classical dynamics and fluids p 1 - university of cambridge

Classical Dynamics and Fluids P 1

CLASSICAL DYNAMICS AND FLUIDS

20 Lectures Prof. S.F. Gull

• NOTES — Provisional hardcopy available in advance. Definitive copies

of overheads available on web. Please report all errors and typos.

• SUMMARY SHEETS — 1 page summary of each lecture.

• EXAMPLES — 2 example sheets — 2 examples per lecture.

• WORKED EXAMPLES — Will be available on the web later.

• WEB PAGE — For feedback, additional pictures, movies etc.

http://www.mrao.cam.ac.uk/∼steve/part1bdyn/

There is a link to it from the Cavendish teaching pages.


REVIEW OF NEWTONIAN MECHANICS

• Newtonian Mechanics is:

- Non-relativistic

i.e. velocities v c (speed of light=3 × 108 m s−1.)

- Classical

i.e. Et h (Planck’s constant= 1.05 × 10−34 J s.)

• Assumptions:

- mass independent of velocity, time or frame of reference;

- measurements of length and time are independent of the frame of

reference;

- all parameters can be known precisely.

• Mechanics:

= Statics (absence of motion);

+ Kinematics (description of motion, using vectors for position and

velocity);

+ Dynamics (prediction of motion, and involves forces and/or energy).


BASIC PRINCIPLES OF NEWTONIAN DYNAMICS

• This course contains many applications of Newton’s Second Law.

• Masses accelerate if a force is applied. . .

• The rate of change of momentum (mass × velocity) is equal to the

applied force.

• Vectorially (p = mv):

dp

dt=

d(mv)

dt= F

• Usually m is a constant so mdv

dt= F

• General casemdv

dt+

dm

dtv = F enables you to do rocket science.

v

u

m

dm0

• Rocket of mass m(t) moving with velocity v(t)

expels a mass dm of exhaust gases

backwards at velocity −u0 relative to the rocket.

• In the absence of gravity or other external forces

dmu0 +mdv = 0

where mi,f are the initial and final masses.

• Integrating, we find v = u0 log(mi/mf)

• For a rocket accelerating upwards against gravity

mdv

dt+

dm

dtu0 +mg = 0.


IMPORTANT EXAMPLE — SIMPLE HARMONIC OSCILLATOR

• The simple harmonic oscillator (SHO) occurs many times during the

course.

• Mass m moving in one dimension with coordinate x on a spring with

restoring force F = −kx.

• The constant k is known as the spring constant.

• Newtonian equation of motion: mx = −kx, where x denotesdx

dtetc.

• General solution: x = A cosωt+B sinωt where ω2 = k/m.

• Can also write solution as x = <(Aeiωt), where A is complex.

• We can integrate the equation of motion to get a conserved quantity —

the energy.

• Multiplying the equation of motion by x (a good general trick) we get

mxx+ kxx = 0 ⇒ 12mx

2 + 12kx

2 = E = constant

• Here the quantity T ≡ 12mx

2 is the kinetic energy of the mass and

V ≡ 12kx

2 is the potential energy stored in the spring.

• For many dynamical systems (such as the SHO) the time t does not

appear explicitly in the equations of motion and the total energy

E = T + V is conserved.

• This conserved quantity is also known as the Hamiltonian.

.


THE ENERGY METHOD

• If we know from physical grounds that the energy is conserved, we can

always derive the equations of motion of systems that only have one

degree of freedom (such as the Simple Harmonic Oscillator):

12mx

2 + 12kx

2 = E ⇒ x(mx+ kx) = 0 ⇒ mx = −kx

This works because x need not always be zero.

• We will call this the energy method.

• We can sometimes derive the equations of motion of much more

complicated systems with n degrees of freedom is a similar way. It’s

certainly not rigorous, but works for most of the systems studied in this

course.

• The theoretically more advanced methods of Lagrangian and

Hamiltonian mechanics derive the equations of motion from a

variational principle. They are rigorous, but still use the kinetic energy

T and potential energy V (actually in the combination T − V ).

• We will see later in the course the use of the energy method to derive

the equation of motion of a particle at radius r in a central force:

12mr

2 + Veff(r) = E

Differentiating with respect to time we get

r

(

mr +dVeff

dr

)

= 0

• We strike out the r to obtain the equation of motion.


THE ENERGY METHOD: EXAMPLE

• Ladder leaning against a smooth

wall, resting on a smooth floor

(not recommended practice).

• Newtonian method needs reaction

forces N and R.

• Take moments to get the angular

acceleration. Try it. . .

• The energy method is easier:

(1) Potential energy. This is easy: V = 12mgl cos θ.

(2) Kinetic energy. Write this as the sum of the kinetic energy of the

centre of mass plus the energy of rotation about the centre of mass.

T = 12m(x2 + y2) + 1

2Iθ2, where I = 1

12ml2 is the moment of

inertia of a uniform rod about its centre.

Coordinates of centre of mass: x = 12 l sin θ ; y = 1

2 l cos θ

Work out velocities: x = 12 l cos θ θ ; y = − 1

2 l sin θ θ

T = 12m(x2 + y2) + 1

2Iθ2 = 1

8ml2θ2 + 1

24ml2θ2 = 1

6ml2θ2

(3) Energy method:d(T + V )

dt= 0 ⇒ θ

(13 θml

2 − 12mgl sin θ

)

= 0.

(4) Equation of motion: θ =3g

2lsin θ.


