Symmetries of Galilean Spacetime

Ask why momentum is conserved and you usually get a circular answer: because the internal forces cancel. Emmy Noether gave the real one in 1918, and it is not about forces at all. It is about the fact that the laws of physics do not care where you do the experiment: slide the apparatus a metre to the left and nothing changes, and that indifference, that symmetry, is momentum conservation. Do the experiment an hour later instead: energy. Turn it through an angle: angular momentum. Newtonian spacetime has exactly ten such indifferences: reset the clock (1 way), shift the origin (3 ways), turn the axes (3 ways), and — the family students forget — set the entire laboratory drifting at a constant velocity (3 ways). That last family, the Galilean boosts, is the mathematical face of the principle of relativity: a shipboard experiment cannot reveal that the ship is moving. Together the ten form the Galilei group, the symmetry group of Newtonian mechanics. Ten symmetries must, by Noether's theorem, buy ten conserved quantities. Energy is one; momentum and angular momentum are six more. What conserves the last three? A quantity no first course names: $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ . Its constancy says the centre of mass of an isolated system slides along a straight line at constant speed forever — through collisions, explosions, chemistry, anything. That theorem is no throwaway corollary of momentum conservation; it is the boost's Noether charge. And when you look closely at how boosts and translations combine, a number falls out of the algebra that is not a symmetry generator at all. It is the mass.

Key takeaways

The Galilei group is the 10-parameter group mapping one inertial frame to another while leaving Newton's laws form-invariant: 1 time translation + 3 space translations + 3 rotations + 3 boosts.
Noether's theorem turns each symmetry into a conserved charge: time translation gives energy, space translation momentum, rotation angular momentum — and the boost gives $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ .
$\mathbf{G}$ is the centre-of-mass theorem: $\mathrm{d}\mathbf{G}/\mathrm{d}t = 0$ says the centre of mass drifts uniformly forever, and it survives collisions that destroy kinetic energy.
The boost is only a quasi-symmetry: it changes the Lagrangian by the total derivative $\mathrm{d}(M\mathbf{X}_{\rm cm})/\mathrm{d}t$ , and that boundary term subtracts $M\mathbf{X}_{\rm cm}$ from the charge.
Boosts and space translations commute as maps of spacetime, but their charges do not: $\{K_i, P_j\} = M\delta_{ij}$ — a number, not a generator. This mass central extension (Bargmann) is a structure the Poincaré group lacks.
Key idea: ten indifferences of Newtonian spacetime, ten conservation laws. The unfamiliar tenth family — the boosts — conserves $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ , and its algebra hands you the mass as a central charge.

The core insight

Every continuous symmetry of the action is a conservation law (Noether, 1918), and Galilean spacetime has ten: $1 + 3 + 3 + 3 = 10$ . Nine charges are familiar — energy, momentum, angular momentum. The tenth family belongs to the boosts and conserves $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ : the centre of mass drifts uniformly, forever. In the algebra of these charges the boost and the momentum fail to commute by exactly the total mass, $\{K_i,P_j\} = M\delta_{ij}$ — the fingerprint of absolute time.

Where this lives today

The Galilean Noether charges are tested hardest inside numerical integrators. Long-term Solar System integrations — asking whether Mercury's orbit is stable for the next five billion years — run for $10^{12}$ steps, and any scheme that quietly leaks energy or momentum manufactures a fictitious secular drift long before it produces an answer. Symplectic integrators are built to respect the symmetries rather than the trajectory, conserving momentum and angular momentum to machine precision by construction; a slow drift in the centre-of-mass charge $\mathbf{G}$ is the standard diagnostic that an N-body or molecular-dynamics code has a bug. The central extension is no ornament either. Bargmann's superselection rule is a live constraint in atom interferometry, where the boost phase $e^{imvx/\hbar}$ is exactly what an atom acquires when the laboratory — or a laser frequency — is Doppler-shifted: the mass in the commutator of a boost with a translation is the mass whose fringes are being counted.

