Lecture Notes on Classical Mechanics

advanced classical mechanics lecture notes and classical mechanics and general properties of matter and classical mechanics for beginners pdf free download
Dr.NavneetSingh Profile Pic
Dr.NavneetSingh,India,Teacher
Published Date:21-07-2017
Your Website URL(Optional)
Comment
Lecture Notes on Classical Mechanics (A Work in Progress) Daniel Arovas Department of Physics University of California, San Diego May 8, 2013Chapter 1 Introduction to Dynamics 1.1 Introduction and Review Dynamics is the science of how things move. A complete solution to the motion of a system means that we know the coordinates of all its constituent particles as functions of time. For a single point particle moving in three-dimensional space, this means we want to know its position vector r(t) as a function of time. If there are many particles, the motion is described by a set of functionsr (t), wherei labels which particle we are talking about. So i generally speaking, solving for the motion means being able to predict where a particle will be at any given instant of time. Of course, knowing the functionr (t) means we can take i its derivative and obtain the velocityv (t) =dr/dt at any time as well. i i The complete motion for a system is not given to us outright, but rather is encoded in a set of differential equations, called the equations of motion. An example of an equation of motion is 2 dx m =−mg (1.1) 2 dt with the solution 1 2 x(t)=x +v t− gt (1.2) 0 0 2 where x and v are constants corresponding to the initial boundary conditions on the 0 0 position and velocity: x(0) =x , v(0) =v . This particular solution describes the vertical 0 0 motion of a particle of mass m moving near the earth’s surface. In this class, we shall discuss a general framework by which the equations of motion may be obtained, and methods for solving them. That “general framework” is Lagrangian Dy- namics, which itself is really nothing more than an elegant restatement of Isaac Newton’s Laws of Motion. 34 CHAPTER 1. INTRODUCTION TO DYNAMICS 1.1.1 Newton’s laws of motion Aristotle held that objects move because they are somehow impelled to seek out their natural state. Thus, a rock falls because rocks belong on the earth, and flames rise because fire belongs in the heavens. To paraphrase Wolfgang Pauli, such notions are so vague as to be “not even wrong.” It was only with the publication of Newton’s Principia in 1687 that a theory of motion which had detailed predictive power was developed. Newton’s three Laws of Motion may be stated as follows: I. A body remains in uniform motion unless acted on by a force. II. Force equals rate of change of momentum: F =dp/dt. III. Any two bodies exert equal and opposite forces on each other. Newton’s First Law states that a particle will move in a straight line at constant (possibly zero) velocity if it is subjected to no forces. Now this cannot be true in general, for suppose we encounter such a “free” particle and that indeed it is in uniform motion, so thatr(t)= r +v t. Now r(t) is measured in some coordinate system, and if instead we choose 0 0 to measure r(t) in a different coordinate system whose origin R moves according to the functionR(t), then in this new “frame of reference” the position of our particle will be ′ r (t)=r(t)−R(t) =r +v t−R(t) . (1.3) 0 0 2 2 IftheaccelerationdR/dt isnonzero,thenmerelybyshiftingourframeofreferencewehave apparentlyfalsifiedNewton’sFirstLaw–afreeparticledoesnot moveinuniformrectilinear motion whenviewed from an accelerating frame of reference. Thus, together withNewton’s Lawscomesanassumptionabouttheexistenceofframesofreference–calledinertial frames ′ – in which Newton’s Laws hold. A transformation from one frameK to another frameK which moves at constant velocityV relative toK is called a Galilean transformation. The equations of motion of classical mechanics are invariant (do not change) under Galilean transformations. At first, the issue of inertial and noninertial frames is confusing. Rather than grapple with this, we will try to build some intuition by solving mechanics problems assuming we are in an inertial frame. The earth’s surface, where most physics experiments are done, is not an inertial frame, due to the centripetal accelerations associated with the earth’s rotation about its own axis and its orbit around the sun. In this case, not only is our coordinate system’s origin – somewhere in a laboratory on the surface of the earth – accelerating, but the coordinate axes themselves are rotating with respect to an inertial frame. The rotation of the earth leads to fictitious “forces” such as the Coriolis force, which have large-scale consequences. For example, hurricanes, when viewed from above, rotate counterclockwise inthenorthernhemisphereandclockwiseinthesouthernhemisphere. Lateroninthecourse we will devote ourselves to a detailed study of motion in accelerated coordinate systems. Newton’s “quantity of motion” is the momentum p, defined as the productp = mv of a particle’s mass m (how much stuff there is) and its velocity (how fast it is moving). In1.1. INTRODUCTION AND REVIEW 5 order to convert the Second Law into a meaningful equation, we must know how the force F depends on the coordinates (or possibly velocities) themselves. This is known as a force law. Examples of force laws include: Constant force: F =−mg Hooke’s Law: F =−kx 2 Gravitation: F =−GMmrˆ/r v Lorentz force: F =qE+q ×B c Fluid friction (v small): F =−bv . Note that for an object whose mass does not change we can write the Second Law in the 2 2 familiar formF =ma, wherea =dv/dt =dr/dt is the acceleration. Most of our initial efforts will lie in using Newton’s Second Law to solve for the motion of a variety of systems. The Third Law is valid for the extremely important case of central forces which we will discussingreat detail lateron. Newtoniangravity –theforcewhichmakestheplanetsorbit the sun – is a central force. One consequence of the Third Law is that in free space two isolated particles will accelerate in such a way thatF =−F and hence the accelerations 1 2 are parallel to each other, with a m 1 2 =− , (1.4) a m 2 1 where the minus sign is used here to emphasize that the accelerations are in opposite directions. We can also conclude that the total momentum P =p +p is a constant, a 1 2 result known as the conservation of momentum. 1.1.2 Aside : inertial vs. gravitational mass In addition to postulating the Laws of Motion, Newton also deduced the gravitational force law, which says that the forceF exerted by a particle i by another particle j is ij r −r i j F =−Gmm , (1.5) ij i j 3 r −r i j where G, the Cavendish constant (first measured by Henry Cavendish in 1798), takes the value −11 2 2 G=(6.6726±0.0008)×10 N·m /kg . (1.6) Notice Newton’s Third Law in action: F +F = 0. Now a very important and special ij ji feature of this “inverse square law” force is that a spherically symmetric mass distribution has the same force on an external body as it would if all its mass were concentrated at its6 CHAPTER 1. INTRODUCTION TO DYNAMICS center. Thus, for a particle of mass m near the surface of the earth, we can take m =m i ˆ and m =M , withr −r ≃R r and obtain e e j i j F =−mgrˆ≡−mg (1.7) 2 2 whererˆisaradialunitvector pointingfromtheearth’scenterandg =GM /R ≃9.8m/s e e is the acceleration