Energy conservation from time-translation invariance

Derivation · the atom

Energy conservation from time-translation invariance

D-002 noether-time-translation Home PU-201 Threads symmetry · energy Depends on D-001 verified

Statement

If the Lagrangian has no explicit time dependence, a specific combination of coordinates and velocities is constant along the motion — and for ordinary systems that combination is the total energy.

Why it matters

Energy conservation is usually presented as an empirical law. It is not. It is a consequence of the statement that the laws of physics are the same today as they were yesterday. This derivation is the smallest complete instance of Noether's theorem, and it is worth doing in full before the general case, because the general case obscures how simple the mechanism is.

Assumptions

∂L/∂t = 0.The Lagrangian contains no explicit t. Drop this and the conserved quantity acquires a source term equal to −∂L/∂t — which is precisely how a time-dependent driving force pumps energy into a system.

The equations of motion hold.The result is an on-shell statement. It says nothing about arbitrary paths, only about the ones the system actually follows.

Coordinates do not depend explicitly on time.If the transformation to generalised coordinates is itself time-dependent (a rotating frame, say), the conserved quantity still exists but is no longer the energy. This is the single most common source of confusion in the topic.

Derivation

dL/dt = Σ_i ( ∂L/∂q_i · q̇_i + ∂L/∂q̇_i · q̈_i ) + ∂L/∂t

Total time derivative of L along the path, by the chain rule. The last term is the explicit dependence, and the assumption sets it to zero. A

∂L/∂q_i = d/dt ( ∂L/∂q̇_i )

Substitute the Euler–Lagrange equation (D-001). This is the step that makes the result on-shell: we are now restricted to physical paths. A

dL/dt = Σ_i [ d/dt ( ∂L/∂q̇_i ) · q̇_i + ∂L/∂q̇_i · q̈_i ]

Insert step 2 into step 1. The bracket is now visibly the product rule expanded. B

dL/dt = d/dt ( Σ_i ∂L/∂q̇_i · q̇_i )

Recognise the bracket as a single derivative and pull it out. Nothing has been assumed beyond steps 1 and 2 — the collapse is pure algebra. B

d/dt ( Σ_i ∂L/∂q̇_i · q̇_i − L ) = 0

Move dL/dt across. A quantity whose time derivative vanishes is conserved — and it has appeared without being looked for. C

Result

H ≡ Σ_i p_i q̇_i − L = constant, p_i ≡ ∂L/∂q̇_i

Reading. The conserved quantity is not defined as the energy; it emerges from the symmetry and then turns out to equal T + V under the conditions below. The object H is the Hamiltonian, arriving here as a by-product before it is given its own role in D-004.

Units check. p q̇ has units of [energy]·[time]/[q] × [q]/[time] = [energy], matching L.

When is H actually the energy?

If the kinetic energy is a homogeneous quadratic function of the velocities — which holds whenever the coordinate transformation is time-independent — then by Euler's theorem on homogeneous functions Σ p_i q̇_i = 2T. With L = T − V this gives H = 2T − (T − V) = T + V. Both conditions are required. A bead on a wire being spun at fixed angular velocity has a conserved H that is not the energy, and the energy is not conserved — the motor is doing work.

Limiting cases

Time-independent Lagrangian, time-independent coordinates → H = T + V, ordinary energy conservation.
Time-independent Lagrangian, rotating (time-dependent) coordinates → H conserved, but H ≠ energy.
Explicit time dependence → the conserved law degrades into dH/dt = −∂L/∂t, the pumping relation.

Breaks when

The Lagrangian carries explicit time dependence — then dH/dt = −∂L/∂t ≠ 0.
The coordinate transformation is time-dependent — H is conserved but is not the energy.
Dissipation is present — no Lagrangian exists in the required form.

Failure modes

Assuming H = T + V always. It requires two separate conditions, both easy to violate.
Confusing ∂L/∂t with dL/dt. Along the motion L generally changes with time even when it has no explicit time dependence.
Believing energy conservation is more fundamental than the symmetry. It is downstream of it — which is why energy is not globally conserved in an expanding universe, where time-translation invariance fails.

