Analytical Mechanics: Lecture Notes (R. Egerton)

Fowles & Cassiday Chapter 6

6.1 Universal gravitation as a vector equation (p.203)

Value of G (accuracy, changing with time?)

Newton: moon and apple. Earth/moon system in dynamic equilibrium.

Meeting of Newton with Halley & Wren.

Inverse square law leads to elliptical orbit; proof required Principia.

6.2 Attraction between a sphere and a particle (Newton's historical problem)

Involves distance between centers.

Proof on pp. 208-209; first consider segments of a spherical shell, then sum over shells.

Note: Eq.(6.2.7) holds for P inside shell but with integration limits R-r to R-r, giving F=0.

Equation assumes inverse-square law so absence of force is a sensitive test of its validity.

6.3 Kepler worked with Tycho Brahe, particularly on the Mars orbit.

1. Elliptical orbits (ellipse = one of the conic sections, p.217)

2. Equal areas in equal time (planets travel faster when nearer the sun)

3. T2 proportional to a3 (a = semimajor axis: see p. 216)

Kepler's laws predate Newton, but can be derived from his mechanics.

6.4 Second Law (easiest to prove)

Define L = r ´ p (direction is perpendicular to plane of orbit)

dL/dt = r ´ F = 0 so angular momentum is conserved (for a central force)

dA = 1/2 (r dr sinq ) = 1/2 ½ r ´ dr½ = 1/2 ½ r ´ vdt½ = (1/2m) L dt = const.

Therefore second law is equivalent to conservation of angular momentum, and does not depend of inverse-square character of the force (unlike the 1st and 3rd laws).

6.5 First Law (just outline proof)

Start with Newton's law as vector eqn: mr'' = f(r) er where er is a unit vector

Equivalent to two scalar equations; using polar coordinates these are:

Radial component: mr'' - m r w 2 = f(r) as given on p.30

[2nd term on LHS = centripetal force, needed even for circular motion where r'' = 0 ]

Tangential: mrw ' + 2mr'w = 0 [second term on LHS is the Coriolis force]

These equations involve time, but we don't need t-dependence to describe orbit.

On p. 213, u = 1/r is used as the parameter, leading to a differential equation of the orbit, Eq. (6.5.10) in which time is eliminated; f(r) can be any central force.

On p. 215, it is shown that an inverse-square law for f(r) produces an elliptical orbit as one option; others are parabola and hyperbolas - conic sections (p.217).

6.6 Third Law

dA/dt = L/2m. Integrate over 1 period gives A = (L/2m) ò dt = (L/2m) T

But A = p ab (compare with circle) where b = (1-e 2)1/2 a = (a /a)1/2 a µ a1/2

Where a = "latus rectum" and is related to L and G (Eq. 6.5.19 of p.216) and is independent of the shape of the ellipse (eccentricity e ), so T µ A µ a3/2

Note (p.222) that most of the planets have nearly circular orbits (small e ).

P. 223: Use of gravitational force law in astronomy:

Irregularities in the motion of Uranus suggested an outer planet, whose approximate coordinates could be computed from the force law. In 1846, that planet was discovered and named Neptune.

Similarly, irregularities in Neptune's orbit led to the discovery of Pluto.

Irregularities in Mercury's orbit suggested an inner planet "Vulcan", but were explained by Einstein on the basis of his General Theory of Relativity. In the absence of GR effects, Newton's law appears to be valid everywhere in the universe.

Spiral galaxies: assume circular motion and most mass in core. The for stars outside core,

v = 2 p r/T where Tµ r3/2 by analogy with planets, so v µ 1/r1/2 ("Keplerian rotation").

But observed tangential speeds do not fall off as fast as this; see Fig. 6.6.1 (p.224), suggesting the presence of additional mass at large r, the so-called dark matter.

6.7 Gravitational potential

By analogy with electrostatics, f(r) = G m M/r2 between m and M

suggests the existence of a potential energy: V(r) = - G m M/r of the system,

and a gravitational potential F (r) = - G M/r at the location of m, due to mass M.

(the minus signs indicate attraction, similar to + and – charges)

To change radius (e.g. of orbit) from r1 to r2 , work done is W = -G M m (1/r2 - 1/r1)

This is independent of the path taken, so inverse-square force is conservative.

p.228 gravitational field intensity: a generalization and vectorization of g due to earth

6.8 Central field

Inverse-square force is just one example of a central force (e.g. 1/rn).

