Chapter 10 Notes: Lagrangian Mechanics

Chapter 10 Notes: Lagrangian Mechanics Thus far we have solved problems by using Newton s Laws (a vector approach) or energy conservation (a scalar approach.) In this chapter we approach the subject in a very different fashion, and one that initially seems far from evident. We will still use kinetic and potential energies, but will define the Lagrangian, L = T V from which we can generate the equations of motion by doing simple derivatives The power of this approach will become evident in examples. The development of this approach spanned the work of Wilhelm von Leibniz(1646-1716), Johann Bernoulli (1667-1748), Jean LeRond D Alembert (1717-1783), and Joseph Louis de Lagrange(1736-1813). We begin with William Rowan Hamilton s (1805-1865) Variational Principle espoused subsequent to the work of those previously mentioned. When the Lagrangian method works it is very slick. So why use Newton s methods at all? The Lagrangian method is particularly powerful when dealing with a conservative system (and indeed that is the only application we will do.) If what we want is an equation of motion, Lagrange works well when we also want to find a constraint force such as a normal force, Lagrangian life may be tougher than Newtonian life. When non-conservative forces or velocity dependent forces are present Newtonian methods are generally preferred. 1 Hamilton s Variational Principle Consider a conservative system. Define the Lagrangian of a system as L = T V (1) the difference between kinetic and potential energies. To do this we must already know expressions for the energies, and this has developed from the Newtonian perspective. Next suppose that the system evolves from time to t 2 along a particular trajectory, that is a particular y(t), ẏ(t) where for simplicity I am considering 1D motion. The starting and 1

ending points are fixed, but we can imagine that there are many paths that can be taken between these fixed points as shown in Figure 10.1.1. Hamilton s Variational Principle states that the integral of the Lagrangian along the actual path of motion is an extremum (maximum or minimum) along the actual path taken. In symbols we evaluate the integral and say that J = L dt (2) δj = δ L dt = 0 (3) The notation may seem a little odd. Recall that if we have a function of a variable, F (x), its extrema are found by df/dx = 0. We are now expressing something slightly different in Equation 3: the integral is a function of two parameters, y, ẏ (position and velocity) represented in 1D by paths on the y t, ẏ t graphs. A change in the parameter, i.e. a change in the path taken between fixed start and end points, will change the result of the integral, and this integral is an extremum (maximum or minimum) for the actual path traveled. E.g. Consider the case of an object falling near the earth in 1D motion. We will show first that Hamilton s Principle leads to the usual differential equation. We have kinetic energy T = mẏ 2 /2 and potential energy V = mgy so L = 1 2 mẏ2 mgy (4) To evaluate the variation we consider δy, δẏ to be slight variations in position and time from the extremum value. Thus 1 [ ] 1 t2 δj = δ 2 mẏ2 mgy dt = [m ẏ δẏ mg δy] dt (5) where this equation illustrates how to bring the variation inside the integral. Now δẏ = d(δy) dt We evaluate the first term using integration by parts as m ẏ δẏ dt = mẏ δy t t2 2 1 This is sort of like a total differential. df = F F dẏ + dy. ẏ y (6) mÿ δy dt (7) 2

Since the endpoints are fixed, with no variation, the first term on the right must be zero. Hence δj = Since δy is arbitrary, this variation can be zero only if This is our usual differential equation of motion. E.g Come again? How does the δy stuff work? [ mÿ mg] δy dt = 0 (8) mÿ mg = 0 (9) If we have initial conditions of t = 0, y = 0, ẏ = 0 the solution to the differential equation is y(t) = gt 2 /2, ẏ = gt. Now we write the variation from this solution as y(α, t) = y(0, t) + αη(t) (10) Here η(t) represents the variation, and is any function of time that (a) has a continuous first derivative (velocity) on the time interval [, t 2 ], and (b) has values η( ) = 0 = η(t 2 ) so that the starting and ending points are fixed. The parameter α is the strength of the variation. When α = 0 we have our true solution. From Equation 10 we have and η( ) = 0 = η(t 2 ) ẏ(α, t) = ẏ(0, t) + α η(t) (11) Using these expressions we can get the kinetic and potential energies and hence the Lagrangian. T = 1 2 m [ẏ(0, t) + α η(t)]2 = 1 2 m [ gt + α η(t)]2 = 1 2 mg2 t 2 mg t α η(t) + 1 2 mα2 η 2 (t) (12) V = mg [y(0, t) + αη(t)] = 1 2 mg2 t 2 + mg α η(t) (13) { L = m g 2 t 2 αg [t η(t) + η(t)] + 1 } 2 α2 η 2 (t) (14) Now evaluate the integral of the Lagrangian. First evaluate the term linear in α, integrating by parts. [t η(t) + η(t)] dt = tη(t) t t2 2 η(t)dt + η(t)dt = 0 (15) This means that J(α) = 1 3 g2 (t 3 2 t 3 1) + 1 2 α2 3 η 2 (t)dt (16)

