Visual walkthrough — Rigid body — definition, degrees of freedom

We build every idea from zero. First we agree on what a "coordinate" even is, then what a "constraint" removes, then we assemble the body one point at a time.

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Step 1 — What a single number ("coordinate") buys you

WHAT. A coordinate is just one number that answers one "how far along?" question. To pin a dot on a line, you say one number: how far from the start. To pin a dot on a floor, you need two: how far right, how far forward. To pin a dot in a room, you need three: right, forward, up.

WHY. Space has three independent directions that can't be built from each other — you can't reach "up" by combining "right" and "forward". Each independent direction demands its own number. We call these numbers $x$, $y$, $z$.

PICTURE. Look at the three coloured rulers below. The blue dot sits at the meeting of "5 units right, 3 units forward, 4 units up". Cover any one ruler and you lose the ability to locate the dot — proof that all three are needed and none is spare.

$\underset{\text{how far right}}{\underbrace{(x}},\underset{\text{how far forward}}{\underbrace{y}},\underset{\text{how far up}}{\underbrace{z}})\Rightarrow 3\text{ free numbers}$

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Step 2 — Many free dots: freedoms just add up

WHAT. Put $N$ dots in the room, each free to sit anywhere, none caring about the others. Count the numbers you must supply.

WHY. "Independent" means choosing dot 1's position tells you nothing about dot 2's. So the total count is simply 3 added to itself $N$ times.

PICTURE. Three free dots below, each with its own little $(x,y,z)$ tag. No strings between them — total freedom.

$\text{total coordinates}=\underset{\text{per dot}}{\underbrace{3}}\times \underset{\text{number of dots}}{\underbrace{N}}=3N$

Here $3N$ is our starting budget of freedom. Everything that follows spends from this budget by adding rules that tie dots together.

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Step 3 — What one "distance rule" (constraint) removes

WHAT. Now glue an invisible rigid stick between dot A and dot B: their separation must stay a fixed length $d$. That single rule is one constraint.

WHY this tool — an equation, not a picture-erasure. A constraint is a written equation linking coordinates: $\underset{\text{squared distance between A and B}}{\underbrace{(x_B-x_A{)}^2+(y_B-y_A{)}^2+(z_B-z_A{)}^2}}=\underset{\text{fixed, chosen once}}{\underbrace{d^2}}$ One equation lets you solve for one unknown in terms of the others. So one independent constraint removes exactly one degree of freedom. This is the master accounting rule: $\boxed{\text{DOF}=(\text{total coordinates})-(\text{independent constraints})}$

PICTURE. Dot A is free anywhere. Dot B is no longer free everywhere — the stick forces it onto the surface of a sphere of radius $d$ centred on A. It went from "anywhere in 3D" (3 numbers) to "anywhere on a 2D skin" (2 numbers). One freedom spent.

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Step 4 — Build the body: point 1 (the anchor)

WHAT. We now assemble the whole rigid body cleverly. Claim: to freeze an entire rigid body in space, it is enough to freeze 3 points that don't lie on one straight line (three non-collinear points). Pin those three and every other particle is dragged into place by its fixed distances. We add them one at a time and count.

Point 1 is placed first, with nothing yet to obey.

WHY. With no earlier point to be a fixed distance from, point 1 roams the whole room freely.

PICTURE. A single yellow anchor dot floating free, tagged with its three numbers.

$\text{point 1 gives}\underset{x_1,y_1,z_1}{\underbrace{3}}\text{DOF}$

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Step 5 — Point 2 (lives on a sphere) and Point 3 (lives on a circle)

WHAT. Attach point 2 at a fixed distance from point 1 — that's 1 constraint. Then attach point 3 at a fixed distance from point 1 and from point 2 — that's 2 constraints.

WHY, geometry-first.

• Point 2: one distance rule ⇒ it sits on the sphere around point 1 ⇒ $3-1=2$ DOF (roam the sphere's surface).
• Point 3: two distance rules. Fixed distance from point 1 puts it on one sphere; fixed distance from point 2 puts it on another sphere. Two spheres cross in a circle. A point on a circle needs just $1$ number (how far around) ⇒ $3-2=1$ DOF.

PICTURE. Point 2 slides on the blue sphere; point 3 slides on the pink circle where the two spheres meet. Watch the freedom shrink: room → sphere → circle.

