Generalized Momenta

Ask a speedometer for a particle's momentum and it will tell you $m\mathbf v$ . Ask the Lagrangian and you may get a different answer entirely. For a pendulum angle the "momentum" comes out as angular momentum, in units of $\mathrm{kg\,m^2/s}$ . For a charge moving through a magnetic field it comes out as $m\dot{\mathbf r}+q\mathbf A$ — carrying a piece of the field along with the motion. And when the dust settles in quantum mechanics, it is this strange, coordinate-bred momentum, not $m\mathbf v$ , that becomes the operator $-i\hbar\nabla$ . All of these are instances of one definition: the generalized momentum conjugate to a coordinate $q_i$ is $p_i=\partial L/\partial\dot q_i$ — the slope of the Lagrangian in the velocity direction. The definition looks innocuous, but it is doing two jobs at once. Dynamically, it is the quantity whose time derivative the Euler–Lagrange equation controls, which is why a coordinate that never appears in $L$ hands you a conservation law on the spot. Structurally, it is the new variable that replaces $\dot q_i$ when mechanics moves from configuration space to phase space: promote each $p_i$ to an independent coordinate and the $n$ second-order equations of Lagrange become the $2n$ first-order equations of Hamilton. You already own the machinery this article leans on — the Euler–Lagrange equations and the Legendre transformation that formalizes the velocity-for-slope trade. What it adds is fluency: how to read a conjugate momentum for any coordinate, why it differs from the kinetic $m\mathbf v$ exactly when it does, and how the momentum zoo — linear, angular, canonical-electromagnetic, relativistic — is one animal seen from four angles.

Key takeaways

The generalized momentum conjugate to a coordinate $q_i$ is defined as $p_i = \partial L/\partial \dot q_i$ , the partial derivative of the Lagrangian with respect to the corresponding generalized velocity.
Conjugate momenta need not equal $m\dot q$ : for a rotational coordinate $p$ is angular momentum, and in a magnetic field $p$ contains a vector-potential term, so it is a 'canonical' rather than 'kinetic' momentum.
If a coordinate is cyclic (absent from $L$ ), its conjugate momentum is conserved — this is the Lagrangian face of Noether's theorem for translations and rotations.
The momenta are the new independent variables that replace velocities when passing from configuration space $(q,\dot q)$ to phase space $(q,p)$ via a Legendre transformation.
Invertibility of the map $\dot q \mapsto p$ requires the Hessian $\partial^2 L/\partial \dot q_i\partial \dot q_j$ to be nonsingular; this is precisely what makes the Hamiltonian formulation possible.
Key idea: Define $p_i=\partial L/\partial\dot q_i$ and promote it to an independent variable: the velocity $\dot q_i$ is eliminated in favour of $p_i$ , turning configuration space into phase space and second-order dynamics into first-order canonical equations.

Where this lives today

Canonical quantization is the definition's afterlife: quantum mechanics promotes precisely the conjugate pair $(q_i,p_i)$ to operators with $[\hat q_i,\hat p_j]=i\hbar\,\delta_{ij}$ , which is why a magnetic Schrödinger equation contains $(-i\hbar\nabla-q\mathbf A)^2$ — the kinetic momentum reconstructed from the canonical operator. The same minimal-coupling shift $p\to p-qA$ , promoted to field theory, is the interaction structure of QED and of the entire Standard Model. In condensed matter the distinction earns its keep daily: crystal momentum, Berry-phase corrections to it, and the magnetic translation algebra of the quantum Hall effect are all statements about canonical rather than kinetic momentum. And accelerator physicists steer beams through storage rings in canonical coordinates — tracking codes and beam-optics Hamiltonians are written in $(q,p)$ precisely so that symplectic integrators conserve the phase-space structure over billions of turns.

