kepler geometry of iho elliptical orbits · kepler geometry of iho (isotropic harmonic oscillator)...

Kepler Geometry of IHO (Isotropic Harmonic Oscillator) Elliptical Orbits(Ch. 9 and Ch. 11 of Unit 1)

Constructing 2D IHO orbits by phasor plotsReview of phasor “clock” geometry (From Lecture 7)Integrating IHO equations by phasor geometry

Constructing 2D IHO orbits using Kepler anomaly plotsMean-anomaly and eccentric-anomaly geometryCalculus and vector geometry of IHO orbitsA confusing introduction to Coriolis-centrifugal force geometry

Some Kepler’s “laws” for central (isotropic) force F(r)Angular momentum invariance of IHO: F(r)=-k·r with U(r)=k·r2/2 (Derived rigorously) Angular momentum invariance of Coulomb: F(r)=-GMm/r2 with U(r)=-GMm·/r (Derived later)Total energy E=KE+PE invariance of IHO: F(r)=-k·r (Derived rigorously)Total energy E=KE+PE invariance of Coulomb: F(r)=-GMm/r2 (Derived later)

Brief introduction to matrix quadratic form geometry

Lecture 8 Revised 12.21.12 from 9.13.2012

1Friday, December 21, 2012

0 12

3

4

5

6

78-7-6

-5

-4

-3

-2-1

01

2

3 4 5

6

78

-7-6

-5

-4-3

-2-1

(2,4)(3,5)(4,6)

(5,7)

(6,8)

(7,-7)

x-position

x-velocity vx/ω

position

x

velocity vx/ω

(a) 1-D Oscillator Phasor Plot

(b) 2-D Oscillator Phasor Plot

Phasor goes

clockwise

by angle ω t

ω t

(8,-6)(9,-5)

y-velocity

vy/ω

(x-Phase 45°

behind

the y-Phase)y-position

φ counter-clockwise

if y is behind x

clockwise

orbit

if x is behind y

(1,3) Left-

handed

Right-

handed

(0,2)

Fx

v0Fy

F=-r(a) (b)

Isotropic Harmonic Oscillator phase dynamics in uniform-body

Unit 1Fig. 9.10


0 12

3

4

5

6

78-7-6

-5

-4

-3

-2-1

0123 4 5

67

8-7

-6-5-4-3-2-1

(0,4) (1,5)(2,6)(3,7)

(4,8)(5,9)

x-position

x-velocity vx/ω(a) Phasor Plots

for

2-D Oscillator

or

Two 1D Oscillators

(x-Phase 90° behind

the y-Phase)

a

b b

a

a

0 12

3

4

5

6

78-7-6

-5

-4

-3

-2-1

0123 4 5

67

8-7

-6-5-4-3-2-1

(0,0)(1,1)

(5,5)

(2,2)(3,3)

(4,4)

x-position

a

b b

a

a

y-position

y-velocity

vy/ω

x-velocity vx/ω

(b)

x-Phase 0° behind

the y-Phase

(In-phase case)

y-velocity

vy/ω

0 12345

678-7

-6-5-4-3-2-1

(0,4)(1,5)(2,6)(3,7)

b

y-position

b

Fig. 9.11

Unit 1Fig. 9.12

These are more generic exampleswith radius of x-phasor differing from that of the y-phasor.


x

y

121

2

3

4

56

7

8

9

10

121

2

3

4

56

7

8

9

10

11

00°°--1155°°--3300°°--4455°°++1155°°++3300°°

++4455°°

--6600°°

++6600°°

--7755°°

++7755°°

--9900°°

++9900°°

--110055°°

++110055°°

--112200°°

++112200°°

--113355°°

++113355°°

--115500°°

++115500°°

--116655°°

++116655°°

118800°°

118800°°++227700°° ++224400°°++330000°°336600°° ++221100°°++333300°°

vx /ω

vy /ω

Δαy-x

xphasor

yphasor

x-yspace

x

y

xvx

y

v y

++334455°°--334455°°

...after time advances by 180°:(αx=120°-180°=-60°,αy= +135°-180°=-45°)

r(180°)

Ellipsometry Contact Plotsvs.

