2012 INBRE workshop...4-1 slope=-4 6 0 m p h 1 0 m p h Velocity-velocityPlot 60mph 10mph...

c-v c +vv c

u=4c/5

√(c-v)(c +v)=c√(1-v2/c2)=c sech ρ=c cos σ

v=c tanh ρ= c sin σσ

c sin σ

Multiply segments by cosh ρ = sec σ=1/√(1-v2/c2)to recover dimensions in (ck, ω) plot

Space-Time Geometry

2012 INBRE workshop

Weapons of Math Instruction:Geometry in Elementary and Advanced Physics

2012 INBRE workshop


Bill Harter Tyle Reimer Alvason LiBill Harter Tyle Reimer Alvason Li1

c-v c +vv c

u=4c/5

√(c-v)(c +v)=c√(1-v2/c2)=c sech ρ=c cos σ

v=c tanh ρ= c sin σσ

c sin σ

Multiply segments by cosh ρ = sec σ=1/√(1-v2/c2)to recover dimensions in (ck, ω) plot

Space-Time Geometry

2012 INBRE workshop


2012 INBRE workshop


Bill Harter Tyle Reimer Alvason LiBill Harter Tyle Reimer Alvason LiLogistics and love: Carole Harter

2

SOME PHYSICS YOU CAN DO BETTER WITH GEOMETRY

•Superball collision problems (Discover momentum-energy rules and Lagrange, Hamilton, and Poincare classical mechanics)

•Rutherford-Coulomb scattering orbits and caustics •Runga-Lenz-Lagrange scattering orbits and caustics

•Space-time wave fractal (“quantum carpet”) gives a lesson in fractions that is quite appealing

•Einstein-Lorentz-Minkowskii relativity (Discover relativistic quantum mechanics)

•Accurate pocket sundial that tells time and predicts sunrise, sunset, civil and nautical twilight, and sunburn hazard.


3


•Superball collision problems (Discover momentum-energy rules and Lagrange, Hamilton, and Poincare classical mechanics)

•Rutherford-Coulomb scattering orbits and caustics •Runga-Lenz-Lagrange scattering orbits and caustics

•Space-time wave fractal (“quantum carpet”) gives a lesson in fractions that is quite appealing


•Accurate pocket sundial that tells time and predicts sunrise, sunset, civil and nautical twilight, and sunburn hazard.


4

Ka-bong!Ka-bong! Ka-runch!Ka-runch!

Totallyinelastic

case

Perfectlyelasticcase

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile

1minute(60sec.) (b) Collision!

After collision...what velocities?Before collision.....A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

6600mph1100mph

5


Totallyinelastic

case


??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile



6600mph1100mph

Conventional solution:Get out formulas: Σ mV(before)= Σ mV(after) [momentum conservation]

Σ mV2(before)= Σ mV2(after) [energy conservation]

6


Totallyinelasticcase


??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile



6600mph

Conventional solution:Get out formulas:ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines! ... (and a circle...)

7




??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile



6600mph


But an UNconventional way is quicker and slicker.......... (Just have to draw 2 lines! ... (and a circle...)

..... and most importantly, DERIVE LOGICAL STRUCTURE!

8




??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile



50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

6600mph1100mph

VVeelloocciittyy-vveelloocciittyy PPlloott

6600mph

1100mph


9




??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile



50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

6600mph1100mph


6600mph

1100mph


50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV6600mph

1100mph

MSUVVVSUV+MVWVVVW =constant is AAxxiioomm ##11

4

-1

10




??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1-6 sec.-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile



50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

4

-1

slope =-4

6600mph1100mph


6600mph

1100mph


50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV6600mph

1100mph

= ΔVSUV

ΔVVWMSUV-mVW

4-1 =

MSUVVVSUV+MVWVVVW =constant is AAxxiioomm ##11

11


Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile


After collision...what velocities?Before collision.....

A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph


6600mph

1100mph

Conventional solution:

Get out formulas:

ΣmV(before)=ΣmV(before) [momentum conservation]ΣmV2(before)=ΣmV2(before)[energy conservation]etc.

