40

Suppose we don't know anything about statistical mechanics, not even the existence of atoms.

Why is entropy defined as $$\delta S=\frac{\delta Q}{T}$$ instead of, say, $$\delta S=\frac{\delta Q}{T^2}$$ or any other function that will conserve the 2nd Law of thermodynamics? In a nutshell, is our entropy definition unique?

knzhou
  • 101,976
oamer
  • 565
  • One advantage of the definition of entropy may be related to the following question http://physics.stackexchange.com/questions/275244/is-the-entropy-a-lorentz-invariant/275252#275252 – jim Sep 06 '16 at 07:58

2 Answers2

45

In thermodynamics, the definition of entropy is unique up to redefinitions of temperature.

The Zeroth Law of thermodynamics tells us that a temperature scale exists, but it doesn't specify anything more than that. Therefore, we are free to replace temperature $T$ with any monotonic function $f(T)$, in which case the definition of entropy becomes $$\Delta S = \frac{\Delta Q}{f(T)}.$$ As you've seen, this doesn't upset the Second Law. It does change what "temperature" means, though. The Carnot efficiency, the form of the ideal gas law, etc. all have to be changed.

This might sound puzzling, because nobody ever seems to mention this. The reason is that there are many ways of sneaking the choice $f(T) = T$ in. For example, a standard thermodynamics book could begin with the ideal gas law, which defines temperature by $T = pV/nR$. Then it can be used to derive the efficiency for an ideal gas Carnot engine. Comparing other cycles with this one in turn leads to the Clausius inequality, and hence the usual definition of entropy $\Delta S = \Delta Q / T$.

On the other hand, if one starts with the Carnot engine without specifying the working fluid, then the most one can conclude is that a Carnot engine running between reservoirs of temperature $T_1$ and $T_2$ has an efficiency $\eta$ obeying $$1 - \eta(T_1, T_2) = \frac{g(T_2)}{g(T_1)}$$ as can be shown by considering the composition of two Carnot engines in series. If the book is sloppy, then at some point in this analysis it will implicitly take $g(T) = T$, thereby fixing a temperature scale. This, of course, agrees with the "ideal gas" temperature $T = pV/nR$. But we're also free to take any function $g$, and choosing a nontrivial $g$ is equivalent to choosing a nontrivial $f = g^{-1}$ above.

Luckily, all of this discussion is moot, because there really is a sense that $f(T) = T$ is the best choice. That's because in statistical mechanics, we have a more fundamental definition of entropy, $$S = k_B \log \Omega.$$ This definition is unique, and it forces the choice $f(T) = T$.

knzhou
  • 101,976
  • 5
    I personally prefer the Gibbs entropy formula since it applies more generally than to situations where there are $\Omega$ equally probably outcomes: $$S = -k_B \sum_{i} P_i \ln P_i,$$ where the sum is over the possible states labeled by $i$, and $P_i$ is the probability of state $i$. – Sean E. Lake Sep 05 '16 at 19:47
  • 3
    @SeanLake Yeah, agreed. I prefer to keep equations as short as possible in my answers, but your formula is the more general one, and definitely the fundamental definition of $S$. – knzhou Sep 05 '16 at 19:48
  • Another fun fact: the thermodynamic temperature is also known as the ideal gas temperature because it is what you get from a gas thermometer when it's in the range where $PV=NkT$ applies. – Sean E. Lake Sep 05 '16 at 19:50
  • @knzhou what you are saying cannot be right, it is not only a convention. One can define heat transfer $\delta Q$ independently from the absolute temperature (1st law) and then Clausius' contour integral is zero for any reversible cycle with Kelvin's absolute temperature in the denominator $\oint \frac{\delta Q}{T} =0$. This integral is only scale independent and not some other nonlinear transformation of $T$ – hyportnex Sep 05 '16 at 19:50
  • Does this answer address issues with dimensional analysis? Wouldn't the function $f(T)$ have to have the same units as $T$? – Alex Ortiz Sep 05 '16 at 19:52
  • @hyportnex As I said, the form of the entropy (or equivalent the form of your integral) is fixed, after $T$ is defined. You can perform a redefinition of $T$ first, though, and this must affect the form of the integral. If $\int dQ/T = 0$ and $T' = f(T)$, we don't have $\int dQ/T' = 0$. – knzhou Sep 05 '16 at 19:53
  • @AOrtiz Nope, a redefinition of temperature could also change its units. (For example, if we turned $T$ into $T^2$, the unit of temperature would effectively be the square root of the current unit we have.) My point is that at this very early stage in thermodynamics, where we're still trying to define entropy, temperature is just as undefined! – knzhou Sep 05 '16 at 19:53
  • @knzhou Thanks for the prompt reply. Your answer to my question is counterintuitive. Do you know of any resources that I could use to familiarize myself with the topic at hand? – Alex Ortiz Sep 05 '16 at 19:55
  • @AOrtiz I don't think so. Nobody uses these alternative definitions of temperature, simply because they're not useful. Every formula just gets more complicated. – knzhou Sep 05 '16 at 19:56
  • 1
    A standard thermodynamics book might help, if you watch carefully for where they slip in the assumption $f(T) = T$. Some books do this without mentioning it at all! – knzhou Sep 05 '16 at 19:56
  • 1
    @AOrtiz Bowley and Sanchez has a good discussion exactly along these lines. https://www.amazon.com/Introductory-Statistical-Mechanics-Roger-Bowley/dp/0198505760/ref=sr_1_1?ie=UTF8&qid=1473105349&sr=8-1&keywords=thermodynamics+textbook+bowley – Sean E. Lake Sep 05 '16 at 19:56
  • @knzhou: An elaborate discussion on this along the same line can be also be found in Fermi's lectures. –  Sep 06 '16 at 05:37
  • So to sum it up : It boils down to temperature function convention in which we choose the simplest one $f(T) = T$.

    If another definition for temperature was introduced entropy will still have the same from (to preserve cyclic properties) but then $T=g^{-1}$ must be found which will just make everything more complicated . Am I getting it right ?

     – oamer Sep 06 '16 at 12:03
  • my question would be: isn't this operationally speaking, a gauge symmetry? what deeper meaning would we conclude of considering it as such? – lurscher Nov 09 '17 at 01:17
3

It can be proved that, if the definition is $\delta S=\delta Q_\text{rev}/T$, the entropy is a state function for an ideal gas (a different definition would not ensure "statefunctionness"). That was the reason why Clausius chose $\delta S=\delta Q_\text{rev}/T$, but then he decided that the entropy so defined is a state function for ANY system. The proof he offered, however, was invalid. Since then the statement "entropy is a state function" has been repeated so many times that nowadays nobody would question it. Still it is just as unjustified as in the time of Clausius.

CDCM
  • 1,822
Pentcho Valev
  • 316