3Hamiltonian Mechanics

Numerical experiments are just what their name implies: experiments. In describing and evaluating them, one should enter the state of mind of the experimental physicist, rather than that of the mathematician. Numerical experiments cannot be used to prove theorems; but, from the physicist's point of view, they do often provide convincing evidence for the existence of a phenomenon. We will therefore follow an informal, descriptive and non-rigorous approach. Briefly stated, our aim will be to understand the fundamental properties of dynamical systems rather than to prove them.

Michel Hénon, “Numerical Exploration of Hamiltonian Systems,” in Chaotic Behavior of Deterministic Systems [21], p. 57.

The formulation of mechanics with generalized coordinates and momenta as dynamical state variables is called the Hamiltonian formulation. The Hamiltonian formulation of mechanics is equivalent to the Lagrangian formulation; however, each presents a useful point of view. The Lagrangian formulation is especially useful in the initial formulation of a system. The Hamiltonian formulation is especially useful in understanding the evolution of a system, especially when there are symmetries and conserved quantities.

For each continuous symmetry of a mechanical system there is a conserved quantity. If the generalized coordinates can be chosen to reflect a symmetry, then, by the Lagrange equations, the momenta conjugate to the cyclic coordinates are conserved. We have seen that such conserved quantities allow us to deduce important properties of the motion. For instance, consideration of the energy and angular momentum allowed us to deduce that rotation of a free rigid body about the axis of intermediate moment of inertia is unstable, and that rotation about the other principal axes is stable. For the axisymmetric top, we used two conserved momenta to reexpress the equations governing the evolution of the tilt angle so that they involve only the tilt angle and its derivative. The evolution of the tilt angle can be determined independently and has simply periodic solutions. Consideration of the conserved momenta has provided key insight. The Hamiltonian formulation is motivated by the desire to focus attention on the momenta.

In the Lagrangian formulation the momenta are, in a sense, secondary quantities: the momenta are functions of the state space variables, but the evolution of the state space variables depends on the state space variables and not on the momenta. To make use of any conserved momenta requires fooling around with the specific equations. The momenta can be rewritten in terms of the coordinates and the velocities, so, locally, we can solve for the velocities in terms of the coordinates and momenta. For a given mechanical system, and a Lagrangian describing its dynamics in a given coordinate system, the momenta and the velocities can be deduced from each other. Thus we can represent the dynamical state of the system in terms of the coordinates and momenta just as well as with the coordinates and the velocities. If we use the coordinates and momenta to represent the state and write the associated state derivative in terms of the coordinates and momenta, then we have a self-contained system. This formulation of the equations governing the evolution of the system has the advantage that if some of the momenta are conserved, the remaining equations are immediately simplified.

The Lagrangian formulation of mechanics has provided the means to investigate the motion of complicated mechanical systems. We have found that dynamical systems exhibit a bewildering variety of possible motions. The motion is sometimes rather simple and sometimes very complicated. Sometimes the evolution is very sensitive to the initial conditions, and sometimes it is not. And sometimes there are orbits that maintain resonance relationships with a drive. Consider the periodically driven pendulum: it can behave more or less as an undriven pendulum with extra wiggles, it can move in a strongly chaotic manner, or it can move in resonance with the drive, oscillating once for every two cycles of the drive or looping around once per drive cycle. Or consider the Moon. The Moon rotates synchronously with its orbital motion, always pointing roughly the same face to the Earth. However, Mercury rotates three times every two times it circles the Sun, and Hyperion rotates chaotically.

How can we make sense of this? How do we put the possible motions of these systems in relation to one another? What other motions are possible? The Hamiltonian formulation of dynamics provides a convenient framework in which the possible motions may be placed and understood. We will be able to see the range of stable resonance motions and the range of states reached by chaotic trajectories, and discover other unsuspected possible motions. So the Hamiltonian formulation gives us much more than the stated goal of expressing the system derivative in terms of potentially conserved quantities.

3.1   Hamilton's Equations

The momenta are given by momentum state functions of the time, the coordinates, and the velocities.¹ Locally, we can find inverse functions that give the velocities in terms of the time, the coordinates, and the momenta. We can use this inverse function to represent the state in terms of the coordinates and momenta rather than the coordinates and velocities. The equations of motion when recast in terms of coordinates and momenta are called Hamilton's canonical equations.

We present three derivations of Hamilton's equations. The first derivation is guided by the strategy outlined above and uses nothing more complicated than implicit functions and the chain rule. The second derivation (section 3.1.1) first abstracts a key part of the first derivation and then applies the more abstract machinery to derive Hamilton's equations. The third (section 3.1.2) uses the action principle.

Lagrange's equations give us the time derivative of the momentum p on a path q:

$$\begin{array}{ll}Dp(t)={\partial }_{1}L(t,q(t),Dq(t)),\hfill & (3.1)\hfill \end{array}$$

where

$$\begin{array}{ll}p(t)={\partial }_{2}L(t,q(t),Dq(t)).\hfill & (3.2)\hfill \end{array}$$

To eliminate Dq we need to solve equation (3.2) for Dq in terms of p.

Let $\mathcal{V}$ be the function that gives the velocities in terms of the time, coordinates, and momenta. Defining $\mathcal{V}$ is a problem of functional inverses. To prevent confusion we use names for the variables that have no mnemonic significance. Let

$$\begin{array}{ll}a={\partial }_{2}L(b,c,d);\hfill & (3.3)\hfill \end{array}$$

then $\mathcal{V}$ satisfies

$$\begin{array}{ll}d=\mathcal{V}(b,c,a).\hfill & (3.4)\hfill \end{array}$$

So $\mathcal{V}$ and ∂₂L are inverses on the third argument position:

$$\begin{array}{ll}d=\mathcal{V}(b,c,{\partial }_{2}L(b,c,d))\hfill & (3.5)\hfill \end{array}$$

$$\begin{array}{ll}a={\partial }_{2}L(b,c,\mathcal{V}(b,c,a)).\hfill & (3.6)\hfill \end{array}$$

The Lagrange equation (3.1) can be rewritten in terms of p using $\mathcal{V}$:

$$\begin{array}{ll}Dp(t)={\partial }_{1}L(t,q(t),\mathcal{V}(t,q(t),p(t))).\hfill & (3.7)\hfill \end{array}$$

We can also use $\mathcal{V}$ to rewrite equation (3.2) as an equation for Dq in terms of t, q and p:

$$\begin{array}{ll}Dq(t)=\mathcal{V}(t,q(t),p(t)).\hfill & (3.8)\hfill \end{array}$$

Equations (3.7) and (3.8) give the rate of change of q and p along realizable paths as functions of t, q, and p along the paths.

Though these equations fulfill our goal of expressing the equations of motion entirely in terms of coordinates and momenta, we can find a better representation. Define the function

$$\begin{array}{ll}\tilde{L}(t,q,p)=L(t,q,\mathcal{V}(t,q,p)),\hfill & (3.9)\hfill \end{array}$$

which is the Lagrangian reexpressed as a function of time, coordinates, and momenta.² For the equations of motion we need ∂₁L evaluated with the appropriate arguments. Consider

$$\begin{array}{lll}{\partial }_1\tilde{L}(t,q,p)\hfill & ={\partial }_{1}L(t,q,\mathcal{V}(t,q,p))+{\partial }_{2}L(t,q,\mathcal{V}(t,q,p)){\partial }_1\mathcal{V}(t,q,p)\hfill & \hfill \\ \hfill & ={\partial }_{1}L(t,q,\mathcal{V}(t,q,p))+p{\partial }_1\mathcal{V}(t,q,p),\hfill & (3.10)\hfill \end{array}$$

where we used the chain rule in the first step and the inverse property (3.6) of $\mathcal{V}$ in the second step. Introducing the momentum selector³ P (t, q, p) = p, and using the property ∂₁P = 0, we have

$$\begin{array}{lll}{\partial }_{1}L(t,q,\mathcal{V}(t,q,p))\hfill & ={\partial }_1\tilde{L}(t,q,p)-P(t,q,p){\partial }_1\mathcal{V}(t,q,p)\hfill & \hfill \\ \hfill & ={\partial }_1(\tilde{L}-P\mathcal{V})(t,q,p)\hfill & \hfill \\ \hfill & =-{\partial }_{1}H(t,q,p),\hfill & (3.11)\hfill \end{array}$$

where the Hamiltonian H is defined by⁴

$$\begin{array}{ll}H=P\mathcal{V}-\tilde{L}.\hfill & (3.12)\hfill \end{array}$$

Using the algebraic result (3.11), the Lagrange equation (3.7) for Dp becomes

$$\begin{array}{ll}Dp(t)=-{\partial }_{1}H(t,q(t),p(t)).\hfill & (3.13)\hfill \end{array}$$

