Noether’s theorem and conserved quantities

Start with one free particle

Supply a classical point particle of constant mass m = 3 kg with no applied force. Use a fixed inertial Cartesian plane: x increases right, y increases up, and the origin stays fixed. On 0 ≤ t ≤ 2 s choose x(t) = −1 m + (2 m/s)t and y(t) = 3 m − (1 m/s)t. Its constant velocity is (vₓ, vᵧ) = (2, −1) m/s. Both acceleration components are zero, as required by this force-free model.

Momentum is mass times velocity here: pₓ = 6 kg m/s and pᵧ = −3 kg m/s. “Conserved” means constant along a solution as time passes. The position changes, but each momentum component keeps its value. The question is how a property of the model lets us identify such quantities before solving every possible motion.

Use kinetic energy K = m(vₓ² + vᵧ²)/2 and potential V = 0. Mechanical energy is E = K + V = 7.5 J in this model. The Lagrangian is the energy-valued function L = K − V = m(vₓ² + vᵧ²)/2; the action is its time integral, with units J s. The path, transformations and table can be read first. The short general proof later uses partial derivatives and the product rule.

Test a continuous model symmetry

Translate the whole history by a constant length a in the x direction: X(t) = x(t) + a and Y(t) = y(t). The same a is used at every time. Velocities stay unchanged, and our L has no position dependence, so its value is unchanged for any such translation. This is a symmetry of this model, not an arbitrary time-dependent displacement.

A continuous symmetry belongs to a differentiable family of transformations containing the identity. A finite translation might use a = 2 m. Its infinitesimal version describes the first-order change as a varies near zero. The family must preserve L for all nearby position and velocity arguments; checking only one path or one finite parameter value is not enough.

Here these are active transformations: we compare different histories in the same fixed axes. A translated history can start elsewhere; a rotated history can point elsewhere. We are not moving the axes or adding a force. A symmetry comparison also transforms the endpoint positions; it is different from a small variation required to vanish at fixed endpoints in a stationarity test.

Connect translation to momentum

A partial derivative changes one Lagrangian argument while holding the others fixed. Define pᵢ = ∂L/∂vᵢ. In this Cartesian kinetic model these quantities are the ordinary momentum components. The local Euler–Lagrange equation is dpᵢ/dt = ∂L/∂qᵢ, where qᵢ names a coordinate such as x or y. For our free constant-mass model it gives m aᵢ = 0.

Because x is absent from L, ∂L/∂x = 0. Thus dpₓ/dt = 0 along a solution. A coordinate absent in this way is called cyclic; its velocity can still appear. The same reasoning applies to y for the free particle. Both momentum components are conserved.

For the finite a = 2 m translation, X(t) = 1 m + (2 m/s)t and Y(t) = 3 m − (1 m/s)t. Its momentum is still (6, −3) kg m/s. Conservation is constancy along each solution; it does not require every conserved quantity to have the same value on all symmetry-related solutions.

Connect rotation to angular momentum

A counterclockwise rotation about the fixed origin maps positions to X = x cos φ − y sin φ and Y = x sin φ + y cos φ, and rotates velocities by the same formula. The physical angle is θ = φ rad: φ is its dimensionless numerical radian value. For every φ, the identity cos²φ + sin²φ = 1 makes Vₓ² + Vᵧ² = vₓ² + vᵧ². Thus the free-particle L is invariant under the whole rotation family.

The first-order position change per unit φ at zero is (−y, x). Noether’s conserved expression for this generator is pₓ(−y) + pᵧx = x pᵧ − y pₓ. Call it J_z, the angular-momentum component about the declared origin, with positive sign for counterclockwise orientation and units kg m²/s. The symbol J_z keeps it distinct from the Lagrangian L.

For our original history J_z = (−1 m + (2 m/s)t)(−3 kg m/s) − (3 m − (1 m/s)t)(6 kg m/s) = −15 kg m²/s. A finite 90° rotation gives (X, Y) = (−y, x) and momentum (3, 6) kg m/s, while J_z stays −15 kg m²/s. The translated history instead has J_z = −21 kg m²/s about the same origin; it is still conserved along that history.