Now for any function η(t), η 2 (t) is positive, and the latter integral must be positive, so J(α) must be parabolic in α. The minimum in this function occurs for α = 0 as expected from Hamilton s Principle. E.g. Motion of a particle in force-free region. We have Newton s First Law that says the motion should be a straight line. We will use Hamilton s Principle to show that if we assume that the motion is sinusoidal, the sinusoid must have zero amplitude, and hence be straight line motion between the start and end. We suppose the particle to move from t = 0, x = 0, y = 0 to t = x 1 /v x, x = x 1, y = 0. The components of the position during motion are (Like α in the previous exam- where η, the amplitude of the sinusoid, can be varied. ple.) x = v x t (17) y = ±η sin πv xt x 1 (18) The potential energy V is constant, so L = 1 ( ) ] 2 [v 2 m x 2 ηπvx + cos 2 πv xt V (19) x 1 x 1 Evaluating J(η), and this is a minimum for η = 0. J(η) = mv xx 1 2 V x 1 v x + η 2 mv xπ 2 4x 1 (20) 2 Generalized Coordinates, Degrees of Freedom, Holonomic and Non-Holonomic Constraints Next in the development is a discussion of generalized coordinates. Recall Chapter 1 when we defined various 3D orthogonal coordinate systems, cartesian (x, y, z), cylindrical polar (r, θ, z) and spherical polar (r, θ, φ). These are all orthogonal meaning that ê a ê b = 0 where a and b are two of the unit vectors for a particular coordinate system. Generalized coordinates by contrast do not need to be orthogonal, but are chosen to best represent a complicated system. 4

E.g. A pendulum constrained to move in the x y plane. Initially we might choose cartesian (x, y, z) coordinates, however they do not encapsulate two constraints: the plane of oscillation is x y and the length of the pendulum, r, is fixed: i.e. z = 0 and r 2 x 2 y 2 = 0. It is more convenient to choose cylindrical polar coordinates with z = 0, i.e. plane polar coordinates. The second constraint is r = constant meaning that there is only one variable, θ, and we can describe the motion as one dimensional meaning a single variable describes the motion. Similarly motion of a roller coaster car is one dimensional. Generalized coordinates q i are a set of coordinates that are independent, and just sufficient to uniquely specify the configuration of a system. Degrees of freedom are the number of variables that are needed to describe a system. Thus a single particle in 3D space has 3 degrees of freedom (x, y, z) while two particles in 3D space have 6 degrees of freedom, (x 1, y 1, z 1, x 2, y 2, z 2 ). However a rigid dumbbell composed of two particles separated by a rigid rod has only 5 degrees of freedom. A good choice of generalized coordinates for the dumbbell would be (x, y, z) for the center of mass plus (θ, φ) for the orientation of the rod joining the particles (see Figure 10.2.2). Conditions of constraint are statements about limitations on motion of a system. They are categorized into holonomic and non-holonomic types. Holonomic constraints can be represented by an equality involving coordinates such as d 2 [ (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2] = 0. Being on a surface, or having a fixed separation between two particles are examples. Non-holonomic constraints include those that involve an inequality (being outside the earth) or involving velocity constraints. We will focus on holonomic constraints and leave non-holonomic constraints for an advanced mechanics course. If we have N particles and m holonomic constraints, there are 3N m degrees of freedom and the same number of generalized coordinates. 3 Many Paths to One Answer: Getting T and V in terms of generalized coordinates Here we will look at a more complicated system (one we would NOT want to tackle with just Newton s Laws) and get the same result three different ways. The system consists of two particles. One has mass M and is constrained to move in a straight line on a frictionless surface. Attached to it is a simple pendulum of length r with mass m constrained to swing in a plane. We want to find kinetic and potential energies so that we can construct the Lagrangian. 5

The natural choice for generalized coordinates are the position of the first mass, X, and the angle that the pendulum makes with the vertical, θ. Refer to Figure 10.3.1. Path 1 Start with the cartesian coordinates, (X, Y, Z) and (x, y, z). Constraints can be written Y = Z = z = 0 and (x X) 2 + y 2 r 2 = 0. This means we have two generalized coordinates (duh, we just said that!) In terms of cartesian coordinates, T = 1 2 MẊ2 + 1 2 m(ẋ2 + ẏ 2 ) V = mgy (21) Now we write transformation equations from the cartesian coordinates to the generalized coordinates. Hence and using these in Equations 21 X = X (22) x = X + r sin θ (23) y = r cos θ (24) (25) Ẋ = Ẋ (26) ẋ = Ẋ + r θ cos θ (27) ẏ = r θ sin θ (28) T = 1 2 MẊ2 + 1 [Ẋ2 2 m + (r θ) 2 + 2Ẋr θ ] cos θ (29) (30) V = mgr cos θ (31) Using the coordinate transformations is usually necessary to get the potential energy. There are alternate ways to get the kinetic energy. Notice that The kinetic energy term has a cross term in it. The generalized coordinates are not orthogonal. The potential energy depends on a single generalized coordinate (this is not true in general.) 6

