242.AFirstApproach to Classical Mechanicsdenote the derivative of x with respect to t and we let x denote the secondderivativeof xwithrespect tot.Newton'slawnowtakes theform(2.7)mx(t) = F(x(t),x(t),whereF :Rn× Rn→Rn is some force law, which in general may dependon both the position and velocity of the particle.We begin by considering forces that are independent of velocity, and welook fora conserved energy function in this setting.Proposition2.5 Consider Newton's law (2.7) in the case of a velocityindependentforce:mx(t)=F(x(t)).Thenan energyfunctionoftheform1m [x + V(x)E(x,x) =2is conservedifandonlyifVsatisfies-VV=F.where VV is the gradient of V.Saying that E is "conserved" means that E(x(t),x(t) is independent oft for each solution x(t) of Newton's law.The function V is the potentialenergyofthesystemProof.Differentiating gives"av/(m[x(t)2+ V(x(t))) =m;(t);(t)+ti(t)dtdrj=i=x(t)· [mx(t) +VVl= x(t) · [F(x) + VV(x)]Thus,dE/dtwill alwaysbeequal to zero if and only if we have-VV(x) = F(x)forall x.We now encounter something that did not occur in the one-dimensionalcase. In Ri,any smooth function can be expressed as the derivative of someother function. In Rn, however, not every vector-valued function F(x) canbe expressed as the (negative of) the gradient of some scalar-valued functionV.Definition 2.6 Suppose F is a smooth, Rn-valued function on a domainU C Rn.Then F is called conservative if there erists a smooth, real-valuedfunctionVonUsuchthatF=-VV.If the domain U is simply connected, then there is a simple local conditionthat characterizes conservative functions
24 2. A First Approach to Classical Mechanics denote the derivative of x with respect to t and we let x¨ denote the second derivative of x with respect to t. Newton’s law now takes the form mx¨(t) = F(x(t), x˙(t)), (2.7) where F : Rn × Rn → Rn is some force law, which in general may depend on both the position and velocity of the particle. We begin by considering forces that are independent of velocity, and we look for a conserved energy function in this setting. Proposition 2.5 Consider Newton’s law (2.7) in the case of a velocityindependent force: mx¨(t) = F(x(t)). Then an energy function of the form E(x, x˙) = 1 2 m |x˙| 2 + V (x) is conserved if and only if V satisfies −∇V = F, where ∇V is the gradient of V. Saying that E is “conserved” means that E(x(t), x˙(t)) is independent of t for each solution x(t) of Newton’s law. The function V is the potential energy of the system. Proof. Differentiating gives d dt 1 2 m |x˙(t)| 2 + V (x(t)) = mn j=i x˙ j (t)¨xj (t) +n j=1 ∂V ∂xj x˙ j (t) = x˙(t) · [mx¨(t) + ∇V ] = x˙(t) · [F(x) + ∇V (x)] Thus, dE/dt will always be equal to zero if and only if we have −∇V (x) = F(x) for all x. We now encounter something that did not occur in the one-dimensional case. In R1, any smooth function can be expressed as the derivative of some other function. In Rn, however, not every vector-valued function F(x) can be expressed as the (negative of) the gradient of some scalar-valued function V. Definition 2.6 Suppose F is a smooth, Rn-valued function on a domain U ⊂ Rn. Then F is called conservative if there exists a smooth, real-valued function V on U such that F = −∇V. If the domain U is simply connected, then there is a simple local condition that characterizes conservative functions
2.2 Motion in R"25Proposition 2.7 Suppose U is a simply connected domain in Rn andFis a smooth, Rn-valued function on U. Then F is conservative if and onlyfF satisfiesOFi_OFk=0(2.8)Orkorjat each point in U.When n = 3, it is easy to check that the condition (2.8) is equivalentto the curl ×F of F being zero on U.The hypothesis that U be simplyconnected cannot be omitted:seeExercise7Proof.IfFis conservative,then02VaFkaFarkOrroryOrjokOrjat every point in U.In the other direction, if F satisfies (2.8),V can beobtained by integrating F along paths and using the Stokes theorem toestablish independence of choice of path.See,for example,Theorem 4.3onP.549 of [44] for aproof in the n =3 case.The proof