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Suppose the potential energy is independent of $\dot{q},$ i.e $\frac{\partial V}{\partial\dot{q}}=0$. What is the most general form of the kinetic energy such that the Hamiltonian is the total energy? My thoughts are that since $\frac{d\mathcal{H}}{dt}=0,$ this implies $$\mathcal{H}:=p\dot{q}-\mathcal{L}=E.$$ For a lot of systems, the Lagrangian is simply the difference of the kinetic and potential energies, but I'd rather not make that assumption here. How do I apply the given condition of a velocity independent potential to make the necessary conclusions?

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  • $\begingroup$ In Classical Mechanics, a sufficient condition for $H=E$ is that the system is scleronomous and the potential does not depend on velocities. $\endgroup$ – Diracology Jun 6 '17 at 0:01
  • $\begingroup$ @Diracology well, that's a new word for Scrabble I've never known! I'm guessing it comes from "σκληρός" for "stiff / hard" i.e. unmoving in time, and its opposite rheonomous from the Greek word for "flow" (which I can't remember - but it's "rheo.." something-or-other, since there is "rheology" for fluid mechanics). $\endgroup$ – WetSavannaAnimal Jun 6 '17 at 0:16
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    $\begingroup$ @WetSavannaAnimalakaRodVance I've been looking for the origin of the words scleronomous and rheonomous but I haven't found it. Your guess is very reasonable and I thank for that! $\endgroup$ – Diracology Jun 6 '17 at 0:27
  • $\begingroup$ @Diracology your first comment probably should have been posted as an answer $\endgroup$ – David Z Jun 6 '17 at 6:08
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If the system is scleronomous and the potential does not depend on generalized velocities, then the hamiltonian equals the energy.

By being scleronomous it means that there is no explicit time dependence in the transformations between Cartesian and generalized coordinates, $\vec r_a=\vec r_a(q_1,\ldots,n)$. In this case, the kinetic energy is written as a quadratic form $$T=\frac12\sum_{i,j}A_{ij}(q)\dot q_i\dot q_j,$$ which is a homogeneous functions of degree two on velocities.

Since the potential does not depend on velocities, then $$\sum_i\dot q_i\frac{\partial L}{\partial\dot q_i}=\sum_i\dot q_i\frac{\partial T}{\partial\dot q_i},$$ and by Euler homogeneous function theorem this equals to $2T$. Therefore, $$H=2T-(T-V)=T+V=E.$$

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  • $\begingroup$ is the converse also true? In other words, is it true that if $T=T_2$ (where $T_2$ is the part of the kinetic energy that is homogeneous of degree two on velocities) then the system is scleronomous? $\endgroup$ – grjj3 Feb 24 '18 at 18:16

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