The variable X does not appear. Path 2 Using generalized coordinates at the start to get velocities. The velocity of M is V M = îẋ and the velocity of m is the sum of the velocity of M and the velocity of m relative to M. v m = V M + v m,rel = îẋ + ê θr θ (32) Then since ê θ = cos θ î + sin θ ĵ, ẋ = Ẋ + r θ cos θ and ẏ = r θ sin θ as in Path 1. Path 3 Similar to Path 2 If we write T = 1 2 M V M V M + 1 2 m v m v m (33) then using î ê θ = cos θ we find the expression for kinetic energy in Equation 30. Usually either Path 1 or Path 3 will be helpful. 4 Lagrange s Equations of Motion Now we apply Hamilton s Variational Principle to the situation of a Lagrangian as a function of generalized coordinates (q i, q i, i = 1 N) and generate the Lagrangian Equations of motion. δj = δ L dt = δl dt = The q i are parameters that are functions of time, and we can write i δ q i = d(δq i) dt ( δq i + ) δ q i dt = 0 (34) q i q i We can use this to evaluate the second term of Equation 34 using integration by parts. ( ) δ q i dt = t t2 2 ( ) d δq i δq i dt (36) q i i q i i dt q i i The first term vanishes at the endpoints (δq 1 = δq 2 = 0) and so δ L dt = i [ q i d dt ( q i (35) )] δq i dt = 0 (37) 7

The generalized coordinates are independent, and so each δq i is independent of the other variations. Thus to ensure that the integral is identically zero for all variations the item in square brackets must be zero for each generalized coordinate. The Lagrange equations of motion are thus This can also be written as q i d dt q i = d dt ( ) = 0 i = 1, 2,, N (38) q i ( ) q i i = 1, 2,, N (39) In Section 6 we will define a conjugate generalized momentum for each generalized coordinate, p i = q i and the equations of motion can be written ṗ i dp i dt = q i (40) Equations 38, 39, and 40 are all equivalent. Now we need to see how to use them. The equations are sometimes called the Euler-Lagrange 5 Some Applications of the Lagrange Formulation The general strategy consists of 1. Select a suitable set of generalized coordinates 2. Find transformation equations between the cartesian and generalized coordinates 3. Write the kinetic energy as a function of generalized coordinates using the ideas from Section 3, Path 1 or Path 3 4. Write the potential energy as a function of generalized coordinates 5. Construct the Lagrangian and do derivatives to get the equations of motion (use one of Equations 38, 39, and 40.) 5.1 Using Lagragian to get things we already know E.g. 1 Free particle Yup this is boring, but let s make sure it works for a particle moving in 1D with no forces, therefore no potential energy. There is only one generalized coordinate, x. The Lagrangian is simply L = T = 1 2 mẋ2. 8

Then the partial derivatives are x i = 0 and ẋ i = mẋ and the equation of motion is 0 = d (mẋ) (41) dt i.e. linear momentum is constant, just a restatement of Newton s First Law. E.g. 2 Projectile Motion in Uniform g Imagine motion in the x z plane. The generalized coordinates are just (x, z) and we can quickly write the Lagrangian The partials are L = 1 2 m(ẋ2 + ż 2 ) mgz (42) x = 0 z = mg ẋ i = mẋ ż i = mż (43) so the equations of motion are 0 = d (mẋ) i.e. horizontal momentum is constant (44) dt and these are just what we get from Newton s Laws mg = d (mż) = m z (45) dt E.g. 3 Harmonic Oscillator in 1D We have a single generalized coordinate x representing the displacement of the mass from the unstretched location of the spring. The Lagrangian is L = 1 2 mẋ2 1 2 kx2 and the partials are x = kx and our equation of motion is simply ẋ = mẋ (46) which is just our usual differential equation. kx = mẍ (47) E.g. 4 Central Force in 2D Here we have a constraint, z = 0 so we need (3-1) = 2 generalized coordinates. We choose the polar coordinates (r, θ) and have transformation equations (Path 1) x = r cos θ y = r sin θ ẋ = ṙ cos θ r θ sin θ ẏ = ṙ sin θ + r θ cos θ (48) 9

Since the potential energy comes from a central force, it only depends on r. The kinetic energy can be calculated by Path 1 by (1/2)(ẋ 2 + ẏ 2 ) using the above transformations (cross terms cancel) resulting in the Lagrangian L = 1 2 m (ṙ 2 + r 2 θ2 ) V (r) (49) The text illustrates getting the Lagrangian using Path 3. The partials are ṙ = mṙ r = mr θ 2 V r θ = mr2 θ Recognizing that f(r) = V/ r we write the equations of motion as m r = mr θ 2 + f(r) These are familiar from Chapter 6. θ = 0 (50) d dt (mr2 θ) = 0 (51) Notice that anytime a generalized coordinate (like θ here) does not appear in the Lagrangian, then the conjugate momentum, / q i (yet to be discussed) is constant. In this example the angular momentum, mr 2 θ is constant. 5.2 More difficult problems E.g. 5 The Complete Atwood s Machine Consider a pulley mounted on a fixed, frictionless bearing having moment of inertia I and radius a. A string of length l connects two blocks of masses m 1, m 2 that can only move vertically. We neglect any air resistance and assume an ideal string (massless and non-stretchy) that does not slip on the pulley. This is shown in figure 10.5.1. There are 5 holonomic constraints, y 1 = z 1 = y 2 = z 2 = 0 produce 1D motion, and l = constant for the string length. So we need (3 2 5 = 1) generalized coordinates. (Or if you include the pulley, with three holonomic constraints on position of pulley, (3 3 8 = 1) generalized coordinates.) We use x for the position of m 1 and then the position of m 2 is (l πa x). We also know that since the string does not slip, ω = ẋ/a. The Lagrangian is then L = 1 (m 1 + m 2 + Ia ) 2 2 ẋ 2 + m 1 gx + m 2 g(l πa x) (52) The partials are x = (m 1 m 2 )g 10 (m ẋ = 1 + m 2 + Ia ) 2 ẋ (53)