in higher dimensionsis the same, provided one knows the general version of the Stokes theorem.We may also consider velocity-dependent forces.If, for example, F(x,v)= -v+Fi(x), where is a positive constant, then we will again haveenergy that is decreasing with time.There is another new phenomenon,however, in dimension greater than 1, namely the possibility of having aconserved energy even when the force depends on velocity.Proposition 2.8 Suppose a particle in Rn moves inthe presence of a forceFof the formF(x, v) = -VV(x) + F2(x, v),where V is a smooth function and where F2 satisfies(2.9)v. F2(x, v) = 0for all x and v in IRn. Then the energy function E(x,v) = m v/+V(x)is constant along each trajectory.If, for example, F2 is the force exerted on a charged particle in R3 by amagnetic field B(x), thenF2(x, v) = qv × B(x),where q is the charge of the particle, which clearly satisfies (2.9)Proof.SeeExercise8
2.2 Motion in Rn 25 Proposition 2.7 Suppose U is a simply connected domain in Rn and F is a smooth, Rn-valued function on U. Then F is conservative if and only if F satisfies ∂Fj ∂xk − ∂Fk ∂xj = 0 (2.8) at each point in U. When n = 3, it is easy to check that the condition (2.8) is equivalent to the curl ∇ × F of F being zero on U. The hypothesis that U be simply connected cannot be omitted; see Exercise 7. Proof. If F is conservative, then ∂Fj ∂xk = − ∂2V ∂xk∂xj = − ∂2V ∂xj∂xk = ∂Fk ∂xj at every point in U. In the other direction, if F satisfies (2.8), V can be obtained by integrating F along paths and using the Stokes theorem to establish independence of choice of path. See, for example, Theorem 4.3 on p. 549 of [44] for a proof in the n = 3 case. The proof in higher dimensions is the same, provided one knows the general version of the Stokes theorem. We may also consider velocity-dependent forces. If, for example, F(x, v) = −γv + F1(x), where γ is a positive constant, then we will again have energy that is decreasing with time. There is another new phenomenon, however, in dimension greater than 1, namely the possibility of having a conserved energy even when the force depends on velocity. Proposition 2.8 Suppose a particle in Rn moves in the presence of a force F of the form F(x, v) = −∇V (x) + F2(x, v), where V is a smooth function and where F2 satisfies v · F2(x, v) = 0 (2.9) for all x and v in Rn. Then the energy function E(x, v) = 1 2m |v| 2 + V (x) is constant along each trajectory. If, for example, F2 is the force exerted on a charged particle in R3 by a magnetic field B(x), then F2(x, v) = qv × B(x), where q is the charge of the particle, which clearly satisfies (2.9). Proof. See Exercise 8
262.AFirstApproachtoClassicalMechanics2.3SystemsofParticlesIf we have a system if N particles, each moving in IR", then we denote theposition of thejth particlebyx=(Ti,...,)Thus, in the expression r,, the superscript j indicates the jth particle, whilethe subscript k indicates thekth component.Newton's lawthen takes theformmjx -F'(x,.,xN,xl,.,xN),j=1,2,...,N,where mj is the mass of the jth particle. Here, Fj is the force on the jthparticle, which in general will depend on the position and velocity not onlyof that particle, but also on the position and velocity of the other particles.2.3.1Conservation of EnergyIna system of particles,wecannot expectthatthe energy ofeach individ-ual particle will be conserved, because as the particles interact, they canexchange energy. Rather, we should expect that, under suitable assump-tions on the forces Fj, we can define a conserved energy function for thewhole system (the total energy of the system).Let us consider forces depending only on the position of the particles,and let us assume that the energy function will be of the formN7E(x....,xN,vl,..,vN) =Im, +V(x....xN).(2.10)Wewill nowtrytoseewhatformforV (ifany)will allowEtobeconstantalong eachtrajectory.Proposition 2.9 An energy function of the form (2.10) is constant alongeach trajectory ifViV=-Fj(2.11)for each j, where Vj is the gradient with respect to the variable xProof.WecomputethatNdEZ[mjxi.x+viv.]dtj=1NEx.[m,x+viv]j=iNEx·[Fi+vi],j=i
