Noether's theorem — symmetry ↔ conservation law: Worked examples

This page is a drill. The parent note built the machinery; here we point it at every kind of case you can meet, so no exam or homework question is a stranger. We reuse only three tools, all defined in the parent:

Recall The three tools we lean on (from the parent)

Conjugate momentum $p_{i} \equiv \frac{\partial L}{\partial q ˙ _{i}}$ — the " $L$ -slope in the direction of the velocity $\overset{q}{˙}_{i}$ ".
Cyclic coordinate: if $L$ has no $q_{k}$ in it, then $\frac{\partial L}{\partial q _{k}} = 0$ and $p_{k}$ is frozen.
Noether charge: for a symmetry $q_{i} \to q_{i} + ϵ δ q_{i}$ with $δ L = ϵ \frac{d F}{d t}$ , the constant is $Q = \sum_{i} p_{i} δ q_{i} - F .$ When $F = 0$ this is just $\sum_{i} p_{i} δ q_{i}$ .

Recall Two more results we will quote (also from the parent)

Euler–Lagrange equation (the equation of motion for each coordinate): $\frac{d}{d t} (\frac{\partial L}{\partial q ˙ _{i}}) - \frac{\partial L}{\partial q _{i}} = 0 ⟺ \frac{d p _{i}}{d t} = \frac{\partial L}{\partial q _{i}} .$ Read the right-hand form as: "the rate of change of a momentum equals how strongly $L$ leans on that coordinate." If $L$ does not lean on $q_{i}$ (the coordinate is absent), the momentum cannot change.
Energy function (the time-translation charge): the conserved quantity of $t \to t + ϵ$ is $h = \sum_{i} p_{i} \overset{q}{˙}_{i} - L,$ and its rate obeys the identity $\frac{d h}{d t} = - \frac{\partial L}{\partial t}$ . So $h$ is frozen exactly when $L$ has no explicit time.

Everything below feeds off the vault chain Lagrangian Mechanics → Euler-Lagrange Equation → the big three Conservation of Energy, Conservation of Momentum, Angular Momentum.

The scenario matrix

Think of Noether problems as a grid. Each column is a type of symmetry; each row is a type of trap (a sign, a zero, a degenerate or limiting case). Our job is to hit every cell at least once.

Cell	What makes it tricky	Which example hits it
A. Pure cyclic coordinate	spot " $L$ has no $q$ " instantly	Ex 1
*B. Symmetry present but $L$ looks* like it depends on the coord**	need $δ L = 0$ check, not eyeballing	Ex 2
C. Rotation → angular momentum, sign of the cross product	getting $δ r = \overset{n}{^} \times r$ right, sign of $L_{z}$	Ex 3
D. Symmetry broken → NOT conserved (degenerate case)	explicit time / position dependence kills conservation	Ex 4
E. Boundary term $F \neq = 0$ (Galilean boost)	must subtract $F$ , else wrong charge	Ex 5
F. Canonical $p \neq = m v$ (velocity-dependent term)	charged particle in a field	Ex 6
G. Limiting / zero input	what happens as a parameter $\to 0$	Ex 7
H. Real-world word problem	translate a physical story into a symmetry	Ex 8
I. Exam-style twist (partial symmetry)	only one component is conserved	Ex 9
J. Pathological / edge cases	non-invertible $L$ , singular coordinate maps	Ex 10

Ten examples for ten cells. Read the Forecast and guess before you scroll.

Ex 1 — Cell A: pure cyclic coordinate

Worked example Free particle in polar coordinates

A free particle in a plane, written in polar coordinates $(r, ϕ)$ , has $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) .$ Find a conserved quantity by inspection.

Forecast: Which coordinate is missing from $L$ ? Guess the frozen quantity before reading.

Scan $L$ for a missing coordinate. $L$ contains $r, \overset{r}{˙}, \dot{ϕ}$ — but no bare $ϕ$ . So $ϕ$ is cyclic. Why this step? Apply the Euler–Lagrange equation to $ϕ$ : $\frac{d}{d t} p_{ϕ} = \frac{\partial L}{\partial ϕ}$ . Since $ϕ$ never appears in $L$ , the right side is $0$ , so $\frac{d}{d t} p_{ϕ} = 0$ . Spotting the missing coordinate is spotting the symmetry (rotational).
Compute the conjugate momentum $p_{ϕ}$ . $p_{ϕ} = \frac{\partial L}{\partial ϕ ˙} = m r^{2} \dot{ϕ} .$ Why this step? The Noether charge for the shift $ϕ \to ϕ + ϵ$ is $p_{ϕ} \cdot δ ϕ$ with $δ ϕ = 1$ , i.e. just $p_{ϕ}$ .
State the conservation law. $m r^{2} \dot{ϕ} = const$ — this is the angular momentum $L_{z}$ .

