Lagrangian L = T − V: Worked examples — Physics deep dive

This page is a drill. The parent note showed you the machinery; here we push it into every corner — signs of the potential, curved coordinates, coupled coordinates, coordinates that don't even appear in $L$ , degenerate free particles, and an exam twist. Before we start, one reminder of what each symbol means, because we will earn nothing on trust.

Definition The vocabulary we will use on every line

$q$ — a generalized coordinate: any single number that helps pin down where the system is (a length $x$ , a height $y$ , an angle $θ$ ). See Generalized Coordinates and Constraints.
$\overset{q}{˙}$ — the generalized velocity: how fast that number changes per second. The dot means "time-derivative of the thing under it".
$\overset{q}{¨}$ — the acceleration of that coordinate (two dots = two time-derivatives).
$T$ — kinetic energy, the energy of motion, always $\frac{1}{2} m v^{2}$ where $v$ is the real speed.
$V$ — potential energy, the energy of position (height in gravity, stretch in a spring).
$L = T - V$ — the Lagrangian. Not the total energy.
The Euler–Lagrange equation (the engine, one per coordinate): $\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = 0$ Read it as: "the rate of change of (slope of $L$ in the velocity) equals (slope of $L$ in the position)." Full derivation lives in Euler–Lagrange Equation.

The scenario matrix

Every problem $L = T - V$ can throw belongs to one of these cells. The examples below are chosen so that together they cover all of them — none is left as "you'll figure it out".

Cell	What makes it distinct	Covered by
A. Straight-line, positive-slope $V$	$V$ increases with $q$ (climbing) → restoring force points down/back	Ex 1
B. Straight-line, negative-slope $V$	$V$ decreases with $q$ (a downhill push) → force points forward	Ex 2
C. Curved coordinate ( $T$ has $q$ in it)	$T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ — the velocity term itself depends on position	Ex 3
D. Two coupled coordinates	Two Euler–Lagrange equations that talk to each other	Ex 4
E. Cyclic (ignorable) coordinate	$q$ missing from $L$ → a conserved momentum	Ex 5
F. Degenerate / zero potential	$V = 0$ , the free particle — the limiting case	Ex 6
G. Sign-of-force stress test	Same spring, both sides of equilibrium (positive AND negative $x$ )	Ex 7
H. Real-world word problem + exam twist	Bead on a rotating hoop: a constraint that adds to $T$	Ex 8

The worked examples

Cell A — potential that grows with the coordinate

Worked example Ex 1 · Block sliding up a frictionless ramp of incline

$α$ A block of mass $m$ slides on a ramp tilted at angle $α$ above horizontal. Let $s$ = distance measured up the slope. Find its equation of motion.

Forecast: guess the sign of $\overset{s}{¨}$ before reading. Gravity should pull it back down, so we expect $\overset{s}{¨} < 0$ (a negative, i.e. down-slope, acceleration). Hold that thought.

Figure — the incline triangle.

A blue slope of length $s$ (the hypotenuse) rising over a white horizontal base; the red dashed vertical leg is the height gained, $s sin α$ ; the green arc marks $α$ and the green down-arrow is gravity $m g$ . The picture is the whole point of step 1 — read the height off the triangle.

Height gained climbing distance $s$ is $s sin α$ (look at the red vertical leg in the figure — it is the opposite side of the right triangle whose hypotenuse is $s$ ). Why this step? Potential energy only cares about height, and $sin α$ is exactly "vertical rise per unit slope-distance".
Write the energies: $T = \frac{1}{2} m \overset{s}{˙}^{2}$ , $V = m g s sin α$ . Why this step? Steps 2–3 of the parent recipe: energies in the chosen coordinate.
Form $L = \frac{1}{2} m \overset{s}{˙}^{2} - m g s sin α$ . Why this step? The Lagrangian is defined as $T - V$ ; we must assemble it before the engine can chew on it.
Velocity slope: $\frac{\partial L}{\partial s ˙} = m \overset{s}{˙}$ , then take its time-derivative: $\frac{d}{d t} (\frac{\partial L}{\partial s ˙}) = \frac{d}{d t} (m \overset{s}{˙}) = m \overset{s}{¨}$ . Why this step? This is the " $\frac{d}{d t} \frac{\partial L}{\partial s ˙}$ " half of the engine — first the velocity-slope, then how it changes in time.
Position slope: $\frac{\partial L}{\partial s} = - m g sin α$ . Why this step? This is the second half of the engine; only $V$ contains $s$ , and its slope is the (negative of the) force pushing on $s$ .
Euler–Lagrange: put the two halves into $\frac{d}{d t} \frac{\partial L}{\partial s ˙} - \frac{\partial L}{\partial s} = 0$ , giving $m \overset{s}{¨} - (- m g sin α) = 0 \Rightarrow \overset{s}{¨} = - g sin α .$ Why this step? This is the engine itself — setting rate-of-velocity-slope equal to position-slope is exactly the equation of motion.

