. The kinetic energy of this system is T = T i. m i. Now let s consider how the kinetic energy of the system changes in time. Assuming each.

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1 Chapter 2 Systems of Partcles 2. Work-Energy Theorem Consder a system of many partcles, wth postons r and veloctes ṙ. The knetc energy of ths system s T = T = 2 mṙ2. 2. Now let s consder how the knetc energy of the system changes n tme. Assumng each m s tme-ndependent, we have dt dt = m ṙ r. 2.2 Here, we ve used the relaton d A 2 = 2A da dt dt. 2.3 We now nvoke Newton s 2nd Law, m r = F, to wrte eqn. 2.2 as T = F ṙ. We ntegrate ths equaton from tme t A to t B : T B t B T A = dt t A t B dt dt = dt F ṙ W A B, 2.4 t A where W A B s the total work done on partcle durng ts moton from state A to state B, Clearly the total knetc energy s T = T and the total work done on all partcles s W A B = W A B. Eqn. 2.4 s known as the work-energy theorem. It says that In the evoluton of a mechancal system, the change n total knetc energy s equal to the total work done: T B T A = W A B.

2 2 CHAPTER 2. SYSTEMS OF PARTICLES Fgure 2.: Two paths jonng ponts A and B. 2.2 Conservatve and Nonconservatve Forces For the sake of smplcty, consder a sngle partcle wth knetc energy T = 2 mṙ2. The work done on the partcle durng ts mechancal evoluton s W A B = t B t A dt F v, 2.5 where v = ṙ. Ths s the most general expresson for the work done. If the force F depends only on the partcle s poston r, we may wrte dr = v dt, and then Consder now the force W A B = r B r A dr F r. 2.6 F r = K y ˆx + K 2 xŷ, 2.7 where K,2 are constants. Let s evaluate the work done along each of the two paths n fg. 2.: x B y B W I = K dx y A + K 2 dy x B = K y A x B x A + K 2 x B y B y A 2.8 x A y A x B y B W II = K dx y B + K 2 dy x A = K y B x B x A + K 2 x A y B y A. 2.9 x A y A

3 2.2. CONSERVATIVE AND NONCONSERVATIVE FORCES 3 Note that n general W I W II. Thus, f we start at pont A, the knetc energy at pont B wll depend on the path taken, snce the work done s path-dependent. The dfference between the work done along the two paths s W I W II = K 2 K x B x A y B y A. 2.0 Thus, we see that f K = K 2, the work s the same for the two paths. In fact, f K = K 2, the work would be path-ndependent, and would depend only on the endponts. Ths s true for any path, and not just pecewse lnear paths of the type depcted n fg. 2.. The reason for ths s Stokes theorem: dl F = ds ˆn F. 2. C C Here, C s a connected regon n three-dmensonal space, C s mathematcal notaton for the boundary of C, whch s a closed path, ds s the scalar dfferental area element, ˆn s the unt normal to that dfferental area element, and F s the curl of F : ˆx ŷ ẑ F = det x y z F x F y F z Fz = y F y ˆx + z Fx z F z Fy ŷ + x x F x ẑ. 2.2 y For the force under consderaton, F r = K y ˆx + K 2 xŷ, the curl s F = K 2 K ẑ, 2.3 whch s a constant. The RHS of eqn. 2. s then smply proportonal to the area enclosed by C. When we compute the work dfference n eqn. 2.0, we evaluate the ntegral C dl F along the path γ II γ I, whch s to say path I followed by the nverse of path II. In ths case, ˆn = ẑ and the ntegral of ˆn F over the rectangle C s gven by the RHS of eqn When F = 0 everywhere n space, we can always wrte F = U, where Ur s the potental energy. Such forces are called conservatve forces because the total energy of the system, E = T + U, s then conserved durng ts moton. We can see ths by evaluatng the work done, r B W A B = dr F r r A r B = dr U r A = Ur A Ur B. 2.4 If C s multply connected, then C s a set of closed paths. For example, f C s an annulus, C s two crcles, correspondng to the nner and outer boundares of the annulus.

