Lecture Notes on Classical Mechanics for Physics 106ab ..., Study notes of Physics

Download Lecture Notes on Classical Mechanics for Physics 106ab ... and more Study notes Physics in PDF only on Docsity! Lecture Notes on Classical Mechanics for Physics 106ab Sunil Golwala Revision Date: January 15, 2007 Introduction These notes were written during the Fall, 2004, and Winter, 2005, terms. They are indeed lecture notes – I literally lecture from these notes. They combine material from Hand and Finch (mostly), Thornton, and Goldstein, but cover the material in a different order than any one of these texts and deviate from them widely in some places and less so in others. The reader will no doubt ask the question I asked myself many times while writing these notes: why bother? There are a large number of mechanics textbooks available all covering this very standard material, complete with worked examples and end-of-chapter problems. I can only defend myself by saying that all teachers understand their material in a slightly different way and it is very difficult to teach from someone else’s point of view – it’s like walking in shoes that are two sizes wrong. It is inevitable that every teacher will want to present some of the material in a way that differs from the available texts. These notes simply put my particular presentation down on the page for your reference. These notes are not a substitute for a proper textbook; I have not provided nearly as many examples or illustrations, and have provided no exercises. They are a supplement. I suggest you skim them in parallel while reading one of the recommended texts for the course, focusing your attention on places where these notes deviate from the texts. ii CONTENTS 5.3.3 Motion under the Influence of External Torques . . . . . . . . . . . . . . . . . 313 6 Special Relativity 323 6.1 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 6.1.1 The Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 6.1.2 Transformation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 6.1.3 Mathematical Description of Lorentz Transformations . . . . . . . . . . . . . 333 6.1.4 Physical Implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 6.1.5 Lagrangian and Hamiltonian Dynamics in Relativity . . . . . . . . . . . . . . 346 A Mathematical Appendix 347 A.1 Notational Conventions for Mathematical Symbols . . . . . . . . . . . . . . . . . . . 347 A.2 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348 A.3 Vector and Tensor Definitions and Algebraic Identities . . . . . . . . . . . . . . . . . 349 A.4 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 A.5 Taylor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 A.6 Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 A.7 Legendre Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 B Summary of Physical Results 359 B.1 Elementary Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 B.2 Lagrangian and Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . 363 B.3 Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 B.4 Central Forces and Dynamics of Scattering . . . . . . . . . . . . . . . . . . . . . . . 379 B.5 Rotating Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 B.6 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 v Chapter 1 Elementary Mechanics This chapter reviews material that was covered in your first-year mechanics course – Newtonian mechanics, elementary gravitation, and dynamics of systems of particles. None of this material should be surprising or new. Special emphasis is placed on those aspects that we will return to later in the course. If you feel less than fully comfortable with this material, please take the time to review it now, before we hit the interesting new stuff! The material in this section is largely from Thornton Chapters 2, 5, and 9. Small parts of it are covered in Hand and Finch Chapter 4, but they use the language of Lagrangian mechanics that you have not yet learned. Other references are provided in the notes. 1 CHAPTER 1. ELEMENTARY MECHANICS 1.1 Newtonian Mechanics References: • Thornton and Marion, Classical Dynamics of Particles and Systems, Sections 2.4, 2.5, and 2.6 • Goldstein, Classical Mechanics, Sections 1.1 and 1.2 • Symon, Mechanics, Sections 1.7, 2.1-2.6, 3.1-3.9, and 3.11-3.12 • any first-year physics text Unlike some texts, we’re going to be very pragmatic and ignore niceties regarding the equivalence principle, the logical structure of Newton’s laws, etc. I will take it as given that we all have an intuitive understanding of velocity, mass, force, inertial reference frames, etc. Later in the course we will reexamine some of these concepts. But, for now, let’s get on with it! 1.1.1 The equation of motion for a single particle We study the implications of the relation between force and rate of change of momentum provided by Newton’s second law. Definitions Position of a particle as a function of time: ~r(t) Velocity of a particle as a function of time: ~v(t) = d dt ~r(t). We refer to the magnitude of the velocity, v = |~v|, as the speed. Acceleration of a particle as a function of time: ~a(t) = d dt ~v(t) = d2 dt2 ~r(t). Momentum of a particle: ~p(t) = m(t)~v(t) Newton’s second law In inertial frames, it holds that ~F (t) = d dt ~p(t) (1.1) If the mass is not time-dependent, we have ~F (t) = m d dt ~v(t) = m d2 dt2 ~r(t) (1.2) We use the “dot” shorthand, defining ~̇r = d dt ~r and ~̈r = d2 dt2 ~r, which gives ~F = ~̇p = m~̇v = m~̈r (1.3) Newton’s second law provides the equation of motion, which is simply the equation that needs to be solved find the position of the particle as a function of time. Conservation of Linear Momentum: Suppose the force on a particle is ~F and that there is a vector ~s such that the force has no component along ~s; that is ~F · ~s = 0 (1.4) 2 1.1. NEWTONIAN MECHANICS • Coordinate system: same as before. • Forces along each axis: mẍ = Fg sin θ − Ff mÿ = FN − Fg cos θ We have the additional frictional force acting along −x. • Apply constraints: there is no motion along the y axis, so ÿ = 0, which gives FN = Fg cos θ. Since Ff = µk FN , the equation resulting from the constraint can be used directly to simplify the other equation. • Solve the remaining equations: Here, we simply have the x equation, ẍ = Fg m sin θ − µk Fg m cos θ = g [sin θ − µk cos θ] That is all that was asked for. For θ = 30◦, the numerical result is ẍ = g (sin 30◦ − 0.3 cos 30◦) = 0.24 g Example 1.3 (Thornton Example 2.2) Same as Example 1.1, but now allow for static friction to hold the block in place, with coefficient of static friction µs = 0.4. At what angle does it become possible for the block to slide? • Sketch: Same as before, except the distinction is that the frictional force Ff does not have a fixed value, but we know its maximum value is µs FN . • Coordinate system: same as before. • Forces along each axis: mẍ = Fg sin θ − Ff mÿ = FN − Fg cos θ • Apply constraints: there is no motion along the y axis, so ÿ = 0, which gives FN = Fg cos θ. We will use the result of the application of the constraint below. • Solve the remaining equations: Here, we simply have the x equation, ẍ = Fg m sin θ − Ff m • Since we are solving a static problem, we don’t need to go to the effort of integrating to find x(t); in fact, since the coefficient of sliding friction is usually lower than the coefficient of static friction, the above equations become incorrect as the block begins to move. Instead, we want to figure out at what angle θ = θ′ the block begins to slide. Since Ff has maximum value µs FN = µsmg cos θ, it holds that ẍ ≥ Fg m sin θ − µs FN m 5 CHAPTER 1. ELEMENTARY MECHANICS i.e., ẍ ≥ g [sin θ − µs cos θ] It becomes impossible for the block to stay motionless when the right side becomes positive. The transition angle θ′ is of course when the right side vanishes, when 0 = sin θ′ − µs cos θ′ or tan θ′ = µs which gives θ′ = 21.8◦. Atwood’s machine problems Another class of problems Newtonian mechanics problems you have no doubt seen before are Atwood’s machine problems, where an Atwood’s machine is simply a smooth, massless pulley (with zero diameter) with two masses suspended from a (weightless) rope at each end and acted on by gravity. These problems again require only Newton’s second equation. Example 1.4 (Thornton Example 2.9) Determine the acceleration of the two masses of a simple Atwood’s machine, with one fixed pulley and two masses m1 and m2. • Sketch: • Coordinate system: There is only vertical motion, so use the z coordinates of the two masses z1 and z2. 6 1.1. NEWTONIAN MECHANICS • Forces along each axis: Just the z-axis, but now for two particles: m1z̈1 = −m1g + T m2z̈2 = −m2g + T where T is the tension in the rope. We have assumed the rope perfectly transmits force from one end to the other. • Constraints: The rope length l cannot change, so z1 + z2 = −l is constant, ż1 = −ż2 and z̈1 = −z̈2. • Solve: Just solve the first equation for T and insert in the second equation, making use of z̈1 = −z̈2: T = m1(z̈1 + g) −z̈1 = −g + m1 m2 (z̈1 + g) which we can then solve for z̈1 and T : −z̈2 = z̈1 = −m1 −m2 m1 +m2 g T = 2m1m2 m1 +m2 g It is instructive to consider two limiting cases. First, take m1 = m2 = m. We have in this case −z̈2 = z̈1 = 0 T = mg As you would expect, there is no motion of either mass and the tension in the rope is the weight of either mass – the rope must exert this force to keep either mass from falling. Second, consider m1  m2. We then have −z̈2 = z̈1 = −g T = 2m2 g Here, the heavier mass accelerates downward at the gravitational acceleration and the other mass accelerates upward with the same acceleration. The rope has to have sufficient tension to both counteract gravity acting on the second mass as well as to accelerate it upward at g. Example 1.5 (Thornton Example 2.9) Repeat, with the pulley suspended from an elevator that is ac- celerating with acceleration a. As usual, ignore the mass and diameter of the pulley when considering the forces in and motion of the rope. 7 CHAPTER 1. ELEMENTARY MECHANICS because the two masses balance each other. The tensions in the rope holding the pulley and the elevator cable are determined by the total mass suspended on each. Next, consider the case m1  m2. We have z̈1 = −g z̈2 = g + 2 a T = 0 R = mp (g + a) E = [me +mp] (g + a) m1 falls under the force of gravity. m2 is pulled upward, but there is a component of the acceleration in addition to just g because the rope must unwind over the pulley fast enough to deal with the accelerating motion of the pulley. R and E no longer contain terms for m1 and m2 because the rope holding them just unwinds, exerting no force on the pulley. The mass combination that appears in the solutions, m1m2/(m1 +m2), is the typical form one finds for transforming continuously between these two cases m1 = m2 and m1  m2 (or vice versa), as you will learn when we look at central force motion later. Retarding Forces (See Thornton 2.4 for more detail, but these notes cover the important material) A next level of complexity is introduced by considering forces that are not static but rather depend on the velocity of the moving object. This is interesting not just for the physics but because it introduces a higher level of mathematical complexity. Such a force can frequently be written as a power law in the velocity: ~Fr = ~Fr(v) = −k vn~v v (1.6) k is a constant that depends on the details of the problem. Note that the force is always directed opposite to the velocity of the object. For the simplest power law retarding forces, the equation of motion can be solved analyt- ically. For more complicated dependence on velocity, it may be necessary to generate the solution numerically. We will come back to the latter point. Example 1.6 (Thornton Example 2.4). Find the velocity and position as a function of time for a particle initially having velocity v0 along the +x axis and experiencing a linear retarding force Fr(v) = −k v. • Sketch: 10 1.1. NEWTONIAN MECHANICS • Coordinate system: only one dimension, so trivial. Have the initial velocity ẋ0 be along the +x direction. • Forces along each axis: Just the x-axis. mẍ = −k ẋ • Constraints: none • Solve: The differential equation for x is d dt ẋ = − k m ẋ(t) This is different than we have seen before since the right side is not fixed, but depends on the left side. We can solve by separating the variables and integrating: dẋ ẋ = − k m dt∫ ẋ(t) ẋ0 dy y = − k m ∫ t 0 dt′ log ẋ(t)− log ẋ0 = − k m t ẋ(t) = ẋ0 exp ( − k m t ) That is, the velocity decreases exponentially, going to 0 as t → ∞. The position is easily obtained from the velocity: d dt x = ẋ0 exp ( − k m t ) x(t) = x0 + ẋ0 ∫ t 0 dt′ exp ( − k m t′ ) x(t) = x0 + mẋ0 k [ 1− exp ( − k m t )] The object asymptotically moves a distance mẋ0/k. Example 1.7 (Thornton Example 2.5). Repeat Example 1.6, but now for a particle undergoing vertical motion in the presence of gravity. 