891 lines
50 KiB
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891 lines
50 KiB
Plaintext
Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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CHAPTER I
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NEWTON'S LAWS
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The aim of classical mechanics is to describe and predict the motion of bodies and systems of bodies which are subject to various interactions. Newton's three laws of motion form the basis of this description, so let us begin by reviewing these.
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Newton' s laws'
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Newton 's first law deals with non-interacting bodies. It says that the velocity of an isolated body, one removed from the influence of other bodies, is constant. This law defines a set of preferred coordinate frames, inertial frames , as frames in which Newton's first law holds. Given an inertial frame, we can obtain others by translating the original in space and time, by rotating the original through some angle about some axis, or by giving the original frame a uniform velocity. Unless stated otherwise, we always refer motion to an inertial frame. To a first approximation a coordinate frame attached to the earth is an inertial frame. However, various physical phenomena such as the behavior of a Foucault pendulum, the flight of a ballistic missile, atmospheric and ocean currents, indicate its true non-inertial nature. Better approximations to inertial frames are frames attached to the sun or, even better, to the "fixed stars."
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Newton's second and third laws deal with the effects of interaction between bodies on their motion. Interactions cause the velocities of the bodies to change; the bodies undergo acceleration . Let us consider two otherwise isolated interacting small bodies, "particles." We find that the accelerations a1 and a2 of the particles are oppositely directed, and that their magnitudes are related by
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a1 /a2 = k12 ,
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where the ratio k12 is independent of the nature of the interaction between the particles. It does not depend on whether the interaction arises because the particles are in contact with one another, or because they are connected by a string or by a spring, or because they interact gravitationally, etc. The ratio k12 is thus a quantity which we can associate with the pair of particles themselves, as opposed to the particular interaction they happen to be undergoing. Further, if we consider three particles, a pair at a time, we find that the three acceleration ratios are not independent but are related by
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k12k23k31 = 1.
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This, together with k12 = 1/k21, shows that the acceleration ratio can be written
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k12 = m2 /m1 ,
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1Ernst Mach, The Science of Mechanics, (The Open Court Publishing Co., Chicago, 1893), trans. Thomas J. McCormack.
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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2 Chapter I: Newton's Laws
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where m1 (m2) is a property of, something associated with, particle 1 (particle 2) alone.
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The quantity m1 (m2) is called the inertial mass of the particle. Putting these facts together, we see that the accelerations of two interacting particles are related by
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m1a1 = -m2a2.
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We describe the interaction by saying that particle 2 exerts a force F2on1 on particle 1, and particle 1 exerts a force Flon2 on particle 2, such that
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with
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m1a1 = F2on1 and m2a2 = F lon2
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F2on1 = -Flon2
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This last equation is Newton 's third law : the force which particle 2 exerts on particle 1 is equal and opposite to the force which particle 1 exerts on particle 2. It is another way of stating our conclusions about the acceleration ratio of two interacting particles.
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If we now consider three interacting particles, we find that the acceleration of any one of them, say particle 1, is the vector sum of the acceleration of particle 1 due to particle
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2 alone and the acceleration of particle 1 due to particle 3 alone (Fig. 1.01(a)), and thus
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m1a1 = F2on1 + F3on1
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Ftotalonl
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This is Newton' s second law : the acceleration of a particle is directly proportional to the total force acting on it (obtained by adding vectorially all the individual forces) and is inversely proportional to the mass of the particle. This law should be understood in the following way: we are meant to describe the interaction of our chosen particle with other particles by specifying the force acting on it in terms of the locations and velocities of all the particles. The form this takes depends, of course, on the nature of the interactions.
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m3
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Ftotal on 1
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F1 on 3
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m1 F2 on 1 m2 m 1
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(a)
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(b)
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Fig. 1.01. Forces on three interacting particles
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Free fall 3
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Similar equations apply to particles 2 and 3 (Fig. 1.01(b)),
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m2a2 = F3on2 + Flon2 m3a3 = Flon3 + F2on3
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Now suppose that the forces F3on2 and F2on3 are such that particles 2 and 3 are bound together to form a single particle and move with a common acceleration
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a2= a3 =a23.
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Then, adding the above two equations, we find
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(m2 + m3)a23 = Flon2 + Flon3,
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the internal forces F3on 2 and F203 canceling because of Newton's third law. The bound combination thus behaves as a single particle with mass
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m23 = m2 + m3;
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mass is additive. Further, the force acting on the bound combination can be taken to be the total external force; the internal forces which hold the combination together need not be taken into account.
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To see how to apply these laws, we begin with some simple examples, most of which will already be familiar to you.
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Free fall
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For our first example, consider the motion of a body of mass m dropped near the surface of the earth. The only force acting on the body (ignoring air friction) is the downward gravitational force mg; here g -~ 9.8 m/s2 is the approximately constant gravitational field due to the earth. Newton's second law gives
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ma = mg,
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so, canceling the m, we see that the body falls with constant acceleration g. Indeed, since g is body-independent, all bodies fall with the same constant acceleration. 2 The equation of motion for the falling body takes the form (measuring x downwards)
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d2x dt2 - g
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2For the consequences, see A. Einstein in H. A. Lorentz, A. Einstein, H. Minkowski, and H. Weyl, The Principle of Relativity (Dover Publications, New York, NY, 1923), trans. W. Perrett and G. B. Jeffery, p. 99.
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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4 Chapter 1: Newton's Laws
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Integrating , we find that the velocity is given by dx = vo + gt = v(t) dt
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where vo is the initial velocity. Integrating once again, we find that the position of the body is given by
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x(t) = xo + vot + 2 gt2
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where xo is the initial position. The position is thus determined as a function of the time; the expression also contains two constants of integration, adjustable parameters, which are set by the initial conditions, the initial position and velocity.
