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Classical Mechanics Motion Under Central Forces

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Classical Mechanics Motion Under Central Forces Classical Mechanics Motion Under Central Forces Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400085 September 6, 2016 1 Introduction Central forces occupy a special place in physics because some of the most commonly met forces in nature are of this nature. Examples are Hooke's law force, gravitational force etc. Such forces are derivable from a space dependent potential and depend only on the distance between the object and the source of the force. If the source of the force is taken to be the origin, the force is given by F~ (~r) = −∇V (~r), where V (~r) = V (j r j). Since the @V potential is centrally symmetric, F~ (~r) = − r^. Let us begin by making some qualitative @r comments on the nature of force and consequence thereof: 1. If the central force does not have explicit time dependence, it is conservative in nature and there exists an energy function which is conserved. This can be seen by writing the force equation as F~ = m~v_ = −∇V (~r) Taking a product of both sides with ~v, we have, d m~v_ · ~v = −∇V · ~v = − V dt which gives d mv2 + V = 0 dt 2 mv2 so that the energy E = + V = constant. 2 1 c D. K. Ghosh, IIT Bombay 2 2. Since the force does not have any (θ; ') dependence, it is isotropic in space. A consequence of such rotational symmetry is the conservation of angular momentum which can be trivially seen by observing that the torque about the origin, τ = ~r × F~ (~r) = 0. 3. Suppose the initial velocity of the particle is along some direction ~v. Since the force is along the radial direction, the motion takes place always in the plane described by ~r and ~v. This is actually a consequence of the conservation of angular momentum ~r × ~p = L~ . The motion must therefore take place in a plane perpendicular to L~ . 4. The trajectory of the particle is described by ~r(t) and θ(t). Once can eliminate t to get the equation to the trajectory r(θ). 5. The state of motion of two particles is completely specified by six quantities (E; L;~ r; θ). will see later that the orbit is described in terms of six quantities. 2 The Two Body Problem - Reduced Mass For a two particle system, we need to specify two positions and two velocities. However, if the only interaction between the particles is mutual, the centre of mass does not accel- erate and its position R~ becomes a cyclic coordinate. Define centre of mass and relative coordinate R~ and ~r respectively by ~ m1~r1 + m2~r2 1 R = = (m1~r1 + m2~r2) (1a) m1 + m2 M ~r = ~r2 − ~r1 (1b) where we have used M = m1 + m2. In terms of these coordinates we can express the original coordinates as m ~r = R~ − 2 ~r (2a) 1 M m ~r = R~ + 1 ~r (2b) 2 M The kinetic energy of the pair can be expressed as 1 T = (m (~r_ )2 + ~r~ )2) 2 1 1 2 1 m 2 m 2 = m R~_ − 2 ~r_ + m R~_ + 1 ~r_ 2 1 M 2 M 1 1 m m = (m + m )R_ 2 + 1 2 r_2 2 1 2 2 M 1 1 = MR_ 2 + µr_2 (3) 2 2 c D. K. Ghosh, IIT Bombay 3 where µ is defined as the \reduced mass" and is given by 1 1 1 = + (4) µ m1 m2 [Examples of reduced mass: For planet-Sun system, the reduced mass is nearly equal to the mass of the planet since the mass of the Sun is much greater than the planetary mass. Likewise in the hydrogen atom, the mass of the nucleus being much greater than the electronic mass, the reduced mass is nearly equal to the mass of electron. However, if we consider a positronium atom, where an electron moves round a positron which has the same mass as that of the electron the reduced mass is m=2 where m is the mass of an electron (or positron).] Thus the Lagrangian of the system is given by 1 1 L = MR_ 2 + µr_2 − V (j r j) (5) 2 2 (Note that in (3) and (5), R_ andr _ stand the vector velocities and should not be confused with the time derivatives of the distances) In Lagrangian (5), R~ is a cyclic coordinate. Hence the momentum of the centre of mass is conserved. The centre of mass, therefore, moves with a constant velocity. We can move the origin to the location of the centre of mass and consider the motion of a reduced mass about the centre of mass. This reduces a two body problem to a one body problem. 