26 2. A First Approach to Classical Mechanics 2.3 Systems of Particles If we have a system if N particles, each moving in Rn, then we denote the position of the jth particle by xj = (xj 1,.,xj n). Thus, in the expression xj k, the superscript j indicates the jth particle, while the subscript k indicates the kth component. Newton’s law then takes the form mjx¨j = Fj (x1,., xN , x˙ 1,., x˙ N ), j = 1, 2,.,N, where mj is the mass of the jth particle. Here, Fj is the force on the jth particle, which in general will depend on the position and velocity not only of that particle, but also on the position and velocity of the other particles. 2.3.1 Conservation of Energy In a system of particles, we cannot expect that the energy of each individual particle will be conserved, because as the particles interact, they can exchange energy. Rather, we should expect that, under suitable assumptions on the forces Fj , we can define a conserved energy function for the whole system (the total energy of the system). Let us consider forces depending only on the position of the particles, and let us assume that the energy function will be of the form E(x1,., xN , v1,., vN ) = N j=1 1 2 mj vj 2 + V (x1,., xN ). (2.10) We will now try to see what form for V (if any) will allow E to be constant along each trajectory. Proposition 2.9 An energy function of the form (2.10) is constant along each trajectory if ∇jV = −Fj (2.11) for each j, where ∇j is the gradient with respect to the variable xj . Proof. We compute that dE dt = N j=1 mjx˙ j · x¨j + ∇jV · x˙ j = N j=1 x˙ j · mjx¨j + ∇jV = N j=1 x˙ j · Fj + ∇jV
272.3Systems of ParticlesIf viv =-Fi, then E will be conserved.As in the one-particle case, there is a simple condition for the existenceofapotential function V satisfying (2.11)Proposition 2.10 Suppose a force function F = (F1,..,FN) is definedon a simply connected domain U in RnN. Then there erists a smoothfunction V on U satisfyingviV=-Fifor all j if and only if we haveOFOFl(2.12)arm"arkforall j,k,l,andmProof.ApplyProposition 2.7withnreplaced bynN and withjandkreplaced bythepairs (j,k)and (l,m).2.3.2ConservationofMomentumWe now introduce the notion of the momentum of a particle.Definition 2.1l In anN-particle system, the momentum of the jthparticle, denoted p', is the product of the mass and the velocity of thatparticle:pi=m,xjThe total momentum of the system, denoted p, is defined asNp=pij=1Observethatdpidt=mx=FiThus, Newton's law may be reformulated as saying,"The force is the rateof change of the momentum." This is how Newton originally formulatedhissecondlawNewton's third law says,“For every action, thereis an equal and oppositereaction."This law will apply if all forces are of the"two-particle"varietyand satisfy a natural symmetry property. Having two-particle forces meansthat theforce Fi on the jth particle is a sum of terms Fj.k, j+ k, whereFj.k depends only x and xk. The relevant symmetry property is thatFj.k(x,x)=-Fk.j(x,x); that is, the force exerted by the jth particleon the kth particle is the negative (i.e.,"equal and opposite")of the force
2.3 Systems of Particles 27 If ∇jV = −Fj, then E will be conserved. As in the one-particle case, there is a simple condition for the existence of a potential function V satisfying (2.11). Proposition 2.10 Suppose a force function F = (F1,., FN ) is defined on a simply connected domain U in RnN . Then there exists a smooth function V on U satisfying ∇jV = −Fj for all j if and only if we have ∂Fj k ∂xl m = ∂Fl m ∂xj k (2.12) for all j, k, l, and m. Proof. Apply Proposition 2.7 with n replaced by nN and with j and k replaced by the pairs (j, k) and (l,m). 