Generalized coordinates — choosing them, degrees of freedom: Visual walkthrough

This page is the picture-by-picture companion to Generalized coordinates — choosing them, degrees of freedom. If a word feels undefined, that is on us — we build it below.

Step 1 — One free particle: how many numbers to pin it?

WHAT. Put a single dot (a "particle" — an object so small we treat it as one point) somewhere in empty space. To tell a friend exactly where it is, how many numbers must you say?

WHY this question first. Everything in this chapter is bookkeeping about numbers needed to locate things. Before we subtract anything, we must know the cost of describing a system with nothing restricting it. That cost is our starting budget.

PICTURE. Look at the figure. To reach the red dot from the corner you walk a certain amount rightward ( $x$ ), a certain amount forward ( $y$ ), and a certain amount up ( $z$ ). Three arrows, three numbers. Not two, not four — space itself has three independent directions.

Figure — Generalized coordinates — choosing them, degrees of freedom

So one free particle costs 3 numbers. That "3" is the dimension of the space we live in — hold onto it, it becomes the 3 in the formula.

Step 2 — Many particles: the raw budget $3 N$

WHAT. Now scatter $N$ particles (here $N$ just means "however many you have" — a whole number like 2, or 5, or a million). Each one still needs its own 3 numbers, and no particle's address helps you find another's.

WHY. Independent objects have independent addresses. There is no discount for buying in bulk when the particles are unconstrained — so the total cost is simply the per-particle cost times the number of particles.

PICTURE. Three dots, each with its own little $x, y, z$ tripod of arrows. Count all the arrows: $3 + 3 + 3 = 9$ . In general $3 \times N$ .

For a 2D problem (everything trapped in a flat plane, so $z$ is always 0 and buys us nothing) the raw budget is $2 N$ instead. Same idea, one fewer direction.

Step 3 — A constraint is one equation, and it costs you one number

WHAT. A constraint is a rule that forbids some configurations. The kind we handle here — a holonomic constraint — is a rule you can write as a single equation relating the positions (and maybe time): $f (r_{1}, \dots, r_{N}, t) = 0.$ Here $f$ is just "some formula built from the coordinates" and setting it to $0$ is the rule.

WHY an equation removes exactly one number. An equation is a relationship: it lets you solve for one variable in terms of the others. Once you know all the others, that last one is no longer free — it is forced. So the equation eats exactly one unit of freedom. Not two (one equation can't fix two things), not zero (a genuine rule always fixes something).

PICTURE. Left: a particle free to roam a whole flat sheet — 2 numbers $(x, y)$ . Right: we impose the rule " $y = x$ " (that's our $f = y - x = 0$ ). Now the particle is stuck on a line. You still get to choose $x$ freely, but the moment you do, $y$ is decided. One equation → one line → one number left, not two.

Step 4 — Stack $k$ independent rules: subtract $k$

WHAT. If there are $k$ such holonomic constraints, and they are independent (no rule is secretly a repeat or a consequence of the others), then each one independently kills one number. So they kill $k$ numbers in total.

WHY independence matters. If two rules say the same thing, the second one costs you nothing — the number it would remove was already gone. Only new information reduces freedom. The letter $k$ counts genuinely-new rules.

PICTURE. A particle in 3D space (3 numbers). Rule 1: "stay on this tilted plane" → drops it to a 2D sheet. Rule 2 (independent): "stay on this second plane" → the two planes cross in a line, 1 number left. Each independent plane-rule peeled off one dimension: $3 \to 2 \to 1$ .

Step 5 — The pendulum: watch a "2" become a "1"

WHAT. A mass on a rigid rod of length $ℓ$ swinging in a vertical plane. Raw budget in 2D: $2 N = 2$ . The rod's rigidity is the rule "distance from pivot is always $ℓ$ ": $x^{2} + y^{2} = ℓ^{2}$ , i.e. $k = 1$ . So $n = 2 - 1 = 1$ .

WHY the angle is the clever coordinate. The formula tells us one number suffices, but which number? If we insist on $x$ , we must always remember $y = - ℓ^{2} - x^{2}$ — the rule follows us around like homework. Instead pick the angle $θ$ . Then $x = ℓ sin θ, y = - ℓ cos θ,$ and check: $x^{2} + y^{2} = ℓ^{2} (sin^{2} θ + cos^{2} θ) = ℓ^{2}$ automatically, for every $θ$ . The constraint didn't get "solved" — it vanished, baked into the meaning of the coordinate.

