Noether's Theorem for Continuous Symmetries

Derivation

Noether's Theorem for Continuous Symmetries

D-128 Home PU-201 Threads symmetry · energy · force Depends on The Euler-Lagrange Equation, Noether's Theorem for Time Translation

Statement

For a Lagrangian system L(q, q̇, t) obeying the Euler–Lagrange equations, every one-parameter continuous transformation t → t + ε τ(q,t), q_i → q_i + ε ζ_i(q,t) that leaves the action invariant up to a boundary term (δS = ∫ ε dF/dt dt) has an associated quantity that is constant along every physical trajectory: the Noether charge Q = Σ_i p_i ζ_i − h τ − F, with conjugate momenta p_i = ∂L/∂q̇_i and energy function h = Σ_i p_i q̇_i − L. Spatial translations give linear momentum, rotations give angular momentum, and time translation gives energy as special cases.

Why it matters

Noether's theorem (Emmy Noether, 1918) is the deepest organizing principle in physics: it converts symmetries of the dynamics into conserved quantities, one for each continuous symmetry, with an explicit formula for the conserved charge. Conservation laws stop being separate empirical facts and become consequences of the geometry of the action.

The individual conservation laws derived elsewhere in this unit — momentum from translation invariance, angular momentum from rotational invariance, energy from time-translation invariance — are here shown to be one theorem viewed through three different symmetry groups. The same machinery extends without change to field theory (the Noether current ∂_μj^μ = 0), to gauge theories, and underlies the conserved charges of the Standard Model, making this the single most reused result in theoretical physics.

Assumptions

The trajectory satisfies the Euler–Lagrange equations (on-shell).Noether's theorem is an identity on solutions. Off the physical path the Euler–Lagrange bracket does not vanish and dQ/dt ≠ 0; the charge is conserved only along actual motions, not for arbitrary q(t). The transformation is a genuine symmetry: δS = ∫ ε dF/dt dt (invariance up to a boundary term).If the variation of the action contains a bulk piece that is not a total time derivative, there is no conserved charge — the transformation is not a symmetry, only a change of variables, and Q drifts. The symmetry is continuous — a differentiable one-parameter group with an infinitesimal generator.Discrete symmetries (parity, time reversal, lattice translations) have no ε → 0 limit and therefore no Noether charge; they yield selection rules and multiplicative quantum numbers instead of additively conserved quantities. The generators τ, ζ_i and the Lagrangian are smooth (C¹), and the symmetry is global (ε constant, independent of t).A local (ε = ε(t)) symmetry is a gauge symmetry and gives Noether's second theorem — an off-shell identity among the equations of motion, not an independent conserved charge. Non-smooth generators break the variational manipulation in the derivation.

Derivation

d/dt(∂L/∂q̇_i) − ∂L/∂q_i = 0, p_i ≡ ∂L/∂q̇_i

Start from the Euler–Lagrange equations (prior result) holding on the physical trajectory; define the conjugate momentum p_i. Everything downstream is evaluated on-shell. A

t → t̄ = t + ε τ(q,t), q_i → q̄_i = q_i + ε ζ_i(q,t), δt = ε τ, δq_i = ε ζ_i

Introduce the one-parameter family of transformations to first order in the group parameter ε. τ generates the time part, ζ_i the coordinate part; higher powers of ε are not needed for a conservation law. A

δ̄q_i ≡ q̄_i(t) − q_i(t) = δq_i − q̇_i δt = ε(ζ_i − q̇_i τ), δ̄(q̇_i) = d/dt(δ̄q_i)

Separate the form variation δ̄q_i (change of the function at fixed t) from the total variation δq_i (which also moves the argument by δt). Only the form variation commutes with d/dt, which is what makes the next step legal. B

δS = ∫ [ Σ_i(∂L/∂q_i δ̄q_i + p_i δ̄q̇_i) + d/dt(L δt) ] dt

Vary the action S = ∫L dt. The first bracket is the ordinary chain-rule variation of L at fixed t; the extra d/dt(L δt) is the Jacobian of the time reparametrization dt̄ = (1 + d(δt)/dt) dt together with L evaluated at the shifted endpoint. C

Σ_i(∂L/∂q_i δ̄q_i + p_i δ̄q̇_i) = Σ_i[∂L/∂q_i − d/dt p_i] δ̄q_i + d/dt[ Σ_i p_i δ̄q_i ]