The physical picture

Draw Newtonian spacetime as a stack of pancakes. Each pancake is a copy of Euclidean 3-space labelled by a time $t$ , and absolute time means every observer agrees which pancake an event lives on. A Galilei transformation re-labels the stack without destroying it: renumber the pancakes (time translation), slide one sideways (space translation), spin it about a point (rotation), or slide each pancake further than the one below (a boost — the stack shears). What you may never do is tilt an event onto a different pancake: the one move Galilean spacetime forbids and Minkowski spacetime allows. Count the freedom in each move: $1 + 3 + 3 + 3 = 10$ . All ten leave $\mathbf{F} = m\mathbf{a}$ identical, because acceleration is a second derivative and so blind to a constant shift, a constant velocity, or a relabelled clock.

Figure

The four families of Galilei transformation: 1 time translation · 3 space translations · 3 rotations · 3 boosts = 10 parameters. The boost shears the worldlines but leaves every simultaneity line t = const horizontal — absolute time, drawn.

Now put Noether's theorem to work. It says a continuous symmetry of the action always produces a quantity that never changes, and tells you how to build it from the Lagrangian. Feed it the ten Galilean symmetries and seven answers are old friends. The other three come from the boosts, and they are the surprise.

Figure

The Noether ledger: 1 + 3 + 3 + 3 = 10 symmetries → 10 conserved charges. The row nobody expects is the boost, which conserves G = P t − M X_cm: for the two-body system below, G = −30 kg·m at every instant, unchanged even through the collision at t = 2 s that destroys 15 J.

The bottom strip makes the point concrete. Two particles run at each other, collide, stick, and lose $15\ \mathrm{J}$ of kinetic energy to heat. Kinetic energy is gone; $\mathbf{G}$ is untouched. A conservation law that survives an inelastic collision is no accident of the collision law — it is a symmetry the collision cannot break, because no internal force cares which uniformly-moving frame you watch from.

The core equations

The most general Galilei transformation stacks all four families at once: rotate the axes, set the frame drifting at $\mathbf{v}$ , shift the origin, reset the clock. Ten numbers, no more and no fewer — the only maps sending straight worldlines to straight worldlines while preserving absolute time and Euclidean distance within a time-slice are affine ones of this form.

t' = t + s, \qquad \mathbf{x}' = R\,\mathbf{x} - \mathbf{v}t + \mathbf{a} \qquad (1)

where: $s$ — clock offset (1 parameter); $\mathbf{a}$ — origin shift (3); $R \in SO(3)$ — rotation (3); $\mathbf{v}$ — relative velocity (3). Physical reading: $t' = t+s$ contains no $\mathbf{x}$ , so simultaneity is untouched — absolute time as a constraint on the group, and the one line special relativity will erase. Two such maps compose into a third (a semidirect product), with translation slot $R_2\mathbf{a}_1+\mathbf{a}_2-\mathbf{v}_2 s_1$ : that cross-term says boosting and waiting do not commute. Noether's theorem turns a symmetry into a number. Let an infinitesimal change of coordinates $\delta q_i = \epsilon\, K_i(q,t)$ alter the Lagrangian only by a total time derivative, $\delta L = \epsilon\,\mathrm{d}\Lambda/\mathrm{d}t$ — a quasi-symmetry, in which the action changes by an endpoint term alone and the equations of motion are untouched. One combination is then constant along every solution.

Q \;=\; \sum_i p_i\,K_i \;-\; \Lambda, \qquad \frac{\mathrm{d}Q}{\mathrm{d}t} = 0 \qquad (2)

where: $p_i = \partial L/\partial \dot q_i$ — canonical momenta; $K_i$ — the symmetry's displacement shape; $\Lambda$ — the boundary term it leaves behind. Physical reading: strict symmetries have $\Lambda = 0$ and give the textbook charges; the boost is the one Galilean symmetry with $\Lambda \neq 0$ , and that $\Lambda$ puts the $-M\mathbf{X}_{\rm cm}$ into $\mathbf{G}$ . Run the machine on all ten and this is the ledger — the first three entries taught everywhere, the fourth the one this article owns.

