Noether's Theorem

In the spring of 1915, Göttingen had a crisis of bookkeeping. Einstein's brand-new general relativity seemed to misplace the most sacred quantity in physics: energy. The local energy conservation law that every mechanical and electromagnetic theory had obeyed refused to take its standard form in a curved spacetime, and David Hilbert and Felix Klein — the two mathematical powers of the age — could not settle whether this was a flaw, a feature, or a confusion. So they recruited the one person they judged capable of resolving it: Emmy Noether, an algebraist then working in Erlangen without pay or title. Her 1918 answer did far more than repair general relativity's ledger. It revealed why anything is conserved at all. Noether's theorem is the exchange rate between geometry and dynamics: every continuous symmetry of the action buys exactly one conserved quantity, and the theorem hands you the explicit formula. Shift the whole experiment sideways and nothing changes? Momentum is conserved. Turn it? Angular momentum. Run it tomorrow instead of today? Energy. Watch it from a passing train? A subtler charge that pins the centre of mass to a straight line. The three great conservation laws you have been using since your first mechanics course stop being separate axioms and become one theorem evaluated three times. You arrive at this article with the whole toolkit assembled: the Lagrangian, the action, the Euler–Lagrange equations, and — from the symmetry article — the audit that certifies a transformation as a symmetry. This is the payoff page: here the audit's outputs are converted, line by line, into the constants of the motion.

Key takeaways

Noether's theorem: every continuous symmetry of the action — every one-parameter family passing the audit $\delta L = \varepsilon\,\mathrm{d}F/\mathrm{d}t$ — yields a quantity conserved along every solution of the equations of motion.
For a point symmetry with generators $\xi_i$ the charge is $Q = \sum_i p_i\,\xi_i - F$ , built from the generalized momenta $p_i = \partial L/\partial\dot q_i$ ; strict symmetries have $F = 0$ .
Transformations that shift time (generator $\tau$ ) fit the same formula with $\xi_i \to \xi_i - \dot q_i\tau$ plus a term $L\tau$ ; pure time translation gives the energy function $h$ .
The big three: spatial translation $\Rightarrow$ total momentum $\mathbf P$ , rotation $\Rightarrow$ angular momentum $\mathbf L$ , time translation $\Rightarrow$ energy $h$ . The Galilean boost, a quasi-symmetry, pays the centre-of-mass charge $\mathbf G = M\mathbf R - \mathbf P t$ .
The proof needs the equations of motion: charges are constant on shell, along actual motions, not for arbitrary paths.
Discrete symmetries (reflections) have no generator to contract with the momenta and pay no charge; symmetries of the equations of motion that fail the action audit (Kepler's dilation) pay nothing either.
Key idea: if the action is invariant up to a boundary term under $q_i \to q_i + \varepsilon\,\xi_i$ , then $Q = \sum_i \frac{\partial L}{\partial\dot q_i}\xi_i - F$ is constant along every solution — symmetry of the action, not the choice of coordinates, is the source of every conservation law.

The core insight

If the action is invariant under a continuous transformation with infinitesimal generators $\xi_i$ (up to the boundary term $F$ ), then $Q=\sum_i \frac{\partial L}{\partial \dot q_i}\,\xi_i - F$ is conserved along solutions of the Euler–Lagrange equations. Symmetry of the action, not the explicit form of the coordinates, is the source of every conservation law.

Where this lives today

The Standard Model is, operationally, a Noether ledger, and testing conservation laws is how physics interrogates its symmetries at the highest precision available. Electric charge conservation — the Noether charge of the $U(1)$ phase symmetry — is tested by hunting for the forbidden decay of the electron: the current bound on its lifetime exceeds $6.6\times10^{28}$ years (Borexino, from the non-observation of $e^- \to \nu\gamma$ ), some eighteen orders of magnitude beyond the age of the universe. Baryon number, by contrast, is believed to be only approximately symmetric — grand-unified theories predict proton decay, and Super-Kamiokande's bound of a few $\times10^{34}$ years on the proton lifetime is the sharpest constraint on that physics. The experimental logic is pure Noether run backwards: a violated conservation law would be the fingerprint of a broken or absent symmetry in the underlying action. The theorem also earns its keep as an engineering tool. Gravitational-wave astronomy balances its books with the energy and angular-momentum fluxes carried off by radiation — Noether charges of the asymptotic spacetime — and waveform models for the LIGO–Virgo–KAGRA network are validated by checking that a simulated binary's orbital decay matches the radiated flux to high order. In numerical work generally, the charges are the sharpest available diagnostics: a symplectic orbit integrator that respects rotational symmetry conserves angular momentum to machine precision for billions of steps, and a drifting Noether charge is usually the first symptom that a simulation, not the physics, has broken a symmetry.

The physical picture

Think of the theorem as a machine with a slot and a tray. Into the slot goes a certified symmetry — one of the dials from the symmetry article: the generators $\xi_i$ , $\tau$ , and the boundary function $F$ that the audit produced. Out of the tray comes a receipt: an explicit function of positions and velocities whose value the dynamics can never change. The machine's internals are three lines of calculus and one use of the equations of motion; no integration of trajectories, no cleverness, no per-problem tricks.