The physical picture

Think of the Lagrangian $L(q,\dot q)$ as a landscape over velocity. Its slope in the $\dot q_i$ direction is the momentum $p_i$ . Reading off a slope at each velocity gives a map $\dot q\to p$ ; because kinetic energy curves upward everywhere, the landscape is convex and the map can be inverted — the slope itself becomes a valid label for states. That relabelling, formalized as the Legendre transformation of the previous article, is the entire move into phase space. But the slope depends on what coordinate you differentiate along, and that is where the zoo comes from. Differentiate along a Cartesian direction and the slope is $m\dot x$ . Differentiate along an angle — a coordinate measured in radians, not metres — and the slope must supply the missing length dimensions itself: out comes $mr^2\dot\theta$ , an angular momentum. Add a term to $L$ that is linear in velocity, as a magnetic field does, and the slope acquires a velocity-independent offset $q\mathbf A$ that no speedometer will ever register:

Figure

The momentum zoo — one definition, read along three different coordinates

The word conjugate signals a pairing: $q_i$ and $p_i$ are partners that together carry one degree of freedom. Whatever units $q_i$ has, $p_i$ compensates so that the product $p_i\dot q_i$ always lands in energy and $p_i\,\mathrm{d}q_i$ in action — the same $p\,\mathrm{d}q$ that fills action integrals and adiabatic invariants later in the course. Conservation laws hide in the pairing too: if $L$ does not depend on $q_i$ , the Euler–Lagrange equation forces $\dot p_i=0$ immediately. Translational symmetry conserves linear momentum, rotational symmetry conserves the angle's conjugate angular momentum — the cheapest conservation theorem in physics, and the Lagrangian face of Noether's theorem.

The definition and its immediate consequences

Everything descends from one line. For a system with Lagrangian $L(q,\dot q,t)$ , each generalized velocity is assigned its slope:

p_i \equiv \frac{\partial L}{\partial \dot q_i} \qquad (1)

where: $p_i$ — the momentum conjugate to $q_i$ ; the partial derivative holds all other velocities and all coordinates fixed. Check it on the free particle, $L=\tfrac12 m\dot x^2$ : the slope is $m\dot x$ , and the familiar momentum turns out to have been a derived quantity all along. The definition is coordinate-dependent by design — choose different generalized coordinates and you get different (but equivalent) momenta, which is a feature: the momentum automatically speaks the same geometric language as its coordinate. The reason this particular derivative deserves a name is that the Euler–Lagrange equation is built around it. Substituting (1) into $\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot q_i}=\frac{\partial L}{\partial q_i}$ collapses the equation of motion to a statement Newton would recognize:

\frac{\mathrm{d}p_i}{\mathrm{d}t}=\frac{\partial L}{\partial q_i} \qquad (2)

where: $\partial L/\partial q_i$ — the generalized force conjugate to $q_i$ (for $L=T-V$ with $T$ coordinate-independent it is $-\partial V/\partial q_i$ ). Rate of change of momentum equals force, coordinate by generalized coordinate: Newton's second law, wearing whatever geometry the problem chose.

Derivation AMomentum form of the equations of motion and the cyclic shortcut

We are recasting the Euler–Lagrange equations as momentum-balance statements and extracting the conservation law that follows for free. The strategy is pure substitution — definition (1) into the Euler–Lagrange equation — but the payoff justifies the ceremony: the rewritten equation makes one entire class of conservation laws visible by inspection of the Lagrangian, before any dynamics is solved. Watch the logical direction: nothing here assumes the momentum is $m\dot q$ ; every statement holds for the full zoo, electromagnetic offsets and all.

Step 1. Start from the Euler–Lagrange equation for coordinate $q_i$ (one per degree of freedom):

\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot q_i}-\frac{\partial L}{\partial q_i}=0

Step 2. Recognize the first term as the time derivative of the conjugate momentum (1), giving equation (2): $\dot p_i=\partial L/\partial q_i$ . Why this is not circular: (1) is a definition, the Euler–Lagrange equation is physics (stationary action); the substitution merely names the object the physics was already differentiating.

Step 3. Suppose $q_i$ is absent from the Lagrangian — a cyclic (ignorable) coordinate, $\partial L/\partial q_i=0$ . Equation (2) then reads $\dot p_i=0$ :

\frac{\partial L}{\partial q_i}=0 \quad\Longrightarrow\quad p_i=\text{const}

Why the velocity may still appear: cyclic means the coordinate itself is missing; $\dot q_i$ can and usually does appear in $L$ . The conserved object is the slope $p_i$ , not the velocity — for the planar orbit below, $\dot\theta$ changes constantly while $mr^2\dot\theta$ does not.

