Can Entropy Be Expressed as a Power Series in Terms of Internal Energy?

  • #1
Tio Barnabe
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.
 
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  • #2
Tio Barnabe said:
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##?
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.
Tio Barnabe said:
In other words, in such a case we would have T=(dS/dU)−1=1/a1T=(dS/dU)−1=1/a1T = (dS / dU)^{-1} = 1 / a_1 i.e. constant temperature.
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}##.
Tio Barnabe said:
On the other hand, if UUU takes on considerable values, then the entropy would increase exponentially, like S(U)∝eUS(U)∝eUS(U) \propto e^U, and of course, the temperature would be T∝1/eUT∝1/eUT \propto 1 / e^U.
Why?
 
  • #3
NFuller said:
Why?
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.
 
  • #4
Isn't S a function of two thermodynamic variables, not just U? Doesn't the same go for U?
 
  • #5
Chestermiller said:
Isn't S a function of two thermodynamic variables, not just U? Doesn't the same go for U?
What would be such two variables? I'm assuming that it's a function only of ##U##.
 
  • #6
Tio Barnabe said:
What would be such two variables? I'm assuming that it's a function only of ##U##.
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.
 
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  • #7
Tio Barnabe said:
Because the expression for S(U)S(U)S(U) becomes similar to that of the exponential function, apart from the 1/n!1/n!1/n! and the first term of the latter.
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.
Chestermiller said:
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.
##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.
 
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  • #8
NFuller said:
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.
Yes. I was trying to keep it as simple as possible for the OP.
 
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