REVISION: VECTOR CALCULUS

• In dynamics we use vectors to describe the positions, velocities and

accelerations of particles and other bodies, as well as the forces and

couples that act on them

• We need to revise vectors, vector functions, vector identities and

integral theorems.

• Revise scalar product a·b and vector product a×××××b (preferable to

a∧b).

• There are only a couple of vector identities,

but you MUST LEARN THEM.

(1) Scalar triple product

a·(b×××××c) = (a×××××b)·c (Interchange of dot and cross)

a·(b×××××c) = b·c×××××a = −b·a×××××c (Permutations change sign)

(2) Vector triple product

This is the most important identity.

a×××××(b×××××c) = a·c b − a·b c

Rule: Vector outside bracket appears in both scalar products.

Rule: Outer pair takes the plus sign.

(a×××××b)×××××c = a·c b − b·c a


REVISION: VECTOR CALCULUS

Gradient operator ∇

• grad(Φ); ∇Φ

Vector gradient of a scalar field Φ.

• div(E); ∇·EScalar divergence of a vector field E.

• curl(E); ∇×××××E.

Vector curl of a vector field E.

Divergence Theorem (Gauss’ Theorem)

d = d| |S n S • Relates integral of flux vector E

through closed surface S (n outwards)

to volume integral of ∇·E.

∮

dS·E =

∫

dV ∇·E

Stokes’ Theorem

• Relates line integral vector E

around closed loop l

to surface integral of ∇×××××E.

∮

dl·E =

∫

dS·∇×××××E


VECTOR CALCULUS REVISION

• Further identities

∇×××××(∇Φ) = 0

∇·(∇ × E) = 0

∇×××××(∇×××××E) = ∇(∇·E) −∇2E

• The gradient operator ∇ is a vector differential operator. It obeys

the rules for vectors and differential operators.

• In particular, remember:

(a) Leibniz’ theorem (differentiation of a product);

(b) the scope rules for ∇ (i.e. what ∇ acts on).

Unless otherwise indicated, ∇ acts on the expression to its

immediate right.

Examples: (again taken from electromagnetism)

∇·(E×××××H) = ∇·(E×××××H) + ∇·(E×××××H)

= H ·∇×××××E − E·∇×××××H

p×××××(∇×××××E) =∗

∇ (p·∗

E) −p·∇ E

[The above examples are as difficult as they get. . . We won’t need

anything as complicated in this course.]


MECHANICS — REVIEW OF LAST YEAR’S MATERIAL

• Development from Newton’s Laws.

mara = Fa

a = 1, N for the ath of N particles. A rigid body is a special case.

• Overall motion∑

a

mara =∑

a

Fa =∑

a

Fa0 +∑

a

∑

b

Fab

where Fa0 is the external force on particle a and Fab is the force on a

due to b. Since Fab = −Fba by Newton’s 3rd Law, the∑

a

∑

b

term

above sums to zero.

• Define M ≡∑

a

ma and MR ≡∑

a

mara.

• R is the position of the Centre of Mass (CoM).

• With these definitions

MR =∑

a

Fa0 ≡ F0

• The Centre of Mass moves as if it were a particle of mass M acted

upon by the total external force F0.

• In terms of momentum

pa = Fa; P = F0

where P is the total momentum.


REVIEW OF MECHANICS — MOMENTS

• Couple, torque: G ≡ r×××××F .

• Angular momentum: J ≡ r×××××p.

• Since pa = Fa, we have∑

a

ra×××××pa =∑

a

ra×××××Fa

• Expand RHS:

RHS =∑

a

ra×××××Fa0 +∑

a

∑

b

ra×××××Fab

︸︷︷︸

∑

b

∑

a<b

(ra − rb)×××××Fab︸︷︷︸

= 0

• The latter term is zero since Fab is assumed to be along the line

between a and b. We have again used Newton’s Third Law.

• The LHS for one particle is

Ja =d

dt(ra×××××pa) = ra×××××pa

︸︷︷︸+ra×××××pa

zero, sincemr = p

• For the system of particles

J ≡∑

a

Ja =∑

a

ra×××××pa = RHS =∑

a

ra×××××Fa0 ≡ G0

• G0 is the resultant couple G from all external forces.


REVIEW OF MECHANICS — CHOICE OF ORIGIN

• Suppose we move the origin by a constant a, giving new coordinates

r′ with r = r′ + a.

• Then r = r′ and the overall motion is unaffected.

• What about the angular momentum J? For one particle

Ja = J ′

a + a×××××pa, or for the system

J = J ′ +∑

a

a×××××pa = J ′ + a×××××P

i.e. J depends on the choice of origin unless P = 0.

• Intrinsic angular momentum: J in the frame in which P = 0

(zero-momentum, or Centre of Mass frame). The Intrinsic angular

momentum is independent of origin.

• The Centre of Mass frame is thus special, and should be used

wherever possible.

• Similarly G = G′ + a × F .


REVIEW OF MECHANICS — KINETIC ENERGY

• Work done: force × distance moved ‖ force = change in energy.

• For a single particle

F ·dr = mr·dr = m r·r︸︷︷︸

dt

d

dt(12 r·r)

or

F ·dr = d( 12mv

2)

• Kinetic energy: T ≡ 12mv

2.

• Work done on particle = change in kinetic energy.