Verify: The sign is negative → matches our forecast (slides back down). Limiting checks: at $α = 9 0^{\circ}$ (vertical wall of a ramp), $\overset{s}{¨} = - g$ — pure free fall. ✓ At $α = 0$ (flat table), $\overset{s}{¨} = 0$ — no push, sits still. ✓ Units: $g sin α$ is $m/s^{2}$ . ✓

Cell B — potential that falls with the coordinate

Worked example Ex 2 · Same ramp, coordinate measured

down the slope Now let $d$ = distance measured down the slope (positive means further down). This is Cell B: as $d$ grows, height drops, so $V$ has a negative slope.

Forecast: the block speeds up going down, so we expect $\ddot{d} > 0$ (positive).

Height above the starting point is now $- d sin α$ (going down = negative height). Why this step? We must be honest about sign: down means lower potential.
$T = \frac{1}{2} m \dot{d}^{2}$ , $V = - m g d sin α$ , so $L = \frac{1}{2} m \dot{d}^{2} + m g d sin α$ . Why this step? $L = T - V$ with a negative $V$ becomes a plus; this sign flip is the entire distinction from Ex 1.
Velocity half: $\frac{\partial L}{\partial d ˙} = m \dot{d}$ , then $\frac{d}{d t} (\frac{\partial L}{\partial d ˙}) = \frac{d}{d t} (m \dot{d}) = m \ddot{d}$ . Why this step? Same engine as always — velocity-slope, then its time-derivative.
Position half: $\frac{\partial L}{\partial d} = + m g sin α$ . Why this step? The $+ V$ contribution makes the position-slope positive — the source of the forward push.
EL: $m \ddot{d} - m g sin α = 0 \Rightarrow \ddot{d} = + g sin α .$ Why this step? Feeding both halves into $\frac{d}{d t} \frac{\partial L}{\partial d ˙} - \frac{\partial L}{\partial d} = 0$ delivers the equation of motion.

Verify: Positive, as forecast. And notice it is the exact same physics as Ex 1 — flipping the direction of the coordinate flipped the sign of $V$ 's slope, which flipped the sign of the acceleration. The world doesn't change; only our bookkeeping does. This is the whole reason the Lagrangian is coordinate-agnostic. Units: $g sin α$ is $m/s^{2}$ , an acceleration. ✓

Cell C — the coordinate lives inside the kinetic term

Worked example Ex 3 · Pendulum, re-examined for the

$ℓ$ -dependence Bob of mass $m$ on a rod of length $ℓ$ , angle $θ$ from the downward vertical. This is Cell C because the speed of the bob is $ℓ \dot{θ}$ — the length multiplies the velocity, so $T$ carries $ℓ^{2}$ .

Forecast: small swings should look like a spring ( $\ddot{θ} \propto - θ$ ); large swings should slow the restoring pull (because $sin θ$ bends over).

Figure — pendulum geometry.

A blue rod from the white pivot to a yellow bob; the green arc is the angle $θ$ from the dotted vertical; the green tangent arrow is the bob's speed $ℓ \dot{θ}$ (perpendicular to the rod); the red dashed segment is the height risen above the lowest point, $ℓ (1 - cos θ)$ . Steps 1 and 2 are literally read off this picture.

