Recovery of Newton's Second Law from L = T - V

Statement

For a system of particles subject only to conservative forces derivable from a scalar potential V(q), inserting the Lagrangian L = T − V into the Euler-Lagrange equations d/dt(∂L/∂q̇_i) − ∂L/∂q_i = 0 returns exactly Newton's second law mᷤ_i = −∂V/∂x_i. This fixes the otherwise-undetermined form of the physical Lagrangian.

Why it matters

The Euler-Lagrange equation is a purely mathematical consequence of stationary action; on its own it does not tell you which function L reproduces the physics. This derivation supplies the missing physical input: for conservative mechanics the correct choice is kinetic minus potential energy, not their sum, not some other combination. That single fact is the bridge between the variational and Newtonian pictures.

Because T − V is a scalar, once we know it reproduces Newton in Cartesians we may transform to any generalized coordinates without re-deriving the forces, which is the entire practical payoff of the Lagrangian method.

Assumptions

Derivation

We work with a single particle of mass m in one Cartesian coordinate x; the vector and many-particle cases follow index by index. We take the Euler-Lagrange equation and the definitions T = ½mẋ², V = V(x) as given.

Result

Limiting cases

• Free particle (V = const): ∂V/∂x = 0 gives mẸ = 0, uniform motion — Newton's first law as a special case.
• Uniform field (V = mgx): recovers mẸ = −mg, constant acceleration.
• Linear restoring (V = ½kx²): recovers mẸ = −kx, the harmonic oscillator.
• Small velocity, quadratic T: any T reducing to ½mẋ² in Cartesians reproduces the same Newtonian limit, which is why the choice is robust.

Breaks when

Failure modes

• Using L = T + V: writing the Hamiltonian (total energy) where the Lagrangian belongs. This flips the sign of the force and gives runaway/unphysical motion.
• Treating ẋ as dx/dt inside the partial derivatives: x and ẋ are independent slots of L; differentiating them as if related makes ∂T/∂x spuriously nonzero.
• Forgetting the total derivative in d/dt(∂L/∂ẋ): dropping the chain rule when m or the metric depends on position (curvilinear coordinates) loses the connection/centrifugal terms.
• Assuming ∂V/∂ẋ = 0 for magnetic forces: the vector-potential coupling is genuinely velocity-dependent, so this "obvious" simplification is exactly what breaks.
• Confusing generalized force Q_i with Cartesian force F in non-Cartesian coordinates: Q_θ is a torque, not a force, and has different units.

Discussion

The deep point is that the Euler-Lagrange equation is empty of physics until L is specified. Stationary action δS = 0 holds for any functional; nature's content lies entirely in which functional. This derivation shows that, for conservative Newtonian mechanics, the right functional is T − V. The variational framework does not derive L = T − V from first principles — it is reverse-engineered by demanding agreement with the already-known Newtonian equations.

Why the difference T − V rather than the sum? Structurally, T lives in the velocity slot and V in the position slot, so the two derivatives in the Euler-Lagrange equation pick them off separately: the d/dt(∂/∂ẋ) term extracts mẸ from T, and the −∂/∂x term extracts the force from V. The relative minus sign is precisely what aligns those two extractions into mẸ = F. Add them instead and both terms would carry the force with the wrong sign.

The scalar nature of T − V is what makes the whole machine worthwhile. Newton's second law is a vector statement that must be re-projected onto every new coordinate direction and augmented with constraint forces. The Lagrangian, being a single coordinate-free scalar, needs no such bookkeeping: transform T − V to polar, spherical, or generalized constraint coordinates, turn the Euler-Lagrange crank, and the correct equations of motion drop out with the constraint forces already eliminated (via the assumed d'Alembert reduction).

At a foundational level, L is not unique: adding a total time derivative df(x,t)/dt to L leaves the Euler-Lagrange equations unchanged, and multiplying L by a constant rescales the (irrelevant) overall action. So "L = T − V" is really a representative of an equivalence class of Lagrangians all yielding Newton. This gauge freedom becomes physically important in electromagnetism, where the vector-potential coupling qA·ẋ in the generalized potential is defined only up to exactly such a total derivative — the same gauge freedom as in the fields.

