2A First Approach to ClassicalMechanicsMotion in Rl2.12.1.1Newton'slawWe begin by considering the motion of a single particle in Ri, which maybe thought of as a particle sliding along a wire, or a particle with motionthat just happens to lie in a line. We let r(t) denote the particle's positionas a function of time. The particle's velocity is thenw(t) := z(t),where we use a dot over a symbol to denote the derivative of that quantitywith respect to the time t.The particle's acceleration is thena(t) =i(t) =(t),where denotes the second derivative of with respect to t.We assumethat there is a force acting on the particle and we assume at first that theforce F is a function of the particle's position only. (Later, we will look atthe case of forces that depend also on velocity.)Under these assumptions,Newton's second law(F=ma)takes theformF(r(t) =ma =mi(t),(2.1)where m is the mass of the particle, which is assumed to be positive.We willhenceforthabbreviate Newton's second lawas simply"Newton'slaw,"since19B.C.Hall, Quantum Theory for Mathematicians, GraduateTextsin Mathematics 267, DOI 10.1007/978-1-4614-7116-5_2,Springer Science+Business Media NewYork 2013
2 A First Approach to Classical Mechanics 2.1 Motion in R1 2.1.1 Newton’s law We begin by considering the motion of a single particle in R1, which may be thought of as a particle sliding along a wire, or a particle with motion that just happens to lie in a line. We let x(t) denote the particle’s position as a function of time. The particle’s velocity is then v(t) := ˙x(t), where we use a dot over a symbol to denote the derivative of that quantity with respect to the time t. The particle’s acceleration is then a(t)= ˙v(t)=¨x(t), where ¨x denotes the second derivative of x with respect to t. We assume that there is a force acting on the particle and we assume at first that the force F is a function of the particle’s position only. (Later, we will look at the case of forces that depend also on velocity.) Under these assumptions, Newton’s second law (F = ma) takes the form F(x(t)) = ma = mx¨(t), (2.1) where m is the mass of the particle, which is assumed to be positive. We will henceforth abbreviate Newton’s second law as simply “Newton’s law,” since B.C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, DOI 10.1007/978-1-4614-7116-5 2, © Springer Science+Business Media New York 2013 19
202.AFirst Approach to Classical Mechanicswe will use the second law much more frequently than the others. Since(2.1) is of second order, the appropriate initial conditions (needed to geta unique solution)are theposition and velocity at someinitial timeto.Sowe look for solutions of (2.1) subject to(to) = Toi(to) = Vo.Assuming that F is a smooth function, standard results from the ele-mentary theory of differential equations tell us that there exists a uniquelocal solution to (2.1)for each pair of initial conditions.(A local solutionis one defined for t in a neighborhood of the initial time to-) Since (2.1) isin general a nonlinear equation, one cannot expect that, for a general forcefunction F, the solutions will exist for all t.If, for example, F(r)= r?, thenany solution with positive initial position and positive initial velocity willescape to infinityin finite time.(ApplyExercise 4with V(r)=-r3/3.)For aproof existence and uniqueness, see Example 8.2and Theorem 8.13in [28].Definition 2.1 A solution (t)to Newton's law is called a trajectoryExample2.2(Harmonic Oscillator)If the force isgiven by Hooke'slaw, F(r)--kr, where k is a positive constant, then Newton's law can bewritten asmi+ka=0.Thegeneral solution of this equation isr(t) = acos(wt) +bsin(wt),wherew:=Vk/misthefrequencyof oscillation.The system in Example 2.2 is referred to as a (classical) harmonic os-cillator. This system can describe a mass on a spring, where the force isproportional to thedistance thatthe spring is stretched from its equi-librium position.The minus sign in -kr indicates that theforcepulls theoscillatorback toward equilibrium.Here and elsewherein the book.weuse the “angular" notion of frequency, which is the rate of change of theargument of a sine or cosine function. If w is the angular frequency, thenthe“ordinary" frequency-ie., the number of cycles per unit of time--isw/2π.Saying that has (angular)frequency w means that is periodicwithperiod 2元/w.2.1.2 Conservation of EnergyWe return now to the case of a general force function F(r).We definethe kinetic energy of the system to bem?.We also define thepotentialenergy of the system as the functionV(a) = -F(r) dr,(2.2)
