Many-body 1D collisionsElastic examples: Western buckboard, Bouncing column, Newton’s cradleInelastic examples: “Zig-zag geometry” of freeway crashesSuper-elastic examples: This really is “Rocket-Science”
Geometry of common power-law potentials Geometric (Power) series
“Zig-Zag” exponential geometryProjective or perspective geometry
Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields
Compare mks units of Coulomb Electrostatic vs. GravityGeometry of idealized “Sophomore-physics Earth”
Coulomb field outside Isotropic Harmonic Oscillator (IHO) field insideContact-geometry of potential curve(s)“Crushed-Earth” models: 3 key energy “steps” and 4 key energy “levels”
Earth matter vs nuclear matter: Introducing the “neutron starlet” and “Black-Hole-Earth”
Lecture 6 Thur. 9.12.2013
Topics for Lecture 7
1Thursday, September 12, 2013
Many-body 1D collisionsElastic examples: Western buckboard, Bouncing column, Newton’s cradleInelastic examples: “Zig-zag geometry” of freeway crashesSuper-elastic examples: This really is “Rocket-Science”
2Thursday, September 12, 2013
Western buckboard = ?????
3Thursday, September 12, 2013
Western buckboard = ?????
4Thursday, September 12, 2013
Western buckboard = 3-ball analogy
5Thursday, September 12, 2013
Western buckboard = 3-ball analogy Disaster!
6Thursday, September 12, 2013
(a)mk/mk+1=3
(b)mk/mk+1=7
(c) Bouncingcolumn
(d) Singlepop-up
(+1,-1)
(-1,+1)
(1,0)
(0,1)
mk/mk+1=1
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
Bang!
-1 0 1 2 3
-1
1
2
3
4
5
6
-1 0 1 2 3
-1
1
2
3
4
5
6
4
Bang(2)12
Bang(3)23
Bang(4)34
Bang(2)12
Bang(3)23
Bang(4)34
Bang(1)01 Bang(1)01
Unit 1Fig. 8.2a-b
4-Body IBM GeometryFig. 8.2c-d
4-Equal-Body Geometry
4-Equal-Body “Shockwave” or pulse wave
Dynamics
Opposite of continuous wave dynamicsintroduced in Unit 2
http://www.uark.edu/ua/modphys/testing/markup/BounceItWeb.html
7Thursday, September 12, 2013
Many-body 1D collisionsElastic examples: Western buckboard, Bouncing column, Newton’s cradleInelastic examples: “Zig-zag geometry” of freeway crashesSuper-elastic examples: This really is “Rocket-Science”
8Thursday, September 12, 2013
M1:m2=1:1
0 10 20 30 40 50 6015
10
20
30
““KKaa--rruunncchh!!””
“Ka-runch!”“Ka-runch!”
velocity VM(n...3210) of pileup mass(es)
0 1 2 3 4 5VM(01)=30
0 1 2 3 4 5(VM(0)=60, Vm(1)=0)
0 1 2VM(0123)=15
3
VM(012)=2030 1 2 4 5
4 5
0 1 2VM(01234)=12
3 4 5
0 1 2VM(01235)=10
3 4 5
VM(n...3210)
2:13:1
4:1
40
50
60
Speeding car and five stationary cars
Unit 1Fig. 8.5Pile-up:
One 60mph car hits
five standing cars
9Thursday, September 12, 2013
M1:m2=1:1
0 10 20 30 40 50 6015
10
20
30
““KKaa--rruunncchh!!””
“Ka-runch!”“Ka-runch!”
velocity VM(n...3210) of pileup mass(es)
0 1 2 3 4 5VM(01)=30
0 1 2 3 4 5(VM(0)=60, Vm(1)=0)
0 1 2VM(0123)=15
3
VM(012)=2030 1 2 4 5
4 5
0 1 2VM(01234)=12
3 4 5
0 1 2VM(01235)=10
3 4 5
VM(n...3210)
2:13:1
4:1
40
50
60
Speeding car and five stationary cars
Unit 1
Fig. 8.5Pile-up:
One 60mph car hits
five standing cars
01(VM(1)=60, Vm(0)=0)
2345
01VM(10)=302345
01VM(210)=40
2345
01VM(3210)=45
2345
01VM(43210)=48
2345
01VM(543210)=50
2345
M1:m2=1:1
0 10 20 30 40 50 60
45
10
20
30
““KKaa--rruunncchh!!””
““KKaa--rruunncchh!!””
““KKaa--rruunncchh!!””
velocity VM(n...3210) of pileup
2:13:14:1
40
50
60
40
30
4850Five speeding cars and a stationary car
Fig. 8.6Pile-up:
Five 60mph cars hit
one standing cars
10Thursday, September 12, 2013
M1:m2=1:1
0 10 20 30 40 50 6015
10
20
30
““KKaa--rruunncchh!!””
