Generalized (Conjugate) Momentum

Equivalent forms

p_\theta = m r^2 \dot{\theta}

\mathbf{p} = m\mathbf{v} + q\mathbf{A} \;\text{(charged particle)}

Linear momentum, angular momentum, and field momentum are all the same formula wearing different coordinates.

Symbol	Name	SI unit	Dimension	Range
$p_theta$	Angular conjugate momentumoutput Momentum conjugate to the rotation angle θ	kg·m²/s	ML^2T^-1	0 – 100
$m$	Mass Mass on the rotating arm	kg	M	0.1 – 5
$r$	Radius Distance from the rotation axis	m	L	0.1 – 2
$omega$	Angular velocity Rate of rotation θ̇	rad/s	T^-1	0.1 – 10

Unit systems

Symbol	SI	CGS	Imperial
$p_theta$	kg·m²/s	g·cm²/s	lb·ft²/s
$m$	kg	g	lb
$r$	m	cm	ft
$omega$	rad/s	rad/s	rad/s

Where it holds

Any Lagrangian system; for velocity-dependent potentials the canonical momentum differs from

m\cdot v

and is the one that is conserved and quantized.

Dimensional analysis

p_\theta \to [ML^{2}T^{-1}]

mr^{2}\omega \to [M][L^{2}][T^{-1}] = [ML^{2}T^{-1}]

Angular-momentum units, since

\theta is

dimensionless.

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Discovery

Joseph-Louis Lagrange · 1788

In Mécanique analytique (1788) — a mechanics book famously containing not a single diagram — Lagrange showed that ∂L/∂q̇ plays the role of momentum for any coordinate q, unifying linear momentum, angular momentum, and stranger quantities under one definition.

Try this

A spinning skater pulls her arms in and speeds up — she's conserving a momentum that isn't m·v.

For a mass on a rotating arm, compute the momentum conjugate to the angle: p_θ = mr²ω, and see why shrinking r must spin ω up.

Research status: stable

Real-world applications

Figure skating and diving rotation control
Satellite reaction wheels (exchange of conjugate angular momenta)
Quantization rules: [q, $p] = i$ ħ uses canonical, not kinetic, momentum
Plasma physics: drift motion conserves canonical angular momentum

Common misconceptions

Momentum is always mass $\times$ velocity — only in Cartesian coordinates without magnetic fields
$p_\theta has$ units of $kg\cdot m/s$ — it has units of angular momentum because $\theta is$ dimensionless
Conjugate momentum is a mathematical trick — it is the quantity quantum mechanics promotes $to -i$ ħ $\partial /\partial q$

Experimental verification

Skater spin-up, helium-atom spectroscopy (quantized

p_\theta = n

ħ), and the Aharonov–Bohm effect — which directly confirms that canonical momentum involves the vector potential A.

Derivation

Definition, motivated by the E–L equation: writing

d/dt(\partial L/\partial q

̇

) = \partial L/\partial q

in the form ṗ

= \partial L/\partial q

makes

\partial L/\partial q

̇ the natural 'momentum' whose rate of change is the generalized force.

For

L = \tfrac{1}{2}mr^{2}\theta

̇

^{2} - V(r)

:

p_\theta = \partial L/\partial \theta

̇

= mr^{2}\theta

̇.

Limiting cases

q = x

(Cartesian),

L = \tfrac{1}{2}m

ẋ

^{2} - V(x)

⟶

p = m

ẋOrdinary linear momentum is the simplest case.

q = \theta

(rotation)⟶

p_\theta = mr^{2}\theta

̇

= L_z

Conjugate momentum of an angle is angular momentum.

Charged particle in a magnetic field⟶

p = mv + qA

Canonical momentum ≠ kinetic momentum — the field carries part of it.

What if…

What if the coordinate doesn't appear in L at all?

Its conjugate momentum is exactly conserved — a free conservation law.

What if a magnetic field is on?

Canonical $p = mv + qA$ ; the conserved/quantized momentum includes the field's contribution.

What if I use a weird coordinate like

q = x^{3}

?

Fine — $p_q = \partial L/\partial q$ ̇ just looks unusual; the formalism doesn't care.

1

Mass on a rotating arm

Given ·

m:: 1
r:: 1
omega:: 2

Find · p_theta

Steps

$L = \tfrac{1}{2}mr^{2}\theta$ ̇ $^{2} for$ fixed r
$p_\theta = \partial L/\partial \theta$ ̇ $= mr^{2}\theta$ ̇
$p_\theta = 1 \times 1 \times 2 = 2 kg\cdot m^{2}/s$

Result ·

p_\theta = mr^{2}\omega = 1 \times 1^{2} \times 2 = 2 kg\cdot m^{2}/s

2

Skater pulls arms in

Given ·

m:: 1
r:: 0.5
omega:: 2

Find ·

New \omega

when r halves

(p_\theta

conserved)

Steps

$\theta is$ cyclic (no torque $) \to p_\theta$ conserved
$mr^{2}\omega = mr'^{2}\omega$ '
$\omega ' = \omega (r/r')^{2} = 2 \times (1/0.5)^{2} = 8\,\mathrm{rad/s}$ — $4\times$ faster

Result ·

p_\theta = mr^{2}\omega

fixed

\Rightarrow \omega ' = \omega (r/r')^{2} = 2 \times 4 = 8\,\mathrm{rad/s}

Euler–Lagrange Equation→Noether's Theorem→Angular Momentum→Linear Momentum→The Hamiltonian (Legendre Transform)→