E=\sum_i \tfrac12 m_i\dot{\mathbf{x}}_i^2+V,\quad \mathbf{P}=\sum_i m_i\dot{\mathbf{x}}_i,\quad \mathbf{L}=\sum_i m_i\,\mathbf{x}_i\times\dot{\mathbf{x}}_i,\quad \boxed{\mathbf{G}=\mathbf{P}t-M\mathbf{X}_{\rm cm}} \qquad (3)

where: $M = \sum_i m_i$ ; $\mathbf{X}_{\rm cm} = \big(\sum_i m_i\mathbf{x}_i\big)/M$ ; $V$ — any internal potential built from the separations. Physical reading: $\mathrm{d}\mathbf{G}/\mathrm{d}t = 0$ rearranges to $\mathbf{X}_{\rm cm}(t) = \mathbf{X}_{\rm cm}(0) + (\mathbf{P}/M)t$ — the centre-of-mass theorem, promoted from corollary to fundamental Noether charge with a symmetry of its own. Finally the charges form an algebra under the Poisson bracket, and here Galilean spacetime signs its name: nine brackets close on other generators, as a Lie algebra should, and one does not.

\{K_i,P_j\} = M\,\delta_{ij}, \qquad \{K_i,H\}=P_i, \qquad \{K_i,K_j\}=0 \qquad (4)

where: $K_i = M X_{{\rm cm},i} - P_i t = -G_i$ — the boost generator (the sign is convention, the content is not); $H, P_i$ — energy and momentum generators. Physical reading: the first bracket returns a number, not a generator — the total mass, sitting where a symmetry ought to be. That is the mass central extension, the algebraic residue of the boundary term $\Lambda$ ; it also says that $M$ is the momentum a boost injects, since boosting by $\mathbf{v}$ sends $\mathbf{P}\to\mathbf{P}+M\mathbf{v}$ .

Where the ten conservation laws come from

Derivation ANoether's theorem on the Galilei group, and the boost charge

The goal is to run Noether's theorem across all ten Galilean symmetries of an isolated system and read off the ten conserved charges — in particular to derive, not assert, the one that never appears in a first mechanics course. Take $L=\sum_i \tfrac12 m_i \dot{\mathbf{x}}_i^2 - V(\{\mathbf{x}_i-\mathbf{x}_j\})$ , whose potential depends only on separations. The payoff is $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ , whose constancy is the centre-of-mass theorem.

Step 1. Dispatch the three easy families. Under a clock shift $L$ has no explicit $t$ , so $\delta L = 0$ and Eq. (2) with $\Lambda=0$ returns $E$ . Under a rigid shift $\mathbf{x}_i \to \mathbf{x}_i + \boldsymbol{\epsilon}$ every separation is untouched, so again $\delta L = 0$ and the charge is $\mathbf{P}$ ; under a rigid rotation the kinetic term and every separation length are unchanged, giving $\mathbf{L}$ . Why: the potential is built only from what these symmetries cannot see — durations, separations, lengths — so $L$ is strictly invariant and nothing is subtracted from the charge.

Step 2. Now the boost: displace every particle by $\delta \mathbf{x}_i = \boldsymbol{\epsilon}\, t$ , so every velocity gains the same constant, $\delta\dot{\mathbf{x}}_i = \boldsymbol{\epsilon}$ . The potential still cannot see it — each separation shifts by $\boldsymbol{\epsilon}t - \boldsymbol{\epsilon}t = 0$ — but the kinetic term can: $\delta L = \sum_i m_i \dot{\mathbf{x}}_i\cdot\boldsymbol{\epsilon}$ . Why: kinetic energy is quadratic in velocity, so adding the same velocity to every particle leaves a cross-term. The boost is not a strict symmetry of $L$ .