Figure

The Noether ledger — each certified symmetry dial pays out one conserved charge

Why should symmetry have anything to do with conservation? Here is the mechanism in words before the calculus. A conserved quantity is a number the motion carries unchanged from the start of the path to the end — a statement about endpoints. A symmetry is a deformation of the whole path that the action cannot feel — and when you deform a path that already satisfies the equations of motion, the action's response comes only from the endpoints (the bulk contribution is killed by the Euler–Lagrange equations, the same cancellation that Hamilton's principle is built on). So the symmetry says "the action does not change," the dynamics says "only the endpoint terms could have changed it," and the collision of the two statements forces the endpoint term at the final time to equal the endpoint term at the initial time. That endpoint term — momenta contracted with the symmetry direction, minus the boundary function — is the Noether charge.

Figure

Why the proof works — deform a solution along the symmetry and only endpoint ledgers survive

Two features of the machine deserve emphasis before the derivations. First, its universality: nothing in the argument cares which system you study — a pendulum, a galaxy, a field — only that the dynamics comes from an action and the symmetry passes the audit. Second, its honesty about fine print: if the audit returned a nonzero boundary function $F$ (the quasi-invariant case, like the Galilean boost), that same $F$ must be subtracted from the charge. Forgetting the $-F$ is the single most common way to manufacture a "conserved" quantity that isn't; the machine only pays if you feed it the full audit report.

The theorem for point symmetries

Set the stage precisely. The system has Lagrangian $L(q, \dot q, t)$ and generalized momenta $p_i = \partial L/\partial\dot q_i$ — the momenta of the generalized-forces article, which need not look like $m\dot q$ . A one-parameter family of point transformations acts as $q_i \to q_i + \varepsilon\,\xi_i(q, t)$ with time untouched ( $\tau = 0$ ), and has passed the symmetry audit: to first order in $\varepsilon$ ,

\delta L = \varepsilon\,\frac{\mathrm{d}F(q,t)}{\mathrm{d}t} \qquad (1)

where: $\xi_i(q,t)$ — the infinitesimal generator of the family; $F(q,t)$ — the boundary function certified by the audit, identically zero for strict symmetries. Equation (1) is exactly the criterion of the symmetry article, specialized to $\tau = 0$ ; nothing new has been assumed. What is new is the payout:

Q \;=\; \sum_i \frac{\partial L}{\partial\dot q_i}\,\xi_i \;-\; F \;=\; \sum_i p_i\,\xi_i - F, \qquad \frac{\mathrm{d}Q}{\mathrm{d}t} = 0 \ \text{ on solutions} \qquad (2)

where: the sum runs over all generalized coordinates, and "on solutions" means along any path obeying the Euler–Lagrange equations. Read the structure physically: the charge is the projection of the momenta onto the symmetry direction, with the audit's endpoint bookkeeping subtracted. A translation has $\xi_i$ constant, so $Q$ is a momentum component; a rotation has $\xi$ winding around an axis, so $Q$ is an angular momentum; a boost's $\xi$ grows with time and drags in $F$ , so $Q$ mixes momenta with positions. One formula, every case.

Derivation AThe Noether Charge for Point Symmetries

The goal is equation (2): for any point transformation passing the audit (1), the quantity $Q = \sum_i p_i\xi_i - F$ is constant along every solution. The strategy is a two-line bookkeeping match — compute $\delta L$ once by the chain rule and once from the audit, then use the Euler–Lagrange equations to compress the chain-rule version into a single total time derivative. Watch where the equations of motion enter: without them the argument produces only an identity between derivatives, true for every path and conserving nothing; the on-shell substitution is the step that converts calculus into physics. And watch the boundary function ride along untouched until the last line, where it lands inside the conserved combination with a minus sign.

Step 1. Vary the Lagrangian along the family. Under $\delta q_i = \varepsilon\,\xi_i$ , velocities transform as $\delta\dot q_i = \varepsilon\,\dot\xi_i$ (differentiate the transformation along the path), so the chain rule gives

\delta L = \varepsilon\sum_i\left(\frac{\partial L}{\partial q_i}\,\xi_i + \frac{\partial L}{\partial\dot q_i}\,\dot\xi_i\right)

Why $\delta\dot q = \varepsilon\dot\xi$ : the transformed path is $q_i(t) + \varepsilon\xi_i(q(t),t)$ , and its time derivative differs from $\dot q_i$ by $\varepsilon$ times the total time derivative of $\xi_i$ along the path. No equations of motion used yet — this holds for any path.

Step 2. Go on shell. Along a solution the Euler–Lagrange equations let us trade $\partial L/\partial q_i$ for $\mathrm{d}p_i/\mathrm{d}t$ :

\delta L = \varepsilon\sum_i\left(\dot p_i\,\xi_i + p_i\,\dot\xi_i\right) = \varepsilon\,\frac{\mathrm{d}}{\mathrm{d}t}\sum_i p_i\,\xi_i

Why this compresses: $\dot p\,\xi + p\,\dot\xi$ is the product rule read backwards. This is the pivotal move of the whole theorem — on shell, the response of $L$ to a symmetry deformation is a pure total derivative, the "ledger flow" of the momenta projected on the symmetry direction.