Worked number

For a single particle, L = ½ m q̇² − V(q), so p = ∂L/∂q̇ = m q̇ and p q̇ = m q̇². Then H = m q̇² − (½ m q̇² − V) = ½ m q̇² + V = T + V — the kinetic term is homogeneous quadratic (2T = m q̇²), so Euler's theorem lands exactly on the familiar energy.

Run the check →

Cross-link — the general Noether statement is developed in PU-402/L02.

Discussion

The single most important thing to take from this derivation is that H is not built out of energy — energy is read off H afterwards, and only sometimes. The chain of steps never mentions kinetic or potential energy; it uses only two facts, that the Lagrangian carries no explicit clock (∂L/∂t = 0) and that the path obeys Euler–Lagrange. From those alone a combination of coordinates and velocities freezes. The physical content is therefore a statement about time itself: because the dynamics cannot tell when it is running, one bookkeeping number is the same at every instant. Energy conservation is the name we give that number in the common case, not the reason it exists.

The structure ∑p_iq̇_i − L is not arbitrary either. It is exactly the Legendre transform that trades the velocities q̇_i for the momenta p_i = ∂L/∂q̇_i, which is why the conserved quantity is the Hamiltonian rather than something less useful — the same object reappears as the generator of the dynamics in D-004. The reason H equals T + V only under extra conditions is visible once you split the kinetic energy by its dependence on the velocities, T = T₂ + T₁ + T₀, where the subscript counts powers of q̇. Euler's theorem on homogeneous functions gives ∑p_iq̇_i = 2T₂ + T₁, so H = T₂ − T₀ + V. The friendly identity H = T + V survives only when T₁ = T₀ = 0, which is precisely the case of a time-independent coordinate map. The rotating-wire example below has a non-zero T₀, and that one term is the whole reason its conserved H is not the energy.

Seen from higher up, this is the smallest cell of a much larger organism. The general Noether theorem attaches a conserved current to every continuous symmetry of the action; here the symmetry is the one-parameter group of time translations t → t + ε, and its Noether charge is H. The same machinery gives momentum from spatial translation and angular momentum from rotation, and in field theory it produces the stress–energy tensor T^μν, whose conservation ∂_μT^μν = 0 is the field-theoretic descendant of this one line of algebra. In quantum mechanics the connection is even tighter: the Hamiltonian that emerges here as a passive by-product becomes the operator that generates time evolution, Û = exp(−iĤt/ℏ), so a stationary state is nothing but an eigenstate of the very charge this classical argument uncovered.

The deepest lesson is that energy conservation is contingent, not sacred. It holds exactly to the extent that time-translation invariance holds. In an expanding universe the Friedmann background is genuinely time-dependent — there is no global timelike Killing vector — and the energy of, for example, the cosmic photon gas is not conserved: photons redshift and the energy simply is not book-kept anywhere. This is not a failure of physics but a direct reading of the theorem: no symmetry, no charge. Treating energy conservation as more fundamental than the symmetry it rests on gets the logic exactly backwards.

Common misconceptions.

That H = T + V always. It requires two independent conditions (no explicit time dependence and a time-independent coordinate map); either can fail alone.
Confusing dL/dt (how L changes along the actual motion, almost never zero) with ∂L/∂t (whether the formula for L contains t explicitly, the thing we set to zero).
Believing “H conserved” means “energy conserved.” On a driven turntable H is conserved while the true energy is being pumped in by the motor.
Thinking the canonical momentum p_i = ∂L/∂q̇_i is always mv. For a charged particle it carries an extra qA, and getting H right depends on using the canonical, not the mechanical, momentum.

Worked examples

Example 1 — A bead on a wire spun at fixed angular velocity: H conserved, energy not. A bead of mass m = 0.020 kg slides without friction on a straight horizontal wire that is forced to rotate about a vertical axis through one end at a constant angular velocity Ω = 10 rad/s. Let r be the bead's distance along the wire. Because the wire's angle is fixed externally as φ = Ωt, the only true freedom is r.

L = T = ½m(ṙ² + r²Ω²), ∂L/∂t = 0

Kinetic energy in plane polar coordinates with φ̇ = Ω substituted. The formula contains no explicit t, so the derivation applies and H is conserved. Note the split: T₂ = ½mṙ² and T₀ = ½mr²Ω².

p_r = ∂L/∂ṙ = mṙ

The canonical momentum conjugate to r.

