Constraints using Lagrange multipliers

WHY do we need this method?

The WHY: if you ever need to ask "when does the bead fly off the hoop?" or "what is the tension?", the embedded method can't answer it directly. The multiplier method can.

HOW: deriving it from d'Alembert's principle

We start from the foundation, d'Alembert's principle (virtual work of applied + inertial forces vanishes):

$\sum_k \left(\frac{d}{dt}\frac{\partial T}{\partial \dot q_k}-\frac{\partial T}{\partial q_k}-Q_k\right)\delta q_k = 0.$

For conservative applied forces this becomes, with $L=T-V$ :

$\sum_{k=1}^{n}\left(\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}\right)\delta q_k = 0. \tag{1}$

Why this isn't just "each bracket = 0": if the $q_k$ were independent, each $\delta q_k$ is free, so each bracket vanishes — the ordinary Euler–Lagrange equations. But under a constraint $f(q,t)=0$ , the $\delta q_k$ are not independent. Taking the variation of the constraint at fixed time:

$\delta f = \sum_{k}\frac{\partial f}{\partial q_k}\,\delta q_k = 0. \tag{2}$

So (1) must hold only for those $\delta q$ obeying (2). We can't set each bracket to zero yet.

Multiply (2) by an as-yet-unknown function $\lambda(t)$ and add to (1):

$\sum_{k}\left(\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}-\lambda\frac{\partial f}{\partial q_k}\right)\delta q_k = 0.$

Now choose $\lambda$ to kill the bracket of the one dependent coordinate. The remaining $\delta q_k$ are independent, so every bracket must vanish:

For $m$ constraints $f_a=0$ you sum: the RHS becomes $\sum_{a=1}^{m}\lambda_a \,\partial f_a/\partial q_k$ .

Figure — Constraints using Lagrange multipliers

WORKED EXAMPLE 1 — Bead on a wire / disk rolling, find the constraint force

Setup: A particle of mass $m$ slides on the inside of a vertical hoop of radius $R$ (polar coords $r,\theta$ ). Constraint: $f = r - R = 0$ . Find the normal force $N$ from the hoop.

Lagrangian in plane polar: $L=\tfrac12 m(\dot r^2+r^2\dot\theta^2)-mgr\sin\theta.$

Why polar? Because the constraint is simply $r=R$ , making $\partial f/\partial r=1$ , $\partial f/\partial \theta=0$ — clean.

$r$ -equation: $\frac{d}{dt}(m\dot r)-mr\dot\theta^2+mg\sin\theta = \lambda\cdot 1.$ Why this step? The RHS uses $\partial f/\partial r = 1$ .

Apply constraint $r=R\Rightarrow \dot r=0,\ddot r=0$ : $-mR\dot\theta^2+mg\sin\theta=\lambda.$

Why this is the answer: $\lambda = \partial L$ -imbalance along $r$ = the radial constraint force. So $N=\lambda = mg\sin\theta - mR\dot\theta^2.$ The bead leaves the hoop when $N=0$ , i.e. $mg\sin\theta = mR\dot\theta^2$ . That question was unanswerable with the embedded method — this is the payoff.

WORKED EXAMPLE 2 — Atwood-style: rope tension via multiplier

Setup: Disk of radius $a$ , moment $I$ , mass over a pulley etc. Simplify to: a mass $m$ hangs from a string wound on a cylinder. Let $x$ = mass drop, $\phi$ = cylinder rotation. Constraint of "no slipping": $f = x - a\phi = 0.$

$L=\tfrac12 m\dot x^2+\tfrac12 I\dot\phi^2+mgx.$

$x$ -equation: $\partial f/\partial x = 1$ $m\ddot x - mg = \lambda. \tag{i}$ $\phi$ -equation: $\partial f/\partial \phi = -a$ $I\ddot\phi = -a\lambda. \tag{ii}$

Why these? RHS is $\lambda\,\partial f/\partial q$ .

