Lagrangian Mechanics

Alain J. Brizard Saint Michael's College

1Principle 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;:::; qk)inak-dimensional con¯guration space, with kfunction L(q; q_ ; t)knownastheLagrangian,whichappears in the action integral Z tf A[q]= L(q; q_ ; t) dt; (1) ti where the action integral is a functional of the vector function q(t), which provides a path from the initial point qi = q(ti)tothe¯nal point qf = q(tf). The variational principle ¯ " Ã !# ¯ Z d ¯ tf @L d @L ¯ j 0=±A[q]= A[q + ²±q]¯ = ±q j ¡ j dt; d² ²=0 ti @q dt @q_ where the variation ±q is assumed to vanish at the integration boundaries (±qi =0=±qf), yields the Euler-Lagrange equation for the generalized coordinate qj (j =1; :::; k) Ã ! d @L @L = ; (2) dt @q_j @qj The Lagrangian also satis¯esthesecond Euler equation Ã ! d @L @L L ¡ q_j = ; (3) dt @q_j @t and thus for time-independent Lagrangian systems (@L=@t =0)we¯nd that L¡q_j @L=@q_j is a conserved quantity whose interpretationwill be discussed shortly. The form of the Lagrangian function L(r; r_; t)isdictated 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)bewritten in the Euler-Lagrange form (2). One easily obtains the form m L(r; r_; t)= jr_j2 ¡ U (r;t); (4) 2

1 which is simply the kineticenergyofthe particle minus its potential energy. For a time- independent Lagrangian (@L=@t =0),wealso¯ndthatthe energy function @L m r_ ¢ ¡ L = jr_j2 + 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).

2Examples

The construction ofaLagrangian function for a system of N particles proceeds in three steps asfollows. Step I. De¯ne k generalized coordinates fq1(t);:::;qk(t)g that represent the instanta- neous con¯guration of the system of N particles.

Step II. Construct the position vector ra(q; t)anditsassociated velocity

Xk @ra j @ra va(q; q_ ; t)= + q_ j @t j=1 @q for each particle (a =1;:::; N). Step III. Construct the kinetic energy X 1 2 K(q; q_ ; t)= ma jva(q; q_ ; t)j 2 a and the potentialenergyU (q; t)forthesystem 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.

2.1Example 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

x_(µ;µ_)=` µ_ cosµ andy _(µ;µ_)=` µ_ sin µ:

Hence, the kinetic energy of the pendulum is m K = `2µ_2; 2 and choosing the zero potential energy point when µ =0(seeFigure above), the gravitational potential energy is U = mg` (1 ¡ cos µ): The Lagrangian L = K ¡ U is, therefore, written as m L(µ;µ_)= `2µ_2 ¡ mg` (1 ¡ cosµ); 2 and the Euler-Lagrange equation for µ is Ã ! @L d @L = m`2 µ_ ! = m`2 µÄ @µ_ dt @µ_ @L = ¡ mg` sin µ @µ or 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 aconstantgravitational ¯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.Theposition of the bead is given as

x(µ;t)=R sinµ cos('0 + !0t);

y(µ;t)=R sinµ sin('0 + !0t); z(µ;t)=¡ R cos µ; where '(t)='0 + !0 t and its associated velocity components are ³ ´ _ _ x_(µ;µ; t)=R µ cos µ cos ' ¡ !0 sin µ sin ' ; ³ ´ _ _ y_(µ;µ; t)=R µ cos µ sin' + !0 sin µ cos ' ; z_(µ;µ_; t)=R µ_ sin µ; so that the kinetic energy of the bead is m mR2 ³ ´ K(µ;µ_)= jvj2 = µ_2 + !2 sin2 µ : 2 2 0 The gravitational potential energy is U(µ)=mgR (1 ¡ cos µ); where wechosethezeropotential energy point at µ =0(seeFigure above). The Lagrangian L = K ¡ U is, therefore, written as mR2 ³ ´ L(µ;µ_)= µ_2 + !2 sin2 µ ¡ mgR (1 ¡ cos µ); 2 0