Common misconceptions. (1) That L = T − V is a law of nature or a theorem — it is a physically-motivated ansatz validated by reproducing Newton. (2) That the Lagrangian is the energy — it is not; the energy is the Hamiltonian H = T + V, obtained by a Legendre transform. (3) That ∂T/∂x = 0 always — true only when T is written in inertial Cartesians; in curvilinear coordinates T depends on position and that term generates the centripetal acceleration.

Worked examples

Example 1 — Vertical projectile in uniform gravity. A ball of mass m = 0.20 kg moves vertically under gravity g = 9.81 m·sâ»Â². Take upward x positive.

Example 2 — Bead on a frictionless horizontal wire, spring restoring force. Mass m = 0.50 kg, spring constant k = 8.0 N·mâ»Â¹, displacement x from equilibrium.

Problems

1. A particle of mass m = 2.0 kg moves in a potential V(x) = ax³ with a = 3.0 J·mâ»Â³. Using L = T − V, find the equation of motion and the force at x = 2.0 m.
   Solution

   L = ½mẋ² − ax³. Then ∂L/∂ẋ = mẋ, d/dt gives mẸ; ∂L/∂x = −3ax². Euler-Lagrange: mẸ = −3ax², i.e. 2.0 Ẹ = −9.0x². At x = 2.0 m: F = −3ax² = −3(3.0)(4.0) = −36 N, so Ẹ = −18 m·sâ»Â². Units: ax² = J·mâ»Â³Â·m² = J·mâ»Â¹ = N. ✓
2. Show explicitly that L = T + V for a free-falling body gives the wrong sign, and state the resulting (unphysical) motion.
   Solution

   Take L = ½mẋ² + mgx. Then ∂L/∂ẋ = mẋ, d/dt = mẸ; ∂L/∂x = +mg. Euler-Lagrange: mẸ − mg = 0 ⟹ Ẹ = +g, an upward acceleration. The body would accelerate away from Earth — unphysical. This confirms the minus sign in T − V is not a convention but fixes the physics.
3. A particle moves in the plane with T = ½m(ẋ² + áºÂ²) and V = V(x,y). Show the Euler-Lagrange equations reproduce both Cartesian components of Newton's second law.
   Solution

   L = ½m(ẋ²+áºÂ²) − V. For x: ∂L/∂ẋ = mẋ, d/dt = mẸ; ∂L/∂x = −∂V/∂x ⟹ mẸ = −∂V/∂x. For y: identically mỳ = −∂V/∂y. The two equations are independent because T separates into a sum and the coordinates decouple in the kinetic slot; together they are the vector law ma = −∇V.
4. For a single particle, verify that adding a total time derivative df/dt with f = βx (constant β) to L leaves Newton's equation unchanged.
   Solution

   df/dt = βẋ. New Lagrangian L' = ½mẋ² − V + βẋ. Then ∂L'/∂ẋ = mẋ + β; d/dt = mẸ since β is constant. ∂L'/∂x = −∂V/∂x (the βẋ term has no x-dependence). Euler-Lagrange: mẸ = −∂V/∂x, unchanged. The added term shifts the canonical momentum by a constant but not the dynamics — a gauge freedom.
5. A pendulum bob of mass m = 0.30 kg swings on a rigid massless rod of length ℓ = 1.2 m. Using generalized coordinate θ, write L = T − V and obtain the equation of motion; give the small-angle frequency. Explain why the recovered equation is Newton's law in torque form, not force form.
   Solution

   T = ½mℓ²θ̇², V = −mgâ„“cosθ. L = ½mℓ²θ̇² + mgâ„“cosθ. ∂L/∂θ̇ = mℓ²θ̇ (angular momentum), d/dt = mℓ²θ̈; ∂L/∂θ = −mgâ„“sinθ. Euler-Lagrange: mℓ²θ̈ = −mgâ„“sinθ ⟹ θ̈ = −(g/â„“)sinθ. Small angle: θ̈ ≈ −(g/â„“)θ, ω = √(g/â„“) = √(9.81/1.2) = 2.86 rad·sâ»Â¹. The canonical momentum ∂L/∂θ̇ is the angular momentum and ∂L/∂θ is the torque, so the recovered statement is "rate of change of angular momentum = torque" — Newton's law in its rotational (torque) form, because θ is an angular coordinate. Units: mℓ²θ̈ = kg·m²·sâ»Â² = N·m ✓.