• For a system of particles

dT =∑

a

dTa =∑

a

Fa·dra

=∑

a

Fa0·dra +∑ ∑

a<b

Fab·(dra − drb)

where we have used Fab = −Fba. We can write the ab-term as

−Fab d|ra − rb|, where Fab has magnitude = |Fab| and is positive

if force is attractive, negative if repulsive.


REVIEW OF MECHANICS — POTENTIAL AND TOTAL ENERGY

• Potential energy: U is defined as

dU =∑ ∑

a<b

Fab d|ra − rb|

• Note the zero of U =

∫

dU is undefined. It is often taken with

U = 0 with particles at infinite separation, giving negative U for a

system of particles with attractive forces.

• For a rigid body dU = 0 since |ra − rb| is fixed.

• Total Energy: E = T + U .

• As defined above

dE = dT + dU =∑

a

Fa0·dra

• The RHS term is the work done by external forces; it can be

incorporated into U if desired.


REVIEW OF MECHANICS — GALILEAN TRANSFORMATION

• Go from frame S to S′ with r = r′ + V t; V steady; t = t′.

• Momentump = p′ +mV ; P = P ′ +MV

i.e. P in S and P ′ in S′ change together (or remain steady together if

there if no external force). If P ′ = 0, then S′ is the zero-momentum

or Centre of Mass frame.

• Angular momentum

J =∑

a

(r′

a + V t)×××××(p′

a +maV )

• There are 4 terms. The 4th is V ×××××V = 0. The others give

J = J ′ + V t×××××P ′ +∑

a

r′

a×××××maV

︸︷︷︸∑

a

(mar′

a)×××××V = MR′×××××V

• Thus if S′ is the zero-momentum frame, P ′ = 0 and

J = J ′ +MR′×××××V

︸︷︷︸.

in S intrinsic motion of C of M in S

• Energy

T =∑

a

12mav

2a = 1

2ma(v′

a + V )·(v′

a + V )

= T ′ +∑

a

mav′

a

︸︷︷︸

·V + 12MV 2.

(= 0, if S′

= zero-momentum frame)or

T = KE in zero-momentum frame + 12MV 2


COORDINATE SYSTEMS

• Position vector r has

Cartesian coordinates (x, y, z)

cylindrical polar coordinates (ρ, φ, z)

and spherical polar coordinates (r, θ, φ).

• Relation between coordinate systems:

x = ρ cosφ = r sin θ cosφ

y = ρ sinφ = r sin θ sinφ

z = z = r cos θ

• Other important relationships:

ρ =√

x2 + y2

r =√

x2 + y2 + z2

• We define unit vectors (ex, ey, ez) along x, y, z axes.

• Similarly we define unit vectors (eρ, eφ, eθ) along directions of

increasing (ρ, φ, θ) (unambiguous because these are orthogonal

coordinate systems).

• Position vector r = xex + yey + zez = ρeρ + zez = rer


COORDINATE SYSTEMS — CYLINDRICAL POLARS

• In Cartesians, the equation of motion of a particle is

mr = F ; or mx = Fx etc

where we denote time derivativesdr

dt≡ r,

d2r

dt2≡ r etc.

eρ

eφ• Consider first cylindrical polars;

ignore z-motion for the moment

r = ρeρ

where eρ, eφ and ez are unit vectors

in the directions of increasing ρ, φ, z.

• Note that the direction of the vectors eρ and eφ change as the particle

moves.

r = ρeρ + ρ ˙eρ

• As the particle moves from say P to P ′ in dt, eρ and eφ rotate by dφ.

• Elementary geometry gives deρ = dφ eφ or ˙eρ = φeφ and

similarly ˙eφ = −φeρ giving

r = ρeρ︸︷︷︸

+ ρφeφ︸︷︷︸

radial transverse

• In cylindrical polar coordinates the radial velocity is ρ and the

transverse velocity is ρφ.


ACCELERATION IN POLAR COORDINATES

• Similarly, we can work out the rate of change of velocity:

r = ρeρ + ρ ˙eρ︸︷︷︸

+ρφeφ + ρφeφ + ρφ ˙eφ︸︷︷︸

φeφ − φeρ

= (ρ− ρφ2)︸︷︷︸

eρ + (2ρφ+ ρφ)︸︷︷︸

eφ

radial transverse

• The z-motion is independent: (r)z is just zez since ˙ez = 0.

• The radial acceleration is ρ− ρφ2, the second term being the

centripetal acceleration required to keep a particle in an orbit of

constant radius.

• The transverse acceleration is 2ρφ+ ρφ =1

ρ

d

dt

(

ρ2φ)

and shows

that it is related to the angular momentum per unit mass ρ2φ.

• Spherical polars can be treated by putting r = rer , and expanding r

etc. with ˙er expressed in terms of er , eθ and eφ. We shall not need

this here, as it’s a slightly complicated, but if you have computer

algebra available it’s very useful. . .


CYLINDRICAL POLARS AND THE ARGAND DIAGRAM

eiφieiφ

• The complex plane z ≡ x+ iy = ρeiφhas

the same structure as the two-dimensional

plane r = xex + yey = ρeρ.

• The unit vectors correspond to complex numbers:

eρ ↔ eiφ

eφ ↔ ieiφ

• We can therefore derive the radial and transverse components easily

using the Argand diagram.

• Velocity:d

dt(ρeiφ) = ρeiφ + ρφ ieiφ.