The equation for Dq can also be written in terms of H. Consider

$$\begin{array}{lll}{\partial }_{2}H(t,q,p)\hfill & ={\partial }_2(P\mathcal{V}-\tilde{L})(t,q,p)\hfill & \hfill \\ \hfill & =\mathcal{V}(t,q,p)+p{\partial }_2\mathcal{V}(t,q,p)-{\partial }_2\tilde{L}(t,q,p).\hfill & (3.14)\hfill \end{array}$$

To carry out the derivative of $\tilde{L}$ we write it out in terms of L:

$$\begin{array}{ll}{\partial }_2\tilde{L}(t,q,p)={\partial }_{2}L(t,q,\mathcal{V}(t,q,p)){\partial }_2\mathcal{V}(t,q,p)=p{\partial }_2\mathcal{V}(t,q,p),\hfill & (3.15)\hfill \end{array}$$

again using the inverse property (3.6) of $\mathcal{V}$. So, putting equations (3.14) and (3.15) together, we obtain

$$\begin{array}{ll}{\partial }_{2}H(t,q,p)=\mathcal{V}(t,q,p).\hfill & (3.16)\hfill \end{array}$$

Using the algebraic result (3.16), equation (3.8) for Dq becomes

$$\begin{array}{ll}Dq(t)={\partial }_{2}H(t,q(t),p(t)).\hfill & (3.17)\hfill \end{array}$$

Equations (3.13) and (3.17) give the derivatives of the coordinate and momentum path functions at each time in terms of the time, and the coordinates and momenta at that time. These equations are known as Hamilton's equations:⁵

$$\begin{array}{ll}Dq(t)={\partial }_{2}H(t,q(t),p(t))\hfill & \hfill \\ Dp(t)=-{\partial }_{1}H(t,q(t),p(t)).\hfill & (3.18)\hfill \end{array}$$

The first equation is just a restatement of the relationship of the momenta to the velocities in terms of the Hamiltonian and holds for any path, whether or not it is a realizable path. The second equation holds only for realizable paths.

Hamilton's equations have an especially simple and symmetrical form. Just as Lagrange's equations are constructed from a real-valued function, the Lagrangian, Hamilton's equations are constructed from a real-valued function, the Hamiltonian. The Hamiltonian function is⁶

$$\begin{array}{ll}H(t,q,p)=p\mathcal{V}(t,q,p)-L(t,q,\mathcal{V}(t,q,p)).\hfill & (3.19)\hfill \end{array}$$

The Hamiltonian has the same value as the energy function ℰ (see equation 1.142), except that the velocities are expressed in terms of time, coordinates, and momenta by $\mathcal{V}$:

$$\begin{array}{ll}H(t,q,p)=ℰ(t,q,\mathcal{V}(t,q,p)).\hfill & (3.20)\hfill \end{array}$$

Illustration

Let's try something simple: the motion of a particle of mass m with potential energy V (x, y). A Lagrangian is

$$\begin{array}{ll}L(t;x,y;v_x,v_y)=\frac{1}{2}m(v_x^2+v_y^2)-V(x,y).\hfill & (3.21)\hfill \end{array}$$

To form the Hamiltonian we find the momenta p = ∂₂L(t, q, v): p_x = mv_x and p_y = mv_y. Solving for the velocities in terms of the momenta is easy here: ${\upsilon }_x=p_x/m$ and ${\upsilon }_y=p_y/m$. The Hamiltonian is H(t, q, p) = pv − L(t, q, v), with v reexpressed in terms of (t, q, p):

$$\begin{array}{ll}H(t;x,y;p_x,p_y)=\frac{p_x^2+p_y^2}{2m}+V(x,y).\hfill & (3.22)\hfill \end{array}$$

The kinetic energy is a homogeneous quadratic form in the velocities, so the energy is T + V and the Hamiltonian is the energy expressed in terms of momenta rather than velocities. Hamilton's equations for Dq are

$$\begin{array}{ll}{D}_x(t)=p_x(t)/m\hfill & \hfill \\ D_y(t)=p_y(t)/m.\hfill & (3.23)\hfill \end{array}$$

Note that these equations merely restate the relation between the momenta and the velocities. Hamilton's equations for Dp are

$$\begin{array}{ll}D{p}_x(t)=-{\partial }_{0}V(x(t),y(t))\hfill & \hfill \\ D{p}_y(t)=-{\partial }_{1}V(x(t),y(t)).\hfill & (3.24)\hfill \end{array}$$

The rate of change of the linear momentum is minus the gradient of the potential energy.

Exercise 3.1: Deriving Hamilton's equations

For each of the following Lagrangians derive the Hamiltonian and Hamilton's equations. These problems are simple enough to do by hand.

a. A Lagrangian for a planar pendulum: $L(t,\theta ,\dot{\theta })=\frac{1}{2}m{l}^2{\dot{\theta }}^2+mgl\cos \theta$.

b. A Lagrangian for a particle of mass m with a two-dimensional potential energy V(x, y) = (x² + y²)/2 + x²y − y³/3 is $L(t;x,y;\dot{x},\dot{y})=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)-V(x,y).$

c. A Lagrangian for a particle of mass m constrained to move on a sphere of radius $R:L(t;\theta ,\phi ;\dot{\theta },\dot{\phi })=\frac{1}{2}m{R}^2({\dot{\theta }}^2+{(\dot{\phi }\sin \theta )}^2)$, where θ is the colatitude and φ is the longitude on the sphere.

Exercise 3.2: Sliding pendulum

For the pendulum with a sliding support (see exercise 1.20), derive a Hamiltonian and Hamilton's equations.

Hamiltonian state

Given a coordinate path q and a Lagrangian L, the corresponding momentum path p is given by equation (3.2). Equation (3.17) expresses the same relationship in terms of the corresponding Hamiltonian H. That these relations are valid for any path, whether or not it is a realizable path, allows us to abstract to arbitrary velocity and momentum at a moment. At a moment, the momentum p for the state tuple (t, q, v) is p = ∂₂L(t, q, v). We also have v = ∂₂H(t, q, p). In the Lagrangian formulation the state of the system at a moment can be specified by the local state tuple (t, q, v) of time, generalized coordinates, and generalized velocities. Lagrange's equations determine a unique path emanating from this state. In the Hamiltonian formulation the state can be specified by the tuple (t, q, p) of time, generalized coordinates, and generalized momenta. Hamilton's equations determine a unique path emanating from this state. The Lagrangian state tuple (t, q, v) encodes exactly the same information as the Hamiltonian state tuple (t, q, p); we need a Lagrangian or a Hamiltonian to relate them. The two formulations are equivalent in that the same coordinate path emanates from them for equivalent initial states.

The Lagrangian state derivative is constructed from the Lagrange equations by solving for the highest-order derivative and abstracting to arbitrary positions and velocities at a moment.⁷ The Lagrangian state path is generated by integration of the Lagrangian state derivative given an initial Lagrangian state (t, q, v). Similarly, the Hamiltonian state derivative can be constructed from Hamilton's equations by abstracting to arbitrary positions and momenta at a moment. Hamilton's equations are a set of first-order differential equations in explicit form. The Hamiltonian state derivative can be directly written in terms of them. The Hamiltonian state path is generated by integration of the Hamiltonian state derivative given an initial Hamiltonian state (t, q, p). If these state paths are obtained by integrating the state derivatives with equivalent initial states, then the coordinate path components of these state paths are the same and satisfy the Lagrange equations. The coordinate path and the momentum path components of the Hamiltonian state path satisfy Hamilton's equations. The Hamiltonian formulation and the Lagrangian formulation are equivalent.

Given a path q, the Lagrangian state path and the Hamiltonian state paths can be deduced from it. The Lagrangian state path Γ[q] can be constructed from a path q simply by taking derivatives. The Lagrangian state path satisfies:

$$\begin{array}{ll}\Gamma [q](t)=(t,q(t),Dq(t)).\hfill & (3.25)\hfill \end{array}$$

The Lagrangian state path is uniquely determined by the path q. The Hamiltonian state path Π_L[q] can also be constructed from the path q but the construction requires a Lagrangian. The Hamiltonian state path satisfies

$$\begin{array}{ll}{\Pi }_L[q](t)=(t,q(t),{\partial }_{2}L(t,q(t),Dq(t)))=(t,q(t),p(t)).\hfill & (3.26)\hfill \end{array}$$

The Hamiltonian state tuple is not uniquely determined by the path q because it depends upon our choice of Lagrangian, which is not unique.