due to gravity at the earth’s surface. Newton’s Second Law now says thata =−g, i.e. objects accelerate as they fall to earth. However, it is not a priori clear why the inertial mass which enters into the definition of momentum should be the same as the gravitational mass which enters into the force law. Suppose, for instance, that the ′ gravitational mass took a different value, m. In this case, Newton’s Second Law would predict ′ m a=− g (1.8) m ′ and unless the ratio m/m were the same number for all objects, then bodies would fall with different accelerations. The experimental fact that bodies in a vacuum fall to earth at ′ the same rate demonstrates the equivalence of inertial and gravitational mass, i.e.m =m. 1.2 Examples of Motion in One Dimension To gain some experience with solving equations of motion in a physical setting, we consider some physically relevant examples of one-dimensional motion. 1.2.1 Uniform force With F =−mg, appropriate for a particle falling under the influence of a uniform gravita- 2 2 tional field, we have mdx/dt =−mg, or x¨=−g. Notation: 2 7 dx dx ˙ dx ¨ ¨ x˙≡ , x¨≡ , x¨ = , etc. (1.9) 2 7 dt dt dt With v =x˙, we solve dv/dt =−g: v(t) t Z Z dv = ds(−g) (1.10) 0 v(0) v(t)−v(0) =−gt . (1.11) Note that there is a constant of integration, v(0), which enters our solution.1.2. EXAMPLES OF MOTION IN ONE DIMENSION 7 We are now in position to solve dx/dt =v: x(t) t Z Z dx = dsv(s) (1.12) 0 x(0) t Z   x(t) =x(0)+ ds v(0)−gs (1.13) 0 1 2 =x(0)+v(0)t− gt . (1.14) 2 Note that a second constant of integration, x(0), has appeared. 1.2.2 Uniform force with linear frictional damping In this case, dv m =−mg−γv (1.15) dt which may be rewritten dv γ =− dt (1.16) v+mg/γ m dln(v+mg/γ) =−(γ/m)dt . (1.17) Integrating then gives   v(t)+mg/γ ln =−γt/m (1.18) v(0)+mg/γ   mg mg −γt/m v(t) =− + v(0)+ e . (1.19) γ γ Note that the solution to the first order ODE mv˙ =−mg−γv entails one constant of integration, v(0). One can further integrate to obtain the motion   m mg mg −γt/m x(t) =x(0)+ v(0)+ (1−e )− t . (1.20) γ γ γ The solution to the second order ODE mx¨ =−mg−γx˙ thus entails two constants of integration: v(0) and x(0). Notice that as t goes to infinity the velocity tends towards the asymptotic value v =−v , where v = mg/γ. This is known as the terminal veloc- ∞ ∞ ity. Indeed, solving the equation v˙ =0 gives v =−v . The initial velocity is effectively ∞ “forgotten” on a time scale τ≡m/γ. Electrons moving in solids under the influence of an electric field also achieve a terminal velocity. In this case the force is not F =−mg but rather F =−eE, where−e is the8 CHAPTER 1. INTRODUCTION TO DYNAMICS electron charge (e 0) and E is the electric field. The terminal velocity is then obtained from v =eE/γ =eτE/m . (1.21) ∞ The current density is a product: current density =(number density)×(charge)×(velocity) j =n·(−e)·(−v ) ∞ 2 ne τ = E . (1.22) m The ratio j/E is called the conductivity of the metal, σ. According to our theory, σ = 2 ne τ/m. This is one of the most famous equations of solid state physics The dissipation is caused by electrons scattering off impurities and lattice vibrations (“phonons”). In high purity copper at low temperatures (T 4K), the scattering time τ is about a nanosecond ∼ −9 (τ≈10 s). 1.2.3 Uniform force with quadratic frictional damping At higher velocities, the frictional damping is proportional to the square of the velocity. 