Chapter 4 Central forces If the force between two bodies is directed along the line connecting the (centres of mass of) the two bodies, this force is called a central force. Since most of the fundamental forces we know about, including the gravitational, electrostatic and certain nuclear forces, are of this kind, it is clear that studying central force motion is extremely imortant in physics. Moreover, the motion of a system consisting of only two bodies interacting via a central force is one of the few problems in classical mechanics that can be solved completely (once you add a third body it becomes, in general, unsolveable). Examples of such systems are the motion of planets and comets around a star, or satellites around a planet, or binary stars; and classical scattering of atoms or subatomic particles. The full description of atoms and subatomic particles requires quantum mehcanics, but even here the classical analysis of central forces can yield a great deal of insight. In the two-body problem we start with a description in terms of 6 coordinates, namely the three (cartesian) coordinates of each of the two bodies. We shall see that it is possible to reduce this to just 2, and for some purposes only 1 effective degree of freedom. This reduction will happen in 3 steps: 1. We can treat the relative motion as a 1-body problem. 2. The relative motion is 2-dimensional (planar). 3. We can use angular momentum conservation to treat the radial motion as 1- dimensional motion in an effective potential. 4.1 One-body reduction, reduced mass We start with a system of two particles, with coordinates ~r1 and ~r2. We need six coordinates to describe this system, and this is provided by the three components of ~r1 and the three components of ~r2. However, since we know that the potential energy only depends on the combination r = j~rj = j~r1 − ~r2j, we may want to describe it instead in terms of the three components of the relative coordinate ~r = ~r1 − ~r2 and a second 48 vector, which we can take to be the centre-of-mass vector m ~r + m ~r R~ = 1 1 2 2 : (4.1) m1 + m2 ~ First we need to express ~r1 and ~r2 in terms of the new coordinates R; ~r. We have ~r = ~r1 − ~r2 () ~r1 = ~r + ~r2 (4.2) ~ m1 ~r1 + m2 ~r2 m1(~r + ~r2) + m2 ~r2 m1 R = = = ~r2 + ~r; (4.3) m1 + m2 m1 + m2 m1 + m2 which gives ~ m1 ~r2 = R − ~r (4.4) m1 + m2 ~ m2 ~r1 = ~r + ~r2 = R + ~r (4.5) m1 + m2 We now plug (4.4), (4.5) into the expression for the kinetic energy, 1 1 T = m ~r_2 + m ~r_2 2 1 1 2 2 2 2 2 1 ~_ m2 _ 1 ~_ m1 _ = m1 R + ~r + m2 R − ~r 2 m1 + m2 2 m1 + m2 2 1 ~_ 2 2m2 ~_ _ m2 _2 = m1 R + R · ~r + 2 ~r 2 m1 + m2 (m1 + m2) (4.6) 2 1 ~_ 2 2m1 ~_ _ m1 _2 + m2 R − R · ~r + 2 ~r 2 m1 + m2 (m1 + m2) 2 2 1 ~_ 2 1 m1m2 m1m2 _2 = (m1 + m2)R + 2 + 2 ~r 2 2 (m1 + m2) (m1 + m2) 1 _ 1 = MR~ 2 + µ~r_2 : 2 2 The total lagrangian is therefore 1 _ 1 L = T − V = MR~ 2 + µ~r_2 − V (r) : (4.7) 2 2 We see that the lagrangian splits into two separate parts: one describing the motion of the centre of mass, and one describing the relative motion. We can therefore analyse the relative motion without any reference to the overall motion of the centre of mass. More- over, R~ is cyclic, so its canonical momentum is conserved. The canonical momentum conjugate to R~ is @L _ ~ ~_ Pi = = MRi =) P = MR: (4.8) _ @Ri This is just the total momentum of the system: _ _ ~ ~_ m1 ~r1 + m2 ~r2 _ _ P = MR = (m1 + m2) = m1 ~r1 + m2 ~r2 : (4.9) m1 + m2 49 Therefore, the absolute motion is merely linear motion with constant total momentum. From now on, we will ignore the absolute motion completely, and focus only on the relative motion | ie, we will drop the first term in (4.7). The lagrangian then becomes 1 L = µ~r_2 − V (r) : (4.10) 2 This looks exactly like the lagrangian for a single particle with position ~r in a potential V (r), but with mass m m µ = 1 2 = the reduced mass (4.11) m1 + m2 We have therefore reduced two-body motion to the equivalent motion of a single body with mass µ. It is worthwhile looking more closely at the reduced mass and its relation to the masses m1; m2. We can assume without any loss of generality that m1 ≥ m2 (since we could just swap the labels if