20 2. A First Approach to Classical Mechanics we will use the second law much more frequently than the others. Since (2.1) is of second order, the appropriate initial conditions (needed to get a unique solution) are the position and velocity at some initial time t0. So we look for solutions of (2.1) subject to x(t0) = x0 x˙(t0) = v0. Assuming that F is a smooth function, standard results from the elementary theory of differential equations tell us that there exists a unique local solution to (2.1) for each pair of initial conditions. (A local solution is one defined for t in a neighborhood of the initial time t0.) Since (2.1) is in general a nonlinear equation, one cannot expect that, for a general force function F, the solutions will exist for all t. If, for example, F(x) = x2, then any solution with positive initial position and positive initial velocity will escape to infinity in finite time. (Apply Exercise 4 with V (x) = −x3/3.) For a proof existence and uniqueness, see Example 8.2 and Theorem 8.13 in [28]. Definition 2.1 A solution x(t) to Newton’s law is called a trajectory. Example 2.2 (Harmonic Oscillator) If the force is given by Hooke’s law, F(x) = −kx, where k is a positive constant, then Newton’s law can be written as mx¨ + kx = 0. The general solution of this equation is x(t) = a cos(ωt) + b sin(ωt), where ω := k/m is the frequency of oscillation. The system in Example 2.2 is referred to as a (classical) harmonic oscillator. This system can describe a mass on a spring, where the force is proportional to the distance x that the spring is stretched from its equilibrium position. The minus sign in −kx indicates that the force pulls the oscillator back toward equilibrium. Here and elsewhere in the book, we use the “angular” notion of frequency, which is the rate of change of the argument of a sine or cosine function. If ω is the angular frequency, then the “ordinary” frequency—i.e., the number of cycles per unit of time—is ω/2π. Saying that x has (angular) frequency ω means that x is periodic with period 2π/ω. 2.1.2 Conservation of Energy We return now to the case of a general force function F(x). We define the kinetic energy of the system to be 1 2mv2. We also define the potential energy of the system as the function V (x) = − F(x) dx, (2.2)
2.1 Motion in R121so that F(r) = -dV/dr. (The potential energy is defined only up to addinga constant.)The total energy E of the system is then1.E(m, 0) = mo° + V(n).(2.3)The chief significance of the energy function is that it is conserued, meaningthat its value along any trajectory is constant.Theorem 2.3 Suppose a particle satisfies Newton's law in the form miF(). Let V and E be as in (2.2) and (2.3). Then the energy E is conserved,meaning that for each solution r(t) of Newton's law, E(r(t),i(t) is inde-pendent of t.Proof. We verify this by differentiation, using the chain rule:ddm(i(t)? + V(r(t)E(r(t), (t) =dt,dv=mi(t)(t) +-(t)dr= (t)[m(t) - F(r(t)]This last expression is zero by Newton's law. Thus, the time-derivative ofthe energy along any trajectory is zero, so E(r(t),(t)) is independent oft,as claimed.We may call the energy a conserved quantity (or constant of motion),since the particle neither gains nor loses energy as the particle movesaccording to Newton's law.Let us see how conservation of energy helps us understand the solutionto Newton's law.We may reduce the second-order equation mi=F(r)toa pair of first-order equations, simply by introducing the velocity u as anew variable. That is, we look for pairs of functions (r(t),v(t)) that satisfythefollowing system of equationsdr= u(t)dtdu= 二F(r(t).(2.4)dt=mIf (r(t), u(t)) is a solution to this system, then we can immediately see thatr(t) satisfies Newton's law, just by substituting dr/dt for in the secondequation. We refer to the set of possible pairs of the form (r,) (i.e., R2)as the phase space of the particle in R1.The appropriate initial conditionsfor this first-order system are r(O) = ro and (O) = uo.Once we are working in phase space, we can use the conservation ofenergy to help us.Conservation of energy means that each solution to
2.1 Motion in R1 21 so that F(x) = −dV /dx. (The potential energy is defined only up to adding a constant.) The total energy E of the system is then E(x, v) = 1 2 mv2 + V (x). (2.3) The chief significance of the energy function is that it is conserved, meaning that its value along any trajectory is constant. Theorem 2.3 Suppose a particle satisfies Newton’s law in the form mx¨ = F(x). Let V and E be as in (2.2) and (2.3). Then the energy E is conserved, meaning that for each solution x(t) of Newton’s law, E(x(t), x˙(t)) is independent of t. Proof. We verify this by differentiation, using the chain rule: d dtE(x(t), x˙(t)) = d dt 1 2 m( ˙x(t))2 + V (x(t)) = mx˙(t)¨x(t) + dV dx x˙(t) = ˙x(t)[mx¨(t) − F(x(t))]. This last expression is zero by Newton’s law. Thus, the time-derivative of the energy along any trajectory is zero, so E(x(t), x˙(t)) is independent of t, as claimed. We may call the energy a conserved quantity (or constant of motion), since the particle neither gains nor loses energy as the particle moves according to Newton’s law. Let us see how conservation of energy helps us understand the solution to Newton’s law. We may reduce the second-order equation mx¨ = F(x) to a pair of first-order equations, simply by introducing the velocity v as a new variable. That is, we look for pairs of functions (x(t), v(t)) that satisfy the following system of equations dx dt = v(t) dv dt = 1 m F(x(t)). (2.4) If (x(t), v(t)) is a solution to this system, then we can immediately see that x(t) satisfies Newton’s law, just by substituting dx/dt for v in the second equation. We refer to the set of possible pairs of the form (x, v) (i.e., R2) as the phase space of the particle in R1. The appropriate initial conditions for this first-order system are x(0) = x0 and v(0) = v0. Once we are working in phase space, we can use the conservation of energy to help us. Conservation of energy means that each solution to