• Acceleration:

d2

dt2(ρeiφ) = ρeiφ + 2ρφ ieiφ + ρφ ieiφ − ρφ2eiφ

= (ρ− ρφ2

︸︷︷︸) eiφ + (ρφ+ 2ρφ

︸︷︷︸) ieiφ

radial transverse


FRAMES IN RELATIVE MOTION

• Suppose we have a frame S0 in which mr0 = F , with F ascribed to

known physical causes. What is the apparent equation of motion in a

moving frame S?

• Case 1: Suppose r = r0 − R(t). Suppose the axes in S0 and S

remain parallel and t = t0 (as always in classical physics).

r = r0 − R

• For the special caseR = 0 (i.e. steady motion between frames),

mr = mr0 = F , i.e. the same equation of motion (Galilean

transformation).

• For general R(t)mr = mr0 −mR

• The apparent force in S includes both the actual force mr0 and a

fictitious force −mR.

• Fictitious forces are:

(a) associated with accelerated frames;

(b) proportional to mass.

• Question: Is gravity a fictitious force?

• Answer: (according to Einstein and general relativity). Yes! Gravity is

equivalent to acceleration.


ROTATING FRAMES

• Case 2: Frame S rotates

with angular velocity ω,

so that the unit vectors rotate

with respect to the inertial frame S0.

• The rate of change is given by˙ez = ω×××××ez etc.

ω×××××ez

ez

ey

ex

• Let the frames coincide at t = 0:

r0 = xex + yey + zez = r

r0 = xex + x ˙ex + y and z terms

= v + ω×××××r

where v ≡ xex + yey + zez is the apparent velocity in S.

• The acceleration in S0 is

r0 = xex + 2x ˙ex + x¨ex + y and z terms

= xex + 2(ω×××××ex)x+ ω×××××(ω×××××ex)x+ y and z terms

= a + 2ω×××××v + ω×××××(ω×××××r)

where a = xex + yey + zez is the apparent acceleration in S. We

rewrite the momentum equation mr0 = F in terms of the apparent

quantities r, v and a:

ma = F − 2m(ω×××××v) −mω×××××(ω×××××r)

• The observer in S adds Coriolis and Centrifugal forces — which are

inertial or fictitious forces.


ROTATING FRAMES

• There is an operator approach to rotating frames that is a good aid to

memory (and is just as rigorous).

• For any vector A the rates of change

in frame S0 and in frame S

are related by

[dA

dt

]

S0

=

[dA

dt

]

S

+ω×××××A

• Apply this operator relation twice to r

(r =r0 at t=0):

[d2r0

dt2

]

S0

=

([d

dt

]

S

+ω×××××

)([dr

dt

]

S

+ω×××××r

)

• Expanding and setting

[dr

dt

]

S

=v and

[dv

dt

]

S

=a we recover

mr0 =F =ma+2m(ω×××××v)+mω×××××(ω×××××r)

• General case: Observer moves on a path R(t) and uses a frame

rotating at angular velocity ω(t) which is also changing. From previous

results and, because the time derivative now operates on ω we get the

general formula:

ma=F −2m(ω×××××v)−mω×××××(ω×××××r)−mR−mω×××××r

• The mω×××××r term is called the Euler force.


CENTRIFUGAL AND CORIOLIS FORCES

• Centrifugal Force: −mω×××××(ω×××××r).

−mω×××××(ω×××××r) = m(ω2r − r·ω ω)

• Centrifugal Force mω2ρ outwards.

• Coriolis Force: −2m(ω×××××v) appears if a body is moving with respect

to a rotating frame.

• Coriolis Force is a sideways force, perpendicular both to the rotation

axis and to the velocity.

• Problems involving Coriolis Force are often better done by considering

angular momentum.

• Advice: Do not meddle with the signs or the ordering of the terms.

The minus sign reminds us that these terms came from the other side

of the equation, and ω×××××(ω×××××r) construction reminds us of the

operator relation

[d

dt

]

S0

=

[d

dt

]

S

+ ω×××××.


FICTITIOUS FORCES — TERRESTRIAL APPLICATIONS

• Centrifugal force gives rise to the Earth’s equatorial bulge:

∼ Ω2R

g≈ 1

300.

l

N

You are hereW• Coriolis force due to motion on Earth’s surface:

- Direction is sideways and to the right

in the Northern Hemisphere.

- Independent of direction of travel [NSEW].

F=2mvW lcos

zx

East

• Coriolis force on a falling body.

Starts from rest at time t = 0

v = gt

mx = 2mgtΩ cosλ

x = 13gΩt

3 cosλ

• Foucault pendulum. Precesses at Ω sinλ. Which way?

• Roundabouts and other fairground rides. Rollercoasters.

Low

High

High

High

N

w

Low

High

• Weather patterns, trade winds, jet streams, tornados. . .

• Bathtubs (??).


ORBITS — CENTRAL FORCE FIELD

F

O

P

r

v

• Particle moving in central force field.

Potential U(r) yields radial force

F = −∇U = −dU

drer .

• Motion remains in the plane defined by position vector r and velocity v.

• No couple from central force ⇒ angular momentum is conserved:

mr2φ = J = constant (Kepler II)

• Total energy is conserved:

E = U(r) +m

2(r2 + r2φ2) =

m

2r2 + U(r) +

J2

2mr2

• The effective potential Ueff(r) has a contribution from the angular

velocity.

Ueff(r) = U(r) +J2

2mr2

• The effective potential has a centrifugal repulsive term ∝ 1

r2


NON-CIRCULAR ORBITS IN POWER-LAW FORCE

• We can gain a lot of insight into orbits by studying the force law

F = −Arn.

