The Legendre transform and Hamilton's equations

Derivation · the atom

The Legendre transform and Hamilton's equations

D-004 Home PU-201 Threads energy · symmetry Depends on D-001 verified

Statement

Trading velocity for momentum as the independent variable converts n second-order equations into 2n first-order ones, and the trade is performed by a Legendre transform.

Why it matters

The same operation appears twice in an undergraduate degree and is almost never identified as the same operation. In mechanics it takes L(q, q̇) to H(q, p). In thermodynamics it takes U(S, V) to F(T, V) and G(T, P). Cross-linking these two is one of the clearest wins available on the site — see D-016.

Assumptions

The relation p = ∂L/∂q̇ is invertible for q̇. Requires ∂²L/∂q̇_i∂q̇_j to be non-singular. Where it is singular the system is constrained in the Dirac sense — which is exactly the situation in electromagnetism and in gauge theory generally. L is convex in q̇. Guarantees the transform is single-valued and involutive. For ordinary kinetic energies this is automatic.

Derivation

p_i ≡ ∂L/∂q̇_i, H(q, p, t) ≡ Σ_i p_iq̇_i − L

Define the conjugate momentum and the transform. The right-hand side must be expressed entirely in terms of q and p by inverting the first relation — this inversion is the whole content of the operation. A

dH = Σ_i(q̇_i dp_i + p_i dq̇_i) − Σ_i( (∂L/∂q_i) dq_i + (∂L/∂q̇_i) dq̇_i ) − (∂L/∂t) dt

Take the differential of both sides. Nothing clever yet — just the product rule and the chain rule. A

dH = Σ_i q̇_i dp_i − Σ_i (∂L/∂q_i) dq_i − (∂L/∂t) dt

The dq̇ terms cancel identically, because p_i was defined as ∂L/∂q̇_i. This cancellation is the entire point of the Legendre transform: it removes the old variable from the differential, proving H is genuinely a function of q and p. B

∂L/∂q_i = ṗ_i

From Euler–Lagrange (D-001), since d/dt(∂L/∂q̇_i) is by definition ṗ_i. Again the result becomes on-shell here and not before. B

dH = Σ_i(q̇_i dp_i − ṗ_i dq_i) − (∂L/∂t) dt

Compare coefficient by coefficient with the generic differential dH = Σ(∂H/∂q) dq + Σ(∂H/∂p) dp + (∂H/∂t) dt. Since q, p and t are independent, coefficients may be equated. C

Result

q̇_i = ∂H/∂p_i, ṗ_i = − ∂H/∂q_i, ∂H/∂t = − ∂L/∂t

Reading. Motion becomes a flow in a 2n-dimensional phase space. The state is a point; the Hamiltonian is a function on that space; the dynamics is the vector field it generates. The antisymmetry between the two equations is the seed of the symplectic structure and of D-005.

Units check. ∂H/∂p is [energy]/[momentum] = velocity. ✓

Limiting cases

The same operation in thermodynamics. Start from dU = T dS − P dV, where the natural variables are S and V. To swap S for T, define F = U − TS. Then dF = −S dT − P dV — the dS terms cancel exactly as the dq̇ terms did at step 3. It is the same transform with (q̇, p, L) replaced by (S, T, U), and the sign convention differing only by taste. Continue in D-016.

Breaks when

The momenta are not independent functions of the velocities — gauge systems, where the Legendre transform is degenerate and constrained dynamics is required.
L is non-convex in q̇ — the transform is multivalued.

Failure modes

Leaving q̇ in the expression for H. The transform is not complete until every velocity has been eliminated in favour of momentum; a Hamiltonian containing q̇ is meaningless.
Assuming the conjugate momentum equals mv. For a charged particle in a magnetic field p = mv + qA, and the difference is the whole of the Aharonov–Bohm effect.

