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Lagrangian is a function of

  1. Apr 14, 2014 #1
    Lagrangian is a function of ......

    Since Lagrangian is a function of q, q dot & time, then why in describing the Hamiltonian (H), L does not involve time explicitly????
    as H = (p*q dot) - L (q, q dot).
  2. jcsd
  3. Apr 14, 2014 #2
    It should. The crucial point is that, if the Lagrangian doesn't depend explicitly on time then the Hamiltonian coincides with the energy of the system. However, H is defined also when L explicitly depends on t.
  4. Apr 14, 2014 #3
    astro2cosmos, where did you get that equation from? Formally, the Hamiltonian is generated from the Lagrangian by doing a Legendre transform (If you replacy only some of the generalized coordinates by their conjugate momentums, you get a Routh's function, by the way), see Arnol'd or, for a simpler treatment, Landau/Lifshitz; So what should prevent you from treating additional variables? What is true anyway, is

    [itex]\frac{\partial \mathcal H}{\partial t} = \frac{\mathrm d \mathcal H}{\mathrm d t}[/itex].

    Besides that, in a closed inertial system, time is homogenous.
  5. Apr 14, 2014 #4
    Forgotten: For any parameter, including time, the following relation is true:

    [itex]\left( \frac{\partial \mathcal H}{\partial \lambda} \right)_{p, q} = - \left( \frac{\partial \mathcal L}{\partial \lambda} \right)_{p, q}[/itex]
  6. Apr 14, 2014 #5


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    The Lagrangian and the Hamiltonian both can also be explicitly time dependent. The Lagrangian is a function of [itex]q[/itex], [itex]\dot{q}[/itex], and (sometimes) of time. The Hamiltonian is the Legendre transformation of the Lagrangian wrt. [itex]\dot{q}[/itex] vs. the canonical momentum
    [tex]p=\frac{\partial L}{\partial \dot{q}},[/tex]
    [tex]H=p \cdot \dot{q}-L.[/tex]
    The total differential is
    [tex]\mathrm{d} H=\mathrm{d}p \cdot \dot{q} + p \cdot \mathrm{d} \dot{q}-\mathrm{d} q \cdot \frac{\partial L}{\partial q}-\mathrm{d} \dot{q} \cdot \frac{\partial L}{\partial \dot{q}}-\mathrm{d} t \frac{\partial L}{\partial t}=p \cdot \mathrm{d} \dot{q}-\mathrm{d} q \cdot \frac{\partial L}{\partial q}-\mathrm{d} t \frac{\partial L}{\partial t}.[/tex]
    From this you read off that the "natural variables" for [itex]H[/itex] are indeed [itex]q[/itex], [itex]p[/itex], and [itex]t[/itex], and that the following relations hold
    [tex]\left (\frac{\partial H}{\partial p} \right)_{q,t}=\dot{q}, \quad \left (\frac{\partial H}{\partial q} \right)_{p,t}=-\left (\frac{\partial L}{\partial q} \right )_{\dot{q},t}, \quad \left (\frac{\partial H}{\partial t} \right )_{q,p}=-\left (\frac{\partial L}{\partial t} \right)_{q,\dot{q}}.[/tex]
    It is important to keep in mind that in the latter relations different variables are kept fixed when the partial derivative wrt. to the pertinent variable is taken on both sides of this equation! That's why I put the variables to be hold fixed in the different cases as subscipts of the parantheses around the partial derivative explicitly!
  7. Apr 16, 2014 #6
    for what condition L is independent of time?????
  8. Apr 16, 2014 #7


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  9. Apr 18, 2014 #8
    It depends on the problem. The Lagrangian may depend explicitly on time, for example, in a system where a bead is sliding on a moving ring or something. Which means that it is not a closed system because some other system exchanges energy with it. So no conservation of energy, which you can see from the equations about that the time derivative of the Hamiltonian doesn't vanish.

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