222.AFirstApproachtoClassical Mechanicsthe system (2.4) must lie entirely on a single "level curve" of the energyfunction,thatis,theset[(r,u) ERE(r,u) =E(ro,o)) .(2.5)If Fand thereforealso Vis smooth,then E is a smooth function of rand u. Then as long as (2.5) contains no critical points of E, this set willbe a smooth curvein R?, bythe implicit function theorem.If the level set(2.5)is alsoa simpleclosed curve,thenthesolutionsof(2.5)will simplywind around and around this curve.Thus,the set that thesolutionsto (2.5)trace out in phase space can be determined simply from the conservationof energy.The only thing not apparent at the moment is how this curve isparameterized as a function of time.In mechanics, a conserved quantitysuch as the energy in the one-dimensional version of Newton's law-is often referred to as an "integralofmotion."Thereason forthisisthatalthough Newton'ssecond lawisasecond-order equation in r,the energy depends only on r and and noton.Thus,theequationm(号() + V(t)= Eo,where Eo isthe value of theenergy attime to,is actually a first-orderdifferential equation.We can solve for to put this equation into a morestandard form:2(Eo-V((t))(2.6)(t) = ±mWhat this means is that by using conservation of energy we have turned theoriginal second-order equation into a first-order equation.We have therefore"integrated" the original equation once, that is, changed an equation ofthe form (t) - ... into an equation of the form i(t) = ... . The first-order equation (2.6) is separable and can be solved more-or-less explicitly(Exercise 1).2.1.3SystemswithDampingUp to now, we have considered forces that depend only on position.It iscommon, however, to consider forces that depend on the velocity as wellas the position. In the case of a damped harmonic oscillator, for example,one typically assumes that there is, in addition to the force of the spring,a damping force (friction, say) that is proportional to the velocity. Thus,F=-kr-yi, where k is, as before, the spring constant and where>0is the damping constant.The minus sign in front of reflects that thedamping force operates in the opposite direction to the velocity, causingtheparticletoslowdown.Theequationofmotionforsuchasystemisthenmi++ka=0
22 2. A First Approach to Classical Mechanics the system (2.4) must lie entirely on a single “level curve” of the energy function, that is, the set (x, v) ∈ R2 E(x, v) = E(x0, v0) . (2.5) If F—and therefore also V —is smooth, then E is a smooth function of x and v. Then as long as (2.5) contains no critical points of E, this set will be a smooth curve in R2, by the implicit function theorem. If the level set (2.5) is also a simple closed curve, then the solutions of (2.5) will simply wind around and around this curve. Thus, the set that the solutions to (2.5) trace out in phase space can be determined simply from the conservation of energy. The only thing not apparent at the moment is how this curve is parameterized as a function of time. In mechanics, a conserved quantity—such as the energy in the onedimensional version of Newton’s law—is often referred to as an “integral of motion.” The reason for this is that although Newton’s second law is a second-order equation in x, the energy depends only on x and ˙x and not on ¨x. Thus, the equation m 2 ( ˙x(t))2 + V (x(t)) = E0, where E0 is the value of the energy at time t0, is actually a first-order differential equation. We can solve for ˙x to put this equation into a more standard form: x˙(t) = ± 2(E0 − V (x(t))) m . (2.6) What this means is that by using conservation of energy we have turned the original second-order equation into a first-order equation. We have therefore “integrated” the original equation once, that is, changed an equation of the form ¨x(t) = ··· into an equation of the form ˙x(t) = ··· . The firstorder equation (2.6) is separable and can be solved more-or-less explicitly (Exercise 1). 2.1.3 Systems with Damping Up to now, we have considered forces that depend only on position. It is common, however, to consider forces that depend on the velocity as well as the position. In the case of a damped harmonic oscillator, for example, one typically assumes that there is, in addition to the force of the spring, a damping force (friction, say) that is proportional to the velocity. Thus, F = −kx − γx,˙ where k is, as before, the spring constant and where γ > 0 is the damping constant. The minus sign in front of γx˙ reflects that the damping force operates in the opposite direction to the velocity, causing the particle to slow down. The equation of motion for such a system is then mx¨ + γx˙ + kx = 0.
