Generalized momenta and generalized forces: Question bank

Definition The symbols used on this page (read first)

$q_{j}$ — a generalized coordinate: one number (a length like $r$ , or an angle like $θ$ ) that helps pin down the configuration. $\overset{q}{˙}_{j} = d q_{j} / d t$ is its generalized velocity.
$r_{i}$ — the ordinary Cartesian position of particle $i$ , written as a function $r_{i} (q_{1}, \dots, q_{n}, t)$ .
$T$ — kinetic energy, $T = \frac{1}{2} \sum_{i} m_{i} ∣ v_{i} ∣^{2}$ (always positive, built from velocities).
$V$ — potential energy, a function of positions (and maybe $t$ ); gravity near Earth gives $V = m g y$ .
$L$ — the Lagrangian, defined as $L = T - V$ (note the minus sign, not $T + V$ ).
$δ W$ — the virtual work: the work the applied forces would do during an imagined tiny displacement $δ r_{i}$ made at frozen time, consistent with the constraints. The little $δ$ (not $d$ ) is the flag for "imagined, time held fixed".

Running example coordinate systems: a pendulum uses one angle $θ$ with rod length $ℓ$ (so $r = (ℓ sin θ, - ℓ cos θ)$ ); polar coordinates use radius $r$ and angle $θ$ (so $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ ).

Key objects you'll reason about: the generalized momentum $p_{j} = \partial L / \partial \overset{q}{˙}_{j}$ , the generalized force $Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j}$ , and the Euler–Lagrange statement $\overset{p}{˙}_{j} = \partial L / \partial q_{j}$ .

The three pictures below are your visual scaffolding — every abstract phrase in this bank (" $\partial r_{i} / \partial q_{j}$ ", "figure-skater effect", "virtual displacement") is drawn there. Glance back at them whenever a line feels abstract.

Figure — Generalized momenta and generalized forces

True or false — justify

Generalized momentum always equals mass times velocity,

p_{j} = m \overset{q}{˙}_{j}

False.

p_{j} = \partial L / \partial \overset{q}{˙}_{j}

; for an angular coordinate

p_{θ} = m ℓ^{2} \dot{θ}

has units kg·m²/s. The "

m

" and even the dimensions depend on which coordinate you chose.

The generalized force conjugate to an angle is a torque.

True. Since

Q_{j} δ q_{j}

must be an energy, and

δ θ

is dimensionless (radians),

Q_{θ}

must carry units of energy = N·m = torque.

If a Lagrangian has no explicit dependence on

q_{j}

, then

\overset{q}{˙}_{j}

is constant.

False. It is the momentum

p_{j}

that is conserved, not the velocity. For

p_{θ} = m r^{2} \dot{θ}

constant,

\dot{θ}

changes as

r

changes (the figure-skater effect drawn in figure 3).

For a conservative system with a time-independent potential, the generalized force is entirely captured by

Q_{j} = - \partial V / \partial q_{j}

True as a statement about the applied conservative force alone. But beware: this

Q_{j}

is not always the whole right-hand side of the equation of motion, because

\overset{p}{˙}_{j} = \partial L / \partial q_{j} = \partial T / \partial q_{j} - \partial V / \partial q_{j}

can also carry a

\partial T / \partial q_{j}

term (centrifugal). "Complete generalized force" and "complete driving term" are different claims — keep them separate.

Canonical momentum

p_{j}

is a vector in physical

x y z

space.

False. Each

p_{j}

is a single scalar number attached to coordinate

q_{j}

. It lives in configuration/phase space, not in real 3D space — see Hamiltonian Mechanics.

If two systems have the same kinetic energy

T

, they have the same generalized momenta.

False.

p_{j}

depends on how $T$ varies with each $\overset{q}{˙}_{j}$ , not just

T

's value. Two systems can share a numeric

T

at an instant yet have different

\partial T / \partial \overset{q}{˙}_{j}

A cyclic (ignorable) coordinate means the coordinate never appears anywhere in the problem.

False. It means

q_{j}

is absent from $L$ (so

\partial L / \partial q_{j} = 0

); its velocity

\overset{q}{˙}_{j}

typically still appears in

L

. That's exactly why

p_{j} = \partial L / \partial \overset{q}{˙}_{j}

is meaningful and conserved.

Generalized momentum is always the same as the physical linear momentum in that direction.

False. Only when

q_{j}

is a Cartesian coordinate and

T

has no cross-terms. In a magnetic field,

p_{x} = m \overset{x}{˙} + q A_{x}

— the canonical momentum differs from the kinetic momentum

m \overset{x}{˙}

If a coordinate is cyclic, its conjugate momentum is conserved even when the Lagrangian depends explicitly on time.