Verify: Units: $kg \cdot m^{2} \cdot s^{- 1} = J\cdots$ , correct for angular momentum. Sanity: even though $r$ does appear in $L$ , $r$ is not cyclic (there is an $r^{2} \dot{ϕ}^{2}$ term), so $p_{r}$ is not conserved — only $p_{ϕ}$ . That asymmetry is the whole point.

Ex 2 — Cell B: symmetry hidden behind a coordinate that looks used

Worked example Two particles joined by a spring

$L = \frac{1}{2} m \overset{x}{˙}_{1}^{2} + \frac{1}{2} m \overset{x}{˙}_{2}^{2} - \frac{1}{2} k (x_{1} - x_{2})^{2} .$ Both coordinates $x_{1}$ and $x_{2}$ appear in $L$ . Is anything conserved?

Forecast: Neither coordinate is cyclic on its own. Does that mean nothing is conserved? Guess.

Try the translation symmetry $δ x_{1} = 1, δ x_{2} = 1$ (shift both particles equally). Why this step? The potential depends only on the difference $x_{1} - x_{2}$ . Shifting both by the same amount leaves $x_{1} - x_{2}$ untouched — that is the symmetry, even though each coordinate individually appears.
Check $δ L = 0$ . $δ L = \frac{\partial L}{\partial x _{1}} (1) + \frac{\partial L}{\partial x _{2}} (1) = - k (x_{1} - x_{2}) (1) - k (x_{2} - x_{1}) (1) = 0. ✓$ Why this step? Noether needs $δ L = 0$ (here $F = 0$ ). Eyeballing " $x_{1}$ appears" would wrongly conclude no symmetry; the honest check catches it.
Build the charge $Q = p_{1} δ x_{1} + p_{2} δ x_{2} = m \overset{x}{˙}_{1} + m \overset{x}{˙}_{2} = P_{total}$ .

Verify: $\frac{d P}{d t} = m \overset{x}{¨}_{1} + m \overset{x}{¨}_{2}$ . Equations of motion: $m \overset{x}{¨}_{1} = - k (x_{1} - x_{2})$ , $m \overset{x}{¨}_{2} = - k (x_{2} - x_{1}) = + k (x_{1} - x_{2})$ . Sum $= 0$ . So total momentum is genuinely constant. ✓ (The internal spring cannot push the center of mass — Newton's third law reborn as a symmetry.)

Ex 3 — Cell C: rotation, the cross-product sign

Worked example Central-force particle in 2D

$L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) - V (x^{2} + y^{2}) .$ The potential depends only on distance $r = x^{2} + y^{2}$ . Find the conserved charge from rotational symmetry, being careful with signs.

Forecast: Will the charge be $+ (x \overset{y}{˙} - y \overset{x}{˙})$ or $- (x \overset{y}{˙} - y \overset{x}{˙})$ ? Guess the sign now.

What the figure shows: the dashed lavender circle is one orbit of radius $r$ ; the mint dots are sample points on it; each coral arrow is that point's displacement $δ r$ under a tiny rotation. Notice every arrow points along the circle, in the counter-clockwise sense, and is perpendicular to the slate radius line — that is the geometric meaning of $δ r = \overset{z}{^} \times r = (- y, x)$ .