Relative phase Δα=αx-αyΔα=−15°

(Right-polarized anti-clockwise case)

...after time advances by 180°:(αx=120°-180°=-60°,αy= +135°-180°=-45°)

Initial phase (αx=120°,αy=+135°)

vxv y

vx-vyspace

v(180°)-45°

-60°


121

2

3

4

56

7

8

9

10

11

121

2

3

4

56

7

8

9

10

11

x

y

x

y

vx/ω

vy/ω

vx/ω

vy/ω

Initial position: r(0)=(7.0 3.0)

Inital velocity: v(0)=(-8.0 5.0)


121

2

3

4

56

7

8

9

10

11

121

2

3

4

56

7

8

9

10

11

x

y

x

y

vx/ω

vy/ω

vx/ω

vy/ω




b bsin ω t

ω t

ω ta cos ω t

ay=bsin ω t

x=a cos ω t

ω t

r

φ

b

-b

-b

a

a

-a

ω t

a

y=bsin ω t

x=a cos ω t

r

b

-b

-b

a

a

-a

-a

ω t

Step 1. Draw concentriccircles of radius a and band a radius OA at angleω t

Step 3. Draw horizontial line BRfrom b-circle at ω t to line AX.Intersection is orbit point R.

Step 2. Draw vertical line AXfrom a-circle at ωt to x-axis

O

A

O

A

O

A

X X

BR

abω t

b

-b

-b

a

-a

-a

ab

O X O X

Step 4-NRepeatas oftenas needed

ba

Linear HarmonicForce-FieldOrbits

Unit 1Fig. 11.1

(top 2/3’s)

Kepler’s Mean Anomaly Line(slope angle θ =ωt)

Kepler’s Eccentric Anomaly Line

(slope is polar angle φ=atan[y/x])


b bsin ω t

ω t

ω ta cos ω t

ay=bsin ω t

x=a cos ω t

ω t

r

φ

b

-b

-b

a

a

-a

ω t

a

y=bsin ω t

x=a cos ω t

r

b

-b

-b

a

a

-a

-a

ω t

Step 1. Draw concentriccircles of radius a and band a radius OA at angleω t

Step 3. Draw horizontial line BRfrom b-circle at ω t to line AX.Intersection is orbit point R.

Step 2. Draw vertical line AXfrom a-circle at ωt to x-axis

O

A

O

A

O

A

X X

BR

abω t

b

-b

-b

a

-a

-a

ab

O X O X

Step 4-NRepeatas oftenas needed

ba

Linear HarmonicForce-FieldOrbits

Unit 1Fig. 11.1

Kepler’s Mean Anomaly

Line

Kepler’s Eccentric Anomaly Line

(slope is polar angle φ=atan[y/x])


r(t) φv(t)/ω

90°

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

r(t)φ=ω tv(t)/ω

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

90°

Time frame angle

φ=ω t(Mean Anomaly)

(a) Orbits (b) Tangents

radius vector :r =rxry

⎛

⎝⎜⎜

⎞

⎠⎟⎟= x

y⎛

⎝⎜

⎞

⎠⎟ =

acosω tbsinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

acosφbsinφ

⎛

⎝⎜⎜

⎞

⎠⎟⎟

velocity vector : v =vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

−aω sinω tbω cosω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= dr

dt= r =

acos φ + 2π( )

bsin φ + 2π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

( for ω = 1)

Unit 1Fig. 11.5

Calculus of IHO orbitsTo make velocity vector v just rotate by π/2 or 90°the mean-anomaly φ of position vector r

mean-anomaly φ of position vector r rotated by π/2 or 90° is m.a. of vector v


r(t) φv(t)/ω

90°

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

r(t)φ=ω tv(t)/ω

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

90°

Time frame angle




⎛

⎝⎜⎜

⎞

⎠⎟⎟= x

y⎛

⎝⎜

⎞

⎠⎟ =

acosω tbsinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

acosφbsinφ

⎛

⎝⎜⎜

⎞

⎠⎟⎟


⎛

⎝⎜⎜

⎞

⎠⎟⎟=


⎛

⎝⎜⎜

⎞

⎠⎟⎟= dr

dt= r =

acos φ + 2π( )