3 Lagrangian in Spherical Polar Coordinates We will express (5) in a spherical polar coordinates (r; θ; '), taking the origin at the centre of mass. Note that as the motion will be shown to take place in a plane (taken as the x-y plane) which is conventionally written as the polar (r; θ) plane, we have taken ' as the polar angle and θ as the angle made by the radius vector with the x-axis. c D. K. Ghosh, IIT Bombay 4 z P ϕ y θ r Q x The Lagrangian is given by µ L = r_2 + r2 sin2 'θ_2 + r2'_ 2 − V (r) (6) 2 θ is cyclic. However ' is not cyclic and the corresponding Euler Lagrange equation is d @L @L − = 0 dt @'_ @' which gives, d (µr2'_) − µr2 sin ' cos 'θ_2 = 0 dt This gives, µr2'¨ + 2µrr_'_ − µr2 sin ' cos 'θ_2 = 0 We can choose the initial condition to be ' = π=2 and' _ = 0, so that' ¨ = 0 at t = 0. Since' ¨ = 0 and' _ = 0,' _ cannot change and remains zero. Consequently, ' also cannot change and remains at its initial value π=2. The fact that the motion is confined to a plane (which we conveniently take to be the x-y or the polar r − θ plane) is a consequence of the conservation of angular momentum, which, in turn, is a consequence of the force being central. d~l d = (~r × ~p) dt dt d~r d~p = × ~p + ~r × dt dt = ~v × m~v + ~r × F~ = 0 Since the angular momentum is conserved, the motion takes place in a plane defined by the initial position ~r0 and the initial velocity ~v0. Let this plane be P . By definition, P c D. K. Ghosh, IIT Bombay 5 ~r × ~v is perpendicular to the vectorn ^ = 0 0 . However, since the angular momentum j ~r0 × ~v0 j ~l = m~r × ~v does not change with time, neither does the plane P . Hence the motion takes place in the x-y plane with µ L = (_r2 + r2θ_2) − V (r) (7) 2 Looking at the Lagrangian (7), we notice that since θ is cyclic, the corresponding mo- mentum µr2θ_ is a constant of motion. For the case of planetary motion, this results in Kepler's second law as from the figure below one can see that the rate of sweeping the area by the radius vector is given by dA d 1 1 = r(rθ) = r2θ_ dt dt 2 2 which remains constant. dθ δΑ O Note that since µr2θ_ is constant, θ_ cannot change sign and as a result θ is a monotonic function of time. Let us now look at the Euler-Lagrange equation for r, d @L @L − = 0 dt @r_ @r which gives @V µr¨ − µrθ_2 + = 0 @r @V Using = −F (r) so that we get @r µr¨ = µrθ_2 + F (r) Using the constancy of the angular momentum and denoting it as l = µr2θ_, we get the following differential equation for the radial acceleration l2 µr¨ = + F (r) (8) µr3 This equation can be interpreted by looking at the two terms on the right. The force F (r) is simply the central force with which we started derived from the potential V (r), which in case of planetary motion is the attractive gravitational potential or in case of atomic motion, the attractive Coulomb potential. The first term l2/µr3 can be thought to have c D. K. Ghosh, IIT Bombay 6 l2 been derived from a repulsive potential . This force is always repulsive irrespective 2µr2 whatever be the nature of the central potential. For the case of the attractive Coulomb (or gravitational) potential, this term compensates for the attractive nature. At short distances, this term (known as the centrifugal barrier) dominates and the net effect is repulsive. However, the potential is soon dominated by the attractive term. l 2 2mr 2 V centrifugal r Effective −k/r Coulomb The reason for calling it a centrifugal force term is that as l has the dimension of mr2!, this term as the form of mr!2, the familiar form for the centrifugal force. The radial equation (8) is a second order differential equation and it is possible, in principle, to solve this to get complete solution. However, we can get the first integral of the Lagrangian which gives us a first order differential equation. Define: l2 V = V (r) + eff 2µr2 so that dV l2 @V eff = − + dr µr3 @r so that the radial equation (8) can be written as d l2 µr¨ = − + V (r) (9) dr 2µr2 This is a second order differential equation and it is possible, in principle, to solve this equation. However, we can get the first integral of the Lagrangian which gives us a first order differential equation. (9) can be written as d 1 d l2 µr_2 = − V (r) + dt 2 dt 2µr2 c D.
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