# Why does the Sun rotate?

On its axis, with a 24 day period, that is.

It seems that angular momentum conservation alone would be a completely insufficient mechanism to maintain rotation against molecular friction over millions let alone billions of years:

Observed differential rotation from poles to equator and strong convection currents produce mixing and shear which will dissipate to heat. It seems the conversion from angular momentum to shear is a 2nd order effect, but even if it were 5th,6th,7th order, over millions of years initial angular momentum would vanish.

So why is there continued solar rotation?

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Apr 28, 2020 at 15:23

First, we need to correct a misconception in the question:

It seems that angular momentum conservation alone would be a completely insufficient mechanism to maintain rotation against molecular friction over millions let alone billions of years

For a closed system, both linear and angular momentum are conserved. The action of any internal forces, including friction, must obey this law. Friction cannot decrease the total angular momentum of a closed system; it can only redistribute angular momentum between different components of the system.

For example, a mechanical gyroscope consists of a disk connected via bearings to an outer frame. If we set that disk spinning and leave the gyroscope floating in space, then eventually, friction in the bearings will cause the disk to slow down. In order to slow down the disk, the outer frame must exert a frictional torque on the disk. By Newton's Third Law, this means that the disk also exerts a frictional torque on the outer frame. So, as the disk begins to slow down, the outer frame begins to rotate. Eventually, the gyroscope reaches a state where the disk and the outer frame are rotating at the same angular velocity.

Let's examine the energy and angular momentum balance at the beginning and the end of this process. The disk, of mass $$m_d$$ and radius $$r_d$$, has initial angular velocity $$\omega_0$$. The outer frame, which is a spherical shell with mass $$m_f$$ and radius $$r_f$$, is initially at rest. So the initial total energy will be:

$$E_0=\frac{1}{2}I_d\omega_0^2=\frac{1}{4}m_dr_d^2\omega_0^2$$

and the total angular momentum will be:

$$L_0=I_d\omega_0=\frac{1}{2}m_dr_d^2\omega_0$$

Now, let's apply friction. The force due to friction at some point on the bearing contact is some constant $$f$$ (since the load on the bearing doesn't change as a function of time). Since the force due to friction is constant, the torque on the disk due to friction is also constant (let's call it $$\tau$$). So the angular momentum of the disk as a function of time will be:

$$L_d(t)=L_0-\tau t$$

This is valid until friction brings the disk and outer frame into equlibrium. But thanks to Newton's Third Law, the torque on the disk by the outer frame is accompanied by an equal and opposite torque on the outer frame by the disk. So the angular momentum of the outer frame will be:

$$L_f(t)=\tau t$$

This means that the rotational kinetic energy of the disk $$K_d$$ and frame $$K_f$$ will be:

$$K_d(t)=\frac{1}{2}I_d\omega_d(t)^2=\frac{1}{2}I_d\left(\frac{L_d(t)}{I_d}\right)^2=\frac{(L_0-\tau t)^2}{m_dr_d^2}$$

$$K_f(t)=\frac{1}{2}I_f\omega_f(t)^2=\frac{1}{2}I_f\left(\frac{L_f(t)}{I_f}\right)^2=\frac{3\tau^2t^2}{4m_fr_f^2}$$

If we take the sum of these two energies, we get the total kinetic energy $$K$$ of the system as a function of time:

$$K(t)=\frac{(L_0-\tau t)^2}{m_dr_d^2}+\frac{3\tau^2t^2}{4m_fr_f^2}$$

If we plot this function, we get the following general shape:

So the frictional torque is decreasing the total kinetic energy of the system, by converting some of the initial kinetic energy into heat. However, the total angular momentum as a function of time is:

$$L(t)=L_d(t)+L_f(t)=L_0-\tau t+\tau t=L_0$$

which is constant! So friction subtracts from the kinetic energy while leaving the angular momentum the same.

This is generally true for any closed system. Friction, like all internal forces, obeys angular momentum conservation even as it acts to dissipate kinetic energy. It does this by redistributing kinetic energy from objects at a smaller radius to objects at a larger radius.

If you apply this to a fluid, "objects at a smaller radius" are the inner layers of the rotating fluid, and "objects at a larger radius" are the outer layers. As we saw in the gyroscope example, frictional forces act until the angular velocity at all layers is the same. In other words, frictional forces convert differential rotation into rigid-body rotation, while still conserving angular momentum.

So how can an object maintain differential rotation in the face of dissipative forces like friction? One answer is convection. Convection arises when the temperature gradient in a fluid is large enough to overcome the adiabatic threshold; its effect is to mix some of the inner layers of the fluid with some of the outer layers. The endpoint of friction, rigid-body rotation, requires that the outer layers of the fluid have more rotational kinetic energy than the inner layers (in order for them to have the same angular momentum). Convection generally equalizes the kinetic energy in the region of the fluid in which it's active, so it directly works against friction in this case and maintains differential rotation (since friction is trying to establish an unequal kinetic energy distribution).

However, in the absence of an internal energy source, a temperature gradient cannot last forever. Eventually, as heat gradually diffuses through the fluid, the temperature gradient that generates convection will disappear, the convection region will shrink and disappear as well, and friction will begin converting the differential rotation to rigid-body rotation. But the Sun has an internal energy source, namely, nuclear fusion in the core. This allows it to maintain the temperature gradient necessary for convection, even in the presence of diffusion of heat, and so allows it to maintain differential rotation even in the presence of friction.

