Constraint Forces via Lagrange Multipliers

Derivation

Constraint Forces via Lagrange Multipliers

D-127 Home PU-201 Threads force Depends on The Euler-Lagrange Equation

Statement

For a system whose configuration coordinates q₁,…,q_n are subject to m independent holonomic constraints f_α(q,t)=0, adjoining each constraint to Lagrange's equations with an undetermined multiplier λ_α(t) gives the unreduced equations of motion, and the adjoined terms Q_i^c=Σ_αλ_α ∂f_α/∂q_i are exactly the generalized forces of constraint that would otherwise have to be supplied by hand.

Why it matters

The usual virtue of Lagrangian mechanics is that constraint forces vanish from the equations: choose independent coordinates and the normal reactions, tensions and rail forces never appear. But those forces are frequently the physical quantity you actually want — the tension that snaps a cable, the normal force that decides when a car leaves a hump, the rail load on a track. The multiplier method keeps the constraints as live equations and returns the forces as a by-product, at the price of a few extra unknowns.

It also unifies statics, dynamics and constrained optimization under one theorem: the multipliers here are the same objects as the Karush–Kuhn–Tucker multipliers in optimization and the “shadow prices” of duality. Understanding that the multiplier is a force is the conceptual pivot.

Assumptions

Constraints are holonomicexpressible as f_α(q,t)=0; if dropped for a genuinely non-holonomic (non-integrable velocity) constraint, the coordinates cannot be reduced and the clean “force ∝ ∇f” reading below no longer follows from a configuration gradient.

Constraints are ideal (workless)the constraint forces do no virtual work, Σ_i Q_i^c δq_i=0; if dropped, e.g. sliding friction, the multiplier delivers only the normal reaction and the tangential dissipative force must be added separately as an applied generalized force.

Applied forces are carried by Leither derivable from a potential or supplied as generalized forces on the right; if dropped, any force not in L is silently attributed to the constraint and λ_α is corrupted.

The constraint Jacobian ∂f_α/∂q_i has full rank mthe m×n matrix of constraint gradients is of rank m along the motion; if dropped (redundant or tangentially-meeting constraints, or a singular configuration) the multipliers are non-unique or blow up.

f_α and L are C²the constraint surfaces and Lagrangian are twice differentiable; if dropped at a kink or corner the gradients that define the force direction are undefined there.

Derivation

δS = ∫_t₁^t₂ δL dt = 0, δq_i(t₁)=δq_i(t₂)=0

Hamilton's principle, stated in the full unreduced coordinates q₁,…,q_n before any constraint is used to eliminate variables. A

δL = Σ_i ( ∂L/∂q_i δq_i + ∂L/∂q̇_i δq̇_i )

Chain rule; L=L(q,q̇,t) and the variation commutes with time so δq̇_i=d(δq_i)/dt. A

∫ Σ_i ∂L/∂q̇_i δq̇_i dt = −∫ Σ_i ( d/dt ∂L/∂q̇_i ) δq_i dt

Integration by parts in time; the boundary term [ ∂L/∂q̇_i δq_i ] vanishes because δq_i is fixed at both endpoints. A

∫_t₁^t₂ Σ_i E_i δq_i dt = 0, E_i ≡ d/dt( ∂L/∂q̇_i ) − ∂L/∂q_i

Collecting steps 2–3 and defining the Euler–Lagrange operator E_i (the prior result). This must hold for all admissible variations. B

δf_α = Σ_i ( ∂f_α/∂q_i ) δq_i = 0 (α=1,…,m)

Admissibility: the varied path must also satisfy the constraints, so at each instant the variation is orthogonal to every constraint gradient. Hence the δq_i are not independent — we may not set each E_i=0. B

∫ Σ_α λ_α(t) δf_α dt = ∫ Σ_i ( Σ_α λ_α ∂f_α/∂q_i ) δq_i dt = 0

Multiply each constraint variation by an as-yet-arbitrary function λ_α(t) and sum. By step 5 this integral is identically zero, so it may be added to step 4 without changing anything. C

∫ Σ_i [ E_i − Σ_α λ_α ∂f_α/∂q_i ] δq_i dt = 0

Subtracting step 6 from step 4. Now choose the m multipliers λ_α to annihilate the bracket for the m dependent coordinates — solvable precisely because the Jacobian has rank m. C