Step 4. Read the conservation physically. A missing coordinate means the Lagrangian — the dynamics — is indifferent to translations along that coordinate. Indifference along $x$ conserves linear $p_x$ ; indifference under rotation conserves the angular $p_\theta$ . Which symmetries produce which momenta, and how the statement generalizes beyond coordinate shifts, is the systematic subject of Cyclic Coordinates and Conservation of Conjugate Momenta and, in full generality, Noether's Theorem — here we only need the shortcut itself. Quick check: for $L=\tfrac12 m\dot x^2-V(x)$ equation (2) gives $\dot p=-V'(x)$ — Newton's law in one line — and a constant potential $V=\text{const}$ immediately makes $p=m\dot x$ conserved, as expected.

The momentum–velocity map

For the Lagrangians mechanics actually meets — kinetic energy quadratic in velocities, possibly a term linear in velocities, minus a potential — the definition unpacks into an explicit, affine relation between momenta and velocities. Differentiating $L=\tfrac12\sum_{jk}M_{jk}(q)\,\dot q_j\dot q_k+\sum_j a_j(q,t)\,\dot q_j-V(q)$ termwise:

p_i=\sum_j M_{ij}(q)\,\dot q_j + a_i(q,t) \qquad (3)

where: $M_{ij}$ — the mass (metric) matrix of the kinetic term; $a_i$ — the offset contributed by velocity-linear terms, of which the electromagnetic $q\mathbf A$ is the canonical example. Two structural facts to read off. The momenta are linear in the velocities with a shift — so inverting for $\dot q(q,p)$ is matrix algebra, $\dot q=M^{-1}(p-a)$ . And the existence of that inverse is governed by the same matrix:

\det\!\left(\frac{\partial^2 L}{\partial \dot q_i\,\partial \dot q_j}\right)=\det M\neq 0 \qquad (4)

where: the left side is the velocity Hessian of $L$ — precisely the convexity condition of the Legendre Transformation article, now revealed as the invertibility of the momentum–velocity dictionary. Positive-definite kinetic energy guarantees it; degenerate Lagrangians (gauge theories) violate it and pay in constraints.

coordinate / system	conjugate momentum	is it $m\dot q$ ?
Cartesian $x$ , free or in $V(x)$	$p=m\dot x$	yes
plane polar angle $\theta$	$p_\theta=mr^2\dot\theta$	no — angular momentum
plane polar radius $r$	$p_r=m\dot r$	yes (radial component)
charge in field $(\mathbf A,\phi)$	$\mathbf p=m\dot{\mathbf r}+q\mathbf A$	no — canonical ≠ kinetic
relativistic free particle	$\mathbf p=\gamma m\mathbf v$	no — $\gamma$ correction
any cyclic coordinate	$p_i=\text{const}$	conserved whatever its form

The electromagnetic momentum — canonical versus kinetic

The magnetic force $q\mathbf v\times\mathbf B$ does no work and cannot be written as $-\nabla V$ ; it enters the Lagrangian through the velocity-dependent potential $U=q\phi-q\,\dot{\mathbf r}\cdot\mathbf A$ (built carefully in Velocity-Dependent Potentials). This is the one place in classical mechanics where the offset $a_i$ in (3) is not optional bookkeeping but the whole story.

Go deeper · Derivation B · 1 minConjugate momentum of a charged particle in an electromagnetic field

Derivation BConjugate momentum of a charged particle in an electromagnetic field

The goal is the single most important non-obvious momentum in physics: what $p_i=\partial L/\partial\dot q_i$ delivers for a charge $q$ moving through potentials $(\phi,\mathbf A)$ . The strategy is a one-line differentiation of the standard electromagnetic Lagrangian — the subtlety is entirely in the interpretation, so the steps linger there. Watch which quantity is gauge-dependent and which is measurable: mixing them up is the classic exam disaster.