11 CHAPTER 1. ELEMENTARY MECHANICS • Sketch: • Coordinate system: only one dimension, so trivial. Have the initial velocity ż0 be along the +z direction. • Forces along each axis: Just the z-axis. m z̈ = −mg − k ż • Constraints: none • Solve: The differential equation for z is d dt ż = −g − k m ż(t) Now we have both constant and velocity-dependent terms on the right side. Again, we solve by separating variables and integrating: dż g + k m ż = −dt∫ ż(t) ż0 dy g + k m y = − ∫ t 0 dt′ m k ∫ g+ k m ż(t) g+ k m ż0 du u = − ∫ t 0 dt′ log ( g + k m ż(t) ) − log ( g + k m ż0 ) = − k m t 1 + k mg ż(t) = ( 1 + k mg ) exp ( − k m t ) ż(t) = −mg k + (mg k + ż0 ) exp ( − k m t ) We see the phenomenon of terminal velocity: as t→∞, the second term vanishes and we see ż(t) → −mg/k. One would have found this asymptotic speed by also solving the equation of motion for the speed at which the acceleration vanishes. The position as a function of time is again found easily by integrating, which yields z(t) = z0 − mg k t+ ( m2 g k2 + m ż0 k )[ 1− exp ( − k m t )] 12 1.1. NEWTONIAN MECHANICS Conservation of Angular Momentum Just as we proved that linear momentum is conserved in directions along which there is no force, one can prove that angular momentum is conserved in directions along which there is no torque. The proof is identical, so we do not repeat it here. Choice of origin There is a caveat: angular momentum and torque depend on the choice of origin. That is, in two frames 1 and 2 whose origins differ by a constant vector ~o such that ~r2(t) = ~r1(t)+~o, we have ~L2(t) = ~r2(t)× ~p(t) = ~r1(t)× ~p(t) + ~o× ~p(t) = ~L1(t) + ~o× ~p(t) ~N2(t) = ~r2(t)× ~F (t) = ~r1(t)× ~F (t) + ~o× ~F (t) = ~N1(t) + ~o× ~F (t) where we have used the fact that ~p and ~F are the same in the two frames (~p because it involves a time derivative; ~F via its relation to ~p by Newton’s second law). Thus, while Newton’s second law and conservation of angular momentum certainly hold regardless of choice of origin, angular momentum may be constant in one frame but not in another because a torque that vanishes in one frame may not vanish in another! In contrast, if linear momentum is conserved in one frame it is conserved in any displaced frame. Thus, angular momentum and torque are imperfect analogues to linear momentum and force. Let’s consider this in more detail. We first solve the linear equations of motion for a particle moving in a circle at fixed speed as shown in the previous figure. Choose the origin of the system to be at the center of the circle and the motion to be in the xy plane. Clearly, in this frame, the particle’s position and velocity as a function of time are ~r1(t) = R (x̂ cosωt+ ŷ sinωt) ~v(t) = ωR (−x̂ sinωt+ ŷ sinωt) where we obtained the velocity by simple differentiation. We do not subscript ~v because it is independent of the choice of origin. The mass is fixed so the momentum is just ~p(t) = m~v(t). The force is, by Newton’s second law, ~F (t) = d~p dt = −mω2R (x̂ cosωt+ ŷ sinωt) = −mω2R r̂1(t) = −m v2 R r̂1(t) where r̂1(t) is a unit vector pointing along ~r1(t). Clearly, the force is back along the line to center of the circle, has magnitude F = mv2/R = mω2R, and is perpendicular to the velocity. The velocity, momentum, and force are independent of the choice of origin. Let’s determine the angular momentum and torque. First consider the same coordinate system with position vector ~r1(t). Since ~F points back along ~r1, it holds that the torque ~N = ~r1 × ~F vanishes. The angular momentum vector is ~L1 = ~r1 × ~p = mvR ẑ. Since v is fixed, ~L1 is fixed, as one would expect in the absence of torque. 15 CHAPTER 1. ELEMENTARY MECHANICS Next, consider a frame whose origin is displaced from the center of the particle orbit along the z axis, as suggested by the earlier figure. Let ~r2 denote the position vector in this frame. In this frame, the torque ~N2 is nonzero because ~r2 and ~F are not colinear. We can write out the torque explicitly: ~N2(t) = ~r2(t)× ~F (t) = r2 [sinα(x̂ cosωt+ ŷ sinωt) + ẑ cosα]× F (−x̂ cosωt− ŷ sinωt) = r2F [− sinα(x̂× ŷ cosωt sinωt+ ŷ × x̂ sinωt cosωt) − cosα(ẑ × x̂ cosωt+ ẑ × ŷ sinωt)] = r2 F cosα (x̂ sinωt− ŷ cosωt) where, between the second and third line, terms of the form x̂× x̂ and ŷ × ŷ were dropped because they vanish. Let’s calculate the angular momentum in this system: ~L2(t) = ~r2(t)× ~p(t) = r2 [sinα(x̂ cosωt+ ŷ sinωt) + ẑ cosα]×mv (−x̂ sinωt+ ŷ cosωt) = r2mv [ sinα(x̂× ŷ cos2 ωt− ŷ × x̂ sin2 ωt) + cosα(−ẑ × x̂ sinωt+ ẑ × ŷ cosωt) ] = ẑ mv r2 sinα+mv r2 cosα (ŷ sinωt− x̂ cosωt) So, in this frame, we have a time-varying component of ~L2 in the plane of the orbit. This time derivative of ~L2 is due to the nonzero torque ~N2 present in this frame, as one can demonstrate directly by differentiating ~L2(t) and using F = mv2/R = mv2/(r2 cosα) and v = Rω = r2 ω cosα. The torque is always perpendicular to the varying component of the angular momentum, so the torque causes the varying component of the angular momentum to precess in a circle. One can of course consider even more complicated cases wherein the origin displacement includes a component in the plane of the motion. Clearly, the algebra gets more complicated but none of the physics changes. 1.1.3 Energy and Work We present the concepts of kinetic and potential energy and work and derive the implications of Newton’s second law on the relations between them. Work and Kinetic Energy We define the work done on a particle by a force ~F (t) in moving it from ~r1 = ~r(t1) to ~r2 = ~r(t2) to be W12 = ∫ t2 t1 ~F · d~r (1.9) The integral is a line integral; the idea is to integrate up the projection of the force along the instantaneous direction of motion. We can write the expression more explicitly to make this clear: W12 = ∫ t2 t1 ~F (t) · d~r dt dt = ∫ t2 t1 ~F (t) · ~v(t) dt 16 1.1. NEWTONIAN MECHANICS The value of this definition is seen by using Newton’s second law to replace ~F : W12 = ∫ t2 t1 d~p dt · ~p m dt (1.10) = 1 2m ∫ ~p2 ~p1 d(~p · ~p) = p2 2 2m − p2 1 2m ≡ T2 − T1 where we have defined the kinetic energy T = p2/2m = mv2/2. Thus, the work the force does on the particle tells us how much the particle’s kinetic energy changes. This is kind of deep: it is only through Newton’s second law that we are able to related something external to the particle – the force acting on it – to a property of the particle – its kinetic energy. Note here that, to demonstrate the connection between work and kinetic energy, we have had to specialize to consider the total force on the particle; Newton’s second law applies only to the total force. For example, consider an elevator descending at constant velocity. Two forces act on the elevator: a gravitational force pointing downward and tension force in the elevator cable pointing upward. If the elevator is to descend at constant velocity, the net force vanishes. Thus, no work is done on the elevator, as is evinced by the fact that its speed (and therefore its kinetic energy) are the same at the start and end of its motion. One could of course calculate the work done by the gravitational force on the elevator and get a nonzero result. This work would be canceled exactly by the negative work done by the cable on the elevator. Potential Energy, Conservation of Energy, and Conservative Forces Consider forces that depend only on position ~r (no explicit dependence on t, ~v). Counterex- ample: retarding forces. Furthermore, consider forces for which W12 is path-independent, i.e., depends only on ~r1 and ~r2. Another way of saying this is that the work done around a closed path vanishes: pick any two points 1 and 2, calculate the work done in going from 1 to 2 and from 2 to 1. The latter will be the negative of the former if the work done is path-independent. By Stokes’ Theorem (see Appendix A), we then see that path-independence of work is equivalent to requiring that ~∇× ~F = 0 everywhere. (Do there exist position-dependent forces for which this is not true? Hard to think of any physically realized ones, but one can certainly construct force functions with nonzero curl.) Then it is possible to define the potential energy as a function of position: U(~r) = U(0)− ∫ ~r 0 ~F (~r1) · d~r1 (1.11) The potential energy is so named because it indicates the amount of kinetic energy the particle would gain in going from ~r back to the origin; the potential energy says how much work the force ~F would do on the particle. The offset or origin of the potential energy is physically irrelevant since we can only measure changes in kinetic energy and hence differences in potential energy. That is, U(~r2)− U(~r1) = − ∫ ~r2 ~r1 ~F (~r) · d~r = −W12 = T1 − T2 17 CHAPTER 1. ELEMENTARY MECHANICS Nonconservative Forces, Mechanical vs. Thermal Energy Of course, there are many forces that are not conservative. We consider the example of a particle falling under the force of gravity and air resistance, and launched downward with the terminal velocity. The particle’s velocity remains fixed for the entire fall. Let’s examine the concepts of work, kinetic energy, potential energy, and conservation of energy for this case. • Work: The net force on the particle vanishes because the air resistance exactly cancels gravity. The particle’s speed remains fixed. Thus, no work is done on the particle. • Kinetic Energy: Since the particle’s speed remains fixed, its kinetic energy is also fixed, consistent with the fact that no work is done. • Potential Energy: Clearly, the particle’s potential energy decreases as it falls. • Conservation of Energy: So, where does the potential energy go? Here we must make the distinction between mechanical energy and thermal energy. We demon- strated earlier that the sum of the kinetic and potential energy is conserved when a particle is acted upon by conservative forces. The kinetic and potential energy are the total mechanical energy of the particle. For extended objects consisting of many particles, there is also thermal energy, which is essentially the random kinetic en- ergy of the particles making up the extended object; the velocities of these submotions cancel, so they correspond to no net motion of the object. Of course, we have not con- sidered thermal energy because we have only been talking about motion of pointlike particles under the influence of idealized forces. Even in the presence of nonconserva- tive forces, the sum of the mechanical and thermal energy is conserved. The potential energy lost by the falling particle in our example is converted to thermal energy of the falling particle and the surrounding air. We will be able to rigorously prove the conservation of total energy later when we consider the dynamics of systems of particles. Calculating Motion from the Potential Energy For particles acting under conservative forces, we have seen that mechanical energy is con- served. We can use this fact to deduce the dynamics purely from the potential energy function. • Solving for the motion using the potential energy Conservation of energy tells us that there is a constant E such that E = T + U = 1 2 mv2 + U(x) Rearranging, we have dx dt = v = ± √ 2 m [E − U(x)] Formally, we can integrate to find t− t0 = ∫ x x0 ± dx′√ 2 m [E − U(x)] 20 1.1. NEWTONIAN MECHANICS Given U(x) it is possible to find x(t). In some cases, it is possible to do this analytically, in others numerical integration may be required. But the fundamental point is that U(x) determines the motion of the particle. • Is the motion bounded? T ≥ 0 always holds. But ~v may go through zero and change sign. If this happens for both signs of ~v, then the motion is bounded. Consider the abstract potential energy curve shown in the following figure (Thornton Figure 2.14): c© 2003 Stephen T. Thornton and Jerry B. Marion, Classical Dynamics of Particles and Systems ~v goes to zero when T vanishes. T can vanish if there are points x such that U(x) ≥ E. Thus, if the particle begins at a point x0 such that there are points xa and xb, xa < x0 < xb, such that U(xa) ≥ E and U(xb) ≥ E, then T vanishes at those points and the velocity vanishes. In order to make the velocity change sign, there must be a force continuing to accelerate the particle at these endpoints. The force is ~F = −~∇U = −x̂ dU/dx in this one-dimensional example; i.e., if U has a nonzero derivative with the appropriate sign at xa and xb, then the particle turns around and the motion is bounded. A particle with energy E1 as indicated in the figure has bounded motion. There can be multiple regions in which a particle of a given energy can be bounded. In the figure, a particle with energy E2 could be bounded between xa and xb or xe and xf . Which one depends on the initial conditions. The particle cannot go between the two bounded regions. The motion is of course unbounded if E is so large that xa and xb do not exist. The motion can be bounded on only one side and unbounded on the other. For example, a particle with energy E3 as indicated in the figure is bounded on the left at xg but unbounded on the right. A particle with energy E4 is unbounded on both sides. • Equilibria A point with ~∇U = 0 is an equilibrium