---
title: "Legendre transform"
firstLetter: "L"
publishDate: 2021-02-22
categories:
- Mathematics
- Physics

date: 2021-02-22T21:36:35+01:00
draft: false
markup: pandoc
---

# Legendre transform

The **Legendre transform** of a function $f(x)$ is a new function $L(f')$,
which depends only on the derivative $f'(x)$ of $f(x)$, and from which
the original function $f(x)$ can be reconstructed. The point is,
analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)),
that $L(f')$ contains the same information as $f(x)$, just in a different form.

Let us choose an arbitrary point $x_0 \in [a, b]$ in the domain of
$f(x)$. Consider a line $y(x)$ tangent to $f(x)$ at $x = x_0$, which has
a slope $f'(x_0)$ and intersects the $y$-axis at $-C$:

$$\begin{aligned}
    y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
\end{aligned}$$

The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$
(or sometimes $-C$) for all $x_0 \in [a, b]$,
where $C$ corresponds to the tangent line at $x = x_0$. This yields:

$$\begin{aligned}
    L(f'(x)) = f'(x) \: x - f(x)
\end{aligned}$$

We want this function to depend only on the derivative $f'$, but
currently $x$ still appears here as a variable. We fix that problem in
the easiest possible way: by assuming that $f'(x)$ is invertible for all
$x \in [a, b]$. If $x(f')$ is the inverse of $f'(x)$, then $L(f')$ is
given by:

$$\begin{aligned}
    \boxed{
        L(f') = f' \: x(f') - f(x(f'))
    }
\end{aligned}$$

The only requirement for the existence of the Legendre transform is thus
the invertibility of $f'(x)$ in the target interval $[a,b]$, which can
only be true if $f(x)$ is either convex or concave, i.e. its derivative
$f'(x)$ is monotonic.

Crucially, the derivative of $L(f')$ with respect to $f'$ is simply
$x(f')$. In other words, the roles of $f'$ and $x$ are switched by the
transformation: the coordinate becomes the derivative and vice versa.
This is demonstrated here:

$$\begin{aligned}
    \boxed{
        \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f')
    }
\end{aligned}$$

Furthermore, Legendre transformation is an *involution*, meaning it is
its own inverse. Let $g(L')$ be the Legendre transform of $L(f')$:

$$\begin{aligned}
    g(L') = L' \: f'(L') - L(f'(L'))
    = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x)
\end{aligned}$$

Moreover, the inverse of a (forward) transform always exists, because
the Legendre transform of a convex function is itself convex. Convexity
of $f(x)$ means that $f''(x) > 0$ for all $x \in [a, b]$, which yields
the following proof:

$$\begin{aligned}
    L''(f')
    = \dv{x(f')}{f'}
    = \dv{x}{f'(x)}
    = \frac{1}{f''(x)}
    > 0
\end{aligned}$$

Legendre transformation is important in physics,
since it connects [Lagrangian](/know/concept/lagrangian-mechanics/)
and Hamiltonian mechanics to each other.
It is also used to convert between thermodynamic potentials.



## References
1.  H. Gould, J. Tobochnik,
    *Statistical and thermal physics*, 2nd edition,
    Princeton.