“Ka-runch!”“Ka-runch!”
velocity VM(n...3210) of pileup mass(es)
0 1 2 3 4 5VM(01)=30
0 1 2 3 4 5(VM(0)=60, Vm(1)=0)
0 1 2VM(0123)=15
3
VM(012)=2030 1 2 4 5
4 5
0 1 2VM(01234)=12
3 4 5
0 1 2VM(01235)=10
3 4 5
VM(n...3210)
2:13:1
4:1
40
50
60
Speeding car and five stationary cars
Unit 1
Fig. 8.5Pile-up:
One 60mph car hits
five standing cars
01(VM(1)=60, Vm(0)=0)
2345
01VM(10)=302345
01VM(210)=40
2345
01VM(3210)=45
2345
01VM(43210)=48
2345
01VM(543210)=50
2345
M1:m2=1:1
0 10 20 30 40 50 60
45
10
20
30
““KKaa--rruunncchh!!””
““KKaa--rruunncchh!!””
““KKaa--rruunncchh!!””
velocity VM(n...3210) of pileup
2:13:14:1
40
50
60
40
30
4850Five speeding cars and a stationary car
Fig. 8.6Pile-up:
Five 60mph cars hit
one standing cars
12345Five speeding cars and five stationary cars
1 2 3 4 5
Fig. 8.7Pile-up:
Five 60mph cars hit
five standing cars(Fug-gedda-aboud-dit!!)
11Thursday, September 12, 2013
Many-body 1D collisionsElastic examples: Western buckboard, Bouncing column, Newton’s cradleInelastic examples: “Zig-zag geometry” of freeway crashesSuper-elastic examples: This really is “Rocket-Science”
12Thursday, September 12, 2013
1/2
1/3
1/4
1/51/61/71/81/91/10
1
2
3
4
5
6
7
8
slope
1:1
slope
1:2
slope
1:3
1
110
16
1/5
1/4
1/4
1/4
18
1/8
1/8
19
1/10
1/9
17
1/7
1/7
1/6
1/5
1/9
1/10
1/6
1/5
00ΔV(0)=+1/10Δv
e=-1
11 22 33 44 55pow(0)
66 77 88 99 1100
11 22 33 44 55 66 77 88 99 110000
00 11 22 33 44 55ΔV(1)=+1/9Δv
e=-1
66 77 88 99 1100
00 11 22 33 44 55Δve=-1
66 77 88 99 1100ΔV(2)=+1/8
(a) Harmonic progression
1,1/2,1/3,1/4,1/5,1/6,1/7,...
(b) Harmonic series
1+1/2+1/3+1/4+1/5+1/6+1/7+...
00 11 22 33 44Δve=-1
55ΔV(5)=+1/566 77 88 99 1100
00 33 44 55ΔV(4)=+1/6Δv
e=-1
11 22 66 77 88 99 1100
00 33 44 55Δve=-1
11 22 66 77 88 99 1100ΔV(3)=+1/7
00 11 22 33 44Δve=-1
55ΔV(6)=+1/4
77 88 99 110066
ΔV(7)=+1/388 99 1100
ve(exhaustparticle)
0.5
VM(rocket)
0.5
-1.0
0 1101
213
14
15
16
17
1819
110
pow(0)
slope
1:10
pow(1)
slope
1:9
pow(2)
slope
1:8
pow(3)
slope
1:7
pow(4)
slope
1:6
pow(5)
slope
1:5
pow(6)
slope
1:4
pow(1)
pow(2)
pow(3)
pow(4)
pow(5)
pow(6)
pow(7)
+__=0.336=0.1
ΔV(6)=+1/4
ΔV(5)=+1/5
ΔV(4)=+1/6
ΔV(3)=+1/7
ΔV(2)=+1/8
ΔV(1)=+1/9
ΔV(0)=+1/10
110+__
=1.096
19
18
17
16
14
15
110+__
=0.846
19
18
17
16
15
16
110+__=0.646
19
18
17
17
110+__=0.478
19
18
18
110
19
19
110+__=0.211
1.0
1.0 Unit 1Fig. 8.8a-b
Rocket Science!
m·Δv1+9m·ΔVM(1)=0
m·Δv2+8m·ΔVM(2)=0
m·Δv0+10m·ΔVM(0)=0
m·Δv3+7m·ΔVM(3)=0
m·Δv4+6m·ΔVM(4)=0
m·Δv5+5m·ΔVM(5)=0
m·Δv6+4m·ΔVM(6)=0
m·Δv7+3m·ΔVM(7)=0
13Thursday, September 12, 2013
0th: V(0)=1/10=0.1 1st: V(1)=1/10+1/9=0.211 2nd: V(2)=1/10+1/9+1/8=0.336 3rd: V(3)=V(2)+1/7=0.478 4th: V(4)=V(3)+1/6=0.646 5th: V(5)=V(4)+1/5=0.846 6th: V(6)=V(5)+1/4=1.096 7th: V(7)= V(6)+1/3=1.429 8th: V(8)=V(7)+1/2=1.929
1/2
1/3
1/4
1/51/61/71/81/91/10
1
2
3
4
5
6
7
8
slope
1:1
slope
1:2
slope
1:3
1
110
16
1/5
1/4
1/4
1/4
18
1/8
1/8
19
1/10
1/9
17
1/7
1/7
1/6
1/5
1/9
1/10
1/6
1/5
00ΔV(0)=+1/10Δv
e=-1
11 22 33 44 55pow(0)
66 77 88 99 1100
11 22 33 44 55 66 77 88 99 110000
00 11 22 33 44 55ΔV(1)=+1/9Δv
e=-1
66 77 88 99 1100
00 11 22 33 44 55Δve=-1
66 77 88 99 1100ΔV(2)=+1/8
(a) Harmonic progression
1,1/2,1/3,1/4,1/5,1/6,1/7,...
