# Calculate Temperature of reservoir from microstate probabilities

1. Feb 5, 2013

### opaka

1. The problem statement, all variables and given/known data
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 Pa = exp (-βEa)/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).

2. Relevant equations
the probability of finding a microstate for a given energy Pa = ga exp(-βEa)/Z
the partition function Za = Ʃa ga exp(-βEa)
average energy U= partial derivative with respect to β of ln Za

3. 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 Za 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.

2. Feb 5, 2013

### 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
pi/p0 = exp(-βEi) and we can then find the temperature of the heat bath to be T= (-E)/(k ln (pi/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.

3. Feb 5, 2013

### Mute

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?

4. Feb 10, 2013

### 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.