True. Cyclicity is about

\partial L / \partial q_{j} = 0

only; explicit

t

-dependence does not touch that derivative, so

\overset{p}{˙}_{j} = 0

still holds. (What fails under explicit time-dependence is conservation of energy, not of a cyclic

p_{j}

Spot the error

"The bead on the rotating wire has no applied force, so

Q_{r} = 0

and

m \overset{r}{¨} = 0

Error: the equation of motion is

m \overset{r}{¨} = m r ω^{2}

. The centrifugal term arises from

\partial T / \partial r

(the

q

-dependence of kinetic energy), not from any

V

. "No applied force" does not mean "no force-like term".

p_{θ} = m ℓ^{2} \dot{θ}

, and

θ

is cyclic for the pendulum, so angular momentum is conserved."

Error: for the pendulum

V = - m g ℓ cos θ

depends on

θ

, so

θ

is not cyclic. Hence

\overset{p}{˙}_{θ} = - m g ℓ sin θ \neq = 0

; angular momentum is not conserved.

"To get

Q_{j}

from virtual work I displace the system by

δ r_{i} = \sum_{j} \frac{\partial r _{i}}{\partial q _{j}} δ q_{j} + \frac{\partial r _{i}}{\partial t} δ t

Error: a virtual displacement freezes time (that's what

δ W

is built on — see the definition box), so the

\partial r_{i} / \partial t

term is dropped. Keeping it mixes real (

d

) with virtual (

δ

) displacements and corrupts

Q_{j}

"Since

Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j}

is a dot product of forces, its units are always newtons."

Error: the weight

\partial r_{i} / \partial q_{j}

(drawn as the tangent arrow in figure 2) has units of (length / unit of

q_{j}

). For an angular

q_{j}

this weight is a length, so

Q_{j}

comes out in N·m (torque), not N.

"The Euler–Lagrange equation says

\overset{p}{˙}_{j} = Q_{j}

, so I compute

Q_{j} = - \partial V / \partial q_{j}

and I'm done — no need for

\partial T / \partial q_{j}

Error: the true statement is

\overset{p}{˙}_{j} = \partial L / \partial q_{j} = \partial T / \partial q_{j} - \partial V / \partial q_{j}

. Whenever

T

depends on

q_{j}

(curvilinear coordinates, rotation), the

\partial T / \partial q_{j}

term is essential.

"Conservation of

p_{j}

requires no force at all acting on the system."

Error: it requires only that

L

be independent of that one coordinate

q_{j}

. Forces and potentials can be huge; as long as they don't depend on

q_{j}

p_{j}

stays constant — this is the Noether's Theorem symmetry statement.

"A velocity-dependent force like magnetism can't be handled by a potential, so it has no place in

Q_{j}

Error: magnetic (Lorentz) forces enter through a velocity-dependent generalized potential

U (q, \overset{q}{˙}, t)

, and the generalized force is recovered as

Q_{j} = \frac{d}{d t} \frac{\partial U}{\partial q ˙ _{j}} - \frac{\partial U}{\partial q _{j}}

. They fit the framework — just not the simple

- \partial V / \partial q_{j}

shortcut.

Why questions

Why is

p_{j}

defined as

\partial L / \partial \overset{q}{˙}_{j}

rather than as

m \overset{q}{˙}_{j}

Because that definition makes Euler–Lagrange read

\overset{p}{˙}_{j} = \partial L / \partial q_{j}

— Newton's "rate of change of momentum = force" in any coordinates. It's chosen precisely to preserve that law.

Why does an angular coordinate automatically give angular momentum as its conjugate momentum?

Because

p_{θ} = \partial L / \partial \dot{θ}

and for rotational motion

T \propto r^{2} \dot{θ}^{2}

, so

p_{θ} = m r^{2} \dot{θ}

— which is the definition of Angular momentum. The coordinate's geometry dictates the momentum's identity.

Why do we bother with generalized coordinates instead of just

x, y, z

Because constraints are baked in: a pendulum needs one

θ

instead of two Cartesian coordinates tied by

x^{2} + y^{2} = ℓ^{2}

. Fewer variables, no constraint forces to solve for.

Why does the same formula

Q_{j} δ q_{j} = δ W

yield a force for one coordinate and a torque for another?

Because

Q_{j}

is defined as "virtual work per unit of

q_{j}

". If

q_{j}

is a length you get N; if it's an angle you get N·m. The units of

Q_{j}

auto-adjust so the product

Q_{j} δ q_{j}

is always energy.