Step 1. Start from the electromagnetic Lagrangian in Cartesian coordinates:

L=\tfrac12 m\,\dot{\mathbf r}^2+q\,\dot{\mathbf r}\cdot\mathbf A(\mathbf r,t)-q\,\phi(\mathbf r,t)

Why this L is right: its Euler–Lagrange equations reproduce the Lorentz force $m\ddot{\mathbf r}=q(\mathbf E+\dot{\mathbf r}\times\mathbf B)$ — the verification lives in the velocity-dependent-potentials article and is assumed here.

Step 2. Differentiate with respect to $\dot x$ (the $y,z$ components are identical). The kinetic term gives $m\dot x$ ; the coupling term is linear in $\dot x$ with coefficient $qA_x$ ; the scalar potential term has no velocity in it:

p_x=\frac{\partial L}{\partial \dot x}=m\dot x+qA_x \qquad\Longrightarrow\qquad \mathbf p=m\dot{\mathbf r}+q\mathbf A

Step 3. Name the two pieces. The kinetic momentum $m\dot{\mathbf r}$ is what a detector measures — gauge-invariant, the thing that curves in a magnetic field. The canonical momentum $\mathbf p$ is what the Lagrangian formalism conserves, transforms, and eventually quantizes. They differ by $q\mathbf A$ , a field-configuration term attached to the particle's position. Why the canonical one is gauge-dependent: under $\mathbf A\to\mathbf A+\nabla\chi$ the momentum shifts by $q\nabla\chi$ while the trajectory — hence $m\dot{\mathbf r}$ — is unchanged. Canonical momentum is a coordinate on phase space, not an observable; only gauge-invariant combinations like $\mathbf p-q\mathbf A$ are measurable.

Step 4. Extract the conservation payoff. If the potentials are independent of $x$ (say $\mathbf B$ uniform with a gauge choice respecting that symmetry), then $x$ is cyclic and it is $p_x=m\dot x+qA_x$ — not $m\dot x$ — that is conserved. The kinetic momentum oscillates as the particle gyrates; the canonical combination stays fixed. This is the classical seed of magnetic translation groups and Landau levels. Two quick checks: setting $\mathbf A\to0$ collapses the canonical momentum onto the kinetic one, as it must; and the dimensions work out since $[q\mathbf A]=\mathrm{C\cdot T\cdot m}=\mathrm{kg\,m/s}$ , matching the units of linear momentum.

Figure

Canonical versus kinetic momentum for a gyrating charge — the measurable piece rotates, the conserved combination does not

The cyclic coordinate at work: central forces

The cleanest laboratory for conserved conjugate momenta is the planar central-force problem, where the geometry does all the talking. The result anchors the entire orbits chapter, so it is worth deriving once from the momentum point of view.

Go deeper · Derivation C · 1 minAngular momentum from the cyclic angle in a central force

Derivation CAngular momentum from the cyclic angle in a central force

We are deriving the conservation of angular momentum for a particle in a potential $V(r)$ that depends only on distance — as a two-line application of the cyclic-coordinate shortcut, rather than the torque argument of Newtonian mechanics. The strategy: write the planar Lagrangian in polar coordinates, notice which coordinate is absent, and read off the conserved slope. The instructive moment is dimensional: the "momentum" of an angle arrives in $\mathrm{kg\,m^2/s}$ without being asked.

Step 1. Write the Lagrangian in plane polar coordinates $(r,\theta)$ , using $v^2=\dot r^2+r^2\dot\theta^2$ :

L=\tfrac12 m\big(\dot r^2+r^2\dot\theta^2\big)-V(r)

Step 2. Scan for cyclic coordinates: $\theta$ appears only through $\dot\theta$ — the potential is blind to direction. Its conjugate momentum, from definition (1):

p_\theta=\frac{\partial L}{\partial\dot\theta}=mr^2\dot\theta

Why the $r^2$ : the angle is dimensionless, so the momentum absorbs the moment arm twice — once for the lever, once because the velocity along $\theta$ is $r\dot\theta$ . This is the moment of inertia $mr^2$ times angular velocity: angular momentum.