Worked number

Take the simplest non-trivial Hamiltonian, a free particle in one dimension, H = p²/2m. The first of Hamilton's equations gives q̇ = ∂H/∂p = p/m, and the second gives ṗ = −∂H/∂q = 0.

m = 2 kg, p = 6 kg·m·s⁻¹ ⟹ q̇ = 6 / 2 = 3 m·s⁻¹, ṗ = 0

Momentum is conserved and the velocity recovered from ∂H/∂p matches p/m — the units check delivered on a number. Run the check →

Discussion

The Legendre transform carries no new physics. It contains exactly the information already in the Lagrangian, re-packaged so that a different pair of variables is primary. What changes is the shape of the equations of motion. Euler–Lagrange gives n second-order equations, one per coordinate, and a second-order equation needs two initial data (position and velocity) to be solved. Hamilton's equations replace them with 2n first-order equations, one for each coordinate and one for each momentum, and a first-order system needs one datum per equation — the same total count, 2n, now read as a single point in a 2n-dimensional phase space. The physical picture that follows is the whole payoff: a mechanical state is a point (q,p), the Hamiltonian is a scalar function defined on that space, and the motion is the flow of the vector field that H generates. Time evolution becomes geometry.

The result is structured the way it is because of the sign asymmetry in the pair q̇_i = ∂H/∂p_i, ṗ_i = −∂H/∂q_i. That single minus sign is not cosmetic. It says that q and p are not interchangeable labels but a conjugate pair: each drives the other's rate of change with opposite signs, which is precisely the structure of a rotation, or more generally of an area-preserving flow. Written compactly, ξ̇ = J ∇H with J the antisymmetric J² = −I matrix, and the whole of Hamiltonian mechanics is the study of what this antisymmetric object — the symplectic form — permits. It is the seed of D-005 (Liouville's theorem and conservation of phase-space volume) and of the Poisson-bracket algebra.

The transform's deepest content is duality. Because L is assumed convex in q̇, the Legendre transform is involutive: transforming H back returns L exactly, so no information is created or destroyed and neither description is privileged. This same convex-duality machinery reappears far outside mechanics. In thermodynamics it is the entire tower of potentials — U→F=U−TS, U→H=U+PV, U→G — each swapping an extensive variable for its conjugate intensive one, cancelling a differential term exactly as the dq̇ terms cancelled at step 3 (see D-016). It is the Wick-rotated bridge to the path integral, the relation between a Lagrangian density and a Hamiltonian density in field theory, and, through the identification of p with −iℏ∂_q, the doorway to canonical quantization: you quantize the Hamiltonian, never the Lagrangian, because q and p are the pair that satisfies [q,p] = iℏ.

There is also a subtlety worth stating plainly at degree level. The Legendre transform is only as good as the invertibility of p = ∂L/∂q̇. When the Hessian ∂²L/∂q̇_i∂q̇_j is singular — as it is for any gauge theory, including free electromagnetism — some velocities cannot be solved for. The Legendre transform is then degenerate, momenta obey primary constraints, and Dirac's constrained-Hamiltonian formalism is needed to recover a consistent dynamics. This is not a pathology to be avoided; it is the generic situation for the fundamental fields of nature, and the constraints turn out to be the generators of the gauge symmetry itself.

Common misconceptions. The most persistent error is treating the conjugate momentum as if it were always mass times velocity. It equals mv only for a plain kinetic term; add a velocity-dependent interaction and the two part ways. For a charge q in a magnetic field the canonical momentum is p = mv + qA, and the gap between canonical and kinetic momentum is exactly what the Aharonov–Bohm effect measures. A second error is to imagine the Hamiltonian is always the energy; it equals the energy only when the coordinates are time-independent and the potential is velocity-independent, and it is conserved only when ∂H/∂t = 0. A third, purely procedural, error is to leave a velocity q̇ sitting inside H: the transform is not finished until every velocity has been eliminated in favour of momentum, because H is by construction a function of q and p alone.

Worked examples

Example 1 — Hamiltonian of the simple harmonic oscillator. A mass on a spring, m = 0.20 kg, k = 8.0 N m⁻¹, released from q₀ = 5.0 cm at rest. Build the Hamiltonian by Legendre transform and extract the motion.