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Simple harmonic oscillator In the above example the integration could be done immediately since we knew the
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time dependence of the force: it was constant. Usually, however, the force is not known a priori as a function of time. Rather, it is known as a function of position. Take, for example, a mass m attached to a spring (Fig. 1.02).
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F = -kx Fig. 1.02. Mass attached to a spring It is found (Hooke 's law) that the force F which the spring exerts on the mass is proportional to the amount x the spring is stretched and is directed opposite the stretch,
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F = -kx. The proportionality constant k, a measure of the strength of the spring , is called the spring constant. Newton's second law then gives
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ma = -kx. We cannot integrate this directly as in the free fall case. However, if we multiply the equation by the velocity v, the left-hand side becomes
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dv d may=m v=-(2mv2)
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dt dt '
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Simple harmonic oscillator 5
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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and the right-hand side becomes -kxv= -kxddtx =- ddt (2 1kx2).
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Thus the motion is such that
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2 2 E= mv2 + kx2
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is constant in time. This quantity is called the total energy and is the sum of the kinetic energy T = +mv2 and the potential energy V = -kx2.
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The energy equation can be rearranged to obtain
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X
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dx
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t
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= fodt=t .
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Lk 2
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Xo m - m x
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To do the x-integration, we set
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x = Fk sin dx = 2JcEos4)d4) k
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where 4) is a new variable, the phase. The left-hand side then becomes
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1 f 17 cos4)
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= m Lifd4 =
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2 00 1- sin 2 d^ k o
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so that the phase is a linear function of time,
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^ 4) =+o+wt.
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The rate at which the phase increases with time, co = k/m, is called the angular frequency. The position of the mass as a function of time is thus given by
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x = Asin(wt + 4)o)
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where A = 2E/k is called the amplitude of the motion. The mass oscillates back and
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forth between x = +A and x = -A, going through a complete cycle in a time i = 2n/w, the period of the motion. This very important motion is called simple harmonic motion . Its importance derives from the fact that, except in unusual circumstances, motion near any stable equilibrium point is simple harmonic motion.
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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6 Chapter I: Newton's Laws
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The approach which we have used here to discuss the simple harmonic oscillator can be applied to any one-dimensional conservative system, for which F = - dV/dx ; all we
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have to do is to replace 2 kx2 by the appropriate potential energy V(x). Such systems are thus in principle always integrable.
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Central force
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When we move from one-dimensional problems to three-dimensional problems, the degree of complexity increases enormously. Indeed, most three-dimensional problems cannot be integrated analytically. There is, however, a class of problems which can still be handled moderately easily, namely the motion of a particle acted on by a force F = Fr which is always directed towards (or away from) a fixed point, the force center. This is the so-called central force problem (Fig. 1.03(a)). For such problems the torque r x F on the particle about the force center is zero, and the angular momentum L = r x (mv) is constant. The motion thus lies in a plane L - r = 0 which is perpendicular to L and which passes through the force center , the orbital plane . Further, the motion is such that the magnitude L of the angular momentum about the force center is constant.
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F(t1)
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r/
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0
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Force center Force center
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(a)
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(b)
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Fig. 1.03. (a) Central force, (b) Polar coordinates
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It is convenient to introduce polar coordinates r and 0 in the orbital plane, with the force center the pole (Fig. 1.03(b)). The position of the particle is then given by r = rr, the velocity by
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dr dr „ dr v = =-r+r
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dt dt dt
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= rr + r99 ,
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and the acceleration by
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Central force
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7
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+( a = dv =rr + rdi +r68+ rOO+rO d9 dt dt dt _ ('r- r6 2 )r r6+2i0)0 .
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In deriving these we have used the results
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di = 0a0ndde= -6r
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dt
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dt
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which follow readily from Fig. 1.04.3
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A
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r+dr
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A
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A
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dr=d6
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Ar
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41. A
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d8=-dOr
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%6 0 + de
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Fig. 1.04. Changes in the unit vectors
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Newton's second law then gives the equations of motion
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m(f-r02)=F m(r6 + 2r6) = 0
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The second of these can be written
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1 d (mr26) = 0 r dt
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which shows that L = mr26 is constant in time. The quantity L = (distance r from origin) x (component mr0 of my perpendicular to r) is the magnitude of the angular momentum, so this simply confirms what we already know. Its content can be expressed in a rather picturesque way.
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3Alternate derivations can be found in Daniel Kleppner and Robert J. Kolenkow , An Introduction to Mechanics , (McGraw-Hill Book Company, New York , NY, 1973), pp. 27-38.
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8 Chapter I: Newton's Laws
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Area swept out in time t to t + dt
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Fig. 1.05. Area swept out by radius vector
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From Fig. 1.05 , the radius vector sweeps out an area dA = 4(r)(rodt) in a time dt, so the
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rate at which it sweeps out area is dA/dt = 2 r26 = L/2m. This, as we have seen, is constant. Thus the particle moves along its orbit in such a way that the radius vector sweeps out equal areas in equal times. Applied to the solar system, this is known as
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0 Kepler's second law of planetary motion. We see, however , that it holds for any
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central force , not just for the gravitational force.
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The fact that L is constant can be used to eliminate = L/mr2 from the radial equation of motion. If, further, the central force is conservative, so that F = -dV(r)/dr where V is the potential energy , we can write
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L2 dV d L2 mr = - 2 + V(r)
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mr3 dr dr 2mr
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The radial motion is thus the same as one-dimensional motion (but with r > 0) in an effective potential
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L2 Veff (r) = 2mr2 + V(r).