2 The frictional force is then F =−cv sgn(v), where sgn(v) is the sign of v: sgn(v) = +1 f if v 0 and sgn(v) =−1 if v 0. (Note one can also write sgn(v) = v/v wherev is the absolute value.) Why all this trouble with sgn(v)? Because it is important that the frictional force dissipate energy, and therefore thatF be oppositely directed with respect to f 2 the velocity v. We will assume that v0 always, hence F =+cv . f 2 Notice that there is a terminal velocity, since settingv˙ =−g+(c/m)v =0 givesv =±v , ∞ p where v = mg/c. One can write the equation of motion as ∞ dv g 2 2 = (v −v ) (1.23) ∞ 2 dt v ∞ and using   1 1 1 1 = − (1.24) 2 2 v −v 2v v−v v+v ∞ ∞ ∞ ∞ we obtain dv 1 dv 1 dv = − 2 2 v −v 2v v−v 2v v+v ∞ ∞ ∞ ∞ ∞   1 v −v ∞ = dln 2v v +v ∞ ∞ g = dt . (1.25) 2 v ∞1.2. EXAMPLES OF MOTION IN ONE DIMENSION 9 Assumingv(0) =0, we integrate to obtain   1 v −v(t) gt ∞ ln = (1.26) 2 2v v +v(t) v ∞ ∞ ∞ which may be massaged to give the final result v(t) =−v tanh(gt/v ) . (1.27) ∞ ∞ Recall that the hyperbolic tangent function tanh(x) is given by x −x sinh(x) e −e tanh(x) = = . (1.28) x −x cosh(x) e +e Again, as t→∞ one has v(t)→−v , i.e.v(∞) =−v . ∞ ∞ Advanced Digression: To gain an understanding of the constant c, consider a flat surface of area S moving through a fluid at velocity v (v 0). During a time Δt, all the fluid molecules inside the volume ΔV =S·vΔt will have executed an elastic collision with the moving surface. Since the surface is assumed to be much more massive than each fluid molecule, the center of mass frame for the surface-molecule collision is essentially the frame of the surface itself. If a molecule moves with velocity u is the laboratory frame, it moves with velocity u−v in the center of mass (CM) frame, and since the collision is elastic, its final CM frame velocity is reversed, to v−u. Thus, in the laboratory frame the molecule’s velocity has become 2v−u and it has suffered a change in velocity of Δu=2(v−u). The total momentumchange isobtainedbymultiplyingΔubythetotal massM =̺ΔV, where ̺ is the mass density of the fluid. But then the total momentum imparted to the fluid is ΔP =2(v−u)·̺SvΔt (1.29) and the force on the fluid is ΔP F = =2S̺v(v−u) . (1.30) Δt Now it is appropriate to average this expression over the microscopic distribution of molec- 2 ular velocities u, and since on averagehui = 0, we obtain the resulthFi = 2S̺v , where h···i denotes a microscopic average over the molecular velocities in the fluid. (There is a subtlety here concerning the effect of fluid molecules striking the surface from either side – you shouldsatisfyyourself that thisderivationis sensible) Newton’s ThirdLaw thenstates 2 that the frictional force imparted to the moving surface by the fluid is F =−hFi =−cv , f wherec =2S̺. In fact, our derivation is too crude to properly obtain the numerical prefac- tors, and it is better to write c =μ̺S, where μ is a dimensionless constant which depends on the shape of the moving object. 1.2.4 Crossed electric and magnetic fields Consider now a three-dimensional example of a particle of charge q moving in mutually ˆ perpendicularE andB fields. We’ll throw in gravity for good measure. We takeE =Ex,10 CHAPTER 1. INTRODUCTION TO DYNAMICS B =Bzˆ, andg =−gzˆ. The equation of motion is Newton’s 2nd Law again: q ¨ ˙ mr =mg+qE+ r×B . (1.31) c The RHS (right hand side) of this equation is a vector sum of the forces due to gravity plus the Lorentz force of a moving particle in an electromagnetic field. In component notation, we have qB mx¨=qE+ y˙ (1.32) c qB my¨=− x˙ (1.33) c mz¨=−mg . (1.34) The equations for coordinatesx andy are coupled, while that forz is independent andmay be immediately solved to yield 1 2 z(t) =z(0)+z˙(0)t− gt . (1.35) 2 The remaining equations may be written in terms of the velocities v =x˙ and v =y˙: x y v˙ =ω (v +u ) (1.36) x c y D v˙ =−ω v , (1.37) y c x where ω = qB/mc is the cyclotron frequency and u = cE/B is the