Classical Mechanics: Lagrangian and Hamiltonian Dynamics, Central Forces, Normal Modes, Rigid Bodies and Special Relativity
1. D’Alembert, Lagrange, Hamilton’s principle and the calculus of variations
D’Alembert’s principle states that the applied forces minus the "inertial forces" do no work in any virtual displacement consistent with the constraints: Σ(Fᵢ − ṗᵢ)·δrᵢ = 0. Holonomic constraints (expressible as f(r, t) = 0) reduce 3N coordinates to n generalised coordinates qⱼ, and the ideal constraint forces, doing no virtual work, drop out. The result is the Euler–Lagrange equation d/dt(∂L/∂q̇ⱼ) − ∂L/∂qⱼ = 0 with L = T − V. For a plane pendulum, L = ½ml²θ̇² + mgl cos θ, and the equation is ml²θ̈ + mgl sin θ = 0 — the tension never appears.
Hamilton’s principle reaches the same equations from the other end: the actual path between fixed end-points makes the action S = ∫L dt stationary, δS = 0. This is one case of the calculus of variations: a functional J = ∫f(y, y′, x) dx is stationary when d/dx(∂f/∂y′) − ∂f/∂y = 0. With f = √(1 + y′²) (arc length), ∂f/∂y = 0 gives y′ = constant: the shortest path in a plane is a straight line. When f has no explicit x, the Beltrami identity f − y′∂f/∂y′ = constant gives a first integral directly — the route to the brachistochrone, a cycloid.
2. Symmetry and conservation laws
The generalised momentum conjugate to qⱼ is pⱼ = ∂L/∂q̇ⱼ. If L does not contain qⱼ — the coordinate is cyclic — then ṗⱼ = ∂L/∂qⱼ = 0 and pⱼ is conserved. This is Noether’s theorem in its simplest form: every continuous symmetry of L gives a conserved quantity.
| Symmetry of L | Conserved quantity | Example |
|---|---|---|
| Translation in x | Linear momentum pₓ | free particle; x is cyclic |
| Rotation about an axis | Angular momentum about it | central force: p_θ = mr²θ̇ |
| No explicit time dependence | Energy function h = Σpq̇ − L | h = T + V when T is quadratic in q̇ and constraints are fixed |
3. Central force motion and the Kepler problem
Two bodies interacting through V(r) reduce to one body of reduced mass μ = m₁m₂/(m₁ + m₂). The motion is planar, the angular momentum l = μr²θ̇ is conserved (equal areas in equal times), and the radial motion is one-dimensional in the effective potential V_eff = V(r) + l²/(2μr²). The orbit equation, with u = 1/r, is d²u/dθ² + u = −(μ/(l²u²)) F(1/u). A circular orbit sits at the minimum of V_eff; for a power law F = −krⁿ it is stable only for n > −3.
- Kepler orbits. For V = −k/r the orbit is a conic, r = p/(1 + e cos θ). e < 1 is an ellipse (E < 0), e = 1 a parabola (E = 0), e > 1 a hyperbola (E > 0). For an ellipse r_min = a(1 − e) and r_max = a(1 + e), so e = (r_max − r_min)/(r_max + r_min).
- Energy and period. E = −k/(2a), depending on the semi-major axis alone. Kepler’s third law is T² = 4π²a³/[G(m₁ + m₂)], so doubling a multiplies the period by 23/2 = 2.83.
- Speeds and the virial theorem. Conservation of l gives v_peri r_min = v_apo r_max. For 1/r potentials the time averages obey ⟨T⟩ = −½⟨V⟩, so on a circular orbit E = −T.