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In particular, the total energy
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E = 2 mIvI2 +V(r) = 2 m(r2 + r262) + V(r)
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= 2m
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r2
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L2
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+ 2 2
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mr
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+
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V(r)
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=
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2 mr2
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+
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Veff
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(r)
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6is constant in time. This can be rearranged and integrated to obtain r as a function of t and
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the parameters E, L, and initial radius ro. The result can then be substituted into
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= L/mr2, and this equation integrated to obtain 0 as a function oft and the parameters E, L, ro, and initial angle 00. These four parameters, together with the two required to
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Gravitationalforce: qualitative 9
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specify the orbital plane, specify the orbit (r0 and 00 give the initial position in the orbital plane, and E and L then give the magnitude and direction of the initial velocity).
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Before embarking on the integrations, either analytically or numerically, it is worthwhile to get an overview of the general behavior to be expected, to obtain qualitative pictures of the possible orbits and the ranges of the parameters over which they occur.
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These can be obtained by examining a sketch of V eff (r) versus r. We illustrate this idea by applying it to the important central force, gravitation.
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Gravitational force : qualitative
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The gravitational force between two bodies, which are small compared to the distance r between them, is given by
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F -- k --dV r2 dr
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where V = - k r
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is the gravitational potential . The constant k equals GmM where m and M are the masses of the bodies and G is the gravitational constant. The same expression may be used for the electrostatic force, the Coulomb force , between two electrically charged bodies,
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provided we set k = - q lq 2 where q, and q2 are the electric charges on the bodies. If one
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of the bodies is much lighter than the other, say m << M, as is the case for the planets compared with the sun, or artificial satellites with the earth, or an electron with a nucleus, we can regard the heavy body as providing an approximately fixed force center about which the lighter body orbits. The effective potential is then
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and is shown in Fig. 1.06.
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= L2 - k V eff
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2mr2 r
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V eff (r)
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- E3 -
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r2
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r E Fig. 1.06. Effective potential for the gravitational force
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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10
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Chapter 1: Newton's Laws
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We have seen that the constant total energy E of an orbiting body is the sum of the
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kinetic energy 2 mr2 due to its radial motion and the effective potential Veff . Since the kinetic energy is non-negative, the total energy must be greater than or equal to the energy Eo at the bottom of the Veff potential well (Fig. 1.06). If the energy is E0, then the radius is fixed at ro and the orbit is a circle. If the energy is E1 with Eo < E1 < 0, the radial motion is like that of a particle in a one-dimensional- potential well, the orbit radius oscillating back and forth between an inner turning radius r1 and an outer turning radius
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r2, with E = Veff and r = 0 at r1 and r2. All the while the angle 0 is increasing. We see that the orbit then looks qualitatively like one of those in Fig. 1.07.
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(a)
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(b)
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Fig. 1.07. Qualitative shape of bound orbit
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Since for gravity the force is attractive rather than repulsive at the inner turning radius, the contact there must look like Fig. 1.07(a) rather than like (b). Detailed calculations to follow show that the orbit for this case is in fact an ellipse. If the energy is E2 = 0 or E3 > 0, there is an inner turning radius but no outer turning radius. The orbiting body comes in from infinity, reaches a minimum radius r3, and moves out again to infinity. The orbit looks qualitatively like that in Fig. 1.08.
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Fig. 1.08. Qualitative shape of scattering orbit
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Detailed calculations to follow show that for the gravitational force the orbit is in fact a parabola for energy E = 0 and a hyperbola for energy E > 0.
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Gravitationalforce: quantitative
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11
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The ideas used above to discuss the qualitative shape of the orbits in a gravitational potential can also be applied to an arbitrary central potential. In addition to the situations already considered for gravity , we may then encounter attractive potentials V(r) which blow up faster than -1/r2 as r -b 0. The effective potential Veff then tends to -0o rather than + oo as r -^ 0 and , depending on the energy , there may be no inner turning radius. For example, an orbit with energy E shown in Fig . 1.09(a) spirals in to the force center as in Fig. 1.09(b).
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Veff(r) E
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r
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- (a) (b) Fig. 1.09. (a) A possible effective potential, (b) Qualitative shape of capture orbit
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Gravitational force : quantitative
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Let us now return to the gravitational force and consider the detailed integration. The energy and angular momentum equations
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2
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E = 1 mr2 + L - k and L = mr29 2 2mr2 r
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lead, on integration, to r and 0 as functions of time. Rather than doing these integrations
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immediately, however, it is more useful first to obtain r as a function of 0; that is, to obtain the equation which describes the shape of the orbit. To do this, we set
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r=dr0=dr L dO d9 mr2
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in the energy equation and hence, on rearranging, write
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r
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= dr =('9 dO 0-90.
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r 2
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2mE
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-
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1_2mk
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+
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Jeo
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ro L2 r2 Or
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The r-integration is performed by setting first u = 1/r, du = -dr/r2 to give
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12
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Chapter 1: Newton's Laws
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Then set
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U
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u
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du du
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2mE 2mk _ 22 F 2 2L2E / mk 2
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°o
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LZ
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+
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LZ
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u
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u
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uo
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1+ 2
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mk
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^ UL _ 2
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`
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u
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mk = mk 2 2
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c1o+s22 aLa2nEd_dmuks22i1n L+2aE2 da
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.
