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

1. Sep 12, 2017

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.

Last edited by a moderator: Sep 12, 2017
2. Sep 12, 2017

NFuller

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?

3. Sep 13, 2017

Tio Barnabe

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. Sep 13, 2017

Staff: Mentor

Isn't S a function of two thermodynamic variables, not just U? Doesn't the same go for U?

5. Sep 13, 2017

Tio Barnabe

What would be such two variables? I'm assuming that it's a function only of $U$.

6. Sep 13, 2017

Staff: Mentor

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.

7. Sep 13, 2017

NFuller

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.

8. Sep 13, 2017

Staff: Mentor

Yes. I was trying to keep it as simple as possible for the OP.