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Fermi Acceleration-- A real scientific problem

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James A. Rome

Tennessee Governor's Academy

August 2010

- Enrico Fermi postulated an acceleration method for cosmic rays that has a simple physical model. Imagine a ball bouncing between two massive walls, one of which oscillates back and forth.

The question to be

solved is whether

the ball can be

accelerated to very

large velocities

u

V(t)

Top View

- http://www.dynamical-systems.org/fermi/info.html
- and you can see an animation at
- http://www.dynamical-systems.org/fermi/index.html
- http://books.google.com/books?id=a0AABe1Hve8C&lpg=PA10&dq=velocity%20of%20ball%20bouncing%20off%20moving%20wall&pg=PA10#v=onepage&q=velocity%20of%20ball%20bouncing%20off%20moving%20wall&f=false
- http://en.wikipedia.org/wiki/Fermi_acceleration
- http://en.wikipedia.org/wiki/Fermi%E2%80%93Ulam_model

- Assumptions:
- The ball is a point located at its center
- The ball is perfectly elastic (think superball)
- The Wall is infinitely massive and absorbs no energy in the collision
- The collision is instantaneous

- If energy (mv2/2) is conserved, and if the wall does not move, the speed after the collision must equal the speed before the collision, but the direction is reversed.
- The ball bounces off the wall with velocity = -u
- This is in accord with your experiences.
- Between collisions, the ball moves at a constant velocity (with no external forces)

Wall

u

?

- How do you solve this problem?
- First suppose the ball is sitting still, and it is hit by the wall that moves at V towards the ball
- If you ever played ping pong or tennis, you will know that the ball goes to the right—but how fast?

Wall

V

u = 0

After

u = ?

- Einstein said that physics must be the same when viewed from any system moving at a constant velocity.
- Example: You sit in your moving car and toss a ball up. To you it looks just like it would if you were not in the car. The ball goes straight up and straight down.
- But to someone on the ground, the ball will move in a parabola.
- So what do you see if you sit on the wall and solve this problem?

Wall

?

V

- You see the ball approaching you with velocity V
- I think we have already solved this problem. . .

Wall

V

?

- You see the ball approaching you with velocity V
- I think we have already solved this problem. . .
- The ball bounces off the wall with velocity V in the opposite direction
- But the wall is already moving to the right at V
- So in the Lab frame, the ball goes to the right at 2V
- That is why Fermi thought you could make it get faster!

Wall

V

?

- You see the ball approaching you with velocity V
- I think we have already solved this problem. . .
- The ball bounces off the wall with velocity V in the opposite direction
- But the wall is already moving to the right at V
- So in the Lab frame, the ball goes to the right at 2V
- That is why Fermi thought you could make it get faster!

Wall

V

?

If the ball is moving towards the wall with

speed u, after the collision it moves right

with speed u + 2V

The ball gains energy because the wallis moving, and we assumed it wasinfinitely massive. Whatever is movingthe wall puts energy into the ball.

- A light ball bounces in one dimension between two massive walls. One wall oscillates with a velocity given by
- V(t) = (V/4)[1 - 2{t}]
- where {t} is the fractional part of t (denoted by ft in the code).
- V(t)
- |
- |
- V/4|\ |\ |\ |\ |\ |\
- | \ | \ | \ | \ | \ | \ |
- --|--\--|--\--|--\--|--\--|--\--|--\--|- t
- |0 \ |1 \ |2 \ |3 \ |4 \ |5 \ |6
- | \| \| \| \| \| \|
- |

- The velocity is the time rate of change of the distance (m/s) which means the velocity is the slope of the position of the wall.
- The position is the area under the velocity curve up to time t

- area of little triangle =
- (1/2)*(1/2-{t})*V({t})=
(V/4)[1/2 - {t}]2

V(t)

V/4

t

1

{t}

distance =

area of trapezoid =

area of big triangle -

area of little triangle =

(V/16) – (V/4)[1/2 - {t}]2

-V/4

{t=1/2} area = base * height/2 = (1/2)*(1/2)*(V/4) = V/16 = max oscillation

- The velocity is the time rate of change of the distance (m/s) which means the velocity is the slope of the position of the wall.
- The position is the area under the velocity curve up to time t