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L L mk
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L
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mk
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Integration gives a = 0 - 00, which leads to the orbit equation
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2
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u = 1 = 1+ l+ 2L Ecos(8-8) r LZ mk 2 °
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This has the form of a conic section
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P =1 + e cos(0 - 0°) r
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with
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semi
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-
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latus-rec
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m - LZ = 2L2E to p mk and eccentricity e 1 + 2 . We
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have
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chosen
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mk
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the constant of integration so that 0 = 00 is the direction of pericenter, the point on the
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orbit closest to the force center. The angle a = 0 - 0° from pericenter is called the true anomaly. We can show that:
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mkt for E = E° _ - 2 2 , e = 0 and the orbit is a circle,
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L for E° < E < 0, 0 < e < 1 and the orbit is an ellipse,
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for E = 0, e =1 and the orbit is a parabola, and for 0 < E, 1 < e and the orbit is a hyperbola.
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Let us first consider the bound orbits E < 0, those for which the particle is confined
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to a finite region of space. We show that the above equation with 0 < e < 1 represents an
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ellipse (clearly, the special case e = 0 represents a circle). As the old geometry books put it, an ellipse is the locus of all points for which the sum of the distances to two fixed points is a constant. The two fixed points are called the foci of the ellipse. The sum of the distances is the major axis of the ellipse. We denote it by 2a, so a is the semi-major axis. The ratio of the distance between the foci to the major axis is the eccentricity e of the ellipse.
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Gravitationalforce: quantitative
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13
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pericenter
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Fig. 1.10. Elliptic orbit If we apply the trigonometric cosine law to the triangle in Fig. 1.10, we find
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(2a - r)2 = r2 + 4e2a2 + 4ear cos a,
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which, on rearranging, gives the polar form of the equation of an ellipse, as above, with
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semi-latus-rectum p = a(1- e2) . Applied to the solar system, this is Kepler's first law of planetary motion: the planets travel around the sun on an elliptical orbit with the sun at one focus.
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The semi-major axis of the ellipse can be expressed in terms of the energy and angular momentum,
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p=L= 2a2 sL o 2 a= 1E orIEk=k -. mk mk2 21 El 2a
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The semi-major axis depends only on the energy, not on the angular momentum; alternatively, the energy depends only on the semi-major axis, not on the eccentricity.
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Let us now turn our attention to the time dependence of the variables. According to Kepler's second law, the rate at which the radius vector sweeps out area is
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dA= L =1 /(1e2). dt 2m 2 m
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The time required to complete one orbit, the period -r, is the time to sweep out the
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complete area
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,^__
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A= nab =rat 1-e2
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enclosed by the ellipse (here b = a 1- e2 is the semi-minor axis of the ellipse). It is given by
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14
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Chapter I: Newton's Laws
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na2 / - e2 1 ka 2
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which, on simplifying, yields
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i=2ac mad. k
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For the family of planets orbiting the sun, k/m = GM where M is the mass of the sun. Thus, the period of a planet is proportional to the Y2power of the semi-major axis of the planet's orbit; it does not depend on the mass of the planet or the eccentricity of the orbit. This is Kepler's third law of planetary motion.4 See Fig. 1.11 and note that the slope of the log-log plot is % .
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3
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2
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0
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-1
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0 1 2
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log(a/aE)
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Fig. 1.11. Kepler's third law
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Kepler's second law, plus some geometry, can also give the way the particle moves around the orbit as a function of time.5 However, for modem readers more familiar with calculus than with geometry, it is probably easier to obtain this by integrating the equation for radial motion
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.1 2 L2 k
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- mr + --= E. 2 2mr2 r
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4One sometimes sees a "corrected" version in which GM is replaced by G(M+m). This comes from treating the situation as a two-body problem involving the sun and the planet under consideration . But the solar system is a many body problem, and the solar motions which lead to "corrections" of this type are the result of the interaction of the sun not only with the planet under consideration but with all the planets. A proper treatment would have to take all these into account. 5 See, for example , Forest Ray Moulton , An Introduction to Celestial Mechanics, (Macmillan , New York, NY, 1902, 1914, 1923 ), 2nd ed., p.159.
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Gravitational force: quantitative
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15
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This can be rearranged in the form
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r
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2
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Jro k r 2L mr2
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ft dtfi(tt)
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° m
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If the energy E and angular momentum L are expressed in terms of the semi-major axis a and eccentricity e, the left-hand side becomes
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2a r rdr k a(l-e) a2e2 - ( r - a)2
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Here we have also chosen ro as the pericenter radius a(1- e) ; to is then the time of passage of pericenter. The integration is readily performed by setting
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r - a = -ea cos V and dr = ea sin V dh
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where V is a new variable called the eccentric anomaly , whose geometric significance can be seen from Fig. 1.12.
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circle ellipse
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acosV = ea + rcosct
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a r
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particle
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=(a-r)/e
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V a
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geometric center focus of ellipse of ellipse and circle (force center)
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Fig. 1.12. Eccentric anomaly V The integration then becomes
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and we find
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2a a f (1- a cos ) d4 = Vk a(V - e sin V), k fo
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r -=1-ecosi a
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16
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Chapter I: Newton's Laws
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with
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V - esini _ 2n (t - to) i
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where -u is the period of the motion. This last equation is known as Kepler's equation; it is a relation between the eccentric anomaly , and the time t, or the so-called mean
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( anomaly (23t/t)(t - to). Finally, the relation between the eccentric anomaly , and the
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true anomaly 0 - 00) can be found by eliminating r/a between the above r-V equation and the orbit equation, to obtain
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1-ecos,=
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1-e2
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l+ecos(0-00)
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In principle, the determination of the motion is now complete. If, however, we want r and
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0 in terms of the time, as is often the case, we must first invert Kepler's equation to get V in terms of the time. This is in general difficult. In cases in which the eccentricity e of the orbit is small, however, expansions in powers of e are adequate. We use this approach in the next section to determine the parameters of earth's orbit.
