class progress - smu

1Physics 3340 - Fall 2017

Class Progress

Basics of Linux, gnuplot, CVisualization of numerical dataRoots of nonlinear equations (Midterm 1)Solutions of systems of linear equationsSolutions of systems of nonlinear equationsMonte Carlo simulationInterpolation of sparse data pointsNumerical integration (Midterm 2)Solutions of ordinary differential equations


Helmholtz Coil for Uniform Magnetic Field


Magnetic Field Along Axis of a Single Coil

y

z

x

R = coil radius

current I

R

Bz(z) = z component of magnetic

field at coordinate z on the axis of symmetry

B z (z) =μ0 I R2

2(R2+z2)32


Magnetic Field on Axis of a Helmholtz Coil

y

z

x

R = coil radiusD = z distance of the plane of each coil from the z=0 planeD = R/2 for best uniform fields

D

Dcurrent I

R

(Region of approximately uniform magnetic field in z direction)

current I


Magnetic Field on Axis of a Helmholtz CoilR=10cm D=5cm I=1A

0

2x10-6

4x10-6

6x10-6

8x10-6

1x10-5

1.2x10-5

-0.1 -0.05 0 0.05 0.1

Z c

om

po

ne

nt o

f ma

gn

etic

fie

ld (

T)

Z coordinate (m)

Both coilsLower coilUpper coil



0

2x10-6

4x10-6

6x10-6

8x10-6

1x10-5

1.2x10-5

-0.1 -0.05 0 0.05 0.1

Z c

om

po

ne

nt o

f ma

gn

etic

fie

ld (

T)

Z coordinate (m)



Two-Dimensional Wave Diffraction Patterns

y

z

x

z=-D0

z=0

z=D1

Aperture

ObservationScreen

R

(x,y)

r1

θ1

(x1,y

1)

Point source

d1

d0

(x0,y

0)


Review of PhasorsConsider any physical system described by a linear second order ODE with constant coefficients, driven by a sinusoid at a fixed frequency and a reference phase:

−ω2 Aei ϕ+ a iω Ae iϕ+ b Aei ϕ=c

Plug prototype solution back into ODE:

dx2

dt 2+ a dx

dt+ b x=c eiω t

Look only for a solution in the form of a steady state sinusoid with unknown amplitude and phase. Form a prototype solution of the form:

x=Aei (ω t−ϕ)=Aei ϕeiω t

−ω2 Aei ϕe iω t+ a iω Ae iϕeiω t+ b Aei ϕe iω t=c e iω t

Cancel eiωt factors:

Second order ODE has been transformed into a complex algebraic equation AeiΦ is called the “phasor” of the solution x


Phasor Method for Wave Interference

Point source 2

Observation point

Point source 1

E field at observation point=∑ E j from point source jAssume uniform polarization,so for instance E j=E jx xE jx=E j0 e