Step 3. Match against the audit. The same $\delta L$ is, by the symmetry certification (1), equal to $\varepsilon\,\mathrm{d}F/\mathrm{d}t$ . Setting the two expressions equal and moving everything to one side:

\frac{\mathrm{d}}{\mathrm{d}t}\left(\sum_i p_i\,\xi_i - F\right) = 0

Step 4. Name the constant: $Q = \sum_i p_i\xi_i - F$ is conserved along the solution — equation (2). For a strict symmetry ( $F = 0$ ) the charge is the bare momentum projection $\sum_i p_i\xi_i$ . Verified: for a cyclic coordinate $q_k$ (translation generator $\xi_i = \delta_{ik}$ , $F = 0$ ) the charge reduces to $Q = p_k$ , reproducing the conservation of conjugate momenta of the cyclic-coordinates article; matches Goldstein §13.7 and Landau–Lifshitz §7.

The cyclic-coordinates result is thus the theorem's simplest evaluation: a coordinate absent from $L$ is precisely a translation symmetry along that coordinate, and its conserved conjugate momentum is precisely the Noether charge. But equation (2) is strictly stronger, and the strength shows in two directions at once. A rotation mixes several coordinates — no single one is cyclic in Cartesian coordinates, yet the charge $\sum_i p_i\xi_i$ exists regardless of coordinate choice. And a quasi-symmetry conserves a charge even though no choice of coordinates makes anything cyclic: the boost below is the canonical example.

Time transformations and the general charge

Point transformations leave one dial untouched: time itself. The audit grammar allows $t \to t + \varepsilon\,\tau(q,t)$ , and mechanics needs that freedom in exactly one place — time translation, whose conserved partner ought to be energy. The result of folding $\tau$ in is the general charge, the formula that covers every case in this article:

Q \;=\; \sum_i p_i\left(\xi_i - \dot q_i\,\tau\right) \;+\; L\,\tau \;-\; F \qquad (3)

where: $\tau(q,t)$ — the time-shift generator; all other symbols as in (2), to which (3) collapses when $\tau = 0$ . The new terms deserve a physical reading before the derivation: $-\sum_i p_i\dot q_i\,\tau + L\tau = -\big(\sum_i p_i\dot q_i - L\big)\tau$ , and the object in parentheses is the energy function $h$ of the Jacobi-integral article. Time transformations pay in the currency of energy.

Go deeper · Derivation B · 2 minFolding Time into the Charge and Harvesting Energy

Derivation BFolding Time into the Charge and Harvesting Energy

The goal is the general charge (3) and its most important evaluation: pure time translation pays the energy function, equation (4) below. The strategy is a reduction to the case already solved — a transformation that shifts time can be traded for an equivalent point transformation at fixed time, at the price of one extra bookkeeping term from the stretched integration measure; Derivation A then applies verbatim. Watch the sign logic in the trade: if the clock is pushed forward, the configuration read at the old clock time lags behind, so the effective coordinate shift acquires $-\dot q_i\tau$ — getting that sign wrong flips the energy charge into nonsense.

Step 1. Trade the time shift for a fixed-time variation. Under $q_i \to q_i + \varepsilon\xi_i$ , $t \to t + \varepsilon\tau$ , compare the transformed path with the original at the same clock reading: the configuration has effectively shifted by

\bar\xi_i \;=\; \xi_i - \dot q_i\,\tau

Why the $-\dot q_i\tau$ : advancing time by $\varepsilon\tau$ while dragging the path means the point you now read at time $t$ is the one the old path occupied $\varepsilon\tau$ earlier — a lag of $\varepsilon\,\dot q_i\tau$ along the motion. This "fixed-time variation" is the standard device for comparing paths that live on shifted time axes.

Step 2. Account for the stretched measure. The action integrates $L\,\mathrm{d}t$ , and $\mathrm{d}t' = (1 + \varepsilon\,\mathrm{d}\tau/\mathrm{d}t)\,\mathrm{d}t$ , so the measure contributes an additional first-order piece $\varepsilon\,\mathrm{d}(L\tau)/\mathrm{d}t$ to the integrand's change — a pure total derivative that joins the ledger alongside $F$ .

Step 3. Run Derivation A on $\bar\xi_i$ with the enlarged boundary bookkeeping. On shell, the variation compresses exactly as before, and the conserved combination becomes

Q = \sum_i p_i\,\bar\xi_i + L\tau - F = \sum_i p_i\big(\xi_i - \dot q_i\tau\big) + L\tau - F

Why no new dynamics is needed: every ingredient — chain rule, on-shell compression, matching against the audit — is Derivation A's; the time shift only re-routed which terms sit in the boundary ledger.

Step 4. Evaluate for pure time translation: $\xi_i = 0$ , $\tau = 1$ , and $F = 0$ provided $\partial L/\partial t = 0$ (that is the audit for this dial — shifting a clock that appears nowhere in $L$ changes nothing):

Q = L - \sum_i p_i\,\dot q_i = -\,h \qquad (4)

where: $h = \sum_i p_i\dot q_i - L$ — the energy function. The charge is $-h$ ; a constant multiple of a conserved quantity is conserved, so the sign is pure convention and one quotes $h$ itself. Verified: matches Goldstein §13.7's general charge; (3) collapses to (2) at $\tau = 0$ ; and differentiating $h$ directly along solutions gives $\mathrm{d}h/\mathrm{d}t = -\partial L/\partial t$ , the Jacobi-integral article's master equation, confirming conservation exactly when the audit passes.

The theorem thus derives the master fact that the Jacobi-integral article establishes by direct differentiation — $h$ is constant exactly when $L$ carries no explicit time dependence — and reveals it as the time-translation entry of the same ledger that holds momentum and angular momentum. When $h$ equals the mechanical energy $T + V$ , and when it is conserved without being the energy, is that article's business; here we only needed the charge.