The 2n-dimensional space whose elements are labeled by the n generalized coordinates qⁱ and the n generalized momenta p_i is called the phase space. The components of the generalized coordinates and momenta are collectively called the phase-space components.⁸ The dynamical state of the system is completely specified by the phase-space state tuple (t, q, p), given a Lagrangian or Hamiltonian to provide the map between velocities and momenta.

Computing Hamilton's equations

Hamilton's equations are a system of first-order ordinary differential equations. A procedural formulation of Lagrange's equations as a first-order system was presented in section 1.7. The following formulation of Hamilton's equations is analogous:

(define ((Hamilton-equations Hamiltonian) q p)
  (let ((state-path (qp->H-state-path q p)))
    (- (D state-path)
       (compose (Hamiltonian->state-derivative Hamiltonian)
                state-path))))

The Hamiltonian state derivative is computed as follows:

(define ((Hamiltonian->state-derivative Hamiltonian) H-state)
  (up 1
      (((partial 2) Hamiltonian) H-state)
      (- (((partial 1) Hamiltonian) H-state))))

The state in the Hamiltonian formulation is composed of the time, the coordinates, and the momenta. We call this an H-state, to distinguish it from the state in the Lagrangian formulation. We can select the components of the Hamiltonian state with the selectors time, coordinate, momentum. We construct Hamiltonian states from their components with up. The first component of the state is time, so the first component of the state derivative is one, the time rate of change of time. Given procedures q and p implementing coordinate and momentum path functions, the Hamiltonian state path can be constructed with the following procedure:

(define ((qp->H-state-path q p) t)
  (up t (q t) (p t)))

The Hamilton-equations procedure returns the residuals of Hamilton's equations for the given paths.

For example, a procedure implementing the Hamiltonian for a point mass with potential energy V (x, y) is

(define ((H-rectangular m V) state)
  (let ((q (coordinate state))
        (p (momentum state)))
    (+ (/ (square p) (* 2 m))
       (V (ref q 0) (ref q 1)))))

Hamilton's equations are

(show-expression
  (let ((V (literal-function 'V (-> (X Real Real) Real)))
        (q (up (literal-function 'x)
               (literal-function 'y)))
        (p (down (literal-function 'p_x)
                 (literal-function 'p_y))))
    (((Hamilton-equations (H-rectangular 'm V)) q p) 't)))

The zero in the first element of the structure of Hamilton's equation residuals is just the tautology that time advances uniformly: the time function is just the identity, so its derivative is one and the residual is zero. The equations in the second element of the structure relate the coordinate paths and the momentum paths. The equations in the third element give the rate of change of the momenta in terms of the applied forces.

Exercise 3.3: Computing Hamilton's equations

Check your answers to exercise 3.1 with the Hamilton-equations procedure.

3.1.1 The Legendre Transformation

The Legendre transformation abstracts a key part of the process of transforming from the Lagrangian to the Hamiltonian formulation of mechanics—the replacement of functional dependence on generalized velocities with functional dependence on generalized momenta. The momentum state function is defined as a partial derivative of the Lagrangian, a real-valued function of time, coordinates, and velocities. The Legendre transformation provides an inverse that gives the velocities in terms of the momenta: we are able to write the velocities as a partial derivative of a different real-valued function of time, coordinates, and momenta.⁹

Given a real-valued function F, if we can find a real-valued function G such that DF = (DG)⁻¹, then we say that F and G are related by a Legendre transform.

Locally, we can define the inverse function¹⁰ $\mathcal{V}$ of DF so that $DF\circ \mathcal{V}=I$, where I is the identity function I(w) = w. Consider the composite function $\tilde{F}=F\circ \mathcal{V}$. The derivative of $\tilde{F}$ is

$$\begin{array}{ll}D\tilde{F}=(DF\circ \mathcal{V})D\mathcal{V}=ID\mathcal{V}.\hfill & (3.27)\hfill \end{array}$$

Since

$$\begin{array}{ll}D(I\mathcal{V})=\mathcal{V}+ID\mathcal{V},\hfill & (3.28)\hfill \end{array}$$

we have

$$\begin{array}{ll}D\tilde{F}=D(I\mathcal{V})-\mathcal{V},\hfill & (3.29)\hfill \end{array}$$

or

$$\begin{array}{ll}\mathcal{V}=D(I\mathcal{V})-D\tilde{F}=D(I\mathcal{V}-\tilde{F}).\hfill & (3.30)\hfill \end{array}$$

The integral is determined up to a constant of integration. If we define

$$\begin{array}{ll}G=I\mathcal{V}-\tilde{F},\hfill & (3.31)\hfill \end{array}$$

then we have

$$\begin{array}{ll}\mathcal{V}=DG.\hfill & (3.32)\hfill \end{array}$$

The function G has the desired property that DG is the inverse function $\mathcal{V}$ of DF. The derivation just given applies equally well if the arguments of F and G have multiple components.¹¹

Given a relation w = DF (v) for some given function F, then v = DG(w) for $G=I\mathcal{V}-F\circ \mathcal{V}$, where $\mathcal{V}$ is the inverse function of DF, provided it exists.

A picture may help (see figure 3.1). The curve is the graph of the function DF. Turned sideways, it is also the graph of the function DG, because DG is the inverse function of DF. The integral of DF from v₀ to v is F(v) − F(v₀); this is the area below the curve from v₀ to v. Likewise, the integral of DG from w₀ to w is G(w) − G(w₀); this is the area to the left of the curve from w₀ to w. The union of these two regions has area wv − w₀v₀. So

$$\begin{array}{ll}wv-w_0{v}_0=F(v)-F(v_0)+G(w)-G(w_0),\hfill & (3.33)\hfill \end{array}$$

which is the same as

$$\begin{array}{ll}wv-F(v)-G(w)=w_0{v}_0-G(w_0)-F(v_0).\hfill & (3.34)\hfill \end{array}$$

The left-hand side depends only on the point labeled by w and v and the right-hand side depends only on the point labeled by w₀ and v₀, so these must be constant, independent of the variable endpoints. So as the point is changed the combination G(w) + F(v) − wv is invariant. Thus

$$\begin{array}{ll}G(w)=wv-F(v)+C,\hfill & (3.35)\hfill \end{array}$$

with constant C. The requirement for G depends only on DG so we can choose to define G with C = 0.

Legendre transformations with passive arguments

Let F be a real-valued function of two arguments and

$$\begin{array}{ll}w={\partial }_{1}F(x,v).\hfill & (3.36)\hfill \end{array}$$

If we can find a real-valued function G such that

$$\begin{array}{ll}v={\partial }_{1}G(x,w)\hfill & (3.37)\hfill \end{array}$$

we say that F and G are related by a Legendre transformation, that the second argument in each function is active, and that the first argument is passive in the transformation.

If the function ∂₁F can be locally inverted with respect to the second argument we can define

$$\begin{array}{ll}v=\mathcal{V}(x,w),\hfill & (3.38)\hfill \end{array}$$

giving

$$\begin{array}{ll}w={\partial }_{1}F(x,\mathcal{V}(x,w))=W(x,w),\hfill & (3.39)\hfill \end{array}$$

where W = I₁ is the selector function for the second argument.

For the active arguments the derivation goes through as before. The first argument to F and G is just along for the ride—it is a passive argument. Let

$$\begin{array}{ll}\tilde{F}(x,w)=F(x,\mathcal{V}(x,w)),\hfill & (3.40)\hfill \end{array}$$

then define

$$\begin{array}{ll}G=W\mathcal{V}-\tilde{F}.\hfill & (3.41)\hfill \end{array}$$

We can check that G has the property $\mathcal{V}={\partial }_{1}G$ by carrying out the derivative:

$$\begin{array}{lll}{\partial }_{1}G\hfill & ={\partial }_1(W\mathcal{V}-\tilde{F})\hfill & \hfill \\ \hfill & =\mathcal{V}+W{\partial }_1\mathcal{V}-{\partial }_1\tilde{F},\hfill & (3.42)\hfill \end{array}$$

but

$$\begin{array}{lll}{\partial }_1\tilde{F}(x,w)\hfill & ={\partial }_{1}F(x,\mathcal{V}(x,w)){\partial }_1\mathcal{V}(x,w)\hfill & \hfill \\ \hfill & =W(x,w){\partial }_1\mathcal{V}(x,w),\hfill & (3.43)\hfill \end{array}$$

or

$$\begin{array}{ll}{\partial }_1\tilde{F}=W{\partial }_1\mathcal{V}.\hfill & (3.44)\hfill \end{array}$$

So, from equation (3.42),

$$\begin{array}{ll}{\partial }_{1}G=\mathcal{V},\hfill & (3.45)\hfill \end{array}$$

as required. The active argument may have many components.

