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Does time translational symmetry imply H'=0 or E'=0?

  1. Oct 21, 2015 #1
    The Hamiltonian is not always equal to the total energy. In fact the Hamiltonian for a system of particles could be defined as

    ##H=L-\sum \dot{q_i}\frac{\partial L}{\partial \dot{q_i}}##

    Which is the total energy only if the potential energy is a function of ##q_i## and if the kinetic energy is a homogeneous quadratic function of ##\dot{q_i}##.

    I know how to show that the condition ##\frac{\partial L}{\partial t}=0## implies ##\frac{d}{dt}H=0##.

    But I was left wondering: People always say time-translational symmetry implies conservation of energy, but I don't think this is the case. Time translational symmetry implies the conservation of the Hamiltonian, which may or may not be the total energy.

    So which one is true? Does time translational symmetry imply conservation of the Hamiltonian or of the Energy?

    In my opinion it could imply the energy too, given a good set of coordinates that aren't flying around in space w.r.t to an inertial frame such that it would involve time in your Lagrangian...

    Thanks.
     
    Last edited: Oct 21, 2015
  2. jcsd
  3. Oct 22, 2015 #2

    mfb

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    You can always write the total energy as Hamiltonian. It might be possible to write down a proper Hamiltonian for things that are not the total energy (not sure), but that doesn't change the result of energy conservation.
     
  4. Oct 22, 2015 #3
    Hmm. My book derives ##\frac{d}{dt}(L-\sum \dot{q}_i \frac{\partial L}{\partial \dot{q}_i})=0## from time translational symmetry. Where the quantity in parenthesis is ##-H##. In order to show ##H=K+U## you would need ##U=U(q_i)## and ##\sum \dot{q}_i\frac{\partial K}{\partial \dot{q}_i}=2K## (which is Eulers theorem for homogeneous functions). Also you need that the transformation equations between generalized coordinates and rectangular coordinates don't contain time.

    Which makes sense once you verify those statements (I could post some of this work in case its not too clear). I don't see why total energy would always be the Hamiltonian given the restrictions above. Is there a theorem you could point me to? Something to ponder?

    Thanks.
     
  5. Oct 22, 2015 #4
    Is there a way to circumvent the Hamiltonian expression to derive ##\frac{d}{dt} E =0 ## from ##\frac{\partial L}{\partial t}=0##?
     
  6. Oct 22, 2015 #5
    Also I've noticed many proofs of the typical statements of Noether's theorem aren't quite that general as people try to say. For example, "space translational symmetry implies conservation of linear momentum". Well, that requires that the potential be velocity independent. So its not as general as the sentence in quotations tries to imply. I guess most potentials are velocity independent though...
     
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