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Boltzmann equation/ Statistical Mechanics

  1. Apr 4, 2008 #1
    1. The problem statement, all variables and given/known data

    If we assume entropy is a function of the multiplicity, [tex]\Omega[/tex], (S=k*f([tex]\Omega[/tex])) show that that function f([tex]\Omega[/tex]) is ln([tex]\Omega[/tex]).


    2. Relevant equations



    3. The attempt at a solution

    [tex]\Omega[/tex] can be written as N!/ni!. By using stirling's approximation, this becomes [tex]\Omega[/tex]= ((N/e)^N)/((n1/e)^n1*(n2/e)^n2*...(ni/e)^ni). We know that the probability pi=N/ni so this reduces to W=1/(p1^n1*p2^n2*...*pi^ni). To make this user friendly take the log so ln([tex]\Omega[/tex])=-[tex]\Sigma[/tex]pi*ln(pi).

    I just started down the road of trying to use definition of multiplicity and probabilities and I did get to ln([tex]\Omega[/tex]), but it doesn't seem like I'm really doing a solid proof and I'm not sure what's missing/ how to tie it together.
     
  2. jcsd
  3. Apr 4, 2008 #2
    Consider two systems, with entropies S1 and S2, multiplicities W1 and W2. What can you say about the entropy of the combined system? What can you say about the multiplicity of the combined system?
     
  4. Apr 4, 2008 #3
    the total entropy s=s1+s2 and the multiplicity w=w1*w2. Is the log just out of convenience then?
     
  5. Apr 4, 2008 #4
    No -- it's the only function that would fit the requirement that f(w1)+f(w2)=f(w1*w2).
     
  6. Apr 4, 2008 #5
    Makes sense...thank you. It's been a long week. Nice to finally know where that log came from.
     
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