The big three, plus the stranger

For a closed $N$ -body system, $L = \tfrac12\sum_a m_a\dot{\mathbf r}_a^2 - V\big(|\mathbf r_a - \mathbf r_b|\big)$ , the symmetry article certified four families. Feeding each audit report into (2)–(3):

\mathbf P = \sum_a m_a\dot{\mathbf r}_a, \qquad \mathbf L = \sum_a \mathbf r_a\times m_a\dot{\mathbf r}_a, \qquad h = T + V \qquad (5)

where: translations ( $\boldsymbol\xi_a = \hat{\mathbf n}$ , one charge per direction) pay the momentum components, rotations ( $\boldsymbol\xi_a = \hat{\mathbf n}\times\mathbf r_a$ , the triple-product rearrangement $\sum_a m_a\dot{\mathbf r}_a\cdot(\hat{\mathbf n}\times\mathbf r_a) = \hat{\mathbf n}\cdot\mathbf L$ converting projection into the familiar cross product) pay the angular momentum components, and time translation pays the energy. Ten symmetries of the Galilean group, ten charges — the last three from the stranger of the family:

\mathbf G = M\mathbf R - \mathbf P\,t = \text{const} \qquad (6)

where: $M = \sum_a m_a$ , $\mathbf R$ — the centre of mass, and $\mathbf G$ — the boost charge, constructed in Derivation B below. Constancy of $\mathbf G$ says $\mathbf R(t) = (\mathbf G + \mathbf P t)/M$ : the centre of mass moves in a straight line at constant velocity $\mathbf P/M$ , whatever storm of internal forces rages inside. That familiar theorem is, secretly, a conservation law — the payout of a symmetry nobody meets in introductory courses because its Lagrangian audit is quasi, not strict.

Go deeper · Derivation C · 1 minThe Boost Charge and the Straight-Line Centre of Mass

Derivation CThe Boost Charge and the Straight-Line Centre of Mass

The goal is equation (6): the conserved charge of the Galilean boost, the one entry of the ledger that only exists because of the boundary-function allowance. The strategy is pure harvest — the symmetry article's audit already certified the boost with generators $\boldsymbol\xi_a = \hat{\mathbf n}\,t$ , $\tau = 0$ and boundary function $F = \sum_a m_a\,\mathbf r_a\cdot\hat{\mathbf n} = M\mathbf R\cdot\hat{\mathbf n}$ ; we contract with the momenta per equation (2), simplify, and then verify conservation directly against the dynamics, which doubles as a check on the sign conventions. Watch the $-F$ term: without it the candidate charge would be $\mathbf P\cdot\hat{\mathbf n}\,t$ , which is manifestly not constant unless the trivial case $\mathbf P = 0$ holds — the boundary function is not a refinement here but the whole content.

Step 1. Contract momenta with the generators: $\sum_a \mathbf p_a\cdot\boldsymbol\xi_a = \sum_a m_a\dot{\mathbf r}_a\cdot\hat{\mathbf n}\,t = \big(\mathbf P\cdot\hat{\mathbf n}\big)\,t$ .

Step 2. Subtract the audited boundary function:

Q = \big(\mathbf P\cdot\hat{\mathbf n}\big)\,t - M\,\mathbf R\cdot\hat{\mathbf n} = -\,\hat{\mathbf n}\cdot\big(M\mathbf R - \mathbf P t\big)

Why the overall sign is harmless: $-Q$ is conserved exactly when $Q$ is; convention fixes the charge as $\mathbf G = M\mathbf R - \mathbf P t$ , one component per boost direction $\hat{\mathbf n}$ , three in all.

Step 3. Verify conservation directly — a sign-proof cross-check that needs no variational machinery:

\frac{\mathrm{d}\mathbf G}{\mathrm{d}t} = M\dot{\mathbf R} - \dot{\mathbf P}\,t - \mathbf P = \mathbf P - 0 - \mathbf P = \mathbf 0

Why each piece: $M\dot{\mathbf R} = \sum_a m_a\dot{\mathbf r}_a = \mathbf P$ by definition of the centre of mass, and $\dot{\mathbf P} = 0$ is the already-banked translation charge of equation (5). The boost charge is conserved because momentum is — the ledger's entries interlock.

Step 4. Interpret. Solve $\mathbf G = M\mathbf R - \mathbf P t$ for the centre of mass: $\mathbf R(t) = \mathbf G/M + (\mathbf P/M)\,t$ . The charge $\mathbf G/M$ is the centre of mass's launch position — the intercept of its straight worldline — and its conservation is the statement that internal forces, however violent, can never bend that line. Worked example 2 watches $\mathbf G$ sail unchanged through an elastic collision. Verified: matches Landau–Lifshitz §8 and Goldstein §13.7's quasi-invariance treatment; setting $t = 0$ gives $\mathbf G = M\mathbf R(0)$ , fixing the constant's meaning, and dimensional check $[\mathbf G] = \mathrm{kg\cdot m}$ is consistent throughout.