(b) Harmonic series
1+1/2+1/3+1/4+1/5+1/6+1/7+...
00 11 22 33 44Δve=-1
55ΔV(5)=+1/566 77 88 99 1100
00 33 44 55ΔV(4)=+1/6Δv
e=-1
11 22 66 77 88 99 1100
00 33 44 55Δve=-1
11 22 66 77 88 99 1100ΔV(3)=+1/7
00 11 22 33 44Δve=-1
55ΔV(6)=+1/4
77 88 99 110066
ΔV(7)=+1/388 99 1100
ve(exhaustparticle)
0.5
VM(rocket)
0.5
-1.0
0 1101
213
14
15
16
17
1819
110
pow(0)
slope
1:10
pow(1)
slope
1:9
pow(2)
slope
1:8
pow(3)
slope
1:7
pow(4)
slope
1:6
pow(5)
slope
1:5
pow(6)
slope
1:4
pow(1)
pow(2)
pow(3)
pow(4)
pow(5)
pow(6)
pow(7)
+__=0.336=0.1
ΔV(6)=+1/4
ΔV(5)=+1/5
ΔV(4)=+1/6
ΔV(3)=+1/7
ΔV(2)=+1/8
ΔV(1)=+1/9
ΔV(0)=+1/10
110+__
=1.096
19
18
17
16
14
15
110+__
=0.846
19
18
17
16
15
16
110+__=0.646
19
18
17
17
110+__=0.478
19
18
18
110
19
19
110+__=0.211
1.0
1.0
14Thursday, September 12, 2013
0th: V(0)=1/10=0.1 1st: V(1)=1/10+1/9=0.211 2nd: V(2)=1/10+1/9+1/8=0.336 3rd: V(3)=V(2)+1/7=0.478 4th: V(4)=V(3)+1/6=0.646 5th: V(5)=V(4)+1/5=0.846 6th: V(6)=V(5)+1/4=1.096 7th: V(7)= V(6)+1/3=1.429 8th: V(8)=V(7)+1/2=1.929
1/2
1/3
1/4
1/51/61/71/81/91/10
1
2
3
4
5
6
7
8
slope
1:1
slope
1:2
slope
1:3
1
110
16
1/5
1/4
1/4
1/4
18
1/8
1/8
19
1/10
1/9
17
1/7
1/7
1/6
1/5
1/9
1/10
1/6
1/5
00ΔV(0)=+1/10Δv
e=-1
11 22 33 44 55pow(0)
66 77 88 99 1100
11 22 33 44 55 66 77 88 99 110000
00 11 22 33 44 55ΔV(1)=+1/9Δv
e=-1
66 77 88 99 1100
00 11 22 33 44 55Δve=-1
66 77 88 99 1100ΔV(2)=+1/8
(a) Harmonic progression
1,1/2,1/3,1/4,1/5,1/6,1/7,...
(b) Harmonic series
1+1/2+1/3+1/4+1/5+1/6+1/7+...
00 11 22 33 44Δve=-1
55ΔV(5)=+1/566 77 88 99 1100
00 33 44 55ΔV(4)=+1/6Δv
e=-1
11 22 66 77 88 99 1100
00 33 44 55Δve=-1
11 22 66 77 88 99 1100ΔV(3)=+1/7
00 11 22 33 44Δve=-1
55ΔV(6)=+1/4
77 88 99 110066
ΔV(7)=+1/388 99 1100
ve(exhaustparticle)
0.5
VM(rocket)
0.5
-1.0
0 1101
213
14
15
16
17
1819
110
pow(0)
slope
1:10
pow(1)
slope
1:9
pow(2)
slope
1:8
pow(3)
slope
1:7
pow(4)
slope
1:6
pow(5)
slope
1:5
pow(6)
slope
1:4
pow(1)
pow(2)
pow(3)
pow(4)
pow(5)
pow(6)
pow(7)
+__=0.336=0.1
ΔV(6)=+1/4
ΔV(5)=+1/5
ΔV(4)=+1/6
ΔV(3)=+1/7
ΔV(2)=+1/8
ΔV(1)=+1/9
ΔV(0)=+1/10
110+__
=1.096
19
18
17
16
14
15
110+__
=0.846
19
18
17
16
15
16
110+__=0.646
19
18
17
17
110+__=0.478
19
18
18
110
19
19
110+__=0.211
1.0
1.0
By calculus: M·ΔV=-ve·ΔM or: Integrate:dV = −vedMM
dVVINVFIN∫ = −ve M
dMMIN
MFIN∫
ve known as“Specific Impulse”
15Thursday, September 12, 2013
0th: V(0)=1/10=0.1 1st: V(1)=1/10+1/9=0.211 2nd: V(2)=1/10+1/9+1/8=0.336 3rd: V(3)=V(2)+1/7=0.478 4th: V(4)=V(3)+1/6=0.646 5th: V(5)=V(4)+1/5=0.846 6th: V(6)=V(5)+1/4=1.096 7th: V(7)= V(6)+1/3=1.429 8th: V(8)=V(7)+1/2=1.929
1/2
1/3
1/4
1/51/61/71/81/91/10
1
2
3
4
5
6
7
8
slope
1:1
slope
1:2
slope
1:3
1
110
16
1/5
1/4
1/4
1/4
18
1/8
1/8
19
1/10
1/9
17
1/7
1/7
1/6
1/5
1/9
1/10
1/6
1/5
00ΔV(0)=+1/10Δv
e=-1
11 22 33 44 55pow(0)
66 77 88 99 1100
11 22 33 44 55 66 77 88 99 110000
00 11 22 33 44 55ΔV(1)=+1/9Δv
e=-1
66 77 88 99 1100
00 11 22 33 44 55Δve=-1
66 77 88 99 1100ΔV(2)=+1/8
(a) Harmonic progression
1,1/2,1/3,1/4,1/5,1/6,1/7,...