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Parameters of earth 's orbit
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The earth's orbit around the sun lies in the plane of the ecliptic, which is marked by the apparent path of the sun through the constellations of the zodiac over the course of a
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year. The plane of the earth's equator makes an angle of approximately 23° with the plane of the ecliptic, and the intersection of these two planes gives a direction in space. In September of each year the sun passes through the plane of the earth's equator going from north to south. This is known as the autumnal equinox (AE). The direction from sun to earth at this time provides a convenient reference (the first point of Aries) from
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which to measure the angle 0, so at autumnal equinox 0 = 0. As the year progresses, the midday sun (in the northern hemisphere) moves lower and lower in the sky, until at 0 = 4/2, the winter solstice (WS) in December, it reaches its lowest point. The sun then moves higher in the sky and at 0 = a, the vernal equinox (VE) in March, the sun again passes through the plane of the earth's equator, this time going from south to north. The first point of Aries is thus also the direction from earth to sun at the vernal equinox. The sun continues to move higher in the sky, and at 0 = 34/2 , the summer solstice (SS) in June, the midday sun reaches is highest point. See Fig. 1.13.
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' Parameters of earth s orbit 17
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Fig. 1.13. Earth's orbit
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Table I. Equinoxes, solstices, and seasons for 1994 - 1995
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Equinoxes and solstices
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Date
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Day
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(see 1)
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(see 2)
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(see 3)
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Seasons (see 4)
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AE (94) WS (94) VE (95) SS (95) AE (95)
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23 Sept. 01:19 266.0549 21 Dec. 21:23 355.8910 20 Mar. 21:14 444.8847 21 June 15:34 537.6486 23 Sept. 07:13 631.3007
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89.8361 0.245961 Autumn 88.9938 0.243654 Winter 92.7639 0.253977 Spring 93.6521 0.256408 Summer
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1. Times in EST 2. 1 Jan. 1994 0:00 h = 1.0000 3. In days 4. In fractions of the year [AE (94) to AE (95)] = 365.2458 days
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We wish to use the "observed" times of these seasonal events6 to find the parameters of earth's orbit: the eccentricity e which determines the shape of the ellipse, the angle 00 of perihelion which determines the orientation of the ellipse in the plane of the ecliptic, and the time to of passage of perihelion. The last equation of the previous section
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gives the relation between the eccentric anomaly V and the true anomaly 0 - 00. Expanding the right-hand side of this as a power series in the eccentricity, we find
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cos i, = cos( 6 - 00) + e sine (0 - 00) - e2 cos(6 - 80) sin 2(0 - 00)+•
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which yields
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6A convenient and inexpensive source of data is The Old Farmer's Almanac, (Yankee Publishing Inc., Dublin, NH). A more conventional source is The Astronomical Almanac, (US Government Printing Office, Washington, DC).
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1 8
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Chapter 1: Newton's Laws
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1, = (0 - 00) - esin(9 - 00) + 4 e2 sin 2(9 - 90)+• • •.
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This, when substituted into Kepler's equation, then gives the relation between the observed quantities , the true anomaly and the time (in years),
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(0-00)-2esin(9-90)+ 4e2sin2(9-00)+•••= 2n(t-t0).
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Setting 0 = 0 at autumnal equinox, 0 = a/2 at winter solstice, 0 = a at vernal equinox, and 0 = 3a/2 at summer solstice, we obtain the four equations
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-90 + 2e sin 90 - 4 e2 sin 290 +• • • = 2a(tAE- t0 ) '-00-2ecos0o+-le2sin26o+...=2n(tws-to) n - 00 - 2e sin 00 - a e2 sin 200 +• • • = 2n(tvE - t0 ) 2 - 00 + 2e cos90 + *e2 sin 200 +• • • = 2a(tss - t0)
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These can be combined to give
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2 e
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sin90=-
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1NE
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-
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tAE)
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=
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1
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-
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(Autumn
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+
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Winter)
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n 2
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2
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-
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2e
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1
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cos eo - (tss a 2
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-
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tws)
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=
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12
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-
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(Winter
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+
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Spring)
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4to
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=
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t AE
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+
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t ws
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+
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tvE
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+
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tss
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+
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2eo
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n
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- 3
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2
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-1+ 3sien2 290 = tws tAE + tss - tVE = Autumn + Spring . 2 2a
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The first two equations give the eccentricity and angle of perihelion; the third gives the time of perihelion; and the fourth is a check on the consistency of the data with the assumption of a Keplerian orbit. In particular, using the data in Table I we find
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2e sin90 = 0.010385 a - 2e cos 90 = 0.002369 a
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which yield e = 0.016732 and 00 =102.85° . The third equation then gives
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4t0 =1604.4792 + (2 x 102.85/180 - 1.5) x 365.2458
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which yields to = 368.50 = 3 January 1995,12h. Finally , the fourth equation gives
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Scattering
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19
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z % 0.499942 0.499937.
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The equations we have been using are good to order e2; corrections of order
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e3 = 4.7 x 10-6 are expected to modify the results by a few parts in the last figure quoted.
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The uncertainty in the input times (assumed to be ± 1 minute = ± 1.9 x 10-6 year) also affects the last figure.
