---
title: "Legendre transform"
sort_title: "Legendre transform"
date: 2021-02-22
categories:
- Mathematics
- Physics
layout: "concept"
---

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](/know/concept/hamiltonian-mechanics/) mechanics to each other.
It is also used to convert between [thermodynamic potentials](/know/concept/thermodynamic-potential/).



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