(b) Harmonic series
1+1/2+1/3+1/4+1/5+1/6+1/7+...
00 11 22 33 44Δve=-1
55ΔV(5)=+1/566 77 88 99 1100
00 33 44 55ΔV(4)=+1/6Δv
e=-1
11 22 66 77 88 99 1100
00 33 44 55Δve=-1
11 22 66 77 88 99 1100ΔV(3)=+1/7
00 11 22 33 44Δve=-1
55ΔV(6)=+1/4
77 88 99 110066
ΔV(7)=+1/388 99 1100
ve(exhaustparticle)
0.5
VM(rocket)
0.5
-1.0
0 1101
213
14
15
16
17
1819
110
pow(0)
slope
1:10
pow(1)
slope
1:9
pow(2)
slope
1:8
pow(3)
slope
1:7
pow(4)
slope
1:6
pow(5)
slope
1:5
pow(6)
slope
1:4
pow(1)
pow(2)
pow(3)
pow(4)
pow(5)
pow(6)
pow(7)
+__=0.336=0.1
ΔV(6)=+1/4
ΔV(5)=+1/5
ΔV(4)=+1/6
ΔV(3)=+1/7
ΔV(2)=+1/8
ΔV(1)=+1/9
ΔV(0)=+1/10
110+__
=1.096
19
18
17
16
14
15
110+__
=0.846
19
18
17
16
15
16
110+__=0.646
19
18
17
17
110+__=0.478
19
18
18
110
19
19
110+__=0.211
1.0
1.0
By calculus: M·ΔV=-ve·ΔM or: Integrate:dV = −vedMM
dVVINVFIN∫ = −ve M
dMMIN
MFIN∫
VFIN −VIN = −ve lnMFIN − lnMIN[ ]=ve lnMFIN
MIN⎡⎣
⎤⎦The Rocket Equation:
ve known as“Specific Impulse”
16Thursday, September 12, 2013
Geometry of common power-law potentials Geometric (Power) series
“Zig-Zag” exponential geometryProjective or perspective geometry
Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields
Compare mks units of Coulomb Electrostatic vs. Gravity
17Thursday, September 12, 2013
1
1
2 3 4
2
3
4
s1=1.5
s2=2.25
s3=3.375
s4=5.0625
s0=1.0
s-1=0.667s-2=0.444
1-1-2
y = s·x (s=1.5)y = s·x“Zig-Zag” geometry of a power sequence
Example: s=1.5
The y=
x line
at 45
°
18Thursday, September 12, 2013
1
1
2 3 4
2
3
4
s1=1.5
s2=2.25
s3=3.375
s4=5.0625
s0=1.0
s-1=0.667s-2=0.444
1-1-2
y = s·x (s=1.5)y = s·x“Zig-Zag” geometry of a power sequence
...and exponential function...
Example: s=1.5
The y=
x line
at 45
°
19Thursday, September 12, 2013
1
1
2 3 4
2
3
4
s1=1.5
s2=2.25
s3=3.375
s4=5.0625
s0=1.0
s-1=0.667s-2=0.444
1-1-2
AApppprrooxxiimmaattiinnggyy ==ssxx
y = s·x (s=1.5)y = s·x“Zig-Zag” geometry of a power sequence
...and exponential function...
Example: s=1.5
The y=
x line
at 45
°
20Thursday, September 12, 2013
1
1
2 3 4
2
3
4
s1=1.5
s2=2.25
s3=3.375
s4=5.0625
s0=1.0
s-1=0.667s-2=0.444
1-1-2
AApppprrooxxiimmaattiinngg::xx ==ssyy oorr yy==llooggss xx
AApppprrooxxiimmaattiinnggyy ==ssxx
y = s·x (s=1.5)y = s·x“Zig-Zag” geometry of a power sequence
...and exponential function... ...and
logarithm function...