50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

MSUVVVSUV

+MVWVVVW

=constant is AAxxiioomm ##11

BINGO!slope=+1

12


Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile




50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph


6600mph

1100mph


Get out formulas:


50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

FINAL

Ka-bong!

MSUVVVSUV

+MVWVVVW


BINGO!

DOUBLE

BINGO!

“!hnuR-aK”

13


Totally

inelastic

case

Perfectly

elastic

case

??????????

1060

????????

1060

0-0.2-0.4-0.6-0.8-1

-6 sec.

-12 sec.

-24 sec.

-36 sec.

-48 sec.

1 mile




50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

I

VVSUV

“Ka-Runch!”

V VW

=V SUV

line

FINAL

Ka-runch!

4

-1

slope =-4

6600mph1100mph


6600mph

1100mph


Get out formulas:


50mph20 40 60 90

20

40

60

80

50mph

100mph

100mph

VVVW

0

INITIAL

10 30 70 80

90

10

30

70

F

45°

45°

I

VVSUV

6600mph

1100mph

= ΔVSUV

ΔVVW

MSUV

-mVW

4

-1 =

FINAL

Ka-bong!

MSUVVVSUV

+MVWVVVW


BINGO!

DOUBLE

BINGO!

“!hnuR-aK”

Superball Collision Simulatorhttp://www.uark.edu/rso/modphys/animations/BounceItWeb.html

Superball Collision Simulator

14

http://www.uark.edu/rso/modphys/animations/BounceItWeb.html


A problem in ssppaaccee--ttiimmee : (6600mph Cell-faxing 4ton SUV rear-ends 1100mph 1ton VW)Geometry of Galilean translation (A ssyymmmmeettrryy ttrraannssffoorrmmaattiioonn)

5020 40 60 90

10010 30 70 80

FEarth

10

20

30

40

50

60

70

80

90

100

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10-100-90-80-70-60

-50-40-30-20-10

FCOM

ICOM

IEarth

VSUV

VVW

subtracting

(50,50)

(a) Galileo transforms to COM frame

Fig. 2.5ain Unit 1

If you increase your velocity by 50 mph,...

...the rest of the world appears to be 50 mph sslloowweerr

15

M1=70gmM1=70gm

M0=10kgM0=10kgbounceplatebounceplate

RRrrd

Superballpenetrationdepthr22Rd=

SuperballSuperball

ballpointpen

M2=10gm

ballpointpen

M2=10gm

The X-2pen-

launcher

The X-2pen-

launcher

The X-2 Pen launcher and Superball Collision Simulator*

*SimulatorWebsite:http://www.uark.edu/rso/modphys/animations/BounceItWeb.html16



M1=70gmM1=70gm

M0=10kgM0=10kg

bounce

plate

bounce

plate

RRrr

d

Superball

penetration

depth

r2

2Rd=

SuperballSuperball

ballpoint

pen

M2=10gm

ballpoint

pen

M2=10gm

The X-2pen-

launcher

The X-2pen-

launcherFig. 4.1 and Fig. 4.3

in Unit 1

M1

m2BANG!M1

(BiggerBANG!)

(StillBiggerBANG!)

m2

M1M1

m2

Bang1!

Bang2!M1

m2

(a) Super-elastic 2nd-body bounce (b) 2-Bang Model (c) n-BodySupernovaSuperballs

m2

-1.0

1.0

1.0

0.5

2.0

m2 Velocity axis

Vym2

m1 Velocity axis

Vym1

(0,0)

Bang-1(01)

INIT point at

(-1.0,-1.0)

(a)Bang-1(01)

Fig. 4.4a-bin Unit 1

Bang-1(01)

FINAL point

(+1.0,-1.0)

This 1st bang is a floor-bounce of

M1off very massive plate/Earth M

0

2.02.01st bang:

mass (M0)vs. mass (M

1)

-7

+1

1.0

1.0

0.5

2.0

m1 Velocity axis

Vym1

(0,0)

COM-point at

(0.75,0.75))

-1.0

Bang-2(12)

INIT point at

(+1.0,-1.0)

Bang-2(12)

FINAL point

(0.5,2.5)

(b)Bang-2(12)

2nd bang:

(M1)vs.(M

2)

17

M1=70gmM1=70gm


RRrrd


SuperballSuperball

ballpointpen

M2=10gm

ballpointpen

M2=10gm

The X-2pen-

launcher

The X-2pen-


in Unit 1

M1

m2BANG!M1

(BiggerBANG!)

