The transformation of thermodynamic potentials

1.14 The transformation of thermodynamic potentials

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

$z : = d y ∕ d x$

(1.19)

and

$x = x (z) .$

(1.20)

We find

$y = y (x (z)) = f (z),$

(1.21)

consequently

$y = f (d y ∕ d x),$

(1.22)

respectively

$d y ∕ d x = f^{- 1} (y) .$

(1.23)

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

$y = y (x + c o n s t .) .$

(1.24)

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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