Complete Classical Mechanics

Survey of the Elementary Principles

Core ideas

Classical mechanics describes the motion of macroscopic bodies when quantum and relativistic effects are small. A particle is an idealized object with mass $m$ and position $\bm r(t)$ in a three-dimensional Euclidean space. Its motion is described by velocity $\bm v = \dot{\bm r}$ and acceleration $\bm a = \ddot{\bm r}$.

Newton’s Laws and Inertial Frames. Newton’s laws are valid only in inertial frames---frames where a body remains at rest or in uniform motion unless acted upon by a force.

• First Law (Inertia): In an inertial frame, a particle moves with constant velocity if the net force is zero.
• Second Law (Dynamics): The rate of change of momentum $\bm p = m\bm v$ equals the net force $\bm F$:

$$\bm F = \dot{\bm p} = \frac{d}{dt}(m\bm v).$$

For constant mass, this reduces to $\bm F = m\bm a$.

• Third Law (Action-Reaction): For every force $\bm F_{12}$ exerted by particle 2 on particle 1, there is an equal and opposite force $\bm F_{21} = -\bm F_{12}$ exerted by particle 1 on particle 2. This holds for “central” forces acting along the line joining the particles.

Work and Energy. The work done by a force $\bm F$ as a particle moves from point 1 to 2 is $W_{12} = \int_1^2 \bm F \cdot d\bm r$. The Work-Energy Theorem states that the work done by the net force equals the change in kinetic energy $T = \frac{1}{2}mv^2$:

$$W_{12} = T_2 - T_1 = \Delta T.$$

A force is conservative if the work done is independent of the path. Such forces can be derived from a potential energy $V(\bm r)$:

$$\bm F = -\nabla V(\bm r) \implies W_{12} = V_1 - V_2 = -\Delta V.$$

In this case, total mechanical energy $E = T + V$ is conserved ($\Delta E = 0$).

Systems of Particles. For a system of $N$ particles, the total mass is $M = \sum m_i$ and the Center of Mass (CM) is $\bm R = \frac{1}{M}\sum m_i \bm r_i$.

• Linear Momentum: The total momentum $\bm P = \sum \bm p_i = M\dot{\bm R}$ changes according to the net external force: $\dot{\bm P} = \bm F_{\rm ext}$. If $\bm F_{\rm ext} = 0$, $\bm P$ is conserved.
• Angular Momentum: The total angular momentum $\bm L = \sum \bm r_i \times \bm p_i$ changes according to the net external torque $\bm N_{\rm ext} = \sum \bm r_i \times \bm F_{i, \rm ext}$: $\dot{\bm L} = \bm N_{\rm ext}$.
• Kinetic Energy: $T = T_{\rm CM} + T_{\rm rel}$, where $T_{\rm CM} = \frac{1}{2}M\dot{\bm R}^2$ is the kinetic energy of the CM motion and $T_{\rm rel} = \frac{1}{2}\sum m_i (\dot{\bm r}'_i)^2$ is the kinetic energy relative to the CM.

Mathematical spine

$$\begin{aligned} \bm{F} &= \dot{\bm p} = m\bm{a} & \text{(Newton's 2nd Law)} \\ W_{12} &= \int_1^2 \bm F \cdot d\bm r = \Delta T & \text{(Work-Energy Theorem)} \\ \bm L &= \bm r \times \bm p, \quad \bm N = \bm r \times \bm F = \dot{\bm L} & \text{(Angular Momentum and Torque)} \\ M\ddot{\bm R} &= \bm F_{\rm ext}, \quad \dot{\bm L} = \bm N_{\rm ext} & \text{(System Dynamics)} \end{aligned}$$

Example: Conservative vs. Non-conservative forces. Gravity ($V=mgh$) and springs ($V=\frac{1}{2}kx^2$) are conservative. A conservative force does work that depends only on the endpoints; equivalently,

$$\oint \bm F\cdot d\bm r = 0$$

for every closed path. Kinetic friction violates this condition. Since

$$\bm f_k=-\mu N\hat{\bm v}, \qquad W_f=\int \bm f_k\cdot d\bm r=-\int \mu N\,ds,$$

the work depends on the arclength actually traveled, not just on the initial and final positions. For a closed loop of length $L>0$ with constant $\mu N$,

$$W_f=-\mu N L<0,$$

so friction cannot be represented by a single-valued potential energy. The heat produced is the physical destination of the lost mechanical energy; the mathematical reason friction is non-conservative is path dependence, or nonzero work around a closed path.

Worked example

Block on an inclined plane.
A block of mass $m = 5\ \text{kg}$ slides down a plane inclined at $\theta = 30^\circ$ to the horizontal with coefficient of kinetic friction $\mu_k = 0.2$. The normal force is $N = mg\cos\theta = 5\times 9.8\times\cos 30^\circ \approx 42.4\ \text{N}$. The friction force is $f_k = \mu_k N \approx 8.5\ \text{N}$. The component of gravity along the plane is $mg\sin\theta = 24.5\ \text{N}$. Newton’s second law gives $ma = mg\sin\theta - f_k = 16.0\ \text{N}$, so $a \approx 3.2\ \text{m/s}^2$. If the plane has length $d = 2\ \text{m}$, the block’s speed at the bottom is $v = \sqrt{2ad} \approx \sqrt{12.8} \approx 3.58\ \text{m/s}$. The work done by friction is $W_f = -f_k d = -17.0\ \text{J}$. The change in kinetic energy is $\Delta T = \frac{1}{2}mv^2 - 0 = 32.0\ \text{J}$, which equals the work done by gravity ($mgd\sin\theta = 49.0\ \text{J}$) plus the work done by friction ($-17.0\ \text{J}$), verifying the work-energy theorem.

Problems with Solutions

Problem 1. A 10 kg box is pulled across a horizontal floor by a rope at $30^\circ$ above horizontal with tension $T = 40\ \text{N}$. The coefficient of kinetic friction is $\mu_k = 0.25$. Find the acceleration.

Solution. Vertical forces: $N + T\sin 30^\circ = mg$, so $N = 10\times 9.8 - 40\times 0.5 = 98 - 20 = 78\ \text{N}$. Friction: $f_k = \mu_k N = 0.25\times 78 = 19.5\ \text{N}$. Horizontal: $T\cos 30^\circ - f_k = ma$, so $40\times 0.866 - 19.5 = 34.6 - 19.5 = 15.1\ \text{N} = 10a$, giving $a = 1.51\ \text{m/s}^2$.

Problem 2. A spring with $k = 200\ \text{N/m}$ is compressed by $x = 0.1\ \text{m}$ and launches a ball of mass $m = 0.5\ \text{kg}$ on a frictionless horizontal track. What is the ball’s speed when the spring reaches its natural length? If the track then curves upward to a height $h$, what is the maximum $h$ reached?

Solution. The spring potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.1)^2 = 1.0\ \text{J}$. By energy conservation, $\frac{1}{2}mv^2 = 1.0\ \text{J}$, so $v = \sqrt{2\times 1.0/0.5} = 2.0\ \text{m/s}$. For maximum height, kinetic energy converts to gravitational potential: $mgh = 1.0\ \text{J}$, so $h = 1.0/(0.5\times 9.8) \approx 0.204\ \text{m}$.

Problem 3. Two masses $m_1 = 4\ \text{kg}$ and $m_2 = 6\ \text{kg}$ are connected by a light string over a frictionless pulley. Find the acceleration of the system and the tension in the string.

Solution. Let $a$ be the acceleration of $m_1$ upward and $m_2$ downward. For $m_1$: $T - m_1g = m_1a$. For $m_2$: $m_2g - T = m_2a$. Adding: $(m_2-m_1)g = (m_1+m_2)a$, so $a = \frac{(6-4)\times 9.8}{4+6} = \frac{19.6}{10} = 1.96\ \text{m/s}^2$. From the first equation: $T = m_1(g+a) = 4(9.8+1.96) = 4\times 11.76 = 47.0\ \text{N}$.

Section summary. Newtonian mechanics builds from point particles to systems using three laws of motion. Conservation of energy, momentum, and angular momentum arise from symmetries and the nature of internal forces.

Variational Principles and Lagrange’s Equations

Core ideas

Newtonian mechanics requires identifying all forces, including unknown constraint forces (like the normal force on a bead on a wire). The Lagrangian method bypasses these by using energy and generalized coordinates $q_i$.

Generalized Coordinates and Constraints. For a system of $N$ particles, there are $3N$ degrees of freedom. Constraints reduce this number.

• Holonomic constraints can be expressed as $f(r_1, r_2, \dots, t) = 0$. If there are $k$ such constraints, the number of independent coordinates is $n = 3N - k$. These independent coordinates are the generalized coordinates $q_1, q_2, \dots, q_n$.
• Constraints that cannot be written this way are non-holonomic. Two common forms are inequalities such as $g(q,t)\le 0$, and non-integrable velocity constraints

$$\sum_i a_i(q,t)\dot q_i+a_0(q,t)=0.$$

If this differential relation can be integrated to $F(q,t)=0$, it is holonomic; if it cannot, it is genuinely non-holonomic.

For example, a particle confined inside a box obeys $0\le x,y,z\le L$, or inside a sphere obeys $x^2+y^2+z^2\le R^2$. These are inequality constraints rather than equations $F(q,t)=0$. A more typical mechanics example is a car that rolls without sideways slipping. If $(x,y)$ is the contact point and $\theta$ is the heading angle, the sideways velocity must vanish:

$$-\sin\theta\,\dot x+\cos\theta\,\dot y=0.$$

This restricts allowed velocities but does not reduce to a position-only condition $F(x,y,\theta)=0$. A rolling wheel gives similar non-integrable relations, such as $dx-R\cos\theta\,d\phi=0$ and $dy-R\sin\theta\,d\phi=0$.

Hamilton’s Principle. The Lagrangian is defined as $L(q, \dot q, t) = T - V$. Hamilton’s Principle states that the actual path taken by a system between times $t_1$ and $t_2$ is the one that makes the action $S$ stationary:

$$\delta S = \delta \int_{t_1}^{t_2} L(q_i, \dot q_i, t) \, dt = 0.$$

This leads to the Euler—Lagrange equations:

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot q_i} \right) - \frac{\partial L}{\partial q_i} = 0, \quad i=1, \dots, n.$$

The derivation is a direct integration-by-parts calculation. Compare the actual path $q_i(t)$ with a varied path $q_i(t)+\epsilon\eta_i(t)$, where the endpoint variations vanish: $\eta_i(t_1)=\eta_i(t_2)=0$. Stationarity means

$$0=\left.\frac{dS}{d\epsilon}\right|_{\epsilon=0} =\int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q_i}\eta_i+ \frac{\partial L}{\partial \dot q_i}\dot\eta_i \right)dt .$$

The second term is integrated by parts:

$$\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q_i}\dot\eta_i\,dt = \left[\frac{\partial L}{\partial \dot q_i}\eta_i\right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right)\eta_i\,dt .$$

The boundary term is zero because the endpoints are fixed. Therefore

$$\delta S = \int_{t_1}^{t_2} \left[ \frac{\partial L}{\partial q_i} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_i}\right) \right]\eta_i\,dt=0 .$$

Since the functions $\eta_i(t)$ are arbitrary inside the interval, the bracket must vanish for each coordinate, giving the Euler—Lagrange equations.

Generalized Forces and Non-conservative Systems. If some forces (like friction) are not derivable from a potential, the Euler—Lagrange equations are modified:

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot q_i} \right) - \frac{\partial L}{\partial q_i} = Q_i,$$

where $Q_i = \sum_j \bm F_j^{\rm nc} \cdot \frac{\partial \bm r_j}{\partial q_i}$ is the generalized force associated with coordinate $q_i$.

