Kinetic energy in generalized coordinates: Question bank

Definition Every symbol used on this page, defined here first

So this page stands on its own, here is the whole cast of characters:

$q_{i}$ — a generalized coordinate: one independent number describing the system's configuration (an angle, a distance along a rail). $\overset{q}{˙}_{i} = d q_{i} / d t$ is its generalized velocity.
$r_{a} (q, t)$ — the Cartesian position of particle $a$ , written as a function of the coordinates and (maybe) the clock $t$ .
$\partial_{q_{i}} r_{a} \equiv \frac{\partial r _{a}}{\partial q _{i}}$ — how the position of particle $a$ shifts when you nudge coordinate $q_{i}$ and freeze everything else. It is a vector pointing in the direction that coordinate moves the particle.
$\partial_{t} r_{a} \equiv \frac{\partial r _{a}}{\partial t}$ — how the position moves because the constraint itself is being dragged by the clock, with all $q_{i}$ frozen. Zero for a stationary track.
$T$ — kinetic energy $= \sum_{a} \frac{1}{2} m_{a} ∣ \dot{r}_{a} ∣^{2}$ , split as $T = T_{2} + T_{1} + T_{0}$ .
$M_{ij} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{q_{j}} r_{a}$ — the mass (metric) matrix.
$a_{i} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{t} r_{a}$ — the linear-term coefficient: the overlap (dot product) between the direction coordinate $q_{i}$ pushes the particle and the direction the moving constraint drags it. Nonzero only when both a coordinate-motion and a clock-motion exist and are not perpendicular. Then $T_{1} = \sum_{i} a_{i} \overset{q}{˙}_{i}$ .
$V$ — the potential energy $V (q)$ : the position-only energy stored in forces (gravity, springs). It carries no velocities.
$L = T - V$ — the Lagrangian: kinetic minus potential (see Lagrangian mechanics).
$p_{i} = \frac{\partial L}{\partial q ˙ _{i}}$ — the generalized momentum conjugate to $q_{i}$ (see Generalized momentum and conserved quantities).
$H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L$ — the energy function / Hamiltonian (see Energy function and Hamiltonian).

Before you start, three anchors you must keep in your pocket (all built in the parent note):

Scleronomic = the position map $r_{a} (q)$ has no explicit $t$ — the "track" is not being yanked around. Rheonomic = $r_{a} (q, t)$ does contain $t$ — some constraint moves on its own schedule.
The split $T = T_{2} + T_{1} + T_{0}$ counts how many velocity factors $\overset{q}{˙}$ each piece carries: 2 (quadratic), 1 (linear), 0 (velocity-free).
The mass matrix $M_{ij}$ is symmetric, positive-definite, and generally depends on the coordinates $q$ .

Formula The non-trivial cancellation

$H = T_{2} - T_{0} + V$ — shown, not asserted Several traps below hinge on why $H = T + V$ only for scleronomic systems. Here is the full chain so you never take it on faith.

Start from the definition $H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L$ with $p_{i} = \partial L / \partial \overset{q}{˙}_{i}$ and $L = T - V$ .
Since $V = V (q)$ carries no velocities, $\partial L / \partial \overset{q}{˙}_{i} = \partial T / \partial \overset{q}{˙}_{i}$ , so $\sum_{i} p_{i} \overset{q}{˙}_{i} = \sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i}$ .
Apply the Euler-degree count to $T = T_{2} + T_{1} + T_{0}$ : $T_{2}$ is degree 2 in $\overset{q}{˙}$ , $T_{1}$ degree 1, $T_{0}$ degree 0, so $\sum_{i} \overset{q}{˙}_{i} \frac{\partial T}{\partial q ˙ _{i}} = 2 T_{2} + 1 \cdot T_{1} + 0 \cdot T_{0} = 2 T_{2} + T_{1} .$
Therefore $H = (2 T_{2} + T_{1}) - (T_{2} + T_{1} + T_{0} - V) = T_{2} - T_{0} + V .$
Scleronomic case: $T_{1} = T_{0} = 0$ and $T = T_{2}$ , so $H = T_{2} + V = T + V$ . Rheonomic case: the $- T_{0}$ (and the vanished $T_{1}$ ) spoil the identity, so $H \neq = T + V$ in general.

True or false — justify

Kinetic energy is always a quadratic function of the generalized velocities.

False. Only for scleronomic constraints. When

\partial_{t} r_{a} \neq = 0

the linear piece

T_{1} = \sum_{i} a_{i} \overset{q}{˙}_{i}

and the velocity-free piece

T_{0}

appear, so

T

is a general degree-2 polynomial in

\overset{q}{˙}

, not a homogeneous quadratic.

M_{ij}

can be treated as a constant matrix once the coordinates are chosen.

False. It is

M_{ij} (q)

— e.g.

