3.19 The transformation of thermodynamic potentials

Why not just replace a coordinate by its adjacent force?
We will have a closer look at a 1D example of a function \(y(x)\).
Let

 \begin{equation*} z:=dy/dx \end{equation*}(3.34)

and

 \begin{equation*} x=x(z) \qquad . \end{equation*}(3.35)

We find

 \begin{equation*} y = y(x(z)) = f(z) \qquad , \end{equation*}(3.36)

consequently

 \begin{equation*} y = f( dy/dx ) \qquad , \end{equation*}(3.37)

respectively

 \begin{equation*} dy/dx = f^{-1}(y) \qquad . \end{equation*}(3.38)

This differential equation has the solution \(y(x)\); but there exist more solutions

 \begin{equation*} y = y(x+const.) \qquad . \end{equation*}(3.39)

Simply substituting a coordinate \(x\) by the adjacent force \(z\) leads to a loss of information, since the original function is known only up to a constant. For a function with several parameters this is even more critical since the ”constant” may by a function of all the remaining parameters.


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© J. Carstensen (TD Kin I)