Step 3. Apply the shortcut: $\theta$ cyclic $\Rightarrow$ $p_\theta=mr^2\dot\theta=\ell$ is a constant of the motion, even though $\dot\theta$ itself varies whenever $r$ does — the skater pulling in her arms, the comet whipping through perihelion. Why this is Kepler's second law in disguise: the areal sweep rate is $\mathrm{d}A/\mathrm{d}t=\tfrac12 r^2\dot\theta=\ell/2m$ — constant. Equal areas in equal times is a conserved conjugate momentum wearing seventeenth-century clothes. *Two checks: the dimensions $\mathrm{kg\,m^2/s}$ are correct for angular momentum, and the result agrees with the Newtonian $\boldsymbol\ell=\mathbf r\times m\mathbf v$ restricted to the plane.*

Limiting cases

Case	Result	Meaning
Cartesian coordinate, velocity-independent $V$	$p=m\dot q$	kinetic and canonical momenta coincide
angular coordinate	$p_\theta=mr^2\dot\theta$	conjugate momentum is angular momentum
$\mathbf A\to 0$	$\mathbf p\to m\dot{\mathbf r}$	electromagnetic offset switches off smoothly
relativistic Lagrangian $-mc^2/\gamma$	$\mathbf p=\gamma m\mathbf v$	definition (1) generates the relativistic momentum
cyclic $q_i$	$p_i$ conserved	symmetry ⇒ conservation, by inspection
$\det M\to 0$	map $\dot q\mapsto p$ not invertible	degenerate Lagrangian; constraints $\phi(q,p)=0$ appear

Common trapMomentum is not mass times velocity — and the exceptions are the interesting cases

The reflex $p=m\dot q$ is correct only for Cartesian coordinates with velocity-independent potentials, and treating it as a definition rather than a special case causes three distinct failures. For angular coordinates it gets the units wrong: the momentum conjugate to $\theta$ is $mr^2\dot\theta$ , an angular momentum. In electromagnetic problems it gets the conservation law wrong: when a translation symmetry holds, the conserved object is $m\dot{\mathbf r}+q\mathbf A$ , and the measurable $m\dot{\mathbf r}$ is generally not constant — gyrating particles trade momentum with the field every cycle. And in the Hamiltonian formalism it corrupts everything downstream: Hamilton's equations, Poisson brackets, and canonical quantization all take the canonical $p$ , so writing $H(q,m\dot q)$ for a magnetic system produces wrong dynamics silently. The safe habit: never guess a momentum — differentiate the Lagrangian and let equation (1) tell you.

Worked examples

1 · Pendulum: reading a conjugate momentum, with numbers (confidence builder)

A pendulum bob of mass $m=0.30\ \mathrm{kg}$ swings on a light rod of length $l=0.80\ \mathrm{m}$ . As it passes $\theta=10°$ it has $\dot\theta=2.5\ \mathrm{rad/s}$ . Find the conjugate momentum $p_\theta$ and its instantaneous rate of change.

Step 1. Lagrangian in the angle: $L=\tfrac12 ml^2\dot\theta^2+mgl\cos\theta$ . Definition (1): $p_\theta=\partial L/\partial\dot\theta=ml^2\dot\theta$ — the moment of inertia $ml^2$ times angular velocity, in $\mathrm{kg\,m^2/s}$ .

Step 2. Numbers: $p_\theta=0.30\times(0.80)^2\times2.5=0.30\times0.64\times2.5=\boxed{0.48\ \mathrm{kg\,m^2/s}}$ . Note this is not $m$ times any velocity: the bob's linear momentum is $ml\dot\theta=0.30\times0.80\times2.5=0.60\ \mathrm{kg\,m/s}$ — different number, different units.

Step 3. The momentum-form equation of motion (2): $\dot p_\theta=\partial L/\partial\theta=-mgl\sin\theta$ — the generalized force conjugate to an angle is a torque. At $\theta=10°$ : $\dot p_\theta=-0.30\times9.81\times0.80\times\sin10°=-2.354\times0.1736=\boxed{-0.41\ \mathrm{N\,m}}$ .