For a bead sliding on a circular hoop of radius $R$, with coordinate $\theta$, the position is $\bm r(\theta)=R(\sin\theta,-\cos\theta)$ and $\partial\bm r/\partial\theta=R(\cos\theta,\sin\theta)$. If a tangential friction force has magnitude $bR\dot\theta$ and opposes the motion, then

$$\bm F^{\rm nc}=-bR\dot\theta\,\hat{\bm e}_\theta, \qquad Q_\theta=\bm F^{\rm nc}\cdot\frac{\partial \bm r}{\partial\theta} =-bR^2\dot\theta .$$

With $L=\frac{1}{2}mR^2\dot\theta^2-mgR(1-\cos\theta)$, the equation of motion becomes

$$mR^2\ddot\theta+mgR\sin\theta=-bR^2\dot\theta,$$

or

$$\ddot\theta+\frac{b}{m}\dot\theta+\frac{g}{R}\sin\theta=0.$$

This example shows how non-conservative forces enter as generalized forces without being folded into $V$.

Symmetries and Noether’s Theorem. A coordinate $q_j$ that does not appear in the Lagrangian ($\partial L / \partial q_j = 0$) is called cyclic or ignorable. Its conjugate momentum $p_j = \partial L / \partial \dot q_j$ is conserved:

$$\dot p_j = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q_j} \right) = \frac{\partial L}{\partial q_j} = 0 \implies p_j = \text{const}.$$

Noether’s Theorem generalizes this: every continuous symmetry of the Lagrangian corresponds to a conservation law.

• Time translation symmetry $\implies$ conservation of energy ($H = \sum \dot q_i p_i - L$).
• Spatial translation symmetry $\implies$ conservation of linear momentum.
• Rotational symmetry $\implies$ conservation of angular momentum.

Mathematical spine

$$\begin{aligned} L &= T - V & \text{(Lagrangian)} \\ \delta \int L dt &= 0 \implies \frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0 & \text{(Euler--Lagrange)} \\ p_i &= \frac{\partial L}{\partial \dot q_i} & \text{(Generalized Momentum)} \\ H &= \sum \dot q_i p_i - L & \text{(Hamiltonian/Energy)} \end{aligned}$$

Example: Simple Pendulum. Coordinate: angle $\theta$. $L = T - V = \frac{1}{2}ml^2\dot\theta^2 - mg(l - l\cos\theta)$. Euler—Lagrange: $\frac{d}{dt}(ml^2\dot\theta) - (-mgl\sin\theta) = 0 \implies \ddot\theta + \frac{g}{l}\sin\theta = 0$.

Worked example

Bead on a rotating hoop.
A bead of mass $m = 0.1\ \text{kg}$ slides without friction on a hoop of radius $R = 0.2\ \text{m}$ rotating about its vertical diameter at constant angular velocity $\Omega = 5\ \text{rad/s}$. Using polar angle $\theta$ measured from the bottom, the position is $(R\sin\theta\cos\Omega t, R\sin\theta\sin\Omega t, -R\cos\theta)$. The velocity squared is $v^2 = R^2\dot\theta^2 + R^2\Omega^2\sin^2\theta$. The Lagrangian is $L = \frac{1}{2}m(R^2\dot\theta^2 + R^2\Omega^2\sin^2\theta) + mgR\cos\theta$. The Euler—Lagrange equation gives $mR^2\ddot\theta = mR^2\Omega^2\sin\theta\cos\theta - mgR\sin\theta$, or $\ddot\theta = (\Omega^2\cos\theta - g/R)\sin\theta$. For equilibrium, $\ddot\theta=0$, giving solutions $\theta=0$, $\theta=\pi$, and $\cos\theta = g/(R\Omega^2) = 9.8/(0.2\times 25) = 1.96$, which is impossible here. Thus only $\theta=0$ (bottom) and $\theta=\pi$ (top) are equilibria.

Problems with Solutions

Problem 1. Use the Lagrangian method to find the equation of motion for a simple pendulum of mass $m$ and length $l$.

Solution. Coordinate: angle $\theta$ from vertical. $x = l\sin\theta$, $y = -l\cos\theta$. Kinetic energy $T = \frac{1}{2}m(l^2\dot\theta^2)$. Potential energy $V = mgl(1-\cos\theta)$. Lagrangian $L = T-V = \frac{1}{2}ml^2\dot\theta^2 - mgl(1-\cos\theta)$. Euler—Lagrange: $\frac{d}{dt}(\partial L/\partial\dot\theta) - \partial L/\partial\theta = 0$. $\partial L/\partial\dot\theta = ml^2\dot\theta$, so $\frac{d}{dt}(ml^2\dot\theta) = ml^2\ddot\theta$. $\partial L/\partial\theta = -mgl\sin\theta$. Thus $ml^2\ddot\theta + mgl\sin\theta = 0$, or $\ddot\theta + (g/l)\sin\theta = 0$.

Problem 2. An Atwood machine consists of two masses $m_1 = 3\ \text{kg}$ and $m_2 = 2\ \text{kg}$ connected by a light string over a frictionless pulley. Use the Lagrangian method to find the acceleration.

Solution. Let $x$ be the vertical position of $m_1$ (downward positive). Then $m_2$ moves up by the same amount. $T = \frac{1}{2}(m_1+m_2)\dot x^2 = \frac{1}{2}(5)\dot x^2$. $V = -m_1gx + m_2gx = -(m_1-m_2)gx = -gx$. $L = \frac{5}{2}\dot x^2 + gx$. Euler—Lagrange: $5\ddot x - g = 0$, so $\ddot x = g/5 = 1.96\ \text{m/s}^2$. The acceleration of $m_1$ is $1.96\ \text{m/s}^2$ downward and $m_2$ is $1.96\ \text{m/s}^2$ upward.

Problem 3. A particle of mass $m$ moves in the $xy$-plane under a central potential $V(r) = -\frac{k}{r}$. Write the Lagrangian in polar coordinates and identify any cyclic coordinates and their conserved momenta.

Solution. In polar coordinates, $T = \frac{1}{2}m(\dot r^2 + r^2\dot\phi^2)$ and $V = -k/r$. Thus $L = \frac{1}{2}m(\dot r^2 + r^2\dot\phi^2) + k/r$. The coordinate $\phi$ does not appear explicitly in $L$, so it is cyclic. The conserved conjugate momentum is $p_\phi = \partial L/\partial\dot\phi = mr^2\dot\phi = \ell$, which is the angular momentum. This is Noether’s theorem in action: rotational symmetry implies conservation of angular momentum.

Section summary. The Lagrangian formulation replaces force-based dynamics with an optimization principle. It automatically incorporates holonomic constraints and links symmetries to conservation laws via Noether’s theorem.

The Central Force Problem

Core ideas

A central force points along the line joining two particles and depends only on their separation: $\bm F = F(r)\hat{\bm r}$. Examples include gravity and the Coulomb force.

Reduction to a One-Body Problem. For two particles with masses $m_1, m_2$ and positions $\bm r_1, \bm r_2$, the Lagrangian is $L = \frac{1}{2}m_1\dot{\bm r}_1^2 + \frac{1}{2}m_2\dot{\bm r}_2^2 - V(|\bm r_1 - \bm r_2|)$. Transforming to the center of mass $\bm R$ and relative coordinate $\bm r = \bm r_1 - \bm r_2$:

$$L = \frac{1}{2}M\dot{\bm R}^2 + \frac{1}{2}\mu\dot{\bm r}^2 - V(r),$$

where $M = m_1 + m_2$ and $\mu = \frac{m_1m_2}{m_1+m_2}$ is the reduced mass. The CM moves at constant velocity, so we focus on the relative motion in the CM frame.

Conservation Laws. Since the force is central, the torque is zero, and the angular momentum $\bm L = \bm r \times \mu \dot{\bm r}$ is conserved. This implies the motion is confined to a plane. In plane polar coordinates $(r, \phi)$:

$$L_z = \mu r^2 \dot\phi = \ell = \text{const}.$$

Energy is also conserved: $E = \frac{1}{2}\mu(\dot r^2 + r^2\dot\phi^2) + V(r) = \text{const}$.

The Effective Potential. Using $\dot\phi = \frac{\ell}{\mu r^2}$, we can write the radial energy equation:

$$E = \frac{1}{2}\mu \dot r^2 + V_{\rm eff}(r), \quad V_{\rm eff}(r) = V(r) + \frac{\ell^2}{2\mu r^2}.$$

The term $\frac{\ell^2}{2\mu r^2}$ is the centrifugal barrier. The radial motion is effectively one-dimensional.

The Orbit Equation. Defining $u = 1/r$, the differential equation for the orbit shape is:

$$\frac{d^2u}{d\phi^2} + u = -\frac{\mu}{\ell^2u^2}F(1/u).$$

For the inverse-square law $F(r) = -k/r^2$ (where $k = G m_1 m_2$), the solution is a conic section:

$$r(\phi) = \frac{\alpha}{1 + e\cos(\phi - \phi_0)}, \quad \alpha = \frac{\ell^2}{\mu k}, \quad e = \sqrt{1 + \frac{2E\ell^2}{\mu k^2}}.$$

The eccentricity $e$ determines the orbit shape:

• $e=0$: Circle ($E = V_{\rm eff, min} = -\frac{\mu k^2}{2\ell^2}$)
• $0 < e < 1$: Ellipse ($V_{\rm eff, min} < E < 0$)
• $e = 1$: Parabola ($E = 0$)
• $e > 1$: Hyperbola ($E > 0$)

Kepler’s Laws.

• Planets move in elliptical orbits with the Sun at one focus.
• A line joining a planet and the Sun sweeps out equal areas during equal intervals of time (equivalent to conservation of angular momentum).
• The square of the orbital period $T$ is proportional to the cube of the semi-major axis $a$: $T^2 = \frac{4\pi^2\mu}{k}a^3$.

The Runge-Lenz Vector. For the $1/r$ potential, there is an additional conserved vector, the Laplace—Runge—Lenz vector:

$$\bm A = \bm p \times \bm L - \mu k \hat{\bm r}.$$

This vector points along the major axis and its conservation explains why orbits in a $1/r$ potential are closed and do not precess.

Mathematical spine

$$\begin{aligned} V_{\rm eff}(r) &= V(r) + \frac{\ell^2}{2\mu r^2} & \text{(Effective Potential)} \\ \frac{d^2u}{d\phi^2} + u &= \frac{\mu k}{\ell^2} & \text{(Orbit Equation for $1/r^2$ force)} \\ r(\phi) &= \frac{\alpha}{1 + e\cos\phi} & \text{(Conic Section Solution)} \end{aligned}$$

Example: Circular Orbit. For a circular orbit at radius $r_0$, the force must equal the centripetal requirement: $|F(r_0)| = \frac{\mu v^2}{r_0} = \frac{\ell^2}{\mu r_0^3}$. This corresponds to the minimum of $V_{\rm eff}(r)$.

Worked example

Geostationary orbit.
A satellite orbits Earth at radius $r$ with period $T = 24\ \text{h} = 86400\ \text{s}$. For a circular orbit, centripetal force equals gravity: $mv^2/r = GMm/r^2$, so $v^2 = GM/r$. With $v = 2\pi r/T$, we get $(2\pi r/T)^2 = GM/r$, or $r^3 = GMT^2/(4\pi^2)$. Using $GM = gR_E^2 \approx 9.8\times(6.37\times10^6)^2 \approx 3.98\times10^{14}\ \text{m}^3/\text{s}^2$, we find $r^3 = 3.98\times10^{14}\times(86400)^2/(4\pi^2) \approx 7.54\times10^{22}\ \text{m}^3$, giving $r \approx 4.22\times10^7\ \text{m} \approx 6.6R_E$. The altitude is $h = r - R_E \approx 3.58\times10^4\ \text{km}$. The orbital speed is $v \approx 3.07\ \text{km/s}$.

Problems with Solutions

Problem 1. Calculate the escape velocity from the surface of Earth ($R_E = 6.37\times10^6\ \text{m}$, $g = 9.8\ \text{m/s}^2$).

Solution. Escape requires total energy $E \ge 0$. At the surface, $E = \frac{1}{2}mv^2 - GMm/R_E = 0$, so $v_{\rm esc} = \sqrt{2GM/R_E} = \sqrt{2gR_E} = \sqrt{2\times 9.8\times 6.37\times10^6} \approx \sqrt{1.25\times10^8} \approx 1.12\times10^4\ \text{m/s} = 11.2\ \text{km/s}$.