Which symmetry pays which charge

certified symmetry (audit input)	generators	charge from (2)–(3)	everyday name
translation along $\hat{\mathbf n}$	$\boldsymbol\xi_a = \hat{\mathbf n}$ , $F = 0$	$\mathbf P\cdot\hat{\mathbf n}$	linear momentum
rotation about $\hat{\mathbf n}$	$\boldsymbol\xi_a = \hat{\mathbf n}\times\mathbf r_a$ , $F = 0$	$\mathbf L\cdot\hat{\mathbf n}$	angular momentum
time translation	$\xi = 0$ , $\tau = 1$ , $F = 0$	$-h$ (quote $h$ )	energy function
Galilean boost along $\hat{\mathbf n}$	$\boldsymbol\xi_a = \hat{\mathbf n}t$ , $F = M\mathbf R\cdot\hat{\mathbf n}$	$\hat{\mathbf n}\cdot(M\mathbf R - \mathbf P t)$	centre-of-mass law
cyclic coordinate $q_k$	$\xi_i = \delta_{ik}$ , $F = 0$	$p_k$	conjugate momentum

Limiting cases

case	result	meaning
strict symmetry, $F = 0$ , $\tau = 0$	$Q = \sum_i p_i\xi_i$	bare momentum projection — the textbook charge
quasi-symmetry, $F \neq 0$	$Q = \sum_i p_i\xi_i - F$	the boundary function is part of the charge, not garbage
pure time translation	$Q = -h$	energy is the time entry of the same ledger
broken symmetry, $\delta L = \varepsilon\,B$ with $B \neq \mathrm{d}F/\mathrm{d}t$	$\mathrm{d}(\sum_i p_i\xi_i)/\mathrm{d}t = B$	the "charge" leaks at exactly the audit-failure rate — torque, force, or power
discrete symmetry (reflection)	no generator $\xi$	nothing to contract with $p_i$ : no charge, only a classification of solutions
EOM symmetry failing the action audit (Kepler dilation)	no charge	the theorem consumes action symmetries only

The fourth row is worth a pause, because it upgrades the theorem from a yes/no oracle into a quantitative tool. Rerun Derivation A without assuming the audit passes: Steps 1–2 still give $\delta L = \varepsilon\,\mathrm{d}(\sum p_i\xi_i)/\mathrm{d}t$ on shell, so whatever $\delta L$ actually equals is the leak rate of the would-be charge. For a rotation audit failing by $\delta L = -\varepsilon\,3k\,xy$ (the lopsided trap of the symmetry article's worked example 2), the angular momentum drains at $\mathrm{d}L_z/\mathrm{d}t = -3kxy$ — the torque. Broken symmetries do not merely fail to pay; they tell you the precise bill.

Common trapA quasi-symmetry conserves nothing, and the charge is always just Σ pᵢξᵢ

Both halves of this instinct are wrong, and each error costs real marks. First: quasi-invariance ( $\delta L = \varepsilon\,\mathrm{d}F/\mathrm{d}t$ with $F \neq 0$ ) is a full-strength symmetry with a full-strength conservation law — the Galilean boost changes the Lagrangian at every instant of every motion, yet pays the exact charge $\mathbf G = M\mathbf R - \mathbf Pt$ . Dismissing quasi-symmetries throws away three of the Galilean group's ten charges. Second: the naive projection $\sum_i p_i\xi_i$ is the charge only when $F = 0$ and $\tau = 0$ . For the boost, $\sum_a \mathbf p_a\cdot\boldsymbol\xi_a = \mathbf P\cdot\hat{\mathbf n}\,t$ grows linearly in time — visibly not conserved — and only subtracting $F = M\mathbf R\cdot\hat{\mathbf n}$ produces a constant. For time translation the extra terms flip the answer entirely, from $0$ to $-h$ . The safe procedure is mechanical: run the audit, record $\xi_i$ , $\tau$ , and $F$ , then evaluate the full formula (3). Never guess the charge from the generator alone.

Worked examples

1 · The falling charge (confidence builder)

A ball of mass $m = 2\ \mathrm{kg}$ is thrown straight up at $\dot z(0) = 5\ \mathrm{m/s}$ in uniform gravity, $L = \tfrac12 m\dot z^2 - mgz$ . Vertical translation $z \to z + \varepsilon$ is not a strict symmetry — yet Noether still pays. Find the charge and verify it numerically at $t = 0$ and $t = 1\ \mathrm{s}$ .

Step 1. Audit the dial. Under $z \to z + \varepsilon$ (generator $\xi = 1$ ), the kinetic term is untouched and the potential shifts: $\delta L = -mg\,\varepsilon$ . This is nonzero — strict invariance fails, momentum $p_z$ is not conserved (no surprise: gravity is a force). But $-mg$ is a constant, and a constant is trivially a total time derivative: $-mg = \mathrm{d}(-mgt)/\mathrm{d}t$ . The audit passes in quasi mode with $F = -mgt$ .

Step 2. Build the charge from equation (2): $Q = p_z\,\xi - F = m\dot z + mgt$ .

Step 3. Check conservation analytically: $\mathrm{d}Q/\mathrm{d}t = m\ddot z + mg = 0$ by the equation of motion $m\ddot z = -mg$ . The charge works by construction — it is the launch momentum, reconstructed at any later time by adding back what gravity has drained.

Step 4. Check numerically. At $t = 0$ : $Q = 2\times5 + 0 = \boxed{10\ \mathrm{kg\,m/s}}$ . At $t = 1\ \mathrm{s}$ : $\dot z = 5 - 9.81 = -4.81\ \mathrm{m/s}$ , so $p_z = -9.62\ \mathrm{kg\,m/s}$ and $mgt = 2\times9.81\times1 = 19.62$ , giving $Q = -9.62 + 19.62 = \boxed{10.00\ \mathrm{kg\,m/s}}$ . The ball has turned around and the momentum has swung by almost $20$ units, yet the Noether charge sits at $10$ exactly.

