Visual walkthrough — Lagrangian L = T − V

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Step 0 — The one idea, before any symbols

Look at the figure. The red curve is the path nature actually takes. The thin black curves are imaginary competitors — routes the object could have taken but didn't. Our whole job is to find a rule that singles out the red one from the crowd.

We will link this to Generalized Coordinates and Constraints later; for now think "$q$ = position."

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Step 1 — Give each path ONE number: the action

WHAT. To compare paths we need a scoreboard: one number per curve, so we can ask "which curve scores best?" That number is the action $S$.

WHY this and not something simpler? A single instant's energy is not enough — a path is a whole history, so its score must gather information from every moment along the way. The mathematical tool that "adds up a quantity over a continuous stretch" is the integral $\int$. That is exactly why an integral appears and not just a sum of a few points: time is continuous, so we need the continuous version of addition.

PICTURE. At every instant compute the number $L$ (we will say what $L$ is in Step 2), plot $L$ against time, and shade the area underneath. That shaded area is the action.

Different paths give differently-shaped $L$-curves, hence different areas, hence different scores. Good — now we can compete them.

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Step 2 — What is $L$? Moving-energy minus stored-energy

WHAT. The height we integrate is $L=T-V$: kinetic energy minus potential energy, measured at that instant on that path.

WHY these two energies, why the minus? We do not justify the minus yet — that is the payoff of the whole derivation (Step 8). For now just hold the definition. Kinetic energy $T$ measures how fast the object moves; potential energy $V$ measures how much stored push it sits in (height in gravity, stretch in a spring).

PICTURE. Two stacked bars at one frozen instant: a $T$ bar and a $V$ bar. The value of $L$ at that instant is the top of $T$ minus the top of $V$.

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Step 3 — The competing paths: wiggle the true one

WHAT. Take the real path $q(t)$ and build a nearby fake path by adding a tiny wiggle $\delta q(t)$. The fake path is $q(t)+\delta q(t)$.

WHY. "Stationary" (our target condition) means: nudge the path a little and the score barely moves. Just like the bottom of a valley — step sideways and your height barely changes. To test that, we need a controlled small nudge, and $\delta q$ is exactly that nudge.

WHY fixed endpoints? We are comparing routes between the same start event and the same end event — the question is which way you go, not where you end up. So the wiggle must be zero at both ends: $\delta q(t_1)=0,\delta q(t_2)=0.$

PICTURE. The red true path, a black wiggled path hugging it, and the vertical gap $\delta q$ between them — pinned shut at both ends.

The valley analogy is the load-bearing intuition: at the very bottom of a valley the ground is flat — first-order change is zero in every direction. Stationary action = standing at the bottom of the action-valley over the space of all paths.

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Step 4 — How the score changes: vary $S$

WHAT. Compute the change $\delta S$ = (score of wiggled path) − (score of true path), keeping only first order in $\delta q$.

WHY. Because the whole game is $\delta S=0$. We must find a formula for $\delta S$ before we can set it to zero.

The wiggle changes both the position ($q\to q+\delta q$) and the velocity ($\dot{q}\to \dot{q}+\delta \dot{q}$). A tiny change in an ingredient changes $L$ by (rate $L$ responds to that ingredient) × (size of the change) — that "rate of response" is a partial derivative $\partial L/\partial q$. This is why partial derivatives enter: they measure how sensitively $L$ reacts to each of its knobs separately.

PICTURE. Two side-by-side sensitivity slopes: how $L$ climbs as you push $q$, and how $L$ climbs as you push $\dot{q}$.

The two terms are a nuisance: one holds $\delta q$, the other holds $\delta \dot{q}$. We cannot factor out a common $\delta q$ yet. Fixing that is Step 5.

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Step 5 — Integration by parts: make every term carry $\delta q$

WHAT. Rewrite the second term so it also multiplies plain $\delta q$ instead of $\delta \dot{q}$.

WHY this tool? $\delta \dot{q}$ is the derivative of $\delta q$. Integration by parts is precisely the tool that moves a derivative off one factor and onto the other inside an integral — it is the reverse of the product rule. That is exactly the surgery we need: it strips the time-derivative off $\delta q$ so we can factor $\delta q$ out of everything.

