# Can spin angular momentum get converted to orbital angular momentum?

• Kaguro
A day on Mercury is exactly two years, which gives it a sidereal rotation rate of almost 57 Earth days and an orbital period (the year) of about 85 1/4 Earth days.

#### Kaguro

I know that in QM, there is LS coupling. So the interaction is there. But is such an interaction possible in macroscopic objects like a planet?

Delta2
Kaguro said:
Summary:: For macroscopic objects like planets orbiting a star, is it possible for it to slow down rotating and make revolutions faster or vice versa while conserving total angular momentum?
There is no fundamental difference between orbital and spin AM in classical mechanics. The spin AM is precisely the orbital AM of all the particles of a planet about the planet's axis of rotation.

Take a look at:

https://en.wikipedia.org/wiki/Tidal_locking

Kaguro, vanhees71 and etotheipi
That's pretty interesting. Thanks for pointing out the article.

Kaguro said:
Summary:: For macroscopic objects like planets orbiting a star, is it possible for it to slow down rotating and make revolutions faster or vice versa while conserving total angular momentum?
...
But is such an interaction possible in macroscopic objects like a planet?
For a simple example, the angular momentum of Earth's spin is constantly being transferred to the moon, resulting in the lengthening of the day and the steady increase of the moon's orbit. Angular momentum (say about the barycenter of the two) is conserved. The process generates considerable excess heat which must be radiated away.

Kaguro said:
Summary:: For macroscopic objects like planets orbiting a star, is it possible for it to slow down rotating and make revolutions faster or vice versa while conserving total angular momentum?

I know that in QM, there is LS coupling. So the interaction is there. But is such an interaction possible in macroscopic objects like a planet?
Yes. That is the reason, for example, that the same side of the moon always faces Earth. And the same side of the planet Mercury always faces the sun.

Delta2 and Kaguro
Halc said:
For a simple example, the angular momentum of Earth's spin is constantly being transferred to the moon, resulting in the lengthening of the day and the steady increase of the moon's orbit. Angular momentum (say about the barycenter of the two) is conserved. The process generates considerable excess heat which must be radiated away.
Even now? But why, the Earth-Moon system is already in tidal lock, isn't it?

Kaguro said:
Even now? But why, the Earth-Moon system is already in tidal lock, isn't it?
The moon is tide locked with Earth (it more or less faces one side to us at all times), but not v-v. If Earth was locked to the moon, the moon would hover permanently over/near one location on Earth (just like Earth hovers near one location from any spot on the moon that can see it). Instead it still goes around once a month. Earth has a lot of spin to lose (it used to be under 10 hours), and the momentum of its spin causes the moon to recede by 3-4 cm per year. Earth also recedes from the sun due to solar tides, as do all the planets. It is one of the 4 things (two positive, two negative) altering the average orbital distance of Earth over time.

Delta2, Kaguro and PeroK
Halc said:
The moon is tide locked with Earth (it more or less faces one side to us at all times), but not v-v. If Earth was locked to the moon, the moon would hover permanently over/near one location on Earth (just like Earth hovers near one location from any spot on the moon that can see it). Instead it still goes around once a month. Earth has a lot of spin to lose (it used to be under 10 hours), and the momentum of its spin causes the moon to recede by 3-4 cm per year. Earth also recedes from the sun due to solar tides, as do all the planets. It is one of the 4 things (two positive, two negative) altering the average orbital distance of Earth over time.
Ah! I see. That is very interesting. Glad I asked the question.

Now I can picture it. The equilibrium will be attained when a guy on Moon and a guy on Earth will be able to hold the ends of a taut rope passing through the barycentre (if extrapolated).

Mister T said:
That is the reason, for example, that the same side of the moon always faces Earth. And the same side of the planet Mercury always faces the sun.
Mercury is tide locked in a strange harmonic, but it doesn't face one side to the sun. A day on Mercury is exactly two years, which gives it a sideread rotation rate of almost 57 Earth days and an orbital period (the year) of about 85 1/4 Earth days.

Kaguro said:
The equilibrium will be attained when a guy on Moon and a guy on Earth will be able to hold the ends of a taut rope passing through the barycentre.
Yes. The day will at that time be about 60 times what it is now.