bsin φ + 2π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

( for ω = 1)

accelerationor force vector : Fm= a =

axay

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

−aω 2 cosω t

−bω 2 sinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= dv

dt= v = r = d2r

dt2=

acos φ + 22π( )

bsin φ + 22π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

Unit 1Fig. 11.5

or changeof velocity



m.a. φ+π/2 of vector v rotated byanother π/2 is m.a. of vector a


r(t) φv(t)/ω

90°

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

r(t)φ=ω tv(t)/ω

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

90°

Time frame angle




⎛

⎝⎜⎜

⎞

⎠⎟⎟= x

y⎛

⎝⎜

⎞

⎠⎟ =

acosω tbsinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

acosφbsinφ

⎛

⎝⎜⎜

⎞

⎠⎟⎟


⎛

⎝⎜⎜

⎞

⎠⎟⎟=


⎛

⎝⎜⎜

⎞

⎠⎟⎟= dr

dt= r =

acos φ + 2π( )

bsin φ + 2π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

( for ω = 1)


axay

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

−aω 2 cosω t

−bω 2 sinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= dv

dt= v = r = d2r

dt2=

acos φ + 22π( )

bsin φ + 22π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

jerk or changeof acceleration : j=jxjy

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

+aω 3sinω t

−bω 3cosω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= da

dt= a = v = r = d3r

dt3=

acos φ + 23π( )

bsin φ + 23π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

Unit 1Fig. 11.5

or changeof velocity or changeof velocity




...and so forth...


r(t) φv(t)/ω

90°

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

r(t)φ=ω tv(t)/ω

a(t)/ω2

j(t)/ω3

acceleration

jerk

velocity

position

90°

90°

Time frame angle




⎛

⎝⎜⎜

⎞

⎠⎟⎟= x

y⎛

⎝⎜

⎞

⎠⎟ =

acosω tbsinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

acosφbsinφ

⎛

⎝⎜⎜

⎞

⎠⎟⎟


⎛

⎝⎜⎜

⎞

⎠⎟⎟=


⎛

⎝⎜⎜

⎞

⎠⎟⎟= dr

dt= r =

acos φ + 2π( )

bsin φ + 2π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

( for ω = 1)


axay

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

−aω 2 cosω t

−bω 2 sinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= dv

dt= v = r = d2r

dt2=

acos φ + 22π( )

bsin φ + 22π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

jerk or changeof acceleration : j=jxjy

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

+aω 3sinω t

−bω 3cosω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= da

dt= a = v = r = d3r

dt3=

acos φ + 23π( )

bsin φ + 23π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

inaugurationor changeof jerk :i =ixiy

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

+aω 4 cosω t

+bω 4 sinω t

⎛

⎝⎜⎜

⎞

⎠⎟⎟= dj

dt= j= a = v = r = d4r

dt4=

acos φ + 24π( )

bsin φ + 24π( )

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

Unit 1Fig. 11.5

or changeof velocity or changeof velocity or changeof velocity




...and so forth...

...and so on...


F = -kr

orbital velocity=V

(b) “Carnival kid” orbiting inspace attached to a spring

centrifugalforce=+kr=+mω2r

ω t

centripetalforce=

(due to spring)

Carnival kidsays:

“This is awful!I can hardlyhold ontothis darnspring.”

F = -kr

orbital velocity=V

(a) “Earthronaut” orbitingtunnel inside Earth

centrifugalforce=+kr=+mω2r

ω t

centripetalforce=

(due to gravity)

Earthronautsays:

“This is great!I’m weightless.”

apogee(x=a, y=0)aphelion=a

perigee(x=0,y=b)

θperhelion=bmass gaining speed

as it falls

VelocityV

θVelocityV centripetal force F=-kr

Negative power( F•V=|F||V|cos θ <0)

Positive power( F•V=|F||V|cos θ >0)

mass losing speedas it rises

Unit 1Fig. 11.2

Unit 1Fig. 11.3

(Radius r decreasing)(Radius r increasing)


Velocity

V

centripetal force F=-kr

centrifugal force

Total inertial force F=+kr

Coriolis force

centrifugal force

Velocity

along

radial

path

Coriolis force

(depends on

radial path

speed)