2.3.2 Conservation of Momentum We now introduce the notion of the momentum of a particle. Definition 2.11 In an N-particle system, the momentum of the jth particle, denoted pj , is the product of the mass and the velocity of that particle: pj = mjx˙ j . The total momentum of the system, denoted p, is defined as p = N j=1 pj . Observe that dpj dt = mjx¨j= Fj . Thus, Newton’s law may be reformulated as saying, “The force is the rate of change of the momentum.” This is how Newton originally formulated his second law. Newton’s third law says, “For every action, there is an equal and opposite reaction.” This law will apply if all forces are of the “two-particle” variety and satisfy a natural symmetry property. Having two-particle forces means that the force Fj on the jth particle is a sum of terms Fj,k, j = k, where Fj,k depends only xj and xk. The relevant symmetry property is that Fj,k(xj , xk) = −Fk,j (xk, xj ); that is, the force exerted by the jth particle on the kth particle is the negative (i.e., “equal and opposite”) of the force
282.AFirstApproach to Classical Mechanicsexerted by the kth particle on the jth particle.If the forces are assumedalso to be conservative, then the potential energy of the system will be oftheformV(x,x2,..,xN) =vi.k(xi - xk).(2.13)j<kOneimportant consequenceof Newton's third lawis conservation of thetotal momentumof the systemProposition2.12Supposethatforeachj,theforceonthejthparticlesoftheformFi(xl,x2,...,xN) =Fj.(x),xk),kjfor certain functions Fj.k. Suppose also that we have the "equal andopposite"conditionFi.k(x),xk) =-Fkj(xj,xh).Thenthetotal momentumof the systemis conserved.Note that since the rate of change of pj is Fi, the force on the jthparticle, the momentum of each individual particle is not constant in time,except in the trivial case of a noninteracting system (one in which all forcesare zero),Proof.DifferentiatinggivesNRdp _ZFi -ZEFik(x,x)dtdtj=1j=1jk+iBy theequal and opposite condition,Fj.k(x,xk)cancels with Fk.j (x,xk)so dp/dt=0.Let us consider, now, a more general situation in which we have con-servative forces, but not necessarily of the“two-particle"form.It is stillpossibletohave conservation of momentum, as thefollowing result shows.Proposition2.13 If a multiparticle systemhas a forcelaw comingfroma potential V, then the total momentum of the system is conserved if andonlyifV(xl +a,x? +a....,xN +a) = V(x,x?.....xN)(2.14)forallaeRnProof.Apply (2.14)with a =tek,where ex is the vector with a 1 in thekth spot and zeros elsewhere. Differentiating with respect to t at t - 0givesNNdpkNavdpk-7F=->0=>dtdtarkj=1j=1=
28 2. A First Approach to Classical Mechanics exerted by the kth particle on the jth particle. If the forces are assumed also to be conservative, then the potential energy of the system will be of the form V (x1, x2,., xN ) = j<k V j,k(xj − xk). (2.13) One important consequence of Newton’s third law is conservation of the total momentum of the system. Proposition 2.12 Suppose that for each j, the force on the jth particle is of the form Fj (x1, x2,., xN ) = k=j Fj,k(xj , xk), for certain functions Fj,k. Suppose also that we have the “equal and opposite” condition Fj,k(xj , xk) = −Fk,j (xj , xk). Then the total momentum of the system is conserved. Note that since the rate of change of pj is Fj , the force on the jth particle, the momentum of each individual particle is not constant in time, except in the trivial case of a noninteracting system (one in which all forces are zero). Proof. Differentiating gives dp dt = N j=1 dpj dt = N j=1 Fj = j k=j Fj,k(xj , xk). By the equal and opposite condition, Fj,k(xj , xk) cancels with Fk,j (xj , xk), so dp/dt = 0. Let us consider, now, a more general situation in which we have conservative forces, but not necessarily of the “two-particle” form. It is still possible to have conservation of momentum, as the following result shows. Proposition 2.13 If a multiparticle system has a force law coming from a potential V, then the total momentum of the system is conserved if and only if V (x1 + a, x2 + a,., xN + a) = V (x1, x2,., xN ) (2.14) for all a ∈ Rn. Proof. Apply (2.14) with a = tek, where ek is the vector with a 1 in the kth spot and zeros elsewhere. Differentiating with respect to t at t = 0 gives 0 = N j=1 ∂V ∂xj k = − N j=1 Fj k = − N j=1 dpj k dt = −dpk dt