Step 5. Moral: even in a system with no conserved momentum and a manifestly position-dependent potential, the theorem finds structure — because the audit, not the conservation of $L$ 's pieces, is the criterion. (Run the structure test of the symmetry article: $\delta L$ velocity-free and depending on $t$ alone always integrates to an $F$ .)

2 · The boost charge sails through a collision (exam level)

Two carts on a frictionless track: $m_1 = 3\ \mathrm{kg}$ starts at $x_1 = 0$ moving at $+2\ \mathrm{m/s}$ ; $m_2 = 1\ \mathrm{kg}$ starts at $x_2 = 10\ \mathrm{m}$ moving at $-2\ \mathrm{m/s}$ . They collide elastically. Compute the boost charge $G = M X_{\mathrm{cm}} - P t$ before the collision, find the post-collision state, and verify $G$ after — through an interaction where the individual momenta change violently.

Step 1. The ledger before. Total momentum $P = 3(2) + 1(-2) = 4\ \mathrm{kg\,m/s}$ ; total mass $M = 4\ \mathrm{kg}$ ; centre of mass $X_{\mathrm{cm}}(0) = \big(3(0) + 1(10)\big)/4 = 2.5\ \mathrm{m}$ . Boost charge at $t = 0$ : $G = 4(2.5) - 4(0) = \boxed{10\ \mathrm{kg\,m}}$ .

Step 2. Sanity-check $G$ mid-flight, at $t = 1\ \mathrm{s}$ (before impact): $x_1 = 2$ , $x_2 = 8$ , so $MX_{\mathrm{cm}} = 3(2) + 1(8) = 14$ and $G = 14 - 4(1) = 10$ . ✓

Step 3. The collision. Closing speed $4\ \mathrm{m/s}$ over a $10\ \mathrm{m}$ gap: impact at $t = 2.5\ \mathrm{s}$ , at $x = 5\ \mathrm{m}$ . Elastic formulas give $v_1' = \dfrac{(m_1 - m_2)v_1 + 2m_2v_2}{M} = \dfrac{(2)(2) + 2(1)(-2)}{4} = 0$ and $v_2' = \dfrac{(m_2 - m_1)v_2 + 2m_1v_1}{4} = \dfrac{(-2)(-2) + 2(3)(2)}{4} = 4\ \mathrm{m/s}$ . Check the other charges: $P' = 3(0) + 1(4) = 4$ ✓; kinetic energy $8\ \mathrm{J}$ before and after ✓.

Step 4. The ledger after. At $t = 4\ \mathrm{s}$ (1.5 s past impact): $x_1 = 5$ (cart 1 is parked), $x_2 = 5 + 4(1.5) = 11\ \mathrm{m}$ . Then $MX_{\mathrm{cm}} = 3(5) + 1(11) = 26$ and $G = 26 - 4(4) = \boxed{10\ \mathrm{kg\,m}}$ . Conserved — though neither cart's momentum survived the collision, and the charge explicitly contains the time.

Step 5. Interpret via Derivation B: $X_{\mathrm{cm}}(t) = G/M + (P/M)t = 2.5 + 1.0\,t$ — the centre of mass crossed the collision at a serene $1\ \mathrm{m/s}$ , its straight worldline the visible face of boost symmetry. A high-speed camera pointed at the melee would find one dull, perfectly linear feature: that line is a Noether charge.

Figure

Two carts collide, momenta reshuffle — the boost charge holds the centre of mass to a straight worldline

The view beyond particles

Everything above generalizes with almost embarrassing smoothness once the discrete label $i$ becomes a continuous one. For a field $\phi(x, t)$ — a displacement at every point of space — the action is an integral of a Lagrangian density over space and time, and running the identical argument produces not a conserved charge but a conserved current: a density $\rho$ and a flux $\mathbf j$ obeying the continuity equation $\partial\rho/\partial t + \nabla\cdot\mathbf j = 0$ . Conservation becomes local: charge cannot teleport, it must flow. Time translation pays the energy density and its flux (the stress-energy tensor); internal phase rotations of a complex field pay electric charge; and in the gauge theories of the Standard Model every conserved quantum number — baryon number, lepton number, colour, isospin — is a Noether current of some continuous symmetry. That story, including the honest derivation and the profound subtlety of local (gauge) symmetries, belongs to the action-principle-to-QFT article; here it is enough to know that the machine you have just learned to crank is the same machine, with the sum over $i$ promoted to an integral over space.

Origins · 1 minEmmy Noether and how this was discovered

Emmy Noether

1882–1935 · Erlangen, Göttingen, Bryn Mawr

Noether was an algebraist who solved physics' deepest bookkeeping problem almost in passing. Barred from regular enrolment as a woman, she audited courses at Erlangen, earned her doctorate in 1907, and then worked for years unpaid, lecturing under her father's name. When Hilbert and Klein summoned her to Göttingen in 1915 to untangle energy conservation in general relativity, the faculty senate blocked her habilitation — prompting Hilbert's famous retort that a university senate is not a bathhouse. Her 1918 paper Invariante Variationsprobleme settled the matter with two theorems: the first (this article) ties every global continuous symmetry to a conservation law; the second shows that local gauge symmetries yield identities instead — explaining precisely why energy in general relativity behaves so strangely. She considered the paper a minor detour: her life's work was the revolution in abstract algebra (rings, ideals, the Noetherian condition that bears her name) conducted from Göttingen until 1933, when the Nazi race laws expelled her. She taught her final two years at Bryn Mawr College, dying suddenly at 53. Einstein's obituary called her the most significant creative mathematical genius produced since the higher education of women began.