Can show (p.231) that any central force is conservative (test is that curl F =0).

This allows the force to be expressed in terms of a potential f(r) = - dV/dr

6.9 Energy equation (6.9.3) of orbit is equivalent to the force equation (6.5.10).

6.10 For inverse-square force, get Eq.(6.10.7) describing a conic section (see p. 216).

Value of total energy E determines the eccentricity.

If e < 0, have ellipse and a closed (repeating) orbit; occurs if v2 <GM/r.

If e < 0, have parabolic orbit (distance e a between foci = 0); occurs if v2 =GM/r.

If e < 0, have hyperbolic orbit (internal and external foci); occurs if v2 >GM/r.

6.11 Since l º L/m = mv r /m = r2 w = r2 q ' , energy equation of orbit becomes (6.11.1a), equivalent to m(dr/dt)2/2 + U(r) =E = KE + PE

so U(r) = effective potential (energy) = V(r) + (m/2) (l2/r2)

The extra component ml2/2r2 is the centifugal potential (energy), which accounts for the rotational component of motion.

This equation reduces the 2-dimensional orbit to a 1-dimensional problem.

Fig. 6.11.1 gives U(r) for inverse-square force, i.e. V(r) = - k/r = [- GM/r for gravity]

Fig.6.11.2 is a 3D plot with U(r) vertical and orbit (x,y) in horizontal plane.

The centrifugal term appears as a potential barrier, preventing close approach to center.

The limits (maximum and minimum values) or r correspond to dr/dt = 0, i.e. U(r)=E

For V(r) = - k/r : (m/2) (l2/r2) - k/r = E giving Eq.(6.11.4): -2Er2 -2kr +ml2 = 0

r = (-1/2E) [k2 ± (k2 + 2Eml2)1/2]

There can be two roots, the maximum and minimum values of r .

For E < 0, there are two positive roots (elliptical motion; see Fig. 6.11.1)

except if E = - k2/2ml2 in which case a single root indicates circular motion.

If E > 0 , there is only one real root, corresponding to a hyperbolic trajectory (e.g. comet) as illustrated in Fig. 6.11.2 .

--------------------------------------

If force is electrostatic, the situation could represent a hydrogen atom.

Proper treatment is by wave mechanics; however, the radial Schroedinger equation has a similar form - see for example Eq.(7.37) on p.281 of Serway et al. (Modern Physics).

For s-states, L = 0 and there is not centrifugal barrier; charge density is maximum at r=0.

Would correspond to linear motion through nucleus on semi-classical Bohr model.

For p-states, orbital charge density avoids the center (Serway Figs. 7.12 and 7.13)

Bohr model: potential barrier allows elliptical motion (p.130 of Serway et al.)

This behaviour has implications for spectroscopy (delayed absorption edges).

-----------------------------------------

Rutherford scattering, originally of alpha-particles

For SI units, replace Qq by Qq/4p e 0 in Eqs. (6.12.1) to (6.12.14)

Differential equation of the "orbit" is (6.12.2)

Path is actually a hyperbola, since E = KE + PE and PE > 0 (repulsive force)

Point or closest approach to nucleus (charge Q) corresponds to q = q 0; see Fig. 6.12.1

Eq.(6.12.4) gives distance r to origin (nucleus) as function of q - q 0

R is minimum when denominator = maximum, cos term = 1 and q = q 0

Particle distant and approaching corresponds r ~ ¥ , denom=0, and q = 0 (see Fig.)

Particle distant and receding corresponds to r ~ ¥ and same value of cos term: q = 2q 0

For zero denominator, (1 + 2Em l2Q-2q-2) cos2q 0 = 1 = sin2q 0 + cos2q 0

i.e. (2Em)1/2 l(Qq)-1 = tan q 0 = tan (p /2 - q s/2) = cot (q s /2)

As seen from Fig. 6.12.1, scattering angle is q s = p - 2q 0 , as in Eq.(6.12.7)

Scattering angle depends on distance of closest approach to nucleus

(rmin = b/sinq 0 , see Fig. 6.12.1),

which is related to the angular momentum about the nucleus:

(at r=¥ ) L = lm= b(mv0) where b (= l/v0 ) is the impact parameter of the collision

For v0 << c (as before), E = (m/2)v02 at r ~ ¥ (since PE=0) and v0 = (2E/m)1/2

Replacing l by bv0 in (6.12.7): cot(q s/2) = (2Em)1/2 bv0(Qq)-1 = bmv02/(Qq) = 2bE/(Qq)

as in Eq.(6.12.9) on p.242.