4 4 CHAPTER 2. SYSTEMS OF PARTICLES The work-energy theorem then gves T B T A = Ur A Ur B, 2.5 whch says E B = T B + Ur B = T A + Ur A = E A. 2.6 Thus, the total energy E = T + U s conserved Example : ntegratng F = U If F = 0, we can compute Ur by ntegratng, vz. Ur = U0 r 0 dr F r. 2.7 The ntegral does not depend on the path chosen connectng 0 and r. For example, we can take x,0,0 x,y,0 x,y,z Ux,y,z = U0,0,0 dx F x x,0,0 dy F y x,y,0 dz F z x,y,z ,0,0 x,0,0 z,y,0 The constant U0,0,0 s arbtrary and mpossble to determne from F alone. As an example, consder the force F r = ky ˆx kxŷ 4bz 3 ẑ, 2.9 where k and b are constants. We have F x = Fz y F y = z F y = Fx z F z = x F z = Fy x F x = 0, 2.22 y so F = 0 and F must be expressble as F = U. Integratng usng eqn. 2.8, we have x,0,0 x,y,0 x,y,z Ux,y,z = U0,0,0 + dx k 0 + dy kxy + dz 4bz ,0,0 x,0,0 z,y,0 = U0,0,0 + kxy + bz

5 2.3. CONSERVATIVE FORCES IN MANY PARTICLE SYSTEMS 5 Another approach s to ntegrate the partal dfferental equaton U = F. Ths s n fact three equatons, and we shall need all of them to obtan the correct answer. We start wth the ˆx-component, = ky x Integratng, we obtan Ux,y,z = kxy + fy,z, 2.26 where fy, z s at ths pont an arbtrary functon of y and z. The mportant thng s that t has no x-dependence, so f/ x = 0. Next, we have y Fnally, the z-component ntegrates to yeld We now equate the frst two expressons: = kx = Ux,y,z = kxy + gx,z z = 4bz3 = Ux,y,z = bz 4 + hx,y kxy + fy,z = kxy + gx,z Subtractng kxy from each sde, we obtan the equaton fy, z = gx, z. Snce the LHS s ndependent of x and the RHS s ndependent of y, we must have fy,z = gx,z = qz, 2.30 where qz s some unknown functon of z. But now we nvoke the fnal equaton, to obtan bz 4 + hx,y = kxy + qz. 2.3 The only possble soluton s hx,y = C + kxy and qz = C + bz 4, where C s a constant. Therefore, Ux,y,z = C + kxy + bz Note that t would be very wrong to ntegrate / x = ky and obtan Ux,y,z = kxy + C, where C s a constant. As we ve seen, the constant of ntegraton we obtan upon ntegratng ths frst order PDE s n fact a functon of y and z. The fact that fy,z carres no explct x dependence means that f/ x = 0, so by constructon U = kxy + fy, z s a soluton to the PDE / x = ky, for any arbtrary functon fy,z. 2.3 Conservatve Forces n Many Partcle Systems T = U = 2 m ṙ V r + <j v r r j. 2.34

6 6 CHAPTER 2. SYSTEMS OF PARTICLES Here, V r s the external or one-body potental, and vr r s the nterpartcle potental, whch we assume to be central, dependng only on the dstance between any par of partcles. The equatons of moton are wth m r = F ext + F nt, 2.35 F ext = V r 2.36 r F nt = v r r j F nt j r j j Here, F nt j s the force exerted on partcle by partcle j: F nt j = v r r j = r r j r r r j v r r j Note that F nt j = F nt j, otherwse known as Newton s Thrd Law. It s convenent to abbrevate r j r r j, n whch case we may wrte the nterpartcle force as F nt j = ˆr j v r j Lnear and Angular Momentum Consder now the total momentum of the system, P = p. Its rate of change s dp dt = ṗ = F ext + F nt j +F nt j =0 {}}{ j F nt j = F ext tot, 2.40 snce the sum over all nternal forces cancels as a result of Newton s Thrd Law. We wrte P = m ṙ = MṘ 2.4 M = m total mass 2.42 R = m r center-of-mass m Next, consder the total angular momentum, L = r p = m r ṙ. 2.44