4 and the Euler-Lagrange equation for µ is Ã ! @L d @L = mR2 µ_ ! = mR2 µÄ @µ_ dt @µ_ @L = ¡ mgR sin µ @µ 2 2 + mR !0 cos µ sin µ or µ ¶ g µÄ +sinµ ¡ !2 cos µ =0 R 0

2.3Example III: Rotating Pendulum

Consider apendulum of mass m and length b attached to the edge of a disk of radius a rotating at angular velocity ! in aconstantgravitational ¯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_ sinµ y_ = ¡ a! sin !t + b µ_ cos µ and the squared velocity is v2 = a2!2 + b2µ_2 +2ab! µ_ sin(µ ¡ !t): Setting the zero potential energy at x =0,thegravitational potential energy is U = ¡ mg x = mga sin !t ¡ mgb cos µ:

5 The Lagrangian L = K ¡ U is, therefore, written as m h i L(µ;µ_; t)= a2!2 + b2µ_2 +2ab ! µ_ sin(µ ¡ !t) 2 ¡ mga sin !t + mgb cos µ; (5) and the Euler-Lagrange equation for µ is @L = mb2 µ_ + mab! sin(µ ¡ !t) ! _ @µÃ ! d @L = mb2 µÄ + mab!(µ_ ¡ !)cos(µ ¡ !t) dt @µ_ and @L = mab!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)a2!2]and[¡ mga sin !t]intheLagrangian (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)isanarbitrary function, lead to the same Euler-Lagrange equations. In the presentcase, 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,andm3) attached by two massless ropes through two massless pulleys in a constant gravitational ¯eld with acceleration g.Thetwogeneralized 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,andm3 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 m2 m3 K = x_ 2 + (_y ¡ x_ )2 + (_x +_y)2 : 2 2 2 Placing the zero potential energy at thetopof the ¯rst pulley, the total gravitational potential energy, on the other hand, can be written as

U = ¡ gx (m1 ¡ m2 ¡ m3) ¡ gy (m2 ¡ m3) ; where constant terms were omitted. The Lagrangian L = K ¡ U is, therefore, written as

m1 m2 m3 L(x; x;_ y; y_)= x_ 2 + (_x ¡ y_)2 + (_x +_y)2 2 2 2 + gx (m1 ¡ m2 ¡ m3)+gy (m2 ¡ m3) : The Euler-Lagrange equation for x is @L =(m1 + m2 + m3)_x +(m3 ¡ m2)_y ! @x_Ã ! d @L =(m + m + m )Äx +(m ¡ m )Äy dt @x_ 1 2 3 3 2 @L = g (m ¡ m ¡ m ) @x 1 2 3

7 while the Euler-Lagrange equation for y is @L =(m3 ¡ m2)_x +(m2 + m3)_y ! @y_Ã ! d @L =(m ¡ m )Äx +(m + m )Äy dt @y_ 3 2 2 3 @L = g (m ¡ m ) @y 2 3 or

(m1 + m2 + m3)Äx +(m3 ¡ m2)Äy = g (m1 ¡ m2 ¡ m3)

(m3 ¡ m2)Äx +(m2 + m3)Äy = g (m2 ¡ m3)

3SymmetriesandConservation Laws

The Noether theorem states that for each symmetry of the Lagrangian there corresponds aconservation 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,theNoether theorem states that energy is conserved, dE=dt =0,where dq @L E = ¢ ¡ L dt @q_ de¯nes the energy invariant. When the Lagrangian is invariant under spatial translations or rotations, ® ! ® + ±®,theNoether theorem states that the component @L @q @L p = = ¢ ® @®_ @® @q_ of the canonical momentum p = @L=@q_ is conserved, dp®=dt =0.Notethatif® is a linear spatial coordinate, p® denotes a component of the linear momentum while if ® is an angular spatialcoordinate, p® = L³ denotes a component of the angular momentum (with ³ denoting the axis ofsymmetry about which ® is measured).