Step 3. Recognise that change as a total time derivative: $\sum_i m_i \dot{\mathbf{x}}_i = \mathrm{d}\big(\sum_i m_i\mathbf{x}_i\big)/\mathrm{d}t$ , so $\delta L = \boldsymbol{\epsilon}\cdot\mathrm{d}\Lambda/\mathrm{d}t$ with $\Lambda = M\mathbf{X}_{\rm cm}$ . Eq. (2) then applies with $\Lambda \neq 0$ , and the displacement shape is $\mathbf{K}_i = t$ , so $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ . Why: the boost survives only "up to a total derivative", and the price is the $-M\mathbf{X}_{\rm cm}$ that $\Lambda$ subtracts. Drop $\Lambda$ and you keep $\mathbf{P}t$ , which is not conserved.

Step 4. Verify directly: $\dot{\mathbf{G}} = \dot{\mathbf{P}}t + \mathbf{P} - \sum_i m_i\dot{\mathbf{x}}_i$ . The last two terms cancel identically because $\mathbf{p}_i \equiv m_i\dot{\mathbf{x}}_i$ , and $\dot{\mathbf{P}} = 0$ because internal forces cancel in pairs — so $\mathrm{d}\mathbf{G}/\mathrm{d}t = 0$ , which rearranges to $\mathbf{X}_{\rm cm}(t) = \mathbf{X}_{\rm cm}(0) + (\mathbf{P}/M)\,t$ . Why: the centre-of-mass theorem, taught as a throwaway consequence of $\sum \mathbf{F}_{\rm int} = 0$ , is really the tenth conservation law of Newtonian spacetime — the relativity principle as a number. *Verified symbolically in verify_aq-sr-symmetries-galilean-spacetime.py: $\delta L = \epsilon\,\mathrm{d}(\sum m_i x_i)/\mathrm{d}t$ , and $\mathrm{d}\mathbf{G}/\mathrm{d}t=0$ for arbitrary internal $V$ .*

Go deeper · Derivation B · 2 minThe mass central extension: why boosts and translations refuse to commute

Derivation BThe mass central extension: why boosts and translations refuse to commute

Here is the payoff, and the place where the Galilei group announces that it is not the Poincaré group of special relativity in disguise. The question is childishly simple: does it matter whether you boost the laboratory and then move it three metres, or move it three metres and then boost it? On the coordinates, no. On the charges that generate the moves, and on the quantum state, yes — and the discrepancy is exactly the mass. This is Bargmann's central extension, reached without leaving Newtonian physics.

Step 1. Check the coordinates. A boost sends $(\mathbf{x},t) \to (\mathbf{x}+\mathbf{v}t,\,t)$ and a translation sends $(\mathbf{x},t)\to(\mathbf{x}+\mathbf{a},\,t)$ ; compose in either order and you get $(\mathbf{x}+\mathbf{a}+\mathbf{v}t,\,t)$ . They commute, flatly and exactly. Why: say it out loud, because the extension is often mis-taught as "boosts and translations don't commute". As maps of spacetime points, they do. The failure lives one level up.

Step 2. Ask the charges the same question. With $K_i = MX_{{\rm cm},i} - P_i t$ and $\{x_{a,i},\,p_{b,j}\} = \delta_{ab}\delta_{ij}$ , the $-P_it$ piece brackets to zero against $P_j$ while $\sum_a m_a x_{a,i}$ gives $\sum_a m_a\delta_{ij}$ , so $\{K_i,P_j\} = M\,\delta_{ij} \neq 0$ — a multiple of the identity, where a Lie bracket ought to return a generator. Why: the transformations commute yet their charges do not, and the obstruction is a plain number — the total mass. An algebra closing only after such a constant is adjoined is a central extension, and the constant is its central charge: here, the mass. That is the Bargmann algebra.