E_i − Σ_α λ_α ∂f_α/∂q_i = 0 for every i=1,…,n

With the m dependent brackets killed, the remaining n−m variations are independent, so their coefficients vanish too. The fundamental lemma of the calculus of variations then forces every bracket to zero. C

d/dt( ∂L/∂q̇_i ) − ∂L/∂q_i = Σ_α=1^m λ_α ∂f_α/∂q_i

Rearranging step 8: the symbols only. Together with the m constraints f_α=0 this is a closed set of n+m equations for the n+m unknowns {q_i,λ_α}. B

d/dt( ∂L/∂q̇_i ) − ∂L/∂q_i = Q_i^c ⇒ Q_i^c = Σ_α λ_α ∂f_α/∂q_i

Identification: in the unreduced coordinates the applied forces sit inside L, so any force needed to close Newton's law that is not in L is the constraint force Q_i^c. Comparison with step 9 makes the adjoined term equal to it — the claim. B

Result

Q_i^c = Σ_α=1^m λ_α(t) ∂f_α/∂q_i ( in Cartesian: F_a^c = Σ_α λ_α ∇_af_α )

Reading. The generalized force of constraint on coordinate q_i is a superposition of the constraint-surface normals ∂f_α/∂q_i, each weighted by its own multiplier λ_α. The multiplier is therefore the strength of that reaction: a single scalar per constraint that measures how hard the surface must push to keep f_α=0. Because the force points along ∇f_α it is normal to the constraint manifold, which is exactly why an ideal constraint does no work on motions tangent to that manifold.

Units check. Q_i^c is conjugate to q_i, so it carries units of [energy]/[q_i]. The gradient ∂f_α/∂q_i carries [f_α]/[q_i], hence λ_α carries [energy]/[f_α]. For a length-type constraint f=r−ℓ ([f]=m) this gives [λ]=J/m=N — the multiplier is literally a force in newtons.

Limiting cases

Single constraint (m=1). Q_i^c=λ ∂f/∂q_i; the one multiplier is the signed magnitude of the single normal reaction (string tension, rail force).
Force-free direction. If ∂f_α/∂q_k=0 for all α, then Q_k^c=0: the constraint exerts nothing along q_k and motion there is free.
Statics. Setting accelerations to zero recovers ∂V/∂q_i=Σ_αλ_α ∂f_α/∂q_i — the method of virtual work, with λ the equilibrium reaction.
Scleronomic constraint (∂f_α/∂t=0). The constraint force does zero work over the motion, so mechanical energy is conserved even though Q^c≠0.

Breaks when

Genuinely non-holonomic constraints. A non-integrable velocity constraint Σ_ia_αi(q,t)q̇_i+b_α=0 has no f_α(q,t) to differentiate; multipliers can still be introduced on the Pfaffian form, but the coordinates cannot be reduced and the reaction is no longer the gradient of any configuration function (rolling coin, skate).
Non-ideal constraints. With friction the constraint force has a tangential component that does work; the multiplier returns only the workless normal part, so λ ∇f underestimates the true reaction unless the friction is added by hand.
Rank-deficient Jacobian. Redundant or degenerately-meeting constraints (Jacobian rank <m, or a singular configuration) leave the multipliers undetermined or divergent — the linear system in step 7 is not invertible.
Impulsive / unilateral constraints. At impact or when a one-sided constraint (f≥0) is about to release, λ can jump or must change sign; the smooth-equality treatment fails and one needs complementarity conditions (λ≥0, λf=0).

Failure modes

Reduce-and-multiply double count. Eliminating a constraint by choosing independent coordinates and still keeping its multiplier — once the constraint is used up its force is gone; you cannot get it back from the reduced description.
Assuming λ is constant. The multiplier is a function of time (and generally of the state); freezing it turns a correct reaction into a wrong one except in trivial steady motion.
Reading ∂f/∂q_i as the physical force. It is only the direction; the newton-valued force is λ times that gradient, summed over all constraints and converted to Cartesian.
Sign of the tension. Forgetting that Q_r^c=λ can be negative (inward pull) and quoting a negative “tension” instead of |λ|.
Assuming [λ]=N always. The multiplier's units depend on the units of f_α; for an angle- or area-type constraint it is not a force in newtons.
Claiming the constraint does work. Computing Q^c·v and getting nonzero without noticing the constraint is rheonomic (∂f/∂t≠0) — only the explicit time dependence carries energy in or out.

