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Home , Central force, Circular motion, Constant of motion, Force, Velocity

Classical Under Central

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 because some of the most commonly met forces in nature are of this nature. Examples are Hooke’s law , gravitational force etc. Such forces are derivable from a dependent and depend only on the 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 (| r |). Since the ∂V potential is centrally symmetric, F~ (~r) = − rˆ. Let us begin by making some qualitative ∂r comments on the nature of force and consequence thereof:

1. If the does not have explicit dependence, it is conservative in nature and there exists an 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, ϕ) dependence, it is isotropic in space. A consequence of such rotational is the conservation of angular which can be trivially seen by observing that the about the origin, τ = ~r × F~ (~r) = 0.

3. Suppose the initial 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 ~r × ~p = L~ . The motion must therefore take place in a plane perpendicular to L~ .

4. The 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 is described in terms of six quantities.

2 The Two Body Problem - Reduced

For a two particle system, we need to specify two positions and two . However, if the only between the particles is mutual, the centre of mass does not accel- erate and its 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 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 - 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 , the mass of the nucleus being much greater than the electronic mass, the reduced mass is nearly equal to the mass of . 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 (| r |) (5) 2 2 (Note that in (3) and (5), R˙ andr ˙ stand the vector velocities and should not be confused with the time of the ) 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, ϕ), 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 ϕ as the polar and θ as the angle made by the 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 ϕθ˙2 + r2ϕ˙ 2 − V (r) (6) 2 θ is cyclic. However ϕ is not cyclic and the corresponding Euler Lagrange equation is d ∂L ∂L − = 0 dt ∂ϕ˙ ∂ϕ which gives, d (µr2ϕ˙) − µr2 sin ϕ cos ϕθ˙2 = 0 dt This gives, µr2ϕ¨ + 2µrr˙ϕ˙ − µr2 sin ϕ cos ϕθ˙2 = 0 We can choose the initial condition to be ϕ = π/2 andϕ ˙ = 0, so thatϕ ¨ = 0 at t = 0. Sinceϕ ¨ = 0 andϕ ˙ = 0,ϕ ˙ cannot change and remains zero. Consequently, ϕ 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 , 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 ˆ = 0 0 . However, since the angular momentum | ~r0 × ~v0 | ~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 = (r ˙2 + 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 law as from the figure below one can see that the rate of sweeping the 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 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

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 term is that as l has the of mr2ω, this term as the form of mrω2, 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 . However, we can get the first 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. K. Ghosh, IIT Bombay 7

which gives 1 l2 µr˙2 + V (r) + = E = constant (10) 2 2µr2 so that s 2  l2  r˙ = E − V (r) − (11) µ 2µr2 which is a first order equation. For physical solution, the quantity within the square root sign cannot be negative. The time to go from r = r0 at t = 0 to r at t = t is given by

Z t Z r dr t = dt = (12) s 2 0 r0 2  l  E − V (r) − µ 2µr2

ldt Using the constancy of µr2θ˙, we can write dθ = , which when used with (12) gives, µr2 Z l Z r ldr θ = dt = (13) µr2 s 2 r0  l  r2 2µ E − V (r) − 2µr2

Let us look at some qualitative aspects of the solution. We started with a particle (mass µ) having three degrees of freedom. We thus had three second order differential equations, solution of which require six constants. We have L~ and E as four conserved quantities, which leave us with two constants of integration r0 and θ0.