Step 3. Watch it act on a quantum state. A Galilean boost does not merely relabel $x$ ; it multiplies the wavefunction by a mass-dependent phase, $(\hat B_v\psi)(x,t) = e^{\,i m(vx-\tfrac12 v^2 t)/\hbar}\,\psi(x-vt,\,t)$ — the operator mapping free solutions to free solutions, turning a plane wave of momentum $\hbar k$ into one of $\hbar k + mv$ . Translation acts plainly, $(\hat T_a\psi)(x,t)=\psi(x-a,t)$ ; compose in both orders and the results differ by a constant phase, $\hat T_a\hat B_v = e^{-imva/\hbar}\,\hat B_v\hat T_a$ . Why: the group is represented only up to a phase — a projective, or ray, representation. That phase is invisible overall, but between two branches of a superposition it is physical, and it carries the mass.

Step 4. Collect the consequence and the contrast. Because that phase carries $m$ , states of different mass cannot be coherently superposed in non-relativistic quantum mechanics without acquiring an unphysical frame-dependence: Bargmann's mass superselection rule. The relativistic contrast is sharp — there the analogous bracket closes on the Hamiltonian, an existing generator, so no central charge is needed. Why: two groups, the same ten parameters, and yet the Galilean one carries an extra scar — the fingerprint of absolute time, the assumption special relativity removes. *Verified symbolically in verify_aq-sr-symmetries-galilean-spacetime.py: boosts and translations commute on $(x,t)$ ; the ten-generator algebra closes with $\{K_i,P_j\} = M\delta_{ij}$ and $\{K_i,H\}=P_i$ ; and $\hat T_a\hat B_v = e^{-imva/\hbar}\hat B_v\hat T_a$ .*

Limiting cases

Case	Result	Meaning
No boosts ( $\mathbf{v}=0$ )	7-parameter Euclidean group plus the clock shift	Only $E$ , $\mathbf{p}$ , $\mathbf{L}$ survive — $\mathbf{G}$ is what the relativity principle is worth
Frame with $\mathbf{P}=0$	$\mathbf{G} = -M\mathbf{X}_{\rm cm}$ , fixed	The charge degenerates into "the centre of mass never moves"
Non-central internal force	$\mathbf{L}$ dies, $\mathbf{P}$ survives	The symmetry, not the force law, buys the conservation law
$c\to\infty$	Galilei group $=$ İnönü–Wigner contraction of the Poincaré group	Newtonian spacetime is the infinite- $c$ shadow of the relativistic one; the replacements appear under The Lorentz Transformation and Minkowski Spacetime

Common trapThe centre-of-mass theorem is not just momentum conservation said twice

The commonest reflex on meeting $\mathbf{G} = \mathbf{P}t - M\mathbf{X}_{\rm cm}$ is to shrug: if $\mathbf{P}$ is conserved then of course the centre of mass drifts uniformly. Not so. Momentum conservation says $\dot{\mathbf{X}}_{\rm cm}$ is constant; it does not fix the integration constant, and it is that constant $\mathbf{G}$ pins down. Counting is the honest test: ten group parameters demand ten charges, and energy plus momentum plus angular momentum is only seven. If the boosts gave nothing new you would be three short and Noether's theorem would be false. The missing three are the components of $\mathbf{G}$ , each carrying what momentum cannot — where the centre of mass was. Nor does $\mathbf{G}$ survive merely "because nothing pushes the system": below, an inelastic collision destroys $15\ \mathrm{J}$ and $\mathbf{G}$ does not budge.

Worked examples

1 · The boost charge through an inelastic collision

Two particles slide on a frictionless line. At $t=0$ : $m_1 = 2\ \mathrm{kg}$ at $x_1 = 0$ with $u_1 = +4\ \mathrm{m/s}$ ; $m_2 = 3\ \mathrm{kg}$ at $x_2 = 10\ \mathrm{m}$ with $u_2 = -1\ \mathrm{m/s}$ . They meet, collide, and stick. Show that $G = Pt - \sum_i m_i x_i$ is the same number before, at, and after the collision.

Step 1. Get the constants: $P = 2(4) + 3(-1) = 5\ \mathrm{kg\,m/s}$ , $M = 5\ \mathrm{kg}$ , and $X_{\rm cm}(0) = [2(0)+3(10)]/5 = 6\ \mathrm{m}$ — so the centre of mass must move at $P/M = 1\ \mathrm{m/s}$ forever.

