Noether's Theorem

Noether's theorem tells us that conserved quantities come from symmetries of physical systems. For example, momentum is conserved because the laws of physics are translation invariant.
This deep insight is helpful for understanding when quantities should be conserved. A mass falling off of a building is allowed to gain momentum because the system is not translation invariant - as you move vertically, the gravitational potential changes. However, a train moving along its tracks should conserve momentum because no relevant physical quantity changes as you move around the surface of the earth.

Proving Noether's Theorem

In the system of Hamiltonian Mechanics, the proof of Noether's theorem is surprisingly simple and elegant. First, we need to set up some machinery. Recall Hamilton's equations of motion \[\begin{aligned} \dot p_i &= \frac{\partial H} {\partial q_i}\\ -\dot q_i &= \frac{\partial H} {\partial p_i} \end{aligned}\] Hamilton's equations define a vector field $X_H = (\dot q, \dot p)$ on phase space that describes how a particle evolves over time. The trajectory of a particle starting at position $q$ with momentum $p$ is the integral curve of $X_H$ passing through point $(q,p)$.
We can express Hamilton's equations more simply using a symplectic form. A symplectic form is a closed, nondegenerate differential 2-form. Using $\Omega$, Hamilton's equations become \[ dH = \iota_{X_H} \Omega \] Where $d$ is the exterior derivative and $\iota_{X_H} \Omega$ is the interior product, a one-form defined by $(\iota_{X_H} \Omega)(X_1) = \Omega(X_H, X_1)$.
Aside: We call $X_H$ the ``symplectic gradient'' $H$. Given a metric $g$, the regular gradient of a function $f$ can be defined by $df = \iota_{\text{grad}\;f} \; g$. The definition of the symplectic gradient is the same as the definition normal gradient, except we use the symplectic form instead of the metric.
Now, let $X_G$ be an infinitesimal symmetry transformation. Then $\mathcal{L}_{X_G}H = 0$. That is to say, if we move space a small amount in the $X_G$ direction, the Hamiltonian stays the same. This is exactly what we mean by a symmetry. Furthermore, let $X_G$ be the symplectic gradient of some potential function $U_G$. i.e. $dU_G = \iota_{X_G} \Omega$. Then \[\begin{aligned} 0 &= \mathcal{L}_{X_G} H\\ &= \iota_{X_G} dH + d \iota_{X_G} H &&\text{Cartan's magic formula}\\ &= \iota_{X_G} dH + 0 &&H\;\text{doesn't take arguments, so}\; \iota_{X_G}H = 0\\ &= \iota_{X_G} \iota_{X_H} \Omega &&\text{definition of}\;X_H \\ &= \Omega(X_H, X_G) && \text{definition of the}\; \iota \; \text{operation}\\ &= -\Omega(X_G, X_H) && \Omega\;\text{is antisymmetric}\\ &= -\iota_{X_H} \iota_{X_G} \Omega && \text{definition of the}\;\iota\;\text{operation}\\ &= -\iota_{X_H} dU_G && \text{definition of}\; X_G\\ &= -\iota_{X_H} dU_G + d\iota_{X_H} U_G &&U_G\;\text{doesn't take arguments, so}\; \iota_{X_H}U_G = 0\\ &= -\mathcal{L}_{X_H} U_G && \text{Cartan's magic formula}\\ \end{aligned}\] Therefore, the quantity $U_G$ does not change when we flow along the vector field $X_H$. But flow along $X_H$ is time evolution! So $U_G$ is a conserved quantity over time!
Aside: In the above derivation, we used Cartan's magic formula. It's a super useful identity described on Wikipedia here. It's also called the Cartan's homotopy formula since it can be viewed as the statement that the function $\mathcal{L}_x$ is null-homotopic on the de Rham complex. I hope to write a post describing it more at some point in the future.


Translation in One Dimension

Suppose we have a one-dimensional physical system. Furthermore, suppose our Hamiltonian is invariant under translation. The vector field that infinitesimally moves things in the $x$ direction is the vector field that points in the $x$ direction. We need to express this vector field as a symplectic gradient. So we want a function $U(x, p)$ that satisfies \[\begin{aligned} \frac{\partial U(x, p)} {\partial x} &= 0\\ \frac{\partial U(x, p)} {\partial p} &= -1 \end{aligned}\] Clearly, $U(x, p) = -p$. Satisfies this condition. So the quantity $-p$ (and therefore $p$ as well), is conserved in this physical system! Just like that, we have shown the conservation of momentum!

Rotation in Two Dimensions

Suppose we have a two-dimensional physical system that is invariant under rotation. A rotation by an angle $\theta$ is given by the matrix \[\begin{pmatrix} \cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{pmatrix}\] For very small $\theta$, this matrix is approximately \[\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\] Rotation affects both position and momentum the same way. So an infinitesimal rotation is given by the transformation $\dot x = -y, \dot y = x, \dot p_x = -p_y, \dot p_y = p_x$. Now, to express this vector field as a symplectic gradient, we need a function $U(x, y, p_x, p_y)$ satisfying \[\begin{aligned} \frac {\partial U} {\partial x} &= \dot p_x = -p_y & \frac {\partial U} {\partial y} &= \dot p_y = p_x\\ \frac {\partial U} {\partial p_x} &= -\dot x = y & \frac {\partial U} {\partial p_y} &= -\dot y = -x \end{aligned}\] To satisfy these conditions, we pick $U(x, y, p_x, p_y) = yp_x - xp_y$, which is the angular momentum!

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