We have l2 V = V (r) + eff 2µr2 k Since E is conserved, E −V ≥ 0. For an inverse square law attractive force F (r) = − eft r2 k which is derived from V (r) = − r k l2 V = − + eff r 2µr2 The centrifugal term dominates at small distances beyond which the inverse square law takes over. Since the centrifugal term goes to infinity as 1/r2, the mass does not fall onto the centre unless the attractive term goes to −∞ faster than the way the centrifugal term goes to +∞. Note that according to (11), we must have

l2 r2V (r) + < Er2 2µ c D. K. Ghosh, IIT Bombay 8

l2 1 Thus r → 0 only if r2V (r) < − , i.e. if V (r) ∼ with n > 2. 2µ rn Let us look at the effective potential, which has the shape shown in the figure. Consider a particle having an energy E1. The particle can approach the potential barrier from large distances till it reaches r = rmin at which its velocity becomes zero (E1 = Veff ). However at this point −dV/dr is positive and hence the particle turns back and goes to infinity. The trajectory of the particle is a hyperbola.

Hyperbolic trajectory

Veff E 1

Turning point r

E2 An Open Orbit E r r r 3 min min max

Consider a particle having an energy E2. The motion is confined between two limits rmin and rmax (shown in red). The orbit may or may not be closed. For the attractive inverse square law, the orbit is an ellipse.

Finally for a particle having an energy E3, the value of r remains constant so that the particle moves in a . The condition for the path to be closed is that the value of the angle θ, calculated from r = rmin to r = rmax should be an integral multiple of m/n, where m and n are integers, so that after n periods, the particle would go back to its initial position, having completed m revolutions. In general, such paths are open . Only two specific instances are known which give closed orbit, the inverse square law force and the force −kr for a 3d-oscillator.

4 Kepler’s Problem:

Let α V (r) = − r The effective potential is given by

α l2 V = − + eff r 2µr2 Define l2 M 2 = (14) 2µ c D. K. Ghosh, IIT Bombay 9

dV In terms of this the minimum of the effective potential is at (set = 0) dr 2M 2 r = r = (15) min α at which the value of the potential has the value α2 V min = − (16) eff 4M 2 1 Substitute u = , in terms of which r 2 2 Veff = −αu + M u

Using this, the expression (13) for the angle θ is given by Z ldr θ = 2p r 2µ(E − Veff (r)) Z Mdu = − √ (17) E + αu − M 2u2 To integrate, complete the square in the denominator. Define α2 α L2 = E + ; z = Mu − ; dz = Mdu 4M 2 2M (17) can be written as Z Mdu Z dz  z  θ = − √ = √ = cos−1 − C L2 − z2 L2 − z2 L We can absorb the const ant of integration into the angle θ and write the solution as L  α  u = cos θ + M 2ML Reverting to the variable r, the equation to the trajectory is given by M r = α (18) L cos θ + 2M This is the equation to a conic section of the form p r = (19) 1 +  cos θ where the semi-latus rectum p and the eccentricity  are given by 2M 2 p = (20a) α 2ML  = (20b) α c D. K. Ghosh, IIT Bombay 10

p 2b ε a

2a Note that the choice of the origin for the angle (θ = 0) in the equation to the conic section taken above is p r = = a(1 − ) 1 +  which is the shortest distance from the focus to the position of the particle or the perihelion distance. The orbit is an ellipse for negative energy E as  < 1. We can calculate the time period by using the fact that the rate of sweeping the area l 2µ is constant and is equal to . The time period is therefore given by A, where A is 2µ l the area of the ellipse which is given by πab. Using the geometry of the ellipse, one can express the area of the ellipse as πa3/2p1/2. Thus the time period is given by 2µ T = πa3/2p1/2 l 2µ 2M 2 1/2 = πa3/2 l α s 2µ 2l2 = πa3/2 l 2µα rµ = 2 πa3/2 α r µ = πα (21) 2 | E |3 Thus the square of the time period is proportional to the cube of the major axis. Further, the time period only depends on the energy of the particle. In the last step of (21), we have used the following expression for the semi-major axis of an ellipse: p p 2M 2 1 a = a = = 1 − 2 1 − 2 α 2ML2 1 − α 2M 2 1 = α 4M 2L  α2  1 − × E + α2 4M 2 α = 2 | E | c D. K. Ghosh, IIT Bombay 11

