Angular Momentum

Prerequisites: Newtonian Mechanics. Full path (5 pages)

• Intuitive
• Concrete
• Abstract
• Why is it interesting?
• FAQ

Intuitive

A figure skater who spins with her arms stretched out and then pulls them in starts to spin much faster. Nothing pushed her. What stays fixed during the move is her angular momentum, a measure of how much rotational motion a body carries about some point or axis.

Angular momentum grows with the mass that moves, with its speed, and with its distance from the axis. When the skater pulls her arms in, the mass sits closer to the axis, so the rotation has to speed up to keep the angular momentum the same.

The same rule explains why a planet moves fastest when it is closest to the Sun, and why a gyroscope on which no outside twist acts keeps its axis pointing in one direction. Angular momentum can only change when something exerts a torque, that is, a force that acts off-centre and twists the body. The gravitational pull of the Sun points straight at the Sun, so it exerts no torque about the Sun and the planet's angular momentum stays constant.

Angular momentum is conserved because the laws of physics do not single out any direction in space: an experiment rotated as a whole gives the same result. In quantum mechanics angular momentum comes in discrete steps fixed by Planck's constant, and particles carry an intrinsic amount of it called spin.

Concrete

For a point particle with position $\vec r$ and momentum $\vec p = m\vec v$, measured from a chosen origin, the angular momentum is

$$\vec L = \vec r \times \vec p .$$

It depends on the choice of origin. Its time derivative is

$$\frac{d\vec L}{dt} = \dot{\vec r} \times m\dot{\vec r} + \vec r \times \dot{\vec p} = \vec r \times \vec F \equiv \vec \tau ,$$

because the cross product of $\vec v$ with itself vanishes. So $\vec L$ is constant whenever the torque $\vec\tau$ vanishes. This is the case for every central force $\vec F = f(r)\,\hat r$, such as gravity of a fixed Sun or the Coulomb force of a nucleus.

Example: Kepler's second law. For motion in a central force, $\vec L$ is constant, so the orbit stays in the plane perpendicular to $\vec L$. In polar coordinates in that plane, $|\vec L| = m r^2 \dot\varphi$. The area swept by the radius vector in time $dt$ is $dA = \frac{1}{2} r^2\, d\varphi$, hence

$$\frac{dA}{dt} = \frac{|\vec L|}{2m} = \text{const}.$$

Rigid bodies. For a body rotating with angular velocity $\vec\omega$ about a fixed point, $\vec L = \sum_a m_a \vec r_a \times (\vec\omega \times \vec r_a)$, which is linear in $\vec\omega$: $L_i = I_{ij}\,\omega_j$ with the inertia tensor $I_{ij} = \sum_a m_a\left(r_a^2\delta_{ij} - r_{a,i} r_{a,j}\right)$. In general $\vec L$ is not parallel to $\vec \omega$. It is parallel when $\vec\omega$ lies along a principal axis, and then $L = I\omega$. The skater from the intuitive layer has $L = I\omega$ fixed, so a smaller moment of inertia $I$ forces a larger $\omega$.

Connection to symmetry. Take the Lagrangian $L = \frac{m}{2}\dot{\vec r}^{\,2} - V(|\vec r|)$. An infinitesimal rotation about the unit vector $\vec n$ is $\delta\vec r = \epsilon\, \vec n \times \vec r$, and it leaves $|\vec r|$ and $\dot{\vec r}^{\,2}$ unchanged. Noether's theorem then gives the conserved quantity

$$Q = \frac{\partial L}{\partial \dot{\vec r}} \cdot (\vec n \times \vec r) = \vec p \cdot (\vec n \times \vec r) = \vec n \cdot (\vec r \times \vec p),$$

the component of $\vec L$ along $\vec n$. Rotational invariance about every axis gives conservation of all three components.