Write the infinitesimal rotation. A rotation by small angle $ϵ$ about $\overset{z}{^}$ sends $r \to R r$ , giving $δ r = \overset{z}{^} \times r$ . In components (each coral arrow in the figure): $δ x = - y, δ y = + x .$ Why this step? We need the shape $δ q_{i}$ of the symmetry. The cross product $\overset{z}{^} \times r = (- y, x)$ is the velocity of a point being spun counter-clockwise — that is what "rotate the whole picture" means locally.
Check $δ L = 0$ . Kinetic part: $δ \overset{x}{˙} = - \overset{y}{˙}, δ \overset{y}{˙} = \overset{x}{˙}$ , so $m (\overset{x}{˙} δ \overset{x}{˙} + \overset{y}{˙} δ \overset{y}{˙}) = m (- \overset{x}{˙} \overset{y}{˙} + \overset{y}{˙} \overset{x}{˙}) = 0$ . Potential part: $r$ is unchanged by rotation, so $δ V = 0$ . Total $δ L = 0$ . ✓ Why this step? Rotational invariance is only real if both pieces are invariant. Distance-only potentials pass automatically.
Build the charge. $Q = p_{x} δ x + p_{y} δ y = m \overset{x}{˙} (- y) + m \overset{y}{˙} (x) = m (x \overset{y}{˙} - y \overset{x}{˙}) = L_{z} .$ So the sign is positive $m (x \overset{y}{˙} - y \overset{x}{˙})$ .

Verify: Take a particle at $r = (1, 0)$ moving with $v = (0, 1)$ (counter-clockwise), $m = 1$ . Then $L_{z} = 1 \cdot (1 \cdot 1 - 0 \cdot 0) = + 1 > 0$ , correctly positive for counter-clockwise motion — matching the arrow direction in the figure. Units $kg\cdotm^{2} / s$ . ✓

Ex 4 — Cell D: symmetry BROKEN, nothing conserved (degenerate case)

Worked example Driven oscillator — the classic "energy not conserved" trap

$L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2} + x F_{0} cos (ω t) .$ A spring plus an external push that varies with time. Is energy conserved?

Forecast: Intro physics says "energy is always conserved." Is it here? Guess before checking.

Test time-translation symmetry $t \to t + ϵ$ . This is a symmetry only if $L$ has no explicit $t$ . Why this step? The energy function $h = \sum_{i} p_{i} \overset{q}{˙}_{i} - L$ (recalled in the box at the top) is the Noether charge of time-translation, and its rate is $\frac{d h}{d t} = - \frac{\partial L}{\partial t}$ . No time-symmetry ⇒ no reason for $h$ to be frozen.
Look for explicit $t$ . The term $x F_{0} cos (ω t)$ contains $t$ directly. So $\frac{\partial L}{\partial t} = - x F_{0} ω sin (ω t) \neq = 0$ . Why this step? This single non-zero partial derivative is the whole verdict.
Compute the actual rate of change of energy. Using the identity from the recall box, $\frac{d h}{d t} = - \frac{\partial L}{\partial t} = + x F_{0} ω sin (ω t) \neq = 0.$ Energy is not conserved: the driver pumps energy in and out.

Verify: Turn the drive off, $F_{0} = 0$ . Then $\partial L / \partial t = 0$ and $d h / d t = 0$ — energy conservation returns. This is the correct limiting behaviour: the broken symmetry heals exactly when the time-dependent term vanishes. ✓

Ex 5 — Cell E: boundary term $F \neq = 0$ (Galilean boost)

Worked example Free particle under a velocity boost

$L = \frac{1}{2} m \overset{x}{˙}^{2} .$ Apply the boost $x \to x + ϵ t$ (give the whole system a small extra velocity $ϵ$ ). Find the conserved charge — the naive Noether formula will lie unless you include $F$ .

Forecast: Guess: will the naive charge $p δ x$ be conserved by itself, or do we need a correction?

Read off $δ x$ . $x \to x + ϵ t$ means $δ x = t$ (and $δ \overset{x}{˙} = 1$ , since $\frac{d}{d t} (ϵ t) = ϵ$ ). Why this step? The "shape" of a boost grows with time — that is what distinguishes it from a plain shift.
Compute $δ L$ — it is NOT zero. $δ L = \frac{\partial L}{\partial x ˙} δ \overset{x}{˙} = m \overset{x}{˙} \cdot 1 = m \overset{x}{˙} = \frac{d}{d t} (m x) .$ So $δ L = \frac{d F}{d t}$ with $F = m x$ . Why this step? The action changes only by a boundary term, so the symmetry is still valid — but the parent's formula demands we subtract $F$ .
Build the corrected charge. $Q = p δ x - F = (m \overset{x}{˙}) (t) - m x = m (\overset{x}{˙} t - x) .$

Verify: For a free particle $\overset{x}{˙} =$ const $\Rightarrow x = x_{0} + \overset{x}{˙} t$ . Then $Q = m (\overset{x}{˙} t - x_{0} - \overset{x}{˙} t) = - m x_{0} = const$ . ✓ It literally encodes the starting position $x_{0}$ — the boost symmetry conserves "where the center of mass was at $t = 0$ ". Had we forgotten $F$ , the naive $p δ x = m \overset{x}{˙} t$ would grow linearly and be nonsense.

