# Partial Differential with independent quantities held constant meaning?

$$\mu_{JT}=\left(\frac{\partial T}{\partial P}\right)_H= \frac{V}{C_p}(\alpha T -1)$$ and $$\left(\frac{\partial T}{\partial P}\right)_H \left(\frac{\partial H}{\partial T}\right)_P \left(\frac{\partial P}{\partial H}\right)_T =-1$$ and for the speed of sound we have $$v^2=\frac{d P}{d \rho}=\left(\frac{\partial P}{\partial \rho}\right)_s$$ I believed that for partials all other independent variables are held constant but they are constantly being stated as held constant in thermodynamics. Is this to do with inexact differentials? how can one convert a full differential to a partial as in the last example without knowing what all the independent variables are?

The physicists' wording "with constant..." is misleading in a way. I learned in a lecture given by a mathematician that writing down something like $$\left(\frac{\partial E}{\partial S}\right)_{N,V}$$ actually means: There is a function $$\widehat{E}$$ such that in the independent variables $$S,N,V$$ it holds $$E=\widehat{E}(S,N,V)$$ and the partial derivative above means $$\frac{\partial \widehat{E}}{\partial S}\left(S,N,V\right)$$. In a different set of variables you would then use another function, different from $$\widehat{E}$$.

So the "with constant ..." is really a statement about the set of independent variables chosen for a function describing the quantity being modeled.

• thanks makes sense. I was wondering how one would go about a change of basis? as seen above we have Pressure a function of entropy and enthalpy instead of Volume Commented Nov 8, 2022 at 14:42
• "Basis" isn't the word to use here, I think. But anyway, to change between different sets of variables you use rules for differentiation, like the chain rule in the answer by basics. Or for the particular case you ask about: Given a non-zero differential, one will often use the en.wikipedia.org/wiki/Inverse_function_rule, i.e. here $\left(\frac{\partial P}{\partial H}\right)_T = 1/\left(\frac{\partial H}{\partial P}\right)_T$. Commented Nov 8, 2022 at 15:02
• thanks thats great. For a zero differential what is the trick? Commented Nov 8, 2022 at 15:49
• I'm not sure there is one. I guess it'll be one of those exceptions the physicist is far less interested in than the mathematician...^^ Commented Nov 8, 2022 at 21:08
• im actually a chemeng :/ still some of the issues with the natural logs with negative numbers not 0 to infinity got me thinking about the base and whats hiding in plain sight Commented Nov 8, 2022 at 21:11

The thermodynamic state of a system is determined by a finite number of independent state variables.

Most of the mess coming from derivatives in thermodynamics could be solved if one:

• clearly states a function and its independent variables;
• applies the fundamental rules of differentiation of composite functions.

As an example, we can write the internal energy $$e$$ as a function of the density and the entropy $$(\rho, s)$$, or as a function of density and temperature $$(\rho, T)$$. Obviously, we can write the expression of entropy as a function of density and temperature $$(\rho, T)$$. If we pay attention at the functions and their independent variables, we can write the internal energy as

$$e =\tilde{e}(\rho, s) = \tilde{e}(\rho, s(\rho,T)) = \hat{e}(\rho, T)$$,

where I introduced the "tilde"-function for the expression of the energy as a function of density and entropy, and the "hat"-function for the expression of the energy as a function of density and temperature. These two functions are different since they have different independent variables, but represent the same physical quantity.

Now, we could compute the partial derivative of the energy w.r.t. the temperature, keeping fixed values of the density, i.e.

$$\left(\dfrac{\partial e}{\partial T} \right)_\rho = \dfrac{\partial }{\partial T}\bigg|_\rho \hat{e}(\rho, T) = \dfrac{\partial }{\partial T}\bigg|_\rho \tilde{e}(\rho, s(\rho, T)) = \dfrac{\partial \tilde{e}}{\partial s}\bigg|_{\rho} \dfrac{\partial s}{\partial T}\bigg|_{\rho}$$.