# Calculate Temperature of reservoir from microstate probabilities

• opaka
In summary, the conversation discusses a network of non-interacting particles in a 1D lattice with an external magnetic field and how to compute the probability of finding the system in each of its allowed microstates, the average energy of the system, and the temperature of the reservoir in units of ε. The partition function and partial derivative are used in the calculations. The conversation also mentions the possibility of all spins up or all spins down being the ground state.
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 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).

## Homework 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

## 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.

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.

opaka said:
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?

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.

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.

## 1. How do you calculate the temperature of a reservoir from microstate probabilities?

To calculate the temperature of a reservoir from microstate probabilities, you can use the Boltzmann distribution formula: T = E/k ln(W), where T is the temperature, E is the average energy of the system, k is the Boltzmann constant, and W is the number of microstates. You will need to know the energy levels and probabilities of each microstate in order to plug them into this formula.

## 2. What is the importance of calculating the temperature of a reservoir from microstate probabilities?

Calculating the temperature of a reservoir from microstate probabilities is important because it allows scientists to understand the thermodynamic properties of a system. It can also help in predicting the behavior and stability of the system, as well as in designing new materials and processes.

## 3. Can the temperature of a reservoir be calculated from macrostate probabilities?

No, the temperature of a reservoir cannot be calculated from macrostate probabilities alone. Macrostate probabilities only provide information about the overall behavior of the system, while microstate probabilities are needed to accurately calculate the temperature.

## 4. How accurate is the calculation of temperature from microstate probabilities?

The accuracy of the calculation of temperature from microstate probabilities depends on the accuracy of the energy levels and probabilities used in the calculation. If these values are known with high precision, then the calculated temperature will also be accurate.

## 5. Are there any limitations to using microstate probabilities to calculate temperature?

One limitation of using microstate probabilities to calculate temperature is that it assumes that the system is in thermal equilibrium. This means that all energy levels are equally likely to be occupied. In reality, this may not always be the case, especially in complex systems. Additionally, this method may not be applicable to systems that are not in isolation or are undergoing rapid changes.

Replies
1
Views
1K
• Thermodynamics
Replies
3
Views
997
Replies
3
Views
2K
Replies
3
Views
952
• Thermodynamics
Replies
15
Views
1K
Replies
1
Views
942
• Thermodynamics
Replies
5
Views
2K
Replies
1
Views
924