Example: s=1.5The
y=x l
ine at
45°
21Thursday, September 12, 2013
Geometry of common power-law potentials Geometric (Power) series
“Zig-Zag” exponential geometryProjective or perspective geometry
Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields
Compare mks units of Coulomb Electrostatic vs. Gravity
22Thursday, September 12, 2013
y=2x
y=− x12
y=−2x
y= x12
y=x
y=-x
y=x
y=-x
s3
s2
s
1/s1/s21/s3-1/s3-1/s2
-1/s
-s
1/s1/s2 s s2
-1/s2
-1/s
-1/s3
-s
ss2s3
1/s
-s3-s2
1/s 1/s2s
1
1
1
-1
-1
0
0 1
(a) Slope factor s=2 (b) s=1/2
90°90°
90°90°
Unit 1Fig. 9.2
“Zig-Zags” give perspective geometry(1D-vanishing point )
23Thursday, September 12, 2013
y=2x
y=− x12
y=−2x
y= x12
y=x
y=-x
y=x
y=-x
s3
s2
s
1/s1/s21/s3-1/s3-1/s2
-1/s
-s
1/s1/s2 s s2
-1/s2
-1/s
-1/s3
-s
ss2s3
1/s
-s3-s2
1/s 1/s2s
1
1
1
-1
-1
0
0 1
(a) Slope factor s=2 (b) s=1/2
90°90°
90°90°
Unit 1Fig. 9.2
“Zig-Zags” give perspective geometry(1D-vanishing point )
1st-day-of-schoolperspective of 1st-grader
1st-day-of-schoolperspective of 12th-grader
24Thursday, September 12, 2013
Geometry of common power-law potentials Geometric (Power) series
“Zig-Zag” exponential geometryProjective or perspective geometry
Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields
Compare mks units of Coulomb Electrostatic vs. Gravity
25Thursday, September 12, 2013
Each y=x2 parabola point found by just one“Zig-Zag”1. Pick an (x=?)-line
x=?x=1
2. “Zig” from its y=xintersection to x=1 line
x=?
x=1
y=x y=x
3. “Zag” from originback to (x=?)-line
x=?
x=1
y=x“Zag” line is y=(?)·xand hits (x=?)-line aty=(?)·(?)=(?)2
26Thursday, September 12, 2013
Each y=x2 parabola point found by just one“Zig-Zag”1. Pick an (x=?)-line
x=?x=1
2. “Zig” from its y=xintersection to x=1 line
x=?
x=1
y=x y=x
3. “Zag” from originback to (x=?)-line
x=?
x=1
y=x“Zag” line is y=(?)·xand hits (x=?)-line aty=(?)·(?)=(?)2
Unit 1Fig. 9.1
21x=0-1-227Thursday, September 12, 2013
Each y=x2 parabola point found by just one“Zig-Zag”1. Pick an (x=?)-line
x=?x=1
2. “Zig” from its y=xintersection to x=1 line
x=?
x=1
y=x y=x
3. “Zag” from originback to (x=?)-line
x=?
x=1
y=x“Zag” line is y=(?)·xand hits (x=?)-line aty=(?)·(?)=(?)2
Unit 1Fig. 9.1
21x=0-1-228Thursday, September 12, 2013
Each y=x2 parabola point found by just one“Zig-Zag”1. Pick an (x=?)-line
x=?x=1
2. “Zig” from its y=xintersection to x=1 line
x=?
x=1
y=x y=x
3. “Zag” from originback to (x=?)-line
x=?
x=1
y=x“Zag” line is y=(?)·xand hits (x=?)-line aty=(?)·(?)=(?)2
Unit 1Fig. 9.1
21x=0-1-229Thursday, September 12, 2013
1
y = x
(a) Oscillator potential U(x)=x2
F(-1.5)
(b)Hooke-Law FFoorrccee FF((xx)) ==--22xx
x=0 1 2
FF((--00..55)) ==11
FF((--11)) ==22
FF((00)) ==00
-2 -1
FF((00..55)) ==--11
F(-1.25)
F(-1.0)
FF((11)) ==--22
F(+1.25)
F(+1.0)U=1
U=2
U=3
x=0 1 2-2 -1
F(2.0)F(-2.0)
F(2.4)F(-2.4)
F(1.5)
U=4
U=5
FF((xx))−ΔUU
UU((xx))Δxx
−ΔUUFF((xx))11 ==______
Δxx
12
12
12
12
12
12
Each y=x2 parabola point found by just one“Zig-Zag”1. Pick an (x=?)-line
x=?x=1
2. “Zig” from its y=xintersection to x=1 line
x=?
x=1
y=x y=x
3. “Zag” from originback to (x=?)-line
x=?
x=1
y=x“Zag” line is y=(?)·xand hits (x=?)-line aty=(?)·(?)=(?)2
Unit 1Fig. 9.1
30Thursday, September 12, 2013
A more conventional parabolic geometry…(uses focal point)
(a) Parabolic Reflector y=x2
y=1
y=2
y=3
y=4
y=5
(b)Parabolic geometry
λ
λ
Directrix
Latusrectum
reflectsintofocus
Verticalincomingray
DistancetoFocus
Distance =to
directrix
Mλ/2λ/2
Unit 1Fig. 9.3
31Thursday, September 12, 2013
A more conventional parabolic geometry...