Step 4. Sanity checks. $\theta$ is not cyclic (it appears in $\cos\theta$ ), so $p_\theta$ is rightly not conserved — the torque above is exactly its drain. Remove gravity ( $g\to0$ ) and $\theta$ becomes cyclic: $p_\theta$ freezes and the rod spins forever at constant $\dot\theta$ , as (2) demands.

2 · Proton in a magnetic field: canonical is half of kinetic (exam level)

A proton ( $m=1.673\times10^{-27}\ \mathrm{kg}$ , $q=+1.602\times10^{-19}\ \mathrm{C}$ ) gyrates on a circle of radius $r=1.0\ \mathrm{mm}$ in a uniform field $B=0.50\ \mathrm{T}$ along $z$ , described in the symmetric gauge $\mathbf A=\tfrac12\mathbf B\times\mathbf r$ . Find (a) its speed and kinetic angular momentum, (b) the canonical angular momentum $p_\varphi$ , and (c) reconcile the two.

Step 1. Cyclotron kinematics first. The gyration frequency is $\omega_c=qB/m=(1.602\times10^{-19}\times0.50)/1.673\times10^{-27}=4.79\times10^{7}\ \mathrm{rad/s}$ , so the speed is $v=\omega_c r=4.79\times10^{7}\times10^{-3}=4.79\times10^{4}\ \mathrm{m/s}$ . Kinetic angular momentum about the orbit centre: $|mr^2\dot\varphi|=mvr=1.673\times10^{-27}\times4.79\times10^{4}\times10^{-3}=\boxed{8.0\times10^{-26}\ \mathrm{kg\,m^2/s}}$ . Direction check: for $\mathbf B=+B\hat z$ a positive charge gyrates clockwise seen from $+z$ , so $\dot\varphi=-\omega_c$ and the kinetic angular momentum is $mr^2\dot\varphi=-8.0\times10^{-26}\ \mathrm{kg\,m^2/s}$ .

Step 2. Build the canonical momentum. In the symmetric gauge $\mathbf A=\tfrac12 Br\,\hat{\boldsymbol\varphi}$ , so the planar Lagrangian is $L=\tfrac12 m(\dot r^2+r^2\dot\varphi^2)+q\,(r\dot\varphi)\big(\tfrac12 Br\big)=\tfrac12 m(\dot r^2+r^2\dot\varphi^2)+\tfrac12 qBr^2\dot\varphi$ . Definition (1) in the angle:

p_\varphi=\frac{\partial L}{\partial\dot\varphi}=mr^2\dot\varphi+\tfrac12 qBr^2

Step 3. Numbers. The field term: $\tfrac12 qBr^2=\tfrac12\times1.602\times10^{-19}\times0.50\times(10^{-3})^2=4.0\times10^{-26}\ \mathrm{kg\,m^2/s}$ . Then $p_\varphi=-8.0\times10^{-26}+4.0\times10^{-26}=\boxed{-4.0\times10^{-26}\ \mathrm{kg\,m^2/s}}$ — the canonical angular momentum is exactly half the kinetic one, with the same sense.

Step 4. Reconcile. $\varphi$ is cyclic in this gauge, so $p_\varphi$ is a constant of the motion — and indeed every quantity in Step 3 is constant on the circular orbit. The general-orbit statement is stronger: for any motion in the uniform field, $p_\varphi=mr^2\dot\varphi+\tfrac12 qBr^2$ stays fixed while the two pieces trade against each other as $r$ changes. The "missing half" sits in the electromagnetic field's own angular momentum — a classical preview of why Landau levels in quantum mechanics carry the degeneracy they do.