Constraint: $\ddot x = a\ddot\phi \Rightarrow \ddot\phi=\ddot x/a$ . Substitute (ii): $\frac{I}{a}\frac{\ddot x}{a} = -a\lambda \;\Rightarrow\; \lambda = -\frac{I\ddot x}{a^2}.$ Plug into (i): $m\ddot x - mg = -\frac{I\ddot x}{a^2}\Rightarrow \ddot x = \frac{mg}{m+I/a^2}.$ And the tension is its constraint force magnitude: $T=|\lambda| = \frac{I\ddot x}{a^2}=\frac{mg\,(I/a^2)}{m+I/a^2}.$ Why $T=|\lambda|$ : $\lambda\,\partial f/\partial x = \lambda$ is the generalized force on $x$ from the string = $-T$ (string pulls up). Sign just tracks direction.

WORKED EXAMPLE 3 — Sphere rolling down an incline (find friction)

Constraint (rolling, no slip): $f = x - R\theta = 0$ . With $L = \tfrac12 M\dot x^2 + \tfrac12 I\dot\theta^2 + Mg x\sin\alpha$ :

$x$ : $M\ddot x - Mg\sin\alpha = \lambda$
$\theta$ : $I\ddot\theta = -R\lambda$

Same algebra: $\lambda = -f_{\text{friction}}$ . Solid sphere $I=\tfrac25 MR^2$ gives $\ddot x = \tfrac57 g\sin\alpha$ and friction $|\lambda| = \tfrac27 Mg\sin\alpha$ . The multiplier is the static friction force.

Common mistake Steel-manning the common errors

Error 1: "I'll use independent coordinates AND multipliers." Feels right because fewer coordinates seems simpler. Why it's wrong: multipliers require the constraint to still be present in your coordinates. If you already eliminated it, $\partial f/\partial q=0$ and $\lambda$ is meaningless. Fix: keep the redundant coordinate when you want the constraint force.

Error 2: Thinking $\lambda$ has fixed units or is dimensionless. Why tempting: it "looks like a number." Truth: $\lambda\,\partial f/\partial q_k$ must have units of generalized force, so $\lambda$ 's units depend on $f$ . If $f$ has length units, $\lambda$ is a force. Fix: always read units from $\lambda\,\partial f/\partial q$ .

Error 3: Forgetting the constraint equation itself. People write $n$ EL equations and panic that $\lambda$ is undetermined. Fix: you have $n+1$ unknowns and need the constraint $f=0$ as the $(n+1)$ th equation.

Error 4: Using multipliers for non-holonomic velocity constraints by writing $f(q)=0$ . Non-holonomic constraints $\sum a_k \dot q_k=0$ aren't integrable; you must use the velocity form $\frac{d}{dt}\frac{\partial L}{\partial\dot q_k}-\frac{\partial L}{\partial q_k}=\sum_a \lambda_a a_{ak}$ , not a gradient of some $f$ .

Recall Feynman: explain to a 12-year-old

Imagine you're rolling a marble inside a bowl. The bowl pushes the marble to keep it on the surface — that's the "constraint force." Usually in physics we're lazy and just say "the marble stays on the bowl," and we never figure out how hard the bowl pushes. The Lagrange-multiplier trick is like adding a tiny honest helper number $\lambda$ to our equations whose only job is to tell us how hard the bowl is pushing at every moment. When that push drops to zero, the marble jumps off the surface!

Active Recall

Why are virtual displacements not independent under a holonomic constraint?

Because

\delta f=\sum_k(\partial f/\partial q_k)\delta q_k=0

links them; one

\delta q

is determined by the others.

What is the modified Lagrange equation with one constraint?

\frac{d}{dt}\frac{\partial L}{\partial\dot q_k}-\frac{\partial L}{\partial q_k}=\lambda\,\frac{\partial f}{\partial q_k}

, plus

f=0

Physical meaning of the Lagrange multiplier

\lambda

It scales the generalized constraint force

Q_k^{(c)}=\lambda\,\partial f/\partial q_k

(tension, normal force, friction).

For

m

constraints, what is the RHS of the EL equation?

