Hamiltonian as Legendre transformation?

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SUMMARY

The discussion centers on the Legendre transformation and its relationship to the Hamiltonian formulation in classical mechanics. The Legendre transform is defined as f*(p) = max_x(xp - f(x)), where max_x indicates the maximum value of the expression over x. The Hamiltonian is expressed as H(q,p) = p·q̇ - L(q, q̇), which is derived from the Legendre transformation, yet the maximization aspect is not explicitly stated in its definition. The connection between momentum p and the Lagrangian L is established through the equation p = ∂L/∂q̇.

PREREQUISITES
  • Understanding of Legendre transformations in mathematical physics
  • Familiarity with Hamiltonian mechanics
  • Knowledge of Lagrangian mechanics and the Lagrangian function L(q, q̇)
  • Basic calculus, particularly differentiation and maximization techniques
NEXT STEPS
  • Study the derivation of the Hamiltonian from the Lagrangian using Legendre transformations
  • Explore the implications of the maximum condition in Legendre transformations
  • Learn about the relationship between momentum and kinetic energy in classical mechanics
  • Investigate applications of Hamiltonian mechanics in modern physics
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in classical mechanics, as well as students and researchers interested in the mathematical foundations of mechanics and the applications of Legendre transformations.

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

[tex]f^*(p)=\max_x\left(xp-f(x)\right)[/tex]

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

[tex]H(q,p)=p\dot{q}-L(q,\dot{q})[/tex]

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

[tex]g(x)=xp-f(x)[/tex]


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

[tex]p=\frac{\partial L}{\partial\dot{q}}[/tex]

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

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