2.4.3 Principle of least action and conservation laws

Newton's second law of motion in the three-dimensional space can be written as the vector equation

$$\frac d{dt}(m\dot q)=-{U}_{q}$$ (2.39)

where $q=(x,y,z)^T$ is the vector of coordinates, $\dot q={dq}/{dt}$ is the velocity vector, and $U=U(q)$ is the potential; consequently, $m\dot q$ is the momentum and $-{ U}_{ q}$ is the force. Note that we are only considering situations where the force is conservative, i.e., corresponds to the negative gradient of some potential function. Planar motion is obtained as a special case by dropping the $z$ -coordinate.

It turns out that there is a direct relationship between (2.40) and the Euler-Lagrange equation. This is difficult to see right now because the notation in (2.40) is very different from the one we have been using. So, let us modify our earlier notation to better match (2.40). First, let us write $t$ instead of $x$ for the independent variable. Second, let us write $q$ instead of $y$ for the dependent variable. Then also $y'$ becomes $\dot q$ and we have $L=L(t,q,\dot q)$ . In the new notation, the Euler-Lagrange equation becomes

$$\frac d{dt}{L}_{\dot q}={L}_{q}.$$ (2.40)

Note that since $q\in\mathbb{R}^3$ , we are referring to the multiple-degrees-of-freedom version (2.21) of the Euler-Lagrange equation, with $n=3$ .

Some remarks on the above change of notation are in order (as it is also a preview of things to come).The change from $x$ to $t$ is conceptually significant, because it implies that the curves are parameterized by time and thus describe some dynamic behavior (e.g., trajectories of a moving object). In problems such as Dido's problem or catenary problem, where there is no motion with respect to time, this notation would not be justified. Other variational problems, such as the brachistochrone problem, do indeed deal with paths of a moving particle. (We did not, however, explicitly use time when formulating the brachistochrone problem, and it would not make sense to just relabel $x$ as $t$ in our earlier formulation of that problem. Instead, we would need to reparameterize the $(x,y)$ -trajectories with respect to time, which yields a different Lagrangian; we will do this in Section 3.2.) In mechanics, as well as in control theory, time is the default independent variable. Accordingly, when we come to the optimal control part of the book, we will make this change of notation permanent. For now, we adopt it just temporarily while we discuss applications of calculus of variations in mechanics. As for the (time-)dependent coordinate variables, it is natural to denote them by $x$ and $y$ for planar curves and by $x$ , $y$ and $z$ for spatial curves. In general, the choice of a label for the coordinate vector is just a matter of preference and convention; the mechanics literature typically favors $q$ or $r$ , while in control theory it is customary to use $x$ .

Let us now compare (2.41) with (2.40). Is there a choice of the Lagrangian $L$ that would make these two equations the same? The answer is yes, and the reader will have no difficulty in seeing that the following Lagrangian does the job:

$$L:=\frac12{m\big(\dot x^2+\dot y^2+\dot z^2\big)}-U(q)$$ (2.41)

which is the difference between the kinetic energy $T:=\frac12m(\dot q\cdot\dot q)$ and the potential energy. We conclude that Newton's equations of motion can be recovered from a path optimization problem. This important result is known as Hamilton's principle of least action: Trajectories of mechanical systems are extremals of the functional

$$\int_{t_0}^{t_1} (T-U)dt$$

which is called the action integral. In general, these extremals are not necessarily minima. However, they are indeed minima--hence the term ``least action" is accurate--if the time horizon is sufficiently short; this will follow from the second-order sufficient condition for optimality, to be derived in Section 2.6.2.

For example, if the potential is 0, then the trajectories are the extremals of $\int_{t_0}^{t_1}\frac12{m(\dot q\cdot \dot q)} dt$ , which are straight lines. We saw in Example 2.2 that straight lines arise as extremals when the Lagrangian is the arclength; the kinetic energy gives the same extremals. In the presence of gravity, the paths along which the action integral is minimized can be viewed as ``straight lines" (shortest paths, or geodesics) in a curved space whose metric is determined by gravitational forces. This view of mechanics forms the basis for Einstein's theory of general relativity.