(StillBiggerBANG!)

m2

M1M1

m2

Bang1!

Bang2!M1

m2


m2

Fig. 4.5ain Unit 1Start at

(1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0 C

L

P

L is 15::1

P is 7::1C is 4::1

Line CPLis elastic collisionfinal pt. locus fordifferentmomentumslopesormassratiosM1::M2

M2 Velocity axisV2

M1 Velocity axis V1(0,0)

3.0

Bang-2(12)FINALpoints

(a)

18

M1=70gmM1=70gm


RRrrd


SuperballSuperball

ballpointpen

M2=10gm

ballpointpen

M2=10gm

The X-2pen-

launcher

The X-2pen-


in Unit 1

M1

m2BANG!M1

(BiggerBANG!)

(StillBiggerBANG!)

m2

M1M1

m2

Bang1!

Bang2!M1

m2


m2

Fig. 4.5a-bin Unit 1Start at

(1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0 C

L

P

L is 15::1

P is 7::1C is 4::1

Line CPLis elastic collisionfinal pt. locus fordifferentmomentumslopesormassratiosM1::M2

1.0

1.0

-1.0

0.5

2.0 C

L

P

L is 15::1

P is 7::1C is 4::1

II iiss 11::::00oorr ∞∞::::11II

MM iiss 33::::11

MM

3.0M2 Velocity axisV2

M2 Velocity axisV2

M1 Velocity axis V1 M1 Velocity axis V1(0,0) (0,0)

3.0

D is 2::1

Uis 1::1

Bang-2(12)FINALpoints

(a) (b)

-1.0

U2

U1

19

Geometric “Integration” (Converting Velocity data to Spacetime)

Start at

(-1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0

Bang-2(12)

Bang-3(20)

Bang-1(01)

Floor at y=0

Bang-1(01)

Bang-2(12)

Bang-3(20)

M2

slope

+2.5

M1

slope

+0.5

Height y

y=3

y=1

Ceiling at y=7

Bang-4(12)

??

Vy1

Vy2

??

(a)

Time t

(b)

M2

slope

-2.5

20


Start at

(-1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0

Bang-2(12)

Bang-3(20)

Bang-1(01)

Floor at y=0

Bang-1(01)

Bang-2(12)

Bang-3(20)

M2

slope

+2.5

M1

slope

+0.5

Height y

y=3

y=1

Ceiling at y=7

Bang-4(12)

??

Vy1

Vy2

??

(a)

Time t Bang-1(01)

Bang-2(12)

Bang-3(20)

Bang-4(12)

Bang-5(20)

Bang-6(12)

Bang-7(01)

Bang-8(20)

Bang-9(12)

Time t

Floor at y=0

Ceiling at y=7

y

Start at

(-1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0

Bang-2(12)

Bang-3(20)

Bang-4(12)

Bang-1(01)

Bang-5(20)

Bang-6(12)

Bang-7(01)

Vy1

Vy2

(b)

(c)

(d)

M2

slope

-2.5

Fig. 4.7a-din Unit 1

Kinetic Energy Ellipse

KE = 12

M1V12 + 1

2M 2V2

2 =12

+ 72= 4

1 = V12

2KE /M1

+V2

2

2KE /M 2

=x1

2

a12 +

x22

a22

21


Start at

(-1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0

Bang-2(12)

Bang-3(20)

Bang-1(01)

Floor at y=0

Bang-1(01)

Bang-2(12)

Bang-3(20)

M2

slope

+2.5

M1

slope

+0.5

Height y

y=3

y=1

Ceiling at y=7

Bang-4(12)

??

Vy1

Vy2

??