4. Small oscillations, coupled oscillators and normal modes
Expand about a stable equilibrium: T = ½ q̇ᵀMq̇ and V = ½ qᵀKq with constant symmetric matrices. Trying q = a eiωt gives (K − ω²M)a = 0, and non-trivial motion requires det(K − ω²M) = 0. Its roots are the squares of the normal-mode frequencies; each eigenvector is a pattern in which every coordinate oscillates at the same frequency. Any small motion is a superposition of normal modes, and normal coordinates make T and V simultaneously diagonal.
| System | Normal-mode frequencies | Mode shapes |
|---|---|---|
| wall–k–m–k–m–k–wall | √(k/m), √(3k/m) | in phase (1, 1); out of phase (1, −1) |
| two pendula (l, m) joined by a spring k | √(g/l), √(g/l + 2k/m) | spring unstretched; spring working |
| linear triatomic m–k–M–k–m | 0, √(k/m), √[(k/m)(1 + 2m/M)] | translation; symmetric stretch (M still); antisymmetric stretch |
5. Rigid bodies: inertia tensor, Euler angles and the torque-free symmetric top
Angular momentum and angular velocity are related by the inertia tensor, L = Iω, with Iᵢⱼ = Σ m(r²δᵢⱼ − xᵢxⱼ): diagonal entries are moments of inertia, e.g. Iₓₓ = Σm(y² + z²), and off-diagonal entries are products of inertia, Iₓᵧ = −Σmxy. Worked: 1 kg at (1, 1, 0) m and 2 kg at (0, 1, 1) m give Iₓₓ = 1(1) + 2(1 + 1) = 5 kg m² and Iₓᵧ = −(1·1·1 + 2·0·1) = −1 kg m². The tensor is real and symmetric, so an orthogonal transformation to principal axes diagonalises it; there, T = ½(I₁ω₁² + I₂ω₂² + I₃ω₃²).
The orientation of the body frame is fixed by three Euler angles — φ about the space z-axis, θ about the new x-axis (the line of nodes), ψ about the body z-axis — a product of three rotation matrices, each orthogonal with determinant +1. In the principal frame the torque-free motion obeys Euler’s equations, I₁ω̇₁ = (I₂ − I₃)ω₂ω₃ and cyclically. For a symmetric top (I₁ = I₂ ≠ I₃), ω₃ is constant and the perpendicular part of ω rotates about the symmetry axis in the body frame at Ω = (I₃ − I₁)ω₃/I₁; seen from space, the symmetry axis and ω precess about the fixed L. Kinetic energy and L are constant throughout.
6. Hamilton’s equations, canonical transformations and Poisson brackets
The Hamiltonian is the Legendre transform H(q, p, t) = Σpq̇ − L, written in q and p. For L = ½mq̇² − ½kq², p = mq̇ and H = p²/(2m) + ½kq². Hamilton’s equations q̇ = ∂H/∂p, ṗ = −∂H/∂q replace n second-order equations with 2n first-order ones in phase space, and dH/dt = ∂H/∂t, so H is conserved when it has no explicit time.
- Poisson bracket {f, g} = Σ(∂f/∂q ∂g/∂p − ∂f/∂p ∂g/∂q). Then df/dt = {f, H} + ∂f/∂t, so f is a constant of motion when it has zero bracket with H and no explicit time.
- Fundamental brackets {qᵢ, pⱼ} = δᵢⱼ, {qᵢ, qⱼ} = {pᵢ, pⱼ} = 0; for angular momentum {Lₓ, L_y} = L_z and cyclically. Quantum mechanics replaces { , } with [ , ]/(iħ).
- Canonical transformation (q, p) → (Q, P) preserves the form of Hamilton’s equations; the test is {Q, P}q,p = 1. Q = p, P = −q passes; Q = p, P = q gives {Q, P} = −1 and fails. Generating functions such as F₂(q, P) produce them systematically.
7. Special relativity: Lorentz transformations, kinematics and E = mc²
For frames in standard configuration with relative velocity v along x, the Lorentz transformation is x′ = γ(x − vt), t′ = γ(t − vx/c²), with γ = 1/√(1 − v²/c²). It keeps the interval c²t² − x² − y² − z² invariant. Its consequences: a moving clock runs slow (Δt = γΔτ, time dilation); a moving rod is shorter along the motion (L = L₀/γ, length contraction); events simultaneous in one frame are not in another; and velocities add as u = (u′ + v)/(1 + u′v/c²), so 0.5c on 0.5c is 0.8c.
