Alain J. Brizard Saint Michael's College Lagrangian Mechanics 1 Principle of Least Action The con¯guration of a mechanical system evolving in an n-dimensional space, with coordinates x = (x1 ; x2 ; :::; xn), may be described in terms of generalized coordinates q = (q1 ; q2; :::; q k ) in a k-dimensional con¯guration space, with k < n. The Principle of Least Action (also known as Hamilton's principle) is expressed in terms of a function L(q; q; _ t) known as the Lagrangian, which appears in the action integral A[q] = Z tf _ t) dt; L(q; q; (1) ti where the action integral is a functional of the vector function q(t), which provides a path from the initial point q i = q(ti) to the ¯nal point qf = q(tf ). The variational principle ¯ " à !# Z tf ¯ d @L d @L ¯ j 0 = ±A[q] = A[q + ² ±q]¯ = ±q ¡ dt; ¯ d² ti @ qj dt @ q_j ²=0 where the variation ±q is assumed to vanish at the integration boundaries (±q i = 0 = ±qf ), yields the Euler-Lagrange equation for the generalized coordinate qj (j = 1; :::; k) d dt à @L @ q_ j ! = @L ; @ qj (2) The Lagrangian also satis¯es the second Euler equation à d @L L ¡ q_j dt @ q_ j ! = @L ; @t (3) and thus for time-independent Lagrangian systems (@L=@t = 0) we ¯nd that L¡ q_ j @L=@ q_ j is a conserved quantity whose interpretation will be discussed shortly. The form of the Lagrangian function L(r; r_ ; t) is dictated by our requirement that Newton's Second Law m Är = ¡ rU(r; t) describing the motion of a particle of mass m in a nonuniform (possibly time-dependent) potential U(r; t) be written in the Euler-Lagrange form (2). One easily obtains the form L(r; r_ ; t) = m 2 j_rj ¡ U (r; t); 2 1 (4) which is simply the kinetic energy of the particle minus its potential energy. For a timeindependent Lagrangian (@L=@t = 0), we also ¯nd that the energy function r_ ¢ m 2 @L ¡ L = j_rj + U (r) = E; @ r_ 2 is a constant of the motion. Hence, for a simple mechanical system, the Lagrangian function is obtained by computing the kinetic energy of the system and its potential energy and then construct Eq. (4). 2 Examples The construction of a Lagrangian function for a system of N particles proceeds in three steps as follows. Step I. De¯ne k generalized coordinates fq1 (t); :::; q k(t)g that represent the instantaneous con¯guration of the system of N particles. Step II. Construct the position vector ra (q; t) and its associated velocity va (q; q; _ t) = k X @r a @ra + q_j @t @q j j=1 for each particle (a = 1; :::; N). Step III. Construct the kinetic energy _ t) = K(q; q; 1 X _ t)j 2 ma jv a (q; q; 2 a and the potential energy U (q; t) for the system and combine them to obtain the Lagrangian L(q; q; _ t) = K (q; q; _ t) ¡ U(q; t); from which the Euler-Lagrange equations (2) are derived. 2.1 Example I: Pendulum Consider a pendulum composed of an object of mass m and a massless string of constant length ` in a constant gravitational ¯eld with acceleration g. 2 Although the motion of the pendulum is two-dimensional, a single generalized coordinate is needed to describe the con¯guration of the pendulum: the angle µ measured from the negative y-axis (see Figure above). Here, the position of the object is given as x(µ) = ` sin µ and y(µ) = ¡ ` cos µ; with associated velocity components _ = ` µ_ cos µ and y(µ; _ = ` µ_ sin µ: x(µ; _ µ) _ µ) Hence, the kinetic energy of the pendulum is K = m 2 _2 ` µ ; 2 and choosing the zero potential energy point when µ = 0 (see Figure above), the gravitational potential energy is U = mg` (1 ¡ cos µ): The Lagrangian L = K ¡ U is, therefore, written as _ = m `2µ_ 2 ¡ mg` (1 ¡ cos µ); L(µ; µ) 2 and the Euler-Lagrange equation for µ is @L d = m`2 µ_ ! _ dt @µ or à ! @L = m`2 µÄ _ @µ @L = ¡ mg` sin µ @µ g µÄ + sin µ = 0 ` 3 2.2 Example II: Bead on a Rotating Hoop Consider a bead of mass m sliding freely on a hoop of radius R rotating with angular velocity !0 in a constant gravitational ¯eld with acceleration g. Here, since the bead of the rotating hoop moves on the surface of a sphere of radius R, we use the generalized coordinates given by the two angles µ (measured from the negative z-axis) and ' (measured from the positive x-axis), where '_ = !0 . The position of the bead is given as x(µ; t) = R sin µ cos('0 + !0t); y(µ; t) = R sin µ sin(' 0 + !0t); z(µ; t) = ¡ R cos µ; where '(t) = '0 + !0 t and its associated velocity components are ³ ´ ³ ´ _ t) = R µ_ cos µ cos ' ¡ !0 sin µ sin ' ; x(µ; _ µ; _ t) = R µ_ cos µ sin ' + !0 sin µ cos ' ; y(µ; _ µ; _ t) = R µ_ sin µ; z(µ; _ µ; so that the kinetic energy of the bead is 2³ ´ _ = m jvj 2 = m R µ_ 2 + ! 