Compare transformed paths and conservation

The upper diagram draws the original, translated and 90°-rotated histories, each over 0 to 2 s. Both spatial axes use the same metre scale and show their true zero. Circles with a solid line, squares with a dashed line, and triangles with a dotted line distinguish the histories. One shared diamond represents the original start and rotated finish at (−1, 3) m. These are comparisons of solutions, not three interacting particles.

The lower panel follows only the original history in time. Its vertical axis is momentum in kg m/s: pₓ = 6 and pᵧ = −3 stay horizontal, although the position above changes. Energy and angular momentum have different units and are kept in the table below, not placed on that momentum axis.

The lines are exact affine histories on the supplied interval. The markers select t = 0, 1 and 2 s; they are not measured observations. A picture of a 90° rotation alone would not prove a continuous symmetry; the all-angle norm calculation above provides that test.

Why the Noether quantity is constant

Here is a sufficient fixed-time version. Let L have continuous partial derivatives through second order near the relevant states. Let the independent coordinates follow a twice continuously differentiable solution of the Euler–Lagrange equations. Use a smooth one-parameter family Qᵢ(s, q) with Qᵢ(0, q) = qᵢ; s is constant throughout each history, and transformed velocities are the time derivatives of Qᵢ. Require exact L invariance for every nearby state and parameter.

Define ξᵢ = ∂Qᵢ/∂s at s = 0, the generator. Differentiating the invariance condition gives 0 = Σᵢ[(∂L/∂qᵢ)ξᵢ + pᵢ dξᵢ/dt]. On a solution substitute ∂L/∂qᵢ = dpᵢ/dt. The product rule makes this d/dt(Σᵢ pᵢξᵢ) = 0. Therefore C = Σᵢ pᵢξᵢ is conserved under these conditions.

The parameter’s unit matters. For an x translation parameter a measured in m, ξ = (1, 0) is dimensionless and C = pₓ has momentum units. For the numerical radian phase φ, ξ = (−y, x) has length units and C = J_z has angular-momentum units. A conserved expression is often called a Noether charge; that name does not mean every such quantity is electric charge.

The solution condition is essential. Try x_trial(t) = −1 m + (2 m/s)t + (0.5 m/s²)t² and y_trial(t) = 3 m − (1 m/s)t in the same free-particle L. Then dpₓ/dt = 3 N, although ∂L/∂x = 0. This arbitrary trial is not a solution, so the model’s translation symmetry does not conserve its pₓ. The symmetry generators above also need not vanish at the endpoints; they compare transformed histories rather than fixed-endpoint test variations.

Connect time independence to energy

Define H = Σᵢ vᵢpᵢ − L. The chain rule for L and product rule for vᵢpᵢ, followed by the Euler–Lagrange equations, give dH/dt = −∂L/∂t along a solution. If L has no explicit time dependence, H is conserved. This autonomous case is the time-translation conservation connection; it does not require the position to stay still.

For our fixed inertial Cartesian kinetic energy and velocity-independent potential, Σᵢ vᵢpᵢ = 2K, so H = 2K − (K − V) = K + V = E. The free history therefore has E = (3 kg)[(2 m/s)² + (−1 m/s)²]/2 = 7.5 J. Other Lagrangians or time-dependent coordinate choices require their own interpretation; H should not automatically be identified with a guessed energy formula.

The conservation statement holds over the modeled interval. A finite recorded time window or specified endpoint data is not itself a statement that shifting the record gives the identical boundary-value problem. Test the model’s explicit time dependence and the intended data separately.

Change the model and retest the symmetry

For a different example supply L = m(vₓ² + vᵧ²)/2 − ky²/2 with m = 3 kg and k = 12 N/m. The potential singles out the y direction. An x translation still leaves L unchanged, and dpₓ/dt = 0. In contrast, dpᵧ/dt = −ky: at y = 0.5 m this is −6 N. Momentum in y is not generally conserved.

