Statistical mechanics/Thermodynamics two spin-1/2 subsystem

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S_Flaherty
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Homework Statement


Consider two spin-1/2 subsystems with identical magnetic moments (μ) in equal fields (B). The first subsystem has a total of NA spins with initially "a" having magnetic moments pointing against the field and (NA - a) pointing along the field, so that its initial energy is UiA = μB(a - (NA - a) = μB(2a - NA). The second subsystem has a total of NB spins with "b" having moments initially pointing against the field so that its initial energy is UiB = μB(2b - NB). Now suppose that the two subsystems are brought together so that they can exchange energy. Assume that B = constant and a, b, NA, NB >> 1. Show that in equilibrium, a0/NA = b0/NB, and this implies that the two subsystems will have the same "magnetization," i.e. total magnetic moment/spin.


Homework Equations


I'm not really sure what equations are useful in this case because I'm having trouble understanding what I need to be doing. I think I need to use multiplicity so
Ω(N,n) = N!/(n!(N-n)!)


The Attempt at a Solution


I think that I have to first figure out the most probable macrostate because that is where the systems would be in equilibrium(?). So do I go along the lines of solving Ω(NA, a) and Ω(NB, b)? I don't even know if that makes sense or what a0 and b0 represent in this question.
 
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You're going in the right direction. Have you seen entropy yet? If so, can you express the conditions for equilibrium in terms of S and U?

My guess is that a0 and b0 are the number of spins pointing against the field at equilibrium.
 
S_Flaherty said:
I think that I have to first figure out the most probable macrostate because that is where the systems would be in equilibrium(?). So do I go along the lines of solving Ω(NA, a) and Ω(NB, b)? I don't even know if that makes sense or what a0 and b0 represent in this question.
exactly. But also, don't forget to constrain the total energy! And yeah, as DrClaude says, the a0 and b0 seem to simply be a and b once equilibrium is achieved.