Ex 6 — Cell F: canonical $p \neq = m v$ (charged particle)

Worked example Charged particle in a uniform magnetic field

With vector potential $A_{x} = - \frac{1}{2} B y, A_{y} = \frac{1}{2} B x$ (uniform field $B \overset{z}{^}$ ), $L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) + q (\overset{x}{˙} A_{x} + \overset{y}{˙} A_{y}) = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) + \frac{q B}{2} (x \overset{y}{˙} - y \overset{x}{˙}) .$ Find the conserved charge from rotational symmetry — and beware what " $p$ " means.

Forecast: Will the conserved quantity be plain $m (x \overset{y}{˙} - y \overset{x}{˙})$ , or something with an extra $B$ -term? Guess.

Check rotation invariance. Under $δ x = - y, δ y = x$ , the kinetic term is invariant (Ex 3) and the term $x \overset{y}{˙} - y \overset{x}{˙}$ is itself the $z$ -component of $r \times v$ , which is rotation-invariant. So $δ L = 0$ . ✓ Why this step? Uniform field along $\overset{z}{^}$ has full rotational symmetry about $\overset{z}{^}$ .
Use the canonical momenta, not $m \overset{x}{˙}$ . $p_{x} = \frac{\partial L}{\partial x ˙} = m \overset{x}{˙} - \frac{q B}{2} y, p_{y} = \frac{\partial L}{\partial y ˙} = m \overset{y}{˙} + \frac{q B}{2} x .$ Why this step? Mistake 2 in the parent: the Noether charge uses $p_{i} = \partial L / \partial \overset{q}{˙}_{i}$ , which here carries a vector-potential piece.
Assemble the charge. $Q = p_{x} δ x + p_{y} δ y = (m \overset{x}{˙} - \frac{q B}{2} y) (- y) + (m \overset{y}{˙} + \frac{q B}{2} x) (x)$ $= m (x \overset{y}{˙} - y \overset{x}{˙}) + \frac{q B}{2} (x^{2} + y^{2}) .$

Verify: Set $q = 0$ : $Q \to m (x \overset{y}{˙} - y \overset{x}{˙}) = L_{z}$ , the ordinary angular momentum — correct limit. With $B \neq = 0$ , the conserved "canonical angular momentum" gains the $\frac{q B}{2} r^{2}$ term, which is why cyclotron orbits have a fixed guiding center. Units: each term is $kg\cdotm^{2} / s$ (note $[q B] = kg/s$ , times $m^{2}$ ). ✓

Ex 7 — Cell G: limiting / zero input

Worked example Cyclic angle in the limit of vanishing torque

A particle in a potential $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - α ϕ$ , where $α$ is a small constant "torque" parameter. Study the conservation of $p_{ϕ}$ as $α \to 0$ .

Forecast: With $α \neq = 0$ the coordinate $ϕ$ appears in $L$ . Is $p_{ϕ}$ conserved? What happens as $α \to 0$ ?

Test whether $ϕ$ is cyclic. $L$ contains the term $- α ϕ$ , so $\frac{\partial L}{\partial ϕ} = - α \neq = 0$ (for $α \neq = 0$ ). Not cyclic. Why this step? Any explicit appearance of the coordinate breaks the exact shift symmetry.
Euler–Lagrange for $ϕ$ . Using $\frac{d}{d t} p_{ϕ} = \frac{\partial L}{\partial ϕ}$ , $\frac{d}{d t} p_{ϕ} (m r^{2} \dot{ϕ}) = - α \Rightarrow \frac{d p _{ϕ}}{d t} = - α .$ Why this step? This quantifies the breakage: $p_{ϕ}$ drifts at constant rate $- α$ , like a constant applied torque.
Take the limit $α \to 0$ . Then $\frac{d p _{ϕ}}{d t} \to 0$ , and $p_{ϕ} = m r^{2} \dot{ϕ}$ becomes conserved again.

