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B Analysis of the entropy S of an arbritary system can be written as a power series?

  1. Sep 12, 2017 #1
    Is it ok to assume that the entropy ##S## of an arbritary system can be written as a power series as a function of the system's internal energy ##U##? Like

    $$S(U) = \sum_{i=1}^{\infty}a_iU^i = a_1 U + a_2 U^2 + \ ...$$ with ##a_i \in \mathbb{R}##.
    What results could be obtained from such expression? For instance, if the temperature could be defined as ##T = (dS / dU)^{-1}##, then we have an interesting result if ##|U| <<1## and we neglect terms higher than power one in the expansion above. In other words, in such a case we would have ##T = (dS / dU)^{-1} = 1 / a_1## i.e. constant temperature.

    On the other hand, if ##U## takes on considerable values, then the entropy would increase exponentially, like ##S(U) \propto e^U##, and of course, the temperature would be ##T \propto 1 / e^U##.

    The ##U## above perhaps could be the normalized original ##U##, i.e. the original internal energy.
     
    Last edited: Sep 12, 2017
  2. jcsd
  3. Sep 12, 2017 #2
    If ##S(E)## is a well defined function then technically yes but it often isn't. Also both ##S## and ##U## must be made dimensionless here for this to make any sense.
    You would have constant temperature near ##U=0## but depending on the value of the a's this could rapidly blow up even after adding a small amount of energy i.e. if ##a_{2}=10^{20}##.
    Why?
     
  4. Sep 13, 2017 at 5:10 AM #3
    Because the expression for ##S(U)## becomes similar to that of the exponential function, apart from the ##1/n!## and the first term of the latter.
     
  5. Sep 13, 2017 at 8:51 AM #4
    Isn't S a function of two thermodynamic variables, not just U? Doesn't the same go for U?
     
  6. Sep 13, 2017 at 8:58 AM #5
    What would be such two variables? I'm assuming that it's a function only of ##U##.
     
  7. Sep 13, 2017 at 9:23 AM #6
    It's not. The thermodynamic equilibrium state of a single phase constant composition material is determined by two independent thermodynamic variables. For S, it could be represented as a function of U and V. At constant U, S changes with V. So, in your expansion, all the a's would be functions of V.
     
  8. Sep 13, 2017 at 11:12 AM #7
    Not necessarily, it depends what the values of ##a_{n}## are. If these coefficients are very different from ##1/n!## then the series will be a very different from an exponential function.
    ##S## is actually a function of three variables; these could be ##S(N,V,E)##, ##S(N,V,T)##, ##S(V,T,\mu)##, ##S(N,P,T)##, or ##S(N,P,H)##. We can always devise a system where two of the variables are fixed and only one is allowed to vary. As you said, this would require the coefficients would take the form ##a_{n}(V,N)## or whatever other two variables are of interest.
     
  9. Sep 13, 2017 at 11:38 AM #8
    Yes. I was trying to keep it as simple as possible for the OP.
     
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