Thus the expression for the time period (21) can be expressed as

rµ α3/2 T = 2 π α (2 | E |)3/2 rµ 1 = πα (22) 2 | E |3/2 α2 Using the expression E = L2 − and the expression for the eccentricity (20b), we 4M 2 have, 4M 2 2 = E + 1 α2 α2 The minimum energy E, is equal to − (as L2 ≥ 0). Thus the minimum value of the 4M 2 eccentricity  is 0, which represents a circle. For α2 µα2 E = − = − 4M 2 2l2 If in this expression, we put l = n~, the energy is given by µα2 µe4 E = − = − 2n2~2 2n2~2 where, we have used α = e2.

5 Laplace-Runge-Lenz (LRL) Vector

We have already shown that for the inverse square law, the energy and the angular momentum are constants of motion. It turns out that one can find yet another vector which is a constant of motion. For historical reasons, this vector goes by the name of Laplace-Runge- Lenz vector. We define LRL vector as

A~ = ~p × ~l − µαrˆ (23) d d~p We have already shown that ~l is a constant of motion. Hence (~p × ~l) = × ~l. We dt dt have, d~p α × ~l = − ~r × (~r × ~p) dt r3 µα = − ~r × (~r × ~r˙) r3 µα h i = − ~r(~r · ~r˙) − ~r˙(r2) (24) r3 We use the following identity 1   1 d ~r · ~r˙ = ~r˙ · ~r = ~r · ~r˙ + ~r˙ · ~r = (r2) = rr˙ 2 2 dt c D. K. Ghosh, IIT Bombay 12

(Note that the last equality results in the product of two scalars r andr ˙, not the product of the corresponding vectors) Substituting this in (24), we get,

d~p µα h i × ~l = − ~r(~r · ~r˙) − ~r˙(r2) dt r3 µα h i = − rr~r˙ − ~r˙(r2) r3 d ~r d = µα = µα (ˆr) dt r dt

Substituting this in (24), conservation of L-R-L vector follows. The vector L,~ A~ and the scalar E constitute 7 constants of motion. However, a two body system has six degrees of freedom. Since, the condition at initial time is unknown, no more than six constants of motion can be there. We have already seen that L~ and E are conserved. Thus, we look for two relationships between the L-R-L vector and the former four. The first relation is obtained by observing that A~ is perpendicular to ~l:

A~ · ~l = [~p × ~l − mαrˆ] · ~l 1 = (~p × ~l) · ~l − mα ~r · (~r × ~p) r = 0 + 0 = 0

Hence, A~ lies in the plane of motion.

L

Α µα p Α pX L P S R pX L µα

p Q p p X L θ µα

Α L-R-L Vector c D. K. Ghosh, IIT Bombay 13

To get a second relationship, consider

Ar cos θ = A~ · ~r = (~p × ~l − µαrˆ) · ~r = ~l · (~r × ~p) − µαr = l2 − mαr

Rearranging, l2 1 r = mα A 1 + cos θ µα A A2 4M 2 Thus the eccentricity is given by  = . Using (20b), we get 2 = = 1 + E . mα m2α2 α2 Substituting for M 2 and simplifying, we get the second realtionship that we have been looking for: A2 = 2µEl2 + µ2α2 Conservation of LRL vector is a consequence of a symmetry in a higher dimensional space, the SO(4) symmetry. We are not in a position to discuss it in this lecture.