it were the other way around). Then we have m 1 m 1 1 ≥ ; 2 ≤ (4.12) m1 + m2 2 m1 + m2 2 m m m m =) 1 ≥ 1 2 ≥ 2 : (4.13) 2 m1 + m2 2 So we see that the reduced mass has a value that lies between half the larger mass and half the smaller mass. Two special cases of particular interest are where the two masses are equal, and where one mass is much larger than the other. The first includes scattering of identical particles (for example two α-particles) as well as some binary stars. The second includes the motion of a planet or comet around the sun, or satellites around a planet. In the first case, m1 = m2 = m, we get that the reduced mass is µ = m=2, ie the reduced mass is half the mass of each body. In the second case, m2 m1, we can rewrite the reduced mass as 1 m m 2 µ = m = m 1 − 2 + 2 + ::: : (4.14) 2 m2 2 1 + m1 m1 m1 If m2 is small enough compared to m1 (for example, for the earth{sun system we have −6 M⊕=M = 3 · 10 ) we can just set µ = m2, ie the reduced mass equals the smaller of the two masses. 4.2 Angular momentum and Kepler's second law Our system is now equivalent to a single particle with mass µ moving in a spherically symmetric potential V (r). Since we have spherical symmetry, the angular momentum L~ = ~r × ~p is conserved, both in magnitude and in direction. We will use this to simplify the problem further. First, we note that since the direction of L~ is conserved, both the position vector ~r and momentum vector ~p = µ~r_ must be in the plane that is orthogonal to this vector, as the 50 cross product between any two vectors is orthogonal to both vectors. We can choose the z-axis of our coordinate system to be pointing in the direction of L~ , ie L~ = `z^, and in this case both ~r and ~p must be in the xy-plane. We can get a more physical understanding of this by noting that as long as ~p = m~r_ remains in the xy-plane, ~r will not move out of this plane, while as long as ~r remains in that plane there is no force that will move ~p our of the plane, since the central force always points towards the centre (or away from it, in the case of a repulsive force), ie along the vector ~r. We have therefore reduced the motion to planar (2-dimensional) motion, and we can use polar coordinates (r; θ) to describe this motion (where θ is the angle with some arbitrarily chosen direction in the plane of motion). The lagrangian for the system, in those coordinates, becomes 1 1 L = µr_2 + µr2θ_2 − V (r) : (4.15) 2 2 Since L does not depend on the angle θ (this is the remaining rotational symmetry), the angular momentum ` = pθ is conserved, ` = µr2θ_ = constant : (4.16) This is exactly equivalent to Kepler's second law for planetary motion, which gives a nice geometrical interpretation of angular momentum conservation. Consider the area dA swept out by the radius vector in a small (infinitesimal) time dt. The angle swept out in that time is dθ = θdt_ , and the length of the arc swept out is ds = rdθ = rθdt_ (see fig. 4.1). If dθ is small we can approximate the area by a triangle with length r and height ds, ie 1 1 dA 1 ` dA = rds = r · rθdt_ () = r2θ_ = = constant : (4.17) 2 2 dt 2 2µ r(t+dt) rdθ dθ r(t) Figure 4.1: The area swept out by a radius vector. Kepler's second law reads, in words, A line joining a planet and the Sun sweeps out equal areas during equal intervals of time. 51 Kepler came to this conclusion through painstaking observation of planetary motions, and the apparatus of newtonian and lagrangian mechanics, which we are using here, was developed long after his time. But from this derivation we can see that this law is valid not just for planetary motion, but for all central force motion, whatever the force is, and is equivalent to conservation of angular momentum. Example 4.1 The planet Mercury orbits the Sun in 87.97 days, in an elliptic orbit with semimajor axis a = 57:91 · 106km and semiminor axis b = 56:67 · 106km. 1. What is the areal velocity of Mercury? 2. Use the relation between areal velocity and angular momentum to find the speed of Mercury (a) at its perihelion (closest to the sun), 46:00 · 106km from the sun; (b) at its aphelion (furthest from the sun), 69:82 · 106km from the sun.