M_{θ θ} = m r^{2}

in polar coordinates grows with

r

. This dependence must be differentiated when you form

\partial T / \partial q

in the Euler–Lagrange equation.

The mass matrix is always symmetric.

True.

M_{ij} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{q_{j}} r_{a}

is built from a dot product, which is symmetric in

i

and

j

, so swapping the indices gives the identical expression.

The mass matrix can have a zero eigenvalue for a physical system.

False.

M

is positive-definite:

\sum_{ij} M_{ij} \overset{q}{˙}_{i} \overset{q}{˙}_{j} = \sum_{a} m_{a} ∣ \dot{r}_{a} ∣^{2} \geq 0

, and it vanishes only if every

\overset{q}{˙}_{i}

produces zero Cartesian velocity — which would mean that coordinate isn't a real degree of freedom.

For a scleronomic system, energy equals

T + V

True. Scleronomic means

T = T_{2}

is a homogeneous quadratic, so

\sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i} = 2 T_{2}

and the chain

H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L

collapses (see the boxed derivation) to

T_{2} + V = T + V

, with

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

and

L = T - V

If a constraint is time-dependent, the total energy

H

still equals

T + V

False. With

T_{1}, T_{0}

present the derivation above gives

H = T_{2} - T_{0} + V \neq = T + V

in general. The moving constraint can feed energy in or out.

T_{1}

being zero guarantees the constraint is scleronomic.

False. The rotating-wire bead has

T_{1} = 0

but

T_{0} = \frac{1}{2} m r^{2} ω^{2} \neq = 0

; it is fully rheonomic.

T_{1}

vanishes whenever

\partial_{q_{i}} r_{a} ⊥ \partial_{t} r_{a}

(their dot product

a_{i} = 0

), which can happen even while time is explicitly in the map.

T_{0}

is a constant (independent of the state).

False.

T_{0} = \frac{1}{2} \sum_{a} m_{a} (\partial_{t} r_{a})^{2}

depends on the coordinates

q

(e.g.

\frac{1}{2} m r^{2} ω^{2}

grows with

r

). It is "constant" only in the sense of being free of velocities

\overset{q}{˙}

, not free of position.

Choosing different generalized coordinates changes the numerical value of

T

at a given physical state.

False.

T

is a physical scalar (

\sum_{a} \frac{1}{2} m_{a} v_{a}^{2}

); its value at a real instant is coordinate-independent. What changes is the formula — the entries

M_{ij}

transform like a metric tensor, but the number they compute stays fixed.

Any constraint can be absorbed into a choice of generalized coordinates.

False. Only holonomic constraints (relations among positions and time,

f (r_{a}, t) = 0

) can be solved to eliminate coordinates. A non-holonomic constraint — one that ties velocities together and cannot be integrated to a position relation, e.g. a rolling coin's

\overset{x}{˙} = R \dot{ϕ} cos ψ

— cannot be removed this way, and the clean

T (q, \overset{q}{˙})

reduction of this whole page assumes it is absent.

Why questions

Why does the chain rule for

\dot{r}_{a}

acquire an explicit

\partial_{t} r_{a}

term, and why only sometimes?

Because

r_{a} (q, t)

can depend on

t

both through the moving

q_{i} (t)

and directly through the clock. The direct term

\partial_{t} r_{a}

survives only when a constraint is imposed on its own time schedule (rheonomic); a scleronomic map has

\partial_{t} r_{a} = 0

Why is grouping

T

by powers of

\overset{q}{˙}

the "natural" classification?

Because

\dot{r}_{a}

is linear in the velocities plus a velocity-free clock term, so squaring it can only yield products with

2

1

, or

0

velocity factors. No other kind of term is possible, and each group has a distinct physical meaning (kinetic, coupling, effective-potential).

Why does the coefficient

a_{i}

measure a "geometric overlap"?

Because

a_{i} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{t} r_{a}

is a dot product between the direction coordinate

q_{i}

pushes the particle and the direction the moving constraint drags it. When these are perpendicular the dot product is zero, so

T_{1}

disappears even though the constraint is moving — exactly the rotating-wire case.

Why does the mass matrix deserve the name "metric on configuration space"?

Because

2 T_{2} = \sum_{ij} M_{ij} \overset{q}{˙}_{i} \overset{q}{˙}_{j}

measures the squared "speed" of the configuration point, exactly as a metric

g_{ij} \overset{q}{˙}^{i} \overset{q}{˙}^{j}

measures distance-rate on a curved space. Motion of the system is geodesic-like motion in this metric (see Configuration space and the metric tensor).

Why does the count

\sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i}

give

2 T_{2} + T_{1}

, not

2 T

Because that sum weights each piece by its degree in

\overset{q}{˙}

T_{2}

(degree 2) contributes

2 T_{2}

T_{1}

(degree 1) contributes

T_{1}

, and

T_{0}

(degree 0) contributes nothing. Only when

T = T_{2}

alone does it equal

2 T

— which is exactly the scleronomic condition behind

H = T + V

Why is

M

guaranteed positive-definite rather than merely non-negative?

