Lesson 1.1 · Foundations

The Principle of Least Action

This is arguably the deepest principle in all of physics. Everything else—Newton's laws, Maxwell's equations, quantum mechanics, general relativity, the Standard Model—can be derived from it. Understanding this principle means understanding how nature "thinks."

Two Ways to See Mechanics

Newton gave us the first picture:

Newton's view: Forces cause acceleration. Given the forces at any instant, we can compute how the system evolves moment by moment.

This is local and differential. We step through time, computing $F = ma$ at each instant.

But there's another way to see the same physics:

The action principle: Nature chooses, among all possible paths from A to B, the one that makes a certain quantity—the action—stationary.

This is global and integral. Instead of asking "what happens next?", we ask "which entire path does nature choose?"

Many paths connect A to B. Nature chooses the one where the action is stationary.

Why should nature care about global properties?

This is strange if you think about it. How does a particle "know" to take the path that minimizes the action over its entire trajectory? It seems to require knowledge of the future. We'll return to this question—it connects to quantum mechanics in a profound way.

Defining the Action

First, the notation

We describe a particle's motion as a function $q(t)$ — the position at each moment in time. Think of it as a movie of where the particle is.

$q(t)$ = position at time $t$
$\dot{q}(t) = \frac{dq}{dt}$ = velocity (how fast position changes)
$q_1 = q(t_1)$ = starting position
$q_2 = q(t_2)$ = ending position

The particle starts at position $q_1$ at time $t_1$ and ends at position $q_2$ at time $t_2$. The function $q(t)$ describes the entire journey between these fixed endpoints.

What is the Lagrangian?

Before defining the action, we need the Lagrangian. This is a single number that captures the "state" of a system at any instant.

For a particle, the Lagrangian is defined as:

L = T - V = \text{kinetic energy} - \text{potential energy}

For a particle of mass $m$ moving in a potential $V(q)$:

L(q, \dot{q}) = \frac{1}{2}m\dot{q}^2 - V(q)

Why kinetic MINUS potential?

This is strange — total energy is $T + V$. Why the minus sign? Intuitively: the Lagrangian measures the "freedom" of the system. High kinetic energy means lots of motion (freedom). High potential energy means the system is constrained (a ball held high wants to fall). The Lagrangian balances these: $L = \text{freedom} - \text{constraint}$.

The deeper answer comes from relativity: the action for a free particle is proportional to proper time, and the $T - V$ form emerges naturally. For now, accept it as the definition that works.

The Action

Now we can define the action. It's the integral of the Lagrangian along the entire path:

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

The notation $S[q]$ emphasizes that the action is a functional — it takes an entire function $q(t)$ as input (the whole path) and returns a single number (how much "action" that path has).

The Principle

Among all possible paths from $q_1$ to $q_2$, nature chooses the one where the action $S$ is stationary (usually a minimum). This single principle generates all of classical mechanics.

Deriving the Euler-Lagrange Equation

We want to find the path $q(t)$ that makes the action stationary. "Stationary" means that small variations in the path produce no first-order change in $S$.

The true path q(t) and its variations. Endpoints stay fixed: η(t₁) = η(t₂) = 0

Step 1: Consider a varied path

Let $q(t)$ be the true path. Consider a nearby path:

$$ \tilde{q}(t) = q(t) + \epsilon \eta(t) $$

where $\eta(t)$ is any smooth function with $\eta(t_1) = \eta(t_2) = 0$ (the endpoints are fixed), and $\epsilon$ is small.

Step 2: Expand the action

The action of the varied path is:

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

Step 3: Taylor expand the Lagrangian $$ L(q + \epsilon\eta, \dot{q} + \epsilon\dot{\eta}) = L(q, \dot{q}) + \epsilon\left( \frac{\partial L}{\partial q}\eta + \frac{\partial L}{\partial \dot{q}}\dot{\eta} \right) + O(\epsilon^2) $$

Step 4: Demand stationarity

For the action to be stationary:

$$ \frac{d}{d\epsilon} S[q + \epsilon\eta] \bigg|_{\epsilon=0} = 0 $$

This gives:

$$ \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q}\eta + \frac{\partial L}{\partial \dot{q}}\dot{\eta} \right) dt = 0 $$

Step 5: Integration by parts

The second term becomes:

$$ \int_{t_1}^{t_2} \frac{\partial L}{\partial \dot{q}}\dot{\eta} \, dt = \left[ \frac{\partial L}{\partial \dot{q}} \eta \right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \eta \, dt $$

The boundary term vanishes because $\eta(t_1) = \eta(t_2) = 0$.

Step 6: The fundamental lemma

We now have:

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

Since this must hold for all valid $\eta(t)$, the integrand must be zero everywhere.

The result is the Euler-Lagrange equation:

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

A Quick Example

For a particle in a potential $V(q)$, we have $L = \frac{1}{2}m\dot{q}^2 - V(q)$.

Computing each term:

$\frac{\partial L}{\partial q} = -\frac{\partial V}{\partial q} = -V'(q)$
$\frac{\partial L}{\partial \dot{q}} = m\dot{q}$
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = m\ddot{q}$

The Euler-Lagrange equation gives:

-V'(q) - m\ddot{q} = 0 \quad \Rightarrow \quad m\ddot{q} = -V'(q) = F

Newton's second law emerges from the action principle.

What Does This Mean?

We've shown that Newton's laws and the action principle are mathematically equivalent for classical mechanics. But the action principle reveals something deeper.

Why This Matters

Generalization: The action principle works in any coordinate system, with constraints, and extends naturally to fields and relativity.
Symmetries: Conservation laws emerge directly from symmetries of the Lagrangian (Noether's theorem—next lesson).
Quantum mechanics: Feynman's path integral formulation shows that quantum particles explore all paths, weighted by $e^{iS/\hbar}$. Classical mechanics emerges when paths near the action-stationary path interfere constructively.

Connection: Why Quantum Mechanics Resolves the "How Does It Know?" Puzzle

The particle doesn't "know" the future. In quantum mechanics, it takes all paths simultaneously. The classical path emerges because nearby paths have similar phases and add up (constructive interference), while other paths cancel out. The action principle is a shadow of quantum mechanics.

Quantum mechanics: all paths contribute with phase e^(iS/ℏ). Near the classical path, phases align and add constructively.

Exercises

Simple harmonic oscillator: Write the Lagrangian for a mass on a spring ($V = \frac{1}{2}kx^2$). Derive the equation of motion using Euler-Lagrange.
Free particle: Show that a free particle ($V = 0$) moves in a straight line at constant velocity. This is the principle of inertia from the action principle.
Geodesics: The shortest path between two points minimizes $S = \int ds$. Show that this gives a straight line in flat space.

Key Takeaways

The action $S = \int L \, dt$ encodes the dynamics of a system
Nature follows paths where the action is stationary
The Euler-Lagrange equation implements this principle
This formulation reveals the deep structure of physics: symmetries, conservation, and the path to quantum mechanics

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