Prove principle of least action. Long question M.Sc

The Principle of Least Action (more accurately, the Principle of Stationary Action) states that the path taken by a physical system between two states (at two specified times) is the one for which the action is stationary (i.e., its variation is zero). This path is the solution to the Euler-Lagrange equations. Let's prove it: Step 1: Define the Action Integral. The action S for a system described by generalized coordinates qᵢ and generalized velocities q̇ᵢ is given by the integral of the Lagrangian L over time, from an initial time t₁ to a final time t₂: S = ∫ from t₁ to t₂ of L(qᵢ, q̇ᵢ, t) dt Step 2: State the Principle of Stationary Action. The actual path taken by the system is the one for which the variation of the action (δS) is zero: δS = 0 Step 3: Consider a varied path. Let the actual path be qᵢ(t). Consider a neighboring, virtual path qᵢ'(t) = qᵢ(t) + δqᵢ(t), where δqᵢ(t) is a small, arbitrary variation. The corresponding velocity variation is δq̇ᵢ(t) = d(δqᵢ)/dt. The variations are zero at the endpoints: δqᵢ(t₁) = 0 and δqᵢ(t₂) = 0. Step 4: Express the variation of the action. The variation in the action δS is given by: δS = ∫ from t₁ to t₂ of [ L(qᵢ + δqᵢ, q̇ᵢ + δq̇ᵢ, t) - L(qᵢ, q̇ᵢ, t) ] dt Using a Taylor expansion for small variations, we get: δS = ∫ from t₁ to t₂ of Σ [ (∂L/∂qᵢ)δqᵢ + (∂L/∂q̇ᵢ)δq̇ᵢ ] dt Step 5: Apply the condition δS = 0. For the action to be stationary, δS must be zero: ∫ from t₁ to t₂ of Σ [ (∂L/∂qᵢ)δqᵢ + (∂L/∂q̇ᵢ)δq̇ᵢ ] dt = 0 Step 6: Integrate the second term by parts. Consider the second term in the sum: ∫ from t₁ to t₂ of (∂L/∂q̇ᵢ)δq̇ᵢ dt. Since δq̇ᵢ = d(δqᵢ)/dt, we can write: ∫ from t₁ to t₂ of (∂L/∂q̇ᵢ) (d(δqᵢ)/dt) dt Using integration by parts (∫ u dv = uv - ∫ v du), with u = (∂L/∂q̇ᵢ) and dv = (d(δqᵢ)/dt) dt: [ (∂L/∂q̇ᵢ)δqᵢ ] from t₁ to t₂ - ∫ from t₁ to t₂ of (d/dt)(∂L/∂q̇ᵢ) δqᵢ dt Step 7: Apply boundary conditions. Since the variations δqᵢ are zero at the endpoints (δqᵢ(t₁) = 0 and δqᵢ(t₂) = 0), the first term vanishes: [ (∂L/∂q̇ᵢ)δqᵢ ] from t₁ to t₂ = (∂L/∂q̇ᵢ)|_(t₂) δqᵢ(t₂) - (∂L/∂q̇ᵢ)|_(t₁) δqᵢ(t₁) = 0 - 0 = 0 Step 8: Combine terms to derive the Euler-Lagrange Equations. Substituting the result from Step 7 back into the expression for δS (from Step 5): ∫ from t₁ to t₂ of Σ [ (∂L/∂qᵢ)δqᵢ - (d/dt)(∂L/∂q̇ᵢ) δqᵢ ] dt = 0 ∫ from t₁ to t₂ of Σ [ (∂L/∂qᵢ) - (d/dt)(∂L/∂q̇ᵢ) ] δqᵢ dt = 0 Since the variation δqᵢ is arbitrary and non-zero between t₁ and t₂, the term in the square brackets must be zero for each generalized coordinate qᵢ. This leads to the Euler-Lagrange equations: d/dt (∂L/∂q̇ᵢ) - ∂L/∂qᵢ = 0 These equations describe the equations of motion for the system, proving that the path for which the action is stationary is indeed the physical path. 3 done, 2 left today. You're making progress. Send me the next one 📸