Verify: Integrate: $p_{ϕ} (t) = p_{ϕ} (0) - α t$ . At $α = 0$ this is a flat line (constant) — the symmetry, and its conservation law, are recovered continuously in the limit. This is the honest way "zero input" behaves: no discontinuity, just the torque switching off. ✓

Ex 8 — Cell H: real-world word problem

Worked example Ice-skater pulling in her arms

A skater spins freely (no friction torque). She pulls her arms in, halving her moment of inertia from $I_{0}$ to $I_{0} /2$ . Starting angular speed $ω_{0} = 3 rad/s$ . What is her final angular speed, and which symmetry guarantees the answer?

Forecast: Guess her final spin rate before doing any algebra.

Identify the symmetry. Ice is frictionless and space is empty around her: nothing in the physics picks out a preferred direction of rotation about the vertical → rotational symmetry about $\overset{z}{^}$ . Why this step? Noether: rotational symmetry ⇒ angular momentum $L_{z}$ conserved. This is the law we will use, not "energy" (she does work pulling arms in, so energy is not conserved).
Write $L_{z} = I ω = const$ . $I_{0} ω_{0} = \frac{I _{0}}{2} ω_{f} .$ Why this step? $L_{z} = I ω$ is the rigid-body version of the single-particle Noether charge $m r^{2} \dot{ϕ}$ from Ex 1, summed over the whole body. Conservation means the value before equals the value after.
Solve for $ω_{f}$ . Cancel $I_{0}$ : $ω_{0} = \frac{1}{2} ω_{f}$ , so $ω_{f} = 2 ω_{0} = 2 \times 3 = 6 rad/s .$ Why this step? Halving the inertia must double the spin to keep the product $I ω$ fixed — the familiar skater speed-up.

Verify: Angular momentum before $= I_{0} \cdot 3 = 3 I_{0}$ ; after $= \frac{I _{0}}{2} \cdot 6 = 3 I_{0}$ — equal. ✓ Sanity on energy: $K E = \frac{1}{2} I ω^{2}$ goes from $\frac{1}{2} I_{0} (3)^{2} = 4.5 I_{0}$ to $\frac{1}{2} \cdot \frac{I _{0}}{2} (6)^{2} = 9 I_{0}$ — it doubled, the extra energy supplied by her muscles. Energy not conserved, angular momentum is. That difference is exactly what Noether predicts (only the symmetric quantity is frozen). ✓

Ex 9 — Cell I: exam twist, only ONE class of component conserved

Worked example Particle in a uniform gravitational field

$L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2}) - m g z, \overset{z}{^} up .$ Which momentum components are conserved, and which are not? (Classic "partial symmetry" trap.)

Forecast: Guess how many of $p_{x}, p_{y}, p_{z}$ are conserved.

What the figure shows: the mint double-arrow is a horizontal shift — sliding the picture left/right does not change the height energy shading (lavender bands), so horizontal translation is a genuine symmetry. The coral vertical double-arrow moves the particle to a different shade of the $m g z$ background, i.e. changes $L$ — so vertical translation is not a symmetry.

Scan each coordinate. $L$ contains $z$ (in $- m g z$ ) but not $x$ or $y$ . Why this step? Cyclic = "coordinate absent". Only the coordinates absent from $L$ give conserved momenta.
Conclude for $x, y$ . Both cyclic ⇒ by $\frac{d}{d t} p_{i} = \frac{\partial L}{\partial q _{i}} = 0$ , $p_{x} = m \overset{x}{˙} = const, p_{y} = m \overset{y}{˙} = const .$ Why this step? Horizontal translational symmetry is unbroken by gravity, so horizontal momentum is conserved.
Conclude for $z$ . $\frac{\partial L}{\partial z} = - m g \neq = 0$ , so $\frac{d p _{z}}{d t} = \frac{\partial L}{\partial z} = - m g \neq = 0.$ Vertical momentum is not conserved — gravity breaks vertical translation symmetry.

Verify: This is projectile motion: $\overset{x}{˙}, \overset{y}{˙}$ constant (flat trajectory in the horizontal plane), while $p_{z} = m \overset{z}{˙}$ decreases at rate $m g$ (that's just $F = - m g$ ). Set $g = 0$ and all three become conserved — the free particle. The number of conserved momenta = number of unbroken translation directions. ✓

Ex 10 — Cell J: pathological / edge cases

Worked example When the machinery itself wobbles

Noether's recipe assumes two quiet things: that $L$ gives well-defined momenta, and that the coordinate map is smooth and reversible. Two short cases show what happens when those assumptions fail.