point because the force vanishes there. Of course, the particle must have zero velocity when it reaches such a point to avoid going past it into a region where the force is nonzero. There are three types of equilibrium points. A stable equilibrium point is an equilibrium point at which d2U/dx2 is positive. The potential energy surface is concave up, so any motion away from the equilibrium 21 CHAPTER 1. ELEMENTARY MECHANICS point pushes the particle back toward the equilibrium. In more than one dimension, this corresponds to (~s · ~∇)2U being positive for all constant vectors ~s regardless of direction. An unstable equilibrium point is an equilibrium point at which d2U/dx2 is negative. The potential energy surface is concave down, so any motion away from the equilibrium point pushes the particle away from the equilibrium. In more than one dimension, this corresponds to (~s·~∇)2U being negative for all constant vectors ~s regardless of direction. A saddle point is an equilibrium point at which d2U/dx2 vanishes. Higher-order derivatives must be examined to determine stability. The point may be stable in one direction but not another. In more than one dimension, a saddle point can occur if there are some directions ~s with (~s · ~∇)2U < 0 and others with (~s · ~∇)2U > 0. For smooth U , this means that there is some direction ~s with (~s · ~∇)2U = 0. Example 1.9 Consider the system of pulleys and masses shown in the following figure. The rope is of fixed length b, is fixed at point A, and runs over a pulley at point B a distance 2d away. The mass m1 is attached to the end of the rope below point B, while the mass m2 is held onto the rope by a pulley between A and B. Assume the pulleys are massless and have zero size. Find the potential energy function of the following system and the number and type of equilibrium positions. Let the vertical coordinates of the two masses be z1 and z2, with the z-axis origin on the line AB and +z being upward. The potential energy is, obviously U = m1 g z1 +m2 g (z2 − c) The relation between z1 and z2 is −z2 = √[ b+ z1 2 ]2 − d2 So the simplified potential energy is U = m1 g z1 −m2 g √[ b+ z1 2 ]2 − d2 −m2 g c 22 1.2. GRAVITATION Note that the relative position vector ~r21 depends on ~r1 and thus varies. Force exerted on an extended mass distribution by and extended mass We can further generalize, allowing m2 to instead be an extended mass distribution. The two distributions are denoted by ρ1(~r) and ρ2(~r). The force between the two mass distributions is now ~F21 = −G ∫ V2 ∫ V1 d3r2 d 3r1 ρ1(~r1) ρ2(~r2) r221 r̂21 Again, note that ~r21 varies with ~r1 and ~r2. The order of integration does not matter. Gravitational vector field Since the gravitational force is proportional to the mass being acted upon, we can define a gravitational vector field by determining the force that would act on a point mass m2: ~g(~r2) = ~F21 m2 = −G ∫ V1 d3r1 ρ(~r1) r221 r̂21 The gravitational field is of course independent of m2. Note that ~g has units of force/mass = acceleration. 25 CHAPTER 1. ELEMENTARY MECHANICS 1.2.2 Gravitational Potential The gravitational force is conservative, gravitational potential energy During our discussion of conservative forces, we argued that, for a force ~F for which the work done in going from ~r1 to ~r2 is independent of path, it holds that one can define a potential energy U(~r) and that ~F = −~∇U . We can easily demonstrate that the gravitational force between two point masses is conservative. For simplicity, place one mass at the origin. The work done on the particle in moving from ~ri to ~rf is W = ∫ ~rf ~ri ~F (~r) · d~r = −Gm1m2 ∫ rf ri dr r2 = Gm1m2 ( 1 ri − 1 rf ) where the line integral has been simplified to an integral along radius because any azimuthal motion is perpendicular to ~F and therefore contributes nothing to the integral. We can therefore define the gravitational potential energy of the two-mass system U(~r21) = −G m1m2 r21 where we have chosen the arbitrary zero-point of the potential energy so that U(~r21) vanishes as ~r21 →∞. Because U is linear in the two masses, we can of course determine the potential energy of a system of two extended masses: U = −G ∫ V2 ∫ V1 d3r2 d 3r1 ρ1(~r1) ρ2(~r2) r21 The gravitational potential Clearly, m2 is just a constant in the above discussion, so we can abstract out the gravita- tional potential Ψ(~r2) = U(~r21) m2 = −G m1 ~r21 If m1 is extended, we have Ψ(~r2) = −G ∫ V1 d3r1 ρ1(~r1) r21 It is obvious that, just in the way that the gravitational vector field ~g(~r) is the force per unit mass exerted by the mass distribution giving rise to the field, the gravitational potential scalar field Ψ(~r) gives the work per unit mass needed to move a test mass from one position to another. 26 1.2. GRAVITATION One of the primary advantages of using the gravitational potential is to greatly simplify calculations. The potential is a scalar quantity and thus can be integrated simply over a mass distribution. The gravtiational field or force is of course a vector quantity, so calculating it for extended mass distributions can become quite complex. Calculation of the potential followed by taking the gradient ~g = −~∇Ψ is usually the quickest way to solve gravitational problems, both analytically and numerically. Newton’s iron sphere theorem Newton’s iron sphere theorem says that the gravitational potential of a spherically symmetric mass distribution at a point outside the distribution at radius R from the center of the distribution is is the same as the potential of a point mass equal to the total mass enclosed by the radius R, and that the gravitational field at a radius R depends only on mass enclosed by the radius R. We prove it here. Assume we have a mass distribution ρ(~r) = ρ(r) that is spherically symmetric about the origin. Let ri and ro denote the inner and outer limits of the mass distribution; we allow ri = 0 and ro → ∞. We calculate the potential at a point P that is at radius R from the origin. Since the distribution is spherically symmetric, we know the potential depends only on the radius R and not on the azimuthal and polar angles. Without loss of generality, we choose P to be at R ẑ. The potential is Ψ(P = R ẑ) = −G ∫ V d3r ρ(r) |R ẑ − ~r| Obviously, we should do the integral in spherical coordinates as indicated in the sketch below. 27 CHAPTER 1. ELEMENTARY MECHANICS The solution is sketched below. Poisson’s Equation A final application of the gravitational potential and field is to prove Poisson’s Equation and Laplace’s Equation. Suppose we have a mass distribution ρ(~r) that sources a gravitational potential Ψ(~r) and gravitational field ~g(~r). Consider a surface S enclosing a volume V ; ρ need not be fully contained by S. Let us calculate the flux of the field through the surface S by doing an area integral over the surface of the projection of ~g along the surface normal vector n̂: Φ = ∫ S d2r n̂(~r) · ~g(~r) = − ∫ S d2r2 n̂(~r2) · ∫ d3r1 Gρ(~r1) r221 r̂21 = −G ∫ d3r1 ρ(~r1) ∫ S d2r2 n̂(~r2) · r̂21 r221 The integrand of the second integral is the solid angle subtended by the area element d2r2 as seen from ~r1 (the r21 in the denominator cancels a similar factor in the numerator to turn area into solid angle). There is a sign defined by whether the surface normal has a positive or negative projection along r̂21; i.e., whether the surface normal at ~r2 points away from or toward ~r1. If ~r1 is inside the surface S, then the integral is just 4π because the surface normal always points away and 4π is the solid angle subtended by an enclosing surface. If ~r1 is outside the surface S, then the integrated solid angle subtended vanishes because positive contributions from the part of the surface with n̂ pointing away from ~r1 are canceled by negative contributions from sections with n̂ pointing toward ~r1. Therefore, we have Φ = −4πG ∫ V d3r1 ρ(~r1) = −4πGMV 30 1.2. GRAVITATION where MV is the mass enclosed by the surface S in the volume V . Gauss’s divergence theorem (see Appendix A.4) tells us that Φ = ∫ S d2r n̂(~r) · ~g(~r) = ∫ V d3r ~∇ · ~g(~r) i.e., that ∫ V d3r ~∇ · ~g(~r) = −4πG ∫ V d3r ρ(~r) The surface S (and enclosed volume V ) we chose were arbitrary, so it holds that ~∇ · ~g(~r) = −4πGρ(~r) ∇2Ψ(~r) = 4πGρ(~r) The latter relation is Poisson’s Equation, which relates the derivatives of the potential to the mass density. Notice that it is a local relation; it was not necessarily obvious that such a relation would hold given the way the potential is defined as a line integral. When ρ(~r) = 0, the relation is called Laplace’s Equation. There are analogous relations in electromagnetism, relating the electric potential to the electric charge density. Summary of Relationships among Gravititational Quantities 31 CHAPTER 1. ELEMENTARY MECHANICS 1.3 Dynamics of Systems of Particles References: • Thornton and Marion, Classical Dynamics of Particles and Systems, Chapter 9 • Goldstein, Classical Mechanics, Section 1.2 • Symon, Mechanics, Chapter 4 • any first-year physics text We introduce Newton’s third law, define center of mass, and explore the concepts of momentum, angular momentum, and energy for systems of particles. Collisions of two-particle systems are considered. 1.3.1 Newtonian Mechanical Concepts for Systems of Particles Newton’s third law We have so far considered motion of a single particle, which required Newton’s second law. In considering systems of particles, we now require Newton’s third law. There are two forms: weak form The forces exerted by two particles a and b on each other are equal in magnitude and opposite in direction. That is, if ~fab is the force exerted on particle a by particle b, then ~fab = −~fba (1.14) strong form In addition to the above, the force between the two particles a and b is a function of only the difference in the two particles’ positions and is directed along the vector between them: ~fab = fab(rab) r̂ab (1.15) where r̂ab = ~rab/|~rab| and ~rab = ~ra − ~rb. That is, the force is a scalar function of the magnitude of the position difference and is directed along r̂ab. The mathematical form may seem like a stronger statement than the verbal form. But such a dependence implies the force must be directed along r̂ab. The remaining dependence on ~rab must therefore be a scalar, and the only nonzero scalar that can be formed from ~rab is rab, so the scalar function fab must only be a function of rab. Both forms are assumptions that must be checked observationally for different forces. For example, the Lorentz force on a charged particle, ~F = q~v × ~B, satisfies the weak form but not the strong form. Forces that satisfy the strong form are called central forces; examples are gravitational and electrostatic forces. 32 1.3. DYNAMICS OF SYSTEMS OF PARTICLES Rocket Motion The rocket problem is a classic application of conservation of momentum. First consider a rocket in the absence of gravity. It exhausts mass at a rate ṁ = dm/dt where ṁ < 0. The exhaust gas is expelled at speed u (as measured in the rocket frame). We can find the motion of the rocket by requiring conservation of momentum: p(t) = p(t+ dt) mv = (m+ ṁ dt)(v + dv) + (−ṁ dt)(v − u) On the right side, the first term is the rocket momentum after losing mass dm = ṁ dt < 0 and gaining speed dv. The second term is the momentum of the expelled gas. The mass of the expelled gas is −ṁ dt because ṁ < 0. The gas is expelled at speed u in the rocket frame, which is speed v − u in fixed frame. We expand out the right side, dropping second order differentials like dt dv: mv = mv + ṁ v dt+mdv − ṁ v dt+ ṁ u dt mdv = −ṁ u dt There are three unknowns in the above: m, v, and t. We can eliminate t by expanding ṁ dt = (dm/dt) dt = dm, which gives dv u = −dm m v − v0 = −u [logm− logm0] v = v0 + u log (m0 m ) The final speed depends on the exhaust speed and the ratio of initial to final mass. Clearly, the less mass left over, the larger the final speed. Note that ṁ does not enter; it does not matter how quickly you expel the mass. This is sensible, as the thing that sets the final momentum is the total momentum of the gas that has been expelled. The rate at which it was expelled is irrelevant. The above differential equation could also have been written m dv dt = −u dm dt = u ∣∣∣∣dmdt ∣∣∣∣ The right side is referred to as the thrust and has units of force. The left side looks more like mass × acceleration. The thrust is the effective force on the rocket due to expulsion of gas at rate ṁ with speed u. Since the final speed of the rocket depends only logarithmically on m0/m, gaining final speed by simply trying to increase this ratio is a losing battle. A better way to do it is to use a multistage rocket. Let m0 = initial mass ma = mass of first stage payload mb = mass of first stage fuel container (empty) m1 = ma +mb v1 = final speed of first stage mc = mass of second stage payload md = mass of second stage fuel container (empty) m2 = mc +md v2 = final speed of second stage 35 CHAPTER 1. ELEMENTARY MECHANICS Then the speeds are all related by v1 = v0 + u log ( m0 m1 ) v2 = v1 + u log ( ma m2 ) = v0 + u [ log ( m0 m1 ) + log ( m1 m2 )] = v0 + u log ( m0ma m1m2 ) Since ma < m1, the advantage is not gained directly by jettisoning the empty fuel container and engine mb. Rather the advantage is gained because now m2 can be much smaller than it would have been possible to make m1, compensating for the lost factor ma/m1. Now, let’s repeat the problem in the presence of simple uniform gravitational field g. In the presence of an external force −mg, our equation of motion for the system is dp dt = F (e) = −mg p(t+ dt)− p(t) = −mg dt mdv + ṁ u dt = −mg dt dv = − [ ṁ m u+ g ] dt dv = − [ u m + g ṁ ] dm where we have eliminated t in favor of m again. We integrate to find (remembering ṁ < 0): v = v0 + g |ṁ| ∫ m m0 dm′ − u ∫ m m0 dm′ m′ = v0 − g |ṁ| (m0 −m) + u log (m0 m ) = v0 − g t+ u log (m0 m ) where in the last step we made use of the fact that the mass loss is constant som0−m = |ṁ t|. So, the solution is pretty straightforward – same as the one in the absence of gravity, but there is an additional deceleration term due to gravity. Angular Momentum, Conservation of Angular Momentum, External and Internal Torques We consider the angular momentum of a system of particles. Let the position of particle a be ~ra = ~R+ ~sa (1.21) where ~R is the position of the center of mass and ~sa is the position relative to the center of mass. Since ~R may experience acceleration, ~sa can be in a noninertial reference system. 