Rotational

velocity

V=ωr

circle

of

curvature

Velocity

V

centrifugal force is


circle

of

curvature

(a) Centrifugal and Coriolis

Forces on Merry-Go-Round

(b) Centrifugal and Coriolis

Forces on Oscillator Orbit

(Falling phase)

(c) Centrifugal and Coriolis

Forces on Oscillator Orbit

(Rising phase)

centrifugal force

Velocity

along

radial

path

Coriolis force


Velocity

V


centrifugal force


Coriolis force

circle

of

curvature

(d) Centrifugal Force

on Oscillator Orbit

(apogee and perigee)

Velocity

V


centrifugal force is


Unit 1Fig. 11.4

a-d

Quite confusing? Discussion of Coriolisforces will be done more elegantly and made more physically intuitive in Ch. 12 of Unit1 and in Unit 6.

Physicist Force (where m wants to go)

Mathematician Force (would hold m back)


1. Area of triangle rv = r × v/2 is constant

r × v = rxvy − ryvx = acosω t ⋅ bω cosω t( )− bsinω t ⋅ −aω sinω t( ) = ab ⋅ω

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba

Unit 1Fig. 11.8

Some Kepler’s “laws” for central (isotropic) force F(r)...and certainly apply to the IHO: F(r)=-k·r with U(r)=k·r2/2

for IHO

vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟=


⎛

⎝⎜⎜

⎞

⎠⎟⎟

rxry

!

"##

$

%&&= x

y

!

"#

$

%& =

acos! tbsin! t

!

"##

$

%&&

(Recall from Lecture 7: k = Gm 4π

3ρ⊕ )



r × v = rxvy − ryvx = acosω t ⋅ bω cosω t( )− asinω t ⋅ −bω sinω t( ) = ab ⋅ω

2. Angular momentum L = mr × v is conserved

L = m |r × v |= m rxvy − ryvx( ) = m ⋅ab ⋅ω

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba

Unit 1Fig. 11.8

Some Kepler’s “laws” that apply to any central (isotropic) force F(r)...and certainly apply to the IHO: F(r)=-k·r with U(r)=k·r2/2

for IHO

for IHO

rv

|r×v| =r·v·sinrv


3ρ⊕ )



r × v = rxvy − ryvx = acosω t ⋅ bω cosω t( )− asinω t ⋅ −bω sinω t( ) = ab ⋅ω2. Angular momentum L = mr × v is conserved

L = m |r × v |= m rxvy − ryvx( ) = m ⋅ab ⋅ω

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba

Unit 1Fig. 11.8


for IHO

for IHO

3. Equal area is swept by radius vector in each equal time interval T

AT =r × dr

20

T

∫ =r × dr

dt2

dt0

T

∫ = r × v2

dt0

T

∫ = L2m

dt0

T

∫ = L2m

T for IHO

by 2.

rdr

|r×dr| =r·dr·sinrdr


3ρ⊕ )




L = mr × v = m rxvy − ryvx( ) = m ⋅ab ⋅ω = m ⋅ab ⋅ 2πτ

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba

Unit 1Fig. 11.8


for IHO

for IHO


AT =r × dr

20

T

∫ =r × dr

dt2

dt0

T

∫ = r × v2

dt0

T

∫ = L2m

dt0

T

∫ = L2m

T for IHO

In one period: τ= 1υ

= 2πω

= 2mAτ

L the area is: Aτ =

Lτ2m

( = ab ⋅π for ellipse orbit)


3ρ⊕ )




L = mr × v = m rxvy − ryvx( ) = m ⋅ab ⋅ω = m ⋅ab ⋅ 2πτ

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba

Unit 1Fig. 11.8


for IHO

for IHO


AT =r × dr

20

T

∫ =r × dr

dt2

dt0

T

∫ = r × v2

dt0

T

∫ = L2m

dt0

T

∫ = L2m

T for IHO

In one period: τ= 1υ

= 2πω

= 2mAτ

L the area is: Aτ =

Lτ2m

( = ab ⋅π for ellipse orbit)


3ρ⊕ )

( Recall from Lecture 7: ω = k /m = Gρ⊕4π / 3 )



r × v = rxvy − ryvx =ab ⋅ Gρ⊕4π / 3 for IHO

a−1/2b GM⊕ for Coul.

