Derivation of Euler-Lagrange equations from D'Alembert's principle: Question bank

Definition Symbols recalled (so this page stands alone)

$r_{k}$ — the position of particle $k$ ; $\dot{r}_{k}$ its velocity, $\ddot{r}_{k}$ its acceleration.
$p_{k} = m_{k} \dot{r}_{k}$ — the momentum of particle $k$ (mass times velocity); $\dot{p}_{k} = m_{k} \ddot{r}_{k}$ is the rate it changes.
$F_{k}^{(a)}$ — the applied force on particle $k$ : the real pushes and pulls (gravity, springs, your hand) that we do care about.
$F_{k}^{(c)}$ — the constraint force: whatever the wire/rail/rod supplies to keep the constraint satisfied (the wire pushing sideways on a bead). We usually don't know it and don't want it.
$q_{i}$ — generalized coordinates (the minimal free variables, like $θ$ for a pendulum); $\overset{q}{˙}_{i}$ their time-rates.
$δ r$ — a virtual displacement: an imagined shift with time frozen, $δ t = 0$ .
$Q_{i}$ — the generalized force for coordinate $q_{i}$ : the applied forces projected onto $q_{i}$ , $Q_{i} = \sum_{k} F_{k}^{(a)} \cdot \frac{\partial r _{k}}{\partial q _{i}}$ . This is not the potential — the potential $V$ is a separate scalar energy from which conservative $Q_{i}$ can be derived as $Q_{i} = - \partial V / \partial q_{i}$ .
$T$ — kinetic energy, $V$ — potential energy, $L = T - V$ — the Lagrangian.

The picture below is worth pinning before you start: it shows why projecting onto the allowed direction deletes the constraint force, and how a virtual displacement differs from the real one.

Figure — Derivation of Euler-Lagrange equations from D'Alembert's principle

And the two engine-room identities the traps refer to — the "cancellation of dots" and the "commuting of $d / d t$ with $\partial / \partial q$ " — are summarized here so you needn't flip back:

Recall The three derivation steps the traps quote (mini-recap)

Step 3 (product-rule-backwards): $m_{k} \ddot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}} = \frac{d}{d t} (m_{k} \dot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}}) - m_{k} \dot{r}_{k} \cdot \frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}}$ . It trades the unwanted acceleration for velocity-type terms.
Lemma 1 (cancellation of dots): $\frac{\partial r ˙ _{k}}{\partial q ˙ _{i}} = \frac{\partial r _{k}}{\partial q _{i}}$ .
Lemma 2 (commuting derivatives): $\frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}} = \frac{\partial r ˙ _{k}}{\partial q _{i}}$ .

True or false — justify

True or false: A virtual displacement

δ r

is the same thing as the real displacement

d r

the particle actually makes in time

d t

False.

δ r

freezes time (

δ t = 0

), so it explores only where the constraint allows you to be right now;

d r

additionally carries the constraint's own time-drift (

\frac{\partial r}{\partial t} d t

). They coincide only when the constraint is time-independent (scleronomic) — that extra drift term is then zero.

True or false: D'Alembert's principle says the applied forces alone do zero virtual work.

False. It says

\sum_{k} (F_{k}^{(a)} - \dot{p}_{k}) \cdot δ r_{k} = 0

. The reason this is true is that the constraint forces

F_{k}^{(c)}

do zero virtual work (they push perpendicular to allowed motion), so they drop out of Newton's equation when we dot with

δ r

— leaving applied forces and inertia, not applied forces alone.

True or false: In

δ r_{k} = \sum_{i} \frac{\partial r _{k}}{\partial q _{i}} δ q_{i}

we may add a

\frac{\partial r _{k}}{\partial t} δ t

term to be safe.

False. A virtual displacement sets

δ t = 0

by definition, so that term is exactly zero. Keeping it would defeat the purpose: it is precisely the absence of the time-drift that makes

\sum_{k} F_{k}^{(c)} \cdot δ r_{k} = 0

hold even for moving constraints.

True or false: Constraint forces never do real work on the system.

False. On a moving constraint (a sliding wall, a driven rotating wire) the real work

F^{(c)} \cdot d r

can be nonzero, because

d r

has a component along

F^{(c)}

coming from the wall's own motion. Only the virtual work

F^{(c)} \cdot δ r

vanishes, because

δ r

has no such time-drift component.

True or false: Once we reach

\sum_{i} [\dots] δ q_{i} = 0

, each bracket must vanish.

True, but only because for a holonomic system the

δ q_{i}

are independent. The logic is: a weighted sum of independent, arbitrary quantities is zero for all choices only if every weight (bracket) is zero. Take away independence (non-holonomic case) and one

δ q_{i}

constrains another, so the brackets can trade off and need not individually vanish.