Problem 2. A comet has a parabolic orbit around the Sun ($M_\odot = 1.99\times10^{30}\ \text{kg}$) with perihelion distance $q = 0.1\ \text{AU}$ ($1\ \text{AU} = 1.496\times10^{11}\ \text{m}$). Find its speed at perihelion.

Solution. For a parabolic orbit, $E=0$ and $e=1$. At perihelion, $r = q$ and $v = v_{\rm max}$. Energy conservation: $\frac{1}{2}v^2 - GM_\odot/q = 0$, so $v = \sqrt{2GM_\odot/q} = \sqrt{2\times 6.67\times10^{-11}\times 1.99\times10^{30}/(0.1\times 1.496\times10^{11})} \approx \sqrt{1.775\times10^{10}} \approx 1.33\times10^5\ \text{m/s} = 133\ \text{km/s}$.

Problem 3. For an elliptical orbit with semi-major axis $a$ and period $T$, derive the relation $T^2 \propto a^3$ using energy and angular momentum conservation.

Solution. From the orbit equation, $r(\phi) = \alpha/(1+e\cos\phi)$ with $\alpha = \ell^2/(\mu k)$. The semi-major axis is $a = \alpha/(1-e^2)$. The energy is $E = -\mu k^2/(2\ell^2)(1-e^2) = -k/(2a)$. Thus $\ell^2 = \mu k a(1-e^2)$. The area swept per unit time is $dA/dt = \ell/(2\mu)$. The total area is $A = \pi a b = \pi a^2\sqrt{1-e^2}$. Integrating, $T = A/(dA/dt) = 2\mu\pi a^2\sqrt{1-e^2}/\ell = 2\mu\pi a^2\sqrt{1-e^2}/\sqrt{\mu k a(1-e^2)} = 2\pi\sqrt{\mu/k}\,a^{3/2}$. Therefore $T^2 = \frac{4\pi^2\mu}{k}a^3$, which is Kepler’s third law.

Section summary. The central force problem reduces to a 1D radial problem using conservation of angular momentum. For inverse-square forces, orbits are conic sections, consistent with Kepler’s Laws.

The Kinematics of Rigid Body Motion

Core ideas

A rigid body is a system of particles where the distance between any two particles remains constant: $|\bm r_i - \bm r_j| = c_{ij}$. It has 6 degrees of freedom (3 for translation of a reference point, 3 for rotation).

Chasles’ Theorem. Any displacement of a rigid body can be decomposed into a translation of a chosen base point plus a rotation about an axis through that point. Typically, the Center of Mass (CM) is chosen as the base point.

Angular Velocity and Rotations. The velocity of any point $P$ in the body is $\bm v = \bm V + \bm \omega \times \bm r$, where $\bm V$ is the velocity of the base point and $\bm \omega$ is the angular velocity vector. Rotations are often described using Euler angles $(\phi, \theta, \psi)$ which represent a sequence of three rotations (e.g., $z$-$x'$-$z''$ convention) to transform from a space-fixed frame to a body-fixed frame.

Why Euler angles in practice. Euler angles provide the minimal set of three independent generalized coordinates for orientation, so that rotational dynamics can be cast directly into Lagrange’s equations without redundant constraints. In the heavy symmetric top, for instance, the choice of $(\phi,\theta,\psi)$ makes $\phi$ and $\psi$ cyclic, immediately yielding two conserved momenta $p_\phi$ and $p_\psi$ and reducing the problem to a one-dimensional motion in $\theta$. The same coordinates are the standard yaw—pitch—roll variables used to describe spacecraft attitude, gyrocompasses, gimbal-mounted sensors, robotic end-effectors, and aircraft orientation.

In numerical practice one must remember the well-known gimbal-lock singularity at $\theta = 0,\pi$, where $\dot\phi$ and $\dot\psi$ become indistinguishable and the kinematic map $(\dot\phi,\dot\theta,\dot\psi)\to\bm\omega$ loses rank. For this reason attitude-control engineers often integrate the equivalent quaternion (Euler—Rodrigues) parameters and convert back to Euler angles only for human-readable output.

The Inertia Tensor. The inertia tensor $\bm I$ is a symmetric $3 \times 3$ matrix that relates angular momentum $\bm L$ to angular velocity $\bm \omega$: $\bm L = \bm I \bm \omega$. Its components in a given basis are:

$$I_{ij} = \sum_n m_n (r_n^2 \delta_{ij} - x_{n,i}x_{n,j}).$$

• Principal Axes: For any point, there exists a set of axes where $\bm I$ is diagonal. The diagonal elements $I_1, I_2, I_3$ are the principal moments of inertia.
• Parallel Axis Theorem: If $I_{\rm cm}$ is the inertia tensor about the CM, the tensor about an axis parallel but shifted by $\bm a$ is $I = I_{\rm cm} + M(a^2 \bm 1 - \bm a \otimes \bm a)$.

Mathematical spine

$$\begin{aligned} \bm v &= \bm V + \bm \omega \times \bm r & \text{(Velocity in Rigid Body)} \\ \bm L &= \bm I \bm \omega, \quad T_{\rm rot} = \frac{1}{2}\bm \omega \cdot \bm L = \frac{1}{2}\bm \omega \cdot \bm I \bm \omega & \text{(Angular Momentum and KE)} \\ I_{ij} &= \int \rho(\bm r) (r^2 \delta_{ij} - x_ix_j) d^3r & \text{(Continuum Inertia Tensor)} \end{aligned}$$

Example: Inertia Tensor of a Cube. For a uniform cube of side $a$ and mass $M$ about a corner, $I_{xx} = \frac{1}{3}Ma^2$, $I_{xy} = -\frac{1}{4}Ma^2$. About the center, the axes are principal and $I_{xx}=I_{yy}=I_{zz} = \frac{1}{6}Ma^2$.

Worked example

Velocity of a point on a rolling wheel.
A bicycle wheel of radius $R = 0.35\ \text{m}$ rolls without slipping at speed $V = 5\ \text{m/s}$. The angular velocity is $\omega = V/R = 5/0.35 \approx 14.3\ \text{rad/s}$. For a point on the rim at angle $\phi$ from the top, measured in the body frame, the position relative to the center is $\bm r = R(\sin\phi, -\cos\phi)$. The velocity is $\bm v = \bm V + \bm\omega\times\bm r = (V, 0) + (0,0,\omega)\times(R\sin\phi, -R\cos\phi, 0) = (V + \omega R\cos\phi, \omega R\sin\phi)$. At the top ($\phi=0$): $\bm v = (V+V, 0) = (10, 0)\ \text{m/s}$. At the bottom ($\phi=\pi$): $\bm v = (V-V, 0) = 0$, consistent with the no-slip condition.

Problems with Solutions

Problem 1. A uniform rod of mass $M = 2\ \text{kg}$ and length $L = 1\ \text{m}$ rotates about an axis through one end and perpendicular to the rod. Find the moment of inertia and the kinetic energy when $\omega = 3\ \text{rad/s}$.

Solution. For a rod about one end, $I = \frac{1}{3}ML^2 = \frac{1}{3}(2)(1)^2 = \frac{2}{3}\ \text{kg}\cdot\text{m}^2$. Rotational kinetic energy is $T = \frac{1}{2}I\omega^2 = \frac{1}{2}\times\frac{2}{3}\times 9 = 3\ \text{J}$.

Problem 2. A rigid body rotates with $\bm\omega = (2, 3, 1)\ \text{rad/s}$. Find the velocity of a point at $\bm r = (0.1, 0.2, 0)\ \text{m}$ relative to the rotation axis.

Solution. $\bm v = \bm\omega\times\bm r = (2,3,1)\times(0.1,0.2,0) = (3\cdot0 - 1\cdot0.2, 1\cdot0.1 - 2\cdot0, 2\cdot0.2 - 3\cdot0.1) = (-0.2, 0.1, 0.1)\ \text{m/s}$. The speed is $|\bm v| = \sqrt{0.04+0.01+0.01} = \sqrt{0.06} \approx 0.245\ \text{m/s}$.

Problem 3. Use the parallel axis theorem to find the moment of inertia of a solid disk of mass $M = 1\ \text{kg}$ and radius $R = 0.2\ \text{m}$ about a tangent in its plane.

Solution. About the center perpendicular to the disk, $I_{\rm cm} = \frac{1}{2}MR^2 = 0.5\times 1\times 0.04 = 0.02\ \text{kg}\cdot\text{m}^2$. For an axis in the plane, by symmetry $I_{x,\rm cm} = I_{y,\rm cm} = \frac{1}{4}MR^2 = 0.01\ \text{kg}\cdot\text{m}^2$. Shifting to a tangent point on the rim, the displacement is $a = R$ perpendicular to the new axis. The parallel axis theorem gives $I = I_{x,\rm cm} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2 = \frac{5}{4}(1)(0.04) = 0.05\ \text{kg}\cdot\text{m}^2$.

Section summary. Rigid body kinematics describes the 6-DOF motion through CM translation and rotation, characterized by the angular velocity vector and the inertia tensor.

The Rigid Body Equations of Motion

Core ideas

Dynamics in a rotating frame requires accounting for the rotation of the basis vectors. For any vector $\bm A$:

$$\left(\frac{d\bm A}{dt}\right)_{\rm space} = \left(\frac{d\bm A}{dt}\right)_{\rm body} + \bm\omega \times \bm A.$$

Euler’s Equations. Applying this to the angular momentum $\bm L$ in the body-fixed frame of principal axes:

$$\bm N = \dot{\bm L}_{\rm body} + \bm\omega \times \bm L.$$

This gives Euler’s equations:

$$\begin{aligned} I_1 \dot\omega_1 - (I_2 - I_3)\omega_2\omega_3 &= N_1 \\ I_2 \dot\omega_2 - (I_3 - I_1)\omega_3\omega_1 &= N_2 \\ I_3 \dot\omega_3 - (I_1 - I_2)\omega_1\omega_2 &= N_3 \end{aligned}$$

Torque-Free Motion. When $\bm N = 0$, both $E_{\rm rot} = \frac{1}{2}\sum I_i \omega_i^2$ and $L^2 = \sum I_i^2 \omega_i^2$ are conserved.

• Stability (Tennis Racket Theorem): Rotation about the principal axes with the largest ($I_{\rm max}$) or smallest ($I_{\rm min}$) moments is stable. Rotation about the intermediate axis is unstable.
• Symmetric Top: If $I_1 = I_2 \neq I_3$, the angular velocity $\bm \omega$ precesses around the body-fixed symmetry axis $z'$ with frequency $\Omega_{\rm body} = \frac{I_3 - I_1}{I_1}\omega_3$.

The Heavy Symmetric Top. A top with $I_1=I_2$ spinning in a gravitational field exhibits complex motion:

• Precession: The symmetry axis rotates around the vertical (gravity) axis.
• Nutation: The symmetry axis bobs up and down between two polar angles $\theta_1$ and $\theta_2$.
• Stability: A “sleeping top” (vertical spin) is stable only if $\omega_3 > \frac{2}{I_3}\sqrt{Mg l I_1}$.

Nutation amplitude and frequency (fast-spin limit). For a top released with non-zero tilt $\theta_0$ and large spin $\omega_3$ the energy/angular-momentum conservation reduces $\theta(t)$ to motion in an effective potential. Linearising about the mean angle one finds small-amplitude nutation at the angular frequency

$$\omega_{\rm nut} \;\approx\; \frac{I_3\,\omega_3}{I_1},$$

i.e.\ the nutation is fast compared with the precession $\omega_{\rm prec}\approx Mgl/(I_3\omega_3)$, and the two satisfy the simple product

$$\omega_{\rm nut}\,\omega_{\rm prec} \;\approx\; \frac{Mgl}{I_1}.$$

The peak-to-peak amplitude of the wobble for a top released from rest at $\theta_0$ is

$$\Delta\theta \;=\; \theta_2 - \theta_1 \;\approx\; \frac{2 Mgl\sin\theta_0}{I_3\,\omega_3^2}\,\frac{I_1}{I_3},$$

which vanishes as $\omega_3^{-2}$ --- a fast-spinning top exhibits steady precession with imperceptible nutation. The sleeping-top threshold above is the same statement at $\theta_0=0$: nutation about the vertical is bounded only when $I_3^2\omega_3^2 > 4 Mgl I_1$.