• The effective potential is then

Ueff(r) =Arn+1

n+ 1+

J2

2mr2

the only exception being n = −1 (Ueff then contains a log r term).

• The potential is qualitatively different for different values of n.


ANALYSIS OF NEARLY CIRCULAR ORBITS


PRECESSION OF ORBITS


ORBITS — SIMPLE HARMONIC WELL

Energy E = m2

(x2 + y2

)+ A

2

(x2 + y2

)

Ang. Mom. J = m (xy − yx)


KEPLER’S LAWS AND THE INVERSE SQUARE LAW OF FORCE

• 150AD Ptolemy - Earth at centre of solar system, with Sun rotating

around it, and the planetary orbits described by a combination of

circles (epicycles).

• Tycho Brahe (1546-1601) made observations of planetary and stellar

positions to an accuracy of 10 arcsec (the resolution of the eye is

1 arcmin).

• Johannes Kepler (1571 -1630) spent 5 years fitting circles to the

Brahe’s data for Mars’s orbit and found differences of the order of

8 arcmin. Rejected model because of known accuracy of Brahe’s

measurements.

• Eventually Kepler concluded that the orbits were ellipses.

• Kepler’s Laws

• First Law: Planetary orbits are ellipses with the Sun at one focus.

• Second Law: The line joining the planets to the Sun sweeps out

equal areas in equal times. (Implies conservation of angular

momentum)

• Third Law: The square of the period of a planet is proportional to

the cube of its mean distance to the Sun (actually it is proportional

to the cube of the length of the orbit’s major axis).


ORBITS IN INVERSE SQUARE LAW FORCE

• Inverse square law force: F = −A

r2.

• Angular momentum: J = mr2φ

• Energy: 12mr

2 +J2

2mr2− A

r= E

• Define r0 =J2

mA, the radius of the circular orbit with the same J .

• Slightly easier to work with u = 1/r:

r =dr

dφφ = −φr2 du

dφ= − J

m

du

dφ

• Substitute into the energy equation(

du

dφ

)2

+ u2 − 2m

J2(E +Au) = 0

(du

dφ

)2

=2mE

J2+

1

r20︸︷︷︸

−(

u− 1

r0

)2

≡ e2

r20du

√

e2

r20−

(

u− 1

r0

)2= dφ⇒ u =

1

r0(1 + e cos(φ− φ0))

• Equation of conic section

r0 = r(1 + e cosφ)

• For a repulsive potential r0 = r(e cosφ− 1)


INVERSE SQUARE LAW — ELLIPTICAL ORBITS (E < 0)


FORMULA FOR TIME IN ORBIT — (DETAILS NON-EXAMINABLE)

• We haven’t so far got a formula for the coordinates (r, φ) as a function

of time. There is an easy way of doing this, which is coming a bit later,

but we can in fact get a formula for t(φ), rather than φ(t).

• It’s just not very pretty. . .

• From the equation of the ellipse and h ≡ J

m= r2φ we get

dφ

(1 + e cosφ)2 =

hdt

r20

• The integral is found as formula 2.551 of Gradshteyn & Ryzhik:

t=r20

h(1−e2)

(

− esinφ

1+ecosφ+

2√1−e2

tan−1

(tan(φ+ 1

4π)+e√1−e2

))

OCA

P

• You can also get a formula for it

by subtracting the area of the

triangle CPO from the sector CPA.


ELLIPTICAL ORBITS — ANOTHER (NON-EXAMINABLE) WAY

• Return to the energy equation: 12mr

2 +J2

2mr2− A

r= E.

• Change the independent variable, defining a radially-scaled “time”

rds = dt so thatdr

dt=

1

r

dr

ds

• We then get

(dr

ds

)2

− 2E

mr2 − 2A

mr + J2 = 0

• Defining Ω2 =−2E

m(remember E is negative) we get the same form

of equation as before, but for r instead of u.

The distances of furthest and closest approach are a(1 ± e), so the

solution is just r = a(1 − e cosΩs)

• We can now integrate r =dt

dsto find a nice parametric form for the

time t t = as− a

ΩsinΩs. The period is T =

2πa

Ω= 2π

√

ma3

A.

• There is also a simple closed form for φ(s)

tan

(√

1 − e2φ

2

)

=

√

1 + e

1 − etan

Ωs

2

• This is related to the “square root” of the Kepler problem, which

transforms the orbit to a central ellipse (a two-dimensional harmonic

oscillator). Many essential features of this solution were known to

Newton, but the procedure is called the “Kustaanheimo-Stiefel”

transformation.


ELLIPTICAL ORBITS — IMPORTANT THINGS TO REMEMBER

OC

P

semi-major axis

semi-latus rectum

fsemi-minor axis b

a

0r

r• The equation of an ellipse

in polar coordinates:

r0 = r (1 + e cosφ)

• The distances of closest and furthest approach follow from this:

rmin =r0

1 + eand rmax =

r01 − e

• The semi-major axis a satisfies 2a = rmin + rmax

⇒ a =r0

1 − e2b =

r0√1 − e2

• The semi-major axis a determines the energy and the period of the

orbit

E = − A

2a; T =

2π

ω; ω2 =

A

ma3

• The semi-latus rectum r0 determines the angular momentum of the

orbitJ2 = Amr0

• If you need to derive any of these formulae in a hurry consider a silly

case. Use the simple balance of forces argument for circular motion:

A

r2= mω2r

.