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The agreement with the almanac values e=0.01673±0.00002 and 00 =102.87° ±0.08°, where the ± indicates the variation over the course of the year due to various perturbations, is excellent. However, our predicted to is 18 hours early. The reason for this is interesting. We have actually been calculating the parameters of the earthmoon barycenter; in particular, our to is the time of barycenter perihelion. What is listed in almanacs, however, is the time of earth perihelion. These differ by approximately 1.3 sin 4)
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days, where 4) is the (angular) phase of the moon near perihelion. In 1995 perihelion occurred about a third the way through the first quarter of the moon, so 4 30° and the correction to our result is approximately + 16 hours, as required.
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Scattering
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One of the primary ways for exploring the nuclear and sub-nuclear world is via
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scattering experiments. A beam of particles, such as electrons, protons, or a-particles, is directed at a thin target which contains the nucleus to be studied. The particles in the beam interact with the target nuclei and are scattered. A detector counts the number of particles per unit time scattered in various directions (Fig. 1.14(a)). This number is proportional to the (small) cross-sectional area AA of the detector and is inversely proportional to the square of the distance R from the target to the detector. That is, the number An of counts per unit time is proportional to the solid angle AQ = AA/R2 subtended by the detector. Further, it is proportional to the intensity I of the incident beam (the number of incident particles per unit area per unit time) and to the number N of target nuclei in the path of the beam. To obtain a quantity from which these details of the experimental arrangement have been removed, we divide the number of counts per unit time by the solid angle subtended by the detector, by the intensity of the incident beam, and by the number of target nuclei. The resulting quantity,
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do 1 do
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d2 NI dO'
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is called the differential scattering cross section . It has units of "area": oa = (da/dQ) AQ is the area an incident particle must strike, per target nucleus, in order to
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scatter into the solid angle AQ.
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20 Chapter 1: Newton's Laws Detector AA
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Target (a)
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Incident particle
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7
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O A
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b
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(b) Target nucleus
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Fig. 1.14. Scattering (a) macroscopic view, (b) microscopic view
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For thin targets the scattering of an incident particle is the result of a single collision between it and an individual target nucleus. We wish to relate the differential scattering cross section to this individual scattering process. The proper way to do this is to use quantum mechanics. It is nevertheless of interest to see how one approaches such a problem using classical mechanics, if only to introduce general ideas and to provide results which can then be compared with the quantum mechanical results. When the incident particle is far from the target nucleus, the force exerted on it by the target nucleus is small,
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and the particle moves along an incoming asymptotic straight line (Fig. 1.14(b)). If
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extended, this straight line would pass by the target nucleus with distance of closest approach b. This distance b is called the impact parameter. It is related to the (constant) angular momentum L and energy E of the incident particle by
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L=my.0b=b 2mE.
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As a result of its interaction with the target nucleus, the incident particle is deflected from its original path, eventually emerging from the interaction region along an outgoing asymptotic straight line which makes an angle O (Os 8:5 it) with the incident direction. This angle O is called the scattering angle . It is a (single-valued) function- 8(b, E) of the impact parameter b and particle energy E. If we choose a spherical polar coordinate system with origin at the target and polar axis in the direction of the incident beam, the scattering angle O is the polar angle of the scattered particle. The azimuthal angle 4 of the particle does not change (except possibly by at). The number of particles per unit time incident with impact parameter in the range b to b + db and with azimuthal angle in the range + to + + d+ is do = I b db d4 . These particles scatter into the specific polar angle range O to O + dO where O is determined by b, and into the azimuthal angle range 4 (+a) to + + d4 (+n). The solid angle which they subtend is thus dQ = sin() dO d+ . Substituting these results into the above definition of the differential scattering cross section then gives
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Coulomb scattering 21
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do b db dQ sin 8 d8 In writing this down we have assumed that the relationship between impact parameter and scattering angle (at fixed energy) is one to one. For some interactions this may not be the case, with more than one impact parameter yielding the same scattering angle. In such situations the above should be replaced by an appropriate sum.7
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Coulomb scattering
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The scattering of a low energy incident charged particle by a nucleus is largely the result of the electrostatic Coulomb force between it and the nucleus. We have seen that the orbit for a particle moving in a Coulomb potential V = - k/r (with k = -glg2 where qi and q2 are the electric charges of the incident and target particles) is a conic section
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p =1+e.cosO r where p = L2/mk is the semi-latus-rectum and e = 1 + 2L2E/mk2 - 1 + (2bE/k)2 is the eccentricity. Also, for convenience we have here taken pericenter in the direction 0 = 0 for an attractive force (unlike charges ), or in the direction 0 = x for a repulsive force (like charges). See Fig. 1.15. repulsive attractive 4
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Fig. 1.15. Coulomb scattering
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7For some of the things that can happen then see: Roger G. Newton, Scattering Theory of Waves and Particles, (McGraw-Hill Book Company , New York, NY, 1966), pp. 129-134; Herbert Goldstein , Classical Mechanics, (Addison-Wesley Publishing Company, Reading, MA, 1980), 2nd. ed., pp. 110-113.
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22
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Chapter I: Newton's laws
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For E positive, e is greater than one, and the orbit is a hyperbola. Far from the nucleus (r -♦ co) the incident particle travels along incoming or outgoing asymptotic straight lines,
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with directions TO. where
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800 = cos-1(- l/e) (x/2 :s 0. s 3L).
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Note that these directions are the same (apart from sign) for both attractive and repulsive Coulomb scattering. The scattering angle is given by
|
|
8=20.-a.
