Legendre Transform

Definition 1 (Legendre Transformation).

Let $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ be a $C^{1}$ function and consider the transformation $x=\left(x_{1},\cdots,x_{n}\right)\longrightarrow y=\left(y_{1},\cdots,y_{n}% \right)=\left(\partial_{1}f(x),x_{2},\cdots,x_{n}\right)$ . Provided it is possible to invert ¹¹The Inverse Mapping Theorem and its implications must be used here; in order to be possible to invert for $x$ , the Jacobian must be different from zero. The Jacobian being $\frac{\partial^{2}f(x)}{\partial x^{2}}$ in this case indicates that $\frac{\partial^{2}f(x)}{\partial x^{2}}\neq 0$ , which means that $f(x)$ must be strictly concave or strictly convex; this seems clear graphically for $x$ , $x=\varphi(y)$ , we define the Legendre Transform 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 written as $g(y)=x\cdot y-f(x)$ , without explicitly indicating that $x$ must be a function of $y$

Remark 2.

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

Remark 3.

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

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

Figure 1: Translated versions 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 its slope at every point. The key point is that we can, nevertheless, determine a curve by knowing its slope and its origin ordinate at every point.

Take a point P on the curve with abscise $x$ -see figure 2-.

Figure 2: Meaning of Legendre 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 is nothing but the origin ordinate of the slope of $f$ at x. It is obvious -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 Legendre Transformation 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 Legendre transform of $g$ induced by the transformation $y\longrightarrow x$ , $y=\varphi^{-1}(x)$ .

∎

Example 1.

In thermodynamics, a thermodynamic system is completely described by knowing 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 a major 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 Legendre transform $F$ 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 hence is a function of the independent variables $T, V$ Analogously, as it happens that $\frac{\partial U}{\partial V}=-P$ , pressure, we can swap $V$ and $P$ and consider the Legendre Transformation $H$ of $U$ induced by the transformation $(S,V)\longrightarrow(S,P)$ :

H=U+PV

which is called Enthaply and hence is a function of the independent variables $S, P$ .

Example 2.

The Lagrangian formalism in Mechanics allows to completely determine the evolution of a general mechanical system by knowledge of the so called Lagrangian, $\mathcal{L}$ which is a function of generalized coordinates $q$ , generalized velocities ²²The customary notation for generalized velocities is $\dot{q}$ ; however this notation is somehow obscure because it is prone to establish a functional relation between $q$ and $\dot{q}$ as variables of L. As variables 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 usual linear momentum. Generalized moments are conserved in time under certain circumstances, so we would like to swap the role of $v$ and $p$ . Thus we consider the Legendre transform $H$ of $L$ induced by the 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 $t$ are independent variables, as we have inverted the mentioned transformation $(q,v,t)\longrightarrow(q,p,t)$

Title	Legendre Transform
Canonical name	LegendreTransform
Date of creation	2013-03-22 17:45:58
Last modified on	2013-03-22 17:45:58
Owner	fernsanz (8869)
Last modified by	fernsanz (8869)
Numerical id	11
Author	fernsanz (8869)
Entry type	Definition
Classification	msc 14R99
Classification	msc 26B10
Related topic	InverseFunctionTheorem
Defines	Legendre Transform
Defines	Legendre transformation