7 2.4. LINEAR AND ANGULAR MOMENTUM 7 The rate of change of L s then dl dt = = { m ṙ ṙ + m r r } r F ext + j r F nt j = r F ext + 2 j r j F nt j =0 { }}{ r r j F nt j = N ext tot Fnally, t s useful to establsh the result T = 2 m ṙ 2 = 2 MṘ2 + 2 m ṙ Ṙ 2, 2.46 whch says that the knetc energy may be wrtten as a sum of two terms, those beng the knetc energy of the center-of-mass moton, and the knetc energy of the partcles relatve to the center-of-mass. Recall the work-energy theorem for conservatve systems, fnal fnal fnal 0 = de = dt + du ntal ntal ntal = T B T A dr F, 2.47 whch s to say T = T B T A = dr F = U In other words, the total energy E = T + U s conserved: E = 2 m ṙ2 + V r + <j v r r j Note that for contnuous systems, we replace sums by ntegrals over a mass dstrbuton, vz. m φ r d 3 r ρrφr, 2.50 where ρr s the mass densty, and φr s any functon.

8 8 CHAPTER 2. SYSTEMS OF PARTICLES 2.5 Scalng of Solutons for Homogeneous Potentals 2.5. Euler s theorem for homogeneous functons In certan cases of nterest, the potental s a homogeneous functon of the coordnates. Ths means U λr,...,λr N = λ k U r,...,r N. 2.5 Here, k s the degree of homogenety of U. Famlar examples nclude gravty, and the harmonc oscllator, U r,...,r N = G U q,...,q n = 2 <j m m j r r j ; k =, 2.52 σ,σ V σσ q σ q σ ; k = The sum of two homogeneous functons s tself homogeneous only f the component functons themselves are of the same degree of homogenety. Homogeneous functons obey a specal result known as Euler s Theorem, whch we now prove. Suppose a multvarable functon Hx,...,x n s homogeneous: Hλx,...,λx n = λ k Hx,...,x n Then d H n H λx dλ,...,λx n = x = k H 2.55 x λ= = Scaled equatons of moton Now suppose the we rescale dstances and tmes, defnng r = α r, t = β t Then The force F s gven by dr dt = α β d r d t F = r U r,...,r N, = α r αk U r,..., r N d 2 r dt 2 = α β 2 d 2 r d t = α k F. 2.58

9 2.5. SCALING OF SOLUTIONS FOR HOMOGENEOUS POTENTIALS 9 Thus, Newton s 2nd Law says If we choose β such that We now demand α β 2 m d 2 r d t 2 = αk F α β 2 = αk β = α 2 k, 2.60 then the equaton of moton s nvarant under the rescalng transformaton! Ths means that f rt s a soluton to the equatons of moton, then so s α r α 2 k t. Ths gves us an entre one-parameter famly of solutons, for all real postve α. If rt s perodc wth perod T, the r t;α s perodc wth perod T = α 2 k T. Thus, T L 2 k =. 2.6 T L Here, α = L /L s the rato of length scales. Veloctes, energes and angular momenta scale accordngly: [ v ] = L T [ E ] = ML 2 T 2 [ L ] = ML 2 T v / v = L T L T = α 2 k 2.62 E E = L L = L 2 / T 2 = α k 2.63 L T L 2 / T L T = α+ 2 k As examples, consder: Harmonc Oscllator : Here k = 2 and therefore q σ t q σ t;α = α q σ t Thus, rescalng lengths alone gves another soluton. Kepler Problem : Ths s gravty, for whch k =. Thus, rt rt;α = α r α 3/2 t Thus, r 3 t 2,.e. L 3 T 2 =, 2.67 also known as Kepler s Thrd Law. L T