Step 2. At $t=0$ , $G = 0 - [m_1x_1+m_2x_2] = \boldsymbol{-30\ \mathrm{kg\,m}}$ , which is $-MX_{\rm cm}(0) = -(5)(6)$ , as it must be. Check again before impact: the $10\ \mathrm{m}$ gap closes at $5\ \mathrm{m/s}$ , so at $t=1\ \mathrm{s}$ the particles sit at $4\ \mathrm{m}$ and $9\ \mathrm{m}$ , giving $G = 5(1) - [8+27] = -30\ \mathrm{kg\,m}$ . Unchanged.

Step 3. They meet at $t = 2\ \mathrm{s}$ at $x = 8\ \mathrm{m}$ . Kinetic energy before is $\tfrac12(2)(4)^2 + \tfrac12(3)(1)^2 = 17.5\ \mathrm{J}$ ; the merged $5\ \mathrm{kg}$ blob leaves at $1\ \mathrm{m/s}$ with $2.5\ \mathrm{J}$ . So $\mathbf{15\ J}$ is destroyed — exactly $\tfrac12\mu\,v_{\rm rel}^2 = \tfrac12(1.2)(5)^2$ , with reduced mass $\mu = 1.2\ \mathrm{kg}$ .

Step 4. After the wreck, at $t = 5\ \mathrm{s}$ the blob is at $x = 8 + 1(5-2) = 11\ \mathrm{m}$ , so $G = 5(5) - 5(11) = \boldsymbol{-30\ \mathrm{kg\,m}}$ . The charge reads $-30\ \mathrm{kg\,m}$ at $t = 0, 1, 2, 3, 5, 10\ \mathrm{s}$ — through a collision that annihilated $15\ \mathrm{J}$ . A symmetry speaking, not an accident of the collision law.

2 · Rotational symmetry and a two-body angular momentum

Two particles are joined by a spring — a central force along the line between them. At $t=0$ : $m_1 = 1\ \mathrm{kg}$ at $\mathbf{r}_1 = (4,0)\ \mathrm{m}$ with $\mathbf{u}_1 = (0,+4)\ \mathrm{m/s}$ ; $m_2 = 2\ \mathrm{kg}$ at $\mathbf{r}_2 = (-2,0)\ \mathrm{m}$ with $\mathbf{u}_2 = (0,-2)\ \mathrm{m/s}$ . Find $L_z$ , and show that rotational symmetry, not the spring, conserves it.

Step 1. Check the set-up: $m_1\mathbf{r}_1 + m_2\mathbf{r}_2 = 0$ , so the centre of mass sits at the origin, and $\mathbf{P} = (0,\ 1(4)+2(-2)) = 0$ , so it stays there.

Step 2. Compute $L_z = \sum_i m_i (x_i \dot y_i - y_i \dot x_i) = 1[(4)(4)] + 2[(-2)(-2)] = \boldsymbol{24\ \mathrm{kg\,m^2/s}}$ . Cross-check via the reduced mass: with $\mathbf{P}=0$ the angular momentum is entirely orbital in the relative coordinate, so $L_z = \mu\,(x\dot y - y\dot x)_{\rm rel} = \tfrac23 (6)(6) = 24\ \mathrm{kg\,m^2/s}$ . Same number, different route.

Step 3. Now the symmetry argument. The rate of change of $\mathbf{L}$ is the total internal torque, $\dot{\mathbf{L}} = (\mathbf{r}_1-\mathbf{r}_2)\times\mathbf{F}_{12}$ once $\mathbf{F}_{21}=-\mathbf{F}_{12}$ is used. A central force is parallel to $\mathbf{r}_1-\mathbf{r}_2$ , so the cross product vanishes and $L_z = 24\ \mathrm{kg\,m^2/s}$ for all time — whatever the spring constant or natural length.

