# Modified Hamilton's Principle overconstraining a system by imposing too many boundary conditions

In Hamiltonian Mechanics, a version of Hamilton's principle is shown to evolve a system according to the same equations of motion as the Lagrangian, and therefore Newtonian formalism. In particular, letting $$\delta$$ indicate a variation of the path through phase space, $$\delta \int_{t_1}^{t_2} \big(p_i \dot{q}_i - H(p_i,q_i,t)\big)dt = 0$$ is shown to generate the same motion equations as those found by the Legendre transform of the Lagrangian equations of motion. Essentially, when we calculate the Euler-Lagrange equations for the above integrand, we find $$\dot{p}_i + \frac{\partial H}{\partial q_i} = 0\text{, } \dot{q}_i - \frac{\partial H}{\partial p_i} = 0.$$ Now, since these are both first-order equations of motion, we require $$2n$$ boundary conditions to obtain a solution, where $$n$$ is the number of particles. This is perfectly consistent with the Lagrangian formalism which had $$2n$$ boundary conditions due to its $$n$$ distinct second-order motion equations.

An important thing to note is that since $$p_i\dot{q}_i - H(p_i,q_i,t)$$ contains no $$\dot{p}_i$$ dependence, the variation in $$p_i(t)$$ need not be zero at the endpoints of the path. This is not the case with $$q_i(t)$$ because the $$\dot{q}_i$$ dependance results in the appearance of boundary terms that must be set to zero in order to obtain the motion equations.

Now, as far as the boundary conditions are concerned, this makes sense. For the path to be uniquely specified given Newtonian mechanics, we require $$2n$$ boundary conditions. These can either be the usual Newtonian choice of initial positions and velocities/momenta, or they can be the initial and final positions. Both are perfectly acceptable mathematically speaking. If we were required, in the modified Hamilton's principle variation, to enforce that the variations in the momenta also be zero at the initial and final times, this would mean $$2n$$ additional boundary conditions, which would generally over constrain a Newtonian system.

However, my confusion is this: when textbooks (Goldstein in particular) consider generating functions, they require that the integrand be invariant to the addition of a total time derivative of a function $$F(q_i,p_i,t)$$ of the phase space coordinates. However, adding such a function will in general add some $$\dot{p}_i$$ dependence to the integrand, which will then add boundary terms to the equations of motion unless we require that the $$p_i(t)$$ variations be zero at the boundaries. This is fine as far as it goes. We can always define the variation however we like. The point is just that it spits out the right motion equations at the end of the day. But my confusion lies in how this doesn't generally over constrain the system. If we specify all the positions and momenta at both the initial and the final times, might not the required path to connect those points in phase space be non-Newtonian? Does this only matter if you're actually trying to use the stationary action principle to find the paths, rather than just using it to find the motion equations?

These are very good questions. Refs. 1 & 2 are not entirely consistent on these issues.

1. Let us analyze the situation. In general a Hamiltonian version of the stationary action principle is of form $$S_H[z]~=~\int_{t_i}^{t_f}\! dt~L_H(z,\dot{z},t),\tag{1}$$ where the $$2n$$-dimensional phase space has (not necessary canonical) coordinates $$(z^1,\ldots,z^{2n})$$. Since the $$2n$$ EL equations should be 1st-order (as opposed to higher-order) ODEs, the integrand $$L_H(z,\dot{z},t)~=~\sum_{I=1}^{2n}A_I(z,t)\dot{z}^I+B(z,t)\tag{2}$$ must be an affine function of $$\dot{z}$$. The infinitesimal variation of the Hamiltonian action $$S_H$$ is of the form $$\delta S_H ~=~ \text{bulk-terms} ~+~ \text{boundary-terms},\tag{3}$$ where $$\text{bulk-terms}~=~\int_{t_i}^{t_f} \! dt ~\sum_{I=1}^{2n}\frac{\delta S_H}{\delta z^I}\delta z^I \tag{4}$$ yield Hamilton's equations, and where $$\text{boundary-terms}~=~\left[\sum_{I=1}^{2n}\frac{\partial L_H}{\partial \dot{z}^I}\delta z^I\right]_{t=t_i}^{t=t_f}~=~0\tag{5}$$ should vanish because of $$n \text{ initial conditions and } n \text{ final conditions.} \tag{6}$$ Since there are $$2\times 2n=4n$$ boundary terms in eq. (5) but only $$2n$$ boundary conditions (BCs) (6), not all affine integrands (2) are consistent. This mismatch is at the core of OP's question$$^1$$.