36 1.3. DYNAMICS OF SYSTEMS OF PARTICLES We have to be careful which coordinates Newton’s second law is applied to. The angular momentum of particle a in the inertial system is ~La = ~ra × ~pa Let us of course write out the total angular momentum: ~L = ∑ a ~La = ∑ a ~ra × ~pa = ∑ a ~ra ×ma ~̇ra = ∑ a ( ~R+ ~sa ) ×ma ( ~̇R+ ~̇sa ) = ∑ a ma [( ~R× ~̇R ) + ( ~R× ~̇sa ) + ( ~sa × ~̇R ) + ( ~sa × ~̇sa )] Consider the middle two terms: ∑ a ma [( ~R× ~̇sa ) + ( ~sa × ~̇R )] = ~R× [∑ a ma ~̇sa ] + [∑ a ma ~sa ] × ~̇R Since ~sa is referenced to the center of mass, by definition the quantity ∑ ama ~sa vanishes.∑ a ma ~sa = ∑ a ma ( ~ra − ~R ) = M ~R−M ~R = 0 So our expression for ~L simplifies to ~L = ∑ a ma [( ~R× ~̇R ) + ( ~sa × ~̇as )] = ~R×M ~̇R+ ∑ a ~sa ×ma ~̇sa ~L = ~R× ~P + ∑ a ~sa ×ma ~̇sa (1.22) Thus, the total angular momentum is the sum of the angular momentum of the center of mass and the angular momentum of the system relative to its center of mass. Remember that the center of mass system is not necessarily inertial, so ma~̇sa, which looks like a linear momentum, may not behave reasonably. It is best to not call it a linear momentum. The next obvious question is – what does Newton’s second law tell us about ~̇L? We know ~̇L = ∑ a ~̇La = ∑ a ~ra × ~̇pa = ∑ a ~ra × ~F (e) a + ∑ b6=a ~fab  = ∑ a ~ra × ~F (e) a + ∑ a,b, b<a [ ~ra × ~fab + ~rb × ~fba ] 37 CHAPTER 1. ELEMENTARY MECHANICS reorganize the sum and simplify by grouping terms for a given particle pair:∑ a,b, b6=a ∫ 2 1 ~fab · d~ra = ∑ a,b, b<a ∫ 2 1 [ ~fab · d~ra + ~fba · d~rb ] = ∑ a,b, b<a ∫ 2 1 ~fab · [d~ra − d~rb] = ∑ a,b, b<a ∫ 2 1 ~fab · d~rab = ∑ a,b, b<a [ Ũab(rab,1)− Ũab(rab,2) ] where in the second step we have made form of the weak form of Newton’s third law and in the last line we have used the strong form to rewrite the line integral as the pairwise potential energy Ũab. It is not obvious at this point that this new pairwise potential energy we have just defined as the line integral of the central force over the difference coordinate ~rab is the same thing as our original single-particle definition of potential energy. But we can see the equivalency by working from the original definition: Ũab(~ra,2, ~rb,2)− Ũab(~ra,1, ~rb,1) = Ũab(~ra,2, ~rb,2)− Ũab(~ra,2, ~rb,1) +Ũab(~ra,2, ~rb,1)− Ũab(~ra,1, ~rb,1) = − ∫ ~rb,2 ~rb,1 ~fba(~rb, ~ra,2) · d~rb − ∫ ~ra,2 ~ra,1 ~fab(~ra, ~rb,1) · d~ra = ∫ ~rb,2 ~rb,1 ~fab(~ra,2, ~rb) · d~rb − ∫ ~ra,2 ~ra,1 ~fab(~ra, ~rb,1) · d~ra = − ∫ ~ra,2−~rb,2 ~ra,2−~rb,1 fab(rab) r̂ab · d~rab − ∫ ~ra,2−~rb,1 ~ra,1−~rb,1 fab(rab) r̂ab · d~rab = − ∫ ~rab,2 ~rab,1 fab(rab) r̂ab · d~rab = Ũab(rab,2)− Ũab(rab,1) In going from the second line to the third line, we make use of the weak form of Newton’s third law. In going from the third line to the fourth line, we make use of the strong form ~fab(~ra, ~rb) = fab(|~rab|) r̂ab and also change variables to ~rab. Going from the fourth line to the fifth line simply makes use of the fact that the lower limit of integration on the first term and the upper limit of integration on the second term are equal, and also we use ~rab = ~ra − ~rb. The final step makes use of the fact that the integral only depend on ~rab. The point made by this derivation is that the pairwise the expression for the potential energy Ũ(~rab,2)− Ũ(~rab,1) is indeed the same as the expression one would expect from the single-particle definition of potential energy as long as the strong form of Newton’s third law holds. So, the total work done is W12 = ∑ a [Ua(~ra,1)− Ua(~ra,2)] + ∑ a,b, b<a [ Ũab(rab,1)− Ũab(rab,2) ] (1.26) ≡ U1 − U2 (1.27) 40 1.3. DYNAMICS OF SYSTEMS OF PARTICLES and so we have T2 − T1 = U1 − U2 E2 = T2 + U2 = T1 + U1 = E1 i.e., we have total energy conservation. We have assumed only that all the forces are conservative and that the internal forces are conservative and central. We separate the potential energy into two terms: U (e) = ∑ a Ua(~ra) (1.28) U (i) = ∑ a,b, b<a Ũab(rab) = 1 2 ∑ a,b, b6=a Ũab(rab) (1.29) In general, U (i) need not be constant in time. We define a rigid body as one for which the distances rab are constant; displacements d~rab during motion are always perpendicular to ~rab. Since we have earlier assumed the strong form of the third law – i.e., central forces – then this immediately implies ~fab · d~rab = 0 for all displacements d~rab. Hence, no work is done by the internal forces and therefore U (i) remains constant. Since constant offsets in potential energy have no effect on motion, the internal potential for such systems can be ignored: all motion is due only to external forces. Example 1.12 A projectile of mass M explodes in flight into three pieces. The first mass m1 = M/2 continues to travel in the same direction as the original projectile. The second mass m2 = M/6 travels in the opposite direction and m3 = M/3 comes to rest. The energy E converted from chemical energy of the explosive to final state mechanical energy is five times the initial kinetic energy of the projectile. What are the velocities of the three pieces? We begin by writing the final velocities in terms of the initial one: ~v1 = k1 ~v ~v2 = −k2 ~v ~v3 = 0 Conservation of linear momentum gives us M ~v = m1 ~v1 +m2 ~v2 +m3 ~v3 M = M 2 k1 − M 6 k2 k2 = 3 k1 − 6 Conservation of energy gives 6 1 2 M v2 = 1 2 m1 v 2 1 + 1 2 m2 v 2 2 + 1 2 m3 v 2 3 6 = 1 2 k2 1 + 1 6 k2 2 Inserting the result for k2 into the conservation of energy equation gives 36 = 3 k2 1 + (3 k1 − 6)2 36 = 3 k2 1 + 9 k2 1 − 36 k1 + 36 0 = k2 1 − 3 k1 41 CHAPTER 1. ELEMENTARY MECHANICS Clearly, the solution is k1 = 3, k2 = 3, so ~v1 = 3~v ~v2 = −3~v ~v3 = 0 Example 1.13 A rope of uniform linear density λ and mass m is wrapped one complete turn around a hollow cylinder of mass M and radius R. The cylinder rotates freely about its axis the rope unwraps. The rope ends are at z = 0 (one fixed, one loose) when point P is at θ = 0. The system is slightly displaced from equilibrium (so motion will occur). Find the angular velocity as a function of the rotation angle θ of the cylinder. We are trading the potential energy of the rope for the kinetic energy of the rope and cylinder. Let α be a parameter that describes the position along the rope, 0 ≤ α ≤ 2π R where 2π R is the length of the rope. α = 0 at the end that is fixed to the cylinder. The z position of the rope as a function of the angle θ and the parameter α is z(θ, α) = { R sin [ 2π − ( θ + α R )] θ + α R < 2π −Rθ + (2π R− α) θ + α R ≥ 2π Some explanation of the above form is necessary. The angle θ gives the angle of the start of the rope. The angle α/R is the angle between the start of the rope and the position α. Therefore, θ + α R gives the angle of the the position α relative to the θ origin, for the part of the rope that is on the cylinder. • The cutoff point between the two forms is where the rope begins to unwind off the cylinder. This occurs for α such that the angle of the position α is 2π. As explained above, the angle of the position α is θ + α R , so the cutoff is at θ + α R = 2π. • Before the cutoff point, the z coordinate of position α is just the z coordinate of a point on a circle of radius R and at angle θ + α R , with an appropriate change of coordinate to a counterclockwise angle. The counterclockwise angle is 2π − (θ + α R). Hence, the z coordinate is R sin [ 2π − (θ + α R) ] . • After the cutoff point, the rope just hangs straight down. The position of the end of the rope is the amount of the rope that has unwound. The amount of rope that has unwound is Rθ because θ is the angle of the start of the rope on the cylinder. Therefore, the z coordinate of the end of the rope is −Rθ. A position α along the rope is at a distance 2π R − α from the end of the rope (recall, 2π R is the length of the rope). So the z coordinate of the position α is −Rθ + (2π R− α). 42 1.3. DYNAMICS OF SYSTEMS OF PARTICLES where ż is the speed of the falling end (and therefore of the entire falling section). So we have T = M g + Ṗ = M g + λ [( b+ z 2 ) z̈ + 1 2 ż2 ] (a) Assume free fall. Since we assume that the falling part of the chain is in free fall, it holds that z = −1 2 g t2 ż = −g t = √ −2 g z z̈ = −g which gives T = M g + λ [ − ( b+ z 2 ) − z ] g = M g + λ g 2 (−b− 3 z) = M g 2 ( −3 z b + 1 ) When the chain is just released, z = 0 and T = M g/2. When the fall is finished and z = −b, we have T = 2M g. The results for the tension are somewhat nonsensical; as we will see, this is due to the assumption of free fall being incorrect. (b) Now we solve by energy methods. Finding the kinetic energy as a function of position gives us ż, we can then differentiate to find z̈ and insert into the above equation for the tension. We find the kinetic energy by requiring conservation of energy. The potential energy of the chain at any given z coordinate is found by integrating the potential energy of small mass elements dm over the chain. Let θ be a parameter that runs from 0 to b, starting at A, that indicates where the mass element is along the chain; θ is independent of how far end B has fallen. Let z̃(θ) be the z coordinate of the element θ. We know that z̃(θ) = { −θ θ < lfixed (θ − lfixed)− lfixed θ > lfixed The contribution of an element dθ at θ to the mass and potential energy are dm = λ dθ dU = dmg z̃(θ) 45 CHAPTER 1. ELEMENTARY MECHANICS so we then have U(z) = ∫ b 0 dθ λ g z̃(θ) = λ g [ − ∫ lfixed 0 dθ θ + ∫ b lfixed dθ (θ − 2 lfixed) ] = λ g [ − lfixed 2 + b2 − l2fixed 2 − 2 lfixed (b− lfixed) ] = λ g [ l2fixed − 2 lfixed b+ b2 2 ] = λ g [( b2 4 − b z 2 + z2 4 ) − ( b2 − b z ) + b2 2 ] = λ g [ −b 2 4 + b z 2 + z2 4 ] The kinetic energy is given by the mass of the part of the chain that is falling and the speed at which it falls: K = 1 2 λ lfall ż 2 = 1 4 λ (b+ z) ż2 Now, we use conservation of energy, noting that K = 0 when z = 0: U(0) = K + U(z) −λ g b 2 4 = 1 4 λ (b+ z) ż2 + λ g [ −b 2 4 + b z 2 + z2 4 ] −g [ 2 b z + z2 ] = (b+ z) ż2 ż2 = −g 2 b z + z2 b+ z Differentiating yields 2 ż z̈ = −g [ 2 b+ 2 z b+ z − 2 b z + z2 (b+ z)2 ] ż = −g [ 2− 2 b z + z2 (b+ z)2 ] ż z̈ = −g [ 1− 1 2 2 b z + z2 (b+ z)2 ] 46 1.3. DYNAMICS OF SYSTEMS OF PARTICLES Now, insert into our previous equation for the tension in terms of z, ż, and z̈ to find T = M g + λ [( b+ z 2 ) (−g) [ 1− 1 2 2 b z + z2 (b+ z)2 ] + 1 2 (−g) 2 b z + z2 b+ z ] = M g − λ g [ 1 2 (b+ z)2 b+ z − 1 4 2 b z + z2 b+ z + 1 2 2 b z + z2 b+ z ] = M g − 1 4 λ g b+ z [ 2 (b+ z)2 − (2 b z + z2) ] = λ g b (b+ z) b+ z − 1 4 λ g b+ z [ 2 b2 + 6 b z + 3 z2 ] = 1 4 λ g b+ z [ 2 b2 − 2 b z − 3 z2 ] = M g 4 1 b(b+ z) [ 2 b2 − 2 b z − 3 z2 ] The two results for the tension are not the same! In the free-fall solution, the tension increases linearly as the chain falls, simply reflecting the fact that the amount of mass under tension increases linearly with the how far the end B has fallen. In the energy solution, the tension becomes infinite as z → −b. Experimentally, it has been determined that the latter solution is closer to reality (Calkin and March, Am. J. Phys, 57: 154 (1989)), though of course the tension does not become infinite (just really large, 25 times the chain weight). The solutions differ because the free-fall solution makes an assumption about the motion of the chain while the energy method does not need to. This is seen by the fact that the relation between ż, ż and z̈ is more complicated in the energy solution. Experimentally, it is seen that the latter relation is closer to reality and that the chain falls faster than free-fall because some tension is communicated through the bend and exerts an additional downward force on the falling part of the chain. In the energy solution, we see this additional force as the second term in z̈, which always makes |z̈| larger (because z < 0). 