Given this, we need to make the following important point: the Sun isn't actually a closed system. In particular, the Sun is continually losing mass and energy due to radiation and, in particular, the solar wind. In addition, tidal forces between the Sun and the planets of the Solar System transfer angular momentum. The magnetic field of the Sun is also responsible for some "magnetic braking", tied to the rotation rate of the Sun. So, it turns out that the Sun is actually slowly losing angular momentum, because some angular momentum is being carried away with the solar wind. However, the rate of mass loss (and hence of angular momentum loss) due to the solar wind is quite small at present (but it was much larger when the Sun was young), and tidal forces are typically also fairly small, and the Sun's rotation rate is relatively slow, so in general, the Sun's angular momentum only changes very slowly. Frictional forces within the Sun are not a mechanism for angular momentum loss. For more details on the angular momentum history of the Sun and the mechanisms of angular momentum loss, see, for example, this article: http://adsabs.harvard.edu/full/2000ASPC..198..353D.

The angular momentum of the Sun is a conserved quantity unless some process causes angular momentum to be transferred away from the Sun by applying a torque.

Processes like convection and viscosity can redistribute angular momentum within the solar interior, but they cannot make the Sun lose angular momentum as a whole.

The Sun was certainly much faster rotating in the past. When it was 100 million years old it likely had a rotation period somewhere between 0.5 and 5 days (observed in solar analogues at that age). The rotation period has slowed because angular momentum was transferred from the solar photosphere out into the solar wind by channelled mass loss of hot, ionised material along magnetic field lines. The field lines are anchored in the photosphere, but at some distance from the Sun (known as the Alfven radius), the field weakens sufficiently for the plasma and field lines to decouple, and specific angular momentum is lost into the solar wind.

• do not the tides also play a role in transfering angular momenta?adsabs.harvard.edu/full/1927PASP...39..228L Commented Apr 28, 2020 at 16:57
• @annav Just thinking intuitively, the tides could only transfer angular momentum to objects in orbit, so we could assign an upper bound on that effect by summing the angular momentum of the rest of the solar system. Dangit... now I'm pondering odd ratios. Pluto is 0.22% of earth's mass, and 48x wider orbit. That means lil' ol' Pluto has a solid 10% of the angular inertia that Earth does! Never would have thought of that. Commented Apr 28, 2020 at 17:15
• @CortAmmon the sun is also in orbit about the barycenter of the solar system. Commented Apr 28, 2020 at 18:32
• @annav No, they don't. Commented Apr 28, 2020 at 19:01
• @RobJeffries I thought on these lines,en.wikipedia.org/wiki/… Commented Apr 29, 2020 at 4:28

Angular momentum and the concept of rotation are inextricably linked, to the point that "it has non-zero angular momentum" is probably the only precise answer to the question "how do you know a body is rotating?"

You are right considering shear and dissipation of kinetic energy into heat. A body can dissipate plenty of kinetic energy this way, and still lose no angular momentum. This dissipation will dampen all differences in velocity, until the whole body is rotating at a single angular frequency. This does not mean however that at the end the body will have stopped rotating.

### A simple model

Two concentric spherical rigid shells are both rotating around a given axis, but at different angular velocities. The total angular momentum is: $$L = I_i\omega_i + I_o\omega_o$$ Now imagine there is some friction in the system, such that each shell will feel some torque, this torque will be some complicated function of the difference $$\omega_i(t)-\omega_o(t)$$ of angular velocities. You will agree that the net effect of this friction will be to reduce this difference, until equilibrium is reached and both shells are rotating at the same angular velocity $$\omega$$. The angular momentum will be $$L' = (I_i+I_o)\omega$$ But if angular momentum conservation holds, then $$L'=L$$ and you can solve for $$\omega$$: $$\omega = \frac{L}{I_i+I_o}$$ From this calculation we conclude that no matter what kind of dissipative forces are at play, this system will keep rotating, as long as it started with some angular momentum.

This is a very simple model, but it captures the essential meaning of the conservation of angular momentum.

### Slightly more general

We can complicate the model by allowing the dynamics to change the shape of the shells, so that their moments of inertia can change. Intuitively, one shell might expand, another contract, or the can deform in differet ways. Still, you will agree that at equilibrium, the two shells will have the same angular velocity, then conservation of angular momentum will imply: $$\omega' = \frac{L}{I_i'+I_o'}$$ Now we see that the body can stop rotating. For example, the new moment of inertia could be huge, so that $$\omega'$$ is very small. Now, if mass is conserved, then you know that the only way to increase the moment of inertia, is to increase the distance from the axis. So the body can stop rotating, provided part of it goes significantly far away from the axis.

### Simple Conclusions

The internal friction of the Sun cannot make it stop rotating. This is because shear stresses will only reduce differentials in velocitites, making some parts rotate faster, while slowing down some other parts.

The Sun could slow down by expelling some amounts of mass. The Sun indeeds expells matter all the time. The fact that it hasn't stopped rotating gives an upper bound on the amount of matter it expells.