Leading to the equation of motion (m 1 + m 2 + Ia 2 ) ẍ = (m 1 m 2 )g (54) meaning that the acceleration is constant with value This should be familiar from Physics 312. (m 1 m 2 )g m 1 + m 2 + I a 2 (55) E. g. 6 The Double Atwood Machine Replace the second object by a second Atwood s machine with masses m 2, m 3. Simplify the problem by making the two moments of inertia zero, and the pulley radii negligible. The string lengths will be l and l. See Figure 10.5.2 in text. The generalized coordinates will be x, the position of m 1 relative to the first pulley, and x, the position of m 2 relative to the moving pulley. The holonomic constraints are for the three masses and the moving pulley, with no motion in the y, z directions (8 constraints) plus constraints on the string lengths (2 constraints.) Verifying, we need (3(4) - 8-2 = 2) generalized coordinates. Path 1 is easy. x 1 = x x p = l x x 2 = x p +x = l x+x x 3 = x p +(l x ) = l+l x x The text writes the constraints as (56) x 1 + x p l = 0 (57) x 2 + x 3 2x p l = 0 (2x 1 + x 2 + x 3 ) (2l + l ) = 0 (58) Here we have defined down as positive x, so the potential energy is V = m 1 gx 1 m 2 gx 2 m 3 gx 3. Hence we get the Lagrangian L = 1 2 m 1ẋ 2 + 1 2 m 2( ẋ+ẋ ) 2 + 1 2 m 3(ẋ+ẋ ) 2 +(m 1 m 2 m 3 )gx+(m 2 m 3 )gx +Constants The partials are x = (m 1 m 2 m 3 )g ẋ = m 1ẋ m 2 ( ẋ + ẋ ) + m 3 (ẋ + ẋ ) 11 (59)

x = (m 2 m 3 )g and the equations of motion are ẋ = m 2 ( ẋ + ẋ ) + m 3 (ẋ + ẋ ) (60) (m 1 + m 2 + m 3 )ẍ + ( m 2 + m 3 )ẍ = (m 1 m 2 m 3 )g (61) ( m 2 + m 3 )ẍ + (m 2 + m 3 )ẍ = (m 2 m 3 )g (62) These simultaneous equations can be easily solved, although the reult is messy with symbols. Try it with numbers. Let m 1 = 5 kg, m 2 = 2 kg, and m 3 = 3 kg. Find the accelerations ẍ and ẍ and then find the acceleration of each mass. Next try reversing the values for m 2, m 3. E. g. 7 Block Sliding on a Frictionless Wedge that is Free to Move on Frictionless Table A wedge of mass M and angle θ can slide freely in 1D on a horizontal frictionless table. A box of mass m slides along the frictionless incline of the wedge. Find the equations of motion from the Lagrangian. As suggested in Figure 10.5.3 we will have two generalized coordinates, x for the horizontal location of the wedge and x for the position of the box relative to the top of the wedge. We will use Path 3 to get the kinetic energy. The velocity of the wedge is V = ẋî and the velocity of the box is v = ẋî + ẋ î. The kinetic energies are T M = 1 2 M V V = 1 2 Mẋ2 T m = 1 2 m v v = 1 2 mẋ2 + 1 2 mẋ 2 + mẋẋ cos θ (63) The potential energy is V = mg x sin θ and the Lagrangian is L = 1 2 Mẋ2 + 1 2 mẋ2 + 1 2 mẋ 2 + mẋẋ cos θ + mg x sin θ (64) with partials x = 0 ẋ = (M + m)ẋ + mẋ cos θ (65) x = mg sin θ ẋ = mẋ + mẋ cos θ (66) With the first partial being zero, this means that (M + m)ẋ + mẋ cos θ = Constant. In a moment we will refer to this a a generalized momentum (conjugate to x) which in this case is conserved. For this problem this means that the horizontal net momentum is constant. The two equations of motion are (m + M)ẍ + (m cos θ)ẍ = 0 ẍ + ẍ cos θ = g sin θ (67) 12

