Is the Moon a Boson? Astronomy & Quantum Spin Explained

[Mentz114]

This could be an astronomically stupid question but I cannot think of a sensible answer. What is the sum of all the quantum spins of all the particles that make up the moon ? If it was a whole number I suppose the answer would be 'yes'.

 

[Blackberg]

The moon? Why the moon?

 

[bahamagreen]

Why the moon? Probably just to attempt to characterize the presented example object as an "isolated" object; but it's not really. No more easy or clear to define the set of particles that make up the moon at any instant than it is to consider the same for an apple or a chair.

One might try to get to the sense of the question by stipulating an object free of influences (shielded from radiation and placed free of gravitational influence in deep space... etc.), or stipulate the particle count for an otherwise compromised object at an instant in time, but I don't think those work either because of uncertainty principle, at least.

 

[Mentz114]

Let me rephrase the question. If we counted up all the spins in one instant, what is probability that it will be 1) and integer 2) an integer + 1/2 3) a real number ?

 

[Jazzdude]

Let me rephrase the question. If we counted up all the spins in one instant, what is probability that it will be 1) and integer 2) an integer + 1/2 3) a real number ?

Neither. The moon, or any other object with sufficiently many interacting degrees of freedom, is not in a spin eigenstate. Therefore the characterisation as boson or fermion doesn't apply.

 

[naima]

Great.
Is it the same for a small saucepan?

 

[Mentz114]

Neither. The moon, or any other object with sufficiently many interacting degrees of freedom, is not in a spin eigenstate. Therefore the characterisation as boson or fermion doesn't apply.

Thank you.

 

[Khashishi]

If you have something like
$\frac{1}{\sqrt{2}} \left(\left|\mathrm{spin 1/2}\right> + \left|\mathrm{spin 3/2}\right>\right)$
that's still a fermion, right?
And
$\frac{1}{\sqrt{2}} \left(\left|\mathrm{spin 2}\right> + \left|\mathrm{spin 3}\right>\right)$
is still a boson, right?
As long as you are in an eigenstate of particle number, the total spin still has to come out either a superposition of integers or a superposition of integer+1/2.
Of course, the moon is constantly exchanging particles with the surroundings, so maybe it's not in a particle number eigenstate.