Hamilton's principle — least action — Maths notes

WHAT is the action?

WHY $T-V$ and not $T+V$ ? Energy conservation ( $T+V$ ) is constant along the true path — it carries no information distinguishing paths. The difference $T-V$ is what gets "balanced" over time: the system trades kinetic for potential and the integral of the trade-off is extremized.

HOW to derive the equation of motion (from scratch)

We want: which $q(t)$ kills $\delta S$ ? Perturb the true path: $q(t) \to q(t) + \varepsilon\,\eta(t), \qquad \eta(t_1)=\eta(t_2)=0,$ where $\eta$ is an arbitrary smooth "wiggle" that vanishes at the fixed endpoints (WHY zero there? endpoints are held fixed — we don't get to vary them).

The action becomes a function of $\varepsilon$ : $S(\varepsilon)=\int_{t_1}^{t_2} L(q+\varepsilon\eta,\ \dot q+\varepsilon\dot\eta,\ t)\,dt.$

Stationary means $\left.\dfrac{dS}{d\varepsilon}\right|_{\varepsilon=0}=0$ . Differentiate under the integral (chain rule):

$\frac{dS}{d\varepsilon}\Big|_{0}=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\,\eta + \frac{\partial L}{\partial \dot q}\,\dot\eta\right)dt.$

Why this step? Each path-component picks up a derivative of $L$ ; $\eta$ multiplies the $q$ -slot, $\dot\eta$ the $\dot q$ -slot.

Now integrate the second term by parts to free $\dot\eta$ into $\eta$ : $\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q}\,\dot\eta\,dt = \underbrace{\left[\frac{\partial L}{\partial \dot q}\,\eta\right]_{t_1}^{t_2}}_{=\,0\ (\eta\text{ vanishes at ends})} - \int_{t_1}^{t_2}\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right)\eta\,dt.$

Why this step? The boundary term dies because $\eta(t_1)=\eta(t_2)=0$ — this is exactly why fixed endpoints matter. Substituting back:

$\frac{dS}{d\varepsilon}\Big|_{0}=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot q}\right)\eta(t)\,dt = 0.$

This must hold for every wiggle $\eta$ . By the Fundamental Lemma of Calculus of Variations (if $\int f\eta\,dt=0$ for all $\eta$ , then $f\equiv 0$ ), the bracket must vanish:

WHY this recovers Newton: for $L=\tfrac12 m\dot x^2 - V(x)$ : $\dfrac{\partial L}{\partial \dot x}=m\dot x$ , so $\dfrac{d}{dt}(m\dot x)=m\ddot x$ ; and $\dfrac{\partial L}{\partial x}=-V'(x)=F$ . Euler–Lagrange gives $m\ddot x - F = 0 \Rightarrow F=m\ddot x$ . Newton's law falls out for free.

Figure — Hamilton's principle — least action

Worked examples

Worked example 3. Pendulum (uses an angle, no force-decomposition needed)

Coordinate $\theta$ . Height $= -\ell\cos\theta$ , speed $=\ell\dot\theta$ . $L=\tfrac12 m\ell^2\dot\theta^2 + mg\ell\cos\theta$ . $\dfrac{\partial L}{\partial\dot\theta}=m\ell^2\dot\theta$ , $\dfrac{\partial L}{\partial\theta}=-mg\ell\sin\theta$ . E–L: $m\ell^2\ddot\theta + mg\ell\sin\theta=0 \Rightarrow \ddot\theta = -\dfrac{g}{\ell}\sin\theta.$ Why this step? No need to resolve tension or components — choosing $\theta$ and writing $T,V$ does all the geometry. This is the practical power of the principle.

Steel-manned mistakes

Recall Feynman: explain to a 12-year-old

Imagine you must walk from your house to a friend's house and you'll be graded by a weird "tiredness score" added up over the whole trip. Nature plays the same game with every moving thing: it adds up a score (the action) along each possible route. The route that the ball, planet, or pendulum actually takes is the one where you couldn't make the score any better by nudging the route a tiny bit — it's perfectly balanced. So instead of asking "what force pushes it now?", you ask "which whole path is the smartest overall?" Both questions give the same answer, but the path-question is often much easier.

Active recall

What is the action

S

of a path?

