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---
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 $$f(x)$$ can be reconstructed.
The point is that $$L(f')$$ contains the same information as $$f(x)$$,
just in a different form,
analogously to e.g. the [Fourier transform](/know/concept/fourier-transform/).
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 slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$y = -C$$:
$$\begin{aligned}
y(x)
&= f'(x_0) (x - x_0) + f(x_0)
\\
&= f'(x_0) \: x - C
\end{aligned}$$
Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$.
We now define the *Legendre transform* $$L(f')$$,
such that for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$
(some authors use $$-C$$ instead).
Renaming $$x_0$$ to $$x$$:
$$\begin{aligned}
L(f'(x))
&= f'(x) \: x - f(x)
\end{aligned}$$
We want this function to depend only on the derivative $$f'$$,
but currently $$x$$ still appears here as a variable.
We solve this 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:
$$\begin{aligned}
\boxed{
L(f')
= f' \: x(f') - f\big(x(f')\big)
}
\end{aligned}$$
The only requirement for the existence of the Legendre transform is thus
the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$,
which is only satisfied if $$f(x)$$ is either convex or concave,
meaning its derivative $$f'(x)$$ is monotonic.
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:
$$\begin{aligned}
\boxed{
\dv{L}{f'}
= f' \dv{x}{f'} + x(f') - f' \dv{x}{f'}
= x(f')
}
\end{aligned}$$
Furthermore, Legendre transformation is an *involution*,
meaning it is its own inverse.
To show this, 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, a Legendre transform is always invertible,
because the transform of a convex function is itself convex.
Convexity of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$,
so a proof is:
$$\begin{aligned}
L''(f')
&= \dv{}{f'} \Big( \dv{L}{f'} \Big)
= \dv{x(f')}{f'}
= \dv{x}{f'(x)}
= \frac{1}{f''(x)}
> 0
\end{aligned}$$
And an analogous proof exists for concave functions where $$f''(x) < 0$$.
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.
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