(a)

Time t Bang-1(01)

Bang-2(12)

Bang-3(20)

Bang-4(12)

Bang-5(20)

Bang-6(12)

Bang-7(01)

Bang-8(20)

Bang-9(12)

Time t

Floor at y=0

Ceiling at y=7

y

Start at

(-1.0,-1.0)

1.0

1.0

-1.0

0.5

2.0

Bang-2(12)

Bang-3(20)

Bang-4(12)

Bang-1(01)

Bang-5(20)

Bang-6(12)

Bang-7(01)

Vy1

Vy2

(b)

(c)

(d)

M2

slope

-2.5

Fig. 4.7a-din Unit 1

Kinetic Energy Ellipse

KE = 12

M1V12 + 1

2M 2V2

2 = 72

+12= 4

1= V12

2KE /M1

+ V22

2KE /M 2

= x12

a12 +

x22

a22

Ellipse radius 1

a1 = 2KE /M1

= 2KE /7

= 8/7 = 1.07

Ellipse radius 2

a2 = 2KE /M1

= 2KE /1

= 8/1 = 2.83

22

Difficult to see high mass-ratio-skinny-ellipse improved byScale transformation M1v1 -> √M1 v1

M1 =49m2 =1

23

Ellipse rescaling geometry and reflection symmetry analysis Convert to rescaled velocity: symmetrize: V1 = v1 ⋅ m1 , V2 = v2 ⋅ m1 , KE =2

1m1v12+2

1m2v22 =2

1 V12 +21 V22Then collisions become reflections and double-collisions become rotations

where:

cosθ sinθsinθ −cosθ

⎛

⎝⎜⎞

⎠⎟

cosθ ≡ m1 − m2

m1 + m2

⎛⎝⎜

⎞⎠⎟

and: sinθ ≡2 m1m2

m1 + m2

⎛

⎝⎜

⎞

⎠⎟ with: m1 − m2

m1 + m2

⎛⎝⎜

⎞⎠⎟

2

+2 m1m2

m1 + m2

⎛

⎝⎜

⎞

⎠⎟

2

= 1

θ=16.26°1

m1 - m2m1+m2 2√m1 m2

m1+m2

4850=

1450=

slope :

−m2

m1

= −17

slope :

+m2

m1

= +17

11

33

77

1177

1199

11331155

55

991111

2211

θ=16.26°V1

V2

[11]-tangent slope-√m1/√m2= -7

22

44

11

33

77

1177

1199

11331155

55

991111

2211

v1

v2

slope1/1

slope-1/1

[11]-tangentslope

-m1/m2= -49

22

44

24

(a) Lagrangian L = L(v1,v2)

v1

v2(b) Estrangian E = E(V1,V2)

V1=√m1v1

(c) Hamiltonian H = H(p1,p2)

p1=m1v1

p2=m2v2

V2=√m2v2

COM Bisectorslope = 1/1

Collision line andCOM tangent slope= -m1/m2 =-16

Collision line andCOM tangent slope=-√m1/√m2=-4

COM Bisector slope= √m2/√m1 =1/4

Collision line andCOM tangent slope

= -1/1

COM Bisector slope= m2/m1 =1/16

slope√m1√m2

=4

slope=1

25

p2=m2v2

p1=m1v1

Hamiltonian plotH(p)=const.=p•M-1•p/2(b)Lagrangian plot

L(v)=const.=v•M•v/2

v2=p2 /m2

L=const = E

v1=p1 /m1

(a)

v v = ∇∇pH=M-1•p

p = ∇∇vL=M•v

p

Lagrangian tangent at velocity vis normal to momentum p

Hamiltonian tangent at momentum pis normal to velocity v

(c) Overlapping plotsv

p

v

p

p

v (d) Less mass

(e) More mass

H=const = E

L=const = E

H=const = E

26

-1

-1

+2

0 +1

+1

p-slope=v

dH/dp=+1.0

-1 0 +1

-1

+1

+2

v-slope=p

dL/dv =+1.6

1.6

1

1.0

1

Lagrangian plot

L(v)= v•p - H(p)(a) Hamiltonian plot

H(p)=p•v-L(v)(b)

v-slope

intercept

-H = -0.6 p-slope

intercept

-L = -1.0

Lagrangian

L(v) = 1.0 Hamiltonian

H(p) = 0.6

momentum

pvelocity

v

velocity

v = 1.0

momentum

p = 1.6

L(v) H(p)

v-slope is momentum p

velocity v is p-slope

Lagrangian L(v) is -p-slope intercept

-v-slope intercept is Hamiltonian H(p)