$\underset{\text{pt 1: free}}{\underbrace{3}}+\underset{\text{pt 2: on sphere}}{\underbrace{2}}+\underset{\text{pt 3: on circle}}{\underbrace{1}}=6$

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Step 6 — Every other particle is forced (0 new DOF)

WHAT. Bring in particle 4, 5, … up to $N$. Each one has a fixed distance to points 1, 2, and 3 (all distances in a rigid body are frozen). Three distances to three known, non-collinear points pin a point uniquely in 3D.

WHY — three spheres meet at a point. Distance to point 1 ⇒ a sphere. Distance to point 2 ⇒ another sphere (they meet in a circle). Distance to point 3 ⇒ a third sphere that stabs that circle at a single spot. No freedom left: 0 new DOF per extra particle.

PICTURE. Three spheres (around the three anchors) intersecting at one lonely dot — particle 4 has nowhere else to be.

$\text{DOF}=\underset{\text{from 3 anchors}}{\underbrace{6}}+\underset{\text{particles }4\dots N}{\underbrace{0+0+\cdots }}=6(\text{for any }N)$

This is the punchline to the mistake "more particles → more DOF": rigidity saturates the count at 6. See Moment of Inertia for how those "extra" particles, though DOF-free, still matter for how hard it is to spin the body.

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Step 7 — Reading the 6 as 3 + 3 (slide vs turn)

WHAT. Split the 6 into what they physically do: 3 translations (slide the body) and 3 rotations (turn the body).

WHY. Take any one reference point of the body — say its centre of mass. Placing that point uses 3 numbers $(x,y,z)$: that's where the body is (pure slide). With that point pinned, the body can still pivot about it — spin about the $x$, $y$, $z$ axes: 3 angles for which way it faces. These three turning freedoms are the input to Rotational Kinematics.

PICTURE. Left panel: three straight arrows (slide along $x,y,z$). Right panel: three curved arrows (roll, pitch, yaw) about the centre.

$6=\underset{\text{translation: }x,y,z}{\underbrace{3}}+\underset{\text{rotation: roll, pitch, yaw}}{\underbrace{3}}$

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Step 8 — Degenerate cases: when the count drops

WHAT. Extra pins (constraints) subtract from 6. Three classic cases, plus the point-mass exception.

WHY & PICTURE. Each pin removes specific freedoms — read them straight off the geometry.

Pin it how	Kills	DOF left	Reason (geometry)
Fix one point	all 3 slides	$6-3=3$	body can only pivot about that point
Fix an axle (a line)	3 slides + 2 turns	$6-5=1$	only the spin angle $\theta$ survives
Confine to a plane	1 slide ($z$) + 2 turns	$6-3=3$	$(x,y)$ + one in-plane angle
Diatomic (2 point atoms)	spin about the bond does nothing	$5$	$3$ slides $+2$ turns; the third turn moves no mass

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The one-picture summary

One figure, whole story: a room (3 numbers) → a sphere (−1) → a circle (−1) → three anchors giving 6 → split into 3 slides + 3 turns, with the degenerate cases branching off.

Recall Feynman: the walkthrough in plain words

Imagine an empty room. Drop in a marble. To tell me where it floats you say three things: how far right, how far forward, how high — 3 numbers.

Now glue a stiff sculpture together from lots of marbles so nothing can shift inside it. I don't need every marble's numbers — I just watch three well-spread marbles. The first one is free: 3 numbers. The second is roped to the first at a fixed length, so it can only slide over an invisible ball's surface around the first: 2 numbers. The third is roped to both, so it's stuck on a ring where two balls overlap: 1 number. Three plus two plus one is six.

Every remaining marble is roped to all three of those, and three ropes to three spread-out points leave a marble exactly one place to be — so the millionth marble adds nothing. That's why a pebble and a planet both have six.

Finally, sort the six into two piles: three are for sliding the whole thing around (left-right, forward-back, up-down) and three are for turning it in place (tilt, spin, roll). 6 = 3 slides + 3 turns. Nail the sculpture to a wall and you snip some ropes: fixed to one spot leaves 3 turns; stuck on an axle leaves just 1 spin. Same recipe every time — start at six, subtract what you pinned.

Recall

From zero freedom-budget, how many numbers to place point 2 of a rigid body relative to point 1? ::: 2 — it lives on a sphere around point 1 ($3-1$). Why does particle number 100 add 0 DOF? ::: Its fixed distances to 3 non-collinear anchors put it at the single crossing point of three spheres — no freedom left. Slide the 6 into physical piles. ::: 3 translations (slide along $x,y,z$) + 3 rotations (turn about $x,y,z$). Fixed axle DOF, and what was removed? ::: 1 DOF; the axle kills all 3 slides and 2 of the 3 turns ($6-5$).