L = ½mq̇² − ½kq², p ≡ ∂L/∂q̇ = mq̇ ⇒ q̇ = p/m

Write the Lagrangian, define the conjugate momentum, and invert for the velocity — the inversion is the transform.

H = pq̇ − L = p·(p/m) − ½m(p/m)² + ½kq² = p²/2m + ½kq²

Substitute and simplify; every q̇ is gone, so the transform is complete. Here H is manifestly the total energy T + V.

q̇ = ∂H/∂p = p/m, ṗ = −∂H/∂q = −kq

Apply Hamilton's equations. Two first-order equations replace the one second-order Newton equation.

q̈ = ṗ/m = −(k/m)q ⇒ ω = √(k/m) = √(8.0/0.20) = √40 = 6.32 rad s⁻¹

Eliminate p between the two to recover the second-order form and read off the angular frequency. Period T = 2π/ω = 0.99 s.

E = ½kq₀² = ½(8.0)(0.050)² = 1.0×10⁻² J; area of orbit = 2πE/ω = 9.9×10⁻³ J s

The phase-space trajectory is the ellipse p²/2m + ½kq² = E; its enclosed area is the action per cycle.

H = p²/2m + ½kq², ω = 6.32 rad s⁻¹

Answer. The oscillator moves on a closed ellipse in phase space with angular frequency ω = 6.32 rad s⁻¹ (period 0.99 s), total energy 1.0×10⁻² J, and phase-space area 9.9×10⁻³ J s per cycle.

Example 2 — Charged particle in a uniform magnetic field: canonical vs kinetic momentum. A particle of charge q and mass m moves in the plane in a field B = Bẑ, described in the Landau gauge A = (−By, 0, 0). Build the Hamiltonian and show a cyclic coordinate delivers a conserved (canonical) momentum that is not the kinetic momentum. Evaluate for an electron in B = 0.10 T.

L = ½m(ẋ²+ẏ²) + qv·A = ½m(ẋ²+ẏ²) − qByẋ

Write the Lagrangian for a charge in a magnetic field. The interaction is velocity-dependent — this is what will break the naive p=mv identification.

p_x = ∂L/∂ẋ = mẋ − qBy, p_y = ∂L/∂ẏ = mẏ

Compute the conjugate momenta. Note p_x ≠ mẋ: the canonical momentum carries the field term qBy.

ẋ = (p_x+qBy)/m, ẏ = p_y/m ⇒ H = (p_x+qBy)²/2m + p_y²/2m

Invert for velocities and form H = p·v − L. Since the magnetic force does no work, H equals the kinetic energy ½m|v|² — but written in canonical variables.

ṗ_x = −∂H/∂x = 0 (x cyclic) ⇒ p_x = mẋ − qBy = const

Because H is independent of x, its conjugate momentum is a constant of motion — a conserved quantity that mixes velocity and position.

mẍ = qBẏ, mÿ = −qBẋ ⇒ ω_c = |q|B/m

The remaining Hamilton equations give coupled first-order relations for the velocities — uniform circular motion at the cyclotron frequency.

ω_c = eB/m_e = (1.60×10⁻¹⁹)(0.10)/(9.11×10⁻³¹) = 1.76×10¹⁰ rad s⁻¹

Substitute the electron values. A 1.0×10⁶ m s⁻¹ electron then orbits at radius r = v/ω_c = 5.7×10⁻⁵ m ≈ 57 μm.

H = (p_x+qBy)²/2m + p_y²/2m, ω_c = 1.76×10¹⁰ rad s⁻¹

Answer. The canonical momentum p_x = mẋ − qBy is conserved because x is cyclic, yet it differs from the kinetic momentum mẋ by the field term. The electron circulates at ω_c = 1.76×10¹⁰ rad s⁻¹, radius ≈ 57 μm for v = 10⁶ m s⁻¹.