- area of little triangle =
- (1/2)*(1/2-{t})*V({t})=
(V/4)[1/2 - {t}]2

V(t)

V/4

For {t} > 1/2, the area under the curve is

the area of the positive triangle – area of

the dotted triangle (note that V < 0) =

V/16 + (1/2)[{t} – (1/2)]*(V/4)[1 – 2{t}] =

V/16 - (V/4)[{t} - (1/2)]2

t

1

{t}

distance =

area of trapezoid =

area of big triangle -

area of little triangle =

(V/16) – (V/4)[1/2 - {t}]2

-V/4

{t=1/2} area = base * height/2 = (1/2)*(1/2)*(V/4) = V/16 = max oscillation

- The velocity is the time rate of change of the distance (m/s) which means the velocity is the slope of the position of the wall.
- The position is the area under the velocity curve up to time t

- area of little triangle =
- (1/2)*(1/2-{t})*V({t})=
(V/4)[1/2 - {t}]2

V(t)

V/4

For {t} > 1/2, the area under the curve is

the area of the positive triangle – area of

the dotted triangle=

V/16 – (1/2)[{t} – (1/2)]*(V/4)[1 – 2{t}] =

V/16 + (V/4)[{t} - (1/2)]2

t

1

{t}

distance =

area of trapezoid =

area of big triangle -

area of little triangle =

(V/16) – (V/4)[1/2 - {t}]2

Sanity check:

d = 0 at {t} = 0, 1

d = V/16 at {t} = ½

It is two parabolas!

-V/4

{t=1/2} area = base * height/2 = (1/2)*(1/2)*(V/4) = V/16 = max oscillation

- Most of the time, the ball is bouncing from the bouncing wall to the fixed wall and back again.
- It will maintain its speed during this process
- The speed changes only when it hits the bouncing wall.

- The velocity of the ball just before the nth collision with the oscillating wall is (un = vball/V)
- The ball leaves the collision with its initial velocity (-un) + 2*Vwall({t})/V
- and remember that Vwall(t) = (V/4)[1 - 2{t}]
- speed = |un + 2*(1/4)[2{tn} – 1]| =>
- un+1 = |un + {tn} – 1/2|

- If we assume the distance between the walls (d) is much greater than the amplitude of the oscillation (V/16), the ball goes a distance 2d between bounces with the moving wall. distance = speed * time, so, remembering that Vun+1 i the speed just before the n+1 collision,
- tn+1 = tn + 2d/(Vun+1) = tn + M/un+1
- where M = 2d/V, and
- {tn+1}= {{tn} + M/un+1}
- So, what do we need to plot?
- u vs {t}
- u vs iteration

- Can we predict the output?
- It is chaotic, so NO

- So, what do we need to plot?

PhaseCanvas

TimeCanvas

How do we do this using Java?

- Make a New Java Desktop Application
- Call it FermiPlot
- As before, remove the status panel, the progress bar, the code after initComponents(), and the variables for the progress bar at the end of the code.
- You can change the name of the project and the other information in the resources.
- Save the project.

- Make a new Class called FermiCalc
- Need calculations in separate thread to keep the GUI responsive

public class FermiCalc extends Thread// Run in a separate thread

{

PhaseCanvas pc;// The left plot area

TimeCanvas tc;// The right plot area

private static final int NCURVE = 7;

private float M = 10.0f;

private static final Color[] colors = {Color.black, Color.blue,

Color.cyan, Color.red, Color.green, Color.magenta, Color.orange};

// The constructor to instantiate the class instance

public FermiCalc(PhaseCanvas pc, TimeCanvas tc)

{

// Get and store the two drawing canvases

this.pc = pc;

this.tc = tc;

}

// Other methods go here

}

- A Thread must have a run() method that does the work

public void run(){

float u[] = new float[2]; // Relative velocity before/after collision

float ft[] = new float[2]; // Fractional parts of t

float nstep[] = new float[2]; // Steps (need float for the plot)

/* Pick interesting starting points (NCURVE=7 of them) */

double[] ustart = {M, M, .4*Math.sqrt((double)M),

1.45*M,1.5*M, 2.0*M, M};

double[] tstart = {.1, .4, .5, .5, 0.0, .5, 0.0};

int[] stepmