Hamiltonian Mechanics

Example 1.1 (Horizontal spring and pendulum). Consider an ideal spring, that is, an idealized system consisting of a point particle of mass \(m\) attached to a spring with stiffness \(k\), sliding on a frictionless surface. Assume that the motion is one-dimensional, along the axis of the spring, and let \(x\) denote the displacement of the system from its equilibrium position, i.e., the position in which the spring is completely at rest: not compressed nor extended. It is convenient to assume that the sliding happens on a flat surface. This allows us to exclude gravitational forces from the picture, making this example as simple as it could be.

Hooke’s law states that the restoring force \(F\) exerted by an ideal spring is directly proportional to its displacement \(x\) from equilibrium. The law is expressed mathematically as \(F(x) = -kx\), where \(k\) is the spring constant, a measure of the spring’s stiffness. The negative sign indicates that the force acts in the opposite direction of the displacement, always seeking to restore the spring to its equilibrium position. See Figure 1.2, left panel. According to (1.4), then, the motion of the point particle is given by

\begin{equation} \label {eq:spring} m \ddot {x} = - k x \qquad \mbox {or}\qquad \ddot {x} = - \omega ^2 x \quad \mbox {where } \omega = \sqrt {k/m}. \end{equation}

The solution, \(x(t)\), is generally described by

\begin{equation} \label {eq:springsol} x(t) = R \cos (\omega t + \phi ), \end{equation}

with the unknowns \(R\in \mathbb {R}\) and \(\phi \in [0,2\pi )\) uniquely prescribed by the initial conditions

\begin{equation} x(0) = R\cos (\phi ) \qquad \mbox {and}\qquad \dot x(0) = -\omega R \sin (\phi ). \end{equation}

Once we know the initial conditions, the full evolution of the solution \(x(t)\) is known.

Consider, now, an ideal pendulum . That is, a point particle of mass \(m\) attached to a pivot on the ceiling via a rigid rod of length \(l\). Assume the motion is frictionless and happening only in a vertical plane. Let \(x\) denote the angle of displacement of the system from its equilibrium position on this plane, i.e., the lowest point on the arc of motion. Without loss of generality, we can assume the angle at equilibrium to be zero, with positive sign on the right hand side of the vertical, and negative on the left. See Figure 1.2, right panel.

The force acting on the pendulum is Earth’s gravitational attraction. According to (1.4), then, the motion of the point particle is given by

\begin{equation} m l \ddot {x} = - m g \sin (x) \qquad \mbox {or}\qquad \ddot {x} = - \omega ^2 \sin (x) \quad \mbox {where } \omega = \sqrt {g/l}. \end{equation}

Here, \(g \approx 9.8 m s^{-2}\) denotes the gravitational acceleration.

Although it is possible to solve the equation of motion of the pendulum explicitl yby means of elliptic integrals [Eur24], this is rather cumbersome. In some cases, when \(\sin x \approx x\), we can simplify our life and, instead, use the equation of the spring as a model for the pendulum oscillation via the so-called small oscillations approximation. We will come back to this later on.

Example 1.2 (Idealized motion of the Earth around the Sun). Let us approximate the Sun with a point particle of mass \(M\) positioned at the origin \(\vb *{0}\in \mathbb {R}^3\) of the Euclidean space. In this model of the solar system, with the Sun fixed at the origin, we will describe the Earth by a point particle of mass \(m\) whose position (and motion) is described by a vector \(\vb *{x}\in \mathbb {R}^3\).

Due to our choice of coordinates, the gravitational attraction of the Sun acts in the direction of \(-\vb *{x}(t)\). Newton’s law of universal gravitation says that such a force is proportional to

\begin{equation} \frac {GmM}{\|\vb *{0}-\vb *{x}\|^2} = \frac {GmM}{\|\vb *{x}\|^2}, \end{equation}

where \(G \sim 6.674 \cdot 10^{-11} \frac {m^3}{s^2\,kg}\) is called the gravitational constant. Newton’s second law (1.4), then, leads to the equation of motion

\begin{equation} \label {eq:keplerex} m \ddot {\vb *{x}} = - \frac {GmM}{\|\vb *{x}\|^2} \frac {\vb *{x}}{\|\vb *{x}\|}. \end{equation}

This is an autonomous second order ordinary differential equation on the configuration space \(\mathbb {R}^3\setminus \big \{\vb *{0}\big \}\). Providing the initial conditions \(\vb *{x}(0)=\vb *{x}_0\) and \(\dot {\vb *{x}}(0)=\vb *{v}_0\), we can try to solve (1.10), and you are welcome to try. As it is often the case, though, things can become easier by taking a step back and looking at the problem differently.

Behind differential equations in classical mechanics lies a surprisingly rich geometrical structure, in which symmetries and conserved quantities play a special role. These can help to obtain extensive information on the solution of classical equations of motion without the need to explicitly solve the equations (which, in general, is impossible). Let’s have an initial brief look at this in the context of the Kepler problem.

Consider the space \(\big (\mathbb {R}^3\setminus \big \{\vb *{0}\big \}\big )\times \mathbb {R}^3\), and define there, for the moment without further justification or explanation, the total energy function

\begin{equation} \label {eq:energyKepler} E(\vb *{x},\vb *{v}) = \frac {1}{2}m\|\vb *{v}\|^2 - \frac {GmM}{\|\vb *{x}\|}, \end{equation}

the angular momentum

\begin{equation} L(\vb *{x},\vb *{v}) = m \vb *{v} \times \vb *{x}, \end{equation}

and the Laplace-Runge-Lenz vector

\begin{equation} A(\vb *{x},\vb *{v}) = m \vb *{v} \times L(\vb *{x}, \vb *{v}) + \frac {G m^2 M^2}{m+M} \frac {\vb *{x}}{\|\vb *{x}\|}. \end{equation}

Along the solutions \(\vb *{x}(t)\) of (1.10), let us define \(E(t) := E(\vb *{x}(t),\dot {\vb *{x}}(t))\) and, similarly, \(L(t)\) and \(A(t)\).

In due time we will show that, \(L(t) = L(0) \neq 0\) for all times, and also \(E(t) = E(0)\) and \(A(t) = A(0)\) are constant for all times. Furthermore, any solution that lives on the subspace defined by

\begin{align} \big \{ (\vb *{x}, \vb *{v})\in \big (\mathbb {R}^3\setminus \big \{\vb *{0}\big \}\big )\times \mathbb {R}^3 \;\mid \; E(\vb *{x},\vb *{v}) = E(\vb *{x}_0, \vb *{v}_0),\; L(\vb *{x},\vb *{v}) = L(\vb *{x}_0, \vb *{v}_0),\; A(\vb *{x},\vb *{v}) = A(\vb *{x}_0, \vb *{v}_0) \big \}, \end{align} must be a conic of the following type

\begin{equation} \begin{split} \mbox {ellipse} \quad \mbox {if} \quad E(t) = E(0) < 0, \\ \mbox {parabola} \quad \mbox {if} \quad E(t) = E(0) = 0, \\ \mbox {hyperbola} \quad \mbox {if} \quad E(t) = E(0) > 0. \end {split} \end{equation}

This is an example of a central force field, and is one of the most prominent and most important examples in this course. The impatient reader can find in [Kna18, Ch. 1] a nice and compact derivation of Kepler’s laws and the invariants above from (1.10).

Example 1.3. A system of two ideal pendulums is described by two position vectors, so \(\vb *{x} = (\vb *{x}_1, \vb *{x}_2)\in \mathbb {R}^{6}\). Notice, however, that to describe their configuration we only need two angular variables, one for each of the pendulums. So, for all practical purposes, the system could be completely described by \(q = (q^1, q^2) \in \mathbb {S}^1\times \mathbb {S}^1 \simeq \mathbb {T}^2\). See Figure 1.3.

The symbol \(\simeq \) usually denotes some kind of equivalence, but which specific kind is context-dependent and not often well explicited. So, let us take this opportunity to clarify the notation. What we wrote above, \(\mathbb {S}^1\times \mathbb {S}^1 \simeq \mathbb {T}^2\), should be read as: the 2-torus, i.e., the surface of a doughnut, is diffemorphic to the product of two circles. Here, diffeomorphic is a concept from the theory of smooth manifolds, and stands for the existence a smooth bijection between the two constructions.

If we then think of a circle \(\mathbb {S}^1\) as the interval \([0, 2\pi )\) with the endpoints identified, the observation above is saying that the torus \(\mathbb {T}^2\) can be thought of as the square \([0, 2\pi )\times [0, 2\pi )\) with the edges identified in the usual way and \(q = (q^1, q^2) \in [0, 2\pi )\times [0, 2\pi ) \subseteq \mathbb {R}^2\).

Such a long story to say that, while in principle this system is 6-dimensional, for all practical purposes it is 2-dimensional.

Example 1.4. The lagrangian of a mechanical system consisting of a point particle with mass \(m > 0\) in a potential \(U : \mathbb {R}^n \to \mathbb {R}\) is

\begin{equation} L(q, v, t) = \frac 12 m \|v\|^2 - U(q), \end{equation}

which is the difference between the so-called kinetic energy of the particle and its so-called potential energy \(U\). We will see where this equation comes from in Section 1.2.1.

Remark 1.1.

• 1. In the finite dimensional setting, Fréchet differentiability corresponds to total differentiability.

• 2. As in the finite dimensional case, Fréchet differentiability implies the continuity of the mapping.

• 3. The analogy with the finite dimensional case goes further. Indeed, the chain rule, mean value theorem, implicit function theorem, inverse mapping theorem and the statements about local extremes with and without constraints, all hold also for Fréchet differentiable functions on Banach spaces.

• 4. Requiring the operator \(A\) above to be bounded is crucial: in \(\infty \)-dimensional normed spaces, linearity does not imply continuity.

• 5. In calculus of variation, the curve \(h\) from the small neighborhood of \(0\in X\), is usually called a variation and denoted \(\delta x\). Don’t confuse this with a Dirac delta function or a small real parameter.

Remark 1.2. Be careful: when we omit the indices \(i\), we use \(\pdv {L}{q}\) to denote the gradient of \(L\) with respect to the position variables \(q\), and \(\pdv {L}{\dot q}\) the gradient of \(L\) with respect to velocity variables \(\dot q\). In particular, \(\dot q\) in \(\pdv {L}{\dot q}\) serves as a label to denote the velocity variable, and not as the derivative of \(q\) with respect to time. We consider \(q\) and \(\dot q\) as curves parametrised by time only when we apply \(\dv {t}\).

Therefore, you should read the Euler-Lagrange equations above as

\begin{equation} \dv {t}\Bigg (\pdv {L(x,v)}{v^i}\Big |_{(x,v) = (q(t),\dot q(t))}\Bigg ) - \pdv {L(x,v)}{x^i}\Big |_{(x,v) = (q(t),\dot q(t))} = 0, \qquad i=1,\ldots n. \end{equation}

This explicit formulation, though precise, illustrates why such notational shortcuts are common and why I chose to adopt them.

Remark 1.3. In general, the solution \(q=q(t)\) of the Euler-Lagrange equations is just a critical point of the \(S\) functional, i.e., it satisfies \(S[q + \delta q] - S[q] = O(\|\delta q\|^2)\). It does not have to be a minimum. However, under some additional conditions it can be proven to be a local minimum. For example, if the matrix of second derivatives

\begin{equation} \label {eq:hsd} \Lambda := \left ( \pdv {L}{\dot q^i}{\dot q^j} \right )_{1\leq i,j\leq n} \end{equation}

is positive definite. This case is thoroughly studied in calculus of variation and in some branches of differential geometry.

Remark 1.4. The lagrangian of a mechanical system is defined only up to total derivatives. That is, the equations of motion remain unchanged if we add a total derivative to the lagrangian function:

\begin{equation} \widetilde L(q,\dot q, t) = L(q, \dot q, t) + \dv {t} f(q,t). \end{equation}

Indeed, the action \(\widetilde S\) of a system with lagrangian \(\widetilde L\) is

\begin{align} \widetilde S[q] & = \int _{t_1}^{t_2} \widetilde L(q, \dot q, t) \,\dd t \\ & = \int _{t_1}^{t_2} L(q, \dot q, t) \,\dd t + \int _{t_1}^{t_2} \dv {t} f(q,t) \,\dd t \\ & = S[q] + f(q_2, t_2) - f(q_1, t_1). \end{align} As the additional \(f\)-dependent part is constant,

\begin{equation} \widetilde S[q+\delta q] - \widetilde S[q] = S[q+\delta q] - S[q] \end{equation}

and thus the critical points of the two actions are the same.

Similarly, the lagrangian of a mechanical system does not change if it is multiplied by a constant factor: \(\widetilde L = \alpha L\).

As an aside, quantum mechanical systems no longer remain invariant under the transformations above. Transformations of the first type lead to rather subtle and interesting effects depending on the topology of the state space, while the number \(\alpha \) is related to Planck’s constant.

Remark 1.5. In view of our previous discussions, the statement above should be read as

the lagrangian of an isolated point particle in an inertial system of coordinates has the form (1.49) up to total derivatives.

Remark 1.6. A lagrangian can always be multiplied by an arbitrary constant without affecting the equations of motion; such multiplication then amounts to a change in the unit of mass. So, the above definition of mass becomes meaningful when we take the additive property into account: the ratios of the masses remain unchanged and it is only these ratios which are physically meaningful.

Exercise 1.1. Prove Newton’s third law, i.e., for each of the \(k\) point particles it holds

\begin{equation} \vb *{F}_k = -\sum _{j\neq k} \vb *{F}_j. \end{equation}

Hint: use the invariance with respect to spatial translations or sneak peek forward to the conservation of total momentum.

Remark 1.7.

• 1. Mechanical systems are time-reversible: they are also invariant with respect to the transformation \(t\mapsto -t\).

• 2. If you consider systems in interaction with the environment, you may end up with lagrangians that are no longer invariant with respect to galilean transformations and can explicitly depend on time.

Example 1.5 (Free fall). Consider the vertical motion of a point particle of mass \(m\) in the external field with potential \(U(z) = m g z\), see also Example 1.1. The natural lagrangian of the system is

\begin{equation} L = \frac {m\dot z^2}2 - mgz, \end{equation}

which corresponds to the equation of motion \(\ddot z = -g\): this is the equation of motion of a point particle in free fall towards Earth. In agreement with Galileo’s law, the acceleration is constant and does not depend on the mass.

Example 1.6. The motion of \(N\) point particles with masses \(m_1, \ldots , m_N\) in their gravitational field, is described by the lagrangian

\begin{equation} L = \sum \frac {m_k \|\dot {\vb *{x}}_k\|^2}{2} + \sum _{k < l} G \frac {m_k m_l}{\|\vb *{x}_k - \vb *{x}_l\|}, \end{equation}

where \(G\) is the gravitational constant.

To study the approximate description of the motion of a planet of mass \(m\) around the Sun, which has mass \(M \gg m\), one can assume the effects of the planet on the Sun to be negligible and ignore the interaction with the other planets.

In such a case, the motion of the planet is described by the lagrangian of a point particle with a Newtonian gravitational potential

\begin{equation} L = \frac {m\|\dot {\vb *{x}}\|^2}{2} + \frac {G M m}{\|\vb *{x}\|}, \end{equation}

whose equation of motion should remind you of (1.10) from Example 1.2.

Remark 1.8. For \(N=1\), Stokes’ theorem implies that the integral above is independent of the path between \(\vb *{x}_0 = \vb *{x}(t_0)\) and \(\vb *{x}_1 = \vb *{x}(t_1)\) if and only if \(\curl \vb *{F} = 0\), which in turn is true if and only if there exists a function \(U:\mathbb {R}^3\to \mathbb {R}\) such that \(\vb *{F} = -\frac {\partial U}{\partial \vb *{x}}\). In fact, this is true in any dimension by using the general version of Stokes’ theorem: the work done by the force depends only on the endpoints of the integral if and only if there is \(U:\mathbb {R}^n\to \mathbb {R}\), unique up to an additive constant, such that \(\vb *{F} = -\frac {\partial U}{\partial \vb *{x}}\). You can read more about this in [Arn89, Chapter 2.5] and [Kna18, Theorem 6.3 and 8.1].

Remark 1.9. Often in physics, the kinetic energy is instead defined as the work required to bring the mass from rest to some speed \(\dot {\vb {x}}\). One can then perform the computation above backwards

\begin{equation} \int _{x_0}^{x^1} \left \langle \vb *{F}(\vb *{x}), \dd {\vb *{x}}\right \rangle = \int _{t_0}^{t_1} \left \langle \dot {\vb *{x}}(t), \vb *{F}(\vb *{x}(t))\right \rangle \; \dd t = \int _{t_0}^{t_1} m \langle \dot {\vb *{x}}(t), \ddot {\vb *{x}}(t)\rangle \; \dd t = \int _{t_0}^{t_1} \dv {t} \frac {m}{2} \langle \dot {\vb *{x}}(t), \dot {\vb *{x}}(t)\rangle \; \dd t, \end{equation}

, leading to the definition of kinetic energy as \(T = \frac {m}{2} \|\dot {\vb {x}}\|^2\).

Example 1.7. Mechanical systems affected by an external magnetic field \(\vb * B\) can be described in the lagrangian formalism by adding a linear term in the velocities to a natural lagrangian (1.59):

\begin{equation} \label {eq:magLag} \widetilde L = L + \frac ec \sum _{k=1}^N \langle \vb * A(\vb *{x}_k), \dot {\vb *{x}}_k\rangle . \end{equation}

Here the constant \(e\) is the electric charge of the point particles and \(c\) is the speed of light in vacuum. The magnetic field is given by \(\vb * B = \curl \vb * A\), the vector field \(\vb * A\) is called magnetic vector potential.

We will not see this in the course, but magnetic phenomenons have a natural description only in the relativistic approach. We see a hint of this fact here, in that (1.68) is not invariant with respect to the galilean transformations.

Note that the magnetic vector potential is not unique! For any function \(f\), the transformation

\begin{equation} \vb * A \mapsto \vb * A + \pdv {f}{\vb *{x}} \end{equation}

will produce the same field. This is known as gauge transformation. Under this transformation the lagrangian is transformed as

\begin{equation} \widetilde L \mapsto \widetilde L + \frac {e}{c} \dv {f}{t} \end{equation}

but we know that the equations of motion remain invariant under the addition of a total derivative to the Lagrangian. The concept of gauge invariance is central in many problems of modern physics.

Remark 1.10. This identity in (1.75) is often expressed as

\begin{equation} y\,\dd y - F(x)\, \dd x = 0. \end{equation}

This may look like a formal manipulation, but it is actually a rigorous identity in terms of differential forms. To know more, you can refer to [Ser20, Equation (5.1) with \(f=y\) and Remark 5.1.3] or any good textbook on differential forms. For simplicity, we will limit ourselves to the formal manipulation.

Example 1.8. Note that Theorem 1.6 applies to Example 1.1.

• • In the case of the spring, also called the harmonic oscillator, \(U(x) = \frac 12 \omega ^2 x^2\) and the integral curves are of the form \(y^2 + \omega ^2 x^2 = C\). Recalling (1.6), \(C = \omega ^2 R^2 \geq 0\) and the curves are ellipses parametrized by \(C\). This will be a very important example throughout the course.

  Use the sliders to see how the phase plane is changing. What happens if you vary the mass?

• • In the case of the pendulum, \(U(x) = -\omega ^2 \cos (x)\) and the integral curves are solutions of \(\frac 12 y^2 - \omega ^2 \cos (x) = C\), \(C \geq -\omega ^2\). Even though we cannot easily give a time parametrization of the motion itself, this formalism allows us to immediately describe the evolution of the system. We will come back to this fact in more generality in the next section.

  Use the sliders to see how the phase plane is changing. What happens if you vary the mass?

Exercise 1.3. Show that at the equilibrium points of the examples above the system is at rest, that is, \(\dot x = \dot y = 0\).

Example 1.9. Let’s consider \(N\) point particles in physical space with natural lagrangian \(L = T - U\) as in (1.59). Then,

\begin{equation} \vb *{p}_k = \frac {\partial L}{\partial \dot {\vb *{x}}_k} = m_k \dot {\vb *{x}}_k, \end{equation}

and therefore

\begin{equation} \sum _{k=1}^{N} \langle \vb *{p}_k, \dot {\vb *{x}}_k\rangle = \sum _{k=1}^{N} m_k \|\dot {\vb *{x}}\|^2 = 2 T \end{equation}

which implies that the energy of the mechanical system is

\begin{equation} \label {eq:energyFromL} E = 2T - T + U = T + U. \end{equation}

We call \(\vb *{p}_k := m_k \dot {\vb *{x}}_k\) the kinetic momentum, replacing it in the equation above we get that

\begin{equation} E = H(q,p) := \frac {1}{2 m_k}\|\vb *{p}_k\|^2 + U \end{equation}

is the sum of kinetic and potential energies, as in (1.11) and Theorem 1.6.

Remark 1.11. This is probably a good point to discuss the question: why does nature want to minimize the action? And why the lagrangian is of the form (1.59)?

Theorem 1.7 tells us that the total energy is conserved, and (1.89) tells us that for closed systems this implies that the energy is transferred back and forth between the kinetic and the potential components. We saw towards the end of Section 1.2.1 that while the kinetic energy measures how much the system is moving around, the potential energy measures the capacity of the system to change: its name can be intended as ‘potential’ in the sense of yet unexpressed possibilities.

Looking at the lagrangian itself, we see that it is minimal when the potential energy is large and maximal when the kinetic energy is large. So, the lagrangian measures in some sense how ‘active’ a system is: the higher the kinetic energy the more active a system is, and the higher the potential energy the less active the system. The principle of least action tells us that nature is lazy: she likes to find a compromise that minimizes its activity over time, i.e., its total action.

You can read a nice historical account on how scientists came up with the principle of least action in [BW05].

Remark 1.12. Integral curves carry a lot more information than what immediately meets the eye. Observe for example that there is a remarkable relation between the period of an oscillation and the area of the integral curve \(H(x,y) = E\): if \(A\) is the area enclosed by the integral curve, then the period \(\mathcal {T}\) can be obtained as

\begin{equation} \mathcal {T} = \frac {\dd A}{\dd E}. \end{equation}

We will see this in more generality once we deal with action–angle coordinates, but let’s sketch the proof in this case.

From \(H(x,y) = E\) we get \(y = \pm \sqrt {2(U(x) - E)}\), where I considered unit mass for notational simplicity. Say that the curve intercepts the \(x\)-axis at the two return points \(x_m < x_M\), then the area inside the curve, is given by

\begin{equation} A = 2\int _{x_m}^{x_M}\sqrt {2(E - U(x))} \dd x. \end{equation}

As we saw above, the period can be computed as

\begin{equation} \mathcal {T} = 2\int _{x_m}^{x_M} \frac {\dd x}{\sqrt {2(E - U(x))}}. \end{equation}

With the above in mind, the result follows by taking the derivative of \(A\) with respect to \(E\).

Example 1.10. Looking back at Example 1.1, the lagrangian of a pendulum of length \(l\) and mass \(m\) is given by

\begin{equation} L = \frac {ml^2 \dot x^2}2 + mgl \cos x, \end{equation}

thus \(E = \frac {ml^2 \dot x^2}2 - mgl \cos x\). As one can see in Figure 1.5, the motion is bounded for \(|E| \leq mgl\). For any such motion, the angle of maximal oscillation corresponds to a solution of \(U(x) = E\), thus is given by \(E = -mgl \cos x^*_E\) for some \(x^*_E\). Finally, (1.96) implies that the period of oscillation is

\begin{equation} \mathcal {T}(E) = 4 \sqrt {\frac l{2g}} \int _0^{x^*_E} \frac {\dd x}{\sqrt {\cos (x) - \cos (x^*_E)}}, \end{equation}

which, as we briefly discussed previously, does not depend on the mass of the pendulum, but only on its length!

It is possible to use the previous formula to express the period of oscillation in terms of elliptic integrals

\begin{equation} \mathcal {T} = 4 \sqrt {\frac lg} K\left (\sin (x^*/2)\right ), \quad K(k):= \int _0^{\pi /2} \frac {\dd \theta }{\sqrt {1 - k^2\sin ^2(\theta )}}. \end{equation}

If the oscillations are small, \(x^* = \epsilon \ll 1\), this form allows for an immediate expansion of the form

\begin{equation} \label {eq:smallosc-preview} \mathcal {T}_\mathrm {p} = 2\pi \sqrt {\frac lg} \left (1 + \frac {\epsilon ^2}{16} + O(\epsilon ^4)\right ). \end{equation}

Equation (1.104) implies that for small oscillations, the period of oscillation is also independent of the amplitude (and thus on the energy \(E\)). Which is the same result that you get by approximating the potential for small \(x\) by a second order polynomial

\begin{equation} U(x) = -mgl \cos x = C + \frac {mgl}2 x^2 + O(x^4) \end{equation}

and study the corresponding system.

Indeed, ignoring the constant \(C\) for conciseness, one finds the harmonic oscillator lagrangian

\begin{equation} L_{\mathrm {ho}} = \frac {ml^2 \dot x^2}{2} - \frac {mgl x^2}{2}, \end{equation}

which we already solved in Example 1.1. In this specific case \(\omega = \sqrt {g/l}\) and the corresponding period of oscillation is

\begin{equation} \mathcal {T}_{\mathrm {ho}} = \frac {2\pi }\omega = 2\pi \sqrt {\frac lg} \sim \mathcal {T}_\mathrm {p}. \end{equation}

This is an example of a more general phenomenon that we will discuss in detail in Section 2.3.

Example 1.11 (Double pendulum). A double pendulum consists of two point particles of masses \(m_1\) and \(m_2\), connected by massless rods of lengths \(l_1\) and \(l_2\). The first pendulum is attached to the ceiling by a fixed pivot while the second one is attached to the point particle of the first pendulum.

the pendulums make with respect to the corresponding vertical, oriented counterclockwise: a value of \(0\) means that the pendulum is vertical with the mass lying below the pivot.

The lagrangian \(L_1 = T_1 - U_1\) of the first particle is the same as for a simple pendulum:

\begin{equation} T_1 = \frac {m_1 l_1^2 \dot x_1^2}2, \quad \mbox {and}\quad U_1(x_1) = -m_1 g l_1 \cos x_1, \end{equation}

where \(g\) is the usual gravitational acceleration.

The lagrangian corresponding to the second particle is a bit more involved. We start by considering the position of the second particle in the \((\chi ,\eta )\)-plane in which the pendulum swings, centered at the pivot of the first pendulum and with \(\eta \) positive in the downward direction:

\begin{equation} \chi = l_1\sin x_1 + l_2\sin x_2, \quad \mbox {and}\quad \eta = l_1\cos x_1 + l_2\cos x_2. \end{equation}

We can substitute these values into the kinetic energy for the second particle in cartesian coordinates \((\chi ,\eta )\) to get

\begin{equation} T_2 = \frac {m_2 (\dot \chi ^2 + \dot \eta ^2)}2 = \frac {m_2}2 \left ( l_1^2 \dot x_1^2 + l_2^2 \dot x_2^2 + 2l_1l_2 \cos (x_1 -x_2)\dot x_1 \dot x_2 \right ), \end{equation}

while the potential energy is given by

\begin{equation} U_2 = -m_2 g \eta = -m_2g (l_1\cos x_1 + l_2\cos x_2). \end{equation}

The full lagrangian is then given by

\begin{align} L & = T_1 + T_2 - U_1 - U_2 \\ & = \frac 12 (m_1 + m_2) l_1^2 \dot x_1^2 + \frac 12 m_2 l_2^2 \dot x_2^2 + m_2l_1l_2 \cos (x_1 -x_2)\dot x_1 \dot x_2 \\ & \quad + (m_1 + m_2) g l_1 \cos x_1 + m_2gl_2\cos x_2. \end{align}

We can now compute the equations of motion using Euler-Lagrange equations and try to solve them. The solutions are complicated and, in fact, after a certain energy threshold the motion is chaotic. You can have some fun looking at the daily simulations of the pendulum bot: https://twitter.com/pendulum_bot. We will come back to this problem after discussing symmetries and small oscillations.

Example 1.12 (Lissajous Figures). Consider the system of independent harmonic oscillators

\begin{equation} \left \lbrace \begin{aligned} \ddot x_1 & = -x_1 \\ \ddot x_2 & = -\omega ^2 x_2 \end {aligned} \right .. \end{equation}

This can be explicitly solved: the equations are completely uncoupled and we know that the phase portrait of \(x_1\) consists of circles while the phase portrait of \(x_2\) consists of ellipses.

Together, the two equations define a 4 dimensional phase space. The curves in the phase space are combinations of closed curves in each phase space and describe a 2-torus.

We know from the previous part of this section that the energy of each oscillator is conserved, as well as the total energy:

\begin{equation} H_1 = \frac 12\dot x_1^2 + \frac 12 x_1^2, \quad H_2 = \frac 12\dot x_2^2 + \frac 12 \omega ^2 x_2^2, \quad H = H_1 + H_2. \end{equation}

The level curves \(H=E\) describe a 3-sphere in \(\mathbb {R}^4\), which are foliated by all the 2-tori of trajectories with energies that sum to \(E\). For the more geometric inclined, you can look up “Hénon-Heiles model”, “Seifert foliation of a three sphere” or “Hopf fibration” to have a glimpse of some related fascinating topics.

On the configuration plane \((x_1, x_2)\), the motion that we observe is

\begin{equation} x(t) = \left ( R_1 \cos (t + \phi _1),\, R_2 \cos (\omega t + \phi _2) \right ). \end{equation}

This is contained in a rectangle \([-R_1,R_1]\times [-R_2,R_2]\). When \(\omega \) is irrational, the trajectory fills the rectangle, whereas they are closed curves inside the rectangle when it is rational. See also [Kna18, Exercise 6.34] and [Arn89, Chapter 2.5, Example 2].

Exercise 1.4 (Huygens problem). Determine the form of even potentials, \(U(x) = U(-x)\), giving rise to isochronal oscillations, i.e., whose period \(\mathcal {T}\) does not depend on \(E\). (Beware, this is hard!)

Use your result to study the following problem. A point particle of mass \(m\) moves without friction under the effect of a gravitational potential with uniform acceleration \(g\) on the curve

\begin{equation} \gamma (s) = (x(s), y(s)) \quad \mbox {such that}\quad \left \lbrace \begin{aligned} & \dot x(s) \neq 0, \\ & x(-s) = -x(s), \; y(-s) = y(s) \end {aligned} \right . \end{equation}

Find the lagrangian of the system and determine the shape of the curve so that the oscillations are isochronal. This is called Huygens problem: he was studying it to win a competition for the best maritime chronometer.
Hint: use the natural parametrization for the curve: \(\dot x^2 + \dot y^2 = 1\).

Exercise 1.5 (The isoperimetric problem). In the \((x,y)\)-plane in \(P\subset \mathbb {R}^3\) you are given a curve \(\gamma :[t_1,t_2]\to P\) connecting the points \((x_1, y_1, 0)\) and \((x_2, y_2, 0)\). Revolve this curve around the \(x\)-axis. For which curve does the corresponding surface of revolution have minimal area?

Exercise 1.6 (The brachistochrone problem). Let \(P\), \(Q\) be two given points in the vertical \((x,y)\)-plane, where \(Q\) lies beneath \(P\). A particle, subject to constant gravitational acceleration pointing downwards, moves from \(P\) to \(Q\), starting at rest. Determine the curve along which this motion takes the shortest time.

Exercise 1.7. Let \(L=L(q,\dot q):TM \to \mathbb {R}\) be a smooth function. Show that the linear momenta, i.e., the derivatives

\begin{equation} p_i := \pdv {L}{\dot q^i}, \quad i=1,\ldots ,n, \end{equation}

transform in “the opposite way as \(\dot q\)” in the sense that:

\begin{equation} \widetilde p_i = \frac {\partial q^k}{\partial {\widetilde q}^i} p_k, \quad i=1,\ldots ,n, \end{equation}

involves the inverse of the jacobian of \(q\mapsto \widetilde q(q)\). Furthermore, show that the matrix of second derivatives with respect to the coordinates \(\dot q\) of the lagrangian transforms as a \((0,2)\)-tensor, i.e.

\begin{equation} \frac {\partial ^2 L}{\partial \dot {\widetilde q}^i \partial \dot {\widetilde q}^j} = \frac {\partial q^k}{\partial {\widetilde q}^i}\frac {\partial q^l}{\partial {\widetilde q}^j} \frac {\partial ^2 L}{\partial \dot {q}^k \partial \dot {q}^l}. \end{equation}

Exercise 1.8. Show that the Euler-Lagrange equations for the functional

\begin{equation} S[q] = \int _{t_1}^{t_2} \frac 12 g_{ij}(q) \dot q^i \dot q^j\; \dd t \end{equation}

have the form

\begin{equation} \label {eq:geodesic} \ddot q^k + \Gamma _{ij}^k(q) \dot q^i \dot q^j = 0, \quad k=1,\ldots , n, \end{equation}

where \(\Gamma _{ij}^k(q)\) are the Christoffel symbols of the Levi-Civita connection associated with the metric \(\dd s^2\):

\begin{equation} \Gamma _{ij}^k(q) = \frac 12 g^{km}\left ( \frac {\partial g_{mj}}{\partial q^i} + \frac {\partial g_{im}}{\partial q^j}-\frac {\partial g_{ij}}{\partial q^m} \right ), \qquad (g^{ij}(q)) = (g_{ij}(q))^{-1}. \end{equation}

Exercise 1.9.

• 1. Derive formula (1.89) for a natural lagrangian on a riemannian manifold \((M,g)\).

• 2. Given any lagrangian of the form \(L = L(q, \dot q)\) on \(TM\), show that the definition of energy (1.84) does not depend on the choice of local coordinates on \(M\).
  Hint: look at Exercise 1.7.

• 3. Consider the following free lagrangian on a riemannian manifold \((M, g)\):

  \(\seteqnumber{0}{1.}{145}\)

  \begin{equation} L = \frac 12 g_{ij}(q)\dot q^i \dot q^j. \end{equation}

  We have already seen in Exercise 1.8 that the solutions of the Euler-Lagrange equations are geodesic curves. Show that geodesic curves have constant speed, i.e., \(\langle \dot q, \dot q\rangle = \mathrm {const}\).

Example 1.13. Let us consider a spherical pendulum of mass \(m\) and length \(l\), i.e., a point particle of mass \(m\) moving without friction on the surface of a sphere of radius \(l\). Let’s ignore gravity for the moment.

The system has two degrees of freedom, its configuration space \(M\) is a 2-sphere \(S^2\) of radius \(l\). Using spherical coordinates \((\theta , \phi )\) – see also Figure 1.8 – and (1.127), we find the free lagrangian

\begin{equation} \label {eq:LsphericalP} L = \frac m2 l^2(\dot \theta ^2 + \dot \phi ^2 \sin ^2\theta ). \end{equation}

Due to the discussion above, we immediately can say that the point particle moves on the geodesics of the sphere.

Exercise 1.10. Start from a free lagrangian in cartesian coordinates. As we are imposing that the particle lives on the surface of the sphere, change variables using the appropriate spherical coordinates to derive (1.147).

Remark 1.13. It is often necessary to deal with mechanical systems in which the interactions between different bodies take the form of constraints. These constraints can be very complicated and hard to incorporate when working directly with the differential equations of motion. It is often easier to construct a lagrangian out of the kinetic and potential energy of the system and derive the correct equations of motion from there. Mechanical lagrangians will often be of the form (1.148).