The four-momentum (E/c, p) has invariant length: E² = p²c² + m²c⁴, with E = γmc², p = γmv, kinetic energy K = (γ − 1)mc² and rest energy mc². For an electron (mc² = 0.511 MeV) with K = 1.022 MeV, γ = 3 and v = √(1 − 1/9) c = 0.943c. The invariant mass of a system is √(E_tot² − p_tot²c²)/c², the same in every frame — the quantity that sets thresholds: two 1 MeV photons colliding head-on have invariant mass 2 MeV/c², enough to create an electron–positron pair (1.022 MeV/c²).
Key takeaways
- L = T − V and d/dt(∂L/∂q̇) = ∂L/∂q; ideal constraint forces never appear. The same equation extremises any functional ∫f(y, y′, x) dx.
- A cyclic coordinate conserves its momentum; time-independent L conserves h, which equals T + V only for fixed constraints.
- Kepler: E = −k/2a, T² ∝ a³, e = (r_max − r_min)/(r_max + r_min), and on a circular orbit E = −T.
- Normal modes solve det(K − ω²M) = 0; guess the in-phase and out-of-phase patterns from symmetry first.
- Canonical means {Q, P} = 1; Poisson brackets give df/dt = {f, H}. Relativistic kinematics runs on E² = p²c² + m²c⁴ and the invariant mass.
Practice questions (15)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A particle moves in a plane under a central potential V(r), with L = ½m(ṙ² + r²θ̇²) − V(r). Which quantity is conserved because θ is cyclic?
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Answer: A — mr²θ̇
θ does not appear in L, so p_θ = ∂L/∂θ̇ = mr²θ̇ is conserved — the angular momentum. mṙ is the radial momentum, which changes because L depends on r; mrθ̇ is a transverse linear momentum, not a conjugate momentum.The functional J = ∫√(1 + y′²) dx between two fixed points in a plane is stationary for:
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Answer: A — A straight line
f has no y, so ∂f/∂y′ = y′/√(1 + y′²) is constant, which makes y′ constant: a straight line, the shortest path. The cycloid extremises the brachistochrone time ∫√[(1 + y′²)/y] dx, and the catenary the potential energy of a hanging chain.For L = ½mq̇² − ½kq², the Hamiltonian is:
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Answer: A — p²/(2m) + ½kq²
p = ∂L/∂q̇ = mq̇, so H = pq̇ − L = p²/m − (p²/(2m) − ½kq²) = p²/(2m) + ½kq². ½mq̇² + ½kq² has the right value but is written in q̇, not p, so it is not the Hamiltonian; p²/m − ½kq² forgets to subtract L’s kinetic term.Which transformation from (q, p) to (Q, P) is canonical?
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Answer: A — Q = p, P = −q
{Q, P} = ∂Q/∂q ∂P/∂p − ∂Q/∂p ∂P/∂q. For Q = p, P = −q this is 0 − (1)(−1) = 1: canonical. Q = p, P = q gives −1; Q = 2q, P = 2p gives 4; Q = q², P = p² gives 4qp — none equals 1.Two 1 kg masses on a frictionless line are joined wall–spring–mass–spring–mass–spring–wall by three identical springs of 100 N/m. The higher normal-mode angular frequency, in rad/s, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 17.32
K = [[2k, −k], [−k, 2k]] with m = 1: eigenvalues k and 3k, so ω = √(k/m) = 10 and √(3k/m) = √300 = 17.32 rad/s. The out-of-phase mode stretches the middle spring twice, so each mass feels k + 2k. √(2k/m) = 14.14 ignores the coupling.A linear symmetric triatomic molecule O–C–O is modelled as masses 16, 12 and 16 (in u) joined by two identical springs. The ratio of the antisymmetric-stretch frequency to the symmetric-stretch frequency, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.91
Symmetric stretch: C at rest, ω₁ = √(k/m). Antisymmetric: ω₃ = √[(k/m)(1 + 2m/M)]. Ratio = √(1 + 2·16/12) = √3.667 = 1.91. Using M/m instead of m/M gives √(1 + 1.5) = 1.58.Satellite B orbits the same planet as satellite A with twice A’s semi-major axis. The ratio of B’s period to A’s, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.83
T² ∝ a³ gives T_B/T_A = 23/2 = 2.83. The ratio does not depend on the eccentricities or on the satellites’ masses (both ≪ the planet’s). T ∝ a would give 2, and T ∝ a² would give 4.A body in an inverse-square attractive field has closest and farthest distances from the centre of force of 1 and 3 (same units). The eccentricity of the orbit is ____.
