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

I would note that the OP's expansion is actually not applicable to the situations you describe and the result on temperature is not generally applicable.In summary, the conversation discusses the possibility of expressing the entropy of an arbitrary system as a power series in terms of its internal energy. The results obtained from such an expression may vary depending on the values of the coefficients and the range of internal energy. It is also noted that the entropy is a function of both the internal energy and another thermodynamic variable, such as volume or temperature. The expansion may not be applicable in all cases and the resulting temperature is not generally applicable.
  • #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.
 

1. What is entropy and why is it important?

Entropy is a measure of the disorder or randomness in a system. It is important because it helps us understand how energy flows and how systems change over time.

2. What does the power series representation of entropy tell us?

The power series representation of entropy tells us that entropy can be expressed as a polynomial function, with each term representing the contribution of different factors to the overall entropy of the system.

3. How is the power series for entropy derived?

The power series for entropy is derived using statistical mechanics and thermodynamics principles. It involves breaking down the system into smaller components and analyzing their individual contributions to the overall entropy of the system.

4. Can the power series for entropy be used to predict the behavior of a system?

Yes, the power series for entropy can be used to make predictions about the behavior of a system. By examining the different factors that contribute to the entropy, we can gain insight into how the system will change over time.

5. Are there any limitations to using the power series for entropy?

While the power series for entropy is a useful tool for understanding and predicting the behavior of a system, it does have limitations. It is based on certain assumptions and may not accurately represent systems with complex interactions or non-equilibrium states.

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