Hamilton's principle of least action

Now sufficient background has been given to return to the dynamics of mechanical systems. The path through the C-space of a system of bodies can be expressed as the solution to a calculus of variations problem that optimizes the difference between kinetic and potential energy. The calculus of variations principles generalize to any coordinate neighborhood of $ {\cal C}$. In this case, the Euler-Lagrange equation is

$\displaystyle \frac{d}{dt} \frac{\partial L}{\partial {\dot q}} - \frac{\partial L}{\partial q} = 0 ,$ (13.124)

in which $ q$ is a vector of $ n$ coordinates. It is actually $ n$ scalar equations of the form

$\displaystyle \frac{d}{dt} \frac{\partial L}{\partial {\dot q}_i} - \frac{\partial L}{\partial q_i} = 0 .$ (13.125)

The coming presentation will use (13.124) to obtain a phase transition equation. This will be derived by optimizing a functional defined as the change in kinetic and potential energy. Kinetic energy for particles and rigid bodies was defined in Section 13.3.1. In general, the kinetic energy function must be a quadratic function of $ {\dot q}$. Its definition can be interpreted as an inner product on $ {\cal C}$, which causes $ {\cal C}$ to become a Riemannian manifold [156]. This gives the manifold a notion of the ``angle'' between velocity vectors and leads to well-defined notions of curvature and shortest paths called geodesics. Let $ K(q,{\dot q})$ denote the kinetic energy, expressed using the manifold coordinates, which always takes the form

$\displaystyle K(q,{\dot q}) = \begin{matrix}\frac{1}{2} \end{matrix} {\dot q}^T M(q) {\dot q},$ (13.126)

in which $ M(q)$ is an $ n \times n$ matrix called the mass matrix or inertia matrix.

The next step is to define potential energy. A system is called conservative if the forces acting on a point depend only on the point's location, and the work done by the force along a path depends only on the endpoints of the path. The total energy is conserved under the motion of a conservative system. In this case, there exists a potential function $ \phi
: W \rightarrow {\mathbb{R}}$ such that $ F = \partial \phi / \partial p$, for any $ p \in W$. Let $ V(q)$ denote the total potential energy of a collection of bodies, placed at configuration $ q$.

It will be assumed that the dynamics are time-invariant. Hamilton's principle of least action states that the trajectory, $ {\tilde q}: T \rightarrow {\cal C}$, of a mechanical system coincides with extremals of the functional,

$\displaystyle \Phi({\tilde q}) = \int_T \Big( K(q(t),{\dot q}(t)) - V(q(t)) \Big) dt ,$ (13.127)

using any coordinate neighborhood of $ {\cal C}$. The principle can be seen for the case of $ {\cal C}=
{\mathbb{R}}^3$ by expressing Newton's second law in a way that looks like (13.124) [39]:

$\displaystyle \frac{d}{dt} (m {\dot q}) - \frac{\partial V}{\partial q} = 0 ,$ (13.128)

in which the force is replaced by the derivative of potential energy. This suggests applying the Euler-Lagrange equation to the functional

$\displaystyle L(q,{\dot q}) = K(q,{\dot q}) - V(q) ,$ (13.129)

in which it has been assumed that the dynamics are time-invariant; hence, $ L(q,{\dot q},t) = L(q,{\dot q})$. Applying the Euler-Lagrange equation to (13.127) yields the extremals.

The advantage of the Lagrangian formulation is that the C-space does not have to be $ {\cal C}=
{\mathbb{R}}^3$, described in an inertial frame. The Euler-Lagrange equation gives a necessary condition for the motions in any C-space of a mechanical system. The conditions can be expressed in terms of any coordinate neighborhood, as opposed to orthogonal coordinate systems, which are required by the Newton-Euler formulation. In mechanics literature, the $ q$ variables are often referred to as generalized coordinates. This simply means the coordinates given by any coordinate neighborhood of a smooth manifold.

Thus, the special form of (13.124) that uses (13.129) yields the appropriate constraints on the motion:

$\displaystyle \frac{d}{dt} \frac{\partial L}{\partial {\dot q}} - \frac{\partia...
...ac{\partial K(q,{\dot q})}{\partial q} + \frac{\partial V(q)}{\partial q} = 0 .$ (13.130)

Recall that this represents $ n$ equations, one for each coordinate $ q_i$. Since $ K(q,{\dot q})$ does not depend on time, the $ d/dt$ operator simply replaces $ {\dot q}$ by $ {\ddot q}$ in the calculated expression for $ \partial K(q,{\dot q})/\partial {\dot q}$. The appearance of $ {\ddot q}$ seems appropriate because the resulting differential equations are second-order, which is consistent with Newton-Euler mechanics.

Example 13..11 (A Falling Particle)   Suppose that a particle with mass $ m$ is falling in $ {\mathbb{R}}^3$. Let $ (q_1,q_2,q_3)$ denote the position of the particle. Let $ g$ denote the acceleration constant of gravity in the $ -q_3$ direction. The potential energy is $ V(q) = m g q_3$. The kinetic energy is

$\displaystyle K(q,{\dot q}) = \begin{matrix}\frac{1}{2} \end{matrix} m {\dot q}...
...atrix}\frac{1}{2} \end{matrix} m ({\dot q}_1^2 + {\dot q}_2^2 + {\dot q}_3^2) .$ (13.131)

The Lagrangian is

$\displaystyle L(q,{\dot q}) = K(q,{\dot q}) - V(q) = \begin{matrix}\frac{1}{2} \end{matrix} m ({\dot q}_1^2 + {\dot q}_2^2 + {\dot q}_3^2) - m g q_3 = 0 .$ (13.132)

To obtain the differential constraints on the motion of the particle, use (13.130). For each $ i$ from $ 1$ to $ 3$,

$\displaystyle \frac{d}{dt} \frac{\partial L}{\partial {\dot q}} = \frac{d}{dt} (m {\dot q}_i) = m {\ddot q}_i$ (13.133)

Since $ K(q,{\dot q})$ does not depend on $ q$, the derivative $ \partial
K/\partial q_i = 0$ for each $ i$. The derivatives with respect to potential energy are

$\displaystyle \frac{\partial V}{\partial q_1} = 0 \hspace*{0.5truein} \frac{\pa...
...V}{\partial q_2} = 0 \hspace*{0.5truein} \frac{\partial V}{\partial q_3} = mg .$ (13.134)

Substitution into (13.130) yields three equations:

$\displaystyle m {\ddot q}_1 = 0 \hspace*{0.5truein} m {\ddot q}_2 = 0 \hspace*{0.5truein} m {\ddot q}_3 + mg = 0 .$ (13.135)

These indicate that acceleration only occurs in the $ -q_3$ direction, and this is due to gravity. The equations are consistent with Newton's laws. As usual, a six-dimensional phase space can be defined to obtain first-order differential constraints. $ \blacksquare$

The ``least'' part of Hamilton's principle is actually a misnomer. It is technically only a principle of ``extremal'' action because (13.130) can also yield motions that maximize the functional.

Steven M LaValle 2012-04-20