Visual walkthrough — Kinetic energy in generalized coordinates

Step 1 — A point pinned to a track: the position map

Read the symbols where they sit:

• $\mathbf{r}$ ::: the arrow from origin to the particle, its real position in the room.
• $q$ ::: the one number we are allowed to change — position along the track.
• $t$ ::: the clock. It appears only if someone is physically moving the track itself. Hold this thought; it is the hero of Steps 5–7.

WHY a map at all? Because energy is born in room-coordinates ($x,y$) but Lagrangian mechanics wants it in track-coordinates ($q$). A map is the translation dictionary between the two. Look at the figure: as $q$ slides, the tip of $\mathbf{r}$ walks along the printed curve.

Step 2 — How fast does the tip move? (the chain rule, drawn)

Now the actual velocity. The tip moves for two independent reasons:

$\dot{\mathbf{r}}=\underset{\text{tangent arrow}}{\underbrace{\frac{\partial \mathbf{r}}{\partial q}}}\underset{\text{how fast }q\text{ changes}}{\underbrace{\dot{q}}}+\underset{\text{track being dragged}}{\underbrace{\frac{\partial \mathbf{r}}{\partial t}}}.$

Term by term:

• $\dot{\mathbf{r}}$ ::: the true velocity arrow in the room (a dot on top means "rate per second").
• $\frac{\partial \mathbf{r}}{\partial q}\dot{q}$ ::: (tangent direction) $\times$ (speed of the coordinate). This is sliding along the track.
• $\frac{\partial \mathbf{r}}{\partial t}$ ::: the tip moving even when $q$ is frozen, because the track itself is being carried. Zero unless the constraint moves.

The figure shows both contributions as two separate arrows, then their sum as the true velocity.

Step 3 — Energy needs speed-squared: enter the dot product

So the total kinetic energy, summed over every particle $a$ (each mass $m_a$), is $T={\sum }_a\frac{1}{2}{m}_a{\dot{\mathbf{r}}}_a\cdot {\dot{\mathbf{r}}}_a.$

• ${\sum }_a$ ::: add up over all particles.
• $\frac{1}{2}{m}_a$ ::: the familiar "half mass".
• ${\dot{\mathbf{r}}}_a\cdot {\dot{\mathbf{r}}}_a$ ::: the speed-squared of particle $a$.

WHY the dot product and not just multiplying numbers? Because velocity is an arrow, not a number. Only the dot product turns two arrows into the single scalar "speed-squared" while automatically summing over $x$ and $y$ (and $z$). The figure shows one velocity arrow and its own length-squared as the shaded square.

Step 4 — Multiply the two brackets: three families of terms appear

Substitute the Step-2 velocity into the dot product. For each particle: ${\dot{\mathbf{r}}}_a\cdot {\dot{\mathbf{r}}}_a=\left({\sum }_i\frac{\partial {\mathbf{r}}_a}{\partial q_i}{\dot{q}}_i+\frac{\partial {\mathbf{r}}_a}{\partial t}\right)\cdot \left({\sum }_j\frac{\partial {\mathbf{r}}_a}{\partial q_j}{\dot{q}}_j+\frac{\partial {\mathbf{r}}_a}{\partial t}\right).$

We now allow several coordinates $q_1,\dots ,q_n$ — that is why one bracket uses index $i$ and the mirror bracket uses index $j$: they run independently, so every $i$ meets every $j$.

The multiplication grid in the figure has four cells; the two off-diagonal cells are identical (that's why $T_1$ carries a $\times 2$ that later cancels the $\frac{1}{2}$). This is the whole secret of the decomposition — pure algebra of $(A+B{)}^2$.

Step 5 — Reading off the three pieces

Collecting by powers of velocity gives the headline: $\boxed{T=T_2+T_1+T_0}$

$T_1={\sum }_i{a}_i{\dot{q}}_i,a_i={\sum }_a{m}_a\frac{\partial {\mathbf{r}}_a}{\partial q_i}\cdot \frac{\partial {\mathbf{r}}_a}{\partial t}$ $T_0=\frac{1}{2}{\sum }_a{m}_a{\left(\frac{\partial {\mathbf{r}}_a}{\partial t}\right)}^2.$

• $a_i$ ::: overlap between a tangent arrow and the drag arrow. Nonzero only when the track moves.
• $T_0$ ::: half-mass times drag-speed-squared. The energy the particle would have even sitting still on the track.

The figure stacks the three terms as three coloured bars whose heights are "power of $\dot{q}$" = 2, 1, 0.

Step 6 — Degenerate case: nobody moves the track (scleronomic)

This is the 95%-of-textbooks case. Because $T$ is now homogeneous of degree 2 in the $\dot{q}$'s, Euler's theorem gives ${\sum }_i{\dot{q}}_i\partial T/\partial {\dot{q}}_i=2T$ — the identity that later makes $H=T+V$ in the energy function and powers the momentum–energy machinery of the Euler-Lagrange equations.

Worked check — plane polar ($x=r\cos \theta ,y=r\sin \theta$), coordinates $(r,\theta )$, no time: $\frac{\partial \mathbf{r}}{\partial r}=(\cos \theta ,\sin \theta ),\frac{\partial \mathbf{r}}{\partial \theta }=(-r\sin \theta ,r\cos \theta ).$

• $M_{rr}=m({\cos }^2\theta +{\sin }^2\theta )=m$.
• $M_{\theta \theta }=m{r}^2({\sin }^2\theta +{\cos }^2\theta )=m{r}^2$.
• $M_{r\theta }=m(\cos \theta )(-r\sin \theta )+m(\sin \theta )(r\cos \theta )=0$ — the two tangent arrows are perpendicular, so no overlap. $\boxed{T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)}.$ The figure draws the radial and angular tangent arrows at right angles — that right angle is the vanishing of $M_{r\theta }$.

Step 7 — The moving-track case: where $T_1$ and $T_0$ actually appear

The lesson in one line: $T_1$ is nonzero exactly when the tangent arrow and the drag arrow overlap ($a_i=$ their dot product). Perpendicular ⇒ $T_1=0$ (bead case). Parallel ⇒ $T_1$ maximal (sliding-frame case). The figure contrasts the two geometries side by side.

The one-picture summary

Everything above is one identity — $(A+B)\cdot (A+B)$ — sorted by how many velocity factors survive. The final figure is the whole flowchart: position map → chain-rule velocity → dot-product square → three sorted piles, with the "freeze the clock" switch that kills two of them.

Active recall