Using $\delta \dot{q}=\frac{d}{dt}\delta q$:

= \underbrace{\left[\frac{\partial L}{\partial \dot q}\,\delta q\right]_{t_1}^{t_2}}_{\text{boundary term}} - \int_{t_1}^{t_2} \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right)\delta q\,dt$$ **The boundary term dies.** It is evaluated only at the two endpoints — and there $\delta q = 0$ (Step 3, the pinned ends). This is *why fixed endpoints mattered*: they are the reason the boundary term vanishes and only the clean integral survives. **PICTURE.** The boundary term shown as two dots at $t_1$ and $t_2$, both crushed to zero by the pinned wiggle; the interior integral survives. ![[deepdives/dd-physics-2.1.04-d2-s06.png]] > [!formula] After parts, with the boundary killed > $$\int_{t_1}^{t_2}\frac{\partial L}{\partial\dot q}\,\delta\dot q\,dt = -\int_{t_1}^{t_2}\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot q}\right)\delta q\,dt$$ > - $\left[\;\cdot\;\right]_{t_1}^{t_2}$ ::: "evaluate at the top end minus the bottom end" — both give $0$ here > - $\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot q}\right)$ ::: the derivative got shoved onto the velocity-sensitivity instead of onto $\delta q$ > - the minus sign ::: the cost of moving the derivative (product-rule bookkeeping) --- ## Step 6 — Factor $\delta q$ and invoke the "any wiggle" lemma **WHAT.** Substitute Step 5 back into Step 4. Now *both* terms multiply $\delta q$, so pull it out: $$\delta S = \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q}\right)\delta q\,dt.$$ **WHY the "arbitrary wiggle" argument?** We demanded $\delta S = 0$ for the *true* path. But $\delta q$ can be **any** tiny pinned bump — a bump near the start, a bump near the middle, anywhere. If the bracket were nonzero somewhere, we could choose a bump concentrated *right there* with the matching sign, making the integral nonzero — contradiction. The only way the area is zero for *every* possible bump is if the bracket itself is zero everywhere. (This is the **fundamental lemma of the calculus of variations**.) **PICTURE.** A spotlight bump placed at one spot; if the bracket (shown as a shaded background sign) is nonzero there, the shaded area is nonzero — so the bracket must be flat zero all along. ![[deepdives/dd-physics-2.1.04-d2-s07.png]] > [!formula] The Euler–Lagrange equation > $$\boxed{\;\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0\;}$$ > - $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot q}$ ::: rate of change of the "velocity-response" — this will become mass × acceleration > - $\dfrac{\partial L}{\partial q}$ ::: the "position-response" — this will become the force > - $=0$ ::: the two must balance at every instant on the true path This is the master result; see [[Euler–Lagrange Equation]] and [[Principle of Least Action]] for its own dedicated treatments. --- ## Step 7 — Feed in $L = T - V$ and watch Newton appear **WHAT.** Take the simplest case: one particle in one dimension, position $x$, kinetic energy $T=\tfrac12 m\dot x^2$, potential $V=V(x)$. So $L=\tfrac12 m\dot x^2 - V(x)$. Compute the two pieces of Euler–Lagrange. **WHY this case first?** It is the cleanest place to check our machine against something we already trust — [[Newton's Second Law]]. If the abstract result reproduces $F=ma$ here, the whole construction is validated. Compute each piece (remember: $x$ and $\dot x$ are independent knobs, Step 2): $$\frac{\partial L}{\partial \dot x} = m\dot x,\qquad \frac{d}{dt}\big(m\dot x\big)= m\ddot x,\qquad \frac{\partial L}{\partial x} = -\frac{dV}{dx} = F.$$ Plug into $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot x} - \dfrac{\partial L}{\partial x} = 0$: $$m\ddot x - F = 0 \quad\Longrightarrow\quad \boxed{F = m\ddot x}.$$ **PICTURE.** The abstract EL boxes on the left morph into the concrete Newton boxes on the right, arrow by arrow. ![[deepdives/dd-physics-2.1.04-d2-s08.png]] > [!formula] Term-by-term dictionary > $$\underbrace{\frac{d}{dt}(m\dot x)}_{\text{= } m\ddot x \text{ (mass}\times\text{accel)}} - \underbrace{\Big(-\frac{dV}{dx}\Big)}_{\text{= force } F} = 0$$ > - $m\dot x$ ::: this is exactly the **generalized momentum** $p=\partial L/\partial\dot x$ (see [[Generalized Momentum and Conservation]]) > - $m\ddot x$ ::: mass times acceleration > - $-dV/dx$ ::: force = minus the slope of the potential hill (things roll downhill) > [!mistake] Why $T-V$ and not $T+V$? (now we can see it) > Redo Step 7 with $L=T+V$. Then $\partial L/\partial x = +dV/dx = -F$, and Euler–Lagrange gives $m\ddot x + F = 0$ — the **wrong sign**, anti-physics. Only the **minus** in $T-V$ lets the force term land with the sign Newton demands. That single minus is the entire reason for the definition. (Contrast with the total energy $E=T+V$ and the [[Hamiltonian H = T + V|Hamiltonian]] — different object, different job.) --- ## Step 8 — Every case: signs, degenerate inputs, limits We must never leave you facing a scenario we didn't show. **Case $V=0$ (free particle).