Paramagnetic system: computing number of microstates

mondeo2015
Messages
1
Reaction score
0

Homework Statement


We are given a paramagnetic system of N distinguishable particles with 1/2 spin where we use N variables
s_k each binary with possible values of ±1 where the total energy of the system is known as:
\epsilon(s) = -\mu H \sum_{k=1}^{N} s_k where \mu is the magnetic moment of the spin and H is the applied magnetic field. We are requested to assume that N is greatly larger than 1 and we are reminded of the definition of arctanh(x) = \frac{1}{2} ln(\frac{1+x}{1-x}). We have the following three parts:
1. We are asked to find the number of microstates for a given energy E \Omega(E) with as many justifiable simplifications as possible.
2. We are asked find \beta = \frac{1}{kT} for a given E and we are asked to use this information to find single-spin magnetization of the system defined as m= \frac{1}{N}\sum_{k=1}^{N} s_k in terms of \beta and H the applied magnetic field. We are advised that m can be treated as a continuous variable where needed.
3. We are told that two systems (N_1,H),(N_2,H) are Temperaturs \beta_1 , \beta_2 and initial magnetizations m_1,m_2 respectively, are brought into contact where only heat can be exchanged and we are asked to use the preceding parts to show that the finial magnetization at equilibrium is given by m = \frac{N_1m_1 + N_2m_2}{N_1+N_2}, noticing the magnetic field is the same for both systems.

Homework Equations


The Stirling approximation N! \approx Nln(N)-N as seen in Wikipedia
https://en.wikipedia.org/wiki/Stirling's_approximation

The Attempt at a Solution


I was thinking for part A the idea is to use the fact that it is the total number of N distinct solutions to an integer equation where the sum of the s' must be some integer m so this is basic combinatorics where it is finding the number of +1 valued s' (denoted l) which determines the sum which is l-(N-l) = 2l-N this is equal to m therefore l must satisfy l = (N+m)/2 so the total number of possibilities is N \choose \frac{N+m}{2} then to get rid of factorials we probably should use the assumption that N is much greater than 1 so a form of Stirling's approximation must be valid and is to be used to simplify it. For part b. I must admit I am stumped as to how to relate this to temperature and for part c I have no clue all I think must happen is thermal equilibrium but how do we reach this nice expression given? This is where I am stuck and need help, I thank all helpers.
 
Last edited:
Physics news on Phys.org
To find the number of microstates of N distinguishable objects with spin up or spin down you need to divide the N spins into two smaller groups of nu and nd. There are N distinguishable states so N! possible arrangements, but there are nu indistinguishable states so there are nu! arrangements of these. By simple combinatory analysis the total number of microstates is,
Ω = N!/ (nu!(N- nu)!).
The total energy is E = nu(-μH) + nd(μH) and from this,
nu = (1/2)(N-E/(μH)) = (N/2)(1-E/(NμH))
nd = N - (1/2)(N-E/(μH)) = (N/2)(1+E/(NμH))
In order to find β and magnetization I suggest you use the basic thermodynamical relations:
(∂S/∂E)H,N = 1/T and (∂S/∂H)E,N = m/T where the entropy is S = kBln(Ω)
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top