10 0 CHAPTER 2. SYSTEMS OF PARTICLES 2.6 Appendx I : Curvlnear Orthogonal Coordnates The standard cartesan coordnates are {x,...,x d }, where d s the dmenson of space. Consder a dfferent set of coordnates, {q,...,q d }, whch are related to the orgnal coordnates x µ va the d equatons q µ = q µ x,...,x d In general these are nonlnear equatons. Let ê 0 = ˆx be the Cartesan set of orthonormal unt vectors, and defne ê µ to be the unt vector perpendcular to the surface dq µ = 0. A dfferental change n poston can now be descrbed n both coordnate systems: d d ds = ê 0 dx = ê µ h µ qdq µ, 2.69 = where each h µ q s an as yet unknown functon of all the components q ν. Fndng the coeffcent of dq µ then gves h µ qê µ = d = µ= x q µ ê 0 ê µ = d M µ ê 0, 2.70 where M µ q = x. 2.7 h µ q q µ The dot product of unt vectors n the new coordnate system s then ê µ ê ν = MM t µν = d x x h µ qh ν q q µ q ν The condton that the new bass be orthonormal s then d x x = h 2 q µ q µqδ µν ν Ths gves us the relaton Note that = = = h µ q = d 2 x q µ ds 2 = = d h 2 µ qdq µ µ= For general coordnate systems, whch are not necessarly orthogonal, we have d ds 2 = g µν qdq µ dq ν, 2.76 µ,ν= where g µν q s a real, symmetrc, postve defnte matrx called the metrc tensor.

11 2.6. APPENDIX I : CURVILINEAR ORTHOGONAL COORDINATES Fgure 2.2: Volume element Ω for computng dvergences Example : sphercal coordnates Consder sphercal coordnates ρ,θ,φ: x = ρ sn θ cos φ, y = ρ sn θ snφ, z = ρ cos θ It s now a smple matter to derve the results h 2 ρ =, h 2 θ = ρ2, h 2 φ = ρ2 sn 2 θ Thus, ds = ˆρ dρ + ρ ˆθ dθ + ρ sn θ ˆφ dφ Vector calculus : grad, dv, curl Here we restrct our attenton to d = 3. The gradent U of a functon Uq s defned by du = q dq + q 2 dq 2 + q 3 dq 3 U ds Thus, = ê h q q + ê2 h 2 q q 2 + ê3 h 3 q q For the dvergence, we use the dvergence theorem, and we appeal to fg. 2.2: dv A = ds ˆn A, 2.82 Ω Ω

12 2 CHAPTER 2. SYSTEMS OF PARTICLES where Ω s a regon of three-dmensonal space and Ω s ts closed two-dmensonal boundary. The LHS of ths equaton s The RHS s LHS = A h dq h 2 dq 2 h 3 dq q +dq q RHS = A h 2 h 3 dq 2 dq 3 + A 2 h h 2+dq 2 q 3 dq dq 3 + A 3 h h +dq 3 2 dq dq 2 q q 2 q [ ] 3 = A h q 2 h 3 + A2 h q h 3 + A3 h 2 q h 2 dq dq 2 dq We therefore conclude A = h h 2 h 3 [ ] A h q 2 h 3 + A2 h q h 3 + A3 h 2 q h To obtan the curl A, we use Stokes theorem agan, ds ˆn A = dl A, 2.86 Σ where Σ s a two-dmensonal regon of space and Σ s ts one-dmensonal boundary. Now consder a dfferental surface element satsfyng dq = 0,.e. a rectangle of sde lengths h 2 dq 2 and h 3 dq 3. The LHS of the above equaton s The RHS s Σ LHS = ê Ah 2 dq 2 h 3 dq q 2 +dq 2 q RHS = A 3 h 3 dq 3 A 2 h 3+dq 3 2 dq 2 q 2 q [ 3 ] = A3 h 3 A2 h q 2 q 2 dq 2 dq Therefore A = h3 A 3 h 2 A h 2 h 3 q 2 q 3 Ths s one component of the full result h ê h 2 ê 2 h 3 ê 3 A = det q h h 2 h q 2 q 3 2 h A h 2 A 2 h 3 A 3 The Laplacan of a scalar functon U s gven by 2 U = U { h2 h 3 = h h 2 h 3 q h + h h 3 q q 2 h 2 + h h 2 q 2 q 3 h }. 2.9 q 3