1.3.2 The Virial Theorem Here we prove the Virial Theorem, which relates the time-averaged kinetic energy for a bounded system to a quantity called the virial, which is just a time-averaged dot product of the force and position of the various particles in the system. In its basic form, the virial theorem does not have a clear intuitive interpretation, though it is certainly useful. When one considers the specific case of conservative forces that depend on particle radius, the virial becomes simply related to the potential energy of the system. Thus, we obtain a time-averaged relation between kinetic and potential energy. This is an incredibly powerful statement because it doesn’t require specific knowledge of the particle orbits. Generic Version Consider an ensemble of particles, whose positions ~ra and momenta ~pa are bounded, meaning that there are upper limits on both. This means that the particles are both confined to a particular region of space and also that they never approach a force center that might impart to them infinite momentum. Define the quantity S = ∑ a ~pa · ~ra 47 CHAPTER 1. ELEMENTARY MECHANICS where U = ∑ a V (ra) is the total potential energy of the system. Thus, the virial theorem reduces to 〈T 〉 = n 2 〈U〉 (1.31) That is, we obtain a very simple relation between the time-averaged kinetic and potential energies of the system. Example 1.16: The Virial Theorem in Astrophysics The virial theorem is used widely in astrophysics because of the dominance of gravity and because it relates directly observable quantities – kinetic energy and temperature – to unobservable quantities – potential energy and mass. We assume gravitational forces, so n = −1. If we divide the virial theorem by the number of particles, we have 1 N 〈T 〉 = 1 2 1 N |〈U〉| (The sign in n has been canceled by the use of the absolute value sign.) That is, the kinetic energy per particle is half the potential energy per particle. We can use this in different ways to measure total masses of systems: • If we are looking at a gas cloud, we can measure the gas temperature Θ by its free-free photon emission.3 That gives us 〈T 〉. The potential energy can be rewritten in terms of the cloud mass M , the typical gas particle mass µ, and the cloud-averaged particle radius. We denote this latter averaged radius as, somewhat uninformatively, the virial radius, Rv. The virial theorem then tells us 3 2 kΘ = 1 N 〈T 〉 = 1 N 1 2 |〈U〉| = 1 2 GM µ 〈 1 r 〉 ≡ 1 2 GM µ 1 Rv 3 kΘ = GM µ Rv Note that the averaging is done on 1/r, not on r. A typical application would be to use the virial theorem to measure the cloud mass. One has to assume that the cloud is spherically symmetric and optically transparent to its own free-free emission; one can then infer from the observed photon radial distribution the shape (but not the normalization!) of the cloud’s density profile. From the shape of unnormalized profile, one can calculate the virial radius. The gas is almost always mostly ionized hydrogen, so µ is known. That leaves the cloud mass as the only unknown. Thus, one can infer from only the photon emission and the virial theorem the cloud mass without any absolute knowledge of normalization of the photon emission in terms of the density. That’s rather remarkable! • If we are looking at a galaxy, we can measure the line-of-sight velocity of a subset of stars by redshift of known spectral lines. The same technique works for galaxies orbiting in a galaxy clusters. Assuming isotropy of the object, the line-of-sight velocity 3Free-free emission is just the process of electrons scattering via the Coulomb force off ions in a plasma, a gas that is hot enough that the bulk of the atoms are ionized. Since the electrons are accelerated in these scattering events, they emit light in the form of a photon. The typical photon energy depends on the plasma temperature; for the very hot plasma in galaxy clusters, which is at millions of degrees K, the photons are keV-energy X-rays. In our own galaxy, the emission is usually in the radio, with wavelength of 1 cm and longer. 50 1.3. DYNAMICS OF SYSTEMS OF PARTICLES and the velocity transverse to the line of sight will be equal on average (up to a √ 2). Assuming all the orbiting objects in the larger object are of roughly equal mass, the kinetic energy per particle is simply related to the rms of the measured line-of-sight velocity: 1 N 〈T 〉 = 1 2 mv2 3d,rms = 3 2 mv2 1d,rms where we relate the full 3-dimensional rms velocity to the measured one-dimension rms velocity assuming isotropy. We can do the same kind of thing as we did for the gas cloud, except now m will drop out: 3 2 mv2 1d,rms = 1 N 〈T 〉 = 1 N 1 2 |〈U〉| = 1 2 GM m 〈1r〉 ≡ 1 2 GM m 1 Rv 3 v2 1d,rms = GM Rv Since the test particles whose velocities we measure are discrete objects, we can just make a plot of their number density as a function of radius and from that calculate the virial radius. We can thus determine M only from our observations of the test particle positions and line-of-sight velocities! 1.3.3 Collisions of Particles One useful application of the concepts we have described above for systems of particles is in the study of collisions of particles. For an isolated system of particles with no external forces acting on it, the total linear momentum is conserved. If the masses of the particles are fixed, the velocity of the center of mass is constant. The reference frame that moves with the center of mass is therefore inertial, and so we can use Newtonian mechanics in this reference frame. Frequently, working in this frame is much easier than in other frames because the algebra is reduced (no need to carry around an extra velocity) and symmetries are more easily apparent. Transforming between Lab and Center-of-Mass Reference Frames Start with some notation: m1,m2 = masses of particles ~ui, ~vi = initial and final velocities of particle i in lab system ~u′i, ~v ′ i = initial and final velocities of particle i in cm system T0, T ′ 0 = total kinetic energy in lab and cm systems Ti, T ′ i = kinetic energy of particle i in lab and cm systems ~V = velocity of cm system with respect to lab system ψi = deflection angle of particle i in lab system (angle between initial velocity of particle 1 and the final velocity of particle i, cos(ψi) = ~vi · ~u1) θ = deflection angle in cm system (same for both particles by definition of cm system) Qualitatively, scattering of two particles in the lab and center-of-mass systems looks as follows (we choose m2 to be initially at rest in the lab system to provide the most extreme difference between lab and cm systems): 51 CHAPTER 1. ELEMENTARY MECHANICS Because the total momentum vanishes in the cm system, the two particles must exit the collision with collinear velocity vectors. Thus, the final state is parameterized by only one angle θ. When transformed back to the lab system, there are now two angles ψ1 and ψ2 reflecting the degree of freedom of the value of the total velocity ~V . Note that the vector ~V fully describes the components of ~v1 and ~v2 that are not along the line between the two particles since ~vi = ~V + ~v′i and we know the ~v′i are collinear. Let’s consider the problem quantitatively. The center of mass satisfies m1 ~r1 +m2 ~r2 = M ~R m1 ~u1 +m2 ~u2 = M ~V In the case ~u2 = 0, which we can always obtain by working in the rest frame of m2, we have ~V = m1 ~u1 m1 +m2 This is the velocity at which the center of mass moves toward m2. Since the center-of-mass is stationary in the center-of-mass frame, we have ~u′2 = −~V . We also have ~u′1 = ~u1 − ~V = m2 ~u1 m1 +m2 Elastic Collisions: Kinematics We now specialize to elastic collisions, wherein the internal kinetic and potential energies of the colliding bodies are unchanged. We also assume there are no external potential energies, so that the only energies are the “external” kinetic energies of the problem, Because we assume the internal energies are unchanged, conservation of energy implies conservation of mechanical energy. Because we assume no external potentials, conservation of mechanical energy implies conservation of “external” kinetic energy. For our above two-particle collision problem, conservation of linear momentum and energy in the center-of mass frame yield: 0 = m1 ~u ′ 1 +m2 ~u ′ 2 = m1 ~v ′ 1 +m2 ~v ′ 2 1 2 m1 u ′2 1 + 1 2 m2 u ′2 2 = 1 2 m1 v ′2 1 + 1 2 m2 v ′2 2 52 1.3. DYNAMICS OF SYSTEMS OF PARTICLES – By calculus: we saw before that ψ1 stays in the first quadrant for m1 > m2, so maximizing ψ1 corresponds to maximizing tanψ1. Let’s take the derivative and set to zero: d dθ tanψ1 = cos θ m1 m2 + cos θ + sin2 θ( m1 m2 + cos θ )2 = m1 m2 cos θ + cos2 θ + sin2 θ m1 m2 + cos θ = 1 + m1 m2 cos θ m1 m2 cos θ Requiring the above to vanish implies cos θ = −m2/m1. We then have tan2 ψ1,max = sin2 θ( m1 m2 + cos θ )2 = 1− m2 2 m2 1 (m1 m2 − m2 m1 )2 = m2 1m 2 2 −m4 2 (m2 1 −m2 2)2 = m2 2 m2 1 −m2 2 Now we use some trigonometric identities to find sinψ1,max: sin2 ψ1,max = 1 1 + cot2 ψ1,max = 1 1 + m2 1−m2 2 m2 2 = m2 2 m2 1 sinψ1,max = m2 m1 The interpretation of this relation is: the larger the mismatch in masses, the more forward-concentrated the scattering is. Elastic Collisions: Energy Recall earlier that we determined that the kinetic energy of each particle is conserved individually in the cm frame, which is convenient. What happens to the kinetic energies in the lab frame? 55 CHAPTER 1. ELEMENTARY MECHANICS First, we know that the initial kinetic energy in the lab and cm frames for ~u2 = 0 are T0 = 1 2 m1 u 2 1 T ′0 = 1 2 [ m1 u ′2 1 +m2 u ′2 2 ] = 1 2 [ m1 m2 2 (m1 +m2)2 u2 1 +m2 m2 1 (m1 +m2)2 u2 1 ] = 1 2 m1m2 m1 +m2 u2 1 = m2 m1 +m2 T0 Note that this implies T ′0 < T0. The final cm energies are T ′1 = 1 2 m1 v ′2 1 = 1 2 m1 u ′2 1 = 1 2 m1 ( m2 m1 +m2 )2 u2 1 = ( m2 m1 +m2 )2 T0 T ′2 = 1 2 m2 v ′2 2 = 1 2 m2 u ′2 2 = 1 2 m2 ( m1 m1 +m2 )2 u2 1 = m1m2 (m1 +m2)2 T0 T ′1 + T ′2 = m2 2 +m1m2 (m1 +m2)2 T0 = m2 m1 +m2 T0 = T ′0 Obviously, we will want to find the final lab frame energies in terms of the initial lab frame energy. We can write T1 T0 = 1 2 m1 v 2 1 1 2 m1 u2 1 = v2 1 u2 1 The law of cosines applied to the figure above relating ~v1, ~v′1, and ~V gives v′21 = v2 1 + V 2 − 2 v1 V cosψ1 which then implies T1 T0 = v2 1 u2 1 = v′21 u2 1 − V 2 u2 1 + 2 v1 V u2 1 cosψ1 = ( m2 m1 +m2 )2 − ( m1 m1 +m2 )2 + 2 v1 V u2 1 cosψ1 = m2 −m1 m1 +m2 + 2 v1 V u2 1 cosψ1 56 1.3. DYNAMICS OF SYSTEMS OF PARTICLES We can rewrite the third term easily. Recall that ~v1 = ~v′1 + ~V . Since ~V has no y component, it therefore holds that v1 sinψ1 = v′1 sin θ. So we get T1 T0 = m2 −m1 m1 +m2 + 2 v′1 V u2 1 sin θ sinψ1 cosψ1 = m2 −m1 m1 +m2 + 2 m2 m1 +m2 m1 m1 +m2 sin θ tanψ1 = m2 −m1 m1 +m2 + 2 m2 m1 +m2 m1 m1 +m2 sin θ m1 m2 + cos θ sin θ = m2 −m1 m1 +m2 + 2 m1m2 (m1 +m2)2 ( m1 m2 + cos θ ) = m2 2 −m2 1 + 2m2 1 (m1 +m2)2 + 2 m1m2 (m1 +m2)2 cos θ = 1− 2m1m2 (m1 +m2)2 + 2 m1m2 (m1 +m2)2 cos θ T1 T0 = 1− 2m1m2 (m1 +m2)2 (1− cos θ) We can write T1/T0 in terms of the lab-frame angle also. To do this, we need to express cos θ in terms of ψ1. Recall from earlier that tanψ1 = sin θ m1 m2 + cos θ For notational convenience, let r = m1/m2. Let’s manipulate the above: tan2 ψ1 = sin2 θ (r + cos θ)2 (r + cos θ)2 tan2 ψ1 = 1− cos2 θ cos2 θ (tan2 ψ1 + 1) + 2 r tan2 ψ1 cos θ + r2 tan2 ψ1 − 1 = 0 cos2 θ cos−2 ψ1 + 2 r tan2 ψ1 cos θ + r2 tan2 ψ1 − 1 = 0 cos2 θ + 2 r sin2 ψ1 cos θ + r2 sin2 ψ1 − cos2 ψ1 = 0 Apply the quadratic formula to find cos θ = 1 2 [ −2 r sin2 ψ1 ± √ 4 r2 sin4 ψ1 − 4 ( r2 sin2 ψ1 − cos2 ψ1 )] = −r sin2 ψ1 ± √ r2 sin4 ψ1 − r2 sin2 ψ1 + 1− sin2 ψ1 = −r sin2 ψ1 ± √ (r2 sin2 ψ1 − 1)(sin2 ψ1 − 1) = −r sin2 ψ1 ± cosψ1 √ 1− r2 sin2 ψ1 1− cos θ = 1 + r [ sin2 ψ1 ± cosψ1 √ r−2 − sin2 ψ1 ] 57 CHAPTER 1. ELEMENTARY MECHANICS Notes: • In the special case m1 = m2, we have T1/T0 = cos2 ψ1 and T2/T0 = cos2 ψ2 = sin2 ψ1 because ψ1 + ψ2 = π/2 for this case. • The center-of-mass scattering angle θ and the ratio of input to output kinetic energies are very simply related. Example 1.17 A particle of mass m1 elastically scatters from a particle of mass m2 at rest. At what lab- frame angle should one place a detector to detect m1 if it loses one-third of its momentum? Over what range m1/m2 is this possible? Calculate the scattering angle for m1/m2 = 1. Our condition on the final state speed is m1 v1 = 2 3 m1 u1 The energy ratio between output and input kinetic energy is T1 T0 = 1 2 m1 v 2 1 1 2 m1 u2 1 = v2 1 u2 1 = 4 9 We equate this to our formula for the lab-frame scattering angle 4 9 = 1− 2m1m2 (m1 +m2)2 (1− cos θ) 1− cos θ = − 5 18 (m1 +m2)2 m1m2 cos θ = 1− 5 18 (m1 +m2)2 m1m2 ≡ y We want the lab angle, though, so let’s use the relation between cm and lab angle: tanψ = sin θ m1 m2 + cos θ = √ 1− y2 m1 m2 + y Because tanψ must be real, we require 1−y2 ≥ 0. Since y ≤ 1 by