Fig. 6.12.1: a plane through nucleus and perpendicular to the incident particle can be mapped as a target area. For impact parameters in the range b to b+db, the target is an annular ring of radius b and thickness db, and its area (2p b)db is the differential cross section, denoted s (q s) on page 242 but more often as ds /dW ; it is a measure of the amount of scattering in the angular cone of solid angle dW = 2p sin q s dq s

For a target foil with n nuclei per unit area (perpendicular to the incident particles), the number dN of the N incident particles which are scattered through angles in the range q s to q s+dq s (and impact parameters in the range b to b-db) is given by:

dN = n N ds = n N (-2p b)db

Therefore: ds /dW = (2p b)(-db/dW ) = 2p b (-db/dq s) (dq s/dW ) = (b/sinq s) (-db/dq s)

Differentiating Eq. (6.12.9) gives: (2E/qQ) (db/dq s) = (1/2) [-sin2(q s/2)]-1

ds /dW = b [sinq s]-1 (qQ/4E) [sin2(q s/2)]-1

= (Qq/2E)cot(q s/2) [2sin(q s/2)cos(q s/2)]-1 (qQ/4E) [sin2(q s/2)]-1

= (Qq/4E)2 [sin4(q s/2)]-1

This famous formula, derived by Rutherford, provided good agreement with the angular distribution of alpha-particles scattered through large angles. An identical formula can we derived by applying wave mechanics.

At small angles, however, there is a problem: the differential cross section goes to infinity at q s ~ 0 (more seriously, the integrated cross section is also infinite). This is a result of the long-range Coulomb force: there is no impact parameter at which the scattering angle actually becomes zero. In practice, however, the nuclear electrostatic field is screened by atomic electrons and the infinite cross section is avoided.

The above formula also holds for negatively charged incident particles such as electrons; the trajectory is still a hyperbola but the particle is attracted toward the nucleus, which is at the internal focus of the hyperbola.

Cross sections and angular distributions of scattering are of general interest in particle physics.

6.13 Stability of orbits

The self-repeating orbits we have considered for central forces are possible solutions of the equation of motion, but are they the only ones?

We will explore this question for the case of a a mass m in a circular orbit, radius a, when it is suddenly displaced by a small amount x (first-order perturbation).

The radial equation of motion, Eq.(6.13.1), can then be rewritten as Eq.(6.13.4).

Using a Taylor-series expansion of (x+a)n gives Eq(6.13.5), where higher-order terms have been ignored (for small x).

Making use of Eq.(6.13.2) for the unperturbed orbit gives Eq.(6.13.6),

which is of the same form as Eq.(3.2.4a) on p.73 which defines SHM,

provided the force constant k (square-bracket coefficient of x) is positive

(force kx is restoring, towards x = 0).

This requires f(a) + (a/3) f’(a) < 0 ,

which for a power-law central force: f(r) = - c rn give n > -3

Inverse-square force therefore allows stable orbits,

as does an object whirled around on an elastic string or a ball rolling around a parabolic well(n = +1).

Inverse-cube law is unstable, as could be proved by including higher powers of x ;

in fact, the orbit is a spiral, as shown in Problem 6.10

6.14 Apsidal angle y is the angle between maximum and minimum r , for a perturbed orbit.

Period T of the instability is 2p (m/k)1/2 as given by Eq.(6.14.1)

and y = (T/2) (dq /dt)

For pure circular motion, (dq /dt) = L / ma2 = l / a2 = [- f(a)/ma] from Eq.(6.13.2)

so (for small perturbation) y is given by Eq.(6.14.3),

which for a power-law force becomes Eq.(6.14.4).

For inverse-square law (n=-2), y = p so the perturbed circular motion is self-repeating; the SHM corresponds to r = a + A sin(w t) ~ a + A sin q ~ a / [1 – (A/a) sinq ] which is the equation of an ellipse.

For n = - 2.0000001612 for example, the perihelion advances ~ 42"/century (see p. 247).

For n= 1 , y = p /2 : again self-repeating elliptical-like motion.

Other values of n may give non-repeating motion (y = non-integer multiple of p )