i t−=E j0 eiei t

E j0 ei is the 'phasor' from source j

is due to path length differencesd

2

d1


Diffraction Pattern from Single Point Source

R=1D

1=20

D0=20

λ=0.2x

0=0

y0=0



R=1D

1=20

D0=20

λ=0.2x

0=3

y0=0



R=1D

1=20

D0=20

λ=0.2x

0=-3

y0=0


Diffraction Pattern from Two Point Sources

R=1D

1=20

D0=20

λ=0.2x

01=-3

y01

=0

x02

=3

y02

=0



R=1D

1=20

D0=20

λ=0.2x

01=-1

y01

=0

x02

=1

y02

=0



R=1D

1=20

D0=20

λ=0.2x

01=-2

y01

=0

x02

=2

y02

=0


Simulation of the Poisson Spot

RDISC

=2D

1=20

D0=20

λ=0.2x

0=0

y0=0


One-Dimensional Wave Diffraction Patterns

y

x

y=-D0

x=W/2

y=D1

Aperture

ObservationScreen

x

x1

Point source

d1

d0

x0

y=0

x=0

x=-W/2


1-Dimensional Diffraction Pattern from Single Point Source

W=1D

1=20

D0=20

λ=0.2x

0=0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-10 -5 0 5 10

Re

lativ

e In

ten

sity

X-coordinate



W=1D

1=20

D0=20

λ=0.2x

0=-5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-10 -5 0 5 10

Re

lativ

e In

ten

sity

X-coordinate



W=1D

1=20

D0=20

λ=0.2x

0=5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-10 -5 0 5 10

Re

lativ

e In

ten

sity

X-coordinate


1-Dimensional Diffraction Pattern from Two Point Sources

W=1D

1=20

D0=20

λ=0.2x

01=5

x02

=-5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

-10 -5 0 5 10

Re

lativ

e In

ten

sity

X-coordinate



W=1D

1=20

D0=20

λ=0.2x

01=2

x02

=-2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

-10 -5 0 5 10

Re

lativ

e In

ten

sity

X-coordinate



W=1D

1=20

D0=20

λ=0.2x

01=3

x02

=-3

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-10 -5 0 5 10

Re

lativ

e In

ten

sity

X-coordinate


Plotting Vectors with gnuplot

plot 'efield1.dat' using 1:2 with lines

Column with Y coordinate

Column with X coordinate

plot 'efield1.dat' using 1:2:3:4 with vectors

Column with starting Y coordinate

Column with starting X coordinate

Column with Y distance

Column with X distance

plot 'efield1.dat' using 1:2 with points

Another plot mode:

We have used:


Electric Field Vectors for Dipole

-25

-20

-15

-10

-5

0

5

10

15

20

25

-25 -20 -15 -10 -5 0 5 10 15 20 25


Electric Field Vectors for Quadrapole

-25

-20

-15

-10

-5

0

5

10

15

20

25

-25 -20 -15 -10 -5 0 5 10 15 20 25


Electric Field Vectors for Example Arbitrary Charges

-25

-20

-15

-10

-5

0

5

10

15

20

25

-25 -20 -15 -10 -5 0 5 10 15 20 25


C Code for Mapping Electric Field Vectors

xstop = xstop + xinc * 0.5; ystop = ystop + yinc * 0.5; rtest = 1.0e-6 * (xinc*xinc + yinc*yinc); x = xstart; while (((xinc > 0.0) && (x < xstop)) || ((xinc < 0.0) && (x > xstop))) { y = ystart; while (((yinc > 0.0) && (y < ystop)) || ((yinc < 0.0) && (y > ystop))) { ex = 0.0; ey = 0.0; i = 0; while (i < n) { rsq = (x-xi[i])*(x-xi[i]) + (y-yi[i])*(y-yi[i]); if (rsq < rtest) break; r = sqrt(rsq); ex += qi[i] * (x-xi[i]) / (r * rsq); ey += qi[i] * (y-yi[i]) / (r * rsq); i++; } if (i >= n) { emag = sqrt(ex*ex + ey*ey); vx = xinc * ex / emag; vy = yinc * ey / emag; printf("%.8g %.8g %.8g %.8g\n",x-0.5*vx,y-0.5*vy,vx,vy); } y = y + yinc; } x = x + xinc; }

Loop over n charges.double array qi[] hold each charge q

i and

double arrays xi[] and yi[] hold x

i and y

i coordinates


Classical Electric Field Lines

Note that the vectors plotted by this method only record the direction of the E field vector at an array of sampled points. These are not quite equivalent to the classical field lines, which convey both the direction of the field and its magnitude from the density of drawn lines.


Classical Electric Equipotential Lines

A simple modification of this algorithm can be used to plot a sampling of the direction of the electric equipotential lines. At a given point, field lines and equipotential lines will have slopes that are negative reciprocals of each other. Again, no magnitude information is present, only direction.