Speed of bob $= ℓ \dot{θ}$ (arc-length rate: radius times angular rate — see the green tangent arrow). Why this step? We need the real speed for $T$ , not just $\dot{θ}$ .
Height above lowest point $= ℓ (1 - cos θ)$ (the red gap in the figure). Why this step? $V$ needs the actual height; $ℓ (1 - cos θ)$ is how far the bob has risen.
$T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ , $V = m g ℓ (1 - cos θ)$ , $L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} - m g ℓ (1 - cos θ)$ . Why this step? Assemble $L = T - V$ ; note $θ$ now appears in $T$ and $V$ — the Cell C feature.
Velocity half: $\frac{\partial L}{\partial θ ˙} = m ℓ^{2} \dot{θ}$ , then $\frac{d}{d t} (\frac{\partial L}{\partial θ ˙}) = m ℓ^{2} \ddot{θ}$ . Why this step? The velocity-slope times its time-derivative — the first half of the engine, now carrying $ℓ^{2}$ .
Position half: $\frac{\partial L}{\partial θ} = - m g ℓ sin θ$ (derivative of $- m g ℓ (1 - cos θ)$ ; the $sin$ appears because $\frac{d}{d θ} (- cos θ) = sin θ$ ; here $T$ has no $θ$ so only $V$ contributes). Why this step? Second half of the engine — the restoring "torque per unit angle".
EL: $m ℓ^{2} \ddot{θ} + m g ℓ sin θ = 0 \Rightarrow \ddot{θ} = - \frac{g}{ℓ} sin θ .$ Why this step? Combine both halves in $\frac{d}{d t} \frac{\partial L}{\partial θ ˙} - \frac{\partial L}{\partial θ} = 0$ to get the equation of motion.

Verify: Small-angle limit $sin θ \approx θ$ gives $\ddot{θ} = - \frac{g}{ℓ} θ$ , simple harmonic with $ω = g / ℓ$ . ✓ At $θ = 0$ (hanging straight down): $\ddot{θ} = 0$ , an equilibrium. ✓ At $θ = π$ (balanced straight up): $sin π = 0$ so $\ddot{θ} = 0$ too — an unstable equilibrium, physically correct. ✓ Units: $g / ℓ$ is $s^{- 2}$ , times the dimensionless $sin θ$ → $rad/s^{2}$ . ✓

Cell D — two coordinates that talk to each other

Worked example Ex 4 · Two masses joined by a spring on a frictionless line

Masses $m$ each at positions $x_{1}$ and $x_{2}$ on a rail, joined by a spring of constant $k$ and natural length $ℓ_{0}$ . Two coordinates → two Euler–Lagrange equations.

Forecast: the spring only cares about the stretch $x_{2} - x_{1}$ , so the two equations should be mirror images pushing the masses toward each other or apart.

$T = \frac{1}{2} m \overset{x}{˙}_{1}^{2} + \frac{1}{2} m \overset{x}{˙}_{2}^{2}$ (each mass carries its own kinetic energy). Why this step? Kinetic energy is additive over independent particles.
Stretch $= (x_{2} - x_{1}) - ℓ_{0}$ , so $V = \frac{1}{2} k [(x_{2} - x_{1}) - ℓ_{0}]^{2}$ . Why this step? A spring stores $\frac{1}{2} k (stretch)^{2}$ , and stretch is relative separation minus natural length.
$L = T - V$ . Why this step? Assemble the Lagrangian; because $V$ mixes $x_{1}$ and $x_{2}$ , the two coordinates will couple.
For $x_{1}$ : velocity half $\frac{\partial L}{\partial x ˙ _{1}} = m \overset{x}{˙}_{1}$ , then $\frac{d}{d t} (\frac{\partial L}{\partial x ˙ _{1}}) = m \overset{x}{¨}_{1}$ ; position half $\frac{\partial L}{\partial x _{1}} = + k [(x_{2} - x_{1}) - ℓ_{0}]$ (chain rule; the inner derivative of $- x_{1}$ gives the $+$ ). Why this step? Remember $q$ and $\overset{q}{˙}$ are independent — when we differentiate w.r.t. $x_{1}$ we leave $\overset{x}{˙}_{1}$ frozen, so only $V$ contributes to the position half.
EL for $x_{1}$ : $m \overset{x}{¨}_{1} = k [(x_{2} - x_{1}) - ℓ_{0}] .$ Why this step? The engine $\frac{d}{d t} \frac{\partial L}{\partial x ˙ _{1}} - \frac{\partial L}{\partial x _{1}} = 0$ gives mass-1's equation.
For $x_{2}$ : identical steps give velocity half $m \overset{x}{¨}_{2}$ and position half $\frac{\partial L}{\partial x _{2}} = - k [(x_{2} - x_{1}) - ℓ_{0}]$ , so EL: $m \overset{x}{¨}_{2} = - k [(x_{2} - x_{1}) - ℓ_{0}] .$ Why this step? Each coordinate gets its own Euler–Lagrange equation; here the sign flips because $x_{2}$ enters the stretch with a $+$ .