Use δ̄q̇_i = d/dt(δ̄q_i) (step 3) and reverse the product rule on p_i d/dt(δ̄q_i) = d/dt(p_iδ̄q_i) − (dp_i/dt) δ̄q_i. This isolates the Euler–Lagrange bracket. C

δS = ∫ d/dt[ Σ_i p_i δ̄q_i + L δt ] dt

On-shell (step 1) the bracket ∂L/∂q_i − dp_i/dt vanishes for every i, so the entire variation of the action collapses to a single total time derivative. This holds for any transformation; it is not yet a conservation law. B

δS = ∫ ε dF/dt dt ⇒ d/dt[ Σ_i p_i δ̄q_i + L δt − ε F ] = 0

Now impose the symmetry: the action changes at most by a boundary term ε F(q,t) (strict invariance is F = 0). Equating the two expressions for δS and moving everything under one d/dt gives a conserved total derivative. B

Σ_i p_i δ̄q_i + L δt = ε[ Σ_i p_iζ_i − (Σ_i p_iq̇_i − L)τ ] = ε[ Σ_i p_iζ_i − h τ ]

Insert δ̄q_i = ε(ζ_i − q̇_iτ) and δt = ετ from step 3 and collect. The combination Σp_iq̇_i − L is exactly the energy function h (the Hamiltonian on the constraint surface). B

d/dt[ Σ_i p_iζ_i − h τ − F ] = 0 ⇒ Q = Σ_i p_iζ_i − h τ − F = const

Divide the conserved derivative of step 7 by the constant ε. The bracket is the Noether charge; its vanishing time derivative is the conservation law. A

Field theory: j^μ = Σ_a (∂ℒ/∂(∂_μφ_a)) Ψ_a − T^μ_ν ξ^ν − F^μ, ∂_μj^μ = 0, T^μ_ν = Σ_a(∂ℒ/∂(∂_μφ_a))∂_νφ_a − δ^μ_νℒ

Repeat steps 1–9 for a Lagrangian density ℒ(φ_a, ∂_μφ_a, x) under x^μ → x^μ + ε ξ^μ, φ_a → φ_a + ε Ψ_a. The total time derivative becomes a four-divergence; the conserved current replaces the charge, and Q = ∫ j⁰ d³x is the constant of motion. T^μ_ν is the canonical stress–energy tensor, the field analogue of h. C

Result

Q = Σ_i p_i ζ_i − h τ − F, dQ/dt = 0, p_i = ∂L/∂q̇_i, h = Σ_i p_iq̇_i − L

Reading. To each continuous symmetry attach its generators (τ for time, ζ_i for the coordinates, F for the boundary term of a quasi-symmetry). Contract the coordinate generator with the conjugate momentum, subtract the energy weighted by the time generator, subtract the boundary term, and the result is a number that does not change as the system evolves. Setting τ = 1, ζ = 0, F = 0 gives Q = −h (energy conservation); a uniform spatial shift ζ_i = constant gives total momentum; a rotation generator ζ = n × r gives the angular-momentum component along n.

Units check. With a dimensionless group parameter ε, the charge has the dimension of action. p_iζ_i = (∂L/∂q̇_i)(δq_i/ε) carries J·s/(q_i/s)·... = J·s, and h τ = J·(δt/ε) = J·s likewise, so every term is J·s = kg·m²·s⁻¹. When ε is chosen to carry dimension (a length for a translation), Q inherits the complementary units: for translation Q = J·s/m = kg·m·s⁻¹ (momentum); for a rotation ε is an angle (dimensionless) so Q = J·s = kg·m²·s⁻¹ (angular momentum); for time translation Q = J·s/s = J (energy). Dimensionally homogeneous in every case.

Limiting cases

Time translation (τ = 1, ζ_i = 0, F = 0): Q = −h. Energy conservation; the prior result noether-time-translation is the τ-only slice of this theorem.
Spatial translation (Cartesian q_i, ζ_i = fixed unit vector e): Q = e·Σ_ip_i = component of total linear momentum along e.
Rotation about axis n (ζ = n × r, τ = 0, F = 0): Q = n·Σ(r × p) = angular-momentum component; isotropy ⇒ all three components conserved.
Galilean boost (ζ = t e, τ = 0): a quasi-symmetry with F = me·r, giving Q = m(e·v t − e·r) = −m e·r_cm(0): the centre of mass moves uniformly.
Cyclic coordinate (∂L/∂q_k = 0): the shift q_k → q_k + ε is a symmetry with ζ = δ_ik, so Q = p_k — Noether reproduces the elementary "conjugate momentum of an ignorable coordinate is conserved".