\sum_{a=1}^{m}\lambda_a\,\partial f_a/\partial q_k

How many equations and unknowns for

n

coords and 1 constraint?

n

EL equations + 1 constraint equation

=n+1

, for

n

coordinates +

\lambda

Why do constraint forces vanish in the embedded (independent-coordinate) method?

Because they do no virtual work (perpendicular to allowed motion), so they drop out when constraints are solved away.

Condition for a bead to leave a circular hoop?

When the normal force

\lambda=mg\sin\theta-mR\dot\theta^2=0

Rolling sphere on incline: what does

\lambda

equal?

Minus the static friction force; magnitude

\tfrac27 Mg\sin\alpha

for a solid sphere.

Why must

\delta f

be taken at fixed time (

\delta

, not

d

Virtual displacements are instantaneous; the explicit

t

-dependence of

f

is frozen during a virtual variation.

What's different for non-holonomic constraints?

Use velocity-form

a_{ak}

coefficients: RHS

=\sum_a\lambda_a a_{ak}

, since

\sum_k a_{ak}\dot q_k=0

isn't an integrable

f(q,t)=0

Connections

Generalized Coordinates and Degrees of Freedom — what we trade when keeping a constraint.
d'Alembert's Principle and Virtual Work — the foundation we derived from.
Euler-Lagrange Equations — the unconstrained special case ( $\lambda=0$ ).
Holonomic vs Non-holonomic Constraints — when this method applies vs needs velocity form.
Normal Force and Tension as Constraint Forces — the physical payoff.
KKT Conditions — the same multiplier idea in optimization/inequality constraints.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Lagrangian mechanics mein do raaste hote hain constraints handle karne ke. Pehla — "embedded" — jahan hum smart coordinates choose karke constraint ko hata dete hain. Isme equations simple ho jaate hain, lekin ek nuksaan hai: jo constraint force thi (tension, normal force, friction), woh equations se gayab ho jaati hai. Agar tumhe pata karna ho ki "bead hoop se kab udd jaayegi?" ya "rassi mein tension kitna hai?", toh ye method jawab nahi de sakta.

Yahaan Lagrange multiplier $\lambda$ aata hai. Hum constraint ko hatate nahi, balki use rakhte hain aur ek unknown number $\lambda$ add kar dete hain. Derivation d'Alembert principle se aati hai: constraint ki wajah se virtual displacements $\delta q$ independent nahi hote ( $\delta f = 0$ unko jodta hai). Isliye hum directly har bracket ko zero nahi kar sakte. Trick: $\lambda$ ko aise choose karo ki dependent coordinate ka bracket cancel ho jaaye. Result milta hai: $\frac{d}{dt}\frac{\partial L}{\partial \dot q_k} - \frac{\partial L}{\partial q_k} = \lambda \frac{\partial f}{\partial q_k}$ , aur saath mein constraint $f=0$ .

Sabse important baat — $\lambda$ ka physical matlab: $\lambda \,\partial f/\partial q_k$ exactly woh generalized constraint force hai. Gradient $\nabla f$ hamesha constraint surface ke perpendicular point karta hai — bilkul wahi direction jahaan normal force lagti hai. Toh $\lambda$ uss force ka magnitude scale hai. Hoop wale example mein $N = mg\sin\theta - mR\dot\theta^2$ , aur jab $N=0$ ho jaata hai tabhi bead udd jaati hai.

Count yaad rakho: $n$ coordinates + $1$ multiplier = $n$ EL equations + $1$ constraint equation. Sab balanced. Galti mat karna — agar tumne coordinate pehle hi eliminate kar diya, toh $\partial f/\partial q = 0$ ho jaayega aur $\lambda$ ka koi matlab nahi rahega. Isliye jab constraint force chahiye, redundant coordinate rakho. Mnemonic: "Gradient times lambda gives the push."

WHY do we need this method?

HOW: deriving it from d'Alembert's principle

WORKED EXAMPLE 1 — Bead on a wire / disk rolling, find the constraint force

WORKED EXAMPLE 2 — Atwood-style: rope tension via multiplier

WORKED EXAMPLE 3 — Sphere rolling down an incline (find friction)

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Analytical Mechanics

Connections