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diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md index 51c003e..c4fdeb4 100644 --- a/source/know/concept/legendre-transform/index.md +++ b/source/know/concept/legendre-transform/index.md @@ -9,76 +9,90 @@ 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. +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 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 -a slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$-C$$: +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 must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$: $$\begin{aligned} - y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C + y(x) + = f'(x_0) (x - x_0) + f(x_0) + = f'(x_0) \: x - C \end{aligned}$$ -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: +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 $$x$$: $$\begin{aligned} - L(f'(x)) = f'(x) \: x - f(x) + L(f'(x)) + = f'(x) \: x - f(x) \end{aligned}$$ -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: +We want this function to depend only on the derivative $$f'$$, +but currently $$x$$ still appears here as a variable. +We fix 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: $$\begin{aligned} \boxed{ - L(f') = f' \: x(f') - f(x(f')) + L(f') + = f' \: x(f') - f(x(f')) } \end{aligned}$$ 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. +the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, +which can only be true if $$f(x)$$ is either convex or concave, +meaning 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: +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: $$\begin{aligned} \boxed{ - \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f') + \dv{L}{f'} + = f' \dv{x}{f'} + x(f') - f' \dv{x}{f'} + = x(f') } \end{aligned}$$ -Furthermore, Legendre transformation is an *involution*, meaning it is -its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$: +Furthermore, Legendre transformation is an *involution*, +meaning it is its own inverse. +To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$: $$\begin{aligned} - g(L') = L' \: f'(L') - L(f'(L')) - = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x) + g(L') + = L' \: f'(L') - L(f'(L')) + = x(f') \: f' - f' \: x(f') + f(x(f')) + = f(x) \end{aligned}$$ -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: +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. |