drift speed for the c D particle. As we shall see, these are the equations for a harmonic oscillator. The solution is  v (t) =v (0) cos(ω t)+ v (0)+u sin(ω t) (1.38) D x x c y c  v (t) =−u + v (0)+u cos(ω t)−v (0) sin(ω t) . (1.39) D D y y c x c Integrating again, the full motion is given by: x(t) =x(0)+A sinδ+A sin(ω t−δ) (1.40) c y(r)=y(0)−u t−Acosδ+A cos(ω t−δ) , (1.41) D c where   q  1 y˙(0)+u 2 −1 D 2 A= x˙ (0)+ y˙(0)+u , δ =tan . (1.42) D ω x˙(0) c Thus, in the full solution of the motion there are six constants of integration: x(0) , y(0) , z(0) , A , δ , z˙(0) . (1.43) Of course instead of A and δ one may choose as constants of integration x˙(0) and y˙(0). 1.3 Pause for Reflection In mechanical systems, for each coordinate, or “degree of freedom,” there exists a cor- responding second order ODE. The full solution of the motion of the system entails two constants of integration for each degree of freedom.Chapter 2 Systems of Particles 2.1 Work-Energy Theorem Consider a system of many particles, with positionsr and velocitiesr˙ . The kinetic energy i i of this system is X X 1 2 T = T = mr˙ . (2.1) i i i 2 i i Now let’s consider how the kinetic energy of the system changes in time. Assuming each m is time-independent, we have i dT i =m r˙ ·r¨ . (2.2) i i i dt Here, we’ve used the relation  d dA 2 A =2A· . (2.3) dt dt ˙ We nowinvoke Newton’s 2ndLaw,mr¨ =F , towriteeqn. 2.2asT =F·r˙ . We integrate i i i i i i this equation from time t to t : A B t B Z dT (B) (A) i T −T = dt i i dt t A t B Z X (A→B) = dtF ·r˙ ≡ W , (2.4) i i i i t A (A→B) where W is the total work done on particle i during its motion from state A to state i P B, Clearly the total kinetic energy is T = T and the total work done on all particles is i i P (A→B) (A→B) W = W . Eqn. 2.4 is known as the work-energy theorem. It says that i i In the evolution of a mechanical system, the change in total kinetic energy is equal to the (B) (A) (A→B) total work done: T −T =W . 1112 CHAPTER 2. SYSTEMS OF PARTICLES Figure 2.1: Two paths joining points A and B. 2.2 Conservative and Nonconservative Forces 1 2 For the sake of simplicity, consider a single particle with kinetic energy T = mr˙ . The 2 work done on the particle during its mechanical evolution is t B Z (A→B) W = dtF·v , (2.5) t A wherev =r˙. This is the most general expression for the work done. If the forceF depends only on the particle’s positionr, we may write dr =vdt, and then r B Z (A→B) W = dr·F(r) . (2.6) r A Consider now the force F(r)=K yxˆ+K xyˆ, (2.7) 1 2 where K are constants. Let’s evaluate the work done along each of the two paths in fig. 1,2 2.1: x y B B Z Z (I) W =K dxy +K dyx =K y (x −x )+K x (y −y ) (2.8) A B A B A B B A 1 2 1 2 x y A A y x B B Z Z (II) W =K dxy +K dyx =K y (x −x )+K x (y −y ) . (2.9) B A B B A A B A 1 2 1 2 x y A A2.2. CONSERVATIVE AND NONCONSERVATIVE FORCES 13 (I) (II) Note that in general W =6 W . Thus, if we start at point A, the kinetic energy at point B will depend on the path taken, since the work done is path-dependent. The difference between the work done along the two paths is (I) (II) W −W =(K −K )(x −x )(y −y ) . (2.10) B A B A 2 1 Thus, we see that if K =K , the work is the same for the two paths. In fact, ifK =K , 1 2 1 2 the work would be path-independent, and would depend only on the endpoints. This is true for any path, and not just piecewise linear paths of the type depicted in fig. 2.1. The reason for this is Stokes’ theorem: I Z dℓ·F = dSn ˆ·∇×F . (2.11) ∂C C Here,C is a connected region in three-dimensional space, ∂C is mathematical notation for 1 the boundary ofC, which is a closed path , dS is the scalar differential area element, n ˆ is the