Hamiltonian as Legendre transformation?

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The Legendre transformation is defined as f*(p) = max_x(xp - f(x)), where max_x indicates the maximum value of the expression over all x. The discussion raises questions about the absence of this maximization process in the Hamiltonian definition, H(q,p) = p·q̇ - L(q, q̇), despite the Hamiltonian being a Legendre transformation. It highlights that maximizing g(x) = xp - f(x) leads to setting g'(x) = 0, which corresponds to p = f'(x). In the context of mechanics, this translates to p being the partial derivative of the Lagrangian with respect to q̇. The conversation emphasizes the connection between the Legendre transformation and mechanics, particularly in the formulation of the Hamiltonian.
pellman
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The definition of a Legendre transformation given on the Wikipedia page http://en.wikipedia.org/wiki/Legendre_transformation is: given a function f(x), the Legendre transform f*(p) is

f^*(p)=\max_x\left(xp-f(x)\right)

Two questions: what does \max_x mean here? And why is it not (explicitly?) included in the definition of the Hamiltonian

H(q,p)=p\dot{q}-L(q,\dot{q})

if the Hamiltonian is a Legendre transformation?
 
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I get it. If we want to maximize

g(x)=xp-f(x)


then we set g'(x)=0 which is the same as putting p=f'(x). In mechanics this amounts to

p=\frac{\partial L}{\partial\dot{q}}

Well.. thanks to anyone who read and at least thought about replying. :-)
 
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