Symmetry First, Conservation Second: Lie Groups and Noether's Theorem — Epoche C2
The order in which mechanics is usually taught suggests a particular story about where conservation laws come from. Energy, momentum and angular momentum arrive one at a time, each with its own demonstrations, as though they were three independent facts about how the world happens to behave. This essay argues that the order is backwards. Given an action principle, each of the three is a theorem, and all three theorems share a single hypothesis: that some continuous family of transformations leaves the action unchanged. The mathematics that carries the argument is the theory of Lie groups, and it works because a Lie group is two things at once. A group that is also a smooth manifold A Lie group is a set carrying two structures simultaneously. It is a group — there is a multiplication, an identity element and an inverse for every element — and it is a smooth manifold, meaning that near any point it looks like ordinary $n$-dimensional space and one may differentiate. The two structures are required to be compatible: multiplication and inversion are smooth maps. Take the rotation group $SO(3)$, the set of $3\times 3$ real matrices $R$ with $R^{\mathsf T}R = I$ and $\det R = 1$, where $R^{\mathsf T}$ is the transpose and $I$ the identity matrix. A general $3\times 3$ matrix has nine entries. The condition $R^{\mathsf T}R = I$ is an equation between symmetric matrices, so it imposes $3(3+1)/2 = 6$ independent scalar conditions, not nine. Subtracting, $9 - 6 = 3$: the rotation group is a three-dimensional smooth surface sitting inside the nine-dimensional space of matrices. That number 3 is the first thing to hold on to. The algebra near the identity Because the group is continuous, its whole local structure is stored at the identity. Consider a one-parameter subgroup — a smooth curve $\gamma(t)$ in the group satisfying $\gamma(s+t) = \gamma(s)\gamma(t)$, so that composing two small transformations gives the transformation with the added parameter. For matrix groups every such curve has the form $\gamma(t) = \exp(tX)$, where $X = \gamma'(0)$ is the velocity of the curve at the identity and $\exp(tX) = \sum_{k \ge 0} t^k X^k / k!$. The set of all such $X$ is the tangent space at the identity, called the Lie algebra $\mathfrak{g}$; it is closed under the bracket $[X,Y] = XY - YX$, which is what makes it an algebra rather than a bare vector space. The dimension count reappears here, and this is a useful check on the picture. Differentiating $\gamma(t)^{\mathsf T}\gamma(t) = I$ at $t = 0$ gives $X^{\mathsf T} + X = 0$. So $\mathfrak{so}(3)$ consists of the antisymmetric $3\times 3$ matrices, which have three independent entries — the same 3 as before. The practical consequence is that an element of the algebra, acting on the coordinates of a mechanical system, is a vector field: an infinitesimal transformation $\delta q = \varepsilon X(q)$, with $\varepsilon$ a small parameter. From an invariant action to a conserved quantity Let a system be described by generalised coordinates $q^i$ and the action $S[q] = \int_{t_1}^{t_2} L(q,\dot q,t)\,dt$, with $L$ the Lagrangian and $\dot q^i$ the time derivative of $q^i$. Suppose the infinitesimal transformation $\delta t = \varepsilon T(q,t)$, $\delta q^i = \varepsilon X^i(q,t)$ leaves $S$ unchanged to first order in $\varepsilon$, for every path and not merely for solutions. Then along any solution of the Euler–Lagrange equations the quantity $$Q \;=\; \sum_i \frac{\partial L}{\partial \dot q^i}\,X^i \;-\; \Big(\sum_i \frac{\partial L}{\partial \dot q^i}\,\dot q^i - L\Big)\,T$$ is constant in time. The formula is not an incantation; it is the boundary term left over when the action is varied along the symmetry direction. Varying $S$ produces a bulk integral plus a boundary term. On a solution the bulk integral vanishes because that is exactly what the Euler–Lagrange equations say, and invariance forces the total to vanish, so the boundary term must take the same value at $t_1$ and at $t_2$ — which is what "conserved" means. Three special cases now follow by choosing $T$ and $X^i$. Time translation. Put $T = 1$ and $X^i = 0$. Then $Q = -\big(\sum_i p_i \dot q^i - L\big) = -H$, where $p_i = \partial L/\partial\dot q^i$ and $H$ is the energy. Energy conservation is the statement that the laws do not care what time it is. Spatial translation. Put $T = 0$ and $X^i = a^i$ with $a$ a fixed direction. Then $Q = \sum_i p_i a^i$, the momentum along $a$. Rotation. Put $T = 0$ and $X = n \times q$, the infinitesimal rotation about the axis $n$. Then $Q = p\cdot(n\times q) = n\cdot(q\times p)$, the angular momentum about $n$. The bookkeeping is exact. The Poincaré group of special relativity has dimension $1 + 3 + 3 + 3 = 10$: one time translation, three space translations, three rotations and three boosts. Noether's theorem accordingly delivers ten conserved quantities, and the last trio, the boosts, is the familiar statement that the centre of energy moves uniformly. Where the argument would fail A theorem is only as informative as its hypotheses, and three of them do real work. First, the symmetry must be a symmetry of the action, not merely of the equations of motion; the two are not the same, and the difference is where scaling symmetries usually fall away. Second, if there is no symmetry there is no law. In a realistic expanding universe — one filled with matter or radiation, so that $a(t)$ is not a coordinate artefact — the metric has no timelike Killing vector — no direction in which the geometry is unchanged by a shift in time — and total energy is correspondingly not conserved. The cosmic microwave background makes this concrete: the radiation temperature was about $3000$ K when the universe became transparent and is $2.725$ K now, so the scale factor has grown by $3000/2.725 \approx 1101$ and every photon has lost that factor in energy. Nothing received it. The local statement $\nabla_\mu T^{\mu\nu} = 0$ survives, where $T^{\mu\nu}$ is the stress-energy tensor