Mathematical spine

$$\begin{aligned} I_1\dot\omega_1 - (I_2 - I_3)\omega_2\omega_3 &= N_1 & \text{(Euler's Equations)} \\ \omega_{\rm prec} &= \frac{Mgl}{L} & \text{(Slow Precession approximation)} \\ E &= \frac{1}{2}I_1(\dot\theta^2 + \dot\phi^2\sin^2\theta) + \frac{1}{2}I_3(\dot\phi\cos\theta + \dot\psi)^2 + Mgl\cos\theta & \text{(Heavy Top Energy)} \end{aligned}$$

Example: Free Precession of the Earth. The Earth is slightly oblate ($I_3 > I_1$). Euler’s equations predict a precession of the spin axis (the “Chandler wobble”) with a period related to the difference in moments of inertia.

Worked example

Stability of a spinning book.
A hardcover book has principal moments $I_1 = 0.05\ \text{kg}\cdot\text{m}^2$, $I_2 = 0.08\ \text{kg}\cdot\text{m}^2$, $I_3 = 0.10\ \text{kg}\cdot\text{m}^2$ (with $I_1 < I_2 < I_3$). According to the tennis-racket theorem, rotation about the $I_1$ (thickness) and $I_3$ (height) axes is stable, while rotation about $I_2$ (width) is unstable. If the book is spun about its intermediate axis with a small perturbation, Euler’s equations give $\dot\omega_1 = \frac{I_2-I_3}{I_1}\omega_2\omega_3 = \frac{-0.02}{0.05}\omega_2\omega_3 = -0.4\omega_2\omega_3$. Linearizing about $\omega_2 \approx \Omega$ shows $\ddot\omega_1 \propto -\Omega^2\omega_1$ with a positive coefficient (since $(I_2-I_3)(I_1-I_2)/(I_1I_3) > 0$), leading to exponential growth: $\omega_1(t) \propto e^{\lambda t}$ with $\lambda = \Omega\sqrt{(I_3-I_2)(I_2-I_1)/(I_1I_3)}$.

Problems with Solutions

Problem 1. A symmetric top has $I_1 = I_2 = 5.0\times10^{-4}\ \text{kg}\cdot\text{m}^2$ and $I_3 = 1.0\times10^{-3}\ \text{kg}\cdot\text{m}^2$. It spins at $\omega_3 = 100\ \text{rad/s}$. Find the body-frame precession rate $\Omega_{\rm body}$.

Solution. For a torque-free symmetric top, $\Omega_{\rm body} = \frac{I_3-I_1}{I_1}\omega_3 = \frac{1.0\times10^{-3} - 5.0\times10^{-4}}{5.0\times10^{-4}}\times 100 = \frac{0.5\times10^{-3}}{5.0\times10^{-4}}\times 100 = 1\times 100 = 100\ \text{rad/s}$. The angular velocity vector precesses around the symmetry axis at $100\ \text{rad/s}$.

Problem 2. A top of mass $M = 0.2\ \text{kg}$ with $I_1 = I_2 = 1.5\times10^{-4}\ \text{kg}\cdot\text{m}^2$ and $I_3 = 3.0\times10^{-4}\ \text{kg}\cdot\text{m}^2$ has its CM a distance $l = 0.03\ \text{m}$ from the pivot. What minimum spin $\omega_3$ is needed for the sleeping-top configuration to be stable?

Solution. The sleeping-top stability condition is $\omega_3 > \frac{2}{I_3}\sqrt{MglI_1}$. Substituting: $\sqrt{MglI_1} = \sqrt{0.2\times 9.8\times 0.03\times 1.5\times10^{-4}} = \sqrt{8.82\times10^{-6}} \approx 2.97\times10^{-3}$. Then $\omega_3 > \frac{2}{3.0\times10^{-4}}\times 2.97\times10^{-3} = \frac{5.94\times10^{-3}}{3.0\times10^{-4}} \approx 19.8\ \text{rad/s}$ (about $3.2$ rev/s).

Problem 3. A cube of side $a = 0.1\ \text{m}$ and mass $M = 0.5\ \text{kg}$ rotates about an axis through its center and perpendicular to one face. Find its angular momentum if $\omega = 20\ \text{rad/s}$.

Solution. For a cube about a central axis perpendicular to a face, $I = \frac{1}{6}Ma^2 = \frac{1}{6}(0.5)(0.1)^2 = \frac{0.005}{6} \approx 8.33\times10^{-4}\ \text{kg}\cdot\text{m}^2$. Angular momentum is $L = I\omega = 8.33\times10^{-4}\times 20 \approx 1.67\times10^{-2}\ \text{kg}\cdot\text{m}^2/\text{s}$.

Section summary. Euler’s equations describe rigid body dynamics in the body frame, revealing the stability of principal axis rotation and the precession/nutation of tops.

Oscillations

Core ideas

Most systems in classical mechanics behave like harmonic oscillators when displaced slightly from a stable equilibrium point.

Small Oscillations around Equilibrium. Consider a system with generalized coordinates $q_i$ and a potential $V(q_1, \dots, q_n)$. Equilibrium occurs where $\frac{\partial V}{\partial q_i} = 0$. If this is a minimum (all eigenvalues of the Hessian $\frac{\partial^2 V}{\partial q_i \partial q_j}$ are positive), the equilibrium is stable. Defining $\eta_i = q_i - q_{i0}$ as the displacement, the Lagrangian for small oscillations is:

$$L \approx \frac{1}{2} \sum_{i,j} (T_{ij} \dot\eta_i \dot\eta_j - V_{ij} \eta_i \eta_j),$$

where $V_{ij} = \left.\frac{\partial^2 V}{\partial q_i \partial q_j}\right|_0$ and $T_{ij}$ is the kinetic energy matrix (usually constant for small displacements).

Normal Modes and the Secular Equation. The equations of motion are $T\ddot{\bm\eta} + V\bm\eta = 0$. Assuming a sinusoidal solution $\bm\eta(t) = \bm a e^{i\omega t}$ leads to the generalized eigenvalue problem:

$$(V - \omega^2 T) \bm a = 0.$$

Nontrivial solutions exist only if the secular equation holds:

$$\det(V - \omega^2 T) = 0.$$

The roots $\omega_k^2$ are the normal frequencies, and the corresponding vectors $\bm a_k$ define the normal modes.

Normal Coordinates. There exists a linear transformation $\bm \eta = \bm A \bm \zeta$ that simultaneously diagonalizes $T$ and $V$. The new coordinates $\zeta_k$ are normal coordinates, and each evolves independently as a simple harmonic oscillator: $\ddot\zeta_k + \omega_k^2 \zeta_k = 0$.

Damping and Driving. Real systems have dissipation ($Q$-factor) and external forces.

• Damping: $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$. Solutions can be underdamped, overdamped, or critically damped.
• Resonance: When driven by $F_0\cos\omega t$, the amplitude is maximized near $\omega \approx \omega_0$. The phase shift $\delta$ changes from 0 to $\pi$ as $\omega$ passes through resonance.

Mathematical spine

$$\begin{aligned} \det(V - \omega^2 T) &= 0 & \text{(Secular Equation)} \\ L &= \frac{1}{2} (\dot{\bm\zeta}^2 - \bm\omega^2 \bm\zeta^2) & \text{(Lagrangian in Normal Coordinates)} \\ Q &= \frac{\omega_0}{2\gamma} & \text{(Quality Factor)} \end{aligned}$$

Example: Two Coupled Pendulums. For two identical pendulums coupled by a spring, there are two modes:

1. In-phase: $\omega_1 = \sqrt{g/l}$ (spring not stretched).
2. Out-of-phase: $\omega_2 = \sqrt{g/l + 2k/m}$ (spring acts as additional restoring force).

Worked example

Driven damped harmonic oscillator.
A mass $m = 0.5\ \text{kg}$ on a spring with $k = 50\ \text{N/m}$ is damped with $b = 1.0\ \text{kg/s}$ and driven by $F(t) = 2\cos(10t)\ \text{N}$. The natural frequency is $\omega_0 = \sqrt{k/m} = 10\ \text{rad/s}$, the damping coefficient is $\gamma = b/(2m) = 1.0\ \text{s}^{-1}$, and the driving frequency is $\omega = 10\ \text{rad/s}$. The steady-state amplitude is $A = F_0/m / \sqrt{(\omega_0^2-\omega^2)^2 + (2\gamma\omega)^2} = 4 / \sqrt{0 + 400} = 4/20 = 0.20\ \text{m}$. Since $\omega = \omega_0$, the system is at resonance and the phase lag is $\delta = \pi/2$. The quality factor is $Q = \omega_0/(2\gamma) = 5$.

Problems with Solutions

Problem 1. Two identical masses $m$ are connected by springs of constant $k$ to each other and to fixed walls. Find the normal-mode frequencies.

Solution. Let displacements be $x_1, x_2$. The equations are $m\ddot x_1 = -kx_1 + k(x_2-x_1) = -2kx_1 + kx_2$ and $m\ddot x_2 = -k(x_2-x_1) - kx_2 = kx_1 - 2kx_2$. Assuming $x_i = A_i\cos\omega t$, the matrix equation is $\begin{pmatrix} 2k-m\omega^2 & -k \\ -k & 2k-m\omega^2 \end{pmatrix}\begin{pmatrix} A_1 \\ A_2 \end{pmatrix} = 0$. The secular equation gives $(2k-m\omega^2)^2 - k^2 = 0$, so $2k-m\omega^2 = \pm k$. Thus $\omega_1^2 = k/m$ (symmetric mode: $A_1=A_2$) and $\omega_2^2 = 3k/m$ (antisymmetric mode: $A_1=-A_2$).

Problem 2. A damped oscillator has $\omega_0 = 5\ \text{rad/s}$ and $\gamma = 3\ \text{s}^{-1}$. Is it underdamped, critically damped, or overdamped? Write the general solution.

Solution. Compare $\gamma$ to $\omega_0$: $\gamma = 3 < 5 = \omega_0$, so the system is underdamped. The damped frequency is $\omega' = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{25-9} = 4\ \text{rad/s}$. The general solution is $x(t) = Ae^{-3t}\cos(4t + \phi)$, where $A$ and $\phi$ are determined by initial conditions.

Problem 3. An undamped oscillator with $m=1\ \text{kg}$, $k=100\ \text{N/m}$ is driven by $F(t) = 5\cos(9t)\ \text{N}$. Find the steady-state amplitude and the average power supplied by the driving force.

Solution. Here $\omega_0 = 10\ \text{rad/s}$ and $\omega = 9\ \text{rad/s}$. The amplitude is $A = F_0/m / |\omega_0^2 - \omega^2| = 5/|100-81| = 5/19 \approx 0.263\ \text{m}$. In steady state, $x(t) = A\cos(9t)$ and $v(t) = -9A\sin(9t)$. The instantaneous power is $P = Fv = -45A\cos(9t)\sin(9t)$, which averages to zero over a cycle because there is no damping and therefore no net energy dissipation.

Section summary. Oscillation theory linearizes motion near equilibrium, reducing complex coupled dynamics to independent normal modes via an eigenvalue problem.

The Classical Mechanics of the Special Theory of Relativity

Core ideas

Special Relativity modifies Newtonian mechanics for high speeds ($v \sim c$), based on the constancy of the speed of light $c$ in all inertial frames.

Lorentz Transformations and Four-Vectors. The transformation between two inertial frames moving at relative velocity $v$ along $x$ is:

$$x' = \gamma(x - vt), \quad t' = \gamma(t - vx/c^2), \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}.$$

Events are points in Minkowski spacetime described by four-vectors $x^\mu = (ct, \bm r)$. The invariant interval is $ds^2 = c^2dt^2 - d\bm r^2$.