and we can solve these simultaneous equations to yield ẍ = g sin θ cos θ (1 + M/m) cos 2 θ ẍ = g sin θ 1 m cos 2 θ/(m + M) (68) Try to think of all limiting cases to check the validity of these equations. 6 Generalized Momenta and Ignorable Coordinates We now have several examples of Lagrangian formulation. In several of the examples the Lagrangian did not include one of the generalized coordinates (look at example 7, x does not appear in L), and in these cases the text refers to the coordinate as ignorable. Whenever we have an ignorable coordinate q, / q = 0 and therefore / q is constant. We define the generalized momentum conjugate to the generalized variable q (conjugate momentum for short) as p q = (69) q and thus can write the equation of motion as ṗ q dp q dt = q If q has units of length, its conjugate momentum has units of momentum, kg m/s. If q is an angle, the conjugate momentum has units of angular momentum, kg m 2 /s. In some cases they clearly represent a momentum or momentum component that we might write down directly (example 1-4, in the latter case p r is the radial component of momentum, p θ is the angular momentum) while in other cases the conjugate momentum is not as obvious (Examples 5-7). Generalized coordinates and their conjugate momenta become central in our definition of the Hamiltonian at the end of this chapter. E.g. 8 Pendulum Attached to Movable Support In Section 3 we began discussing a mass M constrained to move in a straight line on a frictionless surface with a simple pendulum of length r, mass m attached to it. The generalized coordinates were X and θ and the Lagrangian was L = 1 2 MẊ2 + 1 [Ẋ2 2 m + (r θ) 2 + 2Ẋr θ ] cos θ + mgr cos θ (71) X is ignorable so (70) p X = (M + m)ẋ + mr θ cos θ = Constant (72) 13

and the equations of motion are (M + m)ẍ + mr θ cos θ mr θ 2 sin θ = 0 (73) d ( ) r 2 θ + Ẋr cos θ = dt (Ẋr θ + gr) sin θ or θ + Ẍ r cos θ + g sin θ = 0 (74) r The last two equations can be combined to eliminate Ẍ and yield a rather ugly looking second order differential equation in θ. The text makes some simplifying assumptions to see if these equations make sense. Read and understand these E.g. 9 Spherical Pendulum Including Approximations Return to the simple pendulum, a mass m attached to a string of length l that s fixed at the other end, but now let the pendulum be free to move in 3D. This problem is identical to that of a spherical bowl of radius r = l with a frictionless bar of soap free to move in it. There are two degrees of freedom, (θ, φ), the usual spherical polar variables. The velocity (Equation 1.12.12) is v = ê θ l θ + ê φ l sin θ φ (75) If we choose the lowest position for V = 0, the height is y = l(1 cos θ) and the Lagrangian is L = 1 2 ml2 ( θ 2 + φ 2 sin 2 θ) mgl(1 cos θ) (76) Since φ is ignorable we have conservation of the conjugate momentum, p φ = ml 2 φ sin 2 θ = constant ml 2 S (77) φ sin 2 θ = S (78) where we have defined a new constant S = φ sin 2 θ. The remaining equation of motion is then d ( ) ml 2 θ = ml 2 θ = ml 2 dt φ2 sin θ cos θ mgl sin θ or θ + g cos θ sin θ S2 l sin 3 θ = 0 (79) We shall NOT try to solve this in general, but will look at three special cases: (a) simple plane pendulum, φ = 0, (b) Conical pendulum, θ = 0, and (c) slight perturbation from a conical pendulum. 14

1. For the simple plane pendulum, φ = 0, we have S = 0 and the equation of motion becomes θ + g sin θ = 0 (80) l which is just what we have seen before for the plane pendulum. So we gain confidence that our derivation is correct. 2. For a conical pendulum, θ = 0, the bob moves in a circle with the string having a constant angle θ 0 and the equation of motion becomes or, putting in the expression for S g cos θ sin θ S2 l sin 3 θ = 0 φ 2 = g l cos θ = g sec θ (81) l and this gives the required angular velocity φ for the conical motion. 3. Now suppose we disturb the conical pendulum slightly from the above value. To indicate the perfect conical pendulum value we will add a subscript φ 2 0 = Also we will express the ideal conical value of g l cos θ 0 = g l sec θ 0 (82) S 2 0 = g l sin4 θ 0 sec θ 0 (83) Also we will indicate the period of the circular motion of the conical pendulum as T 0 = 2π φ l cos θ 0 = 2π (84) g We assume that the perturbation will not change the value of S significantly (nor will it change the period T 0 ) and write the equation of motion as θ + g ( ) sin θ sin4 θ 0 cos θ l cos θ 0 sin 3 = 0 (85) θ Now consider the expression in parentheses f(θ) = sin θ sin4 θ 0 cos θ 0 cos θ sin 3 θ Do a Taylor series expansion about θ = θ 0 to first order to get (86) f(θ) (3 cos θ 0 + sec θ 0 )(θ θ 0 ) (87) 15

