eljose said:
let be (dS/dt)+(gra(S))^2/2m+(LS)+V(x) where L is the Laplacian Operator and V is the potential...could it be considered as the Hamiltan Jacobi equation for a particle under a potential Vtotal=V(x)+(LS) where S is the action
I assume you mean to equate to 0, i.e.:
\frac{\partial S}{\partial t}\right)+\frac{(\vec\nabla S)^2}{2m}+\vec\nabla S+V(q)=0
If we compare it to the Hamillton-Jacobi equation for the generating function S (a concept more general than the "action")
H\left(q,\frac{\partial S}{\partial q},t\right)+\frac{\partial S}{\partial t}=0
we find they are compatible provided we let
H\left(q,\frac{\partial S}{\partial q},t\right)=\frac{(\vec\nabla S)^2}{2m}+\vec\nabla S+V(q)
Since p_i=\frac{\partial S}{\partial q_i}, we can rewrite it as
H\left(q_i,p_i,t\right)=\frac{(\sum_i p_i)^2}{2m}+\vec\nabla S+V(q)
or
H=T+W
where
W=\sum_i p_i+V(q_i).
Here we see that W=f(p_i,q_i), in other words the "potential" W is not conservative and the meaning of W is that of "virtual work". Is that the source of your doubts?