S[q]=\int_{t_1}^{t_2} L(q,\dot q,t)\,dt

, the time-integral of the Lagrangian over the path.

State Hamilton's principle.

The true path between fixed endpoints makes the action stationary:

\delta S=0

What is the Lagrangian for a classical system?

L=T-V

(kinetic minus potential energy).

Write the Euler–Lagrange equation.

\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot q}-\dfrac{\partial L}{\partial q}=0

Why must the perturbation

\eta(t)

vanish at the endpoints?

Because endpoints are held fixed; this kills the boundary term from integration by parts.

Which lemma converts

\int(\cdots)\eta\,dt=0

into the E–L equation?

The Fundamental Lemma of the Calculus of Variations: if

\int f\eta\,dt=0

for all

\eta

, then

f\equiv0

Show E–L recovers Newton for

L=\tfrac12 m\dot x^2-V

\frac{d}{dt}(m\dot x)+V'(x)=0\Rightarrow m\ddot x=-V'(x)=F

Is "least action" always a minimum?

No — it's stationary; it can be a saddle (e.g. past a conjugate point). Better called "stationary action."

What is the Lagrangian of a simple pendulum (angle

\theta

, length

\ell

L=\tfrac12 m\ell^2\dot\theta^2+mg\ell\cos\theta

, giving

\ddot\theta=-\frac{g}{\ell}\sin\theta

What conserved quantity is built from

L

and equals

T+V

The Hamiltonian

H=\sum_i \dot q_i\,p_i-L

, with

p_i=\partial L/\partial\dot q_i

Connections

Euler–Lagrange equation — the differential consequence of $\delta S=0$ .
Calculus of Variations — the general machinery (functionals, first variation, the fundamental lemma).
Lagrangian Mechanics — generalized coordinates make geometry automatic.
Noether's Theorem — symmetries of $L$ ⇒ conserved quantities.
Hamiltonian Mechanics — Legendre transform of $L$ giving $H=T+V$ .
Fermat's Principle — optics' analogue: light takes the path of stationary time.
Newton's Laws — recovered as a special case.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea bahut elegant hai. Newton bolta hai "har pal force lagao, $F=ma$ solve karo." Hamilton ek doosra angle deta hai: kisi bhi object ke saare possible raaste socho jo start point se end point tak jaate hain. Har raaste ke liye ek number nikaalo jise action $S=\int (T-V)\,dt$ kehte hain — yaani kinetic energy minus potential energy ka time ke upar integral. Asli raasta wahi hota hai jahan ye action stationary ho jaye, matlab thoda sa path hilao to $S$ change na ho (first order me). Isko $\delta S=0$ likhte hain.

Ab WHY $T-V$ ? Kyunki $T+V$ (total energy) to constant rehti hai, woh paths ko distinguish nahi karti. $T-V$ ka time-average balance hota hai, aur usi balance se asli trajectory decide hoti hai. Jab hum is condition ko maths se solve karte hain (path ko thoda perturb karo $q+\varepsilon\eta$ , integrate by parts, endpoints pe wiggle zero hone se boundary term mar jaata hai, phir fundamental lemma), to Euler–Lagrange equation nikalti hai: $\frac{d}{dt}\frac{\partial L}{\partial\dot q}=\frac{\partial L}{\partial q}$ . Aur agar $L=\tfrac12 m\dot x^2-V$ daalo to seedha $m\ddot x=-V'(x)$ , yaani Newton ka law! Toh dono same physics, bas dekhne ka tareeka alag.

Practical fayda: pendulum ya complicated systems me forces aur tensions resolve karna painful hota hai. Lekin agar aap sirf $T$ aur $V$ angle $\theta$ me likh do, geometry apne aap handle ho jaati hai — koi free-body diagram nahi. Ek important warning: "least action" naam se lagta hai hamesha minimum, par actually stationary hota hai, kabhi-kabhi saddle bhi. Isliye exam me "principle of stationary action" bolna safe hai. Yaad rakho mantra: Wiggle karo, integrate by parts, ends phenk do, lemma se bracket zero.

Hamilton's principle — least action

WHAT is the action?

HOW to derive the equation of motion (from scratch)

Worked examples

Steel-manned mistakes

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Advanced Topics (Elite Level)

Connections