Unit 1Fig. 12.3

27

“Monster Mash”classical segue to Heisenberg action relationsExample of very very large M1 ball-walls crushing a poor little m2

How m2 keeps its action An interesting wave analogy: The “Tiny-Big-Bang” [Harter, J. Mol. Spec. 210, 166-182 (2001)],[Harter, Li IMSS (2012)]

A lesson in geometry of fractions: Ford Circles and Farey Sums [Lester. R. Ford, Am. Math. Monthly 45,586(1938)] [John Farey, Phil. Mag.(1816)]

28

2

-2

4

-46

-6

Δx=1.5

Δx=0.75Δx=0.5

Δx=3.0

Δx=0.375

Δx=1.5

Δx=0.75Δx=0.5

Δx=3.0

Δx=0.375

11//11

11//22

11//33

11//4411//5511//6611//77

v2=0

v2=2

v1=1

Time

Space

The Wall

Time

Space

(a) Big ball moves in and traps small ball between it and The Wall

(b) Trajectory geometry exposed

Y Y

Vy2

Vy2

(a) Big space

Low speed

Bang (1)12

Bang (2)20

(b) Decreasing space

Increasing speed

Y

Vy2(c) Small space

High speed

Bang (n)12

Bang (n+1)20

VyY = const.

Y

Vy2

Y

Vy2

Y

Vy2

Unit 1Fig. 6.4

The Classical“Monster Mash”

Classical introduction to

Heisenberg “Uncertainty” Relations

v2 = const.Y

or: Y ⋅v2 = const.

is analogous to: Δx ⋅ Δp = N ⋅

29

-3

-57

-7 8-8

Δx=2.0

Δx=1.5Δx=1.2

Δx=3.0

Δx=1.0

v2=3

Time

v2=-1

5

2

-2

-44

6-6

11//11

11//22

11//33

11//4411//5511//6611//77

9-9

v2=+1

v2=0 v2=0

V1=+1 V1=-1

v2=-0.2

v2=+2.2

Double “Monster Mash”

30

-3

3

The Wall-1

5

3

-1

5

-3

0

4

-26

2

-2

-4

4

6-6

The Wall(a) Galilean shift by V=1 (b)

−vINvFIN=vIN+2V

1

+vIN

V1=1V1=1

V1=1V1=1

V1=1V1=1

V1=1V1=1

V1

V1=1V1=1

V1=1V1=1

B− B0 B+

A−A0 A+

C

O

V1=1V1=1

V1=1V1=1

(a) (b)

(c)

V1V1

vFIN2

+4

+3

+2

+1

0

-1

-2

-3

+5B− B0 B+

Unit 1Fig. 6.6

andFig. 6.7

31

Δm = 9

2Δx = 4 %

-15 -10 -5 0 5 10 15 = m

1/1

0/11/1

2Δx = 4 % 0/1Time t (units of fundamental period ) τ1

Coordinate φ (units of 2π )

0/1

1/10

1/5

3/10

2/5

1/2

3/5

7/10

4/5

9/10

1/1

3/4

1/4

1/21/40-1/4-1/2

1/21/4

0-1/4

-1/2

(Imagine "wrap-around" φ-coordinate) +π /2

−π /2

+π−π

3/4

1/4

0/1

1/2

1/1

Time t

[Harter, J. Mol. Spec. 210, 166-182 (2001)]

32

Δm = 9

2Δx = 4 %

-15 -10 -5 0 5 10 15 = m

1/1

0/1

1/2

1/4

1/61/7

1/3

1/5

2/5

2/7

3/7

1/8

Wave packet starts hereZeros start here Zeros start here

Time t(units of τ1)

Coordinate φ(units of 2π)

1/1

0/10 1/4 1/2-1/2 -1/4

N-level-system and revival-beat wave dynamics (9 or10-levels (0, ±1, ±2, ±3, ±4,..., ±9, ±10, ±11...) excited) Zeros (clearly) and “particle-packets” (faintly) have paths

labeled by fraction sequences like: 07, 17, 27, 37, 47, 57, 67,11

17

27

37

47

57

67

47

57

67

11

01

[Harter, J. Mol. Spec. 210, 166-182 (2001)]