Origins · 1 minJoseph-Louis Lagrange and how this was discovered

Joseph-Louis Lagrange

1736–1813 · Turin, Berlin, Paris

Lagrange built the mechanics in which momenta could finally be defined rather than postulated. Self-taught in Turin, he sent Euler a solution of the isoperimetric problem at nineteen and was refining the calculus of variations into the engine of the Mécanique analytique (1788) — a mechanics text famously without a single diagram — by his fifties. There the quantities $\partial L/\partial\dot q$ appear as the natural currency of the equations of motion, coordinate system be damned: the momentum of an angle, of a constrained bead, of any coordinate whatsoever, all issued by one rule. Napoleon made him a count; the metric system bears his committee work; and when Hamilton and Jacobi rebuilt mechanics around conjugate pairs a generation later, the pairs were Lagrange's.

History

1687

— Newton's Principia treats momentum as mass times velocity, the "quantity of motion"

1788

— Lagrange's Mécanique analytique produces momenta as derivatives of a single function of the coordinates chosen

1834

— Hamilton promotes the conjugate momenta to independent variables, opening phase space

1918

— Noether proves the systematic symmetry–conservation correspondence behind the cyclic-coordinate shortcut

1926

— Schrödinger's quantization takes the canonical momentum, not the kinetic one, to the operator −iℏ∇

1930

— Landau solves the charge-in-field problem quantum mechanically; the gauge structure of canonical momentum sets the level degeneracy

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §2.6 and §8.1 · Landau & Lifshitz, Mechanics, 3rd ed., §7 and §40 · Morin, Introduction to Classical Mechanics, Ch. 6 and Ch. 15 · Taylor, Classical Mechanics, Ch. 13 Applying $p_i=\partial L/\partial\dot q_i$ to the standard Lagrangians gives: free particle $m\dot x$ , angular coordinate $mr^2\dot\theta$ , electromagnetic $m\dot{\mathbf r}+q\mathbf A$ , and relativistic $\gamma m\mathbf v$ . The $\mathrm{d}\dot q$ terms cancel in the Legendre transform, yielding Hamilton's equations, and the Hessian/invertibility statement follows from the inverse function theorem. Worked-example numbers: pendulum $p_\theta=0.48\ \mathrm{kg\,m^2/s}$ , $\dot p_\theta=-0.41\ \mathrm{N\,m}$ ; proton $\omega_c=4.79\times10^7\ \mathrm{rad/s}$ , kinetic angular momentum $8.0\times10^{-26}\ \mathrm{kg\,m^2/s}$ , canonical $-4.0\times10^{-26}\ \mathrm{kg\,m^2/s}$ .

Builds on: The Lagrangian and Euler–Lagrange Equations · Generalized Coordinates · Legendre Transformation Leads to: The Hamiltonian Function · Hamilton's Canonical Equations · Phase Space · Equivalence of Lagrangian and Hamiltonian Formulations

Why should p be promoted to an independent variable at all — the velocity was doing fine?

Because the promotion halves the order of the dynamics and doubles the geometry. With $(q,\dot q)$ you have $n$ second-order equations whose state space has no preferred structure; with $(q,p)$ you get $2n$ first-order equations on a phase space equipped with the symplectic pairing that makes Hamiltonian flow, Liouville's theorem, and canonical transformations possible. The promotion is honest — the Legendre transform is invertible when (4) holds, so no information is gained or lost — but the new packaging exposes structure the velocity description hides. The payoff compounds through the rest of this chapter, starting with The Hamiltonian Function.

Is the canonical momentum measurable, given that it is gauge-dependent?

Not by itself. Under a gauge change $\mathbf A\to\mathbf A+\nabla\chi$ the canonical momentum shifts by $q\nabla\chi$ at fixed physical state, so its value is convention. What is measurable is the kinetic momentum $m\dot{\mathbf r}=\mathbf p-q\mathbf A$ — gauge-invariant, and the thing whose magnitude sets radii of curvature in detectors. This is not a defect: canonical momentum is a coordinate on phase space, and coordinates are always conventions. The physics lives in gauge-invariant combinations, in the equations of motion, and — quantum mechanically — in spectra and phases like the Aharonov–Bohm phase, which cares about $\oint\mathbf A\cdot\mathrm{d}\mathbf r$ precisely through the canonical structure.

When the kinetic momentum of a gyrating charge is not conserved, where does it go?

