Math / integrals

If I have two integrals

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.

Thanks Ultrafilter I should have known that.

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).

http://icg.harvard.edu/~phys151/lectures/Lecture21.pdf

I don’t know if I ever learned that–calculus isn’t really my thing. But maybe one of the weekday folks can help you out more.

It’s not that they are equal for arbitrary F. This defines F (to within some constant).

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).