33

Farey Sum algebra of revival-beat wave dynamicsLabel by numerators N and denominators D of rational fractions N/D


0 1/4 1/2

1/d1

1/d2


1/1

0/1-1/2 -1/4

1/d2

2/d2

3/d2

n2/d2

14/d113/d112/d1

n1/d1(n2-1)/d2

(n1+1)/d1

•••

•••

1/d2

2/d2

n2/d2 path slope is 1/d2

n1/d1 path slope is -1/d1

•••

•••

n1/d1 and n2/d2 pathfractions

numerator/denominator

34

n1+n2d1+d2

tx=


0 1/4 1/2

1/d1

1/d2

n2/d2- t

1/2 -φ= 1/d2


1/1

0/1-1/2 -1/4

n1/d1 - t1/2 -φ = -1/d1

(φx ,tx)

1/d2

2/d2

3/d2

n2/d2

14/d113/d112/d1

n1/d1(n2-1)/d2

(n1+1)/d1

•••

•••

1/d2

2/d2

x

n1/d1 and n2/d2 pathintersection time

(Farey-Sum)

•••

•••

d1n2-n1d2d1+d2

φx=

n1/d1 and n2/d2 pathintersection point

(Ford-Cross)

n2/d2 path slope is 1/d2

n1/d1 path slope is -1/d1

Farey Sum algebra of revival-beat wave dynamicsLabel by numerators N and denominators D of rational fractions N/D

[Lester. R. Ford, Am. Math. Monthly 45,586(1938)] [John Farey, Phil. Mag.(1816)] 35

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

20 3 4 51 7 8 9 106 12 13 14 1511 17 18 19 2016-1-2-3

11

v1=(1,1)v0=(0,1)

(a) 01

Numerator Axis N

DenominatorAxisD

Unit Real Interval Farey Sum related to

vector sumand

Ford Circles1/1-circle has

diameter 1

36

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

20 3 4 51 7 8 9 106 12 13 14 1511 17 18 19 2016-1-2-3

12

11

v1=(1,1)v0=(0,1)

(b) 01

v1=(1,2)

Numerator Axis N

DenominatorAxisD

Unit Real Interval

12

11

01

Farey-Sum Tree

Farey Sum related to

vector sumand

Ford Circles

1/2-circle hasdiameter 1/22=1/4

1/1-circle hasdiameter 1

37

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

20 3 4 51 7 8 9 106 12 13 14 1511 17 18 19 2016-1-2-3

12

13

11

v1=(1,1)v0=(0,1)

(c) 01

v1/2=(1,2)v1/3=(1,3)

Numerator Axis N

DenominatorAxisD

Unit Real Interval

12

11

01

13

23

Farey-Sum Tree


vector sumand

Ford Circles


1/3-circles havediameter 1/32=1/9

38

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

20 3 4 51 7 8 9 106 12 13 14 1511 17 18 19 2016-1-2-3

122

513

14

15

Numerator Axis N

DenominatorAxisD

Unit Real Interval 11

v1=(1,1)v0=(0,1)

(d) 01

v1/2=(1,2)v1/3=(1,3)

v2/5=(2,5)

16

17

27

37

18

38

12

11

01

13

23

14

34

15

25

35

45

16

56

17

27

37

47

57

67

18

38

58

78

Farey-Sum Tree


vector sumand

Ford Circles


1/3-circles havediameter 1/32=1/9

n/d-circles havediameter 1/d2

39

U of A Physics Day 2000(Quantum computer simulation)

That makes an ∞-ly deep “3D-Magic-Eye” picture

40


41

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

46

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3

3

4

Phase

Group

47

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group

Doppler blue-shiftUP by factor e+ρ=2

Doppler red-shiftDN by factor e-ρ=1/2

48

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group



49

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group



50

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group



Group

Phase

Group vector shiftedfrom ϖ(1,0) to ϖ(cosh ρ ,sinh ρ)