definition, this corresponds to requiring 1 + y ≥ 0, or 1 + [ 1− 5 18 (m1 +m2)2 m1m2 ] ≥ 0 −5x2 + 26x− 5 ≥ 0 (−5x+ 1)(x− 5) ≥ 0 60 1.3. DYNAMICS OF SYSTEMS OF PARTICLES The two roots are x = 1/5 and x = 5 and the inequality is satisfied in the range 1/5 < x < 5; i.e., 1 5 ≤ m1 m2 ≤ 5 That is, we can find the solution tanψ given above when the mass ratio is in this range. For m1 = m2, we find y = −1 9 tanψ = √ 1− (−1/9)2 1− 1/9 = √ 80/64 = √ 5/4 ψ ≈ 48.2◦ Inelastic Collisions We conclude with a very brief discussion of inelastic collisions. These are collisions in which linear momentum is conserved but mechanical energy is not because some of the energy goes into internal motions of the colliding objects. Recall that conservation of total linear momentum for a system required only that internal forces obey the weak form of the third law, whereas conservation of mechanical energy required the strong form and that the internal kinetic and potential energies be fixed. We must therefore adjust our law of conservation of energy to be Q+ T1 + T2 = T ′1 + T ′2 where Q represents the amount of energy that will be liberated (Q < 0) or absorbed (Q > 0). A classic inelastic collision is the following one: A ball of putty is thrown at another ball of putty and they stick together, continuing on as one body. Clearly, energy was put into internal motion (heat, etc.) by the collision; the mechanical energy is not conserved. One useful concept to consider when looking at inelastic collisions is that of impulse ~P = ∫ t2 t1 ~Fdt The reason impulse is useful is that, though we usually do not know the details of ~f(t), we can determine the total impulse, which gives the total momentum change. Or, vice versa, we can find the force from the total impulse. Example 1.18 A rope of linear density λ is dropped onto a table. Let part of the rope already have come to rest on the table. As the free end falls a distance z, find the force exerted on the table. During a time dt, a portion of the rope v dt comes to rest, where v = |ż| is the fall speed. The momentum of the rope changes by an amount dp = (λ v dt) v = λ v2 dt This is equal but opposite to the change of momentum of the table. The force on the table due to this impulse is therefore Fimpulse = dp dt = λ v2 = 2λ z g 61 CHAPTER 1. ELEMENTARY MECHANICS where the last step is made by making use of the kinematics of a freely falling object. If the table remains stationary as the rope falls on it, the table must be exerting an equal but opposite force back on the rope. The table thus exerts a total force on the rope F = Fimpulse + Fg = 3λ g z The first part of the force brings the portion of the rope z to rest, while the second term keeps it from falling through the table under the influence of gravity. 62 2.1. THE LAGRANGIAN APPROACH TO MECHANICS 2.1.1 Degrees of Freedom, Constraints, and Generalized Coordinates Degrees of Freedom Obviously, a system of M point particles that are unconstrained in any way has 3M degrees of freedom. There is freedom, of course, in how we specify the degrees of freedom; e.g.: • choice of origin • coordinate system: cartesian, cylindrical, spherical, etc. • center-of-mass vs. individual particles: {~rk} or {~R,~sk = ~rk − ~R} But, the number of degrees of freedom is always the same; e.g., in the center-of-mass system, the constraint ∑ k ~sk = 0 applies, ensuring that {~rk} and {~R,~sk} have same number of degrees of freedom. The motion of such a system is completely specified by knowing the dependence of the available degrees of freedom on time. Example 2.1: In the elliptical wire example, there are a priori 3 degrees of freedom, the 3 spatial coordinates of the point particle. The constraints reduce this to one degree of freedom, as no motion in y is allowed and the motions in z and x are related. The loss of the y degree of freedom is easily accounted for in our Cartesian coordinate system; effectively, a 2D Cartesian sytem in x and z will suffice. But the relation between x and z is a constraint that cannot be trivially accomodated by dropping another Cartesian coordinate. Example 2.2: In the Atwood’s machine example, there are a priori 2 degrees of freedom, the z coordinates of the two blocks. (We ignore the x and y degrees of freedom because the problem is inherently 1-dimensional.) The inextensible rope connecting the two masses reduces this to one degree of freedom because, when one mass moves by a certain amount, the other one must move by the opposite amount. Constraints Constraints may reduce the number of degrees of freedom; e.g., particle moving on a table, rigid body, etc. Holonomic constraints are those that can be expressed in the form f(~r1, ~r2, . . . , t) = 0 For example, restricting a point particle to move on the surface of a table is the holonomic constraint z − z0 = 0 where z0 is a constant. A rigid body satisfies the holonomic set of constraints |~ri − ~rj | − cij = 0 where cij is a set of constants satisfying cij = cji > 0 for all particle pairs i, j. For the curious: it is remarkably hard to find the etymology of holonomic (or holonomy) on the web. I found the following (thanks to John Conway of Princeton): 65 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS I believe it was first used by Poinsot in his analysis of the motion of a rigid body. In this theory, a system is called “holonomic” if, in a certain sense, one can recover global information from local information, so the meaning “entire-law” is quite appropriate. The rolling of a ball on a table is non-holonomic, because one rolling along different paths to the same point can put it into different orientations. However, it is perhaps a bit too simplistic to say that “holonomy” means “entire-law”. The “nom” root has many intertwined meanings in Greek, and perhaps more often refers to “counting”. It comes from the same Indo-European root as our word “number.” Nonholonomic constraints are, obviously, constraints that are not holonomic. Hand and Finch Chapter 1 Appendix A has a nice discussion. We present the highlights here. There are three kinds of nonholonomic constraints: 1. Nonintegrable or history-dependent constraints. These are constraints that are not fully defined until the full solution of the equations of motion is known. Equivalently, they are certain types of constraints involving velocities. The classic case of this type is a vertical disk rolling on a horizontal plane. If x and y define the position of the point of contact between the disk and the plane, φ defines the angle of rotation of the disk about its axis, and θ defines the angle between the rotation axis of the disk and the x-axis, then one can find the constraints ẋ = −r φ̇ cos θ ẏ = −r φ̇ sin θ The differential version of these constraints is dx = −r dφ cos θ dy = −r dφ sin θ These differential equations are not integrable; one cannot generate from the relations two equations f1(x, θ, φ) = 0 and f2(y, θ, φ) = 0. The reason is that, if one assumes the functions f1 and f2 exist, the above differential equations imply that their second derivatives would have to satisfy ∂2f ∂θ ∂φ 6= ∂2f ∂φ ∂θ which is a very unpleasant mathematical condition. Explicitly, suppose f1 existed. Then we would be able to write f1(x, θ, φ) = 0 Let us obtain the differential version of the constraint by differentiating: ∂f1 ∂x dx+ ∂f1 ∂θ dθ + ∂f1 ∂φ dφ = 0 This differential constraint should match the original differential constraint dx = −r dφ cos θ. Identifying the coefficients of the differentials yields ∂f1 ∂x = 1 ∂f1 ∂θ = 0 ∂f1 ∂φ = r cosφ 66 2.1. THE LAGRANGIAN APPROACH TO MECHANICS Taking the mixed second partial derivatives gives ∂2f1 ∂φ ∂θ = 0 ∂2f1 ∂θ ∂φ = −r sinφ which, clearly, do not match. Such constraints are also called nonintegrable because one cannot integrate the dif- ferential equation to find a constraint on the coordinates. Nonintegrability is at the root of the etymology indicated in the quotation above: a differential relation such as the one above is a local one; if the differential relation is integrable, you can ob- tain the constraint at all points in space, i.e., you can find the “entire law”. Clearly, nonintegrability is also related to the fact that the constraint is velocity-dependent: a velocity-dependent constraint is a local constraint, and it may not always be possible to determine a global constraint from it. 2. inequality constraints; e.g., particles required to stay inside a box, particle sitting on a sphere but allowed to roll off 3. problems involving frictional forces Holonomic constraints may be divided into rheonomic (“running law”) and scleronomic (“rigid law”) depending on whether time appears explicitly in the constraints: rheonomic: f({~rk}, t) = 0 scleronomic: f({~rk}) = 0 At a technical level, the difference is whether ∂f ∂t = 0 or not; the presence of this partial derivative affects some of the relations we will derive later. Example 2.1: For the elliptical wire example, the constraint equation is the one we specified ini- tially: ( x a(t) )2 + ( z b(t) )2 = 1 If a and/or b do indeed have time dependence, then the constraint is rheonomic. Otherwise, it is scleronomic. Example 2.2: For the Atwood’s machine, the constraint equation is z1 + z2 + l(t) = 0 where l(t) is the length of the rope (we assume the pulley has zero radius). The signs on z1 and z2 are due to the choice of direction for positive z in the example. The constraint is again rheonomic if l is indeed time dependent, scleronomic if not. Generalized Coordinates In general, if one has j independent constraint equations for a system of M particles with 3M degrees of freedom, then the true number of degrees of freedom is 3M − j. There 67 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS Dot cancellation does not necessarily hold if the constraints are nonholonomic. Suppose ~ri = ~ri({qk}, {q̇k}, t). Then our partial derivative would be ~̇ri = d dt ~ri({qk}, {q̇k}, t) = ∑ k [ ∂~ri ∂qk q̇k + ∂~ri ∂q̇k q̈k ] + ∂~ri ∂t Newton’s second law relates q̈k to qk and q̇k. By definition, if there is a velocity-dependent nonholonomic constraint, then there is a velocity-dependent force. Thus, q̈k will certainly depend directly on q̇k and the second term in the sum will yield additional terms when one does the next step, taking the partial derivative with respect to q̇k. The first and last terms might also yield additional unwanted terms: if ~ri depends on q̇k, then there may still be q̇k-dependent terms left in ∂~ri ∂t , which will survive when the partial derivative with respect to q̇k is taken. Either way, the dot cancellation result would be invalidated. Example 2.1: For the elliptical wire, we have given the relations between the positions x, z and the generalized coordinate α x = a(t) cosα z = b(t) sinα We find the relations between the velocities by differentiating with respect to time: ẋ = ȧ cosα− a sinα α̇ ż = ḃ sinα+ b cosα α̇ Note that we do have terms of the form ∂~ri/∂t, the terms with ȧ and ḃ. Taking the partial derivatives, we find ∂x ∂α = −a sinα ∂z ∂α = b cosα ∂ẋ ∂α̇ = −a sinα ∂ż ∂α̇ = b cosα where the ȧ and ḃ terms have disappeared because α̇ does not appear in them. We see the dot cancellation works. Example 2.2: The Atwood’s machine example is as follows; it is somewhat nontrivial if l is a function of time. Differentiating the relations between z1, z2, and Z gives ż1 = Ż ż2 = −l̇ − Ż So then ∂z1 ∂Z = 1 ∂z2 ∂Z = −1 ∂ż1 ∂Ż = 1 ∂ż2 ∂Ż = −1 Note that dot cancellation works even if l is time dependent. 70 2.1. THE LAGRANGIAN APPROACH TO MECHANICS 2.1.2 Virtual Displacement, Virtual Work, and Generalized Forces Virtual Displacement We define a virtual displacement {δ~ri} as an infinitesimal displacement of the system coordinates {~ri} that satisfies the following criteria (stated somewhat differently than in Hand and Finch, but no different in meaning): 1. The displacement satisfies the constraint equations, but may make use of any remaining unconstrained degrees of freedom. 2. The time is held fixed during the displacement. 