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Hamiltonian V/S Energy!

  1. Jun 8, 2012 #1
    Consider a ball of mass m rotating around an axis Oz (vertical). This ball is on a circle whose center is the same O.
    Given: Angular velocity of ring is d∅/dt = ω.
    Mind explaining it so we can prove that Hamiltonian here is different from Energy?!!
     
  2. jcsd
  3. Jun 8, 2012 #2

    vanhees71

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    The question is, is your assertion true? Let's start from the Lagrangian, using [itex]\phi[/itex] as the generalized coordinate. The motion on a circle is then described by

    [tex]\vec{x}=\begin{pmatrix}
    r \cos \phi \\ r \sin \phi
    \end{pmatrix}
    [/tex]

    with [itex]r=\text{const}.[/itex] The Lagrangian is

    [tex]L=T=\frac{m}{2} r^2 \dot{\phi}^2.[/tex]

    The Hamiltonian is then defined as

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

    with the canonical momentum

    [tex]p=\frac{\partial L}{\partial \dot{\phi}}=m r^2 \dot{\phi}.[/tex]

    The Hamiltonian is thus

    [tex]H(q,p)=\frac{p^2}{2 m r^2}.[/tex]

    Written in terms of [itex]\dot{q}=\partial_p H=p/(m r^2)[/itex] one sees that [itex]H=T[/itex], and thus [itex]H[/itex] is the energy of the system.
     
  4. Jun 8, 2012 #3
    Thank you for your reply, very organized, and this is so true! But why did he ask us to prove that H different from E?
     
  5. Jun 8, 2012 #4
    And why did he mention "Angular velocity of ring is d∅/dt = ω"?
     
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