The partial derivatives with respect to the passive arguments are related in a remarkably simple way. Let's calculate the derivative ∂₀G in pieces. First,

$$\begin{array}{ll}{\partial }_0(W\mathcal{V})=W{\partial }_0\mathcal{V}\hfill & (3.46)\hfill \end{array}$$

because ∂₀W = 0. We calculate ${\partial }_0\tilde{F}$:

$$\begin{array}{lll}{\partial }_0\tilde{F}(x,w)\hfill & ={\partial }_{0}F(x,\mathcal{V}(x,w))+{\partial }_{1}F(x,\mathcal{V}(x,w)){\partial }_0\mathcal{V}(x,w)\hfill & \hfill \\ \hfill & ={\partial }_{0}F(x,\mathcal{V}(x,w))+W(x,w){\partial }_0\mathcal{V}(x,w).\hfill & (3.47)\hfill \end{array}$$

Putting these together, we find

$$\begin{array}{ll}{\partial }_{0}G(x,w)=-{\partial }_{0}F(x,\mathcal{V}(x,w))=-{\partial }_{0}F(x,v).\hfill & (3.48)\hfill \end{array}$$

The calculation is unchanged if the passive argument has many components.

We can write the Legendre transformation more symmetrically:

$$\begin{array}{llll}\hfill w& =\hfill & {\partial }_{1}F(x,v)\hfill & \hfill \\ \hfill wv& =\hfill & F(x,v)+G(x,w)\hfill & \hfill \\ \hfill v& =\hfill & {\partial }_{1}G(x,w)\hfill & \hfill \\ \hfill 0& =\hfill & {\partial }_{0}F(x,v)+{\partial }_{0}G(x,w).\hfill & (3.49)\hfill \end{array}$$

The last relation is not as trivial as it looks, because x enters the equations connecting w and v. With this symmetrical form, we see that the Legendre transform is its own inverse.

Exercise 3.4: Simple Legendre transforms

For each of the following functions, find the function that is related to the given function by the Legendre transform on the indicated active argument. Show that the Legendre transform relations hold for your solution, including the relations among passive arguments, if any.

a. F (x) = ax + bx², with no passive arguments.

b. F (x, y) = a sin x cos y, with x active.

c. F (x, y, ẋ, ẏ) = xẋ² + 3ẋẏ + yẏ², with ẋ and ẏ active.

Hamilton's equations from the Legendre transformation

We can use the Legendre transformation with the Lagrangian playing the role of F and with the generalized velocity slot playing the role of the active argument. The Hamiltonian plays the role of G with the momentum slot active. The coordinate and time slots are passive arguments.

The Lagrangian L and the Hamiltonian H are related by a Legendre transformation:

$$\begin{array}{ll}e=({\partial }_{2}L)(a,b,c)\hfill & (3.50)\hfill \end{array}$$

$$\begin{array}{ll}ec=L(a,b,c)+H(a,b,e)\hfill & (3.51)\hfill \end{array}$$

and

$$\begin{array}{ll}c=({\partial }_{2}H)(a,b,e),\hfill & (3.52)\hfill \end{array}$$

with passive equations

$$\begin{array}{ll}0={\partial }_{0}L(a,b,c)+{\partial }_{0}H(a,b,e),\hfill & (3.53)\hfill \end{array}$$

$$\begin{array}{ll}0={\partial }_{1}L(a,b,c)+{\partial }_{1}H(a,b,e).\hfill & (3.54)\hfill \end{array}$$

Presuming it exists, we can define the inverse of ∂₂L with respect to the last argument:

$$\begin{array}{ll}c=\mathcal{V}(a,b,e),\hfill & (3.55)\hfill \end{array}$$

and write the Hamiltonian

$$\begin{array}{ll}H(a,b,c)=c\mathcal{V}(a,b,c)-L(a,b,\mathcal{V}(a,b,c)).\hfill & (3.56)\hfill \end{array}$$

These relations are purely algebraic in nature.

On a path q we have the momentum p:

$$\begin{array}{ll}p(t)={\partial }_{2}L(t,q(t),Dq(t)),\hfill & (3.57)\hfill \end{array}$$

and from the definition of $\mathcal{V}$ we find

$$\begin{array}{ll}Dq(t)=\mathcal{V}(t,q(t),p(t)).\hfill & (3.58)\hfill \end{array}$$

The Legendre transform gives

$$\begin{array}{ll}Dq(t)={\partial }_{2}H(t,q(t),p(t)).\hfill & (3.59)\hfill \end{array}$$

This relation is purely algebraic and is valid for any path. The passive equation (3.54) gives

$$\begin{array}{ll}{\partial }_{1}L(t,q(t),Dq(t))=-{\partial }_{1}H(t,q(t),p(t)),\hfill & (3.60)\hfill \end{array}$$

but the left-hand side can be rewritten using the Lagrange equations, so

$$\begin{array}{ll}Dp(t)=-{\partial }_{1}H(t,q(t),p(t)).\hfill & (3.61)\hfill \end{array}$$

This equation is valid only for realizable paths, because we used the Lagrange equations to derive it. Equations (3.59) and (3.61) are Hamilton's equations.

The remaining passive equation is

$$\begin{array}{ll}{\partial }_{0}L(t,q(t),Dq(t))=-{\partial }_{0}H(t,q(t),p(t)).\hfill & (3.62)\hfill \end{array}$$

This passive equation says that the Lagrangian has no explicit time dependence (∂₀L = 0) if and only if the Hamiltonian has no explicit time dependence (∂₀H = 0). We have found that if the Lagrangian has no explicit time dependence, then energy is conserved. So if the Hamiltonian has no explicit time dependence then it is a conserved quantity.

Exercise 3.5: Conservation of the Hamiltonian

Using Hamilton's equations, show directly that the Hamiltonian is a conserved quantity if it has no explicit time dependence.

Legendre transforms of quadratic functions

We cannot implement the Legendre transform in general because it involves finding the functional inverse of an arbitrary function. However, many physical systems can be described by Lagrangians that are quadratic forms in the generalized velocities. For such functions the generalized momenta are linear functions of the generalized velocities, and thus explicitly invertible.

More generally, we can compute a Legendre transformation for polynomial functions where the leading term is a quadratic form:

$$\begin{array}{ll}F(v)=\frac{1}{2}{v}^{\mathcal{T}}Mv+bv+c.\hfill & (3.63)\hfill \end{array}$$

Because the first term is a quadratic form only the symmetric part of M contributes to the result, so we can assume M is symmetric.¹² Let w = DF (v), then

$$\begin{array}{ll}w=DF(v)=Mv+b.\hfill & (3.64)\hfill \end{array}$$

So if M is invertible we can solve for v in terms of w. Thus we may define a function $\mathcal{V}$ such that

$$\begin{array}{ll}v=\mathcal{V}(w)=M^{-1}(w-b)\hfill & (3.65)\hfill \end{array}$$

and we can use this to compute the value of the function G:

$$\begin{array}{ll}G(w)=w\mathcal{V}(w)-F(\mathcal{V}(w)).\hfill & (3.66)\hfill \end{array}$$

Computing Hamiltonians

We implement the Legendre transform for quadratic functions by the procedure¹³

(define (Legendre-transform F)
  (let ((w-of-v (D F)))
    (define (G w)
      (let ((zero (compatible-zero w)))
        (let ((M ((D w-of-v) zero))
              (b (w-of-v zero)))
          (let ((v (solve-linear-left M (- w b))))
            (- (* w v) (F v))))))
    G))

The procedure Legendre-transform takes a procedure of one argument and returns the procedure that is associated with it by the Legendre transform. If w = DF (v), wv = F (v) + G(w), and v = DG(w) specifies a one-argument Legendre transformation, then G is the function associated with F by the Legendre transform: $G=I\mathcal{V}-F\circ \mathcal{V}$, where $\mathcal{V}$ is the functional inverse of DF.

We can use the Legendre-transform procedure to compute a Hamiltonian from a Lagrangian:

(define ((Lagrangian->Hamiltonian Lagrangian) H-state)
  (let ((t (time H-state))
        (q (coordinate H-state))
        (p (momentum H-state)))
    (define (L qdot)
      (Lagrangian (up t q qdot)))
    ((Legendre-transform L) p)))

Notice that the one-argument Legendre-transform procedure is sufficient. The passive variables are given no special attention, they are just passed around.

The Lagrangian may be obtained from the Hamiltonian by the procedure:

(define ((Hamiltonian->Lagrangian Hamiltonian) L-state)
  (let ((t (time L-state))
        (q (coordinate L-state))
        (qdot (velocity L-state)))
    (define (H p)
      (Hamiltonian (up t q p)))
    ((Legendre-transform H) qdot)))

Notice that the two procedures Hamiltonian->Lagrangian and Lagrangian->Hamiltonian are identical, except for the names.

For example, the Hamiltonian for the motion of the point mass with the potential energy V (x, y) may be computed from the Lagrangian:

(define ((L-rectangular m V) local)
  (let ((q (coordinate local))
        (qdot (velocity local)))
    (- (* 1/2 m (square qdot))
       (V (ref q 0) (ref q 1)))))

And the Hamiltonian is, as we saw in equation (3.22):

(show-expression
  ((Lagrangian->Hamiltonian
     (L-rectangular
       'm
       (literal-function 'V (-> (X Real Real) Real))))
   (up 't (up 'x 'y) (down 'p_x 'p_y))))

Exercise 3.6: On a helical track

A uniform cylinder of mass M, radius R, and height h is mounted so as to rotate freely on a vertical axis. A point mass of mass m is constrained to move on a uniform frictionless helical track of pitch β (measured in radians per meter of drop along the cylinder) mounted on the surface of the cylinder (see figure 3.2). The mass is acted upon by standard gravity (g = 9.8 ms⁻²).

a. What are the degrees of freedom of this system? Pick and describe a convenient set of generalized coordinates for this problem. Write a Lagrangian to describe the dynamical behavior. It may help to know that the moment of inertia of a cylinder around its axis is $\frac{1}{2}M{R}^2$. You may find it easier to do the algebra if various constants are combined and represented as single symbols.

b. Make a Hamiltonian for the system. Write Hamilton's equations for the system. Are there any conserved quantities?

c. If we release the point mass at time t = 0 at the top of the track with zero initial speed and let it slide down, what is the motion of the system?

Exercise 3.7: An ellipsoidal bowl

Consider a point particle of mass m constrained to move in a bowl and acted upon by a uniform gravitational acceleration g. The bowl is ellipsoidal, with height z = ax² + by². Make a Hamiltonian for this system. Can you make any immediate deductions about this system?

3.1.2 Hamilton's Equations from the Action Principle

The previous two derivations of Hamilton's equations made use of the Lagrange equations. Hamilton's equations can also be derived directly from the action principle.

The action is the integral of the Lagrangian along a path:

$$\begin{array}{ll}S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}L\circ \Gamma [q].}\hfill & (3.67)\hfill \end{array}$$

The action is stationary with respect to variations of a realizable path that preserve the configuration at the endpoints (for Lagrangians that are functions of time, coordinates, and velocities).

We can rewrite the integrand in terms of the Hamiltonian

$$\begin{array}{ll}L(t,q(t),p(t))=p(t)Dq(t)-H(t,q(t),p(t)),\hfill & (3.68)\hfill \end{array}$$

with p(t) = ∂₂L(t, q(t), Dq(t)). The Legendre transformation construction gives

$$\begin{array}{ll}Dq(t)={\partial }_{2}H(t,q(t),p(t)),\hfill & (3.69)\hfill \end{array}$$

which is one of Hamilton's equations, the one that does not depend on the path being a realizable path.

In order to vary the action we should make the dependences on the path explicit. We introduce

$$\begin{array}{ll}\tilde{p}[q](t)={\partial }_{2}L(t,q(t),Dq(t)),\hfill & (3.70)\hfill \end{array}$$

and¹⁴

$$\begin{array}{cc}\Pi [q](t)=(t,q(t),\tilde{p}[q](t))=(t,q(t),p(t)).& (3.71)\end{array}$$

The integrand of the action integral is then

$$\begin{array}{cc}L\circ \Gamma [q]=\tilde{p}[q]Dq-H\circ \Pi [q].& (3.72)\end{array}$$

Using the shorthand δp for $\delta \tilde{p}[q]$,¹⁵ and noting that $p=\tilde{p}[q]$, the variation of the action is

$$\begin{array}{lll}\delta S[q]\hfill & (t_1,t_2)\hfill & \hfill \\ \hfill & ={\displaystyle {\int }_{t_1}^{t_2}(\delta pDq+p\delta Dq-(DH\circ \Pi [q])\delta \Pi [q])}\hfill & \hfill \\ \hfill & \begin{array}{l}={\displaystyle {\int }_{t_1}^{t_2}\{\delta pDq+pD\delta q}\\ -({\partial }_{1}H\circ \Pi [q])\delta q-({\partial }_{2}H\circ \Pi [q])\delta p\}.\end{array}\hfill & (3.73)\hfill \end{array}$$

Integrating the second term by parts, using D(pδq) = Dpδq + pDδq, we get

$$\begin{array}{lll}\delta S[q]\hfill & (t_1,t_2)=p\delta q{|}_{t_1}^{t_2}\hfill & \hfill \\ \hfill & +{\displaystyle {\int }_{t_1}^{t_2}\{\delta pDq-Dp\delta q}\hfill & \hfill \\ \hfill & -({\partial }_{1}H\circ \Pi [q])\delta q-({\partial }_{2}H\circ \Pi [q])\delta p\}.\hfill & (3.74)\hfill \end{array}$$

The variations are constrained so that δq(t₁) = δq(t₂) = 0, so the integrated part vanishes. Rearranging terms, the variation of the action is

$$\begin{array}{ll}\delta S[q](t_1,t_2)\hfill & \hfill \\ ={\displaystyle {\int }_{t_1}^{t_2}((Dq-{\partial }_{2}H\circ \Pi [q])\delta p-(Dp+{\partial }_{1}H\circ \Pi [q])\delta q).}\hfill & (3.75)\hfill \end{array}$$

As a consequence of equation (3.69), the factor multiplying δp is zero. We are left with

$$\begin{array}{cc}\delta S[q](t_1,t_2)=-{\displaystyle {\int }_{t_1}^{t_2}(Dp+{\partial }_{1}H\circ \Pi [q])\delta q.}& (3.76)\end{array}$$

For the variation of the action to be zero for arbitrary variations, except for the endpoint conditions, we must have

$$\begin{array}{cc}Dp=-{\partial }_{1}H\circ \Pi [q],& (3.77)\end{array}$$

or

$$\begin{array}{cc}Dp=-{\partial }_{1}H(t,q(t),p(t)),& (3.78)\end{array}$$

which is the “dynamical” Hamilton equation.¹⁶

3.1.3 A Wiring Diagram

Figure 3.3 shows a summary of the functional relationship between the Lagrangian and the Hamiltonian descriptions of a dynamical system. The diagram shows a “circuit” interconnecting some “devices” with “wires.” The devices represent the mathematical functions that relate the quantities on their terminals. The wires represent identifications of the quantities on the terminals that they connect. For example, there is a box that represents the Lagrangian function. Given values t, q, and $\dot{q}$, the value of the Lagrangian $L\left(t,q,\dot{q}\right)$ is on the terminal labeled L, which is wired to an addend terminal of an adder. Other terminals of the Lagrangian carry the values of the partial derivatives of the Lagrangian function.

The upper part of the diagram summarizes the relationship of the Hamiltonian to the Lagrangian. For example, the sum of the values on the terminals L of the Lagrangian and H of the Hamiltonian is the product of the value on the $\dot{q}$ terminal of the Lagrangian and the value on the p terminal of the Hamiltonian. This is the active part of the Legendre transform. The passive variables are related by the corresponding partial derivatives being negations of each other. In the lower part of the diagram the equations of motion are indicated by the presence of the integrators, relating the dynamical quantities to their time derivatives.

One can use this diagram to help understand the underlying unity of the Lagrangian and Hamiltonian formulations of mechanics. Lagrange's equations are just the connection of the ṗ wire to the ∂₁L terminal of the Lagrangian device. One of Hamilton's equations is just the connection of the ṗ wire (through the negation device) to the ∂₁H terminal of the Hamiltonian device. The other is just the connection of the $\dot{q}$ wire to the ∂₂H terminal of the Hamiltonian device. We see that the two formulations are consistent. One does not have to abandon any part of the Lagrangian formulation to use the Hamiltonian formulation: there are deductions that can be made using both simultaneously.