True or false: The Euler–Lagrange equation

\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{i}} - \frac{\partial L}{\partial q _{i}} = 0

is a new physical law beyond Newton.

False. Each step of the derivation is an equality: it is Newton's

F = m a

with the unknown constraint forces algebraically projected out and the survivors repackaged into energies. Same physics, cleaner bookkeeping — no new postulate is smuggled in.

True or false: We may fold

V

into

L

because

\partial V / \partial \overset{q}{˙}_{i} = 0

True in the usual case

V = V (q, t)

: since

V

has no

\overset{q}{˙}

-dependence, adding

- V

T

changes

\partial L / \partial q_{i}

(giving

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

) but leaves

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

untouched — exactly what we want. For a velocity-dependent potential

U (q, \overset{q}{˙}, t)

the folding still works but you must use the generalized potential form

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

; see the edge case below for the concrete magnetic example.

True or false:

\frac{\partial r ˙ _{k}}{\partial q ˙ _{i}} = \frac{\partial r _{k}}{\partial q _{i}}

(the "cancellation of dots") is just cancelling the dots symbolically.

False in spirit — it's a genuine result. Because

\dot{r}_{k} = \sum_{j} \frac{\partial r _{k}}{\partial q _{j}} \overset{q}{˙}_{j} + \frac{\partial r _{k}}{\partial t}

is linear in the

\overset{q}{˙}_{j}

with coefficients that depend on

q, t

but not on

\overset{q}{˙}

, differentiating in

\overset{q}{˙}_{i}

selects the single coefficient

\frac{\partial r _{k}}{\partial q _{i}}

. The mnemonic looks like cancellation only because the underlying reason really does return that quantity.

Spot the error

"Newton gives

F^{(a)} + F^{(c)} - \dot{p} = 0

; dotting with

δ r

and summing, the applied forces cancel, leaving constraint forces." — where is the slip?

Reversed. It is

\sum F^{(c)} \cdot δ r = 0

(constraint forces perpendicular to allowed motion) that cancels, so the constraint forces vanish and the applied forces survive as the generalized force

Q_{i}

"For the pendulum I'll use Cartesian

x, y

so I have two clean equations." — what's wrong with this choice?

You reintroduce the rod tension (a constraint force) as an unknown and gain a constraint

x^{2} + y^{2} = ℓ^{2}

(rod length

ℓ

). Choosing

q = θ

instead bakes that constraint in automatically, so the tension never appears.

\frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}}

— the position doesn't move in

q_{i}

, so this is zero." — spot the error.

It is not zero. By Lemma 2 (recapped above),

\frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}} = \frac{\partial r ˙ _{k}}{\partial q _{i}}

, which is generally nonzero because

\dot{r}_{k}

depends on the

q

's through its coefficients.

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

always, so I can always write

L = T - V

." — find the hidden assumption.

This needs the applied forces to be conservative (

F^{(a)} = - \nabla V

). Friction, driving forces, or magnetic forces are not captured by a plain scalar

V

; then you keep the raw

Q_{i} = \sum_{k} F_{k}^{(a)} \cdot \frac{\partial r _{k}}{\partial q _{i}}

"In Step 3 we wrote

m_{k} \ddot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}} = \frac{d}{d t} (m_{k} \dot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}})

." — what got dropped?

The correction term

- m_{k} \dot{r}_{k} \cdot \frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}}

. The product rule run backwards has two pieces; dropping the second loses the eventual

- \partial T / \partial q_{i}

term.

T = \frac{1}{2} m \overset{r}{˙}^{2}

for the bead on the rotating wire (radial coordinate

r

, wire spinning at rate

ω

)." — what's missing?

The tangential contribution: the bead is also swept sideways by the wire, so

T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})

. Omitting

r^{2} ω^{2}

kills

\partial T / \partial r

and loses the centrifugal effect.

"Since

δ q_{i}

are infinitesimal, I can set them to specific tiny numbers to solve." — why is this a mistake?

The

δ q_{i}

are arbitrary and independent symbols; the conclusion "each bracket = 0" follows from that arbitrariness (it must hold for all choices), not from any single value. Fixing values would prove nothing.

Why questions

Why do we project onto virtual displacements at all, instead of onto arbitrary directions?

Because virtual displacements point along what the constraints allow, and constraint forces are perpendicular to those — so the dot product

F^{(c)} \cdot δ r

is zero, deleting the forces we don't know (see figure s01).

Why must we use the product-rule-backwards trick in Step 3 rather than differentiate directly?

To trade the unwanted acceleration

\ddot{r}

for time-derivatives of velocity terms, which repackage cleanly into

\frac{d}{d t} \frac{\partial T}{\partial q ˙ _{i}} - \frac{\partial T}{\partial q _{i}}

. Direct differentiation leaves

\ddot{r}

stuck to the constraint geometry with no way to become an energy.