2.2 Motion in Rn23If is small, the solutions to this equation display decaying oscillation,meaning sines and cosines multiplied by a decaying exponential; if islarge, the solutions are pure decaying exponentials (Exercise 5).In the case of the damped harmonic oscillator,there is no longer aconserved energy. Specifically, there is no nonconstant continuous func-tion E on R2 such that E(r(t), (t)) is independent of t for all solutions ofNewton's law. To see this, we simply observe that for > O, all solutionsr(t) have the property that (r(t), (t)) tends to the origin in the plane as ttends to infinity. Thus, if E is continuous and constant along each trajec-tory, the value of E at the starting point has to be the same as the valueat the origin.Wenow considerageneral system with damping.Proposition 2.4 Suppose a particle moves in the presence of a force laugiven by F(r,) =Fi(r) -, with > 0. Define the energy E of thesystemby1mi?+V(z),E(r,)=where dV/dr =-Fi(r). Then along any trajectory r(t), we haved.E(r(t),(t)=-(t)2≤0dtThus, although the energy is not conserved, it is decreasing with time,whichgivesussomeinformationabout thebehaviorofthe systemProof.We differentiate as in the proof of Theorem 2.3, except that nowdV|dr =-Fi(r):是 (),() = ()[m() - ().dtSince F is not the full force function, the quantity in square brackets equalsnot zerobut -.Thus,dE/dt=-?.We can interpret Proposition 2.4as saying that in the presence of friction.the system we are studying gives up some of its energy to heat energy inthe environment, so that the energy of our system decreases with time.We will see that in higher dimensions.it is possible to have conservationof energy in the presence of velocity-dependent forces,provided that theseforces act perpendicularly to the velocity.2.2Motion in RnWe now consider a particle moving in Rn.Theposition x = (ri,..., n)of a particle is now a vector in Rn, as is the velocity v and acceleration a.Weletx= (1,...,in)
2.2 Motion in Rn 23 If γ is small, the solutions to this equation display decaying oscillation, meaning sines and cosines multiplied by a decaying exponential; if γ is large, the solutions are pure decaying exponentials (Exercise 5). In the case of the damped harmonic oscillator, there is no longer a conserved energy. Specifically, there is no nonconstant continuous function E on R2 such that E(x(t), x˙(t)) is independent of t for all solutions of Newton’s law. To see this, we simply observe that for γ > 0, all solutions x(t) have the property that (x(t), x˙(t)) tends to the origin in the plane as t tends to infinity. Thus, if E is continuous and constant along each trajectory, the value of E at the starting point has to be the same as the value at the origin. We now consider a general system with damping. Proposition 2.4 Suppose a particle moves in the presence of a force law given by F(x, x˙) = F1(x) − γx,˙ with γ > 0. Define the energy E of the system by E(x, x˙) = 1 2 mx˙ 2 + V (x), where dV /dx = −F1(x). Then along any trajectory x(t), we have d dtE(x(t), x˙(t)) = −γx˙(t) 2 ≤ 0. Thus, although the energy is not conserved, it is decreasing with time, which gives us some information about the behavior of the system. Proof. We differentiate as in the proof of Theorem 2.3, except that now dV /dx = −F1(x): d dtE(x(t), x˙(t)) = ˙x(t)[mx¨(t) − F1(x(t))]. Since F1 is not the full force function, the quantity in square brackets equals not zero but −γx.˙ Thus, dE/dt = −γx˙ 2. We can interpret Proposition 2.4 as saying that in the presence of friction, the system we are studying gives up some of its energy to heat energy in the environment, so that the energy of our system decreases with time. We will see that in higher dimensions, it is possible to have conservation of energy in the presence of velocity-dependent forces, provided that these forces act perpendicularly to the velocity. 2.2 Motion in Rn We now consider a particle moving in Rn. The position x = (x1,.,xn) of a particle is now a vector in Rn, as is the velocity v and acceleration a. We let x˙ = (˙x1,., x˙ n)