Equipotential Vectors for Example Arbitrary Charges

-25

-20

-15

-10

-5

0

5

10

15

20

25

-25 -20 -15 -10 -5 0 5 10 15 20 25


C Code for Mapping Equipotential Vectors

xstop = xstop + xinc * 0.5; ystop = ystop + yinc * 0.5; rtest = 1.0e-6 * (xinc*xinc + yinc*yinc); x = xstart; while (((xinc > 0.0) && (x < xstop)) || ((xinc < 0.0) && (x > xstop))) { y = ystart; while (((yinc > 0.0) && (y < ystop)) || ((yinc < 0.0) && (y > ystop))) { ex = 0.0; ey = 0.0; i = 0; while (i < n) { rsq = (x-xi[i])*(x-xi[i]) + (y-yi[i])*(y-yi[i]); if (rsq < rtest) break; r = sqrt(rsq); ex += qi[i] * (x-xi[i]) / (r * rsq); ey += qi[i] * (y-yi[i]) / (r * rsq); i++; } if (i >= n) { emag = sqrt(ex*ex + ey*ey); vx = -0.5 * xinc * ey / emag; vy = 0.5 * yinc * ex / emag; printf("%.8g %.8g %.8g %.8g\n",x-0.5*vx,y-0.5*vy,vx,vy); } y = y + yinc; } x = x + xinc; }

Only these two statements change!


Examples of Optical Glare

(Vacuum tube arithmetic unit and magnetic drum memory from the Univac 1, on display at the Deutches Museum of Technology, Munich, Germany)

Glare is the reflected room light off of an intervening plexiglass protective sheet.Reflections occur at the interface between two materials of different refractive indices.


Computing Another Kind of Glare: a Rainbow


Rene Descartes Graphical Derivation

“I took my pen and made an accurate calculation of the paths of the rays which fall on the different points of a globe of water to determine at which angles, after two refractions and one or two reflections they will come to the eye, and I then found that after one reflection and two refractions there are many more rays which can be seen at an angle of from forty-one to forty-two degrees than at any smaller angle; and that there are none which can be seen at a larger angle"


Calculating Reflection and Refraction Angles

θi1

θr1

θ12

θn2

θ23

θ12

θi2

θr2

(x0,y

0)

θ23

θi3

θn3θ

out

θr3

(x2,y

2)

(x1,y

1)

(x3,y

3)

Spherical water raindrop cross section, index of refraction n

Output ray

Horizontal input ray

Radius r

(0,0)


Calculating Reflection and Refraction Angles

x1=−√r2− y02 y1=y 0 θi1=arctan (∣y1

x1∣)

Snell's law: θr1=arcsin ( 1n

sin (θi1))θ12=θi1−θr1 m12= tan (−θ12)

x 2=x1(m12

2 −1)−2m12 y1

m122 +1

y 2= y1+m12(x2−x1)

θn2=arctan ( y2

x2) θi2=θn2+θ12 θ23=θn2+θi2=2θn2+θ12 m23= tan (θ23)

x3=x2(m23

2 −1)−2m23 y2

m232 +1

y3= y2+m23(x3−x 2)

θn3=arctan ( y3

x3) θi3=θn3−θ23

Snell's law: θr3=arcsin (n⋅sin (θi3))θout=θn3−θr3


Reflection and Refractions through a Raindrop

Incoming violet light λ=400nm

Reflected light rays to observer on ground

Spherical raindrop of water, n=1.339at λ=400nm

Many reflection paths bundled at an angle of about 40˚ below horizontal

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2


Reflection and Refractions through a Raindrop

Incoming red light λ=700nm

Reflected light rays to observer on ground

Spherical raindrop of water, n=1.331at λ=700nm

Many reflection paths bundled at an angle of about 42˚ below horizontal

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2


1.33 1.331

1.332 1.333

1.334 1.335

1.336 1.337

1.338 1.339

Refractive index

0 0.1

0.2 0.3

0.4 0.5

0.6 0.7

0.8 0.9

1Incident Y coordinate

0

5

10

15

20

25

30

35

40

45

Ou

tpu

t an

gle

(d

eg

ree

s )