unit normal to that differential area element, and∇×F is the curl ofF:   ˆ ˆ ˆ x y z ∂ ∂ ∂   ∇×F =det ∂x ∂y ∂z F F F x y z       ∂F ∂F ∂F ∂F ∂F ∂F z y x z y x ˆ ˆ ˆ = − x+ − y+ − z . (2.12) ∂y ∂z ∂z ∂x ∂x ∂y For the force under consideration,F(r)=K yxˆ+K xyˆ, the curl is 1 2 ∇×F =(K −K )zˆ, (2.13) 2 1 which is a constant. The RHS of eqn. 2.11 is then simply proportional to the area enclosed H byC. When we compute the work difference in eqn. 2.10, we evaluate the integral dℓ·F C −1 along the path γ ◦γ , which is to say path I followed by the inverse of path II. In this II I ˆ ˆ ˆ case,n=z and the integral ofn·∇×F over the rectangleC is given by the RHS of eqn. 2.10. When∇×F = 0 everywhere in space, we can always writeF =−∇U, where U(r) is the potential energy. Such forces are called conservative forces because the total energy of the system,E =T +U, is then conserved during its motion. We can see this by evaluating the work done, r B Z (A→B) W = dr·F(r) r A r B Z =− dr·∇U r A =U(r )−U(r ) . (2.14) A B 1 If C is multiply connected, then ∂C is a set of closed paths. For example, if C is an annulus, ∂C is two circles, corresponding to the inner and outer boundaries of the annulus.14 CHAPTER 2. SYSTEMS OF PARTICLES The work-energy theorem then gives (B) (A) T −T =U(r )−U(r ) , (2.15) A B which says (B) (B) (A) (A) E =T +U(r )=T +U(r ) =E . (2.16) B A Thus, the total energy E =T +U is conserved. 2.2.1 Example : integrating F =−∇U If∇×F =0, we can compute U(r) by integrating, viz. r Z ′ ′ U(r) =U(0)− dr ·F(r ) . (2.17) 0 The integral does not depend on the path chosen connecting0 andr. For example, we can take (x,0,0) (x,y,0) (x,y,z) Z Z Z ′ ′ ′ ′ ′ ′ U(x,y,z) =U(0,0,0)− dx F (x,0,0)− dy F (x,y,0)− dz F (x,y,z ). (2.18) y z x (0,0,0) (x,0,0) (z,y,0) The constant U(0,0,0) is arbitrary and impossible to determine fromF alone. As an example, consider the force 3 F(r)=−kyxˆ−kxyˆ−4bz zˆ, (2.19) where k and b are constants. We have    ∂F ∂F z y ∇×F = − =0 (2.20) x ∂y ∂z    ∂F ∂F x z ∇×F = − =0 (2.21) y ∂z ∂x    ∂F ∂F y x ∇×F = − =0 , (2.22) z ∂x ∂y so∇×F = 0 andF must be expressible asF =−∇U. Integrating using eqn. 2.18, we have (x,0,0) (x,y,0) (x,y,z) Z Z Z 3 ′ ′ ′ ′ ′ U(x,y,z) =U(0,0,0) + dx k·0 + dy kxy + dz 4bz (2.23) (0,0,0) (x,0,0) (z,y,0) 4 =U(0,0,0)+kxy+bz . (2.24)2.3. CONSERVATIVE FORCES IN MANY PARTICLE SYSTEMS 15 Another approach is tointegrate the partial differential equation∇U =−F. Thisisin fact three equations, and we shall need all of them to obtain the correct answer. We start with ˆ thex-component, ∂U =ky . (2.25) ∂x Integrating, we obtain U(x,y,z) =kxy+f(y,z) , (2.26) wheref(y,z) is at this point an arbitrary function ofy andz. The important thing is that it has nox-dependence, so ∂f/∂x=0. Next, we have ∂U =kx =⇒ U(x,y,z) =kxy+g(x,z) . (2.27) ∂y Finally, the z-component integrates to yield ∂U 3 4 =4bz =⇒ U(x,y,z) =bz +h(x,y) . (2.28) ∂z We now equate the first two expressions: kxy+f(y,z) =kxy+g(x,z) . (2.29) Subtractingkxy from each side, we obtain the equation f(y,z)=g(x,z). Since the LHS is independent of x and the RHS is independent of y, we must have f(y,z) =g(x,z) =q(z) , (2.30) whereq(z) is some unknown function ofz. But now we invoke the final equation, to obtain 4 bz +h(x,y) =kxy+q(z) . (2.31) 4 The only possible solution ish(x,y) =C+kxy andq(z) =C+bz , whereC is a constant. Therefore, 4 U(x,y,z) =C +kxy+bz . (2.32) Note that it would be very wrong to integrate ∂U/∂x =ky and obtain U(x,y,z) =kxy+ ′ ′ C , where C is a constant. As we’ve seen, the ‘constant of integration’ we obtain upon integrating this firstorderPDE isin fact a function ofy andz. Thefact thatf(y,z)carries no explicit x dependence means that ∂f/∂x =0, so by construction U =kxy+f(y,z) is a solution to the PDE ∂U/∂x =ky, for any arbitrary function f(y,z). 