⎧⎨⎪

⎩⎪

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba

Some Kepler’s “laws” that apply to any central (isotropic) force F(r)Apply to IHO: F(r)=-k·r with U(r)=k·r2/2 and Coulomb: F(r)=-GMm/r2 with U(r)=-GMm·/r

for IHO

t = 0 vv

r rrba

v

vrCoulomb:

IHO:

for Coul.





⎧⎨⎪

⎩⎪2. Angular momentum L = mr × v is conserved

L = mr × v = m rxvy − ryvx( ) = m·ab ⋅ Gρ⊕4π / 3 for IHO

m·a−1/2b GM⊕ for Coul.

⎧⎨⎪

⎩⎪

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba


for IHO

for IHO

t = 0 vv

r rrba

v

vrCoulomb:

IHO:

for Coul.

for Coul.





⎧⎨⎪

⎩⎪2. Angular momentum L = mr × v is conserved

L = mr × v = m rxvy − ryvx( ) = m·ab ⋅ Gρ⊕4π / 3 for IHO

m·a−1/2b GM⊕ for Coul.

⎧⎨⎪

⎩⎪

t = 0 t = π/3ω t = π/2ωv=a ωv=b ω

r rr

ba


for IHO

for IHO


τ= 1υ= 2πω

= 2mAτ

L= 2m·ab ⋅π

L=

2m·ab ⋅πm·ab ⋅ Gρ⊕4π / 3

= 2πGρ⊕4π / 3

for IHO

2m·ab ⋅πm·a−1/2b GM⊕

= 2πa−3/2 GM⊕

for Coul.

⎧

⎨

⎪⎪

⎩

⎪⎪

t = 0 vv

r rrba

v

vrCoulomb:

IHO:

for Coul.

for Coul.

In one period:

that is ωIHO

that is ωCoul

Applies toany central

F(r)

Applies toIHO and Coulomb


Kepler laws involve -momentum conservation in isotropic force F(r)Now consider orbital energy conservation of the IHO: F(r)=-k·r with U(r)=k·r2/2

Total energy=KE+PE is constant

KE + PE = 12v iM i v + 1

2r iK i r

= 12

vx vy( )• m 00 m

⎛⎝⎜

⎞⎠⎟•

vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟+ rx ry( )• k 0

0 k⎛⎝⎜

⎞⎠⎟•

rxry

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 12mv2

x + 12mv2

y + 12kr2

x + 12kr2

y

= 12m(−aω sinω t)2 + 1

2m(bω cosω t)2 + 1

2k(acosω t)2 + 1

2k(bsinω t)2

vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟=


⎛

⎝⎜⎜

⎞

⎠⎟⎟

rxry

!

"##

$

%&&= x

y

!

"#

$

%& =

acos! tbsin! t

!

"##

$

%&&



Total IHO energy=KE+PE is constant

KE + PE = 12v iM i v + 1

2r iK i r

= 12

vx vy( )• m 00 m

⎛⎝⎜

⎞⎠⎟•

vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟+ rx ry( )• k 0

0 k⎛⎝⎜

⎞⎠⎟•

rxry

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 12mv2

x + 12mv2

y + 12kr2

x + 12kr2

y

= 12m(−aω sinω t)2 + 1


2k(acosω t)2 + 1

2k(bsinω t)2

= 12ma2ω 2 (sin2ω t) + 1

2mb2ω 2 (cos2ω t)2 + 1

2ka2 (cos2ω t) + 1

2kb2 (sin2ω t)

= 12mω 2 (a2 + b2 ) Given : k = mω 2


= 12ma2ω 2 (sin2ω t) + 1

2mb2ω 2 (cos2ω t)2 + 1

2ka2 (cos2ω t) + 1

2kb2 (sin2ω t)