Observe the difference between the law that the derivative of the momentum equals the force and the principle that the action integral is minimized. The former condition holds pointwise in time, while the latter is a statement about the entire trajectory. However, their relation is not surprising, because if the action integral is minimized then every small piece of the trajectory must also deliver minimal action. In the limit as the length of the piece approaches 0, we recover the differential statement. We already discussed this point in the general context of the Euler-Lagrange equation (see page ).

Now, what is the physical meaning of the Hamiltonian? Substituting $q$ for $y$ in (2.28)-(2.29) and using (2.42), we have

$$H={L}_{\dot q}\cdot\dot q-L=m(\dot q\cdot \dot q)- \left(\frac12{m(\dot q\cdot \dot q)}-U\right)= \frac12{m(\dot q\cdot \dot q)}+U=T+U=E$$

which is the total energy (kinetic plus potential). We see that the Hamiltonian not only enables a convenient rewriting of the Euler-Lagrange equation in the form of the canonical equations (2.30), but also has a very clear mechanical interpretation.

Recall that in Section 2.3.4 we studied two special cases for which we found conserved quantities, i.e., functions that remain constant along extremals. We now revisit these two conservation laws--as well as another related case--in the context of the principle of least action, which permits us to see their physical meaning.

CONSERVATION OF ENERGY. In a conservative system, the potential is fixed and does not change with time. Since the kinetic energy does not explicitly depend on time either, we have $L=L(q,\dot q)$ . In other words, the Lagrangian is invariant under time shifting. In our old notation, this corresponds to the ``no $x$ " case from Section 2.3.4, and we know that in this case the Hamiltonian is conserved. We also just saw that the Hamiltonian is the total energy of the system. Therefore, this is nothing but the well-known principle of conservation of energy.

CONSERVATION OF MOMENTUM. Suppose that no force is acting on the system (i.e., the system is closed). Since the force is given by ${L}_{q}=-{U}_{q}$ , this implies that $U$ must be constant. The kinetic energy $T$ depends on $\dot q$ but not on $q$ . Thus the Lagrangian $L=T-U$ does not explicitly depend on $q$ , which means that it is invariant under parallel translations. This situation corresponds to the ``no $y$ " case from Section 2.3.4, where we saw that the momentum, ${L}_{\dot q}=m\dot q$ in the present notation, is conserved. A more general statement is that for each coordinate $q_i$ that does not appear in $L$ , the corresponding component ${L}_{\dot q_i}=m\dot q_i$ of the momentum is conserved.

CONSERVATION OF ANGULAR MOMENTUM. Consider a planar motion in a central field; in polar coordinates $(r,\theta)$ , this is defined by the property that $U=U(r)$ , i.e., the potential depends only on the radius and not on the angle. This means that no torque is acting on the system, making the Lagrangian $L$ invariant under rotations. Now we can use the fact (noted at the end of Section 2.3.3) that the Euler-Lagrange equation looks the same in all coordinate systems. In particular, in polar coordinates we have

$${L}_{\theta}=\frac d{dt}{L}_{\dot \theta}$$ (2.42)

and the analogous equation for $r$ (which we do not need here). Arguing exactly as before, we can show that in the present ``no $\theta$ " case the corresponding component of the momentum, ${L}_{\dot\theta}$ , is conserved. Converting the kinetic energy from Cartesian to polar coordinates, we have

$$T=\frac12 m(\dot x^2+\dot y^2)=\frac 12m(\dot r^2+r^2\dot\theta^2).$$ (2.43)

Thus the conserved quantity, in polar and Cartesian coordinates, is

$${L}_{\dot\theta}={T}_{\dot\theta}=mr^2\dot\theta=m(x\dot y-y\dot x).$$

These are familiar expressions for the angular momentum.

We remark that all of the above examples are special instances of a general result known as Noether's theorem, which says that invariance of the action integral under some transformation (e.g., time shift, translation, rotation) implies the existence of a conserved quantity.