The Lagrangian Equation INTERMEDIATE MECHANICS WESLEY QUEEN Developing a new perspective Hero of Alexandria (70 A.D.) - light reflections takes the shortest path Pierre de Fermat (1657) - light travels along a path that requires the least time Maupertuis (1747) - action is minimized through the “wisdom of God” Action is a quantity with dimensions of energy x time William Hamilton (1834) – “Of all the possible paths along which a dynamical system may move from one point to another within a specified time interval, the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energies.” The Lagrangian L=T-U T = Kinetic Energy U = Potential Energy Hamilton’s principle asserts that for any given system there is a extremum (maximum or minimum) when the time integral is taken of the difference of Kinetic and Potential Energies. 𝑡2 𝑡1 𝑇 − 𝑈 𝑑𝑡 = Min or Max (Minimum in almost all dynamic systems) Let S = 𝑡2 𝐿 𝑡1 𝑑𝑡 “Action” Calculus of Variations Begun by Newton Developed by Johann Bernoulli, Jakob Bernoulli, and Leonhard Euler Important contributions made by Joseph Lagrange, Hamilton, and Jacobi Leonhard Euler (1707 - 1783) Calculus of Variations Consider a function for any generalized coordinate system 𝑥𝑎 𝑡 = 𝑥0 𝑡 + 𝛼𝛽(𝑡) where 𝑥0 𝑡 produces a min value for S, 𝛼 is a number, and 𝛽(𝑡) is 0 at both end points of our interval. When this is integrated, the t is integrated out and S becomes a number dependent on 𝛼, 𝑡1 , & 𝑡2 . 𝑥 𝑥𝑎 𝑡 = 𝑥0 𝑡 + 𝛼𝛽(𝑡) 𝑡2 How does S depend on 𝛼? 