Relativistic Dynamics. The proper time $\tau$ is the time measured in the particle’s rest frame: $d\tau = dt/\gamma$.

• Four-velocity: $u^\mu = \frac{dx^\mu}{d\tau} = \gamma(c, \bm v)$.
• Four-momentum: $p^\mu = mu^\mu = (E/c, \bm p)$, where $\bm p = \gamma m \bm v$ and $E = \gamma mc^2$.
• Energy-Momentum Relation: The norm of the four-momentum is invariant: $p^\mu p_\mu = (E/c)^2 - p^2 = m^2c^2$, leading to $E^2 = p^2c^2 + m^2c^4$.

Relativistic Lagrangian. The action for a free particle is proportional to its proper time (the “longest” path in Minkowski space):

$$S = -mc \int ds = \int (-mc^2\sqrt{1 - v^2/c^2}) dt.$$

The free particle Lagrangian is $L = -mc^2\sqrt{1 - v^2/c^2}$. For a particle in an electromagnetic field:

$$L = -mc^2\sqrt{1 - v^2/c^2} - q\phi + q\bm A \cdot \bm v.$$

Mathematical spine

$$\begin{aligned} p^\mu &= (E/c, \bm p) & \text{(Four-momentum)} \\ E^2 &= p^2c^2 + m^2c^4 & \text{(Energy-momentum relation)} \\ \bm f &= \frac{d\bm p}{dt} = \frac{d}{dt}(\gamma m \bm v) & \text{(Relativistic force/Minkowski force)} \end{aligned}$$

Example: Relativistic Doppler Effect. The frequency shift for a source moving away is $\nu = \nu_0 \sqrt{\frac{1-v/c}{1+v/c}}$. Unlike the classical case, there is also a transverse Doppler effect ($\nu = \nu_0/\gamma$) due to time dilation.

Worked example

Muon decay and time dilation.
Cosmic-ray muons are created at an altitude of about $h = 10\ \text{km}$ with speed $v = 0.995c$ ($\gamma \approx 10$). In the muon’s rest frame, its mean lifetime is $\tau_0 = 2.2\ \mu\text{s}$. In Earth’s frame, the dilated lifetime is $\tau = \gamma\tau_0 \approx 22\ \mu\text{s}$, during which the muon travels $d = v\tau \approx 0.995c \times 22\ \mu\text{s} \approx 6.6\ \text{km}$. However, muons reach the ground! This is resolved by length contraction: in the muon frame, the atmosphere is contracted to $h' = h/\gamma \approx 1\ \text{km}$, which takes $t' = h'/v \approx 3.4\ \mu\text{s}$ to traverse---less than the mean lifetime. Both frames consistently predict that many muons survive to sea level.

Problems with Solutions

Problem 1. An electron is accelerated to a kinetic energy of $1.0\ \text{MeV}$. Given $m_e c^2 = 0.511\ \text{MeV}$, find its speed and momentum.

Solution. Total energy $E = K + mc^2 = 1.511\ \text{MeV}$. Since $E = \gamma mc^2$, we have $\gamma = 1.511/0.511 \approx 2.957$. Then $v = c\sqrt{1-1/\gamma^2} \approx c\sqrt{1-0.114} \approx 0.941c$. Momentum is $p = \sqrt{E^2 - (mc^2)^2}/c = \sqrt{1.511^2 - 0.511^2}/c \approx \sqrt{2.283 - 0.261}/c \approx 1.423\ \text{MeV}/c$.

Problem 2. A spaceship travels to a star $4.3\ \text{light-years}$ away at $v = 0.8c$. How long does the trip take according to Earth clocks and according to the ship’s clock?

Solution. Earth time: $t = d/v = 4.3/0.8 = 5.375\ \text{years}$. Ship time (proper time): $\tau = t/\gamma = t\sqrt{1-0.8^2} = 5.375 \times 0.6 = 3.225\ \text{years}$. Alternatively, in the ship frame the distance is contracted to $d' = d/\gamma = 4.3 \times 0.6 = 2.58\ \text{ly}$, so $\tau = d'/v = 2.58/0.8 = 3.225\ \text{years}$.

Problem 3. Show that the relativistic kinetic energy $K = (\gamma - 1)mc^2$ reduces to the classical expression $\frac{1}{2}mv^2$ when $v \ll c$.

Solution. For small $x$, $(1-x)^{-1/2} \approx 1 + x/2 + 3x^2/8 + \dots$. With $x = v^2/c^2$, $\gamma \approx 1 + \frac{1}{2}v^2/c^2 + \frac{3}{8}v^4/c^4 + \dots$. Thus $K = mc^2(\gamma - 1) \approx mc^2(\frac{1}{2}v^2/c^2 + \frac{3}{8}v^4/c^4) = \frac{1}{2}mv^2 + \frac{3}{8}mv^4/c^2 + \dots$. The first term is the classical kinetic energy; the rest are relativistic corrections.

Section summary. Relativistic mechanics replaces absolute time with proper time and Newtonian momentum with four-momentum, unified by the invariant mass-shell condition.

The Hamilton Equations of Motion

Core ideas

Hamiltonian mechanics is a reformulation of classical mechanics that emphasizes the symmetry between coordinates $q_i$ and momenta $p_i$. It describes motion as a flow in phase space.

The Legendre Transformation. The Hamiltonian $H(q, p, t)$ is obtained from the Lagrangian $L(q, \dot q, t)$ via a Legendre transformation:

$$H(q, p, t) = \sum_i p_i \dot q_i - L(q, \dot q, t), \quad p_i = \frac{\partial L}{\partial \dot q_i}.$$

If $L = T - V$ and $T$ is a homogeneous quadratic function of $\dot q$, then $H = T + V = E$ (the total energy).

Hamilton’s Equations. Taking the differential of $H$ and using Euler—Lagrange equations leads to Hamilton’s canonical equations:

$$\dot q_i = \frac{\partial H}{\partial p_i}, \quad \dot p_i = -\frac{\partial H}{\partial q_i}.$$

These are $2n$ first-order differential equations, unlike the $n$ second-order Euler—Lagrange equations.

Poisson Brackets. For any two functions $f(q, p, t)$ and $g(q, p, t)$ on phase space, the Poisson bracket is:

$$\{f, g\} = \sum_i \left( \frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q_i} \right).$$

The time evolution of any function $f$ is given by:

$$\frac{df}{dt} = \{f, H\} + \frac{\partial f}{\partial t}.$$

A quantity is conserved if $\{f, H\} = 0$ (and it has no explicit time dependence).

Phase Space and Liouville’s Theorem. A state is a point in the $2n$-dimensional phase space. Liouville’s Theorem states that the density of points in phase space (or the volume of a region of points) is constant along the trajectories of the system.

Mathematical spine

$$\begin{aligned} H(q, p, t) &= \sum p_i \dot q_i - L & \text{(Hamiltonian)} \\ \dot q_i = \frac{\partial H}{\partial p_i}, \quad &\dot p_i = -\frac{\partial H}{\partial q_i} & \text{(Hamilton's Equations)} \\ \dot f = \{f, H\} &+ \frac{\partial f}{\partial t} & \text{(Evolution equation)} \end{aligned}$$

Example: Harmonic Oscillator in Phase Space. $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2$. Hamilton’s equations: $\dot q = p/m$, $\dot p = -m\omega^2 q$. The trajectories in phase space are ellipses.

Worked example

Particle in a uniform gravitational field.
A mass $m = 2\ \text{kg}$ falls vertically in a uniform gravitational field $g = 9.8\ \text{m/s}^2$. Using coordinate $q = y$ (height) and momentum $p = m\dot y$, the Hamiltonian is $H = p^2/(2m) + mgy$. Hamilton’s equations give $\dot y = \partial H/\partial p = p/m$ and $\dot p = -\partial H/\partial y = -mg$. Thus $\ddot y = \dot p/m = -g$, as expected from Newton’s second law. If released from rest at $y_0 = 10\ \text{m}$, then $p(0)=0$ and $y(t) = y_0 - \frac{1}{2}gt^2$, hitting the ground at $t = \sqrt{2y_0/g} \approx 1.43\ \text{s}$ with momentum $p = -mgt \approx -28.0\ \text{kg}\cdot\text{m/s}$.

Problems with Solutions

Problem 1. A 1D harmonic oscillator has Hamiltonian $H = p^2/(2m) + \frac{1}{2}m\omega^2 q^2$. Write Hamilton’s equations and show that they reproduce the familiar second-order equation of motion.

Solution. $\dot q = \partial H/\partial p = p/m$ and $\dot p = -\partial H/\partial q = -m\omega^2 q$. Differentiating the first and substituting the second gives $\ddot q = \dot p/m = -\omega^2 q$, which is the standard harmonic oscillator equation.

Problem 2. For $H = p^2/(2m) - \alpha q$ (a particle in a constant force field), find the time evolution of $q(t)$ and $p(t)$ given $q(0)=0$ and $p(0)=p_0$.

Solution. Hamilton’s equations: $\dot q = p/m$, $\dot p = \alpha$. Integrating, $p(t) = p_0 + \alpha t$ and $q(t) = \int_0^t (p_0 + \alpha t')/m\,dt' = p_0 t/m + \alpha t^2/(2m)$. This is constant-acceleration motion with $a = \alpha/m$.

Problem 3. Compute the Poisson bracket $\{L_z, x\}$ where $L_z = xp_y - yp_x$.

Solution. Using $\{f,g\} = \sum_i(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i})$, only the terms involving $y$ and $p_y$ survive because $x$ depends only on $x$ and $p_x$. We get $\{L_z, x\} = \frac{\partial(xp_y-yp_x)}{\partial y}\frac{\partial x}{\partial p_y} - \frac{\partial(xp_y-yp_x)}{\partial p_y}\frac{\partial x}{\partial y} + \dots = (-p_x)(0) - (x)(0) + \dots$. Actually computing all terms: $\{xp_y-yp_x, x\} = \{xp_y, x\} - \{yp_x, x\}$. The first bracket is $x\{p_y,x\} + p_y\{x,x\} = 0$. The second is $y\{p_x,x\} + p_x\{y,x\} = y(-1) + 0 = -y$. Thus $\{L_z,x\} = 0 - (-y) = y$. (Equivalently, from the standard relation $\{L_i, x_j\} = \epsilon_{ijk}x_k$, we have $\{L_3, x_1\} = \epsilon_{312}x_2 = y$.)

Section summary. Hamiltonian mechanics treats coordinates and momenta as independent variables in phase space, providing a powerful framework for conservation laws and a bridge to quantum mechanics.

Canonical Transformations

Core ideas

A transformation from old coordinates $(q, p)$ to new coordinates $(Q, P)$ is canonical if it preserves the form of Hamilton’s equations. This is equivalent to preserving the Poisson bracket relations:

$$\{Q_i, Q_j\}_{q,p} = 0, \quad \{P_i, P_j\}_{q,p} = 0, \quad \{Q_i, P_j\}_{q,p} = \delta_{ij}.$$

The Symplectic Condition. Defining $\bm \eta = (q_1, \dots, q_n, p_1, \dots, p_n)^T$, Hamilton’s equations can be written as $\dot{\bm\eta} = \bm J \nabla H$, where $\bm J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ is the symplectic matrix. A transformation $\bm\eta \to \bm\xi$ is canonical if the Jacobian $\bm M = \partial \bm\xi / \partial \bm\eta$ satisfies the symplectic condition:

$$\bm M \bm J \bm M^T = \bm J.$$

Generating Functions. Canonical transformations are often derived from a generating function. There are four basic types depending on the choice of independent variables:

• $F_1(q, Q, t)$: $p_i = \frac{\partial F_1}{\partial q_i}, \quad P_i = -\frac{\partial F_1}{\partial Q_i}$.
• $F_2(q, P, t)$: $p_i = \frac{\partial F_2}{\partial q_i}, \quad Q_i = \frac{\partial F_2}{\partial P_i}$.
• $F_3(p, Q, t)$: $q_i = -\frac{\partial F_3}{\partial p_i}, \quad P_i = -\frac{\partial F_3}{\partial Q_i}$.
• $F_4(p, P, t)$: $q_i = -\frac{\partial F_4}{\partial p_i}, \quad Q_i = \frac{\partial F_4}{\partial P_i}$.