Introduce the variable (pronounced ksi) ξ = θ θ 0, with ξ = θ so that the equation of motion is ξ + g l (3 cos θ 0 + sec θ 0 )ξ (88) This is our familiar friend the harmonic oscillator with period l T 1 = 2π g(3 cos θ 0 + sec θ 0 ) (89) So as the pendulum makes its primarily circular motion with period T 0, it bobs up and down with period T 1. E.g. Suppose that l = 60 cm and θ 0 = 35. Then T 0 = 1.407 s and T 1 = 0.811 s. The maxima in θ are a good visual reference to the motion. In the time T 1 between successive maxima, the bob rotates through an angle φ φ 0 T 1 = 2π 3 cos 2 θ 0 + 1 > π (90) and so the point of maximum θ precesses in the direction of increasing φ as shown in Figure 10.6.2. The text integrates the equation of motion once to get an effective potential that it plots versus angle to show the angular turning points. 7 Forces of Constraint: Lagrange Multipliers The previous sections have shown how the Lagrangian lets us determine constants of motion and equations of motion. We may also want to determine the size of the constraint forces such as normal forces or tensions. Rather than return to Newton s Methods, we introduce another new powerful method, Lagrange multipliers. In the Lagrangian method used previously we used the constraints to reduce the number of variables needed to describe the system, and to make all the remaining variables independent. 7.1 Text Development Consider a system with two generalized coordinates, (q 1, q 2 ) connected by a single equation of constraint, f(q 1, q 2, t) = 0 where a time dependent constraint has been allowed. In the Lagrangian approach already used we would use the constraint to reduce the system to a 16

single generalized coordinate. Here we will NOT do this step so that we can determine a force of constraint. Start with Hamilton s Variational Principle from Section 4, [ δ L dt = d ( )] δq i dt = 0 (91) q i dt q i i In Section 4 the constraints had already been used, so the δq s were all independent. In this section we are starting earlier and have not used the constraint, so the δq are NOT independent. We write the variation in the constraint, which at any time must be zero. ( f δf = δq 1 + f ) δq 2 = 0 (92) q 1 q 2 Hence Putting this into Equation 91, [( d ) ( d q 1 dt q 1 q 2 dt ( ) f/ q1 δq 2 = δq 1 (93) f/ q 2 ) ( f/ q1 q 2 f/ q 2 )] δq 1 dt = 0 (94) Now we are varying a single coordinate, and the variation can take any value. To make the integral zero, therefore, the item in square brackets must be zero, or (/ q 1 ) (d/dt)(/ q 1 ) ( f/ q 1 ) = (/ q 2) (d/dt)(/ q 2 ) ( f/ q 2 ) The text states that the left hand side is an expression that is a function of (q 1, q 1, t) with time being either implicitly (via q 1 (t), q 1 (t)) or explicitly involved. Likewise the right hand side is a function of (q 2, q 2, t). For these to be equal for arbitrary choices of the generalized variables, they must both equal a function of time alone, that we call λ(t). Then we can write (95) d + λ(t) f q 1 dt q 1 q 1 = 0 (96) d + λ(t) f q 2 dt q 2 q 2 = 0 (97) These two equations plus the constraint condition provide three equations in the three unknowns (q 1 (t), q 2 (t), λ(t)) that can be solved. The quantities Q i = λ(t) f q i (98) 17

are called the generalized forces and are either a force for a spacelike q or a torque for an angular q. The text generalizes to more than 2 generalized coordinates and more than one constraint. 7.2 Alternate Method from Analytical Mechanics By Louis N. Hand, Janet D. Finch, available as a Google Book We will start with an example that is more straightforward calculus: Consider the function F (x, y) = x 2 +y 2. What values of (x, y) will minimize this function subject to the constraint y = 2x + 1. Geometrically we can see that the constraint describes a straight line on the x y plane, and F is the square of the distance from the origin to the line. Standard calculus would use the constraint equation to eliminate y in F, then df/dx = 0 could be solved for the desired value, x = 0.4. Then y = 0.2. Instead we introduce a Lagrange multiplier λ and write F = x 2 + y 2 + λ(y 2x 1) and get three equations, F / x = 0, F / y = 0 F / λ = 0 or in this case 2x 2λ = 0 (99) 2y + λ = 0 (100) y 2x 1 = 0 (101) Solving the simultaneous equations we get x = λ = 0.4, y = 0.2. Try this one: Let F (x) = 3x 2 + 2y 2 and the constraint be x = cos(2y). Use Lagrange multipliers to minimize F and show that the answers are x = 0.2293, y = 0.6697, λ = 1.3761. Returning to the mechanics realm, and for the case of a single constraint equation f(q 1, q 2, t) = 0, we introduce a modified Lagrangian, Then the equation of motion becomes L = L + λf (102) d = 0 (103) q i dt q i 18

