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Kaguro
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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?
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.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?
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?
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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.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?
Even now? But why, the Earth-Moon system is already in tidal lock, isn't it?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.
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.Kaguro said:Even now? But why, the Earth-Moon system is already in tidal lock, isn't it?
Ah! I see. That is very interesting. Glad I asked the question.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.
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.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.
Yes. The day will at that time be about 60 times what it is now.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.
Such nice calculations.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.
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.Kaguro said:But why is Mercury's orbit so strange? Won't the 1:1 ratio be of the least energy and hence equilibrium?
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,
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:Charles Link said: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.
Yes, spin angular momentum can be converted to orbital angular momentum and vice versa. This is known as spin-orbit coupling and is a fundamental principle in quantum mechanics.
Spin-orbit coupling occurs due to the interaction between the spin of an electron and its motion around an atomic nucleus. This interaction results in the conversion of spin angular momentum to orbital angular momentum.
Spin-orbit coupling is important in understanding the properties of atoms and molecules. It plays a crucial role in determining the energy levels and spectral lines of atoms and is also responsible for the fine structure of atomic spectra.
No, spin-orbit coupling is a quantum phenomenon and can only be observed at the atomic and subatomic level. It is not observable in macroscopic objects due to the large number of particles involved.
No, there are other ways in which spin and orbital angular momentum can be converted, such as through the absorption or emission of photons. However, spin-orbit coupling is the most common and well-studied mechanism for this conversion.