Why does the

\partial T / \partial q_{i}

term matter — isn't kinetic energy "just about speed"?

When the mapping

r_{k} (q, t)

curves or rotates,

T

depends on the coordinate

q

itself (e.g. the

r^{2} ω^{2}

term for the rotating wire, with radial coordinate

r

and rotation rate

ω

), so

\partial T / \partial q \neq = 0

. That term produces centrifugal, Coriolis-like, and geometric forces automatically — see Kinetic energy in generalized coordinates.

Why can we fold a conservative

V

into

L

but not friction?

A conservative force is the gradient of a scalar

V (q)

, so

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

and

\partial V / \partial \overset{q}{˙}_{i} = 0

. Friction is not the gradient of any scalar (it depends on the direction of motion), so no

V

exists to absorb it.

Why is the independence of the

δ q_{i}

specifically a holonomic property?

Holonomic constraints are position-equations, so they can be solved away, leaving coordinates each freely variable at any instant. Non-holonomic constraints tie velocities together and can't be integrated out, so varying one

δ q_{i}

forces others — independence is lost.

Why does Hamilton's principle give the same equation as this force-based derivation?

Both encode the same content: extremizing the action integral is mathematically equivalent to demanding the EL bracket vanish, which is D'Alembert's principle integrated in time. Two doors, one room.

Why do we take

\overset{q}{˙}

and

q

as independent variables when differentiating

L

Inside the partial derivatives

L

is treated as a function on the abstract configuration-velocity space, where

q

and

\overset{q}{˙}

are separate slots. The relation

\overset{q}{˙} = d q / d t

is only imposed afterwards, by the outer total derivative

\frac{d}{d t}

; mixing them mid-step corrupts the partials.

Edge cases

Edge case: a free particle,

V = 0

q = x

. What does EL reduce to and why is that reassuring?

\frac{d}{d t} (m \overset{x}{˙}) - 0 = 0 \Rightarrow m \overset{x}{¨} = 0

— Newton's first law. The formalism must reproduce plain Newton when there are no constraints or potentials.

Edge case: the constraints are time-independent (scleronomic). Do virtual and real displacements now coincide?

Yes to first order, because

\partial r / \partial t = 0

removes the time-drift that separated

δ r

from

d r

. The virtual-real distinction only bites for moving (rheonomic) constraints.

Edge case: a non-holonomic constraint (e.g. rolling without slipping). Can you still set each EL bracket to zero?

No — the

δ q_{i}

are no longer independent, so you must adjoin the constraints with Lagrange multipliers before separating terms.

Edge case:

L

has no explicit dependence on some coordinate

q_{j}

(

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

). What survives?

Then

\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{j}} = 0

, so

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

is conserved — a cyclic coordinate. This is the seed of Noether's theorem: a symmetry of

L

gives a conserved quantity.

Edge case: a charged particle (charge

e

) in an electromagnetic field — the concrete velocity-dependent potential. What replaces "fold in

V

", and does it reproduce the right force?

You use the generalized potential

U = e ϕ - e A \cdot \dot{r}

, where

ϕ

is the scalar potential and

A

the vector potential (both functions of position and time). It does depend on

\dot{r}

, so build

L = T - U = \frac{1}{2} m \dot{r}^{2} - e ϕ + e A \cdot \dot{r}

. Feeding this into

\frac{d}{d t} \frac{\partial L}{\partial r ˙} - \frac{\partial L}{\partial r} = 0

yields, after using

E = - \nabla ϕ - \partial_{t} A

and

B = \nabla \times A

, exactly the Lorentz force

m \ddot{r} = e (E + \dot{r} \times B)

. This is "minimal coupling," and it is the standard example that folding can handle

\overset{q}{˙}

-dependent forces via the generalized-potential form

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

Edge case: what if two chosen generalized coordinates are secretly not independent (a redundant choice)?

The "each bracket = 0" step fails, because the vanishing of

\sum_{i} [\dots] δ q_{i}

no longer forces each coefficient to zero. You must first eliminate the redundancy or use multipliers.

Edge case: the potential depends explicitly on time,

V = V (q, t)

. Does the derivation still hold?

Yes. Time-dependence of

V

leaves

\partial V / \partial \overset{q}{˙}_{i} = 0

intact, so the folding into

L

is valid; only the conserved-energy conclusion (a separate result) is lost.

Recall One-line self-test

If someone hands you a system, which single question decides whether you may write "each bracket = 0"? Are the $δ q_{i}$ independent? ::: Equivalently: are the constraints holonomic? If yes, separate; if no, use multipliers.