# Could temperature have been defined as $-\partial S/\partial U$?

When coming up with a definition of temperature, it's typical to start with an empirical definition that a system with a hotter temperature tends to lose heat to a system with a colder temperature. Combined with the second law of thermodynamics, that leads to the condition that for two systems in thermodynamic equilibrium,

$$\frac{\partial S_1}{\partial U_1} = \frac{\partial S_2}{\partial U_2}$$

(consider it implicit throughout this question that volumes are constant). So temperature has to be some function of $\partial S/\partial U$. Furthermore, we'd like the object that loses the heat to be the one with the hotter temperature, so temperature has to be inversely related to $\partial S/\partial U$.

Of course the conventional choice is $\frac{1}{T} = \frac{\partial S}{\partial U}$, but if we were rederiving thermodynamics and statistical mechanics from scratch, is there any reason we couldn't have chosen

$$B = -\frac{\partial S}{\partial U}$$

instead? Sure, it would lead to many common temperatures being negative, but suppose we're willing to accept that. Is there any other useful property that the standard definition of temperature has which wouldn't be shared by $B$? Would it necessarily lead to nonlinear thermometers, for example?

As Michael Brown pointed out in a comment, this alternative definition is just $B = -k\beta$, so I guess another roughly equivalent way of saying what I'm asking would be, is there a practical reason not to use $\beta$ or $B$ as temperature instead of $T$?

I've already looked at this question and this one and some others, but any close-to-relevant answers there seem to be saying merely that it's conventional to define temperature the way it is defined, not that we couldn't have done it differently.

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Suggestion to the question(v1): Emphasize that the volume $V$ should be kept fixed during the partial differentiation. –  Qmechanic Jan 7 '13 at 0:50
I left that out because I didn't think the fact that it is a constant-volume partial derivative specifically would be relevant to the question. Anyway, anyone familiar with thermodynamics would know that the volume is to be kept constant. But if you think it's important, I'll add it (or you can add it). –  David Z Jan 7 '13 at 0:55
$T'=-1/T=-\beta$, so I guess your new temperature is good for anything $\beta$ is good for. I know several people have suggested calling $\beta$ the "coolness" and negative coolness should be "hotness." So maybe there is some merit in your proposal. Some thermodynamic relations would obviously be a tiny bit more complicated, i.e. $E = -\frac{1}{2T'}$ for equipartition. The upside is that the partition function is really simple: $Z = \sum \exp(T'E)$. I'll let someone who is more of an expert comment on whether anything more than convention is really blocking this $T'$. –  Michael Brown Jan 7 '13 at 1:07
@Michael indeed, I was thinking along the same lines. I'm really looking for that commentary from an expert about whether anything is blocking the use of effectively $\beta$ as the temperature. (I changed notation from $T'$ to $B$ to emphasize that $T'$ has a nontrivial relationship to the temperature we know.) –  David Z Jan 7 '13 at 1:28
More on $S$, $T$ and $U$: physics.stackexchange.com/a/36102/2451 –  Qmechanic Jan 7 '13 at 1:30

If one takes $U$ as the dependent variable and $N, V,$ and $S$ as the independent variables, then one has $U = U(S,V,N)$ with the total derivative of U equal to $$dU = (\partial U/\partial S)_{V,N} dS + (\partial U/\partial V)_{S,N} dV + (\partial U/\partial N)_{S,V} dN.$$ Each of these partial derivatives has a "simple form", with $T = \partial U/\partial S$, $\,-p = \partial U / \partial V$ and $\mu = \partial U/\partial N$. The condition for equilibrium between two systems open to thermal transfer is that the two $T$'s be equal, the condition for two systems open to pressure change is that the two $p$'s be equal, and the same for exchange of particles with the two chemical potentials $\mu$ be the same.

These intensive variables are fundamental for thermodynamics, but could have been defined in any reasonable way, such as $1/T$ ($1/k T$ might have been better) or whatever. So the short of it is that while the derivatives are all important, their actual definition is somewhat arbitrary.

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For a mathematician, the answer is that the customary calibration of temperature---Kelvin's absolute temperature---has enormous advantages over all its rivals (including the one proposed in the above query): it is essentially the only one that leads to the mapping from the pressure-volume plane into the temperature-entropy plane described by the equations of state being area-preserving (recall that this area in both planes can be interpreted as energy). Analytically, this means that if the equations of state have the form $T=f(p,V), S=g(p,V)$ then $f_1g_2-f_2g_1 = 1$ (subscripts denote partial derivatives) and this is equivalent to the Maxwell relations.

We quote from Kelvin's original article: "The characteristic property of the scale which I now propose is, that all degress have the same value; that is, that a unit of heat descending from a body A at the temperature $T$ degrees of this scale, to a body B at the temperature $(T-1)$ degrees would give out the same mechanical effect, whatever the number $T$. This may justly be termed an absolute scale."

One of the central incidents in the history of physics is surely the search for the true scale of temperature (Maxwell's treatise on Heat begins with a lucid discussion of the problems involved) and its solution (the practical one due to Regnault and the theoretical one due to Kelvin---it is no concidence that the latter had worked in the former's laboratory) must be one of the most significant scientific achievements of the 19th century.

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More radically speaking, if the axiomatization of your thermodynamics doesn't explicily mention the intensive variable identified as $\frac{\partial A}{\partial B}$ (with $A,B$ being extensive), and you in turn always define the model you look at by specifying some (extensive) variable (like $B$ here, e.g. you say "My energy is supposed to be ..."), then why intruduce the letter $T$ at all?

What I'm saying is that if you explicitly know $S(U,V)$ and $U(S,V)$, then the quantity which will equilibrate, the "temperature", will always be a function of the parameters stated to exist in your axioms.

Temperature as such is an abstract feature of the theory. For example you might see that $T\propto PV$ or rather $T\propto (-\frac{\partial U}{\partial V})·V$. And then of course, the practical function of the "temperature" is in measuring $V$ and $P$. This question of mine is related.

Of course, if you actually do not introduce a letter for that object, the abstract theory will be loaded with expressions like

$$\Delta S = C_v \ln{\frac{\partial U_2}{\partial S_2} \over \frac{\partial U_1}{\partial S_1}} + R \ln{V_2 \over V_1},$$

where I even cheated as the definition of $C_v$ also depends on a derivative w.r.t. $T$. Or in statistical physics, there your formulas would read like

$$\frac{1}{Z}\exp{(-\frac{1}{k}\frac{\partial S}{\partial E} E \,)}.$$

I want to add that, while you introduce a temperature-less setting in the first sentence of your question and then define $T$ as (function of) a derivative, I think the people actually trying to put phenomenological thermodynamics on an abstract footing are already "using" the real quantity $T$ in their axioms.

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Hm... interesting information, but I'm not sure it quite has the answer I'm looking for, unless you're saying that there's no reason not to use the alternative definition because the theory doesn't care how $\partial S/\partial U$ is abbreviated. But I think I'm asking something on a somewhat more practical level. (By the way, I changed notation from T' to B, in case you'd like to update your answer in response) –  David Z Jan 7 '13 at 1:36
@DavidZaslavsky: Yeah, I'm essentially saying it doesn't matter: You say "we'd like the object that loses the heat to be the one with the hotter temperature" but this is really just making a list, which reduces the properties of a general $f$ in $\f(T)=\frac{\partial S}{\partial U}$. –  NikolajK Jan 7 '13 at 14:44