Breaks when

The symmetry is discrete, not continuous. Parity, time reversal, and lattice translations have no infinitesimal generator (no ε → 0 limit), so the formula Q = Σp_iζ_i − hτ − F is empty. Such symmetries give selection rules and conserved quantum numbers (multiplicative), not additive Noether charges — a different kind of statement entirely.
The action is not invariant (symmetry explicitly broken). An explicit time dependence ∂L/∂t ≠ 0 breaks time-translation invariance, and dh/dt = −∂L/∂t ≠ 0 — energy is not conserved (a driven or time-dependent-mass system). Likewise an external position-dependent field breaks translation/rotation invariance and momentum/angular momentum leak away.
Local (gauge) symmetry. If ε is allowed to depend on t or on x^μ the transformation is a gauge symmetry: Noether's second theorem applies and yields an off-shell identity among the field equations (e.g. the Bianchi identity, ∂_μ∂_νF^μν ≡ 0), not an independent conserved charge. The naive current is then ambiguous up to an improvement term.
Quantum anomalies. A symmetry of the classical action can fail to survive quantization because the path-integral measure is not invariant; the classically conserved current acquires a divergence ∂_μj^μ ≠ 0 (the chiral/ABJ anomaly). The theorem's classical conclusion is exact but does not automatically promote to the quantum theory.

Failure modes

Dropping the −h τ term for transformations that move time. Using the coordinate part Σp_iζ_i alone when τ ≠ 0 (as in a boost combined with a time shift, or a Lorentz boost) loses the energy contribution and gives a non-conserved quantity. The form variation δ̄q = δq − q̇ δt is the origin of this term.
Forgetting the boundary term F. Galilean boosts and many phase symmetries are quasi-symmetries (δL = ε dF/dt, not δL = 0). Setting F = 0 there produces a "charge" that drifts; the centre-of-mass theorem is invisible without F = me·r.
Confusing a symmetry of L with a symmetry of the equations of motion. Scaling and other transformations can map solutions to solutions without leaving the action invariant; only invariance of S (up to a boundary term) yields a Noether charge. A symmetry of the EOM alone does not.
Applying the theorem off-shell. Q is constant only along solutions. Evaluating dQ/dt for an arbitrary trial q(t) and finding it non-zero is not a counterexample — the Euler–Lagrange bracket in step 5 is only zero on the physical path.
Sign of the energy term. Writing Q = Σp_iζ_i + h τ (wrong sign) so that pure time translation appears to conserve +h with the wrong sign convention, or double-counting h when the coordinate generator already contains a q̇ τ piece.

Discussion

The structural surprise buried in the derivation is step 6: on-shell, the variation of the action is always a total time derivative, for any transformation whatsoever. The physical content of Noether's theorem is therefore not in that identity but in the symmetry condition of step 7 — the requirement that this same variation also equal ε dF/dt by virtue of the transformation being a symmetry. Two independent expressions for the boundary term, forced to agree, pin down a combination that must be constant. Symmetry and dynamics meet exactly at that equality.

Each conservation law you have met separately is a projection of this one statement onto a subgroup of the symmetry group of the action. Homogeneity of time → energy; homogeneity of space → momentum; isotropy of space → angular momentum; the Galilean boost → uniform centre-of-mass motion. That these are the ten classical constants of motion of an isolated system is not a coincidence: they are precisely the ten generators of the Galilei group, and Noether attaches one charge to each. In special relativity the same counting over the Poincaré group gives energy–momentum (4) and the relativistic angular momentum / boost tensor M^μν (6), again ten.

On the Hamiltonian side the theorem acquires a converse and a sharper meaning. The Noether charge Q is the generator of its own symmetry through the Poisson bracket: δ(anything) = ε{ · , Q}. Conservation dQ/dt = {Q, H} = 0 is then equivalent to H being invariant under the flow Q generates — symmetry and conservation become the single statement that Q and H Poisson-commute, and the map from symmetries to charges is the moment map of symplectic geometry. This is also why the charges of a non-abelian symmetry close into the same Lie algebra as the generators (e.g. {L_i, L_j} = ε_ijkL_k for rotations).