Discussion

The geometric picture is the cleanest way to hold the result. The constraints cut out a submanifold of configuration space; admissible velocities and virtual displacements are tangent to it, while the constraint gradients ∇f_α span the normal directions. The multiplier equation says the reaction lives entirely in that normal space — it is a covector (a force) built from the constraint gradients, with the λ_α as its components in the gradient basis. This is precisely why ideal constraints drop out when you work in tangent (independent) coordinates: a force normal to the surface has no component along tangential displacements.

There is a genuine engineering trade-off between the two routes. Reducing to n−m independent coordinates gives the smallest system and hides the forces; keeping the multipliers gives n+m equations but hands you every reaction directly. When the reaction is the design quantity — will the cable hold, when does the wheel lift — the multiplier route is not a detour, it is the whole point. Numerically it produces a differential-algebraic system, and stabilization schemes (Baumgarte, index reduction) exist precisely because the algebraic constraint tends to drift.

The deepest reading is that λ_α is a sensitivity. In constrained optimization the multiplier equals the rate of change of the optimum with respect to relaxing the constraint — the shadow price — and the Karush–Kuhn–Tucker conditions are the static, inequality-augmented twin of what we derived. In field theory the same device enforces first-class constraints: in electromagnetism the scalar potential φ enters as the multiplier whose equation of motion is Gauss's law, and Dirac's constraint analysis generalizes the counting of steps 7–8 to infinite dimensions. The undetermined multiplier and the physical reaction are, in every one of these settings, the same mathematical object wearing different clothes.

Common misconceptions. The multiplier is not just a bookkeeping device with no physical content — it is a measurable force (the tension you could read off a spring scale). And “constraint forces do no work” is a statement about virtual displacements consistent with the instantaneous constraint, not about real displacements: a moving (rheonomic) constraint, such as a bead on a wire being cranked, does exchange energy with the system even though it is ideal.

Worked examples

Example 1 — Tension in a plane pendulum. A bob of mass m hangs on a light inextensible string of length ℓ; polar coordinates (r,θ) about the pivot, θ from the downward vertical. Find the string tension.

L = ½m(ṙ²+r²θ̇²) + mgr cosθ, f = r−ℓ = 0

Kinetic energy in polar coordinates; height below pivot is −r cosθ so V=−mgr cosθ. One holonomic constraint. A

m r̈ − m r θ̇² − mg cosθ = λ ∂f/∂r = λ

The r-equation from step 9, with ∂f/∂r=1. B

λ = −m(ℓ θ̇² + g cosθ) ⇒ T = −λ = m(ℓ θ̇² + g cosθ)

On the constraint r=ℓ, ṙ=r̈=0. The tension pulls inward (−r), so T=−λ>0. This is the textbook T=mg cosθ+mv²/ℓ with v=ℓθ̇. B

m=0.5 kg, ℓ=1.2 m, θ=30°, θ̇=2.0 rad/s

Now insert numbers. g=9.81 m/s², cos30°=0.8660. A

T = 0.5 (1.2×2.0² + 9.81×0.8660) = 0.5 (4.80 + 8.496) N

Arithmetic inside the bracket first. A

T = −λ ≈ 6.65 N (inward)

Reading. The multiplier λ≈−6.65 N is the radial reaction; its magnitude is the string tension, larger than the static value mg cosθ=4.25 N because of the centripetal demand.

Units check. m ℓ θ̇² = kg·m·s⁻² = N, and mg cosθ = N. Consistent.

Example 2 — String tension in an Atwood machine. Masses m₁,m₂ hang over an ideal pulley; coordinates x₁,x₂ measured downward. Find the tension.