H = p_rṙ − L = mṙ² − ½m(ṙ² + r²Ω²) = ½mṙ² − ½mΩ²r²

Form the Hamiltonian symbolically. The sign of the second term is the whole story: H = T₂ − T₀, whereas the energy is E = T₂ + T₀.

H = ½(0.020)(0.40)² − ½(0.020)(10)²(0.15)² = 0.0016 − 0.0225 = −0.0209 J

Evaluate at the instant r = 0.15 m, ṙ = 0.40 m/s. Substitute numbers only now.

E = ½m(ṙ² + r²Ω²) = 0.0016 + 0.0225 = 0.0241 J, r̈ = Ω²r > 0

The equation of motion is d(mṙ)/dt = mΩ²r, i.e. r̈ = Ω²r: the bead accelerates outward, so T₀ and hence E grow without bound while H stays fixed. One checks dH/dt = mṙr̈ − mΩ²rṙ = 0.

H = −0.0209 J (conserved), E = 0.0241 J (increasing)

Answer. The conserved Jacobi integral is H = −0.0209 J, while the mechanical energy at this instant is E = +0.0241 J and is not conserved. The difference E − H = 2T₀ = mΩ²r² = 0.045 J is the imprint of the motor: the external torque that keeps Ω fixed does work on the bead, so this is exactly the case flagged in the derivation where a time-dependent coordinate map leaves H conserved but strips it of its identity as the energy.

Example 2 — A bead in a frictionless parabolic bowl: position-dependent inertia, yet H = T + V. A bead of mass m = 0.50 kg slides without friction on the fixed wire y = ½ax² with a = 4.0 m⁻¹, under gravity g = 9.81 m/s². It is released from rest at x₀ = 0.30 m. Find its speed at the bottom. The constraint is time-independent, so we expect H to be the true energy — but the inertia is position dependent, a good test of the machinery.

T = ½m(ẋ² + ẏ²) = ½mẋ²(1 + a²x²), V = mgy = ½mgax²

Use ẏ = axẋ to eliminate ẏ. Here T is homogeneous quadratic in the single velocity ẋ (no T₁ or T₀), even though the coefficient depends on x.

p_x = ∂L/∂ẋ = mẋ(1 + a²x²), H = p_xẋ − L = ½mẋ²(1 + a²x²) + ½mgax² = T + V

Because T = T₂, Euler's theorem gives p_xẋ = 2T, so H = 2T − (T − V) = T + V. This is the energy, and it is conserved.

H = V(x₀) = ½mgax₀² = ½(0.50)(9.81)(4.0)(0.30)² = 0.883 J

Evaluate the conserved H at the release point, where ẋ = 0 so T = 0.

At x = 0: H = ½mv² ⇒ v = √(2H/m) = √(2 × 0.883 / 0.50) = √3.53 = 1.88 m/s

At the bottom x = 0 the metric factor is 1, so the full speed is just ẋ. Setting the whole of H into kinetic energy solves for v.

v = 1.88 m/s

Answer. The bead reaches the bottom at 1.88 m/s. The result agrees with the elementary energy argument v = √(2gy₀) with y₀ = ½ax₀² = 0.18 m — as it must, because here the two conditions of the derivation both hold and H genuinely is the energy, notwithstanding the position-dependent inertia.