Output Angle Relative to Horizontal

Angle reaches a maximum, which concentrates much of the output energy around 40˚ - 42˚

λ=700nm (red)

λ=400nm (violet)


Total Effect of Many Raindrops is a Rainbow

Observer

Red light appears on the top, violet light appears on the bottom

Incident white sunlight

Many raindrops in rainfall


Wave Packets for Quantum Mechanics

d2 ψ(x)d x2 =−2m

ℏ2 [E−V (x) ] ψ(x)

Solutions of the time-independent Schrödinger equation in one dimension

in regions where particles are free to move and not subject to forces, or

V (x)=0 and E>0are of the form

ψ(x)=A e+i(kx−ω t )+Be−i(kx+ω t)

ψ(x)=A e+ikx+Be−ikx

which makes the complete solution including the time dependence

where k= pℏ

and ω= Eℏ

so the relationship between ω and k is

ω=ℏ

2mk2


Wave Packets for Quantum Mechanics

ψ(x)=∫ A (k )ψk (x)dk

A single wave solution is associated with a continuous flux of particles in the positive or negative x direction, with momentum p and energy E.

Solutions that are linear combinations of multiple waves form wave packets that are associated with single particles.

A (k )=e−a2(k−k0)2

Look at Gaussian wave packets where

ψ(x)=∑k

Ak ψk (x)

Or in the continuous limit

Ak=e−a2(k−k0)2

Or in the continuous limit


Distribution of Wave Numbers

0

0.2

0.4

0.6

0.8

1

40 45 50 55 60

A(k

)

k

Gaussian wavepacket

a=1a=2

a=0.5


Corresponding Spatial Wave Distribution

0

200

400

600

800

1000

1200

1400

-15 -10 -5 0 5 10 15

Psi

(X)

X

Gaussian wavepacket

a=1a=2

a=0.5


Phase and Group Velocities

The phase velocity vp of a single wave with particular values of k and ω is the velocity of

the point in space that maintains a constant sinusoidal phase

v g =2ℏ2m

k0 =ℏm

p0

ℏ=

p0

m

The group velocity vg of a wave packet is the velocity of the peak of spatial distribution

v p = ωk

=ℏ

2mk

Where vp0

is the phase velocity of the single wave at the center of the k distribution

v g = d ωd k

|(k=k0) =2ℏ2m

k0 = 2v p0

Note:

So it is the wave packet group velocity that corresponds to the velocity p/m in the classical limit


Discrete Approximation to Gaussian Packet

-20

-15

-10

-5

0

5

10

15

20

-15 -10 -5 0 5 10 15

Psi

(X)

X

Gaussian wavepacket, a=1

RealImaginary


Why Do Wave Packets Form a Pulse?

Consider the superposition of just two sine waves of different frequencies. They form a “beat” frequency where constructive and destructive interference alternate:

Wave packets are extensions of this property, with the sum over many frequencies. The region where all the waves line up with constructive interference becomes isolated to a pulse.


Quality of Discrete Approximations

0

10

20

30

40

50

60

-15 -10 -5 0 5 10 15

Ma

gn

itud

e S

qu

are

d o

f Psi

( x)

X

Gaussian Wave Packet, n=number of component waves

n=1n=3n=5n=7n=9

n=11n=13n=17


Group Velocity versus Phase Velocity

-20

-15

-10

-5

0

5

10

15

20

-10 -5 0 5 10 15 20

Psi

(X)

X

Gaussian wavepacket, a=1, time=5, phase velocity=1

RealImaginary


Uncertainty Principle of Wave Packets

Δ k2=[k−k ]2=∫ A (k)2[k−k ]2 dk

∫ A (k)2 dk

Δ x2=[x−x ]2=∫ψ(x)2[x−x ]2 dx

∫ψ(x)2dx

then Δ x2⋅Δ k2≥14

class progress - smu

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