Legendre Transform

Definition 1 (Legendre Transformation).

Let $f:\\mathbb{R}^{n}\\longrightarrow\\mathbb{R}$ be a $C^{1}$ function and consider thetransformation $x=\\left(x_{1},\\cdots,x_{n}\\right)\\longrightarrow y=\\left(y_{1},\\cdots,y_{n}%\\right)=\\left(\\partial_{1}f(x),x_{2},\\cdots,x_{n}\\right)$ . Provided it ispossible to invert ¹¹The Inverse Mapping Theorem and itsimplications must be used here; in order to be possible to invertfor $x$ , the Jacobian must be different from zero. The Jacobianbeing $\\frac{\\partial^{2}f(x)}{\\partial x^{2}}$ in this case indicatesthat $\\frac{\\partial^{2}f(x)}{\\partial x^{2}}\eq 0$ , which means that $f(x)$ must be strictly concave or strictly convex; this seems cleargraphically for $x$ , $x=\\varphi(y)$ , we define the LegendreTransform of $f$ , $\\mathcal{L}f$ , as the function

	$\\displaystyle\\mathcal{L}f=g:$	$\\displaystyle\\mathbb{R}^{n}$	$\\displaystyle\\longrightarrow\\mathbb{R}$
		$\\displaystyle y$	$\\displaystyle\\longrightarrow g(y)=\\varphi(y)\\cdot y-f(\\varphi(y))$

(here ’ $\\cdot$ ’ denotes the usual scalar product on $\\mathbb{R}^{n}$ ). $\\mathcal{L}$ is called the Legendre Transformation.

Remark 1.

As $x=\\varphi(y)$ , the defining relation is often writtenas $g(y)=x\\cdot y-f(x)$ , without explicitly indicating that $x$ must be a function of $y$

Remark 2.

Note that, in inverting for $x$ , $x=\\varphi(y)$ , we aremaking $y=(y_{1},\\cdots,y_{n})$ the independent variables.This is more an issue related to the Inverse Mapping Theorem, but itis well worth to state it explicitly.

Remark 3.

From the definition we see that the LegendreTransformation allows us to pass from a function $f$ of $(x_{1},\\cdots x_{n})$ to a function in which we have substituted the firstcoordinate by the derivative of $\\partial_{1}f$ . We will deal herewith the case in which just one coordinate is changed but proceedingby induction it is easy to prove the following facts for any numberof variables.

The rationale behind the Legendre transformation is thefollowing. Let’s begin by considering the unidimensional case.Suppose we have the function $x\\to f(x)$ . We could be interested inexpressing the values of $f$ as function of the derivative $m=f_{x}(x)$ instead of as function of $x$ itself without losing anyinformation about $f$ (some examples of this situation will be givenbelow). At first glance one could think of just inverting therelation $m=f_{x}(x)$ for $x$ to write $f(x)=f(f_{x}^{-1}(m))\\equiv g(m)$ . However, this would result in a loss of information becausethere would be infinite functions $f$ which will give rise to thesame $g$ ; namely the family of translated functions $f(x-a)$ for any $a\\in\\mathbb{R}$ will result in the same $g$ . This can be easilyvisualized in the figure.

Figure 1: Translatedversions of the same function have the same relation $f(f_{x}^{-1}(m))$

This is because we can not entirely determine a curve by knowing itsslope at every point. The key point is that we can, nevertheless,determine a curve by knowing its slope and its originordinate at every point.

Take a point P on the curve with abscise $x$ -see figure2-.

Figure 2: Meaning ofLegendre Transformation

Call the origin ordinate of its tangent $\\psi$ and its slope $m$ which is given by

m=\\frac{f(x)-\\psi}{x-0}

Then $\\psi=f-x\\cdot m$ . So, intuitively we see that the Legendre transform isnothing but the origin ordinate of the slope of $f$ at x. It isobvious -at least graphically- that we can recover $f$ knowing $\\psi(m)$ . We now prove it rigourously.

Theorem 1 (Invertibility and duality of Legendre Transformation).

The Legendre Transformation is invertible and the Inverse LegendreTransformation is the Legendre Transformation itself, that is, $\\mathcal{L}\\mathcal{L}^{-1}f=f$ or $\\mathcal{L}=\\mathcal{L}^{-1}$ .

Proof.

Evaluate the function $g$ at point $y=\\varphi^{-1}(x)$ to get

g(\\varphi^{-1}(x))=x\\cdot\\varphi^{-1}(x)-f(x)

this is

f(x)=x\\cdot\\varphi^{-1}(x)-g(\\varphi^{-1}(x))

Now, it is easy to show that $x=\\left(x_{1},\\cdots,x_{n}\\right)=\\left(\\partial_{1}g(y),y_{2},\\cdots,y_{n}\\right)$ . So, according to the definition, this is the Legendretransform of $g$ induced by the transformation $y\\longrightarrow x$ , $y=\\varphi^{-1}(x)$ .

∎

Example 1.

In thermodynamics, a thermodynamic system is completely described byknowing its fundamental equation in energetic form: $\\mathcal{U}=U(S,V)$ where $\\mathcal{U}$ is the energy, $S$ is entropy and $V$ is volume. This relation, although of great theoretical value, has amajor drawback, namely that entropy is not a measurable quantity.However, it happens that $\\frac{\\partial U}{\\partial S}=T$ ,temperature. So, we would like to being able to swap $S$ for $T$ which is an easily measurable quantity. We just take the Legendretransform $F$ of $U$ induced by the transformation $(S,V)\\longrightarrow(T,V)=\\left(\\frac{\\partial U}{\\partial S},V\\right)$ :

F=U-TS

which is called the Helmholtz Potential and henceis a function of the independent variables $T, V$ Analogously, as ithappens that $\\frac{\\partial U}{\\partial V}=-P$ , pressure, we canswap $V$ and $P$ and consider the Legendre Transformation $H$ of $U$ induced by the transformation $(S,V)\\longrightarrow(S,P)$ :

H=U+PV

which iscalled Enthaply and hence is a function of the independentvariables $S, P$ .

Example 2.

The Lagrangian formalism in Mechanics allows to completely determinethe evolution of a general mechanical system by knowledge of the socalled Lagrangian, $\\mathcal{L}$ which is a function of generalizedcoordinates $q$ , generalized velocities ²²The customarynotation for generalized velocities is $\\dot{q}$ ; however thisnotation is somehow obscure because it is prone to establish afunctional relation between $q$ and $\\dot{q}$ as variables of L. Asvariables of L they are just points in $\\mathbb{R}^{n}$ $v$ and time $t$ : $\\mathcal{L}=L(q,v,t$ ). The generalized moments are defined as $p=\\frac{\\partial L}{\\partial v}$ and they play the role of usuallinear momentum. Generalized moments are conserved in time undercertain circumstances, so we would like to swap the role of $v$ and $p$ . Thus we consider the Legendre transform $H$ of $L$ induced bythe transformation $(q,v,t)\\longrightarrow(q,p,t)=\\left(q,\\frac{\\partial L}{\\partial v},t\\right)$ :

H=L-p\\cdot v

which is called the Hamiltonian. As pointed out in Remark 2, $q, p$ and $t$ areindependent variables, as we have inverted the mentionedtransformation $(q,v,t)\\longrightarrow(q,p,t)$