= Conservation of angular momentum
{title2=$d\mathbf L_O/dt=\boldsymbol\tau_O^{\mathrm{ext}}=\mathbf0$}
= Angular momentum conservation
{synonym}
<Conservation of angular momentum> says that total <angular momentum> stays fixed when the net external <torque> about the chosen fixed inertial origin vanishes. Individual <angular momenta> and <angular velocities> can change. The <radius change at constant angular momentum> below makes this distinction visible: pulling two rotating <point masses> inward lowers the <moment of inertia>, raises the <angular velocity>, and supplies <work> that increases the <kinetic energy>.
\Image[/angular-momentum-radius-change.png]
{title=Pulling two point masses inward preserves angular momentum and adds three joules of kinetic energy}
Follow the same constant-<mass> <point masses> throughout, in an <inertial frame>, with all <positions> measured from a fixed origin $O$. If the internal pair <forces> are opposite and central, meaning directed along each pair's separation, then
$$
\mathbf L_O=\sum_i\mathbf r_i\times\mathbf p_i,\qquad
\frac{d\mathbf L_O}{dt}=\boldsymbol\tau_O^{\mathrm{ext}}
=\sum_i\mathbf r_i\times\mathbf F_i^{\mathrm{ext}}.
$$
Thus zero net external <torque> throughout an interval implies constant total <angular momentum> throughout that interval. A vanishing external <torque> component, such as $\tau_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; <angular-momentum flux> must also be accounted for when matter crosses its boundary. https://openstax.org/books/university-physics-volume-1/pages/11-3-conservation-of-angular-momentum[OpenStax University Physics Volume 1, section 11.3: Conservation of Angular Momentum]{ref}
The central qualification is essential. Write $\mathbf F_{ij}$ for the internal <force> on member $i$ from member $j$. Under <Newton's third law>, $\mathbf F_{ji}=-\mathbf F_{ij}$, but the sum of their <torques> is
$$
\mathbf r_i\times\mathbf F_{ij}+\mathbf r_j\times\mathbf F_{ji}
=(\mathbf r_i-\mathbf r_j)\times\mathbf F_{ij}.
$$
This vanishes when $\mathbf F_{ij}$ is parallel to $\mathbf r_i-\mathbf r_j$. Equal-and-opposite <forces> alone do not suffice: two opposite transverse <forces> with separated lines of action form a <couple> with nonzero <torque>. Cancellation of the internal pair <torques>, together with <Newton's second law>, proves the total balance above.
For one <point mass> under a <central force> about $O$, $\mathbf F=f(r)\widehat{\mathbf r}$ gives $\boldsymbol\tau_O=\mathbf r\times\mathbf F=\mathbf0$. The <force> may still change the <speed> and <kinetic energy>; its zero <torque> preserves <angular momentum>. With nonzero $\mathbf L_O$, $\mathbf r\cdot\mathbf L_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot\theta=L_z/m$ governs its angular progress. This removes an unknown from the <central-force radial turning point> calculation and helps reduce the <Kepler orbit> to a radial problem.
There is a historical geometric version of this result. In Book I, Section II, Proposition I of the \i[Principia], Newton proves that motion under a <central force> toward a fixed centre stays in a plane and sweeps equal <areas> in equal <times>, using successive central <impulses> and equal <triangle> <areas>. https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/BookI-II[Newton, Principia, Book I, Proposition I, Motte translation, 1846 edition]{ref} In the modern notation above, constant <areal velocity> is $\dot A=L_z/(2m)$: the old geometric proof and the <angular momentum> calculation express the same invariant.
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