Int from t1 to t2 of [(p[sub]i[/sub]q’[sub]i[/sub] – H(q,p,t)]dt = 0
And
Int from t1 to t2 of [(P[sub]i[/sub]Q’[sub]i[/sub] – K(Q,P,t)]dt = 0
Where ‘ means differentiation with respect to time.
The fact that both integrals equal zero does not mean that the integrands are equal (why is that?) but if I multiply the first integrand by a constant and add dF/dt to second then the integrands are equal? How come?
a*[p[sub]i[/sub]q’[sub]i[/sub] – H] = P[sub]i[/sub]Q’[sub]i[/sub] – K + dF/dt
Suppose you’re in a space where integration is path independent, such as C. Then the integral of any function over a simple closed curve (like the unit circle) is zero.
These integrals are Hamiltonian least action integrals and what they’re doing is developing a canonical transformation from one set of generalized canonical variables to another which significantly reduces the effort needed to solve a classical mechanics problem. Unfortunately I can’t figure out how the generating function F does this. This is from Goldstein’s Classical Mechanics text but it’s also shown here (pdf).
The Hamiltonian dynamics condition (“Hamilton’s principle” is that the functions p(t), q(t) be chosen to make the integral
stationary (usually a minimum or maximum, but occasionally a saddle-point). This is what the “delta” symbol in front of the integrals means, in the PDF you link to (look at Lecture 20 for more details). This is a problem in variational calculus; the solution is just Hamilton’s equations (shown on p.3 of the PDF).
Now adding any constant to the integral won’t change whether it’s stationary. So we add the constant F(t[sub]2[/sub])-F(t[sub]1[/sub]), which we can then write as
(Integral from t[sub]1[/sub] to t[sub]2[/sub]) dF/dt dt.
This is why we can add dF/dt to the integrand without affecting the dynamics (again, more details in Lecture 20). The goal of a canonical transformation is to use this dF/dt to implement a change of variables, from (p,q) to (P,Q).