PICTURE. The rod sweeps a circle. The single arrow that matters is the growing angle $θ$ measured from straight-down. The forbidden directions (in/out along the rod) are drawn faded — you cannot go there, so they cost nothing.

Step 6 — The rigid body: why the count stops at 6

WHAT. A rigid body has infinitely many particles, so naive $3 N = \infty$ . But rigidity ("every pair of points keeps a fixed distance") is a colossal pile of constraints that collapses the count to a flat 6, no matter how big the body.

WHY 6, split as 3 + 3. Freeze one point of the body: that takes 3 numbers to place in space (translation). Now the body can only spin about that point. Its orientation needs 3 more numbers — the Euler angles $(ϕ, θ, ψ)$ . Total $3 + 3 = 6$ . Every extra particle's position is then forced by rigidity, contributing nothing new.

PICTURE. A chalk brick. Blue arrow-triple: where its corner sits (3 translation numbers). Pink curved arrows: three independent ways it can tumble (3 rotation numbers). Six freedoms, full stop.

Step 7 — The degenerate cases: where the picture bends

WHAT. Two "watch out" scenarios where a naive $n = 3 N - k$ misleads.

Case A — the invisible rotation (dumbbell). Two point atoms, one bond-length rule: $3 N - k = 6 - 1 = 5$ . But naively you'd expect $3 + 3 = 6$ (translation + full orientation). The missing one? Spinning about the bond axis moves no point — two points on a line look identical after such a spin. That "rotation" is not a real freedom, so orientation needs only 2 angles, not 3. Result $3 + 2 = 5$ ✓.

Case B — the rolling disk (non-holonomic). A coin rolling without slipping obeys a velocity rule ( $\dot{r}$ tied to the spin) that cannot be squashed into a position-only equation $f (r, t) = 0$ . So it is non-holonomic: it restricts how you move, not where you can be. The set of reachable positions stays large, and $n = 3 N - k$ for position coordinates simply does not apply. You can even roll the coin back to the same spot facing a new direction.

PICTURE. Left: the dumbbell with a faded circular arrow around its axis marked "moves nothing." Right: a rolling coin with its ground track; a green arrow shows the velocity rule, crossed out as "can't integrate to a position rule."

The one-picture summary

Everything above, on one board: the raw budget $3 N$ (or $6$ for a rigid body) sits at the top; each independent holonomic rule is a knife that slices off exactly one number; what survives at the bottom is $n$ , the degrees of freedom — the count of good generalized coordinates you then feed into a Lagrangian and the Euler–Lagrange equations.

Recall Feynman: the whole walkthrough in plain words

Start greedy. Say "I'll describe every dot with three numbers — right, forward, up." Add them all up: that's $3 N$ , your wasteful starting pile. Now read the rules of the problem. Every rule you can write as one equation ("the rod is this long", "the bead stays on the wire") is a knife: it slices exactly one number off your pile, because once you obey the rule, one number is no longer yours to choose. Slice $k$ times for $k$ real, non-repeating rules. Whatever's left in the pile is $n$ — the true number of ways your system can wiggle. Two warnings: (1) a rigid body is a huge pile of rules that always leaves 6, and one "spin" of a two-atom stick moves nothing so it doesn't count; (2) a rolling coin's rule is about how fast, not where, so it's a different animal — don't slice with it. Start raw, subtract the rules, respect the sneaky cases.

Recall

Raw budget for $N$ particles in 3D? ::: $3 N$ numbers. Why does one holonomic constraint remove exactly one DOF? ::: It's one equation, so it lets you solve for one variable in terms of the rest — that variable is no longer free. Why is $θ$ better than $x$ for the pendulum? ::: $x = ℓ sin θ, y = - ℓ cos θ$ satisfies $x^{2} + y^{2} = ℓ^{2}$ automatically, so the constraint vanishes. Why does a two-atom dumbbell have 5 DOF, not 6? ::: Rotation about the bond axis moves no point, so orientation needs 2 angles not 3: $3 + 2 = 5$ . Why can't $n = 3 N - k$ be used for a rolling disk? ::: Its rolling constraint is non-holonomic (a velocity rule that can't be integrated to a position equation), so it doesn't cleanly remove a position coordinate.