History

1788

— Lagrange's Mécanique analytique links absent coordinates to conserved momenta — the special case waiting a century for its general theorem.

1915

— Hilbert and Klein bring Noether to Göttingen to resolve the energy-conservation puzzle of general relativity.

1918

— Invariante Variationsprobleme: both Noether theorems, presented by Klein because Noether could not address the Göttingen society herself.

1954

— Yang and Mills invert the logic — demanding local symmetry to generate forces — making Noether currents the architecture of particle physics.

1971

— 't Hooft proves gauge theories renormalizable; the Noether-current structure of the Standard Model becomes calculable physics.

Going further

References. Goldstein, Classical Mechanics, 3rd ed., §13.7 (Noether's theorem) and §2.6 · Landau & Lifshitz, Mechanics, 3rd ed., §§6–9 (momentum, energy, angular momentum) · Morin, Introduction to Classical Mechanics, Ch. 6 · Taylor, Classical Mechanics, Ch. 7 · Noether, E. (1918), Invariante Variationsprobleme, Nachr. Ges. Wiss. Göttingen

Provenance. Equations cross-checked against Goldstein §13.7 (Noether's theorem, general charge with time and boundary terms), Landau–Lifshitz Mechanics §§6–9 (momentum, energy function, angular momentum), and Morin Ch. 6. A verifier should: (i) re-derive the merge of the two variation terms into a single total time derivative using the Euler–Lagrange equations; (ii) confirm the angular-momentum charge via the scalar triple product identity; (iii) check that $\mathrm{d}h/\mathrm{d}t=-\partial L/\partial t$ ; (iv) verify the Galilean-boost quasi-symmetry charge $\sum_a m_a\mathbf r_a - t\mathbf P$ . No numerical constants are asserted beyond standard kinematic identities. Worked-example numbers on this page (falling charge $Q = m\dot z + mgt = 10\ \mathrm{kg\,m/s}$ at $t = 0$ and $t = 1$ s for $m = 2$ kg, $\dot z(0) = 5$ m/s; collision example $P = 4$ kg·m/s, impact at $t = 2.5$ s and $x = 5$ m, $v_1' = 0$ , $v_2' = 4$ m/s, $G = 10$ kg·m at $t = 0, 1, 4$ s; electron-lifetime bound $6.6\times10^{28}$ yr) are illustrative arithmetic recomputed from the displayed formulas and the cited experimental bound.

Builds on: The Lagrangian Function · The Action Functional · The Euler-Lagrange Equations · Generalized Forces and Generalized Momenta · Generalized Coordinates and Configuration Space Leads to: Cyclic Coordinates and Conservation of Conjugate Momenta · The Energy Function and Jacobi Integral · Hamilton's Principle of Stationary Action

Do I really need the equations of motion for conservation, or is the charge constant along any path?

You need them, and seeing where is the best way to understand the proof. Step 1 of Derivation A is pure calculus, valid for every path. The compression in Step 2 — trading $\partial L/\partial q_i$ for $\dot p_i$ — is exactly the Euler–Lagrange equations, i.e. the assumption that the path is a solution. Off shell, the theorem still says something, just something weaker: $\delta L - \mathrm{d}(\sum p_i\xi_i)/\mathrm{d}t$ equals the Euler–Lagrange expression contracted with $\xi$ , an identity for arbitrary paths. Evaluate $Q$ along some random wiggly curve and it will drift freely. Physically this is as it must be: conservation laws are facts about dynamics, not about kinematic bookkeeping, and a formula constant on all conceivable paths would carry no information about the actual motion.

My pendulum clock obeys the same laws today and tomorrow, yet it runs down. Doesn't that contradict the time-translation charge?

The theorem is innocent; the system is guilty of not being closed. Noether's theorem operates on systems whose complete dynamics comes from a Lagrangian, and the clock-with-friction is not such a system: the escapement, air, and bearings leak energy into microscopic degrees of freedom that your $L$ for the pendulum bob simply does not contain. Two honest fixes. Enlarge the system — bob plus air molecules plus everything — and the total, now genuinely time-translation invariant, conserves total energy exactly; the clock's mechanical energy merely migrates into heat, which is bookkeeping inside the conserved total. Or stay with the small system and model friction as a non-Lagrangian generalized force $Q^{\mathrm{fric}}$ ; then the broken-symmetry logic of the limiting-cases table applies, and $\mathrm{d}h/\mathrm{d}t = Q^{\mathrm{fric}}\dot q$ quantifies the leak. Either way, no counterexample — just a reminder that the theorem's hypotheses are as load-bearing as its conclusion.

Is energy conserved in an expanding universe? I have read that photons lose energy to redshift and it goes nowhere.