Halc said:
Mercury is tide locked in a strange harmonic, but it doesn't face one side to the sun. A day on Mercury is exactly two years, which gives it a sideread rotation rate of almost 57 Earth days and an orbital period (the year) of about 85 1/4 Earth days.

Yes. The day will at that time be about 60 times what it is now.
Such nice calculations.

But why is Mercury's orbit so strange? Won't the 1:1 ratio be of the least energy and hence equilibrium?

Kaguro said:
But why is Mercury's orbit so strange? Won't the 1:1 ratio be of the least energy and hence equilibrium?
The current Mercury orbit is stable due to its eccentricity. The planet is fairly solid compared to larger planets like Earth, so the crust is permanetly bulged in two places corresponding to noon and midnight when the planet is closest to the sun. Any force that tries to slow that spin is met with a torque when the planet is closest that increases the spin, thus making the period a stable equilibrium.

The eccentric orbit causes friction due to librations which results in the same effect as is seen with the moon: it tends to circularize the orbit over time, a lower energy state than the eccentric orbit. Eventually Mercury's orbit will become sufficiently circular that the positive torque effect will no longer be able to cancel the negative friction torque from the planet spinning faster than its orbital period, and Mercury will again begin to slow its spin rate just like is happening with all the other planets.

Venus has retrograde spin, so left alone, it will eventually stop spinning (sideread rate) altogether, and then start spinning the normal way.

Pluto is the only known object that is tide locked with one of its moons (somebody will probably point out another). This is expected since it has by far the largest moon as a mass percentage of its primary. Earth is so proud of its moon, but it's only 1% of Earth's mass, whereas Charon is 12% the mass of Pluto.

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Kaguro
Delta2 and Kaguro
Both of these are so cool!

Halc said:
Mercury is tide locked in a strange harmonic, but it doesn't face one side to the sun. A day on Mercury is exactly two years,

Oh, yes. I'd forgotten about that detail.

See https://en.wikipedia.org/wiki/Einstein–de_Haas_effect for experimental results that show that spin at the atomic level is indeed angular momentum in the macroscopic sense.
But you have to be careful! For electrons there's an additional "gyrofactor" of about ##2## compared to classical (orbital) angular momentum. Einstein couldn't know that and thus predicted a gyrofactor of 1, and de Haas from his measurements got not such a clear value for the gyrofactor but just neglected measurements with the apparently "wrong values" indicating a gyrofactor larger than 1. So he missed a great discovery, which was made a bit later by Barnett:

https://en.wikipedia.org/wiki/Einstein–de_Haas_effect

From a modern perspective after discovering the gyrofactor of about 2 for elementary spin 1/2 particles what was discovered is that ferromagnetism of metals is due to the spins of the electrons.

Kaguro and Delta2

## 1. How is spin angular momentum different from orbital angular momentum?

Spin angular momentum refers to the intrinsic angular momentum of a particle, such as an electron, while orbital angular momentum refers to the angular momentum of a particle in motion around a central point, such as an electron orbiting an atom. Spin angular momentum is a fundamental property of a particle, while orbital angular momentum is dependent on its motion.

## 2. Can spin angular momentum be converted to orbital angular momentum?

No, spin angular momentum cannot be converted to orbital angular momentum. This is because spin is an intrinsic property of a particle and cannot be changed by external factors.

## 3. Can orbital angular momentum be converted to spin angular momentum?

No, orbital angular momentum cannot be converted to spin angular momentum. Again, this is because spin is an intrinsic property of a particle and cannot be changed by external factors.

## 4. How is angular momentum conserved in systems with both spin and orbital angular momentum?

In a closed system, angular momentum is conserved. This means that the total angular momentum, made up of both spin and orbital angular momentum, will remain constant. However, individual components of angular momentum can change within the system.

## 5. What role does spin-orbit coupling play in the conversion of angular momentum?

Spin-orbit coupling refers to the interaction between a particle's spin and its orbital motion. While this interaction can affect the value of both spin and orbital angular momentum, it does not result in the conversion of one to the other. Spin-orbit coupling is responsible for fine structure in atomic spectra, but does not change the total angular momentum of a system.