𝜕 𝑆 𝜕𝛼 𝑥𝑎 𝑡 = 𝜕 𝑡2 𝐿 𝜕𝛼 𝑡1 𝑑𝑡 = 𝑡2 𝜕 𝐿 𝑡1 𝜕𝛼 𝑑𝑡 𝑥𝑎 𝑡 = 𝑥0 𝑡 𝑡1 𝜕 𝑆 𝜕𝛼 𝑥𝑎 𝑡 = 𝑡2 𝜕𝐿 𝜕𝑥𝑎 ( 𝑡1 𝜕𝑥𝑎 𝜕𝛼 + 𝜕𝐿 𝜕 𝑥𝑎 )𝑑𝑡 𝜕𝑥𝑎 𝜕𝛼 Use the Chain rule to separate 𝑡 Calculus of Variations Use substitution: From our initial function: 𝑥𝑎 𝑡 = 𝑥0 𝑡 + 𝛼𝛽(𝑡) 𝜕 𝑆 𝜕𝛼 𝜕 𝑆 𝜕𝛼 𝑥𝑎 𝑡 𝑥𝑎 𝑡 𝜕𝑥𝑎 𝜕𝛼 = 𝑡2 𝜕𝐿 𝜕𝑥𝑎 ( 𝑡1 𝜕𝑥𝑎 𝜕𝛼 = 𝑡2 𝜕𝐿 ( 𝛽 𝑡1 𝜕𝑥𝑎 𝜕𝐿 𝜕 𝑥𝑎 + )𝑑𝑡 𝜕𝑥𝑎 𝜕𝛼 + 𝜕𝐿 𝛽)𝑑𝑡 𝜕 𝑥𝑎 = 𝛽 & 𝜕 𝑥𝑎 𝜕𝛼 =𝛽 Calculus of Variations Use integration by parts: 𝜕 𝑆 𝜕𝛼 𝜕 𝑆 𝜕𝛼 𝜕 𝑆 𝜕𝛼 𝑥𝑎 𝑡 = 𝑥𝑎 𝑡 = 𝑥𝑎 𝑡 = 𝑡2 𝜕𝐿 ( 𝛽 𝑡1 𝜕𝑥𝑎 𝑡2 𝜕𝐿 𝛽𝑑𝑡 𝑡1 𝜕𝑥𝑎 𝑡2 𝜕𝐿 ( 𝑡1 𝜕𝑥𝑎 − + 𝜕𝐿 𝛽)𝑑𝑡 𝜕 𝑥𝑎 𝜕𝐿 𝑡 + 𝛽|𝑡21 𝜕 𝑥𝑎 − 𝑑 𝜕𝐿 )𝛽𝑑𝑡 𝑑𝑡 𝜕 𝑥𝑎 Euler’s Equation: 𝜕𝐿 𝜕𝑥0 = 𝑡2 𝑑 𝜕𝐿 ( )𝛽𝑑𝑡 𝑡1 𝑑𝑡 𝜕 𝑥𝑎 Since 𝑥0 𝑡 produces a stationary value 𝜕 for S, 𝑆 𝑥𝑎 𝑡 = 0 𝜕𝛼 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝑥0 A falling object Consider a falling object (no motion in the x or z dimensions) 1 2 𝑇 = 𝑚𝑦 2 Apply Euler’s Equation 𝑑 −𝑚𝑔 = (𝑚𝑦) 𝑑𝑡 1 2 𝐿 = 𝑚𝑦 2 - 𝑚𝑔𝑦 𝑈 = 𝑚𝑔𝑦 𝜕𝐿 𝜕𝑦 = 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝑦 −𝑚𝑔 = 𝑚𝑦 𝑦 = −g The Lagrangian gives us Newton’s second law with respect to gravity. A projectile Consider a projectile launched in the positive x direction at some angle 𝜃, where 0 < 𝜃 < 90. 1 1 𝑇 = 𝑚𝑥 2 + 𝑚𝑦 2 𝑈 = 𝑚𝑔𝑦 2 2 1 1 2 𝐿 = 𝑚𝑥 + 𝑚𝑦 2 − 𝑚𝑔𝑦 2 2 X dimension 𝜕𝐿 𝜕𝑥 = 0= 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝑥 𝑑 (𝑚𝑥) 𝑑𝑡 Y dimension 𝜕𝐿 𝜕𝑦 = 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝑦 −𝑚𝑔 = 𝑑 (𝑚𝑦) 𝑑𝑡 0 = 𝑚𝑥 −𝑚𝑔 = 𝑚𝑦 𝑥=0 𝑦 = −g An Orbiting body y Consider a body orbiting about a another body with a central force acting on it. Position 𝑥 = 𝑟𝑐𝑜𝑠 𝜃 y = 𝑟𝑠𝑖𝑛 𝜃 Velocity M 𝑥 = 𝑟𝑐𝑜𝑠 𝜃 − 𝑟𝜃sin(𝜃) 𝑦 = 𝑟𝑠𝑖𝑛 𝜃 − 𝑟𝜃𝑐𝑜𝑠(𝜃) The Lagrangian 1 2 𝑇 = 𝑚(𝑥 2 + 𝑦 2 ) 1 2 L = 𝑚(𝑥 2 + 𝑦 2 ) + 1 2 r 𝑈=− 𝐺𝑀𝑚 𝑟 L = 𝑚(𝑟 2 + 𝑟 2 𝜃 2 ) + 𝐺𝑀𝑚 𝑟 𝐺𝑀𝑚 𝑟 m 𝜃 x An Orbiting body y Consider a body orbiting about a another body with a central force acting on it. The Lagrangian 1 