Principle of least action — Hamilton's principle derivation

What is the action?

Hamilton's Principle (the statement)

Deriving the Euler–Lagrange equation from $\delta S = 0$

This is the heart of the chapter. We build everything from scratch.

Setup (WHAT). Take the true path $q(t)$ and perturb it: $q_\varepsilon(t) = q(t) + \varepsilon\,\eta(t)$ where $\eta(t)$ is an arbitrary smooth "wiggle" and $\varepsilon$ is a small number.

Step 1 — Write the action of the varied path. $S(\varepsilon) = \int_{t_1}^{t_2} L\big(q+\varepsilon\eta,\ \dot q+\varepsilon\dot\eta,\ t\big)\, dt$ Why this step? The true path corresponds to $\varepsilon=0$. Stationarity means $S(\varepsilon)$ has a flat tangent there: $\left.\dfrac{dS}{d\varepsilon}\right|_{\varepsilon=0}=0$.

Step 2 — Differentiate under the integral (chain rule). $\frac{dS}{d\varepsilon} = \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}\,\eta + \frac{\partial L}{\partial \dot q}\,\dot\eta\right) dt$ Why this step? $L$ depends on $\varepsilon$ only through $q+\varepsilon\eta$ and $\dot q+\varepsilon\dot\eta$. Chain rule gives factors $\partial q_\varepsilon/\partial\varepsilon=\eta$ and $\partial\dot q_\varepsilon/\partial\varepsilon=\dot\eta$.

Step 3 — Integrate the second term by parts. This is the crucial move: it converts $\dot\eta$ into $\eta$ so we can factor $\eta$ out. $\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q}\,\dot\eta\,dt = \underbrace{\left[\frac{\partial L}{\partial \dot q}\,\eta\right]_{t_1}^{t_2}}_{=\,0} - \int_{t_1}^{t_2}\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right)\eta\,dt$ Why this step? The boundary term vanishes because $\eta(t_1)=\eta(t_2)=0$. That's why endpoint-fixing mattered.

Step 4 — Collect terms. $\frac{dS}{d\varepsilon}\bigg|_{0} = \int_{t_1}^{t_2}\left[\frac{\partial L}{\partial q} - \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q}\right)\right]\eta(t)\, dt = 0$

Step 5 — Apply the Fundamental Lemma of Calculus of Variations. Since $\eta(t)$ is arbitrary, the bracket must vanish at every $t$ (if it were nonzero anywhere, pick an $\eta$ bumped there and the integral wouldn't be zero).

Checking it gives Newton (Feynman-style sanity test)

Take a 1D particle: $T=\tfrac12 m\dot x^2$, $V=V(x)$, so $L=\tfrac12 m\dot x^2 - V(x)$.

• $\dfrac{\partial L}{\partial \dot x} = m\dot x$ ⟹ $\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial \dot x}\right)=m\ddot x$
• $\dfrac{\partial L}{\partial x} = -\dfrac{dV}{dx} = F$

Euler–Lagrange: $m\ddot x - F = 0 \Rightarrow \boxed{m\ddot x = F}$. ✅ Newton's second law falls out. Why this matters: it proves $L=T-V$ was the right choice and that Hamilton's principle is equivalent to Newtonian mechanics.

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

• Lagrangian Mechanics — EL equation generalizes to many coordinates $q_i$.
• Euler–Lagrange Equation — the differential equation this principle produces.
• Calculus of Variations — the math of stationary functionals; same lemma.
• Noether's Theorem — symmetries of $L$ ⟹ conservation laws.
• Hamiltonian Mechanics — Legendre transform of $L$ gives $H=T+V$.
• Fermat's Principle — optics analogue (stationary optical path/time).
• Feynman Path Integral — quantum sum over paths weighted by $e^{iS/\hbar}$.
• Newton's Second Law — equivalent formulation recovered as a check.

Concept Map

Hinglish (regional understanding)