• Some of the $$4n$$ boundary terms (5) could vanish automatically if the integrand $$L_H$$ does not depend on all the dot variables $$(\dot{z}^1,\ldots,\dot{z}^{2n})$$.

The remaining boundary terms (5) must be killed by the BCs (6), which have the following possibilities:

• Essential/Dirichlet BC: $$\quad z^I(t_i)~=~z^I_i\quad\text{and}\quad z^I(t_f)~=~z^I_f.$$

• Natural BC: $$\quad \left.\frac{\partial L_H}{\partial \dot{z}^I}\right|_{t_i}~=~0\quad\text{and}\quad \left.\frac{\partial L_H}{\partial \dot{z}^I}\right|_{t_f}~=~0.$$

• Combinations thereof.

Note that if the remaining terms are more than $$2n$$, then some of the essential & natural BCs must be dependent, i.e. play a double-role$$^2$$.

2. Now let us use canonical coordinates $$(z^1,\ldots,z^{2n})~=~(q^1, \ldots, q^n; p_1,\ldots, p_n).\tag{7}$$ Refs. 1 & 2 originally consider a Hamiltonian Lagrangian of the form $$L_H~=~\sum_{j=1}^np_j\dot{q}^j-H(q,p,t)\tag{8}$$ with $$2n$$ essential/Dirichlet BCs$$^3$$ $$q^j(t_i)~=~q^j_i\qquad\text{and}\qquad q^j(t_f)~=~q^j_f, \tag{9}$$ cf. eq. (8.65) in Ref. 1 and eq. (43.8) in Ref. 2. Note that the Hamiltonian Lagrangian (8) does not depend on $$\dot{p}_j$$. We stress that the momenta $$p_j$$ do not fulfill BCs$$^3$$.

3. Next let us consider canonical transformations (CTs). If we assume that \begin{align} \sum_{j=1}^np_j\dot{q}^j-&H(q,p,t) \cr ~=~&\sum_{k=1}^nP_k\dot{Q}^k-K(Q,P,t)+\frac{dF(q,p;Q,P;t)}{dt} \end{align}\tag{10} holds off-shell, it follows via algebraic manipulations that $$\text{Hamilton's eqs. and Kamilton's eqs. are equivalent.} \tag{11}$$ Refs. 1 & 2 apply a variational argument to deduce (10)$$\Rightarrow$$ (11) by incorrectly$$^4$$ assuming an overcomplete set of $$4n$$ Dirichlet BCs.

4. Nevertheless for CTs of types 1-4 it possible to give a variational proof of (10)$$\Rightarrow$$ (11) by only assuming the $$2n$$ BCs (9). In this related Phys.SE post, the proof for type 1 is explicitly given.

References:

1. H. Goldstein, Classical Mechanics; Sections 8.5 + 9.1.

2. L.D. Landau & E.M. Lifshitz, Mechanics; $$\S43 + \S45$$.

--

$$^1$$ Let us mention that the coherent state path integral famously imposes $$4n$$ real BCs, i.e. the system is overconstrained. In other words, generically there don't exist classical paths! This is related to the overcompleteness of the coherent states, cf. e.g. this Phys.SE post.

$$^2$$ Interestingly, this issue does not arise for Lagrangian theories, where $$4n$$ BCs are just the right number for $$2n$$ 2nd-order ODEs, cf. e.g. this related Phys.SE post.

$$^3$$ After correctly not imposing BCs on the momentum variables in the text before eq. (8.71), Ref. 1 turns around in the text after eq. (8.71) and incorrectly states that one should also impose BCs on the momentum variables! This would lead to an overconstrained system as OP already noted.

$$^4$$ See in the text between eqs. (9.7) & (9.8) in Ref. 1, and in the text under eq. (45.5) in Ref. 2.