2 L = 𝑚(𝑟 2 + 𝑟 2 𝜃 2 ) + 𝐺𝑀𝑚 𝑟 r M Apply Euler’s Equations 𝜕𝐿 𝜕𝜃 𝑑 𝜕𝐿 = 𝑑𝑡 𝜕𝜃 𝑑 𝜕𝐿 𝜕𝑟 𝑑 𝜕𝐿 = 𝑑𝑡 𝜕𝑟 0 = 𝑑𝑡(m𝑟 2 𝜃) 𝑚𝑟𝜃 2 − m𝑟 2 𝜃 = r x P 𝑚𝑟𝜃 2 − 𝑟= Shows conservation of Angular momentum 𝐺𝑀𝑚 𝑟2 𝐺𝑀𝑚 𝑟2 𝐺𝑀 − 𝑟2 + 𝑑 = 𝑑𝑡 (m𝑟) = 𝑚𝑟 𝑟𝜃 2 𝑎𝑐 = 𝑣2 𝑟 = 𝑟 2 𝜃2 𝑟 Indicates the acceleration in the direction r is the central force acceleration + the tangential acceleration m 𝜃 x Charge interaction 𝑞𝑏(𝑥𝑏 ,𝑦𝑏 ) Consider 3 charges interacting in a 2 dimensional plane 1 2 1 2 1 2 𝐿 = 𝑚𝑥𝑎2 + 𝑚𝑦𝑎2 − 𝑈= 𝑘𝑞𝑎 𝑞𝑏 + 2 (𝑥𝑏 −𝑥𝑎 )2 +(𝑦𝑏 −𝑦𝑎 ) 𝑘𝑞𝑎 𝑞𝑏 − 2 (𝑥𝑏 −𝑥𝑎 )2 +(𝑦𝑏 −𝑦𝑎 ) = - 𝑘𝑞𝑎 𝑞𝑐 (𝑥𝑐 −𝑥𝑎 ) ((𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2 )3/2 𝑘𝑞𝑎 𝑞𝑏 2 2 cos(𝜃) 𝑏 −𝑥𝑎 ) +(𝑦𝑏 −𝑦𝑎 ) - (𝑥 𝑘𝑞𝑎 𝑞𝑐 𝑞𝑐(𝑥𝑐 ,𝑦𝑐 ) (𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2 𝜕𝐿 𝜕𝑦𝑎 𝑑 𝜕𝐿 𝑑𝑡 𝜕 𝑥𝑎 𝑘𝑞𝑎 𝑞𝑏 (𝑥𝑏 −𝑥𝑎 ) 2 2 3/2 𝑏 −𝑥𝑎 ) +(𝑦𝑏 −𝑦𝑎 ) ) - ((𝑥 (𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2 𝑞𝑎(𝑥𝑎,𝑦𝑎) Y dimension X dimension 𝜕𝐿 𝜕𝑥𝑎 𝑘𝑞𝑎 𝑞𝑐 - = 𝑚𝑥𝑎 𝑘𝑞𝑎 𝑞𝑐 cos(𝜃) (𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2 - = 𝑚𝑥𝑎 - = y 𝜃 x 1 2 𝑇 = 𝑚𝑥𝑎2 + 𝑚𝑦𝑎2 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝑦𝑎 𝑘𝑞𝑎 𝑞𝑏 (𝑦𝑏 −𝑦𝑎 ) ( 𝑥𝑏 −𝑥𝑎 2 + 𝑦𝑏 −𝑦𝑎 2 )3/2 −( 𝑘𝑞𝑎 𝑞𝑐 (𝑦𝑐 −𝑦𝑎 ) 𝑥𝑐 −𝑥𝑎 2 + 𝑦𝑐 −𝑦𝑎 2 )3/2 = 𝑚𝑦𝑎 𝑘𝑞𝑎 𝑞𝑏 𝑘𝑞𝑎 𝑞𝑐 sin(𝜃) − sin(𝜃) 2 2 𝑥𝑏 −𝑥𝑎 + 𝑦𝑏 −𝑦𝑎 𝑥𝑐 −𝑥𝑎 2 + 𝑦𝑐 −𝑦𝑎 2 = 𝑚𝑦𝑎 A rotating pendulum y Consider a pendulum of length 𝑙 moving on a massless ring of radius r and constant angular velocity ω. r Position 𝑥 = 𝑟𝑐𝑜𝑠 ω𝑡 + 𝑙𝑠𝑖𝑛(θ) y = 𝑟𝑠𝑖𝑛 ω𝑡 − 𝑙𝑐𝑜𝑠(θ) ωt x θ Velocity 𝑙 𝑥 = −𝑟ω𝑠𝑖𝑛 ω𝑡 + 𝑙 𝜃𝑐𝑜𝑠(𝜃) 𝑦 = 𝑟𝜔𝑐𝑜𝑠 ω𝑡 + 𝑙 𝜃𝑠𝑖𝑛(𝜃) The Lagrangian 1 2 𝑇 = 𝑚(𝑥 2 + 𝑦 2 ) 1 2 𝑈 = 𝑚𝑔𝑦 1 2 𝐿 = 𝑚(𝑥 2 + 𝑦 2 ) - 𝑚𝑔𝑦 𝐿 = 𝑚[𝑟 2 𝜔2 + 𝑙 2 𝜃 2 + 2𝑟𝜔𝑙 𝜃𝑠𝑖𝑛(𝜃 − 𝜔𝑡)] - 𝑚𝑔(𝑟𝑠𝑖𝑛 ω𝑡 − 𝑙𝑐𝑜𝑠 θ ) m A rotating pendulum y Consider a pendulum of length 𝑙 moving on a massless ring of radius r and constant angular velocity ω. The Lagrangian r 1 2 𝐿 = 𝑚[𝑟 2 𝜔2 + 𝑙 2 𝜃 2 + 2𝑟𝜔𝑙 𝜃𝑠𝑖𝑛(𝜃 − 𝜔𝑡)] - 𝑚𝑔(𝑟𝑠𝑖𝑛 ω𝑡 − 𝑙𝑐𝑜𝑠 θ ) 𝜕𝐿 𝜕𝜃 = 𝑙 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝜃 = 𝑚𝑟𝜔𝑙 𝜃𝑐𝑜𝑠(𝜃 − 𝜔𝑡) - 𝑚𝑔𝑙𝑠𝑖𝑛 θ 𝑑 [𝑚𝑙 2 𝜃 𝑑𝑡 x θ Apply Euler’s Equation 𝜕𝐿 𝜕𝜃 ωt 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝜃 = 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝜃 = 𝑚𝑙 2 𝜃 + 𝑚𝑟𝜔𝑙 𝜃𝑐𝑜𝑠 (𝜃 − 𝜔𝑡)- 𝑚𝑟𝜔2 𝑙𝑐𝑜𝑠 (𝜃 − 𝜔𝑡) + 𝑚𝑟𝜔𝑙𝑠𝑖𝑛(𝜃 − 𝜔𝑡)] m A rotating pendulum y Consider a pendulum of length 𝑙 moving on a massless ring of radius r and constant angular velocity ω. Apply Euler’s Equation r 𝜕𝐿 𝜕𝜃 = 𝑑 𝜕𝐿 𝑑𝑡 𝜕𝜃 ωt 𝑚𝑟𝜔𝑙𝜃𝑐𝑜𝑠(𝜃 − 𝜔𝑡) − 𝑚𝑔𝑙𝑠𝑖𝑛 θ = 𝑚𝑙2 𝜃 + 𝑚𝑟𝜔𝑙𝜃𝑐𝑜𝑠 (𝜃 − 𝜔𝑡)- 𝑚𝑟𝜔2 𝑙𝑐𝑜𝑠 (𝜃 − 𝜔𝑡) 𝑚𝑙 2 𝜃 = 𝑚𝑟𝜔2 𝑙𝑐𝑜𝑠 (𝜃 − 𝜔𝑡) − 𝑚𝑔𝑙𝑠𝑖𝑛 θ x θ 𝑙 Solve for 𝜃 m 𝜃= 𝑟𝜔2 𝑐𝑜𝑠 (𝜃−𝜔𝑡) 𝑔𝑠𝑖𝑛 θ − 𝑙 𝑙 Reduces to simple pendulum when ω = 0. Lagrangian Benefits Deals with derivatives of kinetic and potential energy differences which are invariant to coordinate transformations. Can greatly simplify complicated systems. Allows us to understand mechanical systems where it is difficult to describe all of the forces explicitly. Provides an alternative way to analyze a mechanical system: rather than seeing only cause and effect, we can now analyze a dynamical system by considering action minimization. Sources http://www.people.fas.harvard.edu/~djmorin/chap6.pdf https://en.wikipedia.org/wiki/Leonhard_Euler (photo) Marion, Thornton. Classical Dynamics of Particles and Systems, 4th edition,1995. Harcourt Brace & Co. http://physicsinsights.org/rotating_polar_1.html