7.3 Examples E.g. 1 Ball rolling down a ramp A ball of radius R and moment of inertia I = mk 2 rolls without slipping down a slope inclined at α to the horizontal. Find the equations of motion and the generalized forces. Choose up the incline to be positive x and positive θ to be consistent with this. The constraint between these variables is x Rθ = 0. The modified Lagrangian is then L = 1 2 mẋ2 + 1 2 I θ 2 mgx sin α + λ(x Rθ) (104) We have two equations of motion that come from this modified Lagrangian, plus the constraint equation, mg sin θ + λ mẍ = 0 (105) λr I θ = 0 (106) x Rθ = 0 (107) These are solved in a trivial fashion to get (using the radius of gyration k) ( ) R 2 ẍ = g sin α R 2 + k ( 2 ) R θ = g sin α R 2 + k 2 ( ) k 2 λ = mg sin α R 2 + k 2 The generalized forces are a friction force F x = λ f ( ) k 2 = mg sin α x R 2 + k 2 and a torque about the center of mass due to friction F θ N = λ f ( ) k 2 = mgr sin α θ R 2 + k 2 (108) (109) (110) (111) (112) E.g. The yo-yo A solid disk, mass m radius a, is used in a yo-yo. Use Lagrangians and Lagrange multipliers to get the acceleration and the tension in the string. 19

Using y (down positive) and φ (clockwise positive) as variables, we can write the Lagrangian L = 1 2 mẏ2 + 1 2 I cm φ 2 + mgy (113) There is a constraint relating y and φ, y aφ = 0 and for a disk I cm = 1 2 ma2. If we use these two facts the Lagrangian becomes with the resulting equation of motion quickly reducing to L = 3 4 mẏ2 + mgy (114) ÿ = 2 3 g (115) Now we solve the problem without using the constraint to eliminate a dependent variable. The constraint is f = y aφ = 0. Start with the modified Lagrangian L = 1 2 mẏ2 + 1 2 I cm φ 2 + mgy + λ(y aφ) (116) Then we get three Lagrange equations using variables y, φ and λ y = d dt ẏ mg + λ = m φ (117) φ = d dt φ λa = I cm φ (118) = d λ dt λ y aφ = 0 (119) Take the second derivative of the last equation, ÿ a φ = 0, insert the moment of inertia, and solve the equations to get λ = 1 mg (120) 3 φ = 2 g 3 a (121) ÿ = 2 3 g (122) 20

The generalized force associated with y is Q y = λ f y = λ = 1 mg (123) 3 This is the tension in the string. The generalized force associated with φ is and this is the torque on the disk. Q φ = λ f φ = λa = 1 mga (124) 3 8 D Alembert s Principle: Generalized Forces This section extends the Lagrangian formulation to systems that might include nonconservative forces. We will not go into the details, but here is what I understand about the meaning of the approach. Newton focuses on vector forces. The Europeans contemporaries of Newton focused on scalar quantities like energy. Bernoulli and D Alembert made the connection between the two approaches by introducing the ideas of virtual work. If the objects in a system are instantaneously displaced from equilibrium by infinitesimal displacements, subject to satisfying constraint equations, then the virtual work done is zero. Since we want to deal with non-equilibrium situations, we move the rate of change of momentum to the force side of the equation and write D Alembert s Principle, ( Fi p ) δ r i = 0 (125) 9 The Hamiltonian Function: Hamilton s Equations Again we will not go into any detail (and really, neither does the text). Define a new function, the Hamiltonian which is H = i q i p i L (126) It can be shown that this function is just the total energy of a system, H = T + V. In the Lagrangian formulation we had L = T (q i, q 1 ) V (q i ) as a function of generalized coordinates and their derivatives. We also defined momenta conjugate to the coordinates, p i = / q i. 21

This equation can be solved (at least in principle) for q i = q i (q 1, p 1 ) and then by using Hamilton s Variational Principle we find Hamilton s canonical equations of motion H p i = q i (127) H q i = ṗ i (128) While Lagrangians for a system of n degrees of freedom yields n second order differential equations, Hamilton s equations yield 2n first order differential equations. E.g. 1D Simple Harmonic Oscillator The energies are T = 1 2 mẋ2 V = 1 2 kx2 (129) hence L = mẋ 2 /2 kx 2 /2. The conjugate momentum is p = / ẋ = mẋ and we invert to get ẋ = p/m. Rewrite the energies in terms of x and p, and add to get H. Hamilton s canonical equations are H = T + V = p2 2m + kx2 2 H p = ẋ H x (130) = ṗ (131) and doing the derivatives, p = ẋ kx = ṗ (132) m The first equation is a restatement of the definition of momentum, and the second is the same as our Newton s Law result. Particle on Cylindrical Surface Subject to a Central Force Imagine an origin surrounded symmetrically by a cylinder of radius R extending in the z-direction. A particle of mass m is constrained to move on the surface of the cylinder. A force F = k r acts on the particle. The constraint is x 2 + y 2 = R 2. The force has an associated potential energy V = 1 2 kr2 = 1 2 k(x2 + y 2 + z 2 ) = 1 2 k(r2 + z 2 ) (133) Symmetry suggests the use of cylindrical coordinates, in which v 2 = ṙ 2 + r 2 θ2 + ż 2, and since r = R = Constant, T = 1 2 m(r2 θ2 + ż 2 ) (134) 22