Problems

(Warm-up.) A block of mass m = 0.40 kg on a spring of stiffness k = 25 N/m has L = ½mẋ² − ½kx². Show from the definition H = pẋ − L that H = ½mẋ² + ½kx², and given amplitude A = 0.10 m find H and the maximum speed.
Solution
p = ∂L/∂ẋ = mẋ, so H = mẋ² − (½mẋ² − ½kx²) = ½mẋ² + ½kx² = T + V. Both conditions hold (no explicit t; T = T₂), so H is the conserved energy. At the turning point ẋ = 0: H = ½kA² = ½(25)(0.10)² = 0.125 J. At x = 0 all of it is kinetic: ½mv_max² = 0.125 ⇒ v_max = √(2×0.125/0.40) = 0.79 m/s.
(Explicit time dependence.) A driven oscillator has L = ½mẋ² − ½kx² + F₀xcosωt. Show the conserved-quantity argument fails and that instead dH/dt = −∂L/∂t. Evaluate the rate at the instant x = 0.050 m, sinωt = 1, with F₀ = 2.0 N, ω = 3.0 rad/s.
Solution
Retracing the derivation, step 1 keeps the explicit term, so dL/dt = d/dt(∑p_iq̇_i) + ∂L/∂t, giving dH/dt = −∂L/∂t. Here ∂L/∂t = −F₀xωsinωt, so dH/dt = +F₀xωsinωt = (2.0)(0.050)(3.0)(1) = 0.30 W. The externally imposed clock in the drive is precisely the source term promised in the assumptions: the driving agent injects energy at this instantaneous rate.
(Euler's theorem in action.) For a particle in a plane written in polar coordinates, T = ½m(ṙ² + r²θ̇²) and L = T − V(r). Show that ∑p_iq̇_i = 2T and hence H = T + V. Verify numerically at m = 1.0 kg, r = 1.5 m, ṙ = 2.0 m/s, θ̇ = 3.0 rad/s.
Solution
p_r = mṙ, p_θ = mr²θ̇. Then ∑p_iq̇_i = mṙ² + mr²θ̇² = 2×½m(ṙ²+r²θ̇²) = 2T, the statement of Euler's theorem for a degree-2 homogeneous function. Hence H = 2T − (T − V) = T + V. Numerically p_r = 2.0 kg·m/s, p_θ = 1×2.25×3.0 = 6.75 kg·m²/s; ∑p_iq̇_i = (2.0)(2.0) + (6.75)(3.0) = 4.00 + 20.25 = 24.25 J, while 2T = 2×½(1.0)(4.0 + 2.25×9.0) = 4.0 + 20.25 = 24.25 J. They match.
(Canonical vs mechanical momentum.) A particle of charge q and mass m in static fields has L = ½mv² + qA·v − qφ. Show p = mv + qA and H = ½mv² + qφ, so the magnetic field drops out of the energy. A proton (m = 1.67×10⁻²⁷ kg, q = 1.60×10⁻¹⁹ C) starts from rest at a point where φ = 0 in a field described by φ = −E₀x, E₀ = 500 V/m, with an arbitrary static B. Find its speed at x = 0.020 m.
Solution
p = ∂L/∂v = mv + qA: the canonical momentum carries the field term. Then H = p·v − L = (mv+qA)·v − ½mv² − qA·v + qφ = ½mv² + qφ. The qA·v terms cancel exactly, which is the Lagrangian statement that the magnetic force does no work. With static φ, A the fields carry no explicit time and H is conserved. From rest at φ = 0, H = 0, so ½mv² = −qφ = qE₀x. Thus v = √(2qE₀x/m) = √(2×1.60×10⁻¹⁹×500×0.020 / 1.67×10⁻²⁷) = √(3.2×10⁻¹⁸/1.67×10⁻²⁷) = √(1.92×10⁹) = 4.4×10⁴ m/s, independent of B.
(Jacobi integral and stability.) A bead of mass m = 0.050 kg is threaded on a frictionless rod that rotates in a horizontal plane at fixed Ω, and is tied to the axis by a spring of stiffness k = 8.0 N/m and zero natural length. Write L, find the conserved H, identify the effective potential V_eff(r), and find the critical angular velocity above which the bead is flung outward. Is H the energy?
Solution
L = ½m(ṙ² + r²Ω²) − ½kr², with ∂L/∂t = 0, so H is conserved. With p_r = mṙ, H = p_rṙ − L = ½mṙ² − ½mΩ²r² + ½kr² = ½mṙ² + V_eff, where V_eff(r) = ½(k − mΩ²)r². The centrifugal term T₀ enters H with a minus sign and combines with the spring. If k > mΩ² the effective potential is an upward parabola: r = 0 is stable and the bead oscillates. If k < mΩ² the coefficient is negative, V_eff is an inverted parabola and the bead accelerates outward. The critical value is Ω_c = √(k/m) = √(8.0/0.050) = √160 = 12.6 rad/s. H is not the energy: the coordinate map is time-dependent (the rod's angle is Ωt), T₀ ≠ 0, and the motor exchanges energy with the bead; the true energy E = ½m(ṙ²+r²Ω²) + ½kr² = H + mΩ²r² is not conserved.