The new Hamiltonian $K$ is related to the old $H$ by $K = H + \frac{\partial F}{\partial t}$.

Goal of Canonical Transformations. The primary use is to transform to a system where all coordinates are cyclic ($\partial K / \partial Q_i = 0$). In such a system, the new momenta $P_i$ are constants of motion, and the problem is trivially solved.

Mathematical spine

$$\begin{aligned} \{Q_i, P_j\} &= \delta_{ij} & \text{(Preservation of Poisson Brackets)} \\ \bm M \bm J \bm M^T &= \bm J & \text{(Symplectic Condition)} \\ K &= H + \frac{\partial F}{\partial t} & \text{(Transformation of the Hamiltonian)} \end{aligned}$$

Example: Identity Transformation. The generating function $F_2(q, P) = \sum q_i P_i$ gives $p_i = P_i$ and $Q_i = q_i$, which is the identity transformation. Changing this to $F_2 = \sum f_i(q) P_i$ generates a point transformation.

Worked example

Transformation to center-of-mass and relative coordinates.
For two particles in 1D with masses $m_1, m_2$, coordinates $q_1, q_2$ and momenta $p_1, p_2$, define $Q_1 = (m_1q_1 + m_2q_2)/(m_1+m_2)$ (CM position), $Q_2 = q_1 - q_2$ (relative coordinate), $P_1 = p_1 + p_2$ (total momentum), and $P_2 = (m_2p_1 - m_1p_2)/(m_1+m_2)$ (relative momentum). The generating function is $F_2 = P_1(m_1q_1+m_2q_2)/(m_1+m_2) + P_2(q_1-q_2)$. One verifies $\{Q_i,P_j\} = \delta_{ij}$ and $\{Q_i,Q_j\} = \{P_i,P_j\} = 0$, confirming the transformation is canonical.

Problems with Solutions

Problem 1. Show that the transformation $Q = \ln(1+\sqrt{q}\cos p)$, $P = 2(1+\sqrt{q}\cos p)\sqrt{q}\sin p$ is canonical by evaluating the Poisson bracket $\{Q,P\}_{q,p}$.

Solution. We need $\{Q,P\} = \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}$. Let $u = 1+\sqrt{q}\cos p$. Then $Q = \ln u$, $P = 2u\sqrt{q}\sin p$. Computing: $\partial Q/\partial q = \cos p/(2\sqrt{q}u)$, $\partial Q/\partial p = -\sqrt{q}\sin p/u$, $\partial P/\partial q = \sin p\cos p/\sqrt{q} + u\sin p/\sqrt{q} = \sin p(1+2\sqrt{q}\cos p)/\sqrt{q}$, $\partial P/\partial p = 2u\sqrt{q}\cos p - 2q\sin^2 p$. After simplification, $\{Q,P\} = 1$, confirming the transformation is canonical.

Problem 2. Find a type-2 generating function $F_2(q,P)$ that produces the point transformation $Q = q^2$ for a 1D system.

Solution. For a point transformation, $p = \partial F_2/\partial q$ and $Q = \partial F_2/\partial P = q^2$. Integrating the second equation, $F_2(q,P) = q^2 P + g(q)$. Then $p = \partial F_2/\partial q = 2qP + g'(q)$. The simplest choice is $g=0$, giving $F_2 = q^2 P$ and $p = 2qP$, or $P = p/(2q)$.

Problem 3. Given the Hamiltonian $H = \frac{p^2}{2} + \frac{q^2}{2}$, show that the transformation $Q = p$, $P = -q$ is canonical and find the new Hamiltonian.

Solution. The Poisson brackets are $\{Q,P\} = \{p,-q\} = -\{p,q\} = 1$, $\{Q,Q\} = \{p,p\} = 0$, $\{P,P\} = \{-q,-q\} = 0$. Thus the transformation is canonical. Since the transformation is time-independent, $K = H$ expressed in new variables: $K = \frac{P^2}{2} + \frac{Q^2}{2}$. This is just a rotation by $90^\circ$ in phase space.

Section summary. Canonical transformations preserve the symplectic structure of phase space, allowing for coordinate changes that simplify the Hamiltonian and reveal constants of motion.

Hamilton—Jacobi Theory and Action-Angle Coordinates

Core ideas

Hamilton—Jacobi (HJ) theory is the ultimate canonical transformation: it seeks a transformation to a frame where the new Hamiltonian is zero, meaning all new coordinates and momenta are constants of motion.

The Hamilton—Jacobi Equation. Using an $F_2(q, P, t)$ generating function, we call the result Hamilton’s Principal Function $S(q, P, t)$. If the new Hamiltonian $K = H + \partial S / \partial t = 0$, then $P_i = \alpha_i$ (constants). The HJ equation is a first-order, non-linear partial differential equation for $S$:

$$H\left(q_1, \dots, q_n, \frac{\partial S}{\partial q_1}, \dots, \frac{\partial S}{\partial q_n}, t\right) + \frac{\partial S}{\partial t} = 0.$$

For time-independent Hamiltonians, we use Hamilton’s Characteristic Function $W$, where $S(q, \alpha, t) = W(q, \alpha) - Et$. Then:

$$H\left(q, \frac{\partial W}{\partial q}\right) = E.$$

Separation of Variables. The HJ equation is often solved by separation of variables: $W(q_1, \dots, q_n) = \sum W_i(q_i)$. This reduces the PDE to $n$ independent ODEs, which can be solved by integration (quadratures).

Action—Angle Variables. For periodic systems, action—angle variables $(J, \theta)$ are the most natural coordinates.

• Action variable $J_i = \frac{1}{2\pi} \oint p_i dq_i$, where the integral is over one cycle of the motion.
• Angle variable $\theta_i$ is the canonical conjugate to $J_i$.
• The Hamiltonian depends only on actions: $H = H(J_1, \dots, J_n)$.
• The equations of motion are trivial: $\dot J_i = 0$, $\dot \theta_i = \frac{\partial H}{\partial J_i} = \nu_i$ (the constant frequency).

Mathematical spine

$$\begin{aligned} H(q, \partial S / \partial q, t) + \partial S / \partial t &= 0 & \text{(Hamilton--Jacobi Equation)} \\ J_i &= \frac{1}{2\pi} \oint p_i dq_i & \text{(Action Variable)} \\ \theta_i(t) &= \nu_i t + \beta_i & \text{(Angle Variable evolution)} \end{aligned}$$

Example: Harmonic Oscillator. $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2 = E$. The action is $J = E/\omega$. The frequency is $\nu = \partial H / \partial J = \omega$. The angle $\theta$ is the phase of the oscillation.

Worked example

Particle in a 1D box via Hamilton—Jacobi.
A particle of mass $m$ moves freely on the interval $[0,L]$ with infinite walls. The Hamiltonian is $H = p^2/(2m) = E$. The HJ equation is $(dW/dx)^2/(2m) = E$, so $dW/dx = \pm\sqrt{2mE}$. Integrating, $W(x) = \pm x\sqrt{2mE}$. The action variable is $J = \frac{1}{2\pi}\oint p\,dx = \frac{1}{2\pi}(2L\sqrt{2mE}) = \frac{L}{\pi}\sqrt{2mE}$. Inverting gives $E = \frac{\pi^2 J^2}{2mL^2}$. The frequency is $\nu = \partial E/\partial J = \frac{\pi^2 J}{mL^2} = \frac{\pi}{L}\sqrt{\frac{2E}{m}} = \frac{v}{2L}$, which is the round-trip frequency as expected.

Problems with Solutions

Problem 1. For a 1D harmonic oscillator with $H = p^2/(2m) + \frac{1}{2}m\omega^2 q^2$, solve the time-independent HJ equation and find Hamilton’s characteristic function $W(q,E)$.

Solution. The HJ equation is $\frac{1}{2m}(\partial W/\partial q)^2 + \frac{1}{2}m\omega^2 q^2 = E$. Solving for $\partial W/\partial q = \sqrt{2mE - m^2\omega^2 q^2}$, we integrate to get $W(q,E) = \int^q \sqrt{2mE - m^2\omega^2 q'^2}\,dq'$. This evaluates to $W = \frac{m\omega}{2}q\sqrt{\frac{2E}{m\omega^2}-q^2} + E\arcsin\left(q\sqrt{\frac{m\omega^2}{2E}}\right)$.

Problem 2. A particle moves in a central potential $V(r) = -k/r$. Using action—angle variables, show that the radial and azimuthal frequencies are equal, confirming that the orbit is closed.

Solution. The actions are $J_\phi = \oint p_\phi\,d\phi = 2\pi\ell$ and $J_r = \oint p_r\,dr$. For the Kepler problem, evaluating the radial action gives $J_r = -J_\phi + \pi k\sqrt{\mu/(-2E)}$. The total energy depends on $J_r + J_\phi$ as $E = -\frac{\mu k^2}{2(J_r+J_\phi)^2}$. The frequencies are $\nu_r = \partial E/\partial J_r$ and $\nu_\phi = \partial E/\partial J_\phi$. Since $E$ depends only on the sum $J_r+J_\phi$, $\nu_r = \nu_\phi$. Equal frequencies mean the radial and angular motions complete a cycle in the same time, producing a closed ellipse.

Problem 3. Find the action variable $J$ for a simple pendulum of length $l$ and mass $m$ in the small-angle approximation.

Solution. For small angles, $H = \frac{p_\theta^2}{2ml^2} + \frac{1}{2}mgl\theta^2 = E$. The motion is harmonic with $\omega = \sqrt{g/l}$ and $E = \frac{1}{2}mgl\theta_0^2$. The phase-space trajectory is an ellipse with semi-axes $p_{\rm max} = ml^2\omega\theta_0 = \sqrt{2mE}\,l$ and $\theta_{\rm max} = \theta_0 = \sqrt{2E/(mgl)}$. The area is $\pi p_{\rm max}\theta_{\rm max} = \pi\sqrt{2mE}\,l \cdot \sqrt{2E/(mgl)} = 2\pi E\sqrt{l/g} = 2\pi E/\omega$. Thus $J = \frac{1}{2\pi}\times(2\pi E/\omega) = E/\omega$.

Section summary. Hamilton—Jacobi theory reduces dynamics to a single PDE, while action-angle variables provide the most efficient description for integrable periodic systems.

Classical Chaos

Core ideas

Chaos refers to the complex, unpredictable behavior that can arise in deterministic nonlinear dynamical systems. It is not due to noise or randomness but is an inherent property of the system’s geometry.

Sensitivity to Initial Conditions. The hallmark of chaos is that two trajectories starting very close together in phase space will diverge exponentially:

$$|\delta x(t)| \approx |\delta x(0)| e^{\lambda t},$$

where $\lambda$ is the Lyapunov exponent. A positive $\lambda$ implies that small uncertainties in measurement grow rapidly, making long-term prediction impossible.

Phase Space and Attractors.

• Phase Portrait: A map of all possible states in phase space. Regular systems have trajectories on circles or tori.
• Strange Attractor: For dissipative chaotic systems, trajectories settle onto a complex, fractal-like structure in phase space.
• Poincaré Section: A way to simplify the analysis by taking a “snapshot” of the system’s state each time it crosses a chosen surface in phase space. This reduces the continuous dynamics to a discrete map.

The KAM Theorem. The Kolmogorov—Arnold—Moser (KAM) theorem addresses what happens to an integrable system when a small nonlinear perturbation is added. It states that many of the original “invariant tori” (regular orbits) survive the perturbation, but they are increasingly destroyed as the perturbation strength increases, leading to “stochastic” or chaotic regions.

Routes to Chaos. Systems often become chaotic through a sequence of bifurcations as a parameter is varied. A common route is period-doubling, where the period of the oscillation doubles repeatedly until it becomes infinite (chaos).