Why is "the coordinate is cyclic" the deepest reason a momentum is conserved?

Because a cyclic coordinate is a symmetry of

L

— shifting

q_{j}

leaves the physics unchanged. Noether's Theorem makes this precise: every continuous symmetry gives a conserved quantity.

Why can

\dot{θ}

increase even when angular momentum

p_{θ}

is conserved?

Because

p_{θ} = m r^{2} \dot{θ}

(see figure 3). If

r

shrinks,

\dot{θ}

must grow to keep the product constant — the figure skater pulling in her arms spins faster.

Why does the "cancellation of dots" identity

\partial v_{i} / \partial \overset{q}{˙}_{j} = \partial r_{i} / \partial q_{j}

hold?

Because

v_{i} = \sum_{j} (\partial r_{i} / \partial q_{j}) \overset{q}{˙}_{j} + \partial r_{i} / \partial t

is linear in each

\overset{q}{˙}_{j}

with coefficient

\partial r_{i} / \partial q_{j}

; differentiating w.r.t.

\overset{q}{˙}_{j}

leaves only that coefficient.

Why does explicit time-dependence in

L

break energy conservation but not cyclic-momentum conservation?

Because energy conservation follows from

L

having no explicit $t$ (

\partial L / \partial t = 0

), while cyclic-momentum conservation follows from

L

having no particular $q_{j}$ (

\partial L / \partial q_{j} = 0

). They are independent symmetries; killing one need not kill the other.

Edge cases

If a system has zero potential (

V = 0

), can there still be a generalized force?

Yes. In curvilinear or rotating coordinates

\partial T / \partial q_{j}

can be nonzero (centrifugal, Coriolis-like terms), producing force-like effects even with

V = 0

What is

p_{j}

for a coordinate that appears in

L

but whose velocity

\overset{q}{˙}_{j}

does not (a non-dynamical coordinate)?

Then

p_{j} = \partial L / \partial \overset{q}{˙}_{j} = 0

. Such a coordinate has no momentum; its Euler–Lagrange equation becomes a constraint

\partial L / \partial q_{j} = 0

, not an equation of motion.

At the instant a pendulum passes

θ = 0

(bottom), what is the generalized force

Q_{θ}

Q_{θ} = - m g ℓ sin 0 = 0

. Gravity provides no restoring torque at the bottom; the bob is momentarily unforced in the

θ

direction even though it's moving fastest.

For a free particle in Cartesian coordinates, do generalized and physical momentum coincide?

Yes — this is the degenerate case.

q = x

L = \frac{1}{2} m \overset{x}{˙}^{2}

, so

p_{x} = m \overset{x}{˙}

exactly. All the "strangeness" of

p_{j}

vanishes precisely when the coordinate is a straight Cartesian length with no velocity cross-terms.

What happens to

Q_{j}

if the applied forces are all perpendicular to

\partial r_{i} / \partial q_{j}

Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j} = 0

. The forces do no work along that coordinate's displacement direction, so they contribute nothing to

Q_{j}

— e.g. a constraint force normal to the wire contributes zero.

In the limit

r \to 0

for a particle in polar coordinates, what happens to

p_{θ} = m r^{2} \dot{θ}

p_{θ} \to 0

\dot{θ}

stays finite. But if

p_{θ}

is conserved and nonzero, then

\dot{θ} \to \infty

r \to 0

— the coordinate system becomes singular at the origin (why polar coordinates fail there).

For a system with an explicitly time-dependent constraint (e.g. the bead on a wire forced to rotate as

θ = ω t

), is mechanical energy

T + V

conserved?

No. The rotating-wire constraint feeds energy in/out through the driving, so

T + V

is not conserved even though

p_{r}

obeys its own equation. The frozen-time virtual displacement is exactly what lets us still define

Q_{r}

cleanly despite the moving constraint.

For a charged particle in a magnetic field, is the kinetic momentum

m \overset{x}{˙}

conserved when

x

is cyclic?

No — it's the canonical momentum

p_{x} = m \overset{x}{˙} + q A_{x}

that is conserved. The magnetic vector potential contribution

q A_{x}

is what makes the conserved quantity differ from plain

m \overset{x}{˙}

Recall One-line summary to carry away

$p_{j} = \partial L / \partial \overset{q}{˙}_{j}$ and $Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j}$ ; both change identity (linear↔angular, force↔torque) with the coordinate, and $\overset{p}{˙}_{j} = \partial L / \partial q_{j}$ is Newton's law in disguise.