EXAMPLE: THE HOHMANN TRANSFER ORBIT

Dv

2

2

Dv1

1

a

a

• The Hohmann transfer orbit is one

half of an elliptic orbit that touches

both the initial orbit and the desired orbit.

• For gravitational case put A = GM .

• In circular orbits T = −E = − 12U ,

for elliptical orbits 〈T 〉 = −〈E〉 = − 12 〈U〉 (Virial Theorem).

• The initial energy is E1 = −GM2a1

, and the velocity has to be

increased until the spacecraft has the energy of the transfer orbit

Et = − GM

a1 + a2.

• The important thing to know is ∆v1, since that determines the amount

of fuel used, but we can work all that out from the energies:

Et = − GM

a1 + a2= −GM

a1+ 1

2v2t1 ⇒ 1

2v2t1 =

GMa2

a1(a2 + a1)

• The rest of relations are easy enough, but not particularly

informative. . .

• The Hohmann transfer is the most fuel efficient orbit, unless there are

other massive bodies in the vicinity, in which case you can use the

gravitational slingshot.


INVERSE SQUARE LAW — UNBOUND ORBITS (E>0)


SQUARE ROOT OF THE KEPLER PROBLEM


SQUARE ROOT OF THE KEPLER PROBLEM II


INVERSE SQUARE LAW ORBITS — AN INTERESTING ALTERNATIVE

• The next two overheads are an alternative derivation of the orbit to the

one given earlier in P29 and P30. The demonstration that the orbit is

an ellipse is interesting, and there is also a proof that the energy is

determined by the semi-major axis.

• Shape of orbit Since the vectors J , v and ˙er have magnitudes

mr2φ,A/mr2 and φ respectively and are mutually perpendicular, we

may write

J×××××v = −A ˙er

the sign is obtained by inspection.

• Since J is constant the equation may be integrated to give

J×××××v +A(er + e) = 0

where e is a vector integration constant.

• Taking the dot-product of this equation with r gives

J×××××v·r︸︷︷︸

+A(r + e·r) = 0

= J ·v×××××r = −J2/m

• Therefore r(1 + e·er) = r(1 + e cosφ) = J2

mA = r0.

which is the polar equation of a conic with focus at r = 0 (Kepler’s 1st

Law).

• The major axis is in the direction of e; e is the eccentricity: 0, < 1, 1 or

> 1 for a circle, ellipse, parabola or hyperbola.


ALTERNATIVE — ENERGY OF THE ORBIT — TIME IN ORBIT

• To get the energy take the scalar product of Ae = −(J×××××v +Aer)

with itself (note that J and v are perpendicular)

A2e2 = J2v2 + 2 J×××××v·er︸︷︷︸

A+A2

= J ·v×××××er = −J2/mrTherefore

A2(e2 − 1) = J2

(

v2 − 2A

mr

)

=2EJ2

m

where E is the total energy. The major axis of the orbit is given by

2a = r0

(1

1 + e+

1

1 − e

)

=2r0

1 − e2= −A

E

i.e. E = −A/2a independent of eccentricity.

• Time in orbit The integral of φ(t) to give φ(t) is insoluble in

elementary functions, but you can get t(φ). It’s not pretty, though. . .

• We get the period, from the area and d(area)/dt = J/2m.

• To get the area, turn the polar equation into Cartesians, to give(x− x0

a

)2

+y2

b2= 1

witha =

r01 − e2

, b =r0√

1 − e2, x0 = ea

The area is πab, so that

Period =πab

J/2m=

πr20(1 − e2)3/2

2m

(r0Am)1/2= 2π

(ma3

A

)1/2

i.e. proportional to a3/2 independent of e; Kepler’s 3rd Law.


GRAVITATIONAL SLINGSHOT

• The escape velocity of a spacecraft from the Solar system at the radius

of the Earth’s orbit is 42 km s−1, which should be compared to its

orbital velocity of 30 km s−1.

• We can use a gravitational ‘slingshot’ around planets to increase

kinetic energy and/or change direction in order to visit other bodies in

the Solar system.

• Voyager 2 made a ‘grand tour’.


GRAVITATIONAL SLINGSHOT — VOYAGER 2


THE TWO-BODY PROBLEM AND REDUCED MASS

• The two masses M1 and M2 orbit the centre of mass.

• Each orbit is an ellipse in a common plane with the centre of mass at

one focus. The ellipses have the same eccentricity and phase.

• The important case is circular motion.

Mass M1 is distanceM2r

M1 +M2from the CoM.

• Balance of forces for M1:

• GM1M2

r2= M1ω

2 M2r

M1 +M2

⇒ ω2 =G(M1 +M2)

r3

• This is the really important result and is usually the best starting point

(e.g. in problem Q10).

• You get the same result by considering the balance of forces for M2.

• We can rearrange the result as µrω2 =GM1M2

r2, in which we have

the true separation r and the actual forceGM1M2

r2, but a modified

reduced mass term µ ≡ M1M2

M1 +M2.

• I would not advocate the use of reduced mass, despite its widespread

use in the textbooks. . .


THE THREE-BODY PROBLEM

• Use a computer. . .

• Some hierarchical systems can be stable indefinitely e.g. Sun, Earth

and the Moon.

• A general 3-body encounter can be very complicated, but a general

feature emerges.

• If 3 bodies are allowed to attract each other from a distance (a), they

will speed up and interact strongly (b). Eventually the interaction is

likely to form a close binary (negative gravitational binding energy)

releasing kinetic energy, which may be enough for the bodies to

escape to infinity (c).

