Boundary conditions for calculus of variations in phase space and under canonical transformations

This might be a stupid question, but I just don't get it. In Hamiltonian mechanics, when examining conditions for a $$(\boldsymbol{q},\boldsymbol{p})\rightarrow(\boldsymbol{Q},\boldsymbol{P})$$ transformation to be canonical, one starts with $$\dot{q}_ip^i-H(\boldsymbol{q},\boldsymbol{p},t)= \dot{Q}_iP^i-\bar{H}(\boldsymbol{Q},\boldsymbol{P},t)+\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t) \ ,$$ where $$\bar{H}$$ is the transformed Hamiltonian, and $$W$$ is the generating function (now a function of $$\boldsymbol{q}$$ and $$\boldsymbol{Q}$$). This term shouldn't break Hamilton's principle, since $$\delta\int_{t_1}^{t_2} dt\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)=\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_2}-\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=0-0=0 \ .$$ But I don't see why the variation of $$W$$ should disappear at the endpoints (say at $$t_1$$). Expanding leads to: $$\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=\left(\frac{\partial W}{\partial q_i}\right)_{t_1}\underbrace{\delta q_i(t_1)}_{=0}+ \left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1)=\left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1) \ .$$ $$\boldsymbol{Q}$$ is itself a function of $$\boldsymbol{q}$$ and $$\boldsymbol{p}$$, so $$\delta Q_i(t_1)=\left(\frac{\partial Q_i}{\partial q_k}\right)_{t_1}\underbrace{\delta q_k(t_1)}_{=0}+\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1)=\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1) \ .$$ It seems as if we also needed the variation of $$\boldsymbol{p}$$ to vanish at the endpoints, and I don't get this because (at least in cartesian coordinates) $$\boldsymbol{p}=m\dot{\boldsymbol{q}}$$ and the velocity can be different along the original and the varied paths even at the endpoints (they can point in totally different directions), so in general $$\delta \dot{\boldsymbol{q}}(t_1)\neq 0$$. What am I doing wrong? Can someone help me with this, please?

• Endpoints are held fixed during the path variation, so the variation of any function at the endpoints is zero. Sep 30, 2018 at 20:56
• Endpoints are fixed indeed ($\delta q(t_1)=\delta q(t_2)=0$), but for some function $f(\boldsymbol{q,p})$, $\delta f(\boldsymbol{q,p})=0$ would require $\delta p(t_1)=\delta p(t_2)=0$ too and I can't see how fixing the endpoints only (and not the derivatives!) guarantees this condition. Sep 30, 2018 at 21:54
• I think you're confusing $\delta p$ and $p$. We're varying both $q$ and $p$ when we work in a variational principle but hold both $q$ and $p$ fixed at the endpoints of the path. $\delta q = \delta p = 0$ at the endpoints, even if the values of $q$ and $p$ are nonzero themselves. Sep 30, 2018 at 22:15
• OK, I think this is, what I actually don't understand: here $\delta p(t_1)=\delta p(t_2)=0$ is the same as $\delta \dot{q}(t_1)=\delta \dot{q}(t_2)=0$ would be in Lagrangian mechanics (again, thinking in cartesian coordinates), but in Lagrangian mechanics we don't have this kind of condition for the velocity. Or do we? Sep 30, 2018 at 22:45

These are very good questions.

1. Let us start with the old phase space variables $$(q^k,p_{\ell})$$. The Hamiltonian action is $$S_H~=~\int_{t_i}^{t_f} \! dt ~L_H, \qquad L_H~:=~\dot{q}^j p_j - H(q,p,t).\tag{A}$$ Its infinitesimal variation reads $$\delta S_H ~=~ \text{bulk-terms} ~+~ \text{boundary-terms},\tag{B}$$ where $$\text{bulk-terms}~=~\int_{t_i}^{t_f} \! dt \left(\frac{\delta S_H}{\delta q^j}\delta q^j + \frac{\delta S_H}{\delta p_j}\delta p_j \right)\tag{C}$$ yield Hamilton's equations, and where $$\text{boundary-terms}~=~\left[p_j\underbrace{\delta q^j}_{=0} \right]_{t=t_i}^{t=t_f}~=~0\tag{D}$$ vanish as they should because of, say$$^1$$, essential/Dirichlet boundary conditions (BCs) $$q^j(t_i)~=~q^j_i\qquad\text{and}\qquad q^j(t_f)~=~q^j_f. \tag{E}$$ Notice that the momenta$$^2$$ $$p_j$$ are unconstrained at the boundary.

2. Next let us consider new phase space variables $$(Q^k,P_{\ell})$$. The action of type 1 reads$$^3$$ $$S_1~:=~\int_{t_i}^{t_f} \! dt ~L_1~=~S_K+\left[ F_1(q,Q,t) \right]_{t=t_i}^{t=t_f}, \qquad S_K~:=~\int_{t_i}^{t_f} \! dt ~L_K,$$ $$L_1~:=~L_K+\frac{dF_1(q,Q,t)}{dt}, \qquad L_K~:=~ \dot{Q}^j P_j - K(Q,P,t),\tag{F}$$ where the old positions $$q^j=q^j(Q,P,t)$$ are implicit functions of the new phase space variables $$(Q^k,P_{\ell})$$. Its infinitesimal variation reads $$\delta S_1 ~=~ \text{bulk-terms} ~+~ \text{boundary-terms},\tag{G}$$ where $$\text{bulk-terms}~=~\int_{t_i}^{t_f} \! dt \left(\frac{\delta S_1}{\delta Q^j}\delta Q^j + \frac{\delta S_1}{\delta P_j}\delta P_j \right)\tag{H}$$ yield Kamilton's equations, and where $$\text{boundary-terms}~=~\left[\underbrace{\left(P_j+\frac{\partial F_1}{\partial Q^j}\right)}_{=0}\delta Q^j +\frac{\partial F_1}{\partial q^i}\underbrace{\delta q^j}_{=0} \right]_{t=t_i}^{t=t_f}~=~0\tag{I}$$ vanish as they should. One drawback is that it is non-trivial how to recast the Dirichlet BCs (E) in the new phase space variables $$(Q^k,P_{\ell})$$.

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$$^1$$ Alternatively, one could impose natural BCs, or perhaps some mixture thereof.

$$^2$$ Note that in QM it would conflict with the HUP to simultaneously impose BCs on a canonical conjugate pair.

$$^3$$ Notation conventions: Kamiltonian $$K\equiv\bar{H}$$ and type 1 generating function $$F_1\equiv G_1\equiv W$$.

• Ah, now I see! It is not $\delta Q^i$ (and thus not $\delta p_k$) which must vanish (and which could not happen anyway), but it's "coefficient", leading you to the well-known canonical conditions for the generating function. Thank you, nice answer :) Oct 1, 2018 at 21:47