(a) Parabolic Reflector y=x2
y=1
y=2
y=3
y=4
y=5
(b)Parabolic geometry
λ
λ
Directrix
Latusrectum
reflectsintofocus
Verticalincomingray
DistancetoFocus
Distance =to
directrix
Mλ/2λ/2
Unit 1Fig. 9.3
Better name† for λ : latus radius†Old term latus rectum is exclusive copyright of
X-Treme Roidrage GymsVenice Beach, CA 90017
radius
32Thursday, September 12, 2013
...conventional parabolic geometry...carried to extremes...
p=λ/2
y =-p
0
p
2p
3p
4p
x =0 p 2p 3p 4p
Parabola4p·y =x2=2λ·y
pdirectrix
Latus rectumλ =2p
slope=1tangent slope=-2
Circleofcurvatureradius=λ
(a)
p
p
slope=1/2y =-p
0
p
2p
3p
4p
x =0 p 2p 3p 4p
Parabola4p·y =x2=2λ·y
tangent slope=-5/2
(b)
Unit 1Fig. 9.4
radius
33Thursday, September 12, 2013
Geometry of common power-law potentials Geometric (Power) series
“Zig-Zag” exponential geometryProjective or perspective geometry
Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields
Compare mks units of Coulomb Electrostatic vs. Gravity
34Thursday, September 12, 2013
x=2.0x=0.5 x=1(0,0)
-0.5
-1
-2
-3
-4
--11//xx
--11//xx22
--11//xx
--11//xx22
2.00.5 x=1(0,0)
-1
-2
-3
-4
UU((xx))==--11//xx
--11//xx
--11//xx22
Step1 : Follow line from origin (0,0)
through (x,-1) intercept to (+1,-1/x) intercept.
Transfer laterally to draw (x,-1/x) point.
Step2 : Follow line from origin (0,0)
through (x,-1/x) point to (+1,-1/x2) intercept.
Transfer laterally to draw (x,-1/x2) point.
FF((xx))==--11//xx22
FF((11))
FF((00..55))
FF((00..77))
FF((xx))
11
Δxx
−ΔUU
UU((xx))
Δxx−ΔUUFF((xx))
11==
______
Step3 :(Optional) Display Force vector
using similar triangle constuction based
on the slope of potential curve.
Unit 1Fig. 9.4
Coulomb geometryForce and Potential F(x)=-1/r2 U(x)=-1/r
x1
1/x
1/x2
x:1=1:(1/x)
1x:(1/x)=1:(1/x2)
35Thursday, September 12, 2013
x=2.0x=0.5 x=1(0,0)
-0.5
-1
-2
-3
-4
--11//xx
--11//xx22
--11//xx
--11//xx22
2.00.5 x=1(0,0)
-1
-2
-3
-4
UU((xx))==--11//xx
--11//xx
--11//xx22
Step1 : Follow line from origin (0,0)
through (x,-1) intercept to (+1,-1/x) intercept.
Transfer laterally to draw (x,-1/x) point.
Step2 : Follow line from origin (0,0)
through (x,-1/x) point to (+1,-1/x2) intercept.
Transfer laterally to draw (x,-1/x2) point.
FF((xx))==--11//xx22
FF((11))
FF((00..55))
FF((00..77))
FF((xx))
11
Δxx
−ΔUU
UU((xx))
Δxx−ΔUUFF((xx))
11==
______
Step3 :(Optional) Display Force vector
using similar triangle constuction based
on the slope of potential curve.
Unit 1Fig. 9.4
Coulomb geometryForce and Potential F(x)=-1/r2 U(x)=-1/r
x1
1/x1/x2
x:1=1:(1/x)
1x:(1/x)=1:(1/x2)
36Thursday, September 12, 2013
Geometry of common power-law potentials Geometric (Power) series
“Zig-Zag” exponential geometryProjective or perspective geometry
Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields
Compare mks units of Coulomb Electrostatic vs. Gravity
37Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
38Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
More precise value for electrostatic constant : 1/4πε0=8.987,551·109Nm2/C2 ~9·109~1010
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+)
quantum of charge: |e|=1.6022·10-19 Coulomb
...but 1 Ampere = 1 Coulomb/sec.
39Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
More precise value for electrostatic constant : 1/4πε0=8.987,551·109Nm2/C2 ~9·109~1010
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+)
quantum of charge: |e|=1.6022·10-19 Coulomb
...but 1 Ampere = 1 Coulomb/sec.
“Fingertip Physics” of Ch. 9 notes that 1 (cm)3 of water (1/38 Mole) has (1/38) 6·1023 molecules (about 6·1023 electrons) That’s about 6·10231.6022·10-19 Coulombor about 10+5 C or 100,000 Coulomb
40Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
Fgrav.(r) = −G mMr2 where :G = 0.000,000,000,067 Newtons ⋅meter ⋅ square
per square Coulomb
2. Gravitational force between m(kilograms) and M(kg.)
More precise value for electrostatic constant : 1/4πε0=8.987,551·109Nm2/C2 ~9·109~1010
More precise value for gravitational constant : G=6.67384(80)·10-11Nm2/C2 ~(2/3)10-10
COMPARE!BIGvssmall
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+)vsAlways Attractive (so far)
quantum of charge: |e|=1.6022·10-19 Coulomb
41Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+) Discussion of repulsive force and PE in Ch. 9...
1(a). Electrostatic potential energy between q(Coulombs) and Q(C.)
U(r) = 14πε0
qQr
where : 14πε0
≅ 9,000,000,000 Jouleper square Coulomb
quantum of charge: |e|=1.6022·10-19 Coulomb
42Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+) Discussion of repulsive force and PE in Ch. 9...
