- #1

pellman

- 684

- 5

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.

[tex]H(x,\frac{\partial S}{\partial x})+\frac{\partial S}{\partial t}=0[/tex]

Now when we say that

[tex]p=\frac{\partial S}{\partial x}[/tex]

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 [tex]p=\frac{\partial S}{\partial x}[/tex] 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

[tex]p(t)=\frac{\partial S}{\partial x}|_{x=x(t)}[/tex]

That is, we can speak of a function [tex]\hat{p}(x,t)\equiv\frac{\partial S}{\partial x}[/tex] and then it is true that

[tex]p(t)=\hat{p}(x(t),t)[/tex].

Let me pause here before proceeding to the question proper. Is what I say so far correct?

[tex]H(x,\frac{\partial S}{\partial x})+\frac{\partial S}{\partial t}=0[/tex]

Now when we say that

[tex]p=\frac{\partial S}{\partial x}[/tex]

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 [tex]p=\frac{\partial S}{\partial x}[/tex] 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

[tex]p(t)=\frac{\partial S}{\partial x}|_{x=x(t)}[/tex]

That is, we can speak of a function [tex]\hat{p}(x,t)\equiv\frac{\partial S}{\partial x}[/tex] and then it is true that

*if x(t),p(t) are solutions to the equations of motions*then for all t[tex]p(t)=\hat{p}(x(t),t)[/tex].

Let me pause here before proceeding to the question proper. Is what I say so far correct?

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