13 2.7. COMMON CURVILINEAR ORTHOGONAL SYSTEMS Common curvlnear orthogonal systems 2.7. Rectangular coordnates In rectangular coordnates x,y,z, we have h x = h y = h z = Thus and the velocty squared s ds = ˆxdx + ŷ dy + ẑ dz 2.93 ṡ 2 = ẋ 2 + ẏ 2 + ż The gradent s The dvergence s The curl s A = U = ˆx x + ŷ y + ẑ z A = A x x + A y y + A z z Az y A y Ax ˆx + z z A z Ay ŷ + x x A x ẑ y The Laplacan s 2 U = 2 U x U y U z Cylndrcal coordnates In cylndrcal coordnates ρ, φ, z, we have ˆρ = ˆx cos φ + ŷ snφ ˆx = ˆρ cos φ ˆφ snφ dˆρ = ˆφ dφ 2.99 ˆφ = ˆx sn φ + ŷ cos φ ŷ = ˆρ sn φ + ˆφ cos φ d ˆφ = ˆρdφ The metrc s gven n terms of h ρ =, h φ = ρ, h z =. 2.0 Thus and the velocty squared s ds = ˆρ dρ + ˆφ ρdφ + ẑ dz 2.02 ṡ 2 = ρ 2 + ρ 2 φ2 + ż

14 4 CHAPTER 2. SYSTEMS OF PARTICLES The gradent s The dvergence s The curl s A = U = ˆρ ρ + ˆφ ρ A = ρ ρa ρ ρ φ + ẑ z ρ A φ φ + A z z A z ρ φ A φ Aρ ˆρ + z z A z ρa φ ˆφ + ρ ρ ρ ρ A ρ ẑ φ The Laplacan s 2 U = ρ ρ + 2 U ρ ρ ρ 2 φ U z Sphercal coordnates In sphercal coordnates r,θ,φ, we have ˆr = ˆxsn θ cos φ + ŷ sn θ sn φ + ẑ sn θ 2.08 ˆθ = ˆxcos θ cos φ + ŷ cos θ sn φ ẑ cos θ 2.09 ˆφ = ˆxsn φ + ŷ cos φ, 2.0 for whch ˆr ˆθ = ˆφ, ˆθ ˆφ = ˆr, ˆφ ˆr = ˆθ. 2. The nverse s ˆx = ˆr sn θ cos φ + ˆθ cos θ cos φ ˆφ sn φ 2.2 ŷ = ˆr sn θ sn φ + ˆθ cos θ sn φ + ˆφ cos φ 2.3 ẑ = ˆr cos θ ˆθ snθ. 2.4 The dfferental relatons are dˆr = ˆθ dθ + sn θ ˆφ dφ 2.5 dˆθ = ˆr dθ + cos θ ˆφ dφ 2.6 d ˆφ = sn θ ˆr + cos θ ˆθ dφ 2.7 The metrc s gven n terms of h r =, h θ = r, h φ = r sn θ. 2.8 Thus ds = ˆr dr + ˆθ r dθ + ˆφr sn θ dφ 2.9

15 2.7. COMMON CURVILINEAR ORTHOGONAL SYSTEMS 5 and the velocty squared s ṡ 2 = ṙ 2 + r 2 θ2 + r 2 sn 2 θ φ The gradent s The dvergence s U = ˆr ρ + ˆθ r θ + ˆφ r sn θ φ. 2.2 The curl s The Laplacan s A = r 2 r 2 A r r A = r sn θ 2 U = r 2 r sn θ Aφ θ + raθ r r r 2 + r + sn θ A θ + A φ r sn θ θ r sn θ φ r 2 sn θ A θ φ A r θ A r sn θ φ ra φ ˆθ r ˆr + r ˆφ snθ + θ θ 2 U r 2 sn 2 θ φ Knetc energy Note the form of the knetc energy of a pont partcle: ds 2 T = 2 m = dt 2 m ẋ 2 + ẏ 2 + ż 2 3D Cartesan 2.25 = 2 m ρ 2 + ρ 2 φ2 2D polar 2.26 = 2 m ρ 2 + ρ 2 φ2 + ż 2 3D cylndrcal 2.27 = 2 m ṙ 2 + r 2 θ2 + r 2 sn 2 θ φ 2 3D polar. 2.28