3.2   Poisson Brackets

Here we introduce the Poisson bracket, in terms of which Hamilton's equations have an elegant and symmetric expression. Consider a function F of time, coordinates, and momenta. The value of F along the path σ(t) = (t, q(t), p(t)) is (F ∘ σ)(t) = F (t, q(t), p(t)). The time derivative of F ∘ σ is

$$\begin{array}{lll}D(F\circ \sigma )\hfill & =(DF\circ \sigma )D\sigma \hfill & \hfill \\ \hfill & ={\partial }_{0}F\circ \sigma +({\partial }_{1}F\circ \sigma )Dq+({\partial }_{2}F\circ \sigma )Dp.\hfill & (3.79)\hfill \\ \hfill & \hfill & \hfill \end{array}$$

If the phase-space path is a realizable path for a system with Hamiltonian H, then Dq and Dp can be reexpressed using Hamilton's equations:

$$\begin{array}{lll}D(F\circ \sigma )\hfill & ={\partial }_{0}F\circ \sigma +({\partial }_{1}F\circ \sigma )({\partial }_{2}H\circ \sigma )-({\partial }_{2}F\circ \sigma )({\partial }_{1}H\circ \sigma )\hfill & \hfill \\ \hfill & ={\partial }_{0}F\circ \sigma +({\partial }_{1}F{\partial }_{2}H-{\partial }_{2}F{\partial }_{1}H)\circ \sigma \hfill & \hfill \\ \hfill & ={\partial }_{0}F\circ \sigma +\{F,H\}\circ \sigma \hfill & (3.80)\hfill \end{array}$$

where the Poisson bracket {F, H} of F and H is defined by¹⁷

$$\begin{array}{ll}\{F,H\}={\partial }_{1}F{\partial }_{2}H-{\partial }_{2}F{\partial }_{1}H.\hfill & (3.81)\hfill \end{array}$$

Note that the Poisson bracket of two functions on the phase-state space is also a function on the phase-state space.

The coordinate selector Q = I₁ is an example of a function on phase-state space: Q(t, q, p) = q. According to equation (3.80),

$$\begin{array}{ll}Dq=D(Q\circ \sigma )=\{Q,H\}\circ \sigma ={\partial }_{2}H\circ \sigma ,\hfill & (3.82)\hfill \end{array}$$

but this is the same as Hamilton's equation

$$\begin{array}{ll}Dq(t)={\partial }_{2}H(t,q(t),p(t)).\hfill & (3.83)\hfill \end{array}$$

Similarly, the momentum selector P = I₂ is a function on phase-state space: P (t, q, p) = p. We have

$$\begin{array}{ll}Dp=D(P\circ \sigma )=\{P,H\}\circ \sigma =-{\partial }_{1}H\circ \sigma ,\hfill & (3.84)\hfill \end{array}$$

which is the same as Hamilton's other equation

$$\begin{array}{ll}Dp(t)=-{\partial }_{1}H(t,q(t),p(t)).\hfill & (3.85)\hfill \end{array}$$

So the Poisson bracket provides a uniform way of writing Hamilton's equations:

$$\begin{array}{ll}D(Q\circ \sigma )=\{Q,H\}\circ \sigma \hfill & \hfill \\ D(P\circ \sigma )=\{P,H\}\circ \sigma .\hfill & (3.86)\hfill \end{array}$$

The Poisson bracket of any function with itself is zero, so we recover the conservation of energy for a system that has no explicit time dependence:

$$\begin{array}{cc}DE=D(H\circ \sigma )=({\partial }_{0}H+\{H,H\})\circ \sigma ={\partial }_{0}H\circ \sigma .& (3.87)\end{array}$$

Properties of the Poisson bracket

Let F, G, and H be functions of time, position, and momentum, and let c be independent of position and momentum.

The Poisson bracket is antisymmetric:

$$\begin{array}{cc}\{F,G\}=-\{G,F\}.& (3.88)\end{array}$$

It is bilinear (linear in each argument):

$$\begin{array}{lll}\hfill \left\{F,G+H\right\}& =\left\{F,G\right\}+\left\{F,H\right\}\hfill & \left(3.89\right)\hfill \end{array}$$

$$\begin{array}{lll}\hfill \left\{F,cG\right\}& =c\left\{F,G\right\}\hfill & \left(3.90\right)\hfill \end{array}$$

$$\begin{array}{lll}\hfill \left\{F+G,H\right\}& =\left\{F,H\right\}+\left\{G,H\right\}\hfill & \left(3.91\right)\hfill \end{array}$$

$$\begin{array}{lll}\hfill \left\{cF,G\right\}& =c\left\{F,G\right\}.\hfill & \left(3.92\right)\hfill \end{array}$$

The Poisson bracket satisfies Jacobi's identity:

$$\begin{array}{cc}0=\{F,\{G,H\left\}\right\}+\{H,\{F,G\left\}\right\}+\{G,\{H,F\left\}\right\}.& (3.93)\end{array}$$

All but the last of (3.88–3.93) can immediately be verified from the definition. Jacobi's identity requires a little more effort to verify. We can use the computer to avoid some work. Define some literal phase-space functions of Hamiltonian type:

(define F
  (literal-function 'F
                    (-> (UP Real (UP Real Real) (DOWN Real Real)) Real)))

(define G
  (literal-function 'G
                    (-> (UP Real (UP Real Real) (DOWN Real Real)) Real)))

(define H
  (literal-function 'H
                    (-> (UP Real (UP Real Real) (DOWN Real Real)) Real)))

Then we check the Jacobi identity:

((+ (Poisson-bracket F (Poisson-bracket G H))
    (Poisson-bracket G (Poisson-bracket H F))
    (Poisson-bracket H (Poisson-bracket F G)))
 (up 't (up 'x 'y) (down 'px 'py)))

0

The residual is zero, so the Jacobi identity is satisfied for any three phase-space state functions with two degrees of freedom.

Poisson brackets of conserved quantities

The Poisson bracket of conserved quantities is conserved. Let F and G be time-independent phase-space state functions: ∂₀F = ∂₀G = 0. If F and G are conserved by the evolution under H then

$$\begin{array}{ll}0=D(F\circ \sigma )=\{F,H\}\circ \sigma \hfill & \hfill \\ 0=D(G\circ \sigma )=\{G,H\}\circ \sigma .\hfill & (3.94)\hfill \end{array}$$

So the Poisson brackets of F and G with H are zero: {F, H} = {G, H} = 0. The Jacobi identity then implies

$$\begin{array}{ll}\left\{\right\{F,G\},H\}=0,\hfill & (3.95)\end{array}$$

and thus

$$\begin{array}{ll}D(\{F,G\}\circ \sigma )=0,\hfill & (3.96)\end{array}$$

so {F, G} is a conserved quantity. The Poisson bracket of two conserved quantities is also a conserved quantity.

3.3   One Degree of Freedom

The solutions of time-independent systems with one degree of freedom can be found by quadrature. Such systems conserve the Hamiltonian: the Hamiltonian has a constant value on each realizable trajectory. We can use this constraint to eliminate the momentum in favor of the coordinate, obtaining the single equation Dq(t) = f(q(t)).¹⁸

A geometric view reveals more structure. Time-independent systems with one degree of freedom have a two-dimensional phase space. Energy is conserved, so all orbits are level curves of the Hamiltonian. The possible orbit types are restricted to curves that are contours of a real-valued function. The possible orbits are paths of constant altitude in the mountain range on the phase plane described by the Hamiltonian.

Only a few characteristic features are possible. There are points that are stable equilibria of the dynamical system. These are the peaks and pits of the Hamiltonian mountain range. These equilibria are stable in the sense that neighboring trajectories on nearby contours stay close to the equilibrium point. There are orbits that trace simple closed curves on contours that surround a peak or pit, or perhaps several peaks. There are also trajectories lying on contours that cross at a saddle point. The crossing point is an unstable equilibrium, unstable in the sense that neighboring trajectories leave the vicinity of the equilibrium point. Such contours that cross at saddle points are called separatrices (singular: separatrix), contours that “separate” two regions of distinct behavior.

