Categories: Mathematics, Physics.

# 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), 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 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.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.