3. The generalized velocities {q̇k} are held fixed during the displacement. A virtual displacement can be represented in terms of position coordinates or generalized coordinates. The advantage of generalized coordinates, of course, is that they automat- ically respect the constraints. An arbitrary set of displacements {δqk} can qualify as a virtual displacement if conditions (2) and (3) are additionally applied, but an arbitrary set of displacements {δ~ri} may or may not qualify as a virtual displacement depending on whether the displacements obey the constraints. All three conditions will become clearer in the examples. Explicitly, the relation between infinitesimal displacements of generalized coordinates and virtual displacements of the position coordinates is δ~ri = ∑ k ∂~ri ∂qk δqk (2.4) This expression has content: there are fewer {qk} than {~ri}, so the fact that δ~ri can be expressed only in terms of the {qk} reflects the fact that the virtual displacement respects the constraints. One can put in any values of the {δqk} and obtain a virtual displacement, but not every possible set of {δ~ri} can be written in the above way. Example 2.1: For the elliptical wire, it is easy to see what kinds of displacements satisfy the first two requirements. Changes in x and z are related by the constraint equation; we obtain the relation by applying the displacements to the constraint equation. We do this by writing the constraint equation with and without the displacements and differencing the two: with displacements: ( x+ δx a(t) )2 + ( z + δz b(t) )2 = 1 without displacements: ( x a(t) )2 + ( z b(t) )2 = 1 difference: ( x+ δx a(t) )2 − ( x a(t) )2 + ( z + δz b(t) )2 − ( z b(t) )2 = 0 x δx [a(t)]2 + z δz [b(t)]2 = 0 ⇒ δx δz = −a(t) b(t) z x 71 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS All terms of second order in δx or δz are dropped because the displacements are infinitesimal.1 The result is that δx and δz cannot be arbitrary with respect to each other and are related by where the particle is in x and z and the current values of a and b; this clearly satisfies the first requirement. We have satisfied the second requirement, keeping time fixed, by treating a and b as constant: there has been no δt applied, which would have added derivatives of a and b to the expressions. If a and b were truly constant, then the second requirement would be irrelevant. The third requirement is not really relevant here because the generalized velocities do not enter the constraints in this holonomic case; but they will enter, for example, the kinetic energy, so it must be kept in mind. The relation between the virtual displacements in the positions and in the general- ized coordinate is easy to calculate: x = a cosα ⇒ δx = −a sinα δα z = b sinα ⇒ δz = b cosα δα =⇒ δx δz = −a b tanα We see that there is a one-to-one correspondence between all infinitesimal displace- ments δα of the generalized coordinate and virtual displacements of the positional coordinates (δx, δz), as stated above. The displacements of the positional coordi- nates that cannot be generated from δα by the above expressions are those that do not satisfy the constraints and are disallowed. Example 2.2: For our Atwood’s machine example, the constraint equation z1 + z2 + l(t) = 0 is easily converted to differential form, giving δz1 + δz2 = 0 Again, remember that we do not let time vary, so l(t) contributes nothing to the differential. This equation is what we would have arrived at if we had started with an infinitesimal displacement δZ of the generalized coordinate Z (holding time fixed according to condition (2)): δz1 = δZ δz2 = −δZ ⇒ δz1 + δz2 = 0 Fixing time prevents appearance of derivatives of l(t). Again, we also see the one- to-one correspondence between infinitesimal displacements of the generalized coor- dinate and virtual displacements of the position coordinates. 1This is one of the first applications of Taylor expansions in this course. This kind of thing will be done regularly, so get used to the technique! 72 2.1. THE LAGRANGIAN APPROACH TO MECHANICS Example 2.2: In the Atwood’s machine example, the constraint forces ~F21 and ~F22 act along the rope. The virtual displacements are also along the rope. Clearly, ~F21 · δ~r1 = F21 δz1 and ~F22 · δ~r2 = F22 δz2 do not vanish. But the sum does: 2∑ i=1 ~Fi2 · δ~ri = F21δz1 + F22δz2 = T (δz1 + δz2) = 0 Notice that, in this case, all the terms pertaining to the particular constraint force have to be summed in order for the result to hold. The virtual work and non- constraint force sum is then δW = 2∑ i=1 1∑ j=1 ~F (nc) ij · δ~ri = 2∑ i=1 ~Fi1 · δ~ri = −g (m1 δz1 +m2 δz2) Note that, unless m1 = m2, δW will in general not vanish, again giving rise to the dynamics of the problem. Generalized Force Our discussion of generalized coordinates essentially was an effort to make use of the con- straints to eliminate the degrees of freedom in our system that have no dynamics. Similarly, the constraint forces, once they have been taken account of by transforming to the gener- alized coordinates, would seem to be irrelevant. We will show how they can be eliminated in favor of generalized forces that contain only the non-constraint forces. To define generalized forces, we combine Equation 2.4, the relation between virtual displace- ments of position coordinates and generalized coordinates, with Equation 2.5, the relation between virtual work and non-constraint forces: δW = ∑ ij ~F (nc) ij · δ~ri = ∑ ij ~F (nc) ij · [∑ k ∂~ri ∂qk δqk ] = ∑ k ∑ ij ~F (nc) ij · ∂~ri ∂qk  δqk ≡ ∑ k Fk δqk where Fk ≡ ∑ ij ~F (nc) ij · ∂~ri ∂qk = δW δqk (2.5) is the generalized force along the kth generalized coordinate. The last form, Fk = δW/δqk, says that the force is simple the ratio of the work done to the displacement when a virtual displacement of only the kth generalized coordinate is performed; it is of course possible to displace only the kth generalized coordinate because the generalized coordinates 75 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS are mutually independent. It is important to remember that the generalized force is found by summing only over the non-constraint forces: the constraint forces have already been taken into account in defining generalized coordinates. The above definition is very sensible. In Section 1.1.3, we defined work to be the line integral of force. Infinitesimally, a force ~F causing a displacement δ~r does work δW = ~F · δ~r. The generalized force is the exact analogue: if work δW is done when the ensemble of forces act to produce a generalized coordinate displacement δqk, then the generalized force Fk doing that work is Fk = δW/δqk. But the generalized force is a simplification because it is only composed of the non-constraint forces. Example 2.1: In the elliptical wire example, the generalized force for the α coordinate (k = 1) is Fα = ~F11 · ∂~r1 ∂α = −mg ẑ · ( x̂ ∂x ∂α + ẑ ∂z ∂α ) = −mg ẑ · (−x̂ a sinα+ ẑ b cosα) = −mg b cosα The constraint force, which acts in both the x and z directions and is α-dependent, does not appear in the generalized force. Example 2.2: In the Atwood’s machine example, the generalized force for the Z coordinate (k = 1 again) is FZ = ~F11 · ∂~r1 ∂Z + ~F21 · ∂~r2 ∂Z = −m1 g ∂z1 ∂Z −m2 g ∂z2 ∂Z = (m2 −m1) g Again, the constraint force (the rope tension) is eliminated. Because Z is just z1 in this case, the generalized force in Z is just the net force on m1 acting in the z1 direction. 2.1.3 d’Alembert’s Principle and the Generalized Equation of Motion The reader is at this point no doubt still wondering, so what? This section gets to the what. d’Alembert’s Principle Our definition of virtual work was δW = ∑ ij ~Fij · δ~ri where the sum includes all (constraint and non-constraint) forces. Assuming our position coordinates are in an inertial frame (but not necessarily our generalized coordinates), New- ton’s second law tells us ∑ j ~Fij = ~pi: the sum of all the forces acting on a particle give the rate of change of its momentum. We may then rewrite δW : δW = ∑ i ∑ j ~Fij · δ~ri = ∑ i ~̇pi · δ~ri 76 2.1. THE LAGRANGIAN APPROACH TO MECHANICS But, we found earlier that we could write the virtual work as a sum over only non-constraint forces, δW = ∑ ij ~F (nc) ij · δ~ri Thus, we may derive the relation ∑ i ∑ j ~F (nc) ij − ~̇pi  · δ~ri = 0 (2.6) The above equation is referred to as d’Alembert’s principle. Its content is that the rate of change of momentum is determined only by the non-constraint forces. In this form, it is not much use, but the conclusion that the rate of change of momentum is determined only by non-constraint forces is an important physical statement. Note that the multiplication by the virtual displacement must be included in order for the statement to hold; the statement ~F (nc) i − ~̇pi = 0 is not in general true; just consider our elliptical wire and Atwood’s machine examples. We may use d’Alembert’s principle to relate generalized forces to the rate of change of the momenta: ∑ k Fk δqk = δW = ∑ k ~̇pi · δ~ri = ∑ i,k ~̇pi · ∂~ri ∂qk δqk Now, unlike the {δ~ri}, the {δqk} are mutually independent. Therefore, we may conclude that equality holds for each term of the sum separately (unlike for Equation 2.6), providing a different version of d’Alembert’s principle:∑ ij ~F (nc) ij · ∂~ri ∂qk = Fk = ∑ i ~̇pi · ∂~ri ∂qk (2.7) This is now a very important statement: the generalized force for the kth generalized coor- dinate, which can be calculated from the non-constraint forces only, is related to a particular weighted sum of momentum time derivatives (the weights being the partial derivatives of the position coordinates with respect to the generalized coordinates). Effectively, we have an analogue of Newton’s second law, but including only the non-constraint forces. This can be a major simplification in cases where the constraint forces are complicated or simply not known. d’Alembert’s principle is not yet useful in the current form because the left side contains generalized forces depending on the generalized coordinates but the right side has derivatives of the momenta associated with the position coordinates. We need time derivatives with respect to the generalized coordinates on the right side so all the dynamics can be calculated in the generalized coordinates. Generalized Equation of Motion Here we perform the manipulation needed to make d’Alembert’s principle useful. We know from Newtonian mechanics that work is related to kinetic energy, so it is natural to expect 77 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS This is an important example of how to convert T from a function of the position velocities to a function of the generalized coordinates and velocities. Now take the prescribed derivatives: d dt ( ∂T ∂α̇ ) − ∂T ∂α = d dt [ mα̇ ( a2 sin2 α+ b2 cos2 α )] − 2mα̇2 ( a2 − b2 ) sinα cosα = mα̈ ( a2 sin2 α+ b2 cos2 α ) In taking the total derivative in the first term, we obtain two terms: the one dis- played in the last line above, and one that exactly cancels the last term in the first line.3 We have the generalized force Fα from before, Fα = −mg b cosα, so the generalized equation of motion is Fα = d dt ( ∂T ∂α̇ ) − ∂T ∂α −mg b cosα = mα̈ ( a2 sin2 α+ b2 cos2 α ) α̈ = − g b cosα a2 sin2 α+ b2 cos2 α Specializing to a = b = r (circular wire), this simplifies to α̈ = − g cosα r One can schematically see what equations of motion one would have obtained if generalized coordinates had not been used. There would be two equations, one for x and one for z. Both would have a component of the constraint force and the z equation would have a gravity term. The constraint equation on x and z would be used to eliminate x, giving two equations for z̈ and the unknown constraint force. The constraint force could be eliminated using one of the equations, resulting in one, rather complicated, equation of motion in z. Clearly, the above equation of motion in α is much simpler! Example 2.2: For the Atwood’s machine example, things are significantly simpler. The kinetic energy is T = 1 2 ( m1 ż 2 1 +m2 ż 2 2 ) Rewriting using the generalized coordinate Z gives T = 1 2 (m1 +m2) Ż2 3This is our first encounter with an equation that has both total time derivatives and partial derivatives with respected to qk and q̇k. One must realize that the total time derivative acts on all variables that have any time dependence. This is why two terms came out of the total time derivative term; one term arising from the total time derivative acting on α̇, which gives α̈, and another arising from the total time derivative acting on sin2 α and cos2 α, which gave the additional term that canceled the ∂T/∂α term. 80 2.1. THE LAGRANGIAN APPROACH TO MECHANICS The kinetic energy derivatives term is d dt ( ∂T ∂Ż ) − ∂T ∂Z = (m1 +m2) Z̈ Using FZ = (m2 −m1) g from earlier, the generalized equation of motion is FZ = d dt ( ∂T ∂Ż ) − ∂T ∂Z (m2 −m1) g = (m1 +m2) Z̈ Z̈ = −m1 −m2 m1 +m2 g which is the same equation of motion obtained for z1 in Example 1.4 in Section 1.1. But, using the formalism of constraints and generalized coordinates, we have no mention of the rope tension in the equations of motion. 2.1.4 The Lagrangian and the Euler-Lagrange Equations For conservative non-constraint forces, we can obtain a slightly more compact form of the general- ized equation of motion, known as the Euler-Lagrange equations. Generalized Conservative Forces Now let us specialize to non-constraint forces that are conservative; i.e., ~F (nc) i = −~∇iU({~rj}) where ~∇i indicates the gradient with respect to ~ri. Whether the constraint forces are conservative is irrelevant; we will only explicitly need the potential for the non-constraint forces. U is assumed to be a function only of the coordinate positions; there is no explicit dependence on time or on velocities, ∂U/∂t = 0 and ∂U/∂~̇ri = 0.4 Let us use this expression in writing out the generalized force: Fk = ∑ i ~F (nc) i · ∂~ri ∂qk = − ∑ i ~∇iU({~rj}) · ∂~ri ∂qk = − ∂ ∂qk U({ql}, t) In the last step we make use of the holonomic constraints to rewrite U as a function of the {ql} and possibly t and realize that the previous line is just the partial derivative of U with respect to qk.5 Thus, rather than determining the equation of motion by calculating the generalized force from the non-constraint forces and the coordinate transformation relations, we can rewrite the potential energy as a function of the generalized coordinates and calculate the general- ized force by gradients thereof. Example 2.1: 4It is possible to consider time-dependent potential energy functions in the following, but we hold off on that discussion until Section 2.1.9. 