• This mechanism is responsible for “evaporation” of stars from star

clusters. (maybe also invalidates the virial theorem?)

• The planet Pluto has a close companion Charon, and has an eccentric

orbit which takes it inside Neptune’s orbit. A 3-body collision amongst

Neptune’s moons is the most likely cause.


HYPERBOLIC ORBITS

O C

P

impact parameter

semi-latus rectum semi-major axis

f f

Path for repulsive force

b

a

0r

r

8

• Attractive potential: all previous formulae still valid, but e > 1 so

a < 0 and energy E = − A

2a=

(e2 − 1)A

2r0> 0.

• Impact parameter b and velocity at infinity v∞ determine angular

momentum J = mbv∞ and energy E = 12mv

2∞

.

• Most problems (e.g. Rutherford scattering Q11) require the total angle

of deflection χ = 2φ∞ − π (χ positive).

• Asymptotes are at ±φ∞; from the equation of the ellipse we have

cosφ∞ = −1/e ⇒ secφ∞ = −e ⇒ tan2 φ∞ = e2 − 1

.• Note that π/2 < φ∞ < π.


HYPERBOLIC ORBITS II

O C

P

impact parameter

semi-latus rectum semi-major axis

f f

Path for repulsive force

b

a

0r

r

8

Attractive potential

• We need to find the eccentricity from the physical parametersE and J .

• From the above we can write (e2 − 1) =2r0E

A=

2J2E

mA2from the

definition of r0. In terms of b and v∞ this means

tan2 φ∞ = e2 − 1 =m2v4

∞b2

A2⇒ tanφ∞ =

mv2∞b

A.Repulsive potential:

• ChangeA

r2→=

B

r2, define J2 = Bmr0 and use other branch.

• Total angle of deflection is now χ = π − 2φ∞ (χ negative).

• The asymptotes are still related to the physical parameters by

tanφ∞ =mv2

∞b

B.


RUTHERFORD SCATTERING


RUTHERFORD SCATTERING II


THE TWO-BODY PROBLEM


UNBOUND INVERSE SQUARE ORBITS

• Hyperbolic orbits e > 1 can occur with an attractive force and E

positive, or a repulsive force and any E, e.g. between two positively

charged nuclei.

• The algebra is similar to that of the ellipse with some changes of sign;

e > 1. b is the impact parameter; if the orbiting body is initially at

infinity with positive speed, b is how close to the central body it would

pass if it were not deviated by the force.

• Important results in practice are

1. J = mr2φ = mbv∞ = mrcvc where suffix c refers to the

moment of closest approach.

2. Energy equation gives rc and vc in terms of b and v∞.

3. The asymptotes occur when r becomes infinite, i.e.

1 + e cosφ∞ = 0. cosφ∞ = −1/e (φ∞ is between π/2 and

π, say π − ψ∞).

4. Geometry of c and e gives |er + e| = tanψ∞ and the

magnitudes of the two terms in r(1 + e cosφ) = r0 give

tanψ∞ =Jv∞A

=mv2

∞b

AThe total deviation during the traverse is π − 2ψ∞.

5. For a repulsive orbit, with force = B/r2, B positive, the angle of

deviation is similarly

π − 2 tan−1

(mv2

∞b

B

)

6. The angle of the asymptotes can also be simply derived from the

x-momentum.


ORBITS IN CENTRAL POWER-LAW FORCE

U

• More here

and here. . .

.

• Let F = −Arn, with A positive, so force is attractive; n = index, with

common cases n = +1 (2D SHM) and n = −2 (gravity,

electrostatics).

Ueff =Arn+1

n+ 1+

J2

2mr2

the only exception being n = −1 (Ueff then contains a log r term).

• Nearly circular orbits are oscillations/perturbations about r0. Taylor

expansion of Ueff gives

Ueff = U0+(r−r0)(

dUeff

dr

)

r0

+ 12 (r−r0)2

(d2Ueff

dr2

)

r0

+ · · ·

• dUeff

dris zero at r0 giving

dUeff

dr= +Arn − J2

mr3= 0 at r0

• The second derivative of Ueff is

d2Ueff

dr2= nArn−1 +

3J2

mr4

=(n+ 3)J2

mr40at r0.


• Therefore we get the SHM equation

mr +(n+ 3)J2

mr40(r − r0)

︸︷︷︸

= 0;

1st-order Taylor expansion of dUeff/dr

i.e. SHM about r0 with angular frequency

ωp =√n+ 3

J

mr20


ORBITS IN POWER-LAW FORCE II

• How does ωp of the perturbation compare with ωc of the circular orbit

at r0? ωc = φ = J/mr20 . Therefore ωp =√n+ 3ωc.

The common cases are

1. n = 1. Force proportional to r, i.e. SHM. ωp = 2ωc, giving a

central ellipse (Lissajous figure).

2. n = −2. Inverse square force. ωp = ωc, giving an ellipse with a

focus at r = 0 (planetary orbit).

General n gives non-commensurate ωp and ωc and non-repeating

orbits.


THE TWO-BODY PROBLEM

M1

M2

2r

1r

CoM

• Two particles of masses M1 and M2 orbiting

each other — positions r1 and r2.

• The energy, angular momentum and

equations of motion can be expressed

in terms of the reduced mass µ

µ ≡ M1M2

M1 +M2and r1 − r2.

• The centre of mass is at R0 =M1r1 +M2r2

M1 +M2.

