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I have the Lagrange function:

$$L=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{-y}}.\tag{1}$$

The energy is then:

$$H=\dot{x}\frac{\partial L}{\partial \dot{x}}+\dot{y}\frac{\partial L}{\partial \dot{y}}-L=0.\tag{2}$$ Can I somehow apply in this case the Hamilton or Hamilton-Jacobi formalism to find the motion?

Qmechanic
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K. Lindy
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    If $H=0$ then everything is super simple, everything is constant (just look at the Hamilton equations). Essentially you've already solved the Hamilton-Jacobi equation. – Javier Jul 10 '18 at 16:50
  • What are the canonical momenta? – DanielC Jul 10 '18 at 16:56
  • Well, but I know that for the Lagrangian L the solution is a motion along a cycloid. So, not everything is constatnt. How to obtain such a solution from the Hamiltonian that is zero? – K. Lindy Jul 11 '18 at 08:22

1 Answers1

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  1. OP's Lagrangian (1) is the square root Lagrangian for a massive relativistic point particle (or equivalently, geodesics) in a curved space with metric $${\bf g}~=~\frac{\mathrm{d}x\odot \mathrm{d}x+ \mathrm{d}y\odot \mathrm{d}y}{-y}, \qquad y~<~0 .\tag{1}$$

  2. The square root Lagrangian (1) has worldline reparametrization invariance, i.e. gauge symmetry. As a result, the Lagrangian energy function $h(x,y,\dot{x},\dot{y})$ [and the Hamiltonian $H(x,y,p_x,p_y)$ in the corresponding Hamiltonian formalism, cf. e.g. this Phys.SE post] vanish.

  3. The easiest way to proceed is to consider the corresponding non-square root Lagrangian, cf. e.g. this Phys.SE post. Then the Legendre transformation, Hamiltonian & Hamilton-Jacobi theory are in principle straightforward to set up.

Qmechanic
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  • OK. But my problem is like follows. The Lagrangian describes motion along cycloid (this can be shown on the basis of Lagrange equations). Can I reproduce this result within the Hamilton-Jacobi formalism? – K. Lindy Jul 11 '18 at 12:06