Problems

Free particle. For L = ½mẋ², construct H, write Hamilton's equations, and solve them. Which quantity is conserved and why?
Solution
p = ∂L/∂ẋ = mẋ, so ẋ = p/m and H = pẋ − L = p²/2m. Hamilton's equations: ẋ = ∂H/∂p = p/m, ṗ = −∂H/∂x = 0. Since H does not contain x (cyclic coordinate), p is conserved: p = p₀. Integrating, x(t) = x₀ + (p₀/m)t — uniform motion, Newton's first law recovered.
Uniform gravitational field. A particle falls under gravity with L = ½mż² − mgz. Find H, write Hamilton's equations, and solve for z(t) with z(0)=h, ż(0)=0. Take g=9.81 m s⁻², h=20 m; when does it land?
Solution
p = mż, so H = p²/2m + mgz. Hamilton's equations: ż = p/m, ṗ = −∂H/∂z = −mg. Thus p(t) = −mgt and ż = −gt, giving z(t) = h − ½gt². Landing (z=0): t = √(2h/g) = √(40/9.81) = 2.02 s. Note H here equals the total mechanical energy and is conserved (no explicit t), while p is not conserved because z is not cyclic.
Plane pendulum. A bob of mass m on a rigid rod of length ℓ has L = ½mℓ²θ̇² + mgℓcosθ. Find the conjugate momentum, the Hamiltonian, Hamilton's equations, and the small-oscillation frequency.
Solution
p_θ = ∂L/∂θ̇ = mℓ²θ̇, so θ̇ = p_θ/mℓ² and H = p_θ²/2mℓ² − mgℓcosθ. Hamilton's equations: θ̇ = ∂H/∂p_θ = p_θ/mℓ², ṗ_θ = −∂H/∂θ = −mgℓsinθ. Combining: θ̈ = −(g/ℓ)sinθ. For small θ, sinθ≈θ, so ω = √(g/ℓ). The phase portrait shows closed orbits (libration) for H < mgℓ and open orbits (rotation) above it, separated by the separatrix through the unstable top.
The transform in thermodynamics (enthalpy). Starting from dU = TdS − PdV, perform the Legendre transform that swaps the volume V for its conjugate pressure P. Identify the natural variables of the new potential and derive the associated Maxwell relation. Compare, term by term, with step 3 of the mechanics derivation.
Solution
Define the enthalpy H ≡ U + PV (the +PV plays the role of the +pq̇ term). Then dH = dU + PdV + VdP = (TdS − PdV) + PdV + VdP = TdS + VdP. The −PdV term cancels against +PdV exactly as the dq̇ terms cancelled in step 3 — the defining property of the transform. The natural variables are (S,P), with T = (∂H/∂S)_P and V = (∂H/∂P)_S. Equality of the mixed second partials, ∂²H/∂P∂S = ∂²H/∂S∂P, gives the Maxwell relation (∂T/∂P)_S = (∂V/∂S)_P. This is literally the same operation as L→H, with (q̇,p,L) ↔ (V,−P,U).
Relativistic free particle (advanced). For a free relativistic particle L = −mc²√(1 − ẋ²/c²). Compute the conjugate momentum, invert it, and show by Legendre transform that H = √(p²c² + m²c⁴). Check the rest-energy and non-relativistic limits.
Solution
p = ∂L/∂ẋ = mẋ/√(1−ẋ²/c²) = γmẋ — the relativistic momentum. Solving for the velocity: p²(1−ẋ²/c²) = m²ẋ² ⇒ ẋ = pc/√(p²+m²c²). Then H = pẋ − L = pẋ + mc²√(1−ẋ²/c²). Substituting ẋ and simplifying (or noting H = γmc² and γ² = 1 + p²/m²c²) gives H = √(p²c² + m²c⁴). Limits: at p→0, H→mc² (rest energy); expanding for p≪mc, H ≈ mc² + p²/2m — the rest energy plus the familiar non-relativistic kinetic term. The convexity of L in ẋ on |ẋ|<c guarantees the transform is single-valued and involutive.