2.3 Conservative Forces in Many Particle Systems X 1 2 T = mr˙ (2.33) i i 2 i X X  U = V(r )+ vr −r . (2.34) i i j i ij16 CHAPTER 2. SYSTEMS OF PARTICLES ′ Here,V(r)istheexternal (orone-body)potential,andv(r−r )istheinterparticle potential, whichweassumetobecentral, dependingonlyonthedistancebetweenanypairofparticles. The equations of motion are (ext) (int) m r¨ =F +F , (2.35) i i i i with ∂V(r ) i (ext) F =− (2.36) i ∂r i  X X ∂vr −r i j (int) (int) F =− ≡ F . (2.37) i ij r i j j (int) Here,F is the force exerted on particle i by particle j: ij   ∂vr −r r −r i j (int) i j ′ F =− =− v r −r . (2.38) i j ij ∂r r −r i i j (int) (int) Note that F =−F , otherwise known as Newton’s Third Law. It is convenient to ij ji abbreviater ≡r −r , in which case we may write the interparticle force as ij i j  (int) ′ F =−rˆ v r . (2.39) ij ij ij 2.4 Linear and Angular Momentum P Consider now the total momentum of the system,P = p . Its rate of change is i i (int) (int) F +F =0 ij ji z X X X dP (ext) (int) (ext) = p˙ = F + F =F , (2.40) tot i i ij dt i i i6=j since the sum over all internal forces cancels as a result of Newton’s Third Law. We write X ˙ P = mr˙ =MR (2.41) i i i X M = m (total mass) (2.42) i i P m r i i i R = P (center-of-mass) . (2.43) m i i Next, consider the total angular momentum, X X L= r ×p = mr ×r˙ . (2.44) i i i i i i i2.4. LINEAR AND ANGULAR MOMENTUM 17 The rate of change ofL is then X  dL = mr˙ ×r˙ +mr ×r¨ i i i i i i dt i X X (ext) (int) = r ×F + r ×F i i i ij i i= 6 j (int) r ×F =0 ij ij z X X (ext) (int) 1 = r ×F + (r −r )×F i i i j ij 2 i i6=j (ext) =N . (2.45) tot Finally, it is useful to establish the result X X  2 2 2 1 1 1 ˙ ˙ T = m r˙ = MR + m r˙ −R , (2.46) i i i i 2 2 2 i i which says that the kinetic energy may be written as a sum of two terms, those being the kinetic energy of the center-of-mass motion, and the kinetic energy of the particles relative to the center-of-mass. Recall the “work-energy theorem” for conservative systems, final final final Z Z Z 0= dE = dT + dU (2.47) initial initial initial Z X (B) (A) =T −T − dr ·F , i i i which is to say Z X (B) (A) ΔT =T −T = dr ·F =−ΔU . (2.48) i i i In other words, the total energy E =T +U is conserved: X X X  1 2 ˙ E = mr + V(r )+ vr −r . (2.49) i i i i j 2 i i ij Note that for continuous systems, we replace sums by integrals over a mass distribution, viz. Z X  3 m φ r −→ drρ(r)φ(r) , (2.50) i i i where ρ(r) is the mass density, and φ(r) is any function.18 CHAPTER 2. SYSTEMS OF PARTICLES 2.5 Scaling of Solutions for Homogeneous Potentials 2.5.1 Euler’s theorem for homogeneous functions Incertaincases ofinterest, thepotential isahomogeneousfunctionofthecoordinates. This means   k U λr ,...,λr =λ U r ,...,r . (2.51) 1 N 1 N Here, k is the degree of homogeneity of U. Familiar examples include gravity, X  m m i j U r ,...,r =−G ; k =−1, (2.52) 1 N r −r i j ij and the harmonic oscillator, X  1 U q ,...,q = V q q ; k =+2 . (2.53) ′ ′ 1 n σσ σ σ 2 ′ σ,σ The sum of two homogeneous functions is itself homogeneous only if the component func- tions themselves are of the same degree of homogeneity. Homogeneous functions obey a special result known as Euler’s Theorem, which we now prove. Suppose a multivariable function H(x ,...,x ) is homogeneous: n 1 k H(λx ,...,λx ) =λ H(x ,...,x ) . (2.54) n n 1 1 Then n X  d ∂H H λx ,...,λx = x =kH (2.55) n 1 i dλ ∂x i i=1 λ=1 2.5.2 Scaled equations of motion Now suppose the we rescale distances and times, defining ˜ ˜ r =αr , t=βt . (2.56) i i Then 2 2 dr α dr˜ d r α d r˜ i i i i = , = . (2.57) 