Phase vector shiftedfrom ϖ(0,1) to ϖ(sinh ρ,cosh ρ)ϖ=2

51

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group



Group

Phase



53

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group



Group

Phase



54

Wavevector ck

Frequency ω

1 300THz

2 600THz

2 900THz

4 1200THz

1 300THz-1 300THz 2-2 1-1 3 4

3

Phase

Group



Group

Phase



55

The Circular Functions “Urban elite” The Hyperbolic Functions “Country-cousins”They’re related by Legendre contact transformation L = p•v-H

tan σ = sinh ρ

Circular Geometry of Lagrangian Functions versus Hyperbolic Geometry of Hamiltonian Functions

59


tan σ = sinh ρ sin σ = tanh ρ


60


tan σ = sinh ρ sin σ = tanh ρcos σ = sech ρ


61


tan σ = sinh ρ sin σ = tanh ρcos σ = sech ρ sec σ = cosh ρ


62

The Circular Functions “Urban elite” The Hyperbolic Functions “Country-cousins”


In spacetime: asimultaneity factor In per-spacetime: momentum velocity u/c group velocity Lorentz contraction -Lagrangian Einstein time dilation Hamiltonian

They’re related by Legendre contact transformation L = p•v-Htan σ = sinh ρ sin σ = tanh ρcos σ = sech ρ sec σ = cosh ρcot σ = csch ρcsc σ = coth ρ

sin σ = tanh ρ = u/csec σ = cosh ρ =1/√1-u2/c2

Old-fashionednotation:

63

Rest energy

Mc2=1

u/c=4/5

Frequency ωorTotal energy

Hamiltonian

H=ω=5/3=coshρ=secσ

Momentum cp or Wavevector ck

cp=ck=4/3=sinhρ=tanσ

-Lagrangian

-L=3/5=sechρ=cosσ

Momentum or Wavevector ck

cp=ck=4/3=sinhρ=tanσPhase velocity

c/u=5/4=cothρ=cscσ

DeBroglie

wavelength λ=3/4=cschρ=cotσ

u/c slope angle υ=atan(u/c)

stellar aberation angle σ=asin(u/c)

DeBroglie

wavelength

λ=3/4=cschρ=cotσ

Phase velocity

c/u=5/4=cothρ=cscσ

Doppler

blue-shift

b=3=e+ρ

Doppler

red-shift

r=1/3=e−ρ

Group velocity

u/c=4/5

=tanhρ=sinσ

Kinetic energy

KE=2/3=coshρ -1=secσ -1

ω

ck

ω´

ck´

67

-2-3 1 2 3 4 5 6 7 8 9

1

2

3

5

6

7

8

9

-1

Phasevector

Groupvector

BHamiltonian

H=Bcosh ρAreaB·ρ/2

per-Space-Time Geometry(frequency vs wavevector)

68

VVW100110120

-120

(60,10)Initial-point

Final“Ka-Bong”-point

a=√=60.21

2·KEMSUV

b=√=120.42

2·KEmVW

ElasticKineticEnergyellipse(KE=7,250)

MomentumPTotal=250line

(50,50)

(40,90)

MSUV=4MSUV=4

mVW=1mVW=1

VSUV

Fig. 3.1 ain Unit 1

69

VVW

100

110

120

-120

(60,10)

Initial-point

Final“Ka-Bong”-point

a=√=60.21

2·KE

MSUV

b=√=120.42

2·KEmVW

Elastic

Kinetic

Energy

ellipse

(KE=7,250)

Momentum

PTotal=250

line

(50,50)

(40,90)

MSUV=4M

SUV=4

mVW=1m

VW=1

VSUV

Fig. 3.1 ain Unit 1

a=√=55.9

2·IE

MSUV

b=√=111.8

2·IEmVW

COM

Energy

ellipse

(ECOM=1,000)

elastic

FIN

IN

inlastic

FIN is

VCOM

point

Inelastic

Kinetic

Energy

ellipse

(IE=6,250)

Fig. 3.1 bin Unit 1

70

2012 INBRE workshop...4-1 slope=-4 6 0 m p h 1 0 m p h Velocity-velocityPlot 60mph 10mph...

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