Numerical answer — type the value.
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Answer: 0.5
r_min = a(1 − e) and r_max = a(1 + e), so e = (3 − 1)/(3 + 1) = 0.5 and a = 2. (r_max − r_min)/r_max = 0.667 and r_min/r_max = 0.333 are the tempting wrong ratios.For a particle in a circular orbit under an inverse-square attractive force, the total energy E and kinetic energy T are related by:
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Answer: A — E = −T
The virial theorem for V ∝ 1/r gives ⟨T⟩ = −½⟨V⟩, i.e. V = −2T on a circular orbit, so E = T + V = −T. Directly: mv²/r = k/r² gives T = k/(2r) and V = −k/r. E = −2T mistakes V for E.Point masses of 1 kg at (1, 1, 0) m and 2 kg at (0, 1, 1) m form a rigid body. The product-of-inertia component Iₓᵧ of its inertia tensor, in kg m², is ____.
Numerical answer — type the value.
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Answer: -1
Iₓᵧ = −Σ m x y = −(1 × 1 × 1 + 2 × 0 × 1) = −1 kg m². The minus sign is part of the definition Iᵢⱼ = Σm(r²δᵢⱼ − xᵢxⱼ); omitting it gives +1. Iₓₓ = Σm(y² + z²) = 5 kg m² is a diagonal component. Type the answer as -1.A symmetric top (I₁ = I₂ ≠ I₃) rotates freely with no external torque. Which statements are correct?
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Answer: A — ω₃, the component along the symmetry axis, is constant; B — The angular momentum L is fixed in space; C — The kinetic energy is constant
Euler’s third equation, I₃ω̇₃ = (I₁ − I₂)ω₁ω₂ = 0, keeps ω₃ constant. With no torque, L is constant in space and T is conserved. The perpendicular part of ω rotates about the symmetry axis at Ω = (I₃ − I₁)ω₃/I₁ in the body frame, so its direction there is not fixed.For the angular-momentum components of a particle, the Poisson bracket {Lₓ, L_y} equals:
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Answer: A — L_z
With Lₓ = yp_z − zp_y and L_y = zpₓ − xp_z, the only surviving terms come from {y, p_y}-type pairs and give xp_y − ypₓ = L_z. iħL_z is the quantum commutator [Lₓ, L_y], which is iħ times the classical bracket.A rocket moves at 0.5c relative to the Earth and fires a probe forward at 0.5c relative to itself. The speed of the probe relative to the Earth, as a fraction of c, is ____.
Numerical answer — type the value.
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Answer: 0.8
u = (u′ + v)/(1 + u′v/c²) = (0.5 + 0.5)/(1 + 0.25) = 0.8c. Galilean addition gives 1.0c, which relativity forbids for any massive body.An electron (rest energy 0.511 MeV) has kinetic energy 1.022 MeV. Its speed as a fraction of c, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.94
K = (γ − 1)mc² gives γ = 1 + 1.022/0.511 = 3, so v/c = √(1 − 1/γ²) = √(8/9) = 0.94. The classical ½mv² = K gives v/c = √(2 × 2) = 2, which is impossible — the sign that relativity is needed.Two photons, each of energy 1 MeV, collide head-on. Which statements are correct?
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Answer: A — The invariant mass of the pair is 2 MeV/c²; B — The collision can create an electron–positron pair; C — The total momentum of the pair is zero in the laboratory
Head-on, the momenta cancel, so M c² = E_tot = 2 MeV, above the 1.022 MeV needed for e⁺e⁻. Photons moving in the same direction have p_tot c = E_tot, so their invariant mass is zero — that is why a single photon cannot pair-produce in empty space.