Visual walkthrough — Generalized momenta and generalized forces

Everything below uses a single running example: a bead free to move on a curved track, whose position we will label by one number $q$. We start with literally nothing but a dot on a wire.

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Step 1 — One dot, one number

PICTURE.

WHY one number and not two. The wire is a curve. A curve is 1-dimensional. Forcing $(x,y)$ onto it means carrying a leftover equation ($y=f(x)$) forever. Choosing $q$ throws that equation away before we do any physics. This is the whole reason Lagrangian Mechanics exists.

The real Cartesian position is a function of that one number: $\vec{r}=\vec{r}(q,t).$

• $\vec{r}$ — the honest position arrow from origin to bead (this is what Newton uses).
• $q$ — our chosen dial.
• $t$ — allowed in case the wire itself is being moved (we will need this later).

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Step 2 — Wiggle the dial: velocity is a rate through the same picture

$\vec{v}=\frac{d\vec{r}}{dt}=\frac{\partial \vec{r}}{\partial q}\dot{q}+\frac{\partial \vec{r}}{\partial t}.$

Read every symbol:

• $\vec{v}$ — the bead's true velocity arrow.
• $\frac{\partial \vec{r}}{\partial q}$ — how far the bead moves in real space per unit dial-turn, holding time frozen. It is a vector tangent to the wire. Call it the "sensitivity arrow."
• $\dot{q}$ — how fast we are turning the dial.
• $\frac{\partial \vec{r}}{\partial t}$ — extra motion if the wire itself is being dragged (zero for a fixed wire).

PICTURE.

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Step 3 — Push the bead: real work, measured along the wire

WHY a dot product and not just $F\delta r$. A force pointing across the wire does no work on a bead that can only move along the wire. The dot product $\vec{F}\cdot \delta \vec{r}$ automatically keeps only the along-wire part of the force. That is exactly the question "how much of this push actually helps the motion?" — and the dot product is the tool that answers it.

$\delta W=\vec{F}\cdot \delta \vec{r}.$

PICTURE.

• $\delta W$ — tiny bit of work (energy) delivered.
• $\vec{F}$ — the real applied force arrow.
• $\delta \vec{r}$ — the tiny allowed step along the tangent.
• $\vec{F}\cdot \delta \vec{r}=|\vec{F}||\delta \vec{r}|\cos \alpha$ — only the component of $\vec{F}$ along the step counts; $\alpha$ is the angle between them.

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Step 4 — Rewrite the nudge in terms of the dial

Now substitute into the work from Step 3: $\delta W=\vec{F}\cdot \frac{\partial \vec{r}}{\partial q}\delta q=\underset{\text{call this }Q}{\underbrace{\left(\vec{F}\cdot \frac{\partial \vec{r}}{\partial q}\right)}}\delta q.$

PICTURE.

Term by term:

• $\delta q$ — tiny dial-turn (the only free thing left).
• $\frac{\partial \vec{r}}{\partial q}$ — sensitivity arrow (Step 2): converts dial-turns to real steps.
• $\vec{F}\cdot \frac{\partial \vec{r}}{\partial q}$ — the bundle that multiplies $\delta q$ to give work. We name it $Q$.

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Step 5 — Why $Q$ becomes a torque when $q$ is an angle

• $q$ a length (metres): $Q$ has units J/m $=$ newtons → an ordinary force.
• $q$ an angle (radians, dimensionless): $Q$ has units J/rad $=$ N·m → a torque.

WHY this is not a coincidence. For an angle, the sensitivity arrow $\partial \vec{r}/\partial q$ has length $\approx r$ (a small angle sweeps an arc $r\delta \theta$). So $Q_{\theta }=\vec{F}\cdot \frac{\partial \vec{r}}{\partial \theta }=(\text{tangential force})\times r,$ which is exactly the schoolbook torque $=$ force $\times$ lever arm. The formula rediscovered torque with no special effort.

PICTURE.

Recall

If $q$ is an angle, the generalized force $Q_{\theta }$ is a ::: torque, with units N·m (joules per radian). The reason its units differ from a plain force is ::: the weight $\partial \vec{r}/\partial q$ has units of length per radian, i.e. a lever arm.

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Step 6 — The momentum side: kinetic energy meets the same tangent arrow

Kinetic energy of one bead is $T=\frac{1}{2}m\vec{v}\cdot \vec{v}$. Differentiate w.r.t. $\dot{q}$ and use the cancel-the-dots identity from Step 2: $\frac{\partial T}{\partial \dot{q}}=m\vec{v}\cdot \frac{\partial \vec{v}}{\partial \dot{q}}=m\vec{v}\cdot \frac{\partial \vec{r}}{\partial q}.$ If $V$ has no $\dot{q}$ (usual case), $p=\partial L/\partial \dot{q}=\partial T/\partial \dot{q}$.

Term by term:

• $m\vec{v}$ — the honest Newtonian momentum arrow.
• $\partial \vec{r}/\partial q$ — the tangent sensitivity arrow (again!).
• $p=(m\vec{v})\cdot (\text{tangent arrow})$ — the projection of true momentum onto the dial's direction.

PICTURE.

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Step 7 — Angle case for momentum, and the cyclic-coordinate payoff

PICTURE.

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The one-picture summary

Everything above is two projections of the same tangent arrow $\partial \vec{r}/\partial q$:

• Project the force onto it → generalized force $Q$ (a torque for angles).
• Project the momentum onto it → generalized momentum $p$ (angular momentum for angles).
• Euler–Lagrange stitches them: $\dot{p}=Q$.

Recall Feynman retelling — the whole walk in plain words

Put a ball on a bent wire. Because it can only move one way, one dial $q$ says where it is. Turn the dial and the ball rolls along a certain direction — that direction (per turn of the dial) is our one magic arrow. Now do two things with it. First, shine your push onto the arrow: whatever fraction of your force lines up with the arrow is the only part that does work, and we call that shadow the generalized force $Q$. Second, shine the ball's momentum onto the same arrow: that shadow is the generalized momentum $p$. If the dial happens to be an angle, both shadows grow by the lever-arm length $r$: the push-shadow becomes a twist (torque) and the momentum-shadow becomes spin (angular momentum) — no new physics, just the arrow being long. Finally, energy bookkeeping (Euler–Lagrange) guarantees the twist equals the rate the spin changes, $\dot{p}=Q$ — Newton's law wearing new clothes. And if the energy doesn't care about the dial's value at all, the spin can never change: that's why angular momentum is conserved.

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