The Lagrangian is L = 1 2 m(r2 θ2 + ż 2 ) 1 2 k(r2 + z 2 ) (135) whence we can get expressions for the generalized momenta p θ = mr 2 θ pz = mż (136) Now we write the Hamiltonian H in terms of variables θ, z, p θ and p z, H = Hamilton s canonical equations are then p2 θ 2mR 2 + p2 z 2m + 1 2 kz2 + 1 2 kr2 (137) ṗ θ = H θ = 0 θ = H p θ = p θ mr 2 ṗ z = H z ż = H p z = p z m = kz (138) (139) In this case, the last two equations just duplicate the definitions of the generalized momenta, angular momentum and linear momentum respectively. The first equation tells us the angular momentum is constant, while the second equation can be written as m z = kz (140) and this is the equation for simple harmonic motion. Thus the particle moves up and down across the z = 0 plane in SHM while rotating around the z-axis at a constant rate. Although Hamilton s canonical equations are of some use in classical mechanics, they play very important roles in both statistical mechanics and quantum mechanics. 10 Another view of Hamilton s Variational Principle (From Classical Dynamics of Particles and Systems, Jerry B. Marion) Here is a bit more rigorous discussion of Euler s equations. We imagine two fixed end points, x 1, x 2 and a function y(x) that describes the path taken between these points. In the Brachistochrone problem this describes the shape of a wire connecting starting and ending points. We define a functional that is evaluated on this path. For the Brachistochrone problem this is the time taken to travel the wire. A useful notation for the functional is f[y(x), y (x); x] (141) 23

This means the functional depends on the function y as well as its derivative dy/dx, and the parametric variable is x. We integrate the functional over the range, J = x2 x 1 f[y, y ; x]dx (142) Our goal in the calculus of variations is to find a function y that will produce a stationary state, that is an extremum for the integral. This desired solution will be called y(0, x). Introduce variations to this desired solution as With fixed endpoints η(x 1 ) = η(x 2 ) = 0. y(α, x) = y(0, x) + αη(x) (143) y (α, x) = y (0, x) + αη (x) (144) A necessary condition for the selection of the proper y(x) is the following. For any variation function η(x), the integral will be a minimum with respect to variations in α, in the limit of α 0. J = 0 (145) α α=0 Now we can do some calculus. From above we can write (notice the total derivative of η in the second equation) y α = η(x) y α = η = dη (146) dx J α = α = = x2 x 1 x2 x 1 x2 f[y, y ; x]dx (147) x ( 1 f y y α + f y ) y dx (148) α ( f y η + f y dη dx ) dx (149) We can evaluate the second term using integration by parts with u = f/ y and dv = dx. Then J dη dx α = x2 x 1 f f η dx + y y η x x2 2 x 1 x 1 At the limits the middle term disappears and J α = x2 x 1 η ( ) d f dx y η dx (150) ( f y d ) f dx (151) dx y 24

The variable α is hidden implicitly in y and y. For this to be true for all variations η(x), f y d ( ) f = 0 (152) dx y This is Euler s Equation. When applied in mechanics to the Lagrangian it is called Lagrange s Equation. Or collectively refer to it as the Euler-Lagrange equation. E.g. The Brachistochrone problem. Imagine two points with coordinates (x 1, y 1 ) and (x 2, y 2 ) connected by a wire of arbitrary shape. A bead can slide along the wire with no friction. Find the shape of the wire (i.e. y(x)) that minimizes the time taken by the bead to move between the points. Choose down as positive-x and to the right as positive-y and assume the bead is at rest at the starting point. We will determine what the functional is that we wish to evaluate, then use Euler s relation. Since the bead is released from rest, x = 0, when it is at x we can use simple kinematics to write the speed, v 2 = 2gx. In a short time the bead will travel a distance ds = ( ) dy 2 dx 2 + dy 2 = 1 + dx = 1 + y dx 2 dx (153) The time taken to travel this distance is Define the functional f[y, y ; x] = Now get the Euler derivatives dt = ds v = 1+y 2 x 1 + y 2 dx 2gx (154) f y = 0 f y = y x(1 + y 2 (155) This means that y x(1 + y 2 ) = Constant 1 2a (156) where we define the constant a for convenience. Squaring, y 2 x(1 + y 2 ) = 1 2a y = dy dx = x 2a x (157) (158) 25

This relation defines the relation between y and x that is required to minimize the time. Now we will show that the relation defines a cycloid. Rearrange and integrate to get y, using initial condition that x = y = 0 at release. y = x dx (159) 2ax x 2 We talked about cycloids last quarter, and had the parameterization x = a(1 cos θ). Inserting this and doing some algebra we get y = a(1 cos θ)dθ = aθ a sin θ (160) which is the parameterized form of y for a cycloid. 10.1 Notation Fowles and Cassiday use δj for the variation if the integral. Marion indicates that this is short hand for the following. With y(α, x) = y(0, x) + αη(x), δj = J x2 ( f α dα = x 1 y d ) f y dα dx (161) dx ẏ α In Fowles and Cassiday notation this is δj = x2 x 1 ( f y d ) f δy dx (162) dx ẏ meaning that J y dα δj α dα δy (163) α 26