** Then $\partial L/\partial x = 0$, so $\dfrac{d}{dt}(m\dot x)=0$: momentum is constant, motion is a straight line at constant speed. On the action picture the $V$ bar vanishes and $L=T$ alone. **Case $F=0$ region but $V\neq 0$ constant.** A flat potential ($V=$const) has zero slope, so $-dV/dx=0$: same straight-line motion. *Adding a constant to $V$ changes nothing* — only the **slope** of $V$ matters, which is why potential energy has an arbitrary zero. **Sign of the force.** If the potential slopes *up* to the right ($dV/dx>0$), then $F=-dV/dx<0$ pushes *left* — downhill. If it slopes *down* to the right, force pushes right. Either way: **toward lower potential.** Both signs shown in the summary figure. **Degenerate wiggle $\delta q \equiv 0$.** Then $\delta S=0$ trivially for *any* path — it tells us nothing. That is why the lemma insists $\delta q$ range over *all* nonzero bumps; the "arbitrary" is essential, not decorative. **Limit of many coordinates.** Nothing in Steps 3–6 used "one dimension." Repeat with an independent wiggle $\delta q_i$ for each coordinate and you get one Euler–Lagrange equation *per* coordinate — they don't interfere because each bump is chosen independently. > [!recall]- Self-check the cases > If $V$ is constant, what is the motion? ::: straight line, constant velocity — constant $V$ has zero slope so zero force. > Which way does force point when $V$ slopes upward to the right? ::: to the left, i.e. toward lower potential ($F=-dV/dx<0$). > Why must $\delta q$ be arbitrary and not one fixed shape? ::: a single fixed wiggle gives one equation; only *all* wiggles force the whole bracket to vanish everywhere. > How many EL equations for $n$ coordinates? ::: exactly $n$ — one independent wiggle each. --- ## The one-picture summary ![[deepdives/dd-physics-2.1.04-d2-s09.png]] Read the flow top to bottom: **many paths** → each gets a **score** (area under $T-V$) → demand the **wiggle changes nothing** → **integration by parts + pinned ends** → **Euler–Lagrange** → plug in $T-V$ → **$F=ma$**. ```mermaid flowchart TD A["Many possible paths q of t"] --> B["Score each path S = integral of L dt"] B --> C["L = T minus V at each instant"] C --> D["Wiggle true path by delta q, ends pinned to zero"] D --> E["First order change delta S in two terms"] E --> F["Integration by parts, boundary term dies"] F --> G["Factor delta q, arbitrary wiggle lemma"] G --> H["Euler Lagrange equation"] H --> I["Insert T minus V, one particle"] I --> J["Newton F = m a recovered"] ``` > [!recall]- Feynman retelling — the whole walkthrough in plain words > Draw a graph of *where the object is* against *time*. The real trip is one curve; imagine a swarm of nearly-identical fake curves crowding around it, all starting and ending at the same two dots. > > Give every curve a **score**: at each moment take "moving-energy minus stored-energy," and add that up across the whole trip — that total is the *action*. Nature's chosen curve is the one sitting at the *bottom of a valley* of scores: nudge it a hair in any direction and the score refuses to change. > > To cash that in, wiggle the real curve by a tiny amount that's glued to zero at the two ends. Work out how the score shifts. One piece of the shift is awkward because it involves the *slope* of the wiggle, not the wiggle itself — so we use a clean trick (integration by parts) to move that slope off, and the glued-down ends make the leftover boundary junk disappear. > > Now the whole shift is (some bracket) times (the wiggle), added over the trip. Since we can wiggle *anywhere* we like and the shift must still be zero, the bracket has to be zero *everywhere*. That bracket-equals-zero **is** the Euler–Lagrange equation. > > Finally, pour in real energies — $\tfrac12 m v^2$ for moving, $V$ for stored — and the equation spits out "mass times acceleration equals minus the slope of the potential," which is just **$F=ma$**. The *minus* in "moving minus stored" is what makes the force come out pushing the right way. That's the whole show. --- ## Connections - [[Euler–Lagrange Equation]] — the boxed result of Step 6, on its own terms. - [[Principle of Least Action]] — the $\delta S = 0$ demand of Steps 3–4. - [[Generalized Coordinates and Constraints]] — what $q$ may be; why constraint forces vanish. - [[Generalized Momentum and Conservation]] — the $m\dot x = \partial L/\partial\dot x$ of Step 7. - [[Hamiltonian H = T + V]] — the $T+V$ object that is *not* the Lagrangian. - [[Newton's Second Law]] — recovered in Step 7 as the sanity check.