The Legendre transformation in 1D

3.20 The Legendre transformation in 1D

We investigate the function \(y(x)\) and \(z:=dy/dx\).
The ”total differential” is

\begin{equation*} dy = z dx \qquad . \end{equation*}

(3.40)

Calculating

\begin{equation*} F(z) = y(x(z)) - z x(z) \label{leg_transformation} \end{equation*}

(3.41)

we find its derivation

\begin{equation*} dF/dz = dy/dx(z) dx/dz(z) - x(z) - z dx/dz(z) = - x(z) \qquad , \end{equation*}

(3.42)

the ”total differential” is

\begin{equation*} dF = -x dz \qquad . \end{equation*}

(3.43)

The transformation in Eq. (3.41) is called Legendre transformation. A coordinate is replaced by its force.
The main advantage of this transformation is the inverse transformation (Legendre transformation of \(F(z)\)).
We find:

\begin{equation*} G(x) = F(z(x))-(-x) z(x) = y(x(z(x)))-z(x) x(z(x)) + x z(x) = y(x) \qquad, \end{equation*}

(3.44)

which is the original function without any loss of information.
Neglecting the additional minus sign of the inverse transformation, the pair

coordinate \(\Leftrightarrow\) force

is absolutely symmetric; e.g. depending on the potential \(-p\) is a force, respectively \(p\) is a coordinate.

Graphical representation of the Legendre transformation \(\longrightarrow\)
Description of the curve by the ”wrapping tangents”
For each \(x\) only one slope \(z\) must exist in order to get a well defined inverse function
What would happen if for a given pressure two possible volumes would exist? (not possible!!, not stable!!)
Thermodynamic functions are always convex and therefore stable

The Legendre transformation allows to transform within a potential from the intensive to the extensive parameter (and vice versa) without loss of information. This calculated potential automatically describes the corresponding thermodynamic contact correctly.
The Legendre transformation can be applied to any coordinate independently.
All contacts can thus be described when knowing one thermodynamic potential of the system for just one thermodynamic contact.