Verify: Add the two equations: $m \overset{x}{¨}_{1} + m \overset{x}{¨}_{2} = 0$ , so the center of mass does not accelerate — momentum conserved, correct for an isolated pair. ✓ Subtract them: with $r = x_{2} - x_{1}$ , $m \overset{r}{¨} = - 2 k (r - ℓ_{0})$ , a clean oscillator of frequency $ω = 2 k / m$ about the natural length. ✓ Mirror symmetry confirmed. Units: $k [(x_{2} - x_{1}) - ℓ_{0}]$ is $(N/m) (m) = N$ , and $m \overset{x}{¨}$ is $kg \cdot m/s^{2} = N$ — both sides are forces. ✓

Cell E — a coordinate that vanishes from $L$

Worked example Ex 5 · Free particle on a plane, polar coordinates

A puck slides freely (no potential) on a horizontal plane. In polar coordinates $(r, ϕ)$ , $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2})$ , $V = 0$ . Notice: $ϕ$ does not appear in $L$ , only $\dot{ϕ}$ does. Such a coordinate is called cyclic or ignorable.

Forecast: if $ϕ$ is missing, its Euler–Lagrange equation should hand us a conserved quantity for free — angular momentum.

$L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2})$ . Why this step? Assemble $L = T - V$ with $V = 0$ ; note $ϕ$ itself is absent — the Cell E feature.
For $ϕ$ (velocity half): the generalized momentum is $p_{ϕ} = \frac{\partial L}{\partial ϕ ˙} = m r^{2} \dot{ϕ}$ . (See Generalized Momentum and Conservation.) Why this step? This is $\frac{\partial L}{\partial ϕ ˙}$ , the quantity whose time-derivative the engine tracks.
For $ϕ$ (position half): $\frac{\partial L}{\partial ϕ} = 0$ because $ϕ$ literally never appears. Why this step? A missing coordinate has zero position-slope — this is what makes it cyclic.
EL for $ϕ$ : $\frac{d}{d t} (\frac{\partial L}{\partial ϕ ˙}) - \frac{\partial L}{\partial ϕ} = \frac{d}{d t} (m r^{2} \dot{ϕ}) - 0 = 0$ , therefore $m r^{2} \dot{ϕ} = constant$ . Why this step? When the position-slope is zero, the engine says the velocity-slope (the momentum) is conserved.
For $r$ (both halves): $\frac{\partial L}{\partial r ˙} = m \overset{r}{˙} \Rightarrow \frac{d}{d t} (m \overset{r}{˙}) = m \overset{r}{¨}$ ; and $\frac{\partial L}{\partial r} = m r \dot{ϕ}^{2}$ . EL: $m \overset{r}{¨} - m r \dot{ϕ}^{2} = 0 \Rightarrow m \overset{r}{¨} = m r \dot{ϕ}^{2} .$ Why this step? The other coordinate still gets its own full Euler–Lagrange equation — the $r$ -equation carries the centrifugal term $m r \dot{ϕ}^{2}$ .

Verify: $m r^{2} \dot{ϕ}$ is exactly the angular momentum $L_{z}$ of a particle — conservation drops out of one missing variable, no torque analysis needed. ✓ The radial equation $m \overset{r}{¨} = m r \dot{ϕ}^{2}$ is the familiar outward centrifugal effect. ✓ Units: $p_{ϕ} = m r^{2} \dot{ϕ}$ has $kg \cdot m^{2} \cdot s^{- 1}$ , the correct dimension of angular momentum. ✓

Cell F — degenerate case: zero potential (free particle)

Worked example Ex 6 · The free particle in 1D — the simplest limit

No forces, no potential: $V = 0$ , $T = \frac{1}{2} m \overset{x}{˙}^{2}$ . This is the degenerate end of the matrix, the sanity floor.