5Some care must be taken with the time dependence. U is not initially a function of t. But the constraints may be rheonomic, so some dependence of U on t may appear when the coordinate transformation is done, and ∂U/∂t may be nonzero. This should not be taken to imply that somehow the potential has become nonconservative – the time-dependence arises purely through the rheonomic constraint. If there is any such confusing circumstance, one should always transform U back to position coordinates to check for time-dependence. And note that U may remain time-independent in some rheonomic constraint cases; see Example 2.5. 81 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS For the elliptical wire example, the potential energy function is due to gravity, U(z) = mg z Rewriting in terms of α gives U(α; t) = mg b(t) sin α The generalized force is then Fα = −∂U(α; t) ∂α = −mg b(t) cosα as obtained before. Note that we may allow b to be a function of time without ruining the conservative nature of the potential energy – U becomes a function of t through the definition of the generalized coordinate but, obviously, if it was initially a conservative potential, a transformation of coordinates cannot change that. Example 2.2: For the Atwood’s machine, the potential energy function is U(z1, z2) = g (m1 z1 +m2 z2) Rewriting in terms of Z gives U(Z; t) = g [(m1 −m2)Z −m2 l(t)] The generalized force is FZ = −∂U(Z; t) ∂Z = g (m2 −m1) as found earlier. Again, l is allowed to be a function of time without ruining the conservative nature of the potential energy. The Euler-Lagrange Equations An even simpler method exists. We may rewrite the generalized equation of motion using the above relation between generalized force and gradient of the potential energy as −∂U ∂qk = d dt ( ∂T ∂q̇k ) − ∂T ∂qk Define the Lagrangian L ≡ T − U (2.11) Since we have assumed holonomic constraints, we have that ∂U/∂q̇k = 0. This lets us replace d/dt (∂T/∂q̇k) with d/dt (∂L/∂q̇k), giving d dt ( ∂L ∂q̇k ) − ∂L ∂qk = 0 (2.12) 82 2.1. THE LAGRANGIAN APPROACH TO MECHANICS 2.1.6 Cyclic Coordinates and Canonical Momenta If the Lagrangian contains q̇k but not qk, we can easily see from the Euler-Lagrange equation that the behavior of that coordinate is trivial: d dt ( ∂L ∂q̇k ) = 0 which implies pk ≡ ∂L ∂q̇k is constant or conserved and is termed the canonical momentum conjugate to qk or the canonically conjugate momentum. The coordinate qk is termed ignorable or cyclic. Once the value of pk is specified by initial conditions, it does not change. In simple cases, the canonical momentum is simply a constant times the corresponding generalized velocity, indicating that the velocity in that coordinate is fixed and the coordinate evolves linearly in time. In more complicated cases, the dynamics are not so trivial, but one does still obtain very useful relations between coordinates and/or velocities that are not generally true. For example, for a particle moving in two dimensions under the influence of a central force like gravity, there is no dependence of L on the azimuthal angle φ, so the angular momentum pφ = mρ2 φ̇ is constant. This tells us the useful fact φ̇ ∝ ρ−2 – as the particle moves inward toward the origin, its angular velocity must increase in a specific way. In general, then, a cyclic coordinate results in a conserved momentum that simplifies the dynamics in the cyclic coordinate. The above definition of canonical momentum holds even when the qk coordinate is not cyclic; we will see its use in the future. Example 2.1: For the elliptical wire, the canonical momentum is pα = ∂L ∂α̇ = mα̇ ( a2 sin2 α+ b2 cos2 α ) The astute reader will notice that, when a = b = r giving a circular wire, pα is the angular momentum about the axis of the circle. α is not cyclic, so the Euler-Lagrange equation for it is not trivial and this momentum is not conserved. If gravity were eliminated, and a = b = r, then α would become cyclic because sin2 α+ cos2 α = 1. Angular momentum, and hence angular velocity, would remain fixed at its initial value: the bead would simply circumnavigate the wire at constant speed. But if a 6= b, then we are left with α dependence in T even if U = 0 and so the Lagrangian is not cyclic in α. Example 2.2: For the Atwood’s machine, the canonical momentum is pZ = ∂L ∂Ż = (m1 +m2) Ż In this case, if U = 0 and l is constant, Z does indeed become cyclic and pZ is conserved. If the blocks are initially at rest, they stay at rest, and if they are initially moving with some speed Ż, that speed is preserved. 85 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS 2.1.7 Summary This section has followed some long logical paths, so we quickly summarize the logic here. degrees of freedom + holonomic constraints ↓ generalized coordinates with dot cancellation generalized coordinates + virtual work + Newton’s second law ↓ d’Alembert’s principle d’Alembert’s principle + generalized force ↓ generalized equation of motion generalized equation of motion + conservative force ↓ Lagrangian, Euler-Lagrange equations Lagrangian + scleronomic constraints ↓ conservation of Hamiltonian Lagrangian + cyclic coordinates ↓ conserved conjugate momenta 2.1.8 More examples In addition to the examples we did alongside the derivation, let us do a few more to further illustrate our results. Example 2.3 Sliding block on sliding inclined plane. See Hand and Finch Sections 1.1 and 1.2 for details; we reproduce some of the more interesting parts here. See Figure 1.1 of Hand and Finch for a sketch. Let b and p subscripts denote the block and the inclined plane. The constraints are that the block cannot move perpendicular to the plane and that the plane cannot move perpendicular to the flat surface it sits on. Therefore, the natural generalized coordinates are X, the horizontal position of the vertical edge of the plane, and d, the distance the block has slid down the plane. (Let h be the height of the plane and α be the angle). Note that d is defined relative to a noninertial reference frame! The constraints can be rewritten as the following transformation equations: ~rp(t) = x̂X ~rb(t) = ~rp(t) + ŷ h+ d [x̂ cosα− ŷ sinα] = x̂ (d cosα+X) + ŷ (h− d sinα) 86 2.1. THE LAGRANGIAN APPROACH TO MECHANICS Note that, though d is a noninertial coordinate, the constraints are still scleronomic because time does not appear explicitly in the transformation relations; rheonomic and noninertial sometimes go hand-in-hand, but not always. The assorted partial derivatives are ∂~rp ∂X = x̂ ∂~rp ∂d = 0 ∂~rb ∂X = x̂ ∂~rb ∂d = x̂ cosα− ŷ sinα The non-constraint forces are ~F (nc) p = −M g ŷ ~F (nc) b = −mg ŷ This last step was very important – we did not have to put in any normal forces, etc., or take any projections. We just blindly put in the non-constraint forces – gravity only in this case – and then we will use the generalized equation of motion to do all the work. The generalized forces are FX = ~F (nc) p · ∂~rp ∂X + ~F (nc) b · ∂~rb ∂X = 0 Fd = ~F (nc) p · ∂~rp ∂d + ~F (nc) b · ∂~rb ∂d = mg sinα Now, we calculate the kinetic energy and the relevant derivatives: T = 1 2 M Ẋ2 + 1 2 m { d dt [x̂ (X + d cosα) + ŷ d sinα]2 } = 1 2 (M +m) Ẋ2 + 1 2 m ( ḋ2 + 2 ḋ Ẋ cosα ) ∂T ∂X = ∂T ∂d = 0 d dt ( ∂T ∂Ẋ ) = d dt ( [m+M ] Ẋ +mḋ cosα ) = [m+M ] Ẍ +md̈ cosα d dt ( ∂T ∂ḋ ) = m [ d̈+ Ẍ cosα ] The generalized equations of motion are then X : 0 = [m+M ] Ẍ +md̈ cosα d : mg sinα = m [ d̈+ Ẍ cosα ] We can solve the X equation for Ẍ: Ẍ = − m M +m d̈ cosα 87 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS where we have assumed the elevator starts at t = 0 with velocity v0 and accelerates with acceleration a. In the inertial frame, the ball has Lagrangian L = 1 2 m ż2 −mg z Rewriting in terms of the generalized coordinate q, we have L = 1 2 m [q̇ + v0 + a t]2 −mg [ q + v0 t+ 1 2 a t2 ] = 1 2 m q̇2 +m q̇ [v0 + a t] + 1 2 m [v0 + a t]2 −mg q −mg [ v0 t+ 1 2 a t2 ] The partial derivatives are ∂L ∂q̇ = m [q̇ + v0 + a t] ∂L ∂q = −mg So the Euler-Lagrange equation is m [q̈ + a] +mg = 0 q̈ = −(g + a) q̇ = u0 − (g + a) t q = u0 t− 1 2 (g + a) t2 If we then transform back to the inertial frame as a check, we find z = (u0 + v0) t− 1 2 g t2 which is as you would expect: if the ball is thrown upward with speed u0 relative to the elevator, its initial velocity in the inertial frame is u0 + v0 and the ball decelerates under gravity as normal. In the noninertial frame, though, the ball decelerates faster, with deceleration g + a, because of the acceleration of the elevator. It is interesting to look at the Hamiltonian because, due to the rheonomic constraints, it is neither the total energy nor is it conserved: H = q̇ ∂L ∂q̇ − L = m q̇ [q̇ + v0 + a t]− L = 1 2 m q̇2 − 1 2 m [v0 + a t]2 +mg q +mg [ v0 t+ 1 2 a t2 ] Clearly, H is time-dependent, and it has no apparent simple relationship to the energy as calculated in the inertial frame, T + U . Example 2.5 Consider a heavy bead sliding on a stiff rotating wire held at a fixed angle, with the bead experiencing the force of gravity and any constraint forces; see Hand and Finch Figure 1.2 for a sketch. Let’s apply our formalism to the problem. 90 2.1. THE LAGRANGIAN APPROACH TO MECHANICS The constraint is that the bead must remain on the wire, and that the wire rotates with angular velocity ω. The natural generalized coordinate is q, the distance that the bead is from the origin along the wire. Let α be the polar angle between the axis of rotation and the wire; it is fixed by assumption. We consider the case shown in Hand and Finch, where the wire is angled upward so that the bead does not fall off. The coordinate transformation relations are ~r(t) = x̂ q sinα cosωt+ ŷ q sinα sinωt+ ẑ q cosα Note that t appears explicitly. Though we have not written the constraint equations ex- plicitly, the presence of t in the transformation from position coordinates to generalized coordinates implies that the constraint is rheonomic. Let’s first go the generalized equation of motion route, so we need to find the partial derivatives of the transformation relation: ∂~r ∂q = x̂ sinα cosωt+ ŷ sinα sinωt+ ẑ cosα The non-constraint force is ~F (nc) = −mg ẑ So the generalized force is Fq = ~F (nc) · ∂~r ∂q = −mg cosα To calculate the kinetic energy, we will need the velocity: ~̇r = ∂~r ∂q q̇ + ∂~r ∂t = x̂ q̇ sinα cosωt+ ŷ q̇ sinα sinωt+ ẑ q̇ cosα −x̂ q sinαω sinωt+ ŷ q sinαω cosωt The kinetic energy is T = 1 2 m~̇r · ~̇r = 1 2 m ( q̇2 + q2 ω2 sin2 α ) where a large amount of algebra has been suppressed. The partial derivatives of T are ∂T ∂q = mq ω2 sin2 α ∂T ∂q̇ = m q̇ The generalized equation of motion therefore is Fq = d dt ( ∂T ∂q̇ ) − ∂T ∂q −mg cosα = d dt (mq̇)−mq ω2 sin2 α q̈ − q ω2 sin2 α = −g cos α 91 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN DYNAMICS We can also employ Lagrangian techniques. The potential energy and the Lagrangian are U = mg z = mg q cosα L = T − U = 1 2 m ( q̇2 + q2 ω2 sin2 α ) −mg q cosα Note that, even though the constraints are rheonomic, U remains time-independent. This is a special case – while there was time dependence in the transformation relations, there was no time dependence in the relation between z and q, and it is only z that appears in the potential energy. The partial derivatives of U are ∂U ∂q = mg cosα ∂U ∂q̇ = 0 Therefore, the Euler-Lagrange equation is d dt ∂ ∂q̇ (T − U)− ∂ ∂q (T − U) = 0 m q̈ −mq ω2 sin2 α+mg cosα = 0 q̈ − ω2 q sin2 α = −g cosα as seen before. The canonical momentum is pq = ∂L ∂q̇ = mq̇ The Hamiltonian is H = q̇ ∂L ∂q̇ − L = m q̇2 − L = 1 2 m q̇2 − 1 2 mq2 ω2 sin2 α+mg q cosα Note that this Hamiltonian is not the total energy of the system. H is conserved because ∂L/∂t vanishes. But, because of the negative sign on the second term, H is not the total energy. This occurs because the constraint was rheonomic. Example 2.6 Another good example would be the ladder sliding against the wall problem that is used as an example in Hand and Finch Chapter 1. Try working it out yourself as we have worked out the examples above. 2.1.9 Special Nonconservative Cases So far we have only considered the Lagrangian when there is a potential energy function that can be derived from conservative forces. There are some special nonconservative cases in which the Lagrangian formalism can continue to be used with some modifications. The discussions of velocity-dependent potentials and general nonconservative forces are based on Goldstein Section 1.5. 92