symmetric.  
> **username_1**: The post-accretion angular momentum loss from the Sun has nothing to do with radiation from the surface. username_4 username_4's comment is also correct. A torque has to be applied to lose angular momentum.  
> **username_6**: I think the most effective angular momentum loss mechanism is the interaction between the solar magnetic field and the solar wind. This is how the solar wind carries away disproportionately more angular momentum.  
> **username_3**: @username_1 So is my comment wrong? Or just irrelevant?  
> **username_1**: @username_3 negligible. We assume the radiation from the Sun is isotropic so I do not see how it can apply a torque to the surface. There will be some tiny second order ($v^2,/c^2$) relativistic aberration effect.  
> **username_5**: @It's relevant. A star loses angular momentum both by magnetic braking and by photonic braking, although the latter apparently is small for the Sun. (It might be more significant for brighter stars.) It's a relativistic effect, akin to Poynting-Robertson drag. See [Poynting-Robertson-like Drag at the Sun’s Surface](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.118.051102). I'll update my answer to reflect this.  
> **username_5**: @username_1 See the comment above.  
> **username_1**: I would read the paper. It says that this tiny relativistic effect is orders of magnitude smaller than magnetic wind braking. And that is now, let alone when the Sun's field was orders of magnitude stronger. "Photon braking" is not why the Sun rotates slowly. Luminous stars rotate very fast. So it's not very effective there either.  

---

### Answer by username_1 on 2024-10-13. Score: 17

Why does the Sun not rotate so fast? Because it has lost the vast majority of the specific angular momentum of the material it has formed from. It has to, otherwise it could not form at all.

The problem of how protostars lose their angular momentum is still not fully solved. Likely mechanisms include jets and winds from the accretion disc, that are coupled to the star by a very strong early stellar magnetic field. This may lock the star and disc into co-rotation, allowing the transfer of angular momentum outwards.

Once the disc cleared, loss of angular momentum continues via the stellar wind and magnetic field. The wind is held in corotation by the magnetic field until it "drops off" the field lines at the Alfven radius (well beyond the orbit of Mercury). The field is then dragged through the wind, applying a torque to the Sun and taking away angular momentum. The classic reference is [Weber & Davis (1967)](https://ui.adsabs.harvard.edu/abs/1967ApJ...148..217W/abstract)

We can be reasonably sure that the young Sun was rotating at 5-50 times faster when it reached the main sequence by observing such stars in young open clusters.

You could reasonably ask why such mechanisms don't operate for Jupiter. Well, the mass loss rate is negligible and it is also empirically observed that lower mass stars and brown dwarfs have *much* longer spin down timescales. Again, for not fully understood reasons.


---

### Answer by username_2 on 2024-10-13. Score: 2

The sun is so big — almost a million miles wide — that it doesn't have to rotate as fast to contain a huge amount of angular momentum.

You are floating outside the space station with a pair of weightlifting barbells. One has 10 pound wights and one has 50 pound weights

If you apply the same amount of energy to make them each spin, the light one will rotate much faster than the heavy one.

Same with Jupiter.

> **Comments:**
> **username_3**: It would be interesting to compare the rotational angular momentum of the Sun with the angular momenta of Jupiter, both rotational and orbital.  
> **username_4**: It doesn't matter if there are a billion particles or just one.  Each individual particle carries its angular momentum with it as it's pulled towards the center.  
> **username_2**: @username_4 Right, as long as the particle remains discrete. But when it merges with the whole, it gives up its individual momentum  and adds its vector to the collective, just like the borg.  

---