• Defineρ1 ≡ r1 −R0 =

M2

M1 +M2(r1 − r2)

ρ2 ≡ r2 −R0 =M1

M1 +M2(r2 − r1)

• Kinetic energy in the centre of mass frame:

T = 12M1ρ

21 + 1

2M2ρ22

= 12

(M1M2

2

(M1+M2)2+

M2

1M2

(M1+M2)2

)

(r1 − r2)2 = µ

2 (r1 − r2)2

• Angular momentum:

J = M1ρ1×××××ρ1 +M2ρ2×××××ρ2 = µ(r1 − r2)×××××(r1 − r2).

• Equations of motion: M1r1 = F 12 ; M2r2 = F 21 = −F 12

r1 − r2 =

(1

M1+

1

M2

)

F 12 =1

µF 12 ; R0 = 0

• Reduced to the one-body problem in the centre of mass frame.


NEWTONIAN EXAMPLE — THE LADDER

• The Cartesian coordinates (x,y) are related to the angle θ by

x= 12 lsinθ y= 1

2 lcosθ

x= 12 lcosθ θ y=− 1

2 lsinθ θ

x= 12 l(cosθ θ−sinθ θ2) y=− 1

2 l(sinθ θ+cosθ θ2)

• The equations of motion are

mx=N ; my=R−mg ; Iθ=Rx−Ny

• Solve for (R,N):

R=mg+ 12ml(sinθ θ+cosθ θ2)

N =− 12ml(cosθ θ−sinθ θ2)

• Evaluate the moment Rx−Ny:

=1

2mlg sinθ− 1

4ml2

(sin

2θ θ+sinθcosθ θ2+cos

2θ θ−sinθcosθ θ2)

=−

1

4ml2θ− 1

2mglsinθ

=1

12ml2θ

• Find the equation of motion: 13ml

2θ= 12mglsinθ


ORBITS

• Ellipse of eccentricity e, Centre at one focus. r0 is semi-latus rectum.

• Cartesian equation r= r0−ex

y2 +x2(1−e2)+2er0x= r20

. Set x′ =x+r0e

1−e2 ⇒y2 +(x′)2(1−e2)=r20

1−e2

(x′)2

a2+y2

b2=1 ; a=

r01−e2 ;b=

r0√1−e2

• Area πab=r20

(1−e2)3/2

• Period T isArea

Rate of sweeping out area.

• Rate of sweeping out area: 12r

2φ=J

2m, hence period

T =2πr20m

J(1−e2)3/2=

√

4π2ma3

A


EXAMPLE

• The Cartesian coordinates (x,y) are related to the angle θ by

x= 12 lsinθ y= 1

2 lcosθ

x= 12 lcosθ θ y=− 1

2 lsinθ θ

x= 12 l(cosθ θ−sinθ θ2) y=− 1

2 l(sinθ θ+cosθ θ2)

• The equations of motion are

mx=−mg+R ; my=N ; Iθ=Ny−Rx

• Solve for (R,N):

R=mg− 12 l(sinθ θ+cosθ θ2)

N = 12 l(cosθ θ−sinθ θ2)

• Evaluate the moment Ny−Rx:

Ny−Rx =1

4l2

(cos

2θ θ−sinθcosθ θ2+sin

2θ θ+sinθcosθ θ2)

=1

4l2θ+

1

2mglsinθ

=1

12ml2θ

• Find the equation of motion 13ml

2θ= 12mglsinθ


ORBITS — SIMPLE HARMONIC OSCILLATOR

• mr =F =−Ar⇒mx=−Ax ; my=−Ay

• Cartesian equation r= r0−ex

y2 +x2(1−e2)+2er0x= r20

. Set x′ =x+r0e

1−e2 ⇒y2 +(x′)2(1−e2)=r20

1−e2

(x′)2

a2+y2

b2=1 ; a=

r01−e2 ;b=

r0√1−e2

• Area πab=r20

(1−e2)3/2

• Period T isArea

Rate of sweeping out area.

• Rate of sweeping out area: 12r

2φ=J

2m, hence period

T =2πr20m

J(1−e2)3/2=

√

4π2ma3

A


THE FOUCAULT PENDULUM

• A very instructive example illustrating most of the general points in this

course.

• Method 1 Let the pendulum have angular frequency Ω2 =g/l. In a

frame rotating with the Earth (angular velocity ΩE) at latitude λ the

Cartesian equations of small oscillation are

x+Ω2x=2ΩE sinλy ; y+Ω2y=−2ΩE sinλx

• These are coupled, second-order equations of the type that we will

meet again later. We seek the solution in matrix form by setting(x(t)

y(t)

)

=

(X

Y

)

eiωt, where X and Y are constant amplitudes.

• We get

(−ω2 +Ω2 −2iωΩE

2iωΩE −ω2 +Ω2

)(X

Y

)

=

(0

0

)

.

• This is a set of linear, homogeneous equations, which only have

non-zero solutions if the determinant of the matrix vanishes:

ω4−2ω2(Ω2 +2ΩE sin2λ)+Ω4 =0

• There are two solutions for ω2 (Normal Modes):

ω2 =Ω2 +2Ω2E sin2λ±2ΩE sinλ

√

Ω2 +Ω2E sin2λ

• For Ω2EΩ these frequencies are approximately ω=Ω±Ω2

Esinλ.

• What are the normal modes? I.e. the relative sizes of X and Y ?

classical dynamics and fluids p 1 - university of cambridge

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