Quantum mechanics. With $\hat{\vec L} = \hat{\vec r}\times\hat{\vec p}$ and $[\hat x_i, \hat p_j] = i\hbar\delta_{ij}$ one finds

$$[\hat L_i, \hat L_j] = i\hbar\, \epsilon_{ijk} \hat L_k .$$

The components do not commute, so only $\hat{\vec L}^2$ and one component, usually $\hat L_z$, have simultaneous eigenstates. Their eigenvalues are $\hbar^2 l(l+1)$ and $\hbar m$ with $l = 0, 1, 2, \ldots$ and $m = -l, -l+1, \ldots, l$. The eigenfunctions in the angles are the spherical harmonics $Y_{lm}(\theta,\varphi)$. The same commutation relations also allow half-integer values; these occur for spin, never for orbital angular momentum.

• What is angular momentum anyway by WIRED magazine

Abstract

Classical mechanics. Let $(M,\omega)$ be a symplectic manifold with a Hamiltonian action of the rotation group $SO(3)$. Angular momentum is the momentum map $\mu : M \to \mathfrak{so}(3)^*$ of this action: for each $\xi \in \mathfrak{so}(3)$ the function $\langle \mu, \xi\rangle$ is the Hamiltonian that generates the flow of the fundamental vector field of $\xi$. For $M = T^*\mathbb R^3$ with the rotations acting on $\vec q$ and $\vec p$ simultaneously, and $\mathfrak{so}(3)^* \cong \mathbb R^3$, one gets $\mu(\vec q, \vec p) = \vec q \times \vec p$. The momentum map is equivariant, which at the level of Poisson brackets reads

$$\{L_i, L_j\} = \epsilon_{ijk} L_k .$$

If the Hamiltonian is $SO(3)$ invariant, $\{H, L_i\} = 0$ and $\mu$ is constant along the flow. This is the Hamiltonian form of Noether's theorem for rotations. Symplectic reduction by this symmetry, $\mu^{-1}(\nu)/SO(3)_\nu$, is the general version of eliminating the angular variables in the Kepler problem.

Quantum mechanics. Rotations act on the Hilbert space $\mathcal H$ by a projective unitary representation of $SO(3)$, equivalently a unitary representation of its double cover $SU(2)$. The angular momentum operators $\hat J_i$ are $\hbar$ times the self-adjoint generators:

$$U\big(R(\theta\,\vec n)\big) = \exp\!\left(-\frac{i}{\hbar}\,\theta\, \vec n\cdot\hat{\vec J}\right),$$

and the commutation relations $[\hat J_i, \hat J_j] = i\hbar\,\epsilon_{ijk}\hat J_k$ are those of $\mathfrak{su}(2)$. Irreducible unitary representations of $SU(2)$ are labelled by $j \in \{0, \frac12, 1, \frac32, \ldots\}$, have dimension $2j+1$, and the Casimir $\hat{\vec J}^2$ acts as $\hbar^2 j(j+1)$. On $L^2(\mathbb R^3)$ the orbital part $\hat{\vec L}$ contains only integer $j = l$, since the action comes from a genuine action of $SO(3)$ on $\mathbb R^3$. Half-integer $j$ require internal degrees of freedom on which a rotation by $2\pi$ acts as $-1$. See SU(2) and representation theory.

Relativistic field theory. Rotations are part of the Lorentz group, and angular momentum becomes part of an antisymmetric tensor of conserved charges. For a conserved, symmetric energy-momentum tensor $T^{\mu\nu}$, the current

$$M^{\lambda\mu\nu} = x^\mu T^{\lambda\nu} - x^\nu T^{\lambda\mu}$$

satisfies $\partial_\lambda M^{\lambda\mu\nu} = T^{\mu\nu} - T^{\nu\mu} = 0$. The charges $J^{\mu\nu} = \int d^3x\, M^{0\mu\nu}$ are the generators of the Lorentz group: $J^{ij}$ are the angular momentum components and $J^{0i}$ are the boost charges. With the canonical (non-symmetric) energy-momentum tensor, $M^{\lambda\mu\nu}$ picks up an extra term from the transformation of the field components, which is the spin current.

Why is it interesting?

Angular momentum is together with energy and ordinary momentum one of the basic fundamentally conserved quantities in physics.

Angular momentum arises as a conserved quantity in every system which is rotational symmetric. This follows from Noether's theorem.

FAQ

Path to this page

1. Calculus
2. Linear Algebra
3. Newton's Second Law
4. Vector Calculus
5. Newtonian Mechanics

Leads to: Spin.