Visual walkthrough — Conservation of angular momentum — conditions

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Step 0 — The two arrows we are allowed to use

Before any formula, we need to agree on what pictures the symbols stand for.

Look at the figure. The amber dot is the pivot (origin). The cyan arrow $\vec{r}$ reaches out to the object; the white arrow $\vec{p}$ shows where it is heading. Nothing else exists yet — no torque, no cross product. We earn those next.

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Step 1 — What does "amount of spin" even mean? Build $\vec{L}$

WHAT. We want one number-with-direction that captures "how much this object is circling the pivot". Call it angular momentum $\vec{L}$.

WHY the cross product? A moving object only circles the pivot if part of its motion is sideways to $\vec{r}$. If it moves straight at (or straight away from) the pivot, it is not orbiting at all. We need a tool that:

• gives zero when $\vec{r}$ and $\vec{p}$ are parallel (pure in/out motion, no circling),
• gives maximum when they are perpendicular (pure sideways motion, fastest circling).

The cross product $\vec{r}\times \vec{p}$ does exactly this. Its size is

$|\vec{L}|=\underset{\text{how far out}}{\underbrace{|\vec{r}|}}\underset{\text{how hard moving}}{\underbrace{|\vec{p}|}}\underset{\begin{array}{c}\text{how sideways}\\ \text{the motion is}\end{array}}{\underbrace{\sin \theta }}$

$\theta$ is the angle between the two arrows. When they line up, $\sin 0=0$ (no spin). When perpendicular, $\sin 90^{\circ }=1$ (all spin). That $\sin \theta$ is why the cross product is the right tool — it automatically extracts the sideways part.

Look at the figure. The shaded parallelogram's area equals $|\vec{r}||\vec{p}|\sin \theta$ — that area is $|\vec{L}|$. The direction of $\vec{L}$ (the little curved amber arrow) is the sense of circling: counter-clockwise here.

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Step 2 — Conservation means "does not change" — so watch it over time

WHAT. "Conserved" means $\vec{L}$ holds still as the clock ticks. In the language of change, the rate of change is zero.

WHY a derivative? The tool that measures "how fast something changes each instant" is the derivative $\frac{d}{dt}$. Asking "is $\vec{L}$ constant?" is exactly asking "is $\frac{d\vec{L}}{dt}=0$?" So we differentiate $\vec{L}$ and see what makes it vanish.

Look at the figure. Two frozen snapshots a tiny time $dt$ apart. The faint arrow is $\vec{L}$ at time $t$, the bright one at $t+dt$. The amber gap between their tips is $d\vec{L}$ — the change we must chase to zero. A flat line means conserved.

$\frac{d\vec{L}}{dt}=\frac{d}{dt}(\vec{r}\times \vec{p})$

Here $\frac{d\vec{L}}{dt}$ = the rate the spin-arrow drifts each second. If it is zero, $\vec{L}$ is frozen — conserved.

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Step 3 — Split the change with the product rule

WHAT. $\vec{L}$ is a product of two things that both change: $\vec{r}$ (the object moves, so its position shifts) and $\vec{p}$ (its velocity can change). So the change of the product has two sources.

WHY the product rule? Whenever a quantity is $A\times B$ and both $A,B$ move, the total change is "change in $A$, times old $B$" plus "old $A$, times change in $B$". This is the product rule, and it holds for cross products too (keeping the order, since $\times$ is not commutative).

$\frac{d\vec{L}}{dt}=\underset{\text{term A: }\vec{r}\text{ changing}}{\underbrace{\frac{d\vec{r}}{dt}\times \vec{p}}}+\underset{\text{term B: }\vec{p}\text{ changing}}{\underbrace{\vec{r}\times \frac{d\vec{p}}{dt}}}$

• $\frac{d\vec{r}}{dt}$ = how fast the position arrow moves = the velocity $\vec{v}$.
• $\frac{d\vec{p}}{dt}$ = how fast the momentum arrow changes.

Look at the figure. Two panels: Term A wiggles the $\vec{r}$ arrow (cyan) while holding $\vec{p}$; Term B wiggles the $\vec{p}$ arrow (white) while holding $\vec{r}$. Their combined effect is the full change of $\vec{L}$.

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Step 4 — Term A dies: a vector cannot circle around itself

WHAT. In term A, replace $\frac{d\vec{r}}{dt}$ with $\vec{v}$. But $\vec{p}=m\vec{v}$, so term A is $\vec{v}\times m\vec{v}$.

WHY it is zero. The cross product measures the sideways angle between two arrows. But $\vec{v}$ and $m\vec{v}$ point the same way — the angle between them is $0$, so $\sin 0=0$.

$\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times (m\vec{v})=m\underset{=0}{\underbrace{(\vec{v}\times \vec{v})}}=0$

Look at the figure. Two arrows lying flat on top of each other. The "parallelogram" they would enclose is squashed to a line — zero area — so this term contributes nothing. Term A is gone.