The theorem has a second half that is routinely conflated with the first. Noether's first theorem (derived here) applies to global symmetries — finitely many parameters — and produces conserved charges. Noether's second theorem applies to local/gauge symmetries — parameters that are arbitrary functions — and produces instead identities among the equations of motion (the field equations become dependent), so the naive conserved current is a total derivative and the associated charge is a boundary integral. This is why the "energy" of the gravitational field is subtle and why electric charge in electromagnetism is a surface flux (Gauss's law) rather than a bulk quantity. Common misconceptions. (1) Noether's theorem is not "symmetry of the Lagrangian" but "symmetry of the action up to a boundary term" — the F term is part of the theorem, not an afterthought. (2) A conserved quantity does not require the Lagrangian itself to be unchanged pointwise; quasi-symmetries with δL = ε dF/dt conserve charges just as well. (3) The theorem runs symmetry → conservation; the converse (every conservation law comes from a symmetry) is true in a suitable Hamiltonian sense but not trivially, and "hidden" symmetries (e.g. the Laplace–Runge–Lenz vector of the Kepler problem, or the extra invariants of the isotropic oscillator) correspond to non-geometric symmetries of phase space.

Worked examples

Rotational symmetry → angular momentum. Particle of mass m in a central potential V(r), plane polar coordinates (r, θ): L = ½m(ṙ² + r²θ̇²) − V(r).

V depends only on r, so L is unchanged by θ → θ + ε: this is the rotational symmetry, with generators τ = 0, ζ_r = 0, ζ_θ = 1, F = 0. B

p_θ = ∂L/∂θ̇ = m r²θ̇; Q = Σ p_iζ_i − h τ − F = p_θ·1 − 0 − 0 = m r²θ̇

Apply the master formula. The conserved charge is the momentum conjugate to the cyclic coordinate θ — the angular momentum ℓ. A

m = 0.20 kg, r = 0.50 m, θ̇ = 3.0 rad·s⁻¹: ℓ = (0.20)(0.50)²(3.0) = 0.20 × 0.25 × 3.0 = 0.15 kg·m²·s⁻¹

Insert numbers into ℓ = m r²θ̇. A

If the particle spirals inward to r′ = 0.30 m at fixed ℓ: θ̇′ = ℓ/(m r′²) = 0.15/(0.20 × 0.090) = 8.33 rad·s⁻¹

ℓ is constant along the motion, so a smaller radius forces a larger angular rate — the conserved charge does physical work for us. A

ℓ = m r²θ̇ = 0.15 kg·m²·s⁻¹ (conserved); r: 0.50 → 0.30 m ⇒ θ̇: 3.0 → 8.33 rad·s⁻¹

Reading. Rotational invariance alone — never solving the radial equation — fixes ℓ and hence the spin-up on infall, the mechanism behind an ice skater pulling in her arms and a collapsing gas cloud accelerating.

Units check. kg·(m)²·(s⁻¹) = kg·m²·s⁻¹ = J·s, the dimension of a Noether charge for a dimensionless (angle) parameter. ✓

Galilean boost → centre-of-mass motion (a quasi-symmetry). Free particle L = ½mẋ² in one dimension; boost x → x + ε t (τ = 0, ζ = t).

Under δx = ε t: δL = mẋ δẋ = mẋ(ε) = ε d/dt(mx). So δL is a total derivative but not zero — a quasi-symmetry with boundary term F = m x. B

p = ∂L/∂ẋ = mẋ; Q = p ζ − h τ − F = (mẋ)(t) − 0 − (m x) = m(ẋ t − x)

Apply the master formula, keeping the F term. Omitting F would leave the non-conserved quantity mẋt. A

Free motion x(t) = x₀ + v t, ẋ = v: Q = m(v t − x₀ − v t) = −m x₀

The explicit t-dependence cancels, confirming Q is constant; its value is minus the mass times the initial position. A

m = 2.0 kg, x₀ = 1.0 m, v = 4.0 m·s⁻¹. At t = 2.0 s: x = 1.0 + 8.0 = 9.0 m, so Q = 2.0(4.0×2.0 − 9.0) = 2.0(−1.0) = −2.0 kg·m

Numerical check at a specific time reproduces Q = −m x₀ = −2.0 kg·m. A

Q = m(ẋ t − x) = −m x₀ = −2.0 kg·m (conserved)

Reading. The boost charge encodes that the centre of mass moves at constant velocity: G(t) = x − vt = x₀ is fixed. The example exists to show the F term is indispensable — without it the "conserved" quantity mẋt grows linearly in time.