Forecast: In each case below, does a conserved charge still exist? Guess before reading.

Case (i): non-invertible ("singular") Lagrangian. $L = \overset{x}{˙} y - \overset{y}{˙} x .$

Compute the momenta. $p_{x} = \frac{\partial L}{\partial x ˙} = y$ , $p_{y} = \frac{\partial L}{\partial y ˙} = - x$ . Why this step? These do not contain $\overset{x}{˙}, \overset{y}{˙}$ at all — you cannot solve them for the velocities. That is exactly what "non-invertible / singular $L$ " means: the map (velocities) $\to$ (momenta) is not one-to-one.
Is there still a symmetry charge? Rotation $δ x = - y, δ y = x$ : check $δ L$ . One finds $δ L = 0$ (the form is rotation-invariant), and the Noether charge is $Q = p_{x} δ x + p_{y} δ y = y (- y) + (- x) (x) = - (x^{2} + y^{2})$ . Why this step? Noether's charge formula still outputs a quantity, but because $L$ is singular the object is a constraint ( $p_{x} - y = 0$ , $p_{y} + x = 0$ ), not an ordinary evolution law. Such systems need Hamiltonian Mechanics with constraints (Dirac's method), not the naive $p = \partial L / \partial \overset{q}{˙}$ story.
Lesson: the formula survives, the interpretation changes — always check that momenta actually depend on velocities before trusting a "conserved" charge.

Case (ii): singular coordinate transformation. Return to polar coordinates from Ex 1. At the origin $r = 0$ , the angle $ϕ$ is undefined and the map $(r, ϕ) \to (x, y)$ is not invertible (its Jacobian $r$ vanishes).

What breaks? The conserved $p_{ϕ} = m r^{2} \dot{ϕ}$ is still perfectly conserved along the motion — but you cannot read off $ϕ$ at $r = 0$ . Why this step? A particle passing exactly through the origin has $r = 0$ , so $p_{ϕ} = m \cdot 0 \cdot \dot{ϕ} = 0$ ; the coordinate singularity forces the charge to $0$ there, consistent with a straight line through the center carrying zero angular momentum.
Lesson: a coordinate singularity does not destroy the conservation law; it only makes that chart a bad place to describe the motion. Switch to Cartesian $x, y$ (smooth everywhere) and $L_{z} = m (x \overset{y}{˙} - y \overset{x}{˙})$ stays finite and conserved.

Verify: Case (i): $\frac{d}{d t} (- (x^{2} + y^{2})) = - 2 (x \overset{x}{˙} + y \overset{y}{˙})$ . The constrained equations of motion of this $L$ give $\overset{x}{˙} = x, \overset{y}{˙} = y$ is not forced; instead the constraints pin $x, y$ so that $x \overset{x}{˙} + y \overset{y}{˙} = 0$ , keeping $Q$ constant — the charge is consistent. Case (ii): at $r = 0$ , $p_{ϕ} = 0$ and the Cartesian $L_{z} = m (x \overset{y}{˙} - y \overset{x}{˙})$ evaluated at $x = y = 0$ is also $0$ ; the two descriptions agree. ✓

Recall Self-test clozes

A driven oscillator conserves energy only in the limit ::: $F_{0} \to 0$ (drive off), restoring time-translation symmetry. For a boost $x \to x + ϵ t$ , the conserved charge is ::: $Q = m (\overset{x}{˙} t - x) = - m x_{0}$ , the initial position. A charged particle's conserved "angular momentum" gains the extra term ::: $\frac{q B}{2} (x^{2} + y^{2})$ , because $p_{i} = \partial L / \partial \overset{q}{˙}_{i}$ includes the vector potential. In uniform gravity the conserved momenta are ::: $p_{x}$ and $p_{y}$ (horizontal), never $p_{z}$ . If the momenta don't contain the velocities (singular $L$ ), the Noether output is ::: a constraint, not an evolution law — use constrained Hamiltonian methods.

Deeper structure of these symmetries lives in Symmetry Groups & Lie Algebras and Gauge Symmetry; the energy charge connects to Hamiltonian Mechanics.