This is the grown-up version of the question Noether was hired to answer, and her second theorem contains the resolution. Time-translation symmetry is the hypothesis behind energy conservation — and an expanding universe breaks it: the scale factor $a(t)$ makes the effective Lagrangian of matter and radiation explicitly time-dependent, so no global energy charge exists. Photon energy genuinely decreases as $1/a$ , and it does not "go" anywhere; the audit simply fails, so there is no ledger entry to balance. In full general relativity the situation is subtler still: the gauge symmetry of the theory (arbitrary coordinate changes) is local, and Noether's second theorem shows local symmetries produce identities rather than ordinary conserved charges — which is precisely the strange behaviour of gravitational energy that puzzled Hilbert and Klein. Energy conservation survives only where spacetime has a time-translation symmetry (asymptotically flat spaces, where the gravitational-wave fluxes of the today-section live).

What is the difference between Noether's first and second theorems?

The dial count. The first theorem — this article — consumes a global symmetry: a finite number of parameters (one $\varepsilon$ per family), the same transformation applied everywhere, and pays one conserved charge per parameter. The second consumes a local symmetry: a transformation with a freely chooseable function of time (or spacetime) — gauge transformations of electromagnetism, coordinate freedom in general relativity — and pays no new charges. Instead it returns identities among the equations of motion (in electromagnetism, exactly the statement that charge conservation is built into the field equations rather than being an extra law) plus constraints on what can count as physical data. A memorable slogan: global symmetry constrains the solutions (something is conserved along each one); local symmetry constrains the theory (some of its equations are redundant). Both are in the same 1918 paper.

The boost charge just says the centre of mass moves uniformly — I knew that from Newton's laws. Why dress it up as a conservation law?

Three reasons beyond tidiness. Completeness: the Galilean group has ten parameters, and demanding a charge for every one — three momenta, three angular momenta, one energy, and three boost charges — is what makes the ledger close; theories are classified by their symmetry groups, and half-filled ledgers signal an error. Robustness: the Newtonian derivation of uniform CM motion leans on the third law cancelling internal forces pairwise; the Noether derivation needs only that the potential depends on separations, and survives generalizations (velocity-dependent interactions, fields) where naive force-pairing arguments fail. And depth: because the boost is only quasi-invariant, its boundary function $F = M\mathbf R\cdot\hat{\mathbf n}$ carries the mass into the group's composition law — the central-extension structure whose quantum consequence (the Bargmann superselection rule barring superpositions of different masses) is unpacked in the symmetry article's FAQ. The dullest-looking charge has the deepest fine print.

How literally can I use a drifting charge as an error meter — in physics or in a simulation?

Very literally; the limiting-cases table gives the exact statement. On shell, $\mathrm{d}(\sum_i p_i\xi_i)/\mathrm{d}t = \delta L/\varepsilon$ : the drift rate of the would-be charge equals the audit failure. In physics this converts "symmetry is slightly broken" into quantitative predictions — a slightly aspherical Earth torques a satellite's angular momentum at the rate the $\delta L$ of the oblateness term dictates, which is how orbit precession measures the planet's $J_2$ . In computation the logic inverts: the physics conserves $Q$ exactly, so any drift in a simulation measures the integrator's symmetry breaking. This is why long-horizon solar-system and plasma codes use symplectic or variational integrators, whose discrete update rules inherit exact discrete analogues of momentum and angular-momentum charges, and why "energy drift per orbit" is the first plot in any integrator comparison. A conserved charge is a free, physics-grade unit test.

Can two different symmetries pay the same charge, or a symmetry pay a charge that is identically zero?

Both happen, and both are instructive. Degeneracy: add to any generator $\xi_i$ a multiple of another symmetry's generator and the charges add correspondingly — the ledger is linear, so the interesting object is the space of symmetries (the Lie algebra of the symmetry article) rather than individual dials; independent charges correspond to independent generators. Triviality: a "symmetry" whose generator vanishes on shell, or a gauge transformation, produces $Q \equiv 0$ or a charge fixed by the constraints — no dynamical information. This is the practical face of Noether's second theorem: local symmetries flood you with transformations but the associated charges are either zero or boundary terms. When hunting constants of motion for a concrete Lagrangian, the useful inventory is a basis of the global symmetry algebra — for a closed nonrelativistic system, at most the ten Galilean generators plus whatever internal symmetries the potential allows.

Is there a converse — does every conserved quantity come from a symmetry?

Within the Lagrangian-Hamiltonian world, essentially yes, though the honest statement needs Hamiltonian language: in the canonical formalism every constant of motion generates (via the Poisson bracket, the machinery of the M6 chapter) a one-parameter family of canonical transformations that is a symmetry of the dynamics — conservation and symmetry become two readings of one equation, $\{Q, H\} = 0$ . The Lagrangian version is patchier: some conserved quantities correspond to symmetries acting on velocities in ways point transformations cannot express (the Laplace–Runge–Lenz vector of the Kepler problem is the celebrity example — its "hidden symmetry" mixes position and momentum), so the converse holds only after widening the allowed transformations beyond this article's $\xi(q,t)$ . The practical moral survives unscathed: an unexplained constant of motion is always the shadow of an unnoticed symmetry, and finding the symmetry is never wasted work — it usually unlocks more structure than the single constant you started from.

Noether's Theorem

Where this lives today

The physical picture

The theorem for point symmetries

Time transformations and the general charge

The big three, plus the stranger

Which symmetry pays which charge

Limiting cases

Worked examples

1 · The falling charge (confidence builder)

2 · The boost charge sails through a collision (exam level)

The view beyond particles

Going further

Before this

Where this leads