2 sin2 µ : K(µ; µ) 0 2 2 The gravitational potential energy is U(µ) = mgR (1 ¡ cos µ); where we chose the zero potential energy point at µ = 0 (see Figure above). The Lagrangian L = K ¡ U is, therefore, written as 2³ ´ _ = m R µ_ 2 + !2 sin2 µ ¡ mgR (1 ¡ cos µ); L(µ; µ) 0 2 4 and the Euler-Lagrange equation for µ is @L d 2 _ = mR µ ! dt @ µ_ or à ! @L = mR2 µÄ _ @µ @L = ¡ mgR sin µ @µ + mR2 !02 cos µ sin µ µ ¶ g 2 Ä µ + sin µ ¡ !0 cos µ = 0 R 2.3 Example III: Rotating Pendulum Consider a pendulum of mass m and length b attached to the edge of a disk of radius a rotating at angular velocity ! in a constant gravitational ¯eld with acceleration g. Placing the origin at the center of the disk, the coordinates of the pendulum mass are x = ¡ a sin !t + b cos µ y = a cos !t + b sin µ so that the velocity components are x_ = ¡ a! cos !t ¡ b µ_ sin µ y_ = ¡ a! sin !t + b µ_ cos µ and the squared velocity is v2 = a2!2 + b2 µ_2 + 2 ab ! µ_ sin(µ ¡ !t): Setting the zero potential energy at x = 0, the gravitational potential energy is U = ¡ mg x = mga sin !t ¡ mgb cos µ: 5 The Lagrangian L = K ¡ U is, therefore, written as h i _ t) = m a 2!2 + b2µ_ 2 + 2 ab ! µ_ sin(µ ¡ !t) L(µ; µ; 2 ¡ mga sin !t + mgb cos µ; (5) and the Euler-Lagrange equation for µ is @L = mb2 µ_ + m ab ! sin(µ ¡ !t) ! @ µ_ à ! d @L = mb2 µÄ + m ab ! (µ_ ¡ !) cos(µ ¡ !t) dt @ µ_ and @L = m ab ! µ_ cos(µ ¡ !t) ¡ mg b sin µ @µ or g a µÄ + sin µ ¡ !2 cos(µ ¡ !t) = 0 b b We recover the standard equation of motion for the pendulum when a or ! vanish. Note that the terms [(m=2) a 2!2] and [¡ mga sin !t] in the Lagrangian (5) play no role in determining the dynamics of the system. In fact, as can easily be shown, a Lagrangian L is always de¯ned up to an exact time derivative, i.e., the Lagrangians L and L0 = L+ df=dt, where f (q; t) is an arbitrary function, lead to the same Euler-Lagrange equations. In the present case, f (t) = [(m=2) a2! 2] t + (mga=!) cos !t and thus this term can be omitted from the Lagrangian (5) without changing the equations of motion. 2.4 Example IV: Compound Atwood Machine Consider a compound Atwood machine composed three masses (labeled m1 , m2, and m3 ) attached by two massless ropes through two massless pulleys in a constant gravitational ¯eld with acceleration g. The two generalized coordinates for this system (see Figure) are the distance x of mass m1 from the top of the ¯rst pulley and the distance y of mass m2 from the top of the second pulley; here, the lengths `a and `b are constants. 6 The coordinates and velocities of the three masses m1 , m2, and m3 are x1 = x ! v1 = x; _ x2 = `a ¡ x + y ! v2 = y_ ¡ x; _ x3 = `a ¡ x + `b ¡ y ! v3 = ¡ x_ ¡ y; _ respectively, so that the total kinetic energy is m1 2 m2 m3 K = x_ + (y_ ¡ x) _ 2 + (x_ + y) _ 2: 2 2 2 Placing the zero potential energy at the top of the ¯rst pulley, the total gravitational potential energy, on the other hand, can be written as U = ¡ g x (m1 ¡ m2 ¡ m3 ) ¡ g y (m2 ¡ m3) ; where constant terms were omitted. The Lagrangian L = K ¡ U is, therefore, written as m1 2 m2 m3 L(x; x; _ y; y) _ = x_ + ( x_ ¡ y) _ 2 + (x_ + y) _ 2 2 2 2 + g x (m1 ¡ m2 ¡ m3 ) + g y (m2 ¡ m3) : The Euler-Lagrange equation for x is @L = (m1 + m2 + m3 ) x_ + (m3 ¡ m2) y_ ! @ x_ à ! d @L = (m1 + m2 + m3) xÄ + (m3 ¡ m2 ) yÄ dt @ x_ @L = g (m1 ¡ m2 ¡ m3 ) @x 7 while the Euler-Lagrange equation for y is @L = (m3 ¡ m2) x_ + (m2 + m3) y_ ! @ y_ à ! d @L = (m3 ¡ m2 ) xÄ + (m2 + m3 ) yÄ dt @ y_ @L = g (m2 ¡ m3) @y or (m1 + m2 + m3 ) xÄ + (m3 ¡ m2) yÄ = g (m1 ¡ m2 ¡ m3 ) (m3 ¡ m2) xÄ + (m2 + m3) yÄ = g (m2 ¡ m3) 3 Symmetries and Conservation Laws The Noether theorem states that for each symmetry of the Lagrangian there corresponds a conservation law (and vice versa). When the Lagrangian L is invariant under a time translation, a space translation, or a spatial rotation, the conservation law involves energy, linear momentum, or angular momentum, respectively. When the Lagrangian is invariant under time translations, t ! t + ±t, the Noether theorem states that energy is conserved, dE=dt = 0, where E = dq @L ¢ ¡ L dt @ q_ de¯nes the energy invariant. When the Lagrangian is invariant under spatial translations or rotations, ® ! ® + ±®, the Noether theorem states that the component p® = @L @q @L = ¢ @ ®_ @® @ q_ of the canonical momentum p = @L=@ q_ is conserved, dp® =dt = 0. Note that if ® is a linear spatial coordinate, p® denotes a component of the linear momentum while if ® is an angular spatial coordinate, p® = L³ denotes a component of the angular momentum (with ³ denoting the axis of symmetry about which ® is measured). 8