Calculate Temperature of reservoir from microstate probabilities

[opaka]

Homework Statement

Consider a network of N = 1006 non-interacting spin 1/2 particles fixed to the sites of a 1D lattice. The network is placed in an external uniform magnetic field so that its total (fixed) energy is given by E = -(N up- N down)ε = -100ε where ε is a positive constant describing how the magnetic moment of each particle couples to the external magnetic field. Now divide the network into a very large component ("the reservoir") and a very small one ("the system"), with the system containing 6 spins and the reservoir containing the remaining 1000.

a. compute the probability of finding the system in each of its allowed 64 microstates
b. compute the average energy of the system
c. use the fact that the probability of finding the system in each of its allowed microstates is given by P_a = exp (-βE_a)/Z where Z is a normalization constant to compute the temperature of the reservoir in units of ε. The simplest way of doing this is to take the ratio of the probabilities of two microstates (i.e. the one having all spins up and the one having all spins down).

Homework Equations

the probability of finding a microstate for a given energy P_a = g_a exp(-βE_a)/Z
the partition function Z_a = Ʃ_a g_a exp(-βE_a)
average energy U= partial derivative with respect to β of ln Z_a

The Attempt at a Solution

No problem with part a, I just used N!/(N-n)!n! to find the degeneracies of each microstate - for all spin up and all spin down that would be 1, and E = ±6ε
And while messy, part b was straightforward, since Z_a would be summing up the 7 individual microstates Z and then taking the partial derivative.
c is where I get stuck: I use P(6 up)/P(6 down) = exp(-β6ε)/exp(β6ε). Unfortunately, too much cancels out, and I end up with neither ε or temperature. Any ideas where I'm approaching this incorrectly.

 

[opaka]

okay, so I found this in another stat mech book:
"the ratio of the probability of being in state i to that of being the ground state (E=0) is
p_i/p₀ = exp(-βE_i) and we can then find the temperature of the heat bath to be T= (-E)/(k ln (p_i/p₀)."

So now my question is, can I justify all spins up or all spins down as a ground state? I would think that 3 spins up and 3 spins down would be the ground state.

 

[Mute]

c is where I get stuck: I use P(6 up)/P(6 down) = exp(-β6ε)/exp(β6ε). Unfortunately, too much cancels out, and I end up with neither ε or temperature. Any ideas where I'm approaching this incorrectly.

Nothing cancels out from this calculation. Take another look: the signs of the exponents are different in the numerator and denominator. Or did you mean something cancels out in a step that you didn't show?

 

[opaka]

Never mind, I must have been sleep deprived or something. I was trying to plug in the probability equations on both sides of the ratio. I figured it out now.

 

[paranormal]

It seems like you are on the right track with your approach. However, instead of using the probabilities for all spin up and all spin down, you should use the probabilities for the two microstates with the lowest and highest energy, respectively, within the system. These would be the microstates with 6 spins all pointing up and 6 spins all pointing down. So you would have P(6 up) = exp(-β6ε)/Z and P(6 down) = exp(β6ε)/Z. Taking the ratio of these probabilities and using the fact that the total energy of the system is fixed at -6ε, you can solve for the temperature in terms of ε. This will give you the temperature of the reservoir in units of ε.