Step 4. Break the symmetry and the law dies. Rotate the internal force by $90^\circ$ , keeping it equal and opposite so momentum is still conserved. The torque is now non-zero, and numerical integration shows $L_z$ drifting far from $24$ within a second while $\mathbf{P}$ stays exactly zero. Rotational invariance was doing the work — not the third law, and not the spring.

Physics of the portrait

Noether's theorem was born not in a mechanics class but in a crisis about energy in general relativity: in 1915 Hilbert and Klein found that the usual conservation law seemed to collapse into an empty identity in Einstein's new field equations, and asked the best algebraist in Göttingen to sort it out. She did far more.

Origins · 1 minEmmy Noether and how this was discovered

Emmy Noether

1882–1935 · Erlangen · Göttingen · Bryn Mawr

Amalie Emmy Noether was the greatest algebraist of her generation and, for most of her career, was not allowed to be paid for it. Born in Erlangen, she audited lectures because women could not matriculate, took her doctorate in 1907, then worked unpaid for seven years. Hilbert and Klein brought her to Göttingen in 1915 precisely because they needed her invariant theory for the new general relativity; the faculty refused her a post, and Hilbert announced lectures under his own name so that she could teach them. Her Invariante Variationsprobleme — presented to the Göttingen Academy on 26 July 1918 and published in its Nachrichten — is the paper physicists know her by, though she thought it a sideline: her real work, the abstract theory of rings, ideals and modules, simply is modern algebra. Dismissed under the Nazi race laws in 1933, she went to Bryn Mawr College and died two years later, aged 53. Einstein wrote to the New York Times that she was "the most significant creative mathematical genius thus far produced since the higher education of women began".

History

1872

— Felix Klein's Erlangen program recasts a geometry as the study of what a transformation group leaves invariant.

1918

— Emmy Noether's Invariante Variationsprobleme, presented to the Göttingen Academy on 26 July and published in its Nachrichten: every continuous symmetry of an action yields a conserved quantity.

1939

— Eugene Wigner classifies the unitary representations of the Poincaré group, labelling particles by mass and spin.

1953

— Erdal İnönü and Eugene Wigner show the Galilei group is the

c\to\infty

contraction of the Poincaré group.

1954

— Valentine Bargmann proves that the Galilei group's ray representations carry a mass central charge — the mass superselection rule, which the Poincaré group does not possess.

Going further

Texts. Landau and Lifshitz, Mechanics, 3rd ed., §§6–9 (the conservation laws from the ten Galilean symmetries, centre-of-mass charge included) · Goldstein, Poole and Safko, Classical Mechanics, 3rd ed., §13.7 (Noether's theorem) · Ballentine, Quantum Mechanics, 2nd ed., Ch. 3 (the Galilei group and mass superselection) · Lévy-Leblond, "Galilei Group and Galilean Invariance", Group Theory and its Applications Vol. 2 (1971).

Provenance. The 10-parameter count, the general transformation, the composition law, the Lie-algebra brackets and the mass central extension are standard (Landau and Lifshitz §§6–9; Ballentine Ch. 3; Bargmann, Ann. Math. 59, 1 (1954); İnönü and Wigner, PNAS 39, 510 (1953)). A verifier should reproduce $\mathrm{d}\mathbf{G}/\mathrm{d}t = 0$ for arbitrary internal $V$ ; the ten-generator Poisson algebra with $\{K_i,P_j\}=M\delta_{ij}$ ; the cocycle $\hat T_a\hat B_v = e^{-imva/\hbar}\hat B_v\hat T_a$ ; and $G=-30\ \mathrm{kg\,m}$ , $15\ \mathrm{J}$ destroyed, $L_z = 24\ \mathrm{kg\,m^2/s}$ . Corrections to the source draft: with $\mathbf{x}'=R\mathbf{x}-\mathbf{v}t+\mathbf{a}$ the composition law's translation slot is $R_2\mathbf{a}_1+\mathbf{a}_2-\mathbf{v}_2 s_1$ , not $+\mathbf{v}_2 s_1$ ; and $[K_i,P_j]=0$ in the unextended algebra is not the group's "tell-tale non-commutativity" — the non-commutation is the extension, $\{K_i,P_j\}=M\delta_{ij}$ .