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Step 5 — Term B becomes torque: name the survivor

WHAT. In term B, $\frac{d\vec{p}}{dt}$ is the rate the momentum changes. Newton's second law says that rate is the force: $\frac{d\vec{p}}{dt}=\vec{F}$. So term B is $\vec{r}\times \vec{F}$.

WHY give it a name? $\vec{r}\times \vec{F}$ is the rotational push — how much a force twists the object about the pivot. We name it torque $\vec{\tau }$.

$\vec{r}\times \frac{d\vec{p}}{dt}=\vec{r}\times \vec{F}=\vec{\tau }$

Its size is $|\vec{\tau }|=|\vec{r}||\vec{F}|\sin \varphi$, where $\varphi$ is the angle between $\vec{r}$ and the force. Just like $\vec{L}$, torque only exists when the force has a sideways component — a straight pull toward/away from the pivot twists nothing.

Look at the figure. The white force arrow splits into a cyan sideways part (twists — makes torque) and a grey along-$\vec{r}$ part (only pushes in/out — no twist). Only the sideways part survives into $\vec{\tau }$.

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Step 6 — The master equation

Put the surviving pieces together: term A $=0$, term B $=\vec{\tau }$.

Reading it as conservation: ${\vec{\tau }}_{ext,net}=0\Rightarrow \frac{d\vec{L}}{dt}=0\Rightarrow \vec{L}=\text{constant}$

Look at the figure. A balance beam: torque $\vec{\tau }$ on one pan sets the "drift speed" of $\vec{L}$ on the other. Empty torque pan ⟹ the $\vec{L}$ pointer is pinned — frozen — conserved. Compare with Torque and its translational twin Conservation of linear momentum.

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Step 7 — Why only EXTERNAL torque? (internal pairs cancel)

WHAT. A real system is many particles pulling on each other. Do those internal pulls change the total $\vec{L}$? We check.

WHY they cancel. By Newton's third law, particle 1 pulls 2 with $\vec{f}$ and particle 2 pulls 1 with $-\vec{f}$ — equal and opposite, along the line joining them. Add their two torques about the same pivot:

${\vec{\tau }}_{int}={\vec{r}}_1\times \vec{f}+{\vec{r}}_2\times (-\vec{f})=\underset{\text{along the joining line}}{\underbrace{({\vec{r}}_1-{\vec{r}}_2)}}\times \vec{f}$

But $\vec{f}$ is also along that joining line (this is what Central forces means). Two parallel arrows ⟹ cross product $=0$.

${\vec{\tau }}_{int}=0$

Look at the figure. The two internal forces (amber) lie exactly on the dashed line linking the particles, so $({\vec{r}}_1-{\vec{r}}_2)$ and $\vec{f}$ are parallel — the parallelogram collapses, zero torque. Only arrows coming from outside the system can twist the total. See Moment of inertia for how many particles combine into $L=I\omega$.

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Step 8 — Edge & degenerate cases (never left hanging)

Look at the figure. Four orbit positions of a planet (quadrants I–IV). At every one the force arrow lies on the $\vec{r}$ line — the twist is zero all the way round. Whether close-and-fast or far-and-slow, $\vec{L}$ never budges.

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The one-picture summary

The whole derivation on one blueprint: start from $\vec{L}=\vec{r}\times \vec{p}$ → differentiate → product rule splits it → term A dies (parallel vectors) → term B is torque → $\frac{d\vec{L}}{dt}={\vec{\tau }}_{ext}$ → set torque to zero → $\vec{L}$ frozen. Internal pairs cancel along the way, so only outside twists matter.

Recall Feynman: the whole walk in plain words

We drew an arrow from a pivot to a moving object ($\vec{r}$) and an arrow for its motion ($\vec{p}$). "Spin around the pivot" ($\vec{L}$) is biggest when the motion is sideways and zero when it is straight in or out — that is why we used the sideways-measuring cross product. To ask "does the spin stay put?" we watched how fast it changes each instant. That change came in two pieces. The first piece asked the object to circle around its own velocity — impossible, so it vanished. The second piece was the twist from a force, which we named torque. So the change of spin is the twist: $\frac{d\vec{L}}{dt}=\vec{\tau }$. No twist from outside ⟹ the spin freezes ⟹ conserved. And the pulls particles exert on each other always cancel in pairs, so only twists from outside the system count. That single sentence contains the skater speeding up, the planet sweeping equal areas, and the putty on the turntable — all of them.

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Connections

• Parent · full conditions & worked examples
• Torque — the right-hand side of the master equation.
• Moment of inertia — how many particles' $\vec{L}$ become $L=I\omega$.
• Newton's third law — why internal torques cancel (Step 7).
• Central forces — the parallel-arrow zero-torque case (Step 8).
• Kepler's laws — equal areas from constant $\vec{L}$.
• Conservation of linear momentum — the translational twin of this whole story.
• Rotational kinetic energy — conserved $L$ does not mean conserved KE.