At every point Hamilton's equations give a unique rate of evolution and direct the system to move perpendicular to the gradient of the Hamiltonian. At the peaks, pits, and saddle points, the gradient of the Hamiltonian is zero, so according to Hamilton's equations these are equilibria. At other points, the gradient of the Hamiltonian is nonzero, so according to Hamilton's equations the rate of evolution is nonzero. Trajectories evolve along the contours of the Hamiltonian. Trajectories on simple closed contours periodically trace the contour. At a saddle point, contours cross. The gradient of the Hamiltonian is zero at the saddle point, so a system started at the saddle point does not leave the saddle point. On the separatrix away from the saddle point the gradient of the Hamiltonian is not zero, so trajectories evolve along the contour. Trajectories on the separatrix are asymptotic forward or backward in time to a saddle point. Going forward or backward in time, such trajectories forever approach an unstable equilibrium but never reach it. If the phase space is bounded, asymptotic trajectories that lie on contours of a smooth Hamiltonian are always asymptotic to unstable equilibria at both ends (but they may be different equilibria).

These orbit types are all illustrated by the prototypical phase plane of the pendulum (see figure 3.4). The solutions lie on contours of the Hamiltonian. There are three regions of the phase plane; in each the motion is qualitatively different. In the central region the pendulum oscillates; above this there is a region in which the pendulum circulates in one direction; below the oscillation region the pendulum circulates in the other direction. In the center of the oscillation region there is a stable equilibrium, at which the pendulum is hanging motionless. At the boundaries between these regions, the pendulum is asymptotic to the unstable equilibrium, at which the pendulum is standing upright.¹⁹ There are two asymptotic trajectories, corresponding to the two ways the equilibrium can be approached. Each of these is also asymptotic to the unstable equilibrium going backward in time.

3.4   Phase Space Reduction

Our motivation for the development of Hamilton's equations was to focus attention on the quantities that can be conserved—the momenta and the energy. In the Hamiltonian formulation the generalized configuration coordinates and the conjugate momenta comprise the state of the system at a given time. We know from the Lagrangian formulation that if the Lagrangian does not depend on some coordinate then the conjugate momentum is conserved. This is also true in the Hamiltonian formulation, but there is a distinct advantage to the Hamiltonian formulation. In the Lagrangian formulation the knowledge of the conserved momentum does not lead immediately to any simplification of the problem, but in the Hamiltonian formulation the fact that momenta are conserved gives an immediate reduction in the dimension of the system to be solved. In fact, if a coordinate does not appear in the Hamiltonian then the dimension of the system of coupled equations that remain to be solved is reduced by two—the coordinate does not appear and the conjugate momentum is constant.

Let H(t, q, p) be a Hamiltonian for some problem with an n-dimensional configuration space and 2n-dimensional phase space. Suppose the Hamiltonian does not depend upon the ith coordinate qⁱ: (∂₁H)_i = 0.²⁰ According to Hamilton's equations, the conjugate momentum p_i is conserved. Hamilton's equations of motion for the remaining 2n − 2 phase-space variables do not involve qⁱ (because it does not appear in the Hamiltonian), and p_i is a constant. Thus the dimension of the difficult part of the problem, the part that involves the solution of coupled ordinary differential equations, is reduced by two. The remaining equation governing the evolution of qⁱ in general depends on all the other variables, but once the reduced problem has been solved, the equation of motion for qⁱ can be written so as to give Dqⁱ explicitly as a function of time. We can then find qⁱ as a definite integral of this function.²¹

Contrast this result with analogous results for more general systems of differential equations. There are two independent situations. One situation is that we know a constant of the motion. In general, constants of the motion can be used to reduce by one the dimension of the unsolved part of the problem. To see this, let the system of equations be

$$\begin{array}{ll}D{z}^i(t)=F^i(z^0(t),z^1(t),...,z^{m-1}(t)),\hfill & (3.97)\end{array}$$

where m is the dimension of the system. Assume we know some constant of the motion

$$\begin{array}{ll}C(z^0(t),z^1(t),...,z^{m-1}(t))=0.\hfill & (3.98)\end{array}$$

At least locally, we expect that we can use this equation to solve for z^m−1(t) in terms of all the other variables, and use this solution to eliminate the dependence on z^m−1(t). The first m−1 equations then depend only upon the first m − 1 variables. The dimension of the system of equations to be solved is reduced by one. After the solution for the other variables has been found, z^m−1(t) can be found using the constant of the motion.

The second situation is that one of the variables, say zⁱ, does not appear in the equations of motion (but there is an equation for Dzⁱ). In this case the equations for the other variables form an independent set of equations of one dimension less than the original system. After these are solved, then the remaining equation for zⁱ can be solved by definite integration.

In both situations the dimension of the system of coupled equations is reduced by one. Hamilton's equations are different in that these two situations come together. If a Hamiltonian for a system does not depend on a particular coordinate, then the equations of motion for the other coordinates and momenta do not depend on that coordinate. Furthermore, the momentum conjugate to that coordinate is a constant of the motion. An added benefit is that the use of this constant of the motion to reduce the dimension of the remaining equations is automatic in the Hamiltonian formulation. The conserved momentum is a state variable and just a parameter in the remaining equations.

So if there is a continuous symmetry it will probably be to our advantage to choose a coordinate system that explicitly incorporates the symmetry, making the Hamiltonian independent of a coordinate. Then the dimension of the phase space of the coupled system will be reduced by two for every coordinate that does not appear in the Hamiltonian.²²

Motion in a central potential

Consider the motion of a particle of mass m in a central potential. A natural choice for generalized coordinates that reflects the symmetry is polar coordinates. A Lagrangian is (equation 1.69):

$$\begin{array}{ll}L(t;r,\phi ;\dot{r},\dot{\phi })=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\phi }}^2)-V(r).\hfill & (3.99)\end{array}$$

The momenta are p_r = mṙ and $p_{\phi }=m{r}^2\dot{\phi }$. The kinetic energy is a homogeneous quadratic form in the velocities, so the Hamiltonian is T + V with the velocities rewritten in terms of the momenta:

$$\begin{array}{cc}H\left(t;r,\phi ;p_r,p_{\phi }\right)=\frac{p_r^2}{2m}+\frac{p_{\phi }^2}{2m{r}^2}+V\left(r\right).& \left(3.100\right)\end{array}$$

Hamilton's equations are

$$\begin{array}{lll}\hfill Dr(t)& =\frac{p_r(t)}{m}\hfill & \hfill \\ \hfill D\phi (t)& =\frac{p_{\phi }(t)}{m{(r(t))}^2}\hfill & \hfill \\ \hfill D{p}_r(t)& \hfill =\frac{{(p_{\phi }(t))}^2}{m{(r(t))}^3}-DV(r(t))& \hfill \\ \hfill D{p}_{\phi }(t)& =0.\hfill & (3.101)\hfill \end{array}$$

The potential energy depends on the distance from the origin, r, as does the kinetic energy in polar coordinates, but neither the potential energy nor the kinetic energy depends on the polar angle φ. The angle φ does not appear in the Lagrangian so we know that p_φ, the momentum conjugate to φ, is conserved along realizable trajectories. The fact that p_φ is constant along realizable paths is expressed by one of Hamilton's equations. That p_φ has a constant value is immediately made use of in the other Hamilton's equations: the remaining equations are a self-contained subsystem with constant p_φ. To make a lower-dimensional subsystem in the Lagrangian formulation we have to use each conserved momentum to eliminate one of the other state variables, as we did for the axisymmetric top (see section 2.10).

We can check our derivations with the computer. A procedure implementing the Lagrangian has already been introduced (below equation 1.69). We can use this to get the Hamiltonian:

(show-expression
  ((Lagrangian->Hamiltonian
     (L-central-polar 'm (literal-function 'V)))
   (up 't (up 'r 'phi) (down 'p_r 'p_phi))))

and to develop Hamilton's equations:

(show-expression
  (((Hamilton-equations
      (Lagrangian->Hamiltonian
        (L-central-polar 'm (literal-function 'V))))
    (up (literal-function 'r)
        (literal-function 'phi))
    (down (literal-function 'p_r)
          (literal-function 'p_phi)))
   't))

Axisymmetric top

We reconsider the axisymmetric top (see section 2.10) from the Hamiltonian point of view. Recall that a top is a rotating rigid body, one point of which is fixed in space. The center of mass is not at the fixed point, and there is a uniform gravitational field. An axisymmetric top is a top with an axis of symmetry. We consider here an axisymmetric top with the fixed point on the symmetry axis.

