# General Form for Kinetic Energy Given Velocity Independent Potential such that $\mathcal{H}=E$

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?

• In Classical Mechanics, a sufficient condition for $H=E$ is that the system is scleronomous and the potential does not depend on velocities. – Diracology Jun 6 '17 at 0:01
• @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). – Selene Routley Jun 6 '17 at 0:16
• @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! – Diracology Jun 6 '17 at 0:27
• @Diracology your first comment probably should have been posted as an answer – David Z Jun 6 '17 at 6:08

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, $$H=2T-(T-V)=T+V=E.$$
• 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? – grjj3 Feb 24 '18 at 18:16