Solving the Paramagnet Entropy Equation

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SUMMARY

The discussion focuses on solving the entropy equation for a paramagnetic system consisting of N fixed particles with spin 1/2 in a magnetic field H along the z-axis. The derived entropy formula is S=k [ (N-E/e)/2 ln( 2N/(N-E/e) ) + (N+E/e)/2 ln( 2N/(N+E/e) ) ]. The solution utilizes the Stirling approximation for the multiplicity Ω and incorporates the relationship between the number of particles in spin states. The key challenge is manipulating the term N*lnN to achieve the desired form involving 2N in the logarithm.

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  • Knowledge of paramagnetic systems and spin 1/2 particles.
  • Proficiency in manipulating logarithmic expressions and algebraic identities.
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Homework Statement



As a model of a paramagnet, consider a system of N fixed particles with spin 1/2 in a magnetic fiels H along z axis. Each particle has an energy e=μH (spin up) or e=-μH

Using S=kln(Ω), show that

S=k [ (N-E/e)/2 ln( 2N/(N-E/e) ) + (N+E/e)/2 ln( 2N/(N+E/e) ) ]


Homework Equations





The Attempt at a Solution



Ω= N!/(N+!N-!)
I used the Stirling approximation
ln(Ω)= NlnN - ( n+ ln(n+) + n- ln(n-) )
Then replaced
n+=1/2 (N+E/e)
n-=1/2(N-E/e)

S= N*lnN + 1/2(N+E/e) ln (2/ (N+E/e)) + 1/2(N-E/e) ln (2/(N-E/e)

Then I don't know what to do with the N*lnN to get the (2N) in the numerator inside the ln .?

Thanks
 
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Note that for any number x,

N*lnN = (1/2)(N+x)lnN + (1/2)(N-x)lnN

Try to choose an appropriate value for x that will lead to the desired result.
 

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