1(a). Electrostatic potential energy between q(Coulombs) and Q(C.)
U(r) = 14πε0
qQr
where : 14πε0
≅ 9,000,000,000 Jouleper square Coulomb
Nuclear size ~ 10-15 m = 1 femtometer =1fm Atomic size ~ 1 Angstrom = 10-10 m Big molecule ~ 10 Angstrom = 10-9 m = 1nanometer=1nm
quantum of charge: |e|=1.6022·10-19 Coulomb
H-atom diameter
1 A=10-1nm
43Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+) Discussion of repulsive force and PE in Ch. 9...
1(a). Electrostatic potential energy between q(Coulombs) and Q(C.)
U(r) = 14πε0
qQr
where : 14πε0
≅ 9,000,000,000 Jouleper square Coulomb
Nuclear size ~ 10-15 m = 1 femtometer =1fm Atomic size ~ 1 Angstrom = 10-10 m Big molecule ~ 10 Angstrom = 10-9 m = 1nanometer=1nm
quantum of charge: |e|=1.6022·10-19 Coulomb
H-atom diameter
1 A=10-1nm
44Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+) Discussion of repulsive force and PE in Ch. 9...
1(a). Electrostatic potential energy between q(Coulombs) and Q(C.)
U(r) = 14πε0
qQr
where : 14πε0
≅ 9,000,000,000 Jouleper square Coulomb
Nuclear size ~ 10-15 m = 1 femtometer =1fm Atomic size ~ 1 Angstrom = 10-10 m Big molecule ~ 10 Angstrom = 10-9 m = 1nanometer=1nm
also:1fm = 10 -13 cm =1Fermi =1Fm
1 Fermi
quantum of charge: |e|=1.6022·10-19 Coulomb
H-atom diameter
1 A=10-1nm
45Thursday, September 12, 2013
Felec.(r) = ± 14πε0
qQr2 where : 1
4πε0
≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb
Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)
Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+) Discussion of repulsive force and PE in Ch. 9...
1(a). Electrostatic potential energy between q(Coulombs) and Q(C.)
U(r) = 14πε0
qQr
where : 14πε0
≅ 9,000,000,000 Jouleper square Coulomb
Nuclear size ~ 10-15 m = 1 femtometer =1fm Atomic size ~ 1 Angstrom = 10-10 m Big molecule ~ 10 Angstrom = 10-9 m = 1nanometer=1nm
also:1fm = 10 -13 cm =1Fermi =1Fm
1 Fermi
nuclear radii are 100,000 to 1,000,000 times smaller than atomic/chemical radii...so nuclear qQ/r energy 100,000 to 1,000,000 times bigger that of atomic/chemical...
quantum of charge: |e|=1.6022·10-19 Coulomb
H-atom diameter
1 A=10-1nm
46Thursday, September 12, 2013
Geometry of idealized “Sophomore-physics Earth” Coulomb field outside Isotropic Harmonic Oscillator (IHO) field insideContact-geometry of potential curve(s)“Crushed-Earth” models: 3 key energy “steps” and 4 key energy “levels”
Earth matter vs nuclear matter: Introducing the “neutron starlet” and “Black-Hole-Earth”
47Thursday, September 12, 2013
d
D
You areHere!
Shell mass element
Shell mass elementM =(solid-angle factor A)D2
Gravity at rdue to shell mass elementsG M - G mD2 d2
D2 - d2
D2 d2( )A = 0
B
Θ
Θ
r
m =(solid-angle factor A) d2
M
m
=(...andweightless!)
You areHere!
OdΩsinΘ
A=
Coulomb force vanishes inside-spherical shell (Gauss-law)
Coulomb force inside-spherical body due to stuff below you, only.
M<
Gravitational force at r< isjust that of planet below r<
r<
M<
Unit 1Fig. 9.6
Cancellation of
48Thursday, September 12, 2013
d
D
You areHere!
Shell mass element
Shell mass elementM =(solid-angle factor A)D2
Gravity at rdue to shell mass elementsG M - G mD2 d2
D2 - d2
D2 d2( )A = 0
B
Θ
Θ
r
m =(solid-angle factor A) d2
M
m
=(...andweightless!)
You areHere!
OdΩsinΘ
A=
Coulomb force vanishes inside-spherical shell (Gauss-law)
Coulomb force inside-spherical body due to stuff below you, only.
M<
Gravitational force at r< isjust that of planet below r<
r<
M<
F inside(r< ) = GmM <
r<2 = Gm 4π
3
M <
4π
3r<3r< = Gm
4π
3ρ⊕r< = mg
r<R⊕
≡ mg ⋅ x
Note:Hooke’s (linear) force law for r< inside uniform body
Unit 1Fig. 9.6
Cancellation of
49Thursday, September 12, 2013
d
D
You areHere!
Shell mass element
Shell mass elementM =(solid-angle factor A)D2
Gravity at rdue to shell mass elementsG M - G mD2 d2
D2 - d2
D2 d2( )A = 0
B
Θ
Θ
r
m =(solid-angle factor A) d2
M
m
=(...andweightless!)
You areHere!
OdΩsinΘ
A=
Coulomb force vanishes inside-spherical shell (Gauss-law)
Coulomb force inside-spherical body due to stuff below you, only.