2 2 2 ˜ ˜ dt β dt β dt dt The forceF is given by i  ∂ F =− U r ,...,r i 1 N ∂r i  ∂ k =− α U r˜ ,...,r˜ 1 N ∂(αr˜) i k−1 ˜ =α F . (2.58) i2.5. SCALING OF SOLUTIONS FOR HOMOGENEOUS POTENTIALS 19 Thus, Newton’s 2nd Law says 2 α d r˜ i k−1 ˜ m =α F . (2.59) i i 2 2 ˜ β dt If we choose β such that We now demand α 1 k−1 1− k 2 =α ⇒ β =α , (2.60) 2 β then the equation of motion is invariant under the rescaling transformation This means  1 k−1 2 that ifr(t) is a solution to the equations of motion, then so is αr α t . This gives us an entire one-parameter family of solutions, for all real positive α. 1 ′ 1− k 2 Ifr(t) is periodic with periodT, ther (t;α) is periodic with periodT =α T. Thus, i 1     1− k ′ ′ 2 T L = . (2.61) T L ′ Here,α =L/L is the ratio of length scales. Velocities, energies and angular momenta scale accordingly:  ′ ′ ′   1 L v L T k 2 v = ⇒ = =α (2.62) T v L T    2 2 2 ′ ′ ′   ML E L T k E = ⇒ = =α (2.63) 2 T E L T   2 2 ′ ′ ′   ML L L T 1 (1+ k) 2 L = ⇒ = =α . (2.64) T L L T As examples, consider: (i) Harmonic Oscillator : Here k =2 and therefore q (t)−→q (t;α) =αq (t) . (2.65) σ σ σ Thus, rescaling lengths alone gives another solution. (ii) Kepler Problem : This is gravity, for which k =−1. Thus,  −3/2 r(t)−→r(t;α) =αr α t . (2.66) 3 2 Thus, r ∝t , i.e.     3 2 ′ ′ L T = , (2.67) L T also known as Kepler’s Third Law.20 CHAPTER 2. SYSTEMS OF PARTICLES 2.6 Appendix I : Curvilinear Orthogonal Coordinates The standard cartesian coordinates arex ,...,x, where d is the dimension of space. 1 d Consider a different set of coordinates,q ,...,q, which are related to the original coor- 1 d dinates x via the d equations μ  q =q x ,...,x . (2.68) μ μ 1 d In general these are nonlinear equations. 0 Leteˆ =xˆ be the Cartesian set of orthonormal unit vectors, and defineeˆ to be the unit μ i i vector perpendicular to the surface dq = 0. A differential change in position can now be μ described in both coordinate systems: d d X X 0 ds = eˆ dx = eˆ h (q)dq , (2.69) i i μ μ μ i=1 μ=1 where each h (q) is an as yet unknown function of all the components q . Finding the μ ν coefficient of dq then gives μ d d X X ∂x i 0 0 h (q)eˆ = eˆ ⇒ eˆ = M eˆ , (2.70) μ μ i μ i μi ∂q μ i=1 i=1 where 1 ∂x i M (q) = . (2.71) μi h (q) ∂q μ μ The dot product of unit vectors in the new coordinate system is then d X  1 ∂x ∂x i i t eˆ ·eˆ = MM = . (2.72) μ ν μν h (q)h (q) ∂q ∂q μ ν μ ν i=1 The condition that the new basis be orthonormal is then d X ∂x ∂x i i 2 =h (q)δ . (2.73) μ μν ∂q ∂q μ ν i=1 This gives us the relation v u   d 2 u X ∂x i t h (q)= . (2.74) μ ∂q μ i=1 Note that d X 2 2 2 (ds) = h (q)(dq ) . (2.75) μ μ μ=1 For general coordinate systems, which are not necessarily orthogonal, we have d X 2 (ds) = g (q)dq dq , (2.76) μν μ ν μ,ν=1 where g (q) is a real, symmetric, positive definite matrix called the metric tensor. μν2.6. APPENDIX I : CURVILINEAR ORTHOGONAL COORDINATES 21 Figure 2.2: Volume element Ω for computing divergences. 2.6.1 Example : spherical coordinates Consider spherical coordinates (ρ,θ,φ): x=ρ sinθ cosφ , y =ρ sinθ sinφ , z =ρ cosθ . (2.77) It is now a simple matter to derive the results 2 2 2 2 2 2 h =1 , h =ρ , h =ρ sin θ . (2.78) ρ θ φ Thus, ˆ ˆ ds =ρˆdρ+ρθdθ+ρ sinθφdφ . (2.79) 2.6.2 Vector calculus : grad, div, curl Here we restrict our attention to d =3. The gradient∇U of a function U(q) is defined by ∂U ∂U ∂U dU = dq + dq + dq 1 2 3 ∂q ∂q ∂q 1 2 3 ≡∇U·ds . (2.80) Thus, eˆ ∂ eˆ ∂ eˆ ∂ 1 2 3 ∇= + + . (2.81) h (q) ∂q h (q) ∂q h (q) ∂q 1 1 2 2 3 3 For the divergence, we use the divergence theorem, and we appeal to fig. 2.2: Z Z dV∇·A = dSn ˆ·A , (2.82) Ω ∂Ω

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.