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|
This, combined with the previous equations , yields the relation between the impact parameter and the scattering angle
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b=Iklcot9. 2E 2
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The scattering angle 8 is a monotonely decreasing function of the impact parameter b,
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decreasing from 8 = n at b = 0 to e = 0 as b -► oo. The differential scattering cross section follows from previous considerations and is given by
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asp
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'
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(
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1
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JL 2
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El
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csc4(\ t el
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This is the famous Rutherford scattering cross section , first derived and used by
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Ernest Rutherford to interpret the experiments of Geiger and Marsden on a-particle scattering from thin metal foils. It led him to the discovery of the nuclear atom. As has already been pointed out, this is really a quantum mechanical problem. Fortunately for the development of atomic physics, however, in this instance classical mechanics, more by accident than anything, leads to the same result as quantum mechanics.8
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Exercises
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|
A particle of mass m moves in one dimension x in a potential well V = Vo tan2 (3rx/2a)
|
|
r where Vo and a are constants . Find, for given total energy E, the position x as a
|
|
function of time and the period of the motion. In particular, examine and interpret the low energy (E << V o) and high energy (E >> Vo) limits of your expressions.
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8See, for example: Kurt Gottfried, Quantum Mechanics: Volume I, (W. A. Benjamin, New York, NY, 1966), pp. 148- 153; Gordon Baym, Lectures on Quantum Mechanics, (W. A. Benjamin, Reading, MA, 1969, 1973), pp. 213-224; J. J. Sakurai, Modern Quantum Mechanics, (Addison-Wesley, Redwood City, CA, 1985), ed. San Fu Tuan, pp. 434-444.
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Exercises 23
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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2. For each of the following central potentials V(r) sketch the effective potential
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L2 Veff (r) = 2 ? 2 + V(r),
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and use your sketch to classify and draw qualitative pictures of the possible orbits. (a) V(r) = 2 kr2 3D isotropic harmonic oscillator (b) V(r) = -V1 for r < a square well
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V (r) = 0 for r > a
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(c) V (r) = - k r2
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(d) Vr = - k r 4
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-ar
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(e) V(r) = -k e t
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Yukawa potential
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Note that the qualitative shape of Veff (r) versus r may depend on L and on the various parameters; consider all cases (but assume that the given parameters are positive).
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3. The first U.S. satellite to go into orbit, Explorer I, which was launched on January 31, 1958 , had a perigee of 360 km and an apogee of 2549 km above the earth's surface. Find: (a) the semi-major axis, (b) the eccentricity, (c) the period, of Explorer I's orbit . The earth 's equatorial radius is 6378 km and the acceleration
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due to gravity at the earth 's surface is g = 9.81 m/s2 .
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4. Mars travels on an approximately elliptical orbit around the Sun. Its minimum distance from the Sun is about 1 . 38 AU and its maximum distance is about 1.67 AU (1 AU = mean distance from Earth to Sun). Find: (a) the semi-major axis, (b) the eccentricity, (c) the period, of Mars' orbit.
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5. The most economical method of traveling from one planet to another, the Hohmann transfer, consists of moving along a (Sun-controlled) elliptical path which is tangent to the (approximately) circular orbits of the two planets. Consider a Hohmann transfer. from Earth (orbit radius 1.00 AU) to Venus (orbit radius 0.72 AU). Find, in units of AU and year: (a) the semi-major axis of the transfer orbit, (b) the time required to go from Earth to Venus, (c) the velocity "kick" needed to place a spacecraft in Earth orbit into the transfer orbit.
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In this problem ignore the effects of the gravitational fields of Earth and Venus on the spacecraft.
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24 Chapter 1: Newton's Laws
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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6. Halley's comet travels around the Sun on an approximately elliptical orbit of eccentricity e = 0.967 and period 76 years. Find: (a) the semi-major axis of the orbit (Ans. 17.9 AU), (b) the distance of closest approach of Halley 's comet to the Sun (Ans. 0.59 AU), (c) the time per orbit that Halley 's comet spends within 1 AU of the Sun (Ans. 78 days).
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7. Define a "season" as a time interval over which the true anomaly increases by t/2. Find the duration of the shortest season for earth . Take the eccentricity of earth's orbit to be 0.0167.
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8. A satellite of mass m moves in a circular orbit of radius ao around the earth. (a) A rocket on the satellite fires a burst radially, and as a result the satellite acquires, essentially instantaneously, a radial velocity u in addition to its angular velocity. Find the semi-major axis, the eccentricity, and the orientation of the elliptical orbit into which the satellite is thrown. (b) Repeat (a), if instead the rocket fires a burst tangentially. (c) In both cases find the velocity kick required to throw the satellite into a parabolic orbit.
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9. Show that the following ancient picture of planetary motion (in heliocentric terms) is in accord with Kepler's picture, if the eccentricity e is small and terms of order e2 and higher are neglected: (a) the earth moves around the sun in a circular orbit of radius a; however, the sun is not at the center of this circle, but is displaced from the center by a distance ea; (b) the earth does not move uniformly around the circle; however, a radius vector from a point which is on a line from the sun to the center, the same distance from and on the opposite side of the center as the sun, to the earth does rotate uniformly.
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10. (a) Show that
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2ro =1 + cos o
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r (the standard form .for a conic section , on setting the eccentricity e =1 and the semilatus-rectum p = 2ro) is the equation of a parabola, by translating it into cartesian coordinates with the origin at the focus and the x-axis through pericenter. (b) A comet travels around the Sun on a parabolic orbit. Show that the distance r of
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the comet from the Sun is related to the time t from perihelion by
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(r + 2ro ) r - ro = 2x t 3 where distances are measured in AU and time is measured in years. (c) If one approximates the orbit of Halley's comet near the Sun by a parabola with
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ro = 0. 59 AU, what does this give for the time Halley's comet spends within 1 AU
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of the Sun? (d) What is the maximum time a comet on a parabolic orbit may spend within 1 AU
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of the Sun?