M<
Gravitational force at r< isjust that of planet below r<
r<
M<
F inside(r< ) = GmM <
r<2 = Gm 4π
3
M <
4π
3r<3r< = Gm
4π
3ρ⊕r< = mg
r<R⊕
≡ mg ⋅ x
Note:Hooke’s (linear) force law for r< inside uniform body
Earth surface gravity acceleration: g = G M⊕
R⊕2 = G M⊕
R⊕3 R⊕ = G 4π
3
M⊕
4π
3R⊕
3R⊕ = G 4π
3ρ⊕R⊕ = 9.8m / s
G=6.67384(80)·10-11Nm2/C2 ~(2/3)10-10
Unit 1Fig. 9.6
Cancellation of
50Thursday, September 12, 2013
d
D
You areHere!
Shell mass element
Shell mass elementM =(solid-angle factor A)D2
Gravity at rdue to shell mass elementsG M - G mD2 d2
D2 - d2
D2 d2( )A = 0
B
Θ
Θ
r
m =(solid-angle factor A) d2
M
m
=(...andweightless!)
You areHere!
OdΩsinΘ
A=
Coulomb force vanishes inside-spherical shell (Gauss-law)
Coulomb force inside-spherical body due to stuff below you, only.
M<
Gravitational force at r< isjust that of planet below r<
r<
M<
F inside(r< ) = GmM <
r<2 = Gm 4π
3
M <
4π
3r<3r< = Gm
4π
3ρ⊕r< = mg
r<R⊕
≡ mg ⋅ x
Note:Hooke’s (linear) force law for r< inside uniform body
Earth surface gravity acceleration: g = G M⊕
R⊕2 = G M⊕
R⊕3 R⊕ = G 4π
3
M⊕
4π
3R⊕
3R⊕ = G 4π
3ρ⊕R⊕ = 9.8m / s
Earthradius :R⊕ = 6.371⋅106m 6.4 ⋅106m
Solarmass :M = 1.9889 × 1030 kg. 2.0 ⋅1030 kg. Earthmass :M⊕ = 5.9722 × 1024 kg. 6.0 ⋅1024 kg. Solar radius :R = 6.955 × 108m. 7.0 ⋅108m.
G=6.67384(80)·10-11Nm2/C2 ~(2/3)10-10
Unit 1Fig. 9.6
Cancellation of
51Thursday, September 12, 2013
Geometry of idealized “Sophomore-physics Earth” Coulomb field outside Isotropic Harmonic Oscillator (IHO) field insideContact-geometry of potential curve(s)“Crushed-Earth” models: 3 key energy “steps” and 4 key energy “levels”
Earth matter vs nuclear matter: Introducing the “neutron starlet” and “Black-Hole-Earth”
52Thursday, September 12, 2013
Example of contacting lineand contact point
directrixdistance
Directrix
Sub-directrix
focal distance =
2.00.5 x=1(0,0)
-1
-2
UU((xx))==--11//xx
FF((11..00))
FF((xx))==--11//xx22((oouuttssiiddee EEaarrtthh))
FF((xx))==-- xx((iinnssiiddee EEaarrtthh))
FF((11..44))FF((22..00))
FF((22..88))
FF((00..88))
FF((00..44))
FF((--11..00))FF((--11..44))
FF((--22..00))
FF((--00..88))
FF((--00..44))
-0.5-1.5
Directrix
Latusrectum
λ
Focus
UU((xx))==((xx22--33))//22
Parabolic potentialinside Earth
Unit 1Fig. 9.7
The ideal “Sophomore-Physics-Earth” model of geo-gravity
53Thursday, September 12, 2013
...conventional parabolic geometry...carried to extremes…
(Review of Lect. 6 p.29)
p=λ/2
y =-p
0
p
2p
3p
4p
x =0 p 2p 3p 4p
Parabola4p·y =x2=2λ·y
pdirectrix
Latus rectumλ =2p
slope=1tangent slope=-2
Circleofcurvatureradius=λ
(a)
p
p
slope=1/2y =-p
0
p
2p
3p
4p
x =0 p 2p 3p 4p
Parabola4p·y =x2=2λ·y
tangent slope=-5/2
(b)
Unit 1Fig. 9.4
radius
54Thursday, September 12, 2013
Geometry of idealized “Sophomore-physics Earth” Coulomb field outside Isotropic Harmonic Oscillator (IHO) field insideContact-geometry of potential curve(s)“Crushed-Earth” models: 3 key energy “steps” and 4 key energy “levels”
Earth matter vs nuclear matter: Introducing the “neutron starlet” and “Black-Hole-Earth”
55Thursday, September 12, 2013
surface gravity: g = −G M⊕
R⊕2
surface potential: PE= −G M⊕
R⊕
surface escape speed: ve= 2G M⊕
R⊕
KE = PE relation: 12mv2
e=GmM⊕
R⊕
-orbit energy: E = −G M⊕
2R⊕
-orbit speed: v = G M⊕
R⊕
The “Three (equal) steps from Hell”
1
2
3
Ground level : PE= −G 3M⊕
2R⊕
Surface level : PE= −G M⊕
R⊕
Orbit level : PE= −G M⊕
2R⊕
Dissociation threshold : PE= 0
56Thursday, September 12, 2013