# Legendre Transform

###### Definition 1 (Legendre Transformation).

Let $f:{\mathbb{R}}^{n}\u27f6\mathbb{R}$ be a ${C}^{1}$ function and consider the
transformation $x=({x}_{1},\mathrm{\cdots},{x}_{n})\u27f6y=({y}_{1},\mathrm{\cdots},{y}_{n})=({\partial}_{1}f(x),{x}_{2},\mathrm{\cdots},{x}_{n})$. Provided it is
possible to invert ^{1}^{1}The 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 $\frac{{\partial}^{2}f(x)}{\partial {x}^{2}}\ne 0$, which means that
$f(x)$ must be strictly concave or strictly convex; this seems clear
graphically for $x$, $x=\phi (y)$, we define the *Legendre
Transform* of $f$, $\mathcal{L}f$, as the function

$\mathcal{L}f=g:$ | ${\mathbb{R}}^{n}$ | $\u27f6\mathbb{R}$ | ||

$y$ | $\u27f6g(y)=\phi (y)\cdot y-f(\phi (y))$ |

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

###### Remark 1.

As $x=\phi (y)$, 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 $x$, $x=\phi (y)$, we are making $y=({y}_{1},\mathrm{\cdots},{y}_{n})$ 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 $({x}_{1},\mathrm{\cdots}{x}_{n})$ 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.

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-.

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,
$\mathrm{L}\mathit{}{\mathrm{L}}^{\mathrm{-}\mathrm{1}}\mathit{}f\mathrm{=}f$ or $\mathrm{L}\mathrm{=}{\mathrm{L}}^{\mathrm{-}\mathrm{1}}$.

###### Proof.

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

$$g({\phi}^{-1}(x))=x\cdot {\phi}^{-1}(x)-f(x)$$ |

this is

$$f(x)=x\cdot {\phi}^{-1}(x)-g({\phi}^{-1}(x))$$ |

Now, it is easy to show that $x=({x}_{1},\mathrm{\cdots},{x}_{n})=({\partial}_{1}g(y),{y}_{2},\mathrm{\cdots},{y}_{n})$. So, according to the definition, this is the Legendre transform of $g$ induced by the transformation $y\u27f6x$, $y={\phi}^{-1}(x)$.

∎

###### 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 $(S,V)\u27f6(T,V)=(\frac{\partial U}{\partial S},V)$:

$$F=U-TS$$ |

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 $(S,V)\u27f6(S,P)$:

$$H=U+PV$$ |

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 ^{2}^{2}The 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 $(q,v,t)\u27f6(q,p,t)=(q,\frac{\partial L}{\partial v},t)$:

$$H=L-p\cdot v$$ |

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 $(q,v,t)\u27f6(q,p,t)$

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 |