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Exercises 25
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11. A particle of mass m moves in a central force field F = -(k/r2 )r. (a) By integrating Newton's second law dp/dt = F, show that the momentum of
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the particle is given by p = po + (mk L)9, where po is a constant vector and L is the magnitude of the angular momentum. (b) Hence show that the orbit in momentum space (the so-called hodograph) is a circle. Where is the center and what is the radius of the circle? (c) Show that the magnitude of po is (mk/L)e, where e is the eccentricity. Sketch the orbit in momentum space for the various cases, e = 0, 0 < e < 1, e =1, e > 1, indicating for the last two cases which part of the circle is relevant. (See: Arnold Sommerfeld, Mechanics, (Academic Press, New York, NY, 1952), trans. Martin 0. Stern, p. 33, 40, 242; Harold Abelson, Andrea diSessa, and Lee Rudolph, "Velocity space and the geometry of planetary orbits," Am. J. Phys. 43, 579-589 (1975).)
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12. Consider the motion of a particle in a central force field with potential V = - k/r. Since the force is central, the angular momentum L = r x p is constant and the orbit lies in a plane passing through the force center and perpendicular to L. (a) Show that for the particular potential V = - k/r there exists an additional vector quantity which is constant, the Laplace-Runge-Lenz vector K=pxL - mkr. Further show that K - L = 0, so that K and L are perpendicular and thus K lies in the orbital plane. (Hint: if you've done exercise 1.11, you need only show that K=po xL).
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(b) By taking the dot product of K with r obtain the equation of the orbit
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a(1- e2) = 1 + ecosO. r
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Hence find a and e in terms of K and L, and also find the direction that K points in the orbital plane.
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2 (c) Express the energy E2=mpr-tkeirnms of K and L.
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13. Consider the motion of a particle of mass m in a central force field with potential
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V=-k+h.
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r r2 (a) Show that the equation for the orbit can be put in the form
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a(1- e2) =1 + ecosa8 , r
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and find a, e, and a in terms of the energy E and angular momentum L of the particle, and the parameters k and h of the potential. (b) Show that this represents a precessing ellipse, and derive an expression for the average rate of precession in terms of the dimensionless quantity q = h/ka. (c) The perihelion of Mercury precesses at the rate of 40 " of arc per century, after all known planetary perturbations are taken into account. What value of q would
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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26 Chapter 1: Newton's Laws
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lead to this result? The eccentricity of Mercury's orbit is 0.206 and its period is 0.24 years. 14. A particle of mass m moves in a 3D isotropic harmonic oscillator potential well
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V. 2 mw2r2 where w, the angular frequency, is a constant.
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= (a) Show that the equation for the orbit has the form L2 1 1+FcE osL22 (8 - 00) mE r2 where E is the energy and L is the angular momentum. (b) Show that this represents an ellipse with geometric center at the force center, and express the energy and angular momentum in terms of the semi-major axis a and eccentricity e of the ellipse. (Ans. E = mw2 (a2 + b2) and L = m(oab where b = a 1- e2 is the semi-minor axis) (c) Show that the period is i = 2x/w independent of the energy and angular momentum, and that the radius is given as a function of time by
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r2= = 1 E- F cE osL 22 w(t - t0) mw2 2
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15. A small meteor approaches the earth with impact parameter b and velocity v. at infinity. Show that the meteor will strike the earth if b<a I1+(vo/vO2 where a is the radius and v0 is the "escape velocity" for the earth.
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16.
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(a) Find the relation between the scattering angle © and the impact parameter b for scattering from a hard sphere of radius a (for which "angle of incidence = angle of reflection"). (b) Use your result to obtain the differential scattering cross section do/do. Integrate to find the total scattering cross section a= f(do/dQ)dQ. where the integration extends over the whole solid angle.
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Lagrangian and Hamiltonian Mechanics Downloaded from www.worldscientific.com by 2603:6010:8600:635:73e9:317c:20e8:d0d on 03/08/24. Re-use and distribution is strictly not permitted, except for Open Access articles.
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Exercises 27
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17. (a) Show that a particle of energy E is refracted in going from a region in which the potential is zero to a region in which the potential is -V1, the angle of incidence 00 and the angle of refraction 01 being related by Snell's law sin 00 n sin 01 where angles are measured from the normal and n = Ji + V1 /E is the index of refraction. (b) Use Snell's law to show that a particle incident at impact parameter b on an attractive square well potential V(x) = -V1 for r < a V(x) = 0 for r > a is scattered through an angle O given by b2 n2 sine O/2 a2 n2 + 1- 2ncos0/2 In particular, show that for small impact parameters (b << a) the scattered particles
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are brought to a focus a distance f -~ ( n ) from the force center. ^n-lA2)
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(c) Find the differential scattering cross section do/dQ.
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18. (a) Show that
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r0 = cos a6 r is the equation of the orbit for a particle moving in a repulsive potential V(r) = k/r2, determining a and r0 in terms of the energy and angular momentum.
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(Ans. a = 1 + 2mk , r0 = aL ) L2 2mE
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(b) Show that the impact parameter b and scattering angle O are related by
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b2 k (^ - e)2 E O(2n - 4)
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(c) Show that the differential scattering cross section is given by da _ n2k it - O dQ E sin O 02 (2r - 0)2
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