6 Determining Force Law from the Orbit Equation:

Starting with l2 ∂V µr¨ = − µr3 ∂r 1 we get, on substituting u = , r du 1 du 1 du dθ = − = − dt u2 dt u2 dθ dt 1 lu2 du l du = − = − u2 µ dθ µ dθ d2r d  l du d  l du dθ = − = − dt2 dt µ dθ dθ µ dθ dt  l d2u l = − u2 µ dθ2 µ l2 d2u = − u2 µ2 dθ2 Thus the differential equation for the orbit becomes, l2 d2u l2 −µ u2 = u3 − F (u) µ2 dθ2 µ i.e. l2  d2u  − u2 + u3 = F (u) (25) µ dθ2 c D. K. Ghosh, IIT Bombay 14

From (25) one can infer the form of F . Example: an The equation to the orbit is given by cos nθ = 1. Determine the force law. rn Solution: Taking logarithm of the orbit equation, we have n ln a − n ln r + ln cos nθ = 0 which, interms of u can be written as n ln a + n ln u + ln cos nθ = 0 Differentiating with respect to u du = u tan nθ dθ The second d2u du = un sec2 nθ + tan nθ = un sec2 nθ + u tan2 nθ dθ2 dθ Thus l2u3 u2n F = − (n + 1) µ a2n Thus the force is proportional to u2n+3 i.e. the force varies as 1/r2n+3. Example: An object moves in a in such a way that the centre of the force is towards a fixed point on the . Find the law of force. Solution: From geometry, we have r = 2R cos θ P

S θ R

Thus the equation to the orbit is 2au cos θ = 1 (This is a special case of the previous example with n = 1. Thus the force is inversely proportional to the fifth of the distance.) We have, u = (1/2R) cos θ. Differentiating, d2u 1 = (sec θ tan2 θ + sec3 θ) = u(tan2 θ + sec2 θ) dθ2 2R . Thus the force is given by l2 d2u l2 l2 l2 F (u) = − u2 = − u3(sec2 θ + tan2θ + 1) = −2 u3 sec2 θ = −8R2 u5 µ2 dθ2 µ2 µ2 µ2 c D. K. Ghosh, IIT Bombay 15

7 Advance of the Perihelion of

As a planet orbits round the Sun, it is found that the location of its perihelion does not remain fixed but advances a bit with every revolution. This is known as the of the orbit. All precess to a lesser or greater degree which can be explained by incorporating the pull due to other planets. The planet Mercury has been an exception because of its proximity to the Sun. As seen from the , Mercury precesses about 574 of an arc (3600seconds = 1◦ over every hundred years. Out this amount 531 seconds of an arc can be explained by gravitational pull of other planets and by the fact that the is made from a non-inertial frame, viz. the Earth. (The balance 43 seconds of arc per century is explained by the General , which along with the of light around the Sun and the gravitational , were proposed by Einstein as test of GTR.)

Sun Mercury P1

P2 P3

1 To understand the advance of perihelion, we add a small repulsive force to the r3 attractive inverse square law, α β V (r) = − + r r2 Note that the added term has the same nature as that of the cenbtrifugal term and as a result, the mathematical manipulations are identical. The expression for the angle (cf. c D. K. Ghosh, IIT Bombay 16

(13)) becomes

Z du θ = −M pE + αu − βu2 − M 2u2 Z du = −M p 2 2 E + αu − (M + β)u  M  Z M 0du = − √ M 0 E + αu − M 02u2 where M 02 = M 2 + β. As before, define

α2 α L2 + E + ; z = M 0u − 4M 02 2M 0 The expression for the angle, after setting the constant of integration to zero, is given by M  z  θ = cos M 0 L Solving for u and reverting back to r, as before, we get

2M 02/α r = 2M 0L 1 + cos γθ α M 0 where γ = . M Thus the eccentricity and the semi-latus rectum in this case are given by

2M 0L  4M 02E 1/2  = = 1 + α α2

2M 0 p = α Clearly, as before  < 1 if E < 0. However, the argument of the cosine, instead of being θ is now γθ. Thus when θ → θ + 2π, cos γθ does not return to its initial value. The shift can be calculated by observing that r M 0 M 2 + β βθ γθ = θ = θ ≈ θ + M M 2 2M 2 Thus the perihelion advances by βθ µβθ ∆θ = = 2M 2 l2 which is proportional to the strength of the repulsive force.