L = ½m₁ẋ₁² + ½m₂ẋ₂² + m₁gx₁ + m₂gx₂, f = x₁+x₂−ℓ = 0

Down-positive, so V=−m₁gx₁−m₂gx₂. The inextensible string fixes x₁+x₂. A

m₁ẍ₁ − m₁g = λ, m₂ẍ₂ − m₂g = λ

Both ∂f/∂x₁=∂f/∂x₂=1, so the same multiplier appears in both equations. B

ẍ₂ = −ẍ₁ ≡ −a ⇒ a = (m₁−m₂)g/(m₁+m₂)

Differentiate the constraint twice for the kinematic link, then subtract the two equations of step 2 to eliminate λ. B

λ = m₁(a−g) = −2m₁m₂g/(m₁+m₂) ⇒ T = −λ = 2m₁m₂g/(m₁+m₂)

Back-substitute a into either equation of step 2. Symbols only — the classic Atwood tension. B

m₁=3 kg, m₂=2 kg, g=9.81 m/s²

Insert numbers: m₁+m₂=5 kg, m₁−m₂=1 kg. A

T = 2(3)(2)(9.81)/5 ≈ 23.5 N, a = (1)(9.81)/5 ≈ 1.96 m/s²

Reading. The single multiplier is the common string tension. It lies between the two weights (m₂g=19.6 N and m₁g=29.4 N), as it must for the lighter mass to accelerate up and the heavier down.

Units check. 2m₁m₂g/(m₁+m₂) has units kg·(m/s²)=N; a has units m/s². Consistent.

Problems

Pendulum tension at the bottom. A bob m=0.8 kg on a string ℓ=0.50 m is released from rest with the string horizontal. Find the tension as it passes the lowest point.
Solution
Energy from horizontal to bottom: drop h=ℓ, so v²=2gℓ. From Example 1, T=m(g+v²/ℓ)=m(g+2g)=3mg. Thus T=3×0.8×9.81=23.5 N. The multiplier is λ=−23.5 N (inward); tension is three times the bob's weight at the bottom of a horizontal release.
Atwood numbers. For m₁=5.0 kg and m₂=3.0 kg, find the acceleration and the string tension.
Solution
a=(m₁−m₂)g/(m₁+m₂)=(2.0)(9.81)/8.0=2.45 m/s². T=2m₁m₂g/(m₁+m₂)=2(5)(3)(9.81)/8=36.8 N (so λ=−36.8 N). Check: m₂g=29.4 N < T < m₁g=49.1 N. ✓
Normal force on a frictionless incline. A block m=2.0 kg rests on a fixed frictionless plane of angle β=30°. Using along-slope and perpendicular coordinates (s,n) with constraint f=n=0, find the multiplier and identify the normal force.
Solution
Height is s sinβ+n cosβ, so L=½m(ṡ²+ṅ²)−mg(s sinβ+n cosβ). The n-equation: mn̈+mg cosβ=λ ∂f/∂n=λ. On the constraint n̈=0, so λ=mg cosβ=N. Numerically N=2.0×9.81×cos30°=17.0 N. The s-equation gives the free slide s̈=−g sinβ=−4.9 m/s², with no multiplier — the constraint exerts nothing along the slope, as expected for a frictionless surface.
Where a particle leaves a sphere. A particle starts at rest at the top of a frictionless sphere of radius R and slides down. Using the constraint f=r−R=0, show the multiplier is the normal reaction and find the angle θ (from the upward vertical) at which the particle leaves the surface.
Solution
With L=½m(ṙ²+r²θ̇²)+mgr cosθ and f=r−R, the r-equation on r=R gives −mRθ̇²−mg cosθ=λ, i.e. the outward normal is N=mg cosθ+λ… more directly, radial Newton: mg cosθ−N=mv²/R with N=−λ. Energy: v²=2gR(1−cosθ). The particle leaves when N=0 (λ=0): mg cosθ=m 2g(1−cosθ) ⇒ 3cosθ=2 ⇒ cosθ=2/3, θ=48.2°. The multiplier vanishing is exactly the physical condition “contact force goes to zero.”
Conical pendulum tension. A bob m=0.30 kg on a string ℓ=1.0 m moves in a steady horizontal circle with the string at α=30° to the vertical. Using the constraint r=ℓ, find the tension (the radial multiplier) and the period.
Solution
For steady conical motion the radial multiplier along the string is −T. Force balance: vertical T cosα=mg ⇒ T=mg/cosα; horizontal T sinα=mω²(ℓ sinα) ⇒ ω²=g/(ℓ cosα). Numbers: T=0.30×9.81/cos30°=2.943/0.8660=3.40 N (so λ=−3.40 N). ω²=9.81/(1.0×0.8660)=11.33, ω=3.37 rad/s, period T_p=2π/ω=1.87 s. The tension exceeds the weight mg=2.94 N by the factor 1/cosα.