pellman
Jan28-09, 09:33 AM
A system with one particle in one dimension x, momentum p, and hamiltonin H(x,p). Hamilton's principal function S(x,t) is a function satisfying.
H(x,\frac{\partial S}{\partial x})+\frac{\partial S}{\partial t}=0
Now when we say that
p=\frac{\partial S}{\partial x}
this is somewhat nonsensical. The LHS, p, is a dynamical variable, independent of x, whereas the RHS is a function of x and t. It makes sense to refer to x(t) and p(t) separately, but not generally of p(x,t).
What p=\frac{\partial S}{\partial x} must mean is that along any particular solution (x(t), p(t)) in the flow of solutions in phase space, it is the case that
p(t)=\frac{\partial S}{\partial x}|_{x=x(t)}
That is, we can speak of a function \hat{p}(x,t)\equiv\frac{\partial S}{\partial x} and then it is true that if x(t),p(t) are solutions to the equations of motions then for all t
p(t)=\hat{p}(x(t),t).
Let me pause here before proceeding to the question proper. Is what I say so far correct?
H(x,\frac{\partial S}{\partial x})+\frac{\partial S}{\partial t}=0
Now when we say that
p=\frac{\partial S}{\partial x}
this is somewhat nonsensical. The LHS, p, is a dynamical variable, independent of x, whereas the RHS is a function of x and t. It makes sense to refer to x(t) and p(t) separately, but not generally of p(x,t).
What p=\frac{\partial S}{\partial x} must mean is that along any particular solution (x(t), p(t)) in the flow of solutions in phase space, it is the case that
p(t)=\frac{\partial S}{\partial x}|_{x=x(t)}
That is, we can speak of a function \hat{p}(x,t)\equiv\frac{\partial S}{\partial x} and then it is true that if x(t),p(t) are solutions to the equations of motions then for all t
p(t)=\hat{p}(x(t),t).
Let me pause here before proceeding to the question proper. Is what I say so far correct?