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I'm trying to compute 1D projectile motion -- basically throwing a ball up and catching it in the same hand. I want to use Lagrangian dynamics and find a numerical solution out of interest.

I started trying to compute a trajectory that minimizes the action functional

$S = \int Ldt = \int((1/2) m \dot{y}^2 - mgy)dt$

using Path Integral Monte Carlo.

Then I realized that the trajectory appears to be the intersection of the Lagrangian surface with a plane. So I started to think level sets.

enter image description here

enter image description here

The slope of the plane appears to be $\frac{dz}{dy} = \frac{-1}{2g}; \frac{dz}{dv}=0$ and passes through the initial condition $(y_0 = 0, v_0 = 5)$. My question is: where does the slope of this plane come from? Any other information about level sets and direct minimization of the action functional would be appreciated also.

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  • $\begingroup$ Comment to the post (v3): Note that the principle of stationary action requires boundary conditions; not initial conditions. $\endgroup$ – Qmechanic Feb 26 '20 at 21:10
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FWIW, the simplest is probably to graph the mechanical energy $(y,\dot{y})\mapsto \frac{m}{2}\dot{y}^2 + mgy $ rather than the Lagrangian. Because of energy conservation the stationary paths would then be level sets.

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  • $\begingroup$ Ah I see. The 2g slope of the apparent plane comes from the difference between the Hamiltonian and the Lagrangian (given that my mass was one). I feel kind of silly for not seeing that. $\endgroup$ – user55937 Feb 27 '20 at 21:32

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