= 12mω 2 (a2 + b2 ) Given : k = mω 2

E = KE + PE = 12mω 2 (a2 + b2 ) = 1

2k(a2 + b2 ) since: ω = k

m = Gρ⊕4π / 3 or: mω 2 = k



KE + PE = 12v iM i v + 1

2r iK i r

= 12

vx vy( )• m 00 m

⎛⎝⎜

⎞⎠⎟•

vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟+ rx ry( )• k 0

0 k⎛⎝⎜

⎞⎠⎟•

rxry

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 12mv2

x + 12mv2

y + 12kr2

x + 12kr2

y

= 12m(−aω sinω t)2 + 1


2k(acosω t)2 + 1

2k(bsinω t)2



We'll see that the Coul. orbits are simpler: (like the period...not a function of b)

E = KE + PE = 12mv2

x +12mv2

y −kr= 1

2mv2

x +12mv2

y −GM⊕m

r= −GM⊕m

a


KE + PE = 12v iM i v + 1

2r iK i r

= 12

vx vy( )• m 00 m

⎛⎝⎜

⎞⎠⎟•

vxvy

⎛

⎝⎜⎜

⎞

⎠⎟⎟+ rx ry( )• k 0

0 k⎛⎝⎜

⎞⎠⎟•

rxry

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 12mv2

x + 12mv2

y + 12kr2

x + 12kr2

y

= 12m(−aω sinω t)2 + 1


2k(acosω t)2 + 1

2k(bsinω t)2

= 12ma2ω 2 (sin2ω t) + 1

2mb2ω 2 (cos2ω t)2 + 1

2ka2 (cos2ω t) + 1

2kb2 (sin2ω t)

= 12mω 2 (a2 + b2 ) Given : k = mω 2

E = KE + PE = 12mω 2 (a2 + b2 ) = 1

2k(a2 + b2 ) since: ω = k

m = Gρ⊕4π / 3 or: mω 2 = k


Quadratic forms and tangent contact geometry of their ellipses

A matrix Q that generates an ellipse by r•Q•r=1 is called positive-definite

r •Q• r = 1

x y( )•1

a20

0 1

b2

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

• xy

⎛

⎝⎜

⎞

⎠⎟ = 1= x y( )•

xa2

yb2

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

= x2

a2+ y2

b2

A inverse matrix Q-1 generates an ellipse by p•Q -1•p=1 called inverse or dual ellipse:

p•Q−1•p = 1

px py( )• a2 0

0 b2

⎛

⎝⎜⎜

⎞

⎠⎟⎟•

pxpy

⎛

⎝⎜⎜

⎞

⎠⎟⎟= 1= px py( )• a2 px

b2 py

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= a2 px

2 + b2 y2


r(φ)

φ=ω t

ab

b-circle

a-circle

r(φ)ab

r(φ)=r(φ+π/2)•

r(φ)•r(φ)

Original ellipse

r•Q•r = r•p = 1

p(φ)

p(φ)

p(φ)= p(φ+π/2)•

p(φ)•

vector p(φ) isperpendicular

to r(φ)•vector p(φ) isperpendicular

to r(φ)

•Inverse ellipse

p•Q-1•p =p•r = 1

p(φ)

(a) Quadratic form ellipse and

Inverse quadratic form ellipse

(b) Ellipse tangents

r(φ)•

φ=ω t

p =Q i r = 1/ a2 00 1/ b2

⎛

⎝⎜

⎞

⎠⎟ •

xy

⎛

⎝⎜

⎞

⎠⎟ =

x / a2

y / b2

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

(1 / a)cosφ(1 / b)sinφ

⎛

⎝⎜

⎞

⎠⎟ where:

x = rx = acosφ = acosω ty = ry = bsinφ = bsinω t

so: p i r = 1

r • p = 0 = rx ry( ) • pxpy

⎛

⎝⎜

⎞

⎠⎟ = −a sinφ b cosφ( ) • (1 / a) cosφ

(1 / b) sinφ⎛⎝⎜

⎞⎠⎟

where:rx = −a sinφry = b cosφ

and:px = (1 / a) cosφ py = (1 / b) sinφ

Note some quadratic form mutual duality relations:unit

mutual projection

p is perpendicular to velocity v=r , a mutual orthogonality•

Unit 1Fig. 11.6


kepler geometry of iho elliptical orbits · kepler geometry of iho (isotropic harmonic oscillator)...

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