Units check. Here the group parameter ε is a velocity (m·s⁻¹), so Q inherits kg·m·s⁻¹ / (m·s⁻¹)·... = kg·m; consistently m·x = kg·m. ✓

Problems

(Easy) A free particle of mass m = 0.50 kg moves in one dimension, L = ½mẋ², with speed ẋ = 6.0 m·s⁻¹. Identify the symmetry associated with translation x → x + ε and use Noether's formula to find the conserved charge and its value.
Solution
Translation generators: τ = 0, ζ = 1, F = 0 (L is unchanged, δL = 0). Q = p ζ − h τ − F = p = mẋ, the linear momentum. Value: Q = 0.50 × 6.0 = 3.0 kg·m·s⁻¹, conserved because L has no x-dependence. Units: kg·m·s⁻¹. ✓
(Easy) For a Lagrangian with no explicit time dependence, ∂L/∂t = 0, state the symmetry generators and the conserved Noether charge, and name it.
Solution
Time-translation t → t + ε: τ = 1, ζ_i = 0, F = 0. Q = Σp_iζ_i − h τ − F = −h = −(Σp_iq̇_i − L). Thus h = Σp_iq̇_i − L is conserved — the energy function (Hamiltonian). For a scleronomic system with velocity-independent V, h = T + V, the total mechanical energy. This is the prior result noether-time-translation as a special case.
(Medium) A particle moves in the plane with L = ½m(ẋ² + ẏ²) − V(√(x²+y²)). Under a rotation the generators are ζ_x = −y, ζ_y = x, τ = 0, F = 0. Find the conserved charge, and evaluate it for m = 1.5 kg at position (x, y) = (2.0, 0) m with velocity (ẋ, ẏ) = (0, 3.0) m·s⁻¹.
Solution
p_x = mẋ, p_y = mẏ. Q = p_xζ_x + p_yζ_y = mẋ(−y) + mẏ(x) = m(x ẏ − y ẋ) = L_z, the angular momentum about the origin. Numerically: Q = 1.5(2.0 × 3.0 − 0 × 0) = 1.5 × 6.0 = 9.0 kg·m²·s⁻¹. Conserved because V is central (rotationally invariant). Units: kg·m²·s⁻¹ = J·s. ✓
(Medium) A 2D isotropic oscillator has L = ½m(ẋ² + ẏ²) − ½k(x² + y²). Show that energy is conserved via time-translation, and compute h for m = 0.40 kg, k = 60 N·m⁻¹, position (0.10, 0) m, velocity (0, 0.50) m·s⁻¹.
Solution
∂L/∂t = 0 ⇒ time translation is a symmetry (τ = 1). h = Σp_iq̇_i − L = m(ẋ² + ẏ²) − [½m(ẋ²+ẏ²) − ½k(x²+y²)] = ½m(ẋ²+ẏ²) + ½k(x²+y²) = T + V. Numerically: T = ½(0.40)(0² + 0.50²) = ½(0.40)(0.25) = 0.050 J; V = ½(60)(0.10² + 0²) = ½(60)(0.010) = 0.30 J; h = 0.35 J, conserved. Units: J. ✓
(Hard) For the same 2D isotropic oscillator (m = 0.40 kg, k = 60 N·m⁻¹), rotational invariance gives the conserved angular momentum L_z = m(xẏ − yẋ). (a) Verify dL_z/dt = 0 using the equations of motion. (b) Evaluate L_z for the state in Problem 4. (c) State why this system has more conserved quantities than the three (energy, L_z) that ordinary geometric symmetries provide.
Solution
(a) EOM: mẍ = −kx, mÿ = −ky. dL_z/dt = m(ẋẏ + xÿ − ẏẋ − yẍ) = m(xÿ − yẍ) = x(−ky) − y(−kx) = −kxy + kxy = 0. Conserved. ✓ (b) L_z = m(xẏ − yẋ) = 0.40(0.10 × 0.50 − 0 × 0) = 0.40 × 0.050 = 0.020 kg·m²·s⁻¹. (c) The isotropic oscillator has a hidden SU(2)/SO(3) symmetry larger than the geometric SO(2) rotations: the symmetric tensor A_ij = (p_ip_j/m + k x_ix_j)/2 is conserved, mixing coordinate and momentum directions in phase space (an "accidental" degeneracy, analogous to the Laplace–Runge–Lenz vector for Kepler). These extra invariants are Noether charges of non-geometric symplectic symmetries, not of point transformations of configuration space — the converse of Noether's theorem in its Hamiltonian form. Units of L_z: kg·m²·s⁻¹. ✓