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Why is Entropy a concave function of internal energy?

  1. Jan 14, 2013 #1
    Hello

    I may well be all wrong about this but I am trying to understand from a microscopic point of view why Entropy is a concave function of internal energy. I found this in the following .pdf:

    http://physics.technion.ac.il/ckfinder/userfiles/files/avron/thermodynamics_potentials.pdf

    I started from this wikipedia article and i understand why, if the particles composing the system have a limited number of available energy levels, then S(E) first increases and then decreases.

    But saying that S(E) is concave should mean:
    - when the temperature is T1, if i give a dE to the system its entropy increases of dS1
    - when the tempereture is T2>T1, if I give the same dE to the system, its Entropy increases only of dS2 < dS1

    I cannot see this with single particles.
    If I have N particles in their lowest energy state there is only one microstate: all the particles are still.
    If I give to this system the tiniest possible amount of energy, it will be taken by just one of the particle, so the possible microstates are N.
    If I add another dE, the possible microstates should be N + N(N-1) = N^2 ... that is or one particle gets both dE or two different particles get it. Every time I add a dE I should increase the power of N.
    Now, if the entropy is somehow proportional to the logarithm of the number of microstates, I should get S proportional to K ln(N^E), that is, something that is proportianl to E... taht is, no concavity

    I am sure I am getting all this wrong... could you please help me understand this?

    Thank You

    Wentu
     
  2. jcsd
  3. Jan 14, 2013 #2

    DrDu

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    Science Advisor

    Where do you get the first N on the left hand side from?
     
  4. Jan 14, 2013 #3
    The first N is for a single particle having 2*dE energy and all other particles ground energy
     
  5. Jan 14, 2013 #4

    DrDu

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    Science Advisor

    Ah, ok. Shouldn't the second term should rather read N(N-1)/2?
     
  6. Jan 14, 2013 #5
    You are right... I was considering distinguishble particles but this isn't enough, so yes, the term should be divided by 2. I wonder if this is enough to change the behaviour from linear to less-than-linear... I think the number of microstates still increases as a power with the increasing of E... but again, i could be all wrong

    W.
     
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