Euler–Lagrange Equation

Equivalent forms

L = T - V

\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x} \;\Rightarrow\; m\ddot{x} = -\frac{dV}{dx}

Coordinates are a choice, physics is not: the equation keeps its form in ANY generalized coordinates.

Symbol	Name	SI unit	Dimension	Range
$omega$	Angular frequencyoutput Oscillation frequency that the E–L equation predicts for L = ½mẋ² − ½kx²	rad/s	T^-1	0 – 50
$m$	Mass Mass in the kinetic term ½mẋ²	kg	M	0.1 – 5
$k$	Spring constant Stiffness in the potential term ½kx²	N/m	MT^-2	1 – 50
$A$	Amplitude Initial displacement of the oscillator	m	L	0.05 – 1

Unit systems

Symbol	SI	CGS	Imperial
$omega$	rad/s	rad/s	rad/s
$m$	kg	g	lb
$k$	N/m	dyn/cm	lbf/ft
$A$	m	cm	ft

Where it holds

Systems with holonomic constraints and forces derivable from a (possibly velocity-dependent) potential; extends to fields, relativity, and — with ħ — quantum theory.

Dimensional analysis

d/dt(\partial L/\partial q

̇

) \to [ML^{2}T^{-2}]/[LT^{-1}]/[T] = [MLT^{-2}]

(for q a length)

\partial L/\partial q \to [ML^{2}T^{-2}]/[L] = [MLT^{-2}]

Both terms are forces (newtons) when q is a length.

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Discovery

Leonhard Euler & Joseph-Louis Lagrange · 1755

19-year-old Lagrange wrote to Euler in 1755 with a purely analytic derivation of the variational equation. Euler, then Europe's greatest mathematician, was so impressed he delayed publishing his own work so the young Lagrange could get credit — and named the field 'calculus of variations' in his honor.

Try this

One equation that re-derives F = ma, the pendulum, orbits, and electric circuits — without ever drawing a force arrow.

Feed L = ½mẋ² − ½kx² into the Euler–Lagrange equation and watch mẍ = −kx (simple harmonic motion) fall out.

Research status: stable

Real-world applications

Robot arm dynamics (every joint angle is a generalized coordinate)
Spacecraft attitude control and multibody simulation
Molecular dynamics force fields
Standard Model of particle physics — written as one Lagrangian

Common misconceptions

It only works in Cartesian coordinates — its whole point is coordinate freedom (angles, arc lengths, anything)
L is always $T - V$ — velocity-dependent potentials (magnetic forces) need generalized potentials
It's just a fancy rewrite of $F = ma$ — it handles constraints without constraint forces, which Newton can't do cleanly

Experimental verification

Reproduces every tested consequence of Newtonian dynamics; constrained systems (beads on wires, rolling bodies, robotic linkages) are routinely engineered with Lagrangian equations and verified to instrument precision.

Derivation

Start from Hamilton's principle

\delta \int L dt = 0

.

For a variation

q \to q + \delta q

with fixed endpoints:

\delta S = \int (\partial L/\partial q \delta q + \partial L/\partial q

̇

\delta q

̇)dt.

Integrate the second term by parts; the boundary term vanishes.

\delta S = \int [\partial L/\partial q - d/dt(\partial L/\partial q

̇

)]\delta q dt = 0

for arbitrary

\delta q

forces the integrand to vanish.

Limiting cases

L = \tfrac{1}{2}m

ẋ

^{2}

(free particle)⟶ mẍ

= 0

No potential → momentum conserved → straight-line motion (Newton's first law).

Cartesian coordinates,

L = T - V

⟶ mẍ

= -\nabla V = F

Newton's second law is the special case.

\partial L/\partial q_i = 0

(cyclic coordinate)⟶

p_i = \partial L/\partial q

̇_i conservedSymmetry ⇒ conservation — the seed of Noether's theorem.

What if…

What if I change coordinates mid-problem?

Nothing breaks — the E–L equation has the same form in every coordinate system. That's its superpower.

What if the Lagrangian has no explicit time dependence?

Energy is conserved (time-translation symmetry, via Noether).

What if there are constraints, like a bead on a wire?

Pick coordinates along the wire; constraint forces vanish from the problem automatically.

1

Harmonic oscillator from L

Given ·

m:: 1
k:: 10

Find · Equation of motion

and \omega

Steps

$L = \tfrac{1}{2}m$ ẋ $^{2} - \tfrac{1}{2}kx^{2}$
$\partial L/\partial$ ẋ $= m$ ẋ $\to d/dt(m$ ẋ $) = m$ ẍ
$\partial L/\partial x = -kx$
E–L: mẍ + $kx = 0 \to x(t) = A \cos (\sqrt{k/m}\cdot t)$

Result · mẍ

= -kx

,

so \omega = \sqrt{k/m} = \sqrt{10} \approx 3.16\,\mathrm{rad/s}

2

Pendulum without force diagrams

Given ·

m:: 1
k:: 0

Find · Equation of motion for a pendulum of length l

Steps

Choose $\theta as$ the generalized coordinate
$L = \tfrac{1}{2}ml^{2}\theta$ ̇ $^{2} + mgl \cos \theta$
$d/dt(ml^{2}\theta$ ̇ $) = -mgl \sin \theta$
$\theta$ ̈ $= -(g/l) \sin \theta$ — tension never appeared

Result ·

\theta

̈

= -(g/l) \sin \theta

Principle of Stationary Action→Generalized (Conjugate) Momentum→The Hamiltonian (Legendre Transform)→Simple Harmonic Motion Period→Newton's Second Law→