Into the field. The particle-plus-field system conserves total momentum; the particle's kinetic momentum oscillates because the field carries the balance, exchanged continuously through the Lorentz force. The canonical momentum $m\dot{\mathbf r}+q\mathbf A$ is the bookkeeping device that makes the particle-side accounting close whenever the potentials have the corresponding symmetry: it is conserved by symmetry of the Lagrangian even though neither of its two pieces separately is. Worked Example 2's constant $p_\varphi$ with both pieces trading against each other is that statement in numbers.

Why does the momentum conjugate to an angle have different units from linear momentum — is that a problem?

It is a feature enforced by the action. The conjugate pair always satisfies $[p_i][q_i]=\mathrm{J\,s}$ : whatever dimensions the coordinate has, the momentum takes the complement. A dimensionless angle therefore gets a momentum in $\mathrm{J\,s}=\mathrm{kg\,m^2/s}$ — angular momentum. This is also why $\sum_i p_i\dot q_i$ can be summed across coordinates of wildly different character in building the Hamiltonian: every term individually lands in energy. If your computed momentum–coordinate product does not have action units, you differentiated along the wrong velocity.

How is the cyclic-coordinate rule related to Noether's theorem — are they the same statement?

The cyclic rule is Noether's theorem's simplest face: invariance of $L$ under translation of one coordinate conserves that coordinate's conjugate momentum. Noether generalizes in two directions the cyclic rule cannot reach: symmetries that mix several coordinates (a rotation acting on $x,y,z$ simultaneously conserves a combination of momenta even though no single Cartesian coordinate is cyclic), and symmetries acting on time (conserving the energy function rather than any $p_i$ ). A practical corollary worth internalizing: whether a symmetry is visible as a cyclic coordinate depends on your coordinate choice — spherical coordinates make rotational symmetry cyclic, Cartesians hide it. Choosing coordinates to expose cyclicity is half the craft of Lagrangian problem-solving.

What exactly breaks when the velocity Hessian is singular?

The dictionary between velocities and momenta loses entries. A zero eigenvalue of $M_{ij}=\partial^2L/\partial\dot q_i\partial\dot q_j$ means some velocity direction leaves all momenta unchanged — so from the momenta alone that velocity cannot be reconstructed, and relations $\phi(q,p)=0$ (primary constraints) hold identically among the would-be phase-space coordinates. The Hamiltonian construction then needs the Dirac–Bergmann constraint analysis. This is not exotic: the Lagrangian of electromagnetism itself is degenerate (the momentum conjugate to $\phi$ vanishes), which is why gauge theories and Hamiltonian mechanics have such an involved marriage.

Does the definition still work for relativistic mechanics?

Perfectly — that is one of its quiet triumphs. The relativistic free particle has $L=-mc^2\sqrt{1-v^2/c^2}$ , which is nothing like $T-V$ , yet definition (1) yields $\mathbf p=\gamma m\mathbf v$ — the correct relativistic momentum, with the Lorentz factor generated by the square root rather than inserted by hand. Adding electromagnetic coupling gives $\mathbf p=\gamma m\mathbf v+q\mathbf A$ , and the Legendre transform of the previous article turns this into $H=\sqrt{(\mathbf p-q\mathbf A)^2c^2+m^2c^4}+q\phi$ . The definition never needed the non-relativistic assumption; it only needed a Lagrangian.

If different gauges give different canonical momenta, which one do I quantize?

Any of them — provided you are consistent. Quantization promotes the canonical pair of whatever gauge you chose to operators; different gauges give unitarily equivalent quantum theories, with wavefunctions differing by the phase $e^{iq\chi/\hbar}$ . What you must never do is mix: computing $\mathbf p$ in one gauge and $\mathbf A$ in another, or replacing the canonical $\mathbf p$ by the kinetic $m\dot{\mathbf r}$ inside the commutation relations. The gauge-covariant structure — kinetic momentum operators $\hat{\boldsymbol\pi}=\hat{\mathbf p}-q\mathbf A$ with $[\hat\pi_x,\hat\pi_y]=i\hbar qB$ — is what carries the physics, and it is gauge-independent by construction.