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One could define temperature as follows: $$T^{-1} = \left(\frac{\partial S}{\partial U}\right)_{N,V}$$ I was reading Schröder, and he says that we can define temperature in another way: $$T = \left(\frac{\partial U}{\partial S}\right)_{N,V}$$ But, it is understood that $$\left(\frac{\partial U}{\partial S}\right)_{N,V} = \left(\frac{\partial S}{\partial U}\right)_{N,V}^{-1}.$$ Doing some reading I found this is not generally true, the partial derivative does not always satisfy this equality. How can we one support Schröder's affirmation?

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    $\begingroup$ Can you give more specifics as to when this is not true? What is your source of this? $\endgroup$ Nov 3, 2020 at 14:34
  • $\begingroup$ Wwll...? This is well know. Knowing you, probably you know this too, so i am assuming that my words is a little confuse. $$\begin{equation} \begin{pmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \phi} \end{pmatrix} =\begin{pmatrix} \frac{\partial \rho}{\partial x} & \frac{\partial \rho}{\partial y} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} \end{pmatrix}^{-1} \tag 2\end{equation}$$ The true relation for the inverse of partial derivatives. $\endgroup$
    – Not
    Nov 3, 2020 at 14:41
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    $\begingroup$ Well, what happens when, e.g., all the other partials are zero in your matrix? $\endgroup$ Nov 3, 2020 at 14:55
  • $\begingroup$ I see what you want to say, so, in this case where N and V are hold constant, the other terms in the matrix "follow it" too? That is, all partial derivatives in this equation is avaliate with N and V constant too? Is yes, so that is it. $\endgroup$
    – Not
    Nov 3, 2020 at 14:58

3 Answers 3

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How can we one support Schröder's affirmation?

The first law, for a closed system ($N$= constant):

$$dU=\delta q -pdV$$

For a reversible process

$$dS=\frac{\delta q_{rev}}{T}$$

and

$$dU=TdS -pdV$$

Then, for a constant volume process, $dV=0$, and

$$\biggr (\frac{\partial U}{\partial S}\biggl )_{N,V}=T$$

Hope this helps

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If a system has a multi-dimensional state space and we consider a change in which all but one of a complete set of parameters is constant, then this suffices to single out a unique direction in the state space. For a small movement along that direction, various properties such as $U$ and $S$ have a unique change, by $dU$ and $dS$. One must then find that for this constrained change, $(dU / dS) = 1 / (dS/dU)$ where it is understood that all the quantities are being evaluated for the given path. It follows that the two definitions in your question must agree with one another for any system with a three-dimensional state space. For such a system with volume $V$ and $N$ particles, an example complete set of parameters is $U,N,V$.

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Omitting (for brevity) the electrochemical potentials, we have the exact differential form

$$ dS(U,V) = {dU + p(U,V)dV \over T} $$

The relation expressing the entropy $ S = S (U, V) $ as a function of the extensive parameters $(U,V)$ is sometimes called the fundamental thermodynamic relationship, because its partial derivatives provide the intensive parameters $(T,p)$ of the thermodynamic system

$$ {\partial S(U,V) \over \partial U} = {1 \over T} \qquad {\partial S(U,V) \over \partial V}= {p \over T} $$

The partial derivatives are positive quantities, so the entropy $S$ is a monotonic function of the energy $U$ and can therefore be inverted with respect to the energy

$$ U =U(S,V) \qquad dU(S,V) = TdS - p dV $$

Therefore we have uniquely

$$ {\partial U(S,V) \over \partial S} = T \qquad {\partial U(S,V) \over \partial V}= -p $$

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