Mathematical spine

$$\begin{aligned} \lambda &= \lim_{t\to\infty} \frac{1}{t} \ln \frac{|\delta x(t)|}{|\delta x(0)|} & \text{(Lyapunov Exponent)} \\ x_{n+1} &= r x_n (1 - x_n) & \text{(Logistic Map - simplest chaos example)} \end{aligned}$$

Concrete example: period-doubling cascade in the logistic map. The discrete map $x_{n+1} = r x_n(1 - x_n)$ on $x\in[0,1]$ shows the period-doubling route to chaos as the parameter $r$ is increased:

• $r < 1$: the only stable fixed point is $x^*=0$ (extinction).
• $1 < r < 3$: a single non-trivial fixed point $x^* = 1 - 1/r$ is stable.
• $r_1 = 3$: first period-doubling bifurcation; a stable 2-cycle appears.
• $r_2 \approx 3.4495$: bifurcation to a stable 4-cycle.
• $r_3 \approx 3.5441$: 8-cycle.
• $r_4 \approx 3.5644$: 16-cycle, and so on.
• $r_\infty \approx 3.56995$: accumulation point --- onset of chaos.

The successive intervals shrink geometrically, and their ratios approach the universal first Feigenbaum constant

$$\delta \;=\; \lim_{n\to\infty} \frac{r_n - r_{n-1}}{r_{n+1} - r_n} \;=\; 4.669\,201\,609\ldots$$

The width of the bifurcating branches scales by a second universal constant $\alpha \approx 2.5029$. Feigenbaum showed in 1978 that $\delta$ and $\alpha$ are independent of the specific map: any smooth one-dimensional map with a single quadratic maximum --- a driven pendulum, a forced oscillator, even a dripping faucet --- displays the same numbers, providing the first experimentally verified case of universality in classical chaos.

Example: Driven Damped Pendulum. A simple pendulum with friction and a periodic driving force. For small drive, it oscillates regularly. As the drive increases, it can undergo period-doubling and eventually move chaotically, never repeating its path.

Worked example

Lyapunov exponent of the logistic map.
For the logistic map $x_{n+1} = r x_n(1-x_n)$ with $r=4$, the Lyapunov exponent is $\lambda = \lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}\ln|f'(x_n)|$ where $f'(x) = 4(1-2x)$. Starting from $x_0 = 0.2$, iteration gives $x_1 = 0.64$, $x_2 = 0.9216$, etc. Numerically averaging $\ln|4(1-2x_n)|$ over $N=10^4$ iterations yields $\lambda \approx \ln 2 \approx 0.693$. Since $\lambda > 0$, the dynamics is chaotic: a tiny initial difference $\delta_0$ grows as $|\delta_n| \approx \delta_0 e^{n\lambda}$, so after $n=10$ iterations the uncertainty has grown by a factor of about $e^{6.93} \approx 10^3$.

Problems with Solutions

Problem 1. Find the fixed points of the logistic map $x_{n+1} = r x_n(1-x_n)$ and determine their stability for $0 < r < 3$.

Solution. Fixed points satisfy $x^* = rx^*(1-x^*)$, giving $x^*=0$ or $x^* = 1-1/r$. Stability requires $|f'(x^*)| < 1$ where $f'(x) = r(1-2x)$. At $x^*=0$, $|f'(0)| = |r|$. For $0<r<1$, this is stable; for $r>1$, unstable. At $x^*=1-1/r$, $|f'(x^*)| = |r(1-2(1-1/r))| = |2-r|$. This is stable when $|2-r|<1$, i.e., $1<r<3$.

Problem 2. A driven damped pendulum obeys $\ddot\theta + 0.5\dot\theta + \sin\theta = 1.5\cos(0.7t)$. Explain why the Poincar’e section is useful and describe what it would show for periodic versus chaotic motion.

Solution. The Poincar’e section records $(\theta, \dot\theta)$ once per drive period $T = 2\pi/0.7$. For periodic motion with period equal to the drive, the section shows a single fixed point. For period-doubling, it shows 2, 4, 8, … points. For chaotic motion, the points never repeat and fill a fractal structure (strange attractor), revealing the underlying order in the chaotic flow.

Problem 3. Estimate the Lyapunov exponent for a system where two nearby trajectories diverge by a factor of $10^6$ in time $t = 10\ \text{s}$.

Solution. Using $|\delta(t)| \approx |\delta(0)|e^{\lambda t}$, we have $10^6 = e^{\lambda(10)}$, so $\lambda = \frac{\ln(10^6)}{10} = \frac{6\ln 10}{10} \approx 1.38\ \text{s}^{-1}$. This positive value indicates chaotic dynamics.

Section summary. Chaos demonstrates that deterministic laws can lead to unpredictable behavior through exponential sensitivity to initial conditions, often visualized through strange attractors and Poincaré sections.

Canonical Perturbation Theory

Core ideas

Many realistic systems are “nearly integrable,” meaning they can be described as an integrable system plus a small perturbation parameterized by a dimensionless number $\epsilon \ll 1$:

$$H(J, \theta) = H_0(J) + \epsilon H_1(J, \theta),$$

where $J=(J_1,\ldots,J_n)$ and $\theta=(\theta_1,\ldots,\theta_n)$ are the action—angle variables of the unperturbed problem.

Time-independent perturbation theory. The goal is to find a near-identity canonical transformation to new variables $(\bar J, \bar \theta)$ such that the new Hamiltonian $K(\bar J)$ depends only on $\bar J$. One uses a type-2 generating function

$$F_2(J,\bar\theta) = \sum_i J_i \bar\theta_i + \epsilon\, W(J,\bar\theta),$$

in which $W$ is the unknown “correction” to the identity. Expanding order by order in $\epsilon$ and demanding that the angle-dependent pieces cancel, the first-order shift in the Hamiltonian is simply the angle-average of the perturbation,

$$K_1(\bar J) = \langle H_1(\bar J, \theta) \rangle_\theta = \frac{1}{(2\pi)^n} \int_0^{2\pi} H_1(\bar J, \theta)\, d^n\theta,$$

and the generating-function correction takes the Fourier form

$$W(\bar J, \bar\theta) = \sum_{\bm n \neq 0} \frac{i\, h_{\bm n}(\bar J)}{\bm n \cdot \bm\omega(\bar J)}\, e^{i \bm n \cdot \bar\theta}, \qquad \bm\omega = \partial H_0/\partial \bar J,$$

where $h_{\bm n}$ are the Fourier coefficients of $H_1$ and $\bm n\in\mathbb{Z}^n\setminus\{0\}$.

Secular terms and the Lindstedt—Poincar’e fix

Naive expansion in $\epsilon$ tends to produce secular terms: terms that grow without bound as a power of $t$ and so spoil the validity of the series after a time of order $1/\epsilon$. The standard cure is the Lindstedt—Poincar’e method, which absorbs the would-be secular contribution into a renormalization of the oscillation frequency.

Worked example: Duffing oscillator. Consider the weakly anharmonic equation

$$\ddot x + x + \epsilon\, x^3 = 0, \qquad x(0)=A,\ \dot x(0)=0.$$

A naive expansion $x = x_0 + \epsilon x_1 + \cdots$ gives $x_0 = A\cos t$ and an inhomogeneous equation for $x_1$ whose driving term contains $\cos t$, producing the resonant secular response

$$x_1(t) \supset -\tfrac{3}{8} A^3\, t \sin t,$$

which diverges with time. To remove it, rescale time by $\tau = \omega t$ with $\omega = 1 + \epsilon\omega_1 + \epsilon^2\omega_2 + \cdots$. The equation becomes

$$\omega^2 x'' + x + \epsilon x^3 = 0,$$

with $x' = dx/d\tau$. Choosing $\omega_1$ to kill the resonant term gives the uniformly valid first-order approximation

$$\omega = 1 + \tfrac{3}{8}\epsilon A^2 + \mathcal{O}(\epsilon^2), \qquad x(t) \approx A\cos(\omega t).$$

The amplitude-dependent frequency shift is the hallmark of nonlinear oscillation.

Resonances and the pendulum normal form

Small denominators. The denominators $\bm n\cdot\bm\omega$ in $W$ become arbitrarily small whenever the unperturbed frequencies are nearly commensurate, $\bm n\cdot\bm\omega \approx 0$. The series then diverges --- the celebrated “problem of small divisors” in celestial mechanics, partially tamed by KAM theory for sufficiently irrational frequency ratios.

Worked resonant example. Consider two degrees of freedom near a $p\!:\!q$ resonance, $p\omega_1 - q\omega_2 \approx 0$, with perturbation

$$H = H_0(J_1,J_2) + \epsilon\, h\, \cos(p\theta_1 - q\theta_2).$$

Average over the fast (non-resonant) angles but retain the slow resonant angle $\psi = p\theta_1 - q\theta_2$. Introducing the canonical transformation generated by $F_2 = \psi P + \theta_2 P_2$ (so that $J_1 = pP$, $J_2 = -qP + P_2$) gives an effective one-degree-of-freedom Hamiltonian

$$H_{\text{eff}}(P,\psi) = H_0(pP, -qP+P_2) + \epsilon h\cos\psi.$$

Expanding $H_0$ about the resonant action $P_*$ where $p\omega_1 = q\omega_2$, one obtains the pendulum form

$$H_{\text{eff}} \approx \tfrac{1}{2} M (P - P_*)^2 + \epsilon h \cos\psi, \qquad M \equiv \frac{\partial^2 H_0}{\partial P^2}\bigg|_{P_*}.$$

Resonant motion thus librates like a pendulum with small-oscillation frequency $\Omega = \sqrt{|M|\,\epsilon h}$ and a separatrix of width $\Delta P \sim \sqrt{\epsilon h/|M|}$ --- the characteristic “resonance island.”

Adiabatic invariants

Concept. If a parameter $\lambda(t)$ of the Hamiltonian changes on a timescale much longer than the orbital period $T$, i.e.\ $\dot\lambda/\lambda \ll 1/T$, then the action

$$J = \frac{1}{2\pi}\oint p\, dq$$

is conserved up to exponentially small corrections in the slowness parameter. $J$ is therefore called an adiabatic invariant.

Worked example: harmonic oscillator with slowly varying frequency. For $H = \tfrac{1}{2}p^2 + \tfrac{1}{2}\omega(t)^2 q^2$ the action of an instantaneous orbit at energy $E$ is the area of a phase-space ellipse with semi-axes $\sqrt{2E}$ and $\sqrt{2E}/\omega$,

$$J = \frac{E}{\omega(t)}.$$

Adiabatic invariance of $J$ therefore implies

$$E(t) = \omega(t)\, J = \omega(t)\, \frac{E_0}{\omega_0},$$

so the energy tracks the frequency: a pendulum whose string is slowly shortened gains energy in proportion to its raised frequency. Historically this is the result Einstein invoked at the 1911 Solvay conference and that Ehrenfest promoted to the quantization rule $J = n\hbar$ of the old quantum theory.

Mathematical spine

$$\begin{aligned} H &= H_0(J) + \epsilon H_1(J,\theta) & \text{(perturbed Hamiltonian)} \\ K_1 &= \langle H_1 \rangle_\theta & \text{(first-order energy shift)} \\ \omega &= \omega_0 + \tfrac{3}{8}\epsilon A^2 + \cdots & \text{(Lindstedt frequency, Duffing)} \\ H_{\text{res}} &\approx \tfrac{1}{2} M\, \delta P^2 + \epsilon h\cos\psi & \text{(pendulum near resonance)} \\ J &= \tfrac{1}{2\pi}\oint p\,dq \approx \text{const} & \text{(adiabatic invariant)} \end{aligned}$$

Astrophysical aside: precession of Mercury’s perihelion. General relativity adds an effective $1/r^3$ correction to the Newtonian $1/r$ potential. Treating it as $\epsilon H_1$ and applying the angle-average prescription yields the celebrated $43''$/century perihelion advance.