Builds on. Inertial Frames of Reference · Galilean Transformations · Absolute Space and Absolute Time · The Principle of Relativity. Leads to. Electromagnetism and the Failure of Galilean Invariance · Einstein's Two Postulates · The Constancy of the Speed of Light.

If momentum is conserved, isn't the centre-of-mass theorem automatic? Why call G a separate law?

Because it contains strictly more information. Momentum conservation says $\dot{\mathbf{X}}_{\rm cm}$ is constant; integrating gives $\mathbf{X}_{\rm cm}(t) = \mathbf{X}_{\rm cm}(0) + (\mathbf{P}/M)t$ with an undetermined integration constant, and $\mathbf{G}$ is that constant, promoted to a charge with a symmetry behind it. The counting settles it: ten symmetries must give ten charges, and $E$ plus $\mathbf{p}$ plus $\mathbf{L}$ is only seven. The three components of $\mathbf{G}$ are the missing three, each remembering where the centre of mass was.

The boost changes the Lagrangian. How can it be a symmetry at all?

It is a quasi-symmetry, and Noether's theorem is built for that case. The boost changes $L$ by a total time derivative, $\delta L = \epsilon\,\mathrm{d}\Lambda/\mathrm{d}t$ with $\Lambda = M\mathbf{X}_{\rm cm}$ , and a total derivative changes the action only by an endpoint term — which contributes nothing to the variational principle, so the equations of motion are exactly invariant even though $L$ is not. The price is that the charge is no longer $\sum_i \mathbf{p}_i\cdot\mathbf{K}_i$ : you must subtract $\Lambda$ , and that gives $\mathbf{G}$ . The leftover $\Lambda$ is no defect — it returns as the mass central charge.

Is the mass central extension purely a quantum effect?

No — that is a common half-truth. The phase $e^{-imva/\hbar}$ is quantum, but the extension is already visible classically: the Poisson bracket $\{K_i,P_j\} = M\delta_{ij}$ is a non-zero number sitting where a generator ought to be. Classically it reads homely — boost an isolated system by $\mathbf{v}$ and its momentum shifts by $M\mathbf{v}$ , so the mass is how much momentum a boost injects. What quantum mechanics adds is consequence: a relative phase between branches of a superposition, hence Bargmann's superselection rule. The structure is classical; the sting is quantum.

Why doesn't special relativity have a mass central charge?

Because there the bracket has somewhere to land. In the Poincaré algebra the bracket of a boost generator with a translation generator closes on the Hamiltonian — an existing generator — so no central element need be adjoined. In the Galilei algebra the same bracket has nothing to close on and must be propped up with a number, the mass. Same ten parameters, but only one group carries this scar, and the scar is the mark of absolute time. What replaces the Galilean structure is set out under The Lorentz Transformation and Minkowski Spacetime; none of it is derived here.

Does Noether's theorem need a Lagrangian, or would the equations of motion be enough?

It needs the action. The theorem is about symmetries of the variational principle, and its proof runs through the variation of the action, so a symmetry of the equations of motion alone will not do. That has teeth here: the boost is an exact symmetry of the equations but only a quasi-symmetry of $L$ , and with the equations alone you would have no $\Lambda$ — and no way to see that $-M\mathbf{X}_{\rm cm}$ belongs in the charge. Dissipative systems, having no Lagrangian, sit outside the theorem entirely.

What does G look like in the centre-of-mass frame?

It degenerates, instructively. Boost to the frame where $\mathbf{P}=0$ ; then $\mathbf{G} = -M\mathbf{X}_{\rm cm}$ , with no $t$ in it at all, and its constancy says simply that the centre of mass never moves. Every internal process — explosion, collision, chemistry — must therefore rearrange the parts around a point nailed to the spot. That is why a firework bursting in still air scatters symmetrically about the point its shell would have reached, and why an astronaut shoving a toolbox recoils so the pair's mass-weighted midpoint stays put.