Because each generalized coordinate is, by construction, an independent degree of freedom, so no nonzero velocity vector

\overset{q}{˙}

can leave every Cartesian velocity zero. Hence

\sum_{a} m_{a} ∣ \dot{r}_{a} ∣^{2} > 0

strictly for

\overset{q}{˙} \neq = 0

Edge cases

A free particle in Cartesian coordinates: what are

T_{2}, T_{1}, T_{0}

T = T_{2} = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2})

with

M_{ij} = m δ_{ij}

(constant, diagonal);

T_{1} = T_{0} = 0

because the map

r = (x, y, z)

has no explicit

t

. This is the degenerate "flat metric" case.

How do you detect that a chosen coordinate is redundant, and what's a concrete example?

Redundancy shows up as a zero eigenvalue of $M$ : some nonzero velocity vector

\overset{q}{˙}

yields zero Cartesian motion, so

det M = 0

. Concretely, describing a particle on a circle by both the angle

θ

and the arc length

s = R θ

gives

\overset{s}{˙} - R \dot{θ} = 0

always producing no independent motion;

M

becomes singular. The fix is to keep only one — drop

s

(or

θ

) so the coordinate count matches the true degrees of freedom.

Rotating wire in the limit

ω \to 0

: does the answer reduce correctly?

Yes —

T_{0} = \frac{1}{2} m r^{2} ω^{2} \to 0

and

T \to \frac{1}{2} m \overset{r}{˙}^{2}

, the scleronomic straight-wire result. The rheonomic terms smoothly switch off as the imposed motion stops, exactly as they should.

Rotating wire with

\overset{r}{˙} = 0

(bead momentarily pinned on the wire): is

T = 0

No —

T = \frac{1}{2} m r^{2} ω^{2}

from

T_{0}

. Zero generalized velocity does not mean zero physical velocity when the constraint itself is moving; the bead is being swept around at speed

r ω

A pendulum of length

ℓ

whose pivot is driven horizontally as

X_{p} = A cos Ω t

— write

T_{2}, T_{1}, T_{0}

explicitly.

With

x = A cos Ω t + ℓ sin θ, y = - ℓ cos θ

, differentiate:

\partial_{θ} r = (ℓ cos θ, ℓ sin θ)

and

\partial_{t} r = (- A Ω sin Ω t, 0)

. So

T_{2} = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}

a_{θ} = m \partial_{θ} r \cdot \partial_{t} r = - m A Ω ℓ cos θ sin Ω t

giving

T_{1} = - m A Ω ℓ cos θ sin Ω t \dot{θ}

, and

T_{0} = \frac{1}{2} m A^{2} Ω^{2} sin^{2} Ω t

. Both survive because the driven pivot puts

t

explicitly in

r (θ, t)

and

\partial_{θ} r

is not perpendicular to

\partial_{t} r

Can

T_{2}

ever be zero for a moving system?

Only instantaneously, when all

\overset{q}{˙}_{i} = 0

; as a form

T_{2}

can never be identically zero because

M

is positive-definite. So a genuinely moving configuration (

\overset{q}{˙} \neq = 0

) always has

T_{2} > 0

If two independent coordinates happen to give orthogonal Cartesian gradients everywhere, what does

M

look like?

It is diagonal,

M_{ij} = 0

for

i \neq = j

, so

T_{2} = \frac{1}{2} \sum_{i} M_{ii} \overset{q}{˙}_{i}^{2}

has no cross terms — the polar case (

M_{r θ} = 0

) is exactly this. Orthogonal coordinates diagonalize the metric.

Does this whole

T = T_{2} + T_{1} + T_{0}

machinery apply to a rolling wheel constrained by

\overset{x}{˙} = R \dot{ϕ}

Not directly — that is a non-holonomic (velocity-level) constraint that cannot be integrated into a position relation, so you cannot eliminate a coordinate to build a plain

T (q, \overset{q}{˙})

. You must instead carry the constraint separately (Lagrange multipliers); the decomposition here presumes only holonomic constraints.

Recall One-line survival summary

"Time in the map ⇒ $T_{1}, T_{0}$ wake up; $M$ is a position-dependent symmetric positive-definite metric; energy $= T + V$ only when the track doesn't move." Trap detector ::: Always ask "Is $t$ explicit in $r_{a} (q, t)$ ?" — that single question decides scleronomic vs rheonomic, and therefore whether $T_{1}, T_{0}$ and the $H = T + V$ shortcut apply.

Question bank — Kinetic energy in generalized coordinates

True or false — justify

Spot the error

Why questions

Edge cases