Worked example

Anharmonic oscillator frequency shift.
Consider a particle of mass $m=1\ \text{kg}$ in a potential $V(x) = \frac{1}{2}x^2 + \epsilon x^4$ (in units where $m=\omega_0=1$) with $\epsilon = 0.01$ and amplitude $A = 0.5\ \text{m}$. Using the Lindstedt—Poincar’e method, the first-order frequency correction is $\omega \approx 1 + \frac{3}{8}\epsilon A^2 = 1 + \frac{3}{8}(0.01)(0.25) = 1 + 0.0009375$. The period shifts from $T_0 = 2\pi\ \text{s}$ to $T \approx 2\pi/1.00094 \approx 6.280\ \text{s}$, a decrease of about $6\ \text{ms}$.

Problems with Solutions

Problem 1. A harmonic oscillator has Hamiltonian $H_0 = \frac{1}{2}(p^2 + q^2)$. A perturbation $H_1 = q^4$ is added with small parameter $\epsilon$. Find the first-order energy shift for a state with action $J$.

Solution. In action—angle variables, $q = \sqrt{2J}\sin\theta$ and the unperturbed frequency is $\omega_0=1$. The first-order shift is the angle average $\langle H_1\rangle = \frac{1}{2\pi}\int_0^{2\pi}(2J)^2\sin^4\theta\, d\theta = \frac{4J^2}{2\pi}\cdot\frac{3\pi}{4} = 3J^2$. Thus $K = J + 3\epsilon J^2$ and the corrected frequency is $\omega = \partial K/\partial J = 1 + 6\epsilon J$.

Problem 2. A pendulum of length $l=1\ \text{m}$ and mass $m=0.5\ \text{kg}$ has its length shortened slowly from $l_0=1\ \text{m}$ to $l_1=0.5\ \text{m}$ over 100 periods. If the initial amplitude is $\theta_0 = 0.1\ \text{rad}$, find the final amplitude and energy.

Solution. For small oscillations, $\omega = \sqrt{g/l}$ and the adiabatic invariant is $J = E/\omega$. Since $E \propto \omega^2 A^2$ for a harmonic oscillator, $J \propto \omega A^2$, so $A^2\omega = \text{const}$ and $A \propto \omega^{-1/2} \propto l^{1/4}$. Thus $A_1 = A_0(l_1/l_0)^{1/4} = 0.1\times(0.5)^{1/4} \approx 0.084\ \text{rad}$. The energy is $E = \frac{1}{2}mgl\theta^2$ (small-angle), so $E_1/E_0 = (l_1/l_0)(\theta_1/\theta_0)^2 = 0.5\times\sqrt{0.5} \approx 0.354$. With $E_0 = \frac{1}{2}(0.5)(9.8)(1)(0.1)^2 \approx 0.0245\ \text{J}$, we get $E_1 \approx 8.7\times10^{-3}\ \text{J}$.

Problem 3. Consider the Duffing equation $\ddot x + x + \epsilon x^3 = 0$ with $x(0)=A$ and $\dot x(0)=0$. Show that the naive perturbation expansion produces a secular term and explain how the Lindstedt—Poincar’e method removes it.

Solution. Writing $x = x_0 + \epsilon x_1 + \dots$, the zeroth order gives $x_0 = A\cos t$. The first-order equation is $\ddot x_1 + x_1 = -x_0^3 = -A^3\cos^3 t = -\frac{3}{4}A^3\cos t - \frac{1}{4}A^3\cos 3t$. The $\cos t$ term drives the harmonic oscillator at resonance, giving a particular solution $x_1 \supset -\frac{3}{8}A^3 t\sin t$, which grows without bound (secular). To fix this, rescale time as $\tau = \omega t$ with $\omega = 1 + \epsilon\omega_1 + \dots$. The equation becomes $\omega^2 x'' + x + \epsilon x^3 = 0$. At $O(\epsilon)$, choosing $\omega_1 = \frac{3}{8}A^2$ cancels the resonant term, leaving only bounded oscillations at the corrected frequency.

Section summary. Canonical perturbation theory systematically organizes the effects of small departures from integrability. Secular terms in the naive series are removed by frequency renormalization (Lindstedt—Poincar’e); near a resonance the same machinery reduces the dynamics locally to a pendulum; and adiabatic invariants such as $E/\omega$ for the slowly varying oscillator capture the robust quantities that survive slow parameter drift.

Introduction to the Lagrangian and Hamiltonian Formulations for Continuous Systems and Fields

Core ideas

Classical mechanics can be extended to systems with an infinite number of degrees of freedom, such as fluids, elastic solids, and electromagnetic fields. These are described by fields $\phi(x, y, z, t)$.

Lagrangian Density. The total Lagrangian $L$ is the spatial integral of a Lagrangian density $\mathcal{L}$:

$$L = \int \mathcal{L}(\phi, \partial_\mu \phi, x^\mu) d^3x.$$

The action is $S = \int L dt = \int \mathcal{L} d^4x$. Hamilton’s Principle ($\delta S = 0$) leads to the Euler—Lagrange equations for fields:

$$\frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \right) = 0,$$

where $\partial_\mu = (\frac{1}{c}\partial_t, \nabla)$ is the four-gradient.

Hamiltonian Density. The conjugate momentum density is $\pi = \frac{\partial \mathcal{L}}{\partial \dot \phi}$. The Hamiltonian density is:

$$\mathcal{H} = \pi \dot \phi - \mathcal{L}.$$

The total Hamiltonian $H = \int \mathcal{H} d^3x$ gives the total energy of the field.

Noether’s Theorem and the Stress-Energy Tensor. Symmetries of the Lagrangian density lead to conserved currents.

• Internal symmetries lead to conserved charges (like electric charge).
• Spacetime symmetries lead to the conservation of the Stress—Energy Tensor $T^{\mu\nu}$:

$$T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial^\nu \phi - g^{\mu\nu} \mathcal{L}.$$

Conservation $\partial_\mu T^{\mu\nu} = 0$ implies conservation of energy and momentum.

Mathematical spine

$$\begin{aligned} \delta \int \mathcal{L} d^4x &= 0 \implies \frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} = 0 & \text{(Field Equations)} \\ \mathcal{H} &= \pi \dot \phi - \mathcal{L} & \text{(Hamiltonian Density)} \\ T^{00} &= \mathcal{H} & \text{(Energy Density)} \end{aligned}$$

Example: The Vibrating String. For a string with tension $\tau$ and linear density $\rho$, $\mathcal{L} = \frac{1}{2}\rho \dot y^2 - \frac{1}{2}\tau (\partial_x y)^2$. The Euler—Lagrange equation yields the wave equation: $\rho \ddot y - \tau y'' = 0$.

Example: The electromagnetic field. The free Maxwell field is described by the gauge potential $A^\mu = (\phi/c, \bm A)$ and the antisymmetric field-strength tensor $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. Its Lorentz-invariant Lagrangian density is

$$\mathcal{L}_{\rm EM} \;=\; -\tfrac{1}{4\mu_0}\, F_{\mu\nu}F^{\mu\nu} \;-\; J^\mu A_\mu,$$

with $J^\mu = (c\rho, \bm J)$ the four-current. The Euler—Lagrange equations applied to $A_\mu$ reproduce the inhomogeneous Maxwell equations $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$, while the homogeneous pair follows automatically from the antisymmetry of $F_{\mu\nu}$. The associated stress—energy tensor $T^{\mu\nu}$ reproduces the energy density $\tfrac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0)$ and the Poynting momentum density $\bm E\times\bm B/(\mu_0 c^2)$.

Example: Non-linear waves --- the sine-Gordon kink. A celebrated nonlinear field theory in $1{+}1$ dimensions is the sine-Gordon model with Lagrangian density

$$\mathcal{L}_{\rm sG} \;=\; \tfrac{1}{2}(\partial_t\phi)^2 - \tfrac{1}{2}c^2(\partial_x\phi)^2 - \frac{m^2 c^4}{\beta^2}\bigl[1 - \cos(\beta\phi)\bigr],$$

which yields the equation of motion $\partial_t^2\phi - c^2\partial_x^2\phi + (m^2 c^4/\beta)\sin(\beta\phi) = 0$. Despite being nonlinear, it admits an exact static kink (soliton) solution interpolating between adjacent vacua $\phi=0$ and $\phi = 2\pi/\beta$:

$$\phi_K(x) \;=\; \frac{4}{\beta}\arctan\!\bigl[\exp(mc\, x)\bigr], \qquad E_{\rm kink} \;=\; \frac{8 m c^3}{\beta^2}.$$

Boosting it gives a localised travelling wave that retains its shape after collisions --- a hallmark of the integrability of the sine-Gordon system. The same equation describes mechanical chains of coupled pendulums and Josephson-junction transmission lines, illustrating how classical field theory unifies wave propagation in seemingly disparate physical systems.

Worked example

Vibrating string with fixed ends.
A uniform string of length $L = 1.0\ \text{m}$, linear density $\rho = 5.0\times10^{-3}\ \text{kg/m}$, and tension $\tau = 20\ \text{N}$ is fixed at both ends. The Lagrangian density is $\mathcal{L} = \frac{1}{2}\rho \dot y^2 - \frac{1}{2}\tau (y')^2$. The wave equation is $\rho \ddot y = \tau y''$, giving wave speed $c = \sqrt{\tau/\rho} = \sqrt{20/0.005} = 63.2\ \text{m/s}$. For fixed ends, normal modes are $y_n(x,t) = A_n \sin(n\pi x/L)\cos(\omega_n t)$ with frequencies $\omega_n = n\pi c/L = n\times 198.7\ \text{rad/s}$ ($n=1,2,\dots$). The fundamental frequency is $f_1 = \omega_1/(2\pi) \approx 31.6\ \text{Hz}$.

Problems with Solutions

Problem 1. A flexible membrane with surface tension $\sigma$ and surface mass density $\mu$ has transverse displacement $z(x,y,t)$. Write the Lagrangian density and derive the wave equation.

Solution. The kinetic energy density is $\frac{1}{2}\mu \dot z^2$ and the potential energy density from stretching is $\frac{1}{2}\sigma[(\partial_x z)^2 + (\partial_y z)^2]$. Thus $\mathcal{L} = \frac{1}{2}\mu \dot z^2 - \frac{1}{2}\sigma[(\partial_x z)^2 + (\partial_y z)^2]$. The Euler—Lagrange equation gives $\mu \ddot z = \sigma(\partial_x^2 z + \partial_y^2 z)$, or $\ddot z = c^2 \nabla^2 z$ with $c^2 = \sigma/\mu$.

Problem 2. For a real scalar field $\phi$ with Lagrangian density $\mathcal{L} = \frac{1}{2}(\partial_\mu \phi)(\partial^\mu \phi) - \frac{1}{2}m^2\phi^2$, find the equation of motion and the Hamiltonian density.

Solution. Using the Euler—Lagrange equation: $\partial_\mu(\partial\mathcal{L}/\partial(\partial_\mu\phi)) = \partial_\mu\partial^\mu\phi = \Box\phi$, and $\partial\mathcal{L}/\partial\phi = -m^2\phi$. The equation of motion is the Klein—Gordon equation: $(\Box + m^2)\phi = 0$. The conjugate momentum is $\pi = \partial\mathcal{L}/\partial\dot\phi = \dot\phi$. The Hamiltonian density is $\mathcal{H} = \pi\dot\phi - \mathcal{L} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2$.

Problem 3. A string of length $L$ has displacement $y(x,t) = A\sin(kx)\cos(\omega t)$. If $y(0,t)=0$ and $y(L,t)=0$, find the allowed values of $k$ and the corresponding frequencies when $c = 100\ \text{m/s}$ and $L = 0.5\ \text{m}$.

Solution. The boundary condition at $x=0$ is satisfied automatically. At $x=L$, $\sin(kL)=0 \implies kL = n\pi$, so $k_n = n\pi/L = 2n\pi\ \text{m}^{-1}$. The dispersion relation is $\omega = ck$, so $\omega_n = c k_n = 200n\pi\ \text{rad/s}$ and $f_n = \omega_n/(2\pi) = 100n\ \text{Hz}$.

Section summary. Field theory generalizes discrete mechanics to continuous media, using Lagrangian densities and local field equations, providing the classical foundation for Electromagnetism and Quantum Field Theory.