At the local state (x, y) = (1, 0.5) m and (vₓ, vᵧ) = (2, 0) m/s, K = 6 J and V = 1.5 J, so L = 4.5 J. Rotate that state by 90°: the speed is unchanged but Y = 1 m, giving V = 6 J and L = 0. This rejects all-angle rotational invariance of this L. It remains time independent, so its standard mechanical energy is conserved along its solutions. These instantaneous checks do not specify a complete spring trajectory.

We used exact invariance of a smooth finite-dimensional L at fixed time. More general Noether formulations can allow changes by a total time derivative with a modified conserved expression. Failing our exact-invariance test does not classify every generalized symmetry. Discrete symmetries alone, arbitrary constraints, dissipative rules and field or gauge theories need their own analysis. The next lesson, Central-force motion and the Kepler problem, applies conservation in a specified force model.

Worked examples

Try it yourself

1. For the original free history, find pₓ, pᵧ and E at t = 2 s.

   Show the answer and reasoning

   pₓ = 6 kg m/s; pᵧ = −3 kg m/s; E = 7.5 J.

   Velocity remains (2, −1) m/s. Multiply by 3 kg for momentum and use m(vₓ² + vᵧ²)/2 for energy; the later position does not alter these free-model values.

2. Instead translate the original history by a = −1 m in x, keeping the origin fixed. What is J_z on that history?

   Show the answer and reasoning

   J_z = −12 kg m²/s, constant along the translated history.

   The shift changes x pᵧ − y pₓ by a pᵧ = (−1 m)(−3 kg m/s) = +3 kg m²/s. Add this to the original −15 kg m²/s. This is an active translation of the history, not a change of origin.

3. After the 90° rotation, what are pₓ, pᵧ, E and J_z?

   Show the answer and reasoning

   pₓ = 3 kg m/s; pᵧ = 6 kg m/s; E = 7.5 J; J_z = −15 kg m²/s.

   The rotation maps velocity to (1, 2) m/s. Momentum components change, but the squared speed and fixed-origin angular-momentum component are preserved by this proper planar rotation.

4. For the arbitrary trial x_trial in the proof section, what are pₓ at 1 s and dpₓ/dt? Does free-model translation symmetry require this trial’s pₓ to be conserved?

   Show the answer and reasoning

   pₓ = 9 kg m/s and dpₓ/dt = 3 N. No: this trial violates the equations of motion.

   Its x velocity is 2 m/s + (1 m/s²)t and x acceleration is 1 m/s². Multiply by 3 kg. The theorem’s solution condition fails, even though the free Lagrangian still has translation symmetry.

5. In the y-only spring model, does a constant x translation preserve L? At y = 0.5 m, which momentum rate vanishes?

   Show the answer and reasoning

   Yes. dpₓ/dt = 0, while dpᵧ/dt = −6 N.

   The potential depends on y, not x. The x coordinate is cyclic; y is not. A continuous x-translation symmetry does not require the y component of momentum to be conserved.

6. For the spring state (1, 0.5) m, a 180° rotation leaves ky²/2 = 1.5 J unchanged. Does this alone establish a continuous rotation symmetry and conserved J_z?

   Show the answer and reasoning

   No. One discrete-angle equality is insufficient; the 90°-rotated state instead has potential 6 J.

   The exact-invariance version needs a continuous family near the identity, valid for the relevant arguments. The 90° counterexample rejects all-angle invariance. Indeed at the original state, dJ_z/dt = x(−ky) = −6 N m, so J_z is not generally conserved for this spring model.

What to carry forward

• A continuous model symmetry is a property of a whole transformation family and the relevant arguments, not one observed path.
• Noether conservation is conditional on sufficient smoothness and the equations of motion.
• Translation and rotation give momentum and fixed-origin angular momentum in the selected Cartesian model, with different units.
• A quantity can be constant on each solution yet have different values on symmetry-related solutions.
• Time independence conserves H = Σvᵢpᵢ − L; identify it with K + V only when the kinetic, potential and coordinate conditions justify that step.