Forecast: with nothing pushing, we expect $\overset{x}{¨} = 0$ — constant velocity, straight-line motion (Newton's first law).

$L = \frac{1}{2} m \overset{x}{˙}^{2}$ . Why this step? Assemble $L = T - V$ with $V = 0$ ; nothing to subtract.
Velocity half: $\frac{\partial L}{\partial x ˙} = m \overset{x}{˙}$ , then $\frac{d}{d t} (m \overset{x}{˙}) = m \overset{x}{¨}$ . Why this step? Velocity-slope then its time-derivative — first half of the engine.
Position half: $\frac{\partial L}{\partial x} = 0$ (no $x$ in $L$ ). Why this step? No potential means no position-slope, so $x$ is cyclic too.
EL: $m \overset{x}{¨} - 0 = 0 \Rightarrow \overset{x}{¨} = 0.$ Why this step? The engine with zero position-slope forces zero acceleration.

Verify: $x$ is also cyclic here, so momentum $p = m \overset{x}{˙}$ is conserved — constant velocity, exactly Newton's first law and the base case the whole framework must reduce to. ✓ Units: $m \overset{x}{¨}$ is $kg \cdot m/s^{2} = N$ ; setting it to zero says "zero net force". ✓ See Newton's Second Law for the $V \neq = 0$ generalization and Principle of Least Action for why the straight line is the stationary path.

Cell G — sign stress test across equilibrium

Worked example Ex 7 · Mass on a spring, checked on BOTH sides of

$x = 0$ $L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2}$ . We already know the EL equation is $m \overset{x}{¨} + k x = 0$ . The point of Cell G is to confirm the force sign is restoring in every case: $x > 0$ , $x < 0$ , and $x = 0$ .

Forecast: the acceleration should always point back toward $x = 0$ .

From the EL equation $m \overset{x}{¨} = - k x$ . To make the sign behaviour concrete and impossible to hand-wave, we plug in specific numbers $k = 4 N/m$ , $m = 1 kg$ — a numeric example is preferable here because the pedagogical claim is about signs across cases, and nothing exposes a sign error faster than watching real numbers flip. Why this step? Choosing clean numbers turns an abstract " $- k x$ " into three checkable arithmetic facts.
Case $x = + 3$ : $\overset{x}{¨} = - 4 (3) = - 12$ — pulls left (negative), back toward origin. ✓ Why this step? Tests the positive-displacement quadrant.
Case $x = - 3$ : $\overset{x}{¨} = - 4 (- 3) = + 12$ — pushes right (positive), back toward origin. ✓ Why this step? Tests the negative-displacement quadrant — the sign must flip.
Case $x = 0$ : $\overset{x}{¨} = 0$ — no force at equilibrium. ✓ Why this step? Tests the degenerate point so no scenario is left unshown.

Verify: In all three cases $sign (\overset{x}{¨}) = - sign (x)$ , the defining property of a restoring force. Frequency $ω = k / m = 4/1 = 2 rad/s$ . ✓ Units: $\overset{x}{¨}$ came out in $m/s^{2}$ (a $N/m$ over $kg$ times $m$ ), and $ω$ in $s^{- 1}$ . ✓

Cell H — real-world word problem with an exam twist

Worked example Ex 8 · Bead on a hoop spinning at fixed rate

$Ω$ (the twist) A bead of mass $m$ slides on a frictionless circular wire of radius $R$ . The hoop is forced to spin about a vertical diameter at constant angular speed $Ω$ . Let $θ$ = angle of the bead from the bottom of the hoop. This is the exam twist: the spin injects an extra kinetic term and creates an effective potential.

Forecast: for slow spin the bead should sit at the bottom ( $θ = 0$ ); for fast enough spin, centrifugal effects should push a new stable angle up the side.

Figure — bead on the spinning hoop.

A blue circle (the hoop) with a white dotted vertical spin-axis; the yellow arrow at top is the forced spin $Ω$ ; the red dot is the bead at angle $θ$ (green arc) from the dotted radius to the bottom; the yellow dashed segment is the bead's distance from the axis, $R sin θ$ , which is why the spin contributes $R sin θ Ω$ to its speed.