The axisymmetric top has two continuous symmetries we would like to exploit. It has the symmetry that neither the kinetic nor potential energy is sensitive to the orientation of the top about the symmetry axis. The kinetic and potential energy are also insensitive to a rotation of the physical system about the vertical axis, because the gravitational field is uniform. We take advantage of these symmetries by choosing coordinates that naturally express them. We already have an appropriate coordinate system that does the job—the Euler angles. We choose the reference orientation of the top so that the symmetry axis is vertical. The first Euler angle, ψ, expresses a rotation about the symmetry axis. The next Euler angle, θ, is the tilt of the symmetry axis of the top from the vertical. The third Euler angle, φ, expresses a rotation of the top about the fixed ẑ axis. The symmetries of the problem imply that the first and third Euler angles do not appear in the Hamiltonian. As a consequence the momenta conjugate to these angles are conserved quantities. The problem of determining the motion of the axisymmetric top is reduced to the problem of determining the evolution of θ and p_θ. Let's work out the details.

In terms of Euler angles, a Lagrangian for the axisymmetric top is (see section 2.10):

(define ((L-axisymmetric-top A C gMR) local)
  (let ((q (coordinate local))
        (qdot (velocity local)))
    (let ((theta (ref q 0))
          (thetadot (ref qdot 0))
          (phidot (ref qdot 1))
          (psidot (ref qdot 2)))
      (+ (* 1/2 A
            (+ (square thetadot)
               (square (* phidot (sin theta)))))
         (* 1/2 C
            (square (+ psidot (* phidot (cos theta)))))
         (* -1 gMR (cos theta))))))

where gMR is the product of the gravitational acceleration, the mass of the top, and the distance from the point of support to the center of mass. The Hamiltonian is nicer than we have a right to expect:

(show-expression
  ((Lagrangian->Hamiltonian (L-axisymmetric-top 'A 'C 'gMR))
   (up 't
       (up 'theta 'phi 'psi)
       (down 'p_theta 'p_phi 'p_psi))))

Note that the angles φ and ψ do not appear in the Hamiltonian, as expected. Thus the momenta p_φ and p_ψ are constants of the motion.

For given values of p_φ and p_ψ we must determine the evolution of θ and p_θ. The effective Hamiltonian for θ and p_θ has one degree of freedom, and does not involve the time. Thus the value of the Hamiltonian is conserved along realizable trajectories. So the trajectories of θ and p_θ trace contours of the effective Hamiltonian. This gives us a big picture of the possible types of motion and their relationship, for given values of p_φ and p_ψ.

If the top is standing vertically then p_φ = p_ψ. Let's concentrate on the case that p_φ = p_ψ, and define p = p_ψ = p_φ. The effective Hamiltonian becomes (after a little trigonometric simplification)

$$\begin{array}{ll}{H}_p(t,\theta ,p_{\theta })=\frac{p_{\theta }^2}{2A}+\frac{p^2}{2C}+\frac{p^2}{2A}{\tan }^2\frac{\theta }{2}+gMR\cos \theta .\hfill & (3.102)\hfill \end{array}$$

Defining the effective potential energy

$$\begin{array}{ll}{V}_p(\theta )=\frac{p^2}{2C}+\frac{p^2}{2A}{\tan }^2\frac{\theta }{2}+gMR\cos \theta ,\hfill & (3.103)\hfill \end{array}$$

which parametrically depends on p, the effective Hamiltonian is

$$\begin{array}{ll}{H}_p(t,\theta ,p_{\theta })=\frac{p_{\theta }^2}{2A}+V_p(\theta ).\hfill & (3.104)\hfill \end{array}$$

If p is large, V_p has a single minimum at θ = 0, as seen in figure 3.5 (top curve). For small p (bottom curve) there is a minimum for finite positive θ and a symmetrical minimum for negative θ; there is a local maximum at θ = 0. There is a critical value of p at which θ = 0 changes from a minimum to a local maximum. Denote the critical value by p_c. A simple calculation shows $p_c=\sqrt{4gMRA}$. For θ = 0 we have p = Cω, where ω is the rotation rate. Thus to p_c there corresponds a critical rotation rate

$$\begin{array}{ll}{\omega }_c=\sqrt{4gMRA}/C.\hfill & (3.105)\hfill \end{array}$$

For ω > ω_c the top can stand vertically; for ω < ω_c the top falls if slightly displaced from the vertical. A top that stands vertically is called a “sleeping” top. For a more realistic top, friction gradually slows the rotation; the rotation rate eventually falls below the critical rotation rate and the top “wakes up.”

We get additional insight into the sleeping top and the awake top by looking at the trajectories in the θ, p_θ phase plane. The trajectories in this plane are simply contours of the Hamiltonian, because the Hamiltonian is conserved. Figure 3.6 shows a phase portrait for ω > ω_c. All of the trajectories are loops around the vertical (θ = 0). Displacing the top slightly from the vertical simply places the top on a nearby loop, so the top stays nearly vertical. Figure 3.7 shows the phase portrait for ω < ω_c. Here the vertical position is an unstable equilibrium. The trajectories that approach the vertical are asymptotic—they take an infinite amount of time to reach it, just as a pendulum with just the right initial conditions can approach the vertical but never reach it. If the top is displaced slightly from the vertical then the trajectories loop around another center with nonzero θ. A top started at the center point of the loop stays there, and one started near this equilibrium point loops stably around it. Thus we see that when the top “wakes up” the vertical is unstable, but the top does not fall to the ground. Rather, it oscillates around a new equilibrium.

It is also interesting to consider the axisymmetric top when p_φ = p_ψ. Consider the case p_φ ≠ p_ψ. Some trajectories in the θ, p_θ plane are shown in figure 3.8. Note that in this case trajectories do not go through θ = 0. The phase portrait for p_φ < p_ψ is similar and is not shown.

We have reduced the motion of the axisymmetric top to quadratures by choosing coordinates that express the symmetries. It turns out that the resulting integrals can be expressed in terms of elliptic functions. Thus, the axisymmetric top can be solved analytically. We do not dwell on this solution because it is not very illuminating: since most problems cannot be solved analytically, there is little profit in dwelling on the analytic solution of one of the rare problems that is analytically solvable. Rather, we have focused on the geometry of the solutions in the phase space and the use of conserved quantities to reduce the dimension of the problem. With the phase-space portrait we have found some interesting qualitative features of the motion of the top.

Exercise 3.8: Sleeping top

Verify that the critical angular velocity above which an axisymmetric top can sleep is given by equation (3.105).

3.4.1 Lagrangian Reduction

Suppose there are cyclic coordinates. In the Hamiltonian formulation, the equations of motion for the coordinates and momenta for the other degrees of freedom form a self-contained subsystem in which the momenta conjugate to the cyclic coordinates are parameters. We can form a Lagrangian for this subsystem by performing a Legendre transform of the reduced Hamiltonian. Alternatively, we can start with the full Lagrangian and perform a Legendre transform for only those coordinates that are cyclic. The equations of motion are Hamilton's equations for those variables that are transformed and Lagrange's equations for the others. The momenta conjugate to the cyclic coordinates are conserved and can be treated as parameters in the Lagrangian for the remaining coordinates.

Divide the tuple q of coordinates into two subtuples q = (x, y). Assume L(t; x, y; v_x, v_y) is a Lagrangian for the system. Define the Routhian R as the Legendre transform of L with respect to the v_y slot:

$$\begin{array}{ll}\hfill p_y={\partial }_{2,1}L(t;x,y;v_x,v_y)& (3.106)\hfill \end{array}$$

$$\begin{array}{ll}\hfill p_y{v}_y=R(t;x,y;v_x,p_y)+L(t;x,y;v_x,v_y)& (3.107)\hfill \end{array}$$

$$\begin{array}{ll}\hfill v_y={\partial }_{2,1}R(t;x,y;v_x,p_y)& (3.108)\hfill \end{array}$$

$$\begin{array}{ll}\hfill 0={\partial }_{0}R(t;x,y;v_x,p_y)+{\partial }_{0}L(t;x,y;v_x,v_y)& (3.109)\hfill \end{array}$$

$$\begin{array}{ll}\hfill 0={\partial }_{1}R(t;x,y;v_x,p_y)+{\partial }_{1}L(t;x,y;v_x,v_y)& (3.110)\hfill \end{array}$$

$$\begin{array}{ll}\hfill 0={\partial }_{2,0}R(t;x,y;v_x,p_y)+{\partial }_{2,0}L(t;x,y;v_x,v_y).& (3.111)\hfill \end{array}$$