The bead's velocity has two perpendicular pieces: along the wire, speed $R \dot{θ}$ ; and around the spin axis, speed $R sin θ \cdot Ω$ (its distance from the axis is $R sin θ$ ). Since they are perpendicular, $v^{2} = R^{2} \dot{θ}^{2} + R^{2} sin^{2} θ Ω^{2}$ , so $T = \frac{1}{2} m (R^{2} \dot{θ}^{2} + R^{2} sin^{2} θ Ω^{2}) .$ Why this step? The forced spin is a constraint that does work in our coordinate, so it lands in $T$ , not $V$ ; we add the two perpendicular speeds via Pythagoras.
Height above the bottom $= R (1 - cos θ)$ , so $V = m g R (1 - cos θ)$ . Why this step? Gravity is a genuine potential; the height is read exactly as in the pendulum.
Form $L = \frac{1}{2} m R^{2} \dot{θ}^{2} + \frac{1}{2} m R^{2} Ω^{2} sin^{2} θ - m g R (1 - cos θ)$ . Why this step? Assemble $L = T - V$ ; the middle term is the spin's gift and will act like a negative potential.
Velocity half: $\frac{\partial L}{\partial θ ˙} = m R^{2} \dot{θ}$ , then $\frac{d}{d t} (\frac{\partial L}{\partial θ ˙}) = m R^{2} \ddot{θ}$ . Why this step? The $Ω$ -term has no $\dot{θ}$ in it, so the velocity-slope is the same as a plain pendulum.
Position half: $\frac{\partial L}{\partial θ} = m R^{2} Ω^{2} sin θ cos θ - m g R sin θ$ . Why this step? Now $θ$ sits in both the spin-kinetic term and $V$ , so both contribute to the position-slope (chain rule on $sin^{2} θ$ gives $2 sin θ cos θ$ , cancelling the $\frac{1}{2}$ ).
EL: $m R^{2} \ddot{θ} - (m R^{2} Ω^{2} sin θ cos θ - m g R sin θ) = 0$ , so $\ddot{θ} = Ω^{2} sin θ cos θ - \frac{g}{R} sin θ = sin θ (Ω^{2} cos θ - \frac{g}{R}) .$ Why this step? The engine $\frac{d}{d t} \frac{\partial L}{\partial θ ˙} - \frac{\partial L}{\partial θ} = 0$ delivers the equation of motion; factor $sin θ$ to expose the equilibria.

Verify: Equilibria where $\ddot{θ} = 0$ : either $sin θ = 0$ (bottom $θ = 0$ ) or $cos θ = \frac{g}{R Ω ^{2}}$ . That second solution only exists when $\frac{g}{R Ω ^{2}} \leq 1$ , i.e. when $Ω^{2} \geq \frac{g}{R}$ — a critical spin rate $Ω_{c} = g / R$ , exactly the forecast. ✓ Below $Ω_{c}$ the bracket $Ω^{2} cos θ - g / R < 0$ near $θ = 0$ , so the bottom is stable. ✓ Units: $Ω^{2}$ and $g / R$ are both $s^{- 2}$ , and $\ddot{θ}$ comes out in $rad/s^{2}$ . ✓

Active recall

Recall Which cell is each new problem in? (cover the answers)

A block sliding down a hill, coordinate down-slope ::: Cell B — negative-slope potential. A double pendulum ::: Cell D — coupled coordinates (two EL equations). A planet orbiting freely, angle coordinate ::: Cell E — cyclic coordinate, angular momentum conserved. A puck with no forces ::: Cell F — degenerate, $V = 0$ , constant velocity. A pendulum on a driven pivot ::: Cell H — constraint that adds to $T$ .

Connections

Euler–Lagrange Equation — the single engine used in every example above.
Generalized Coordinates and Constraints — why $s$ , $θ$ , $(r, ϕ)$ were free to choose.
Generalized Momentum and Conservation — Cell E's conserved $p_{ϕ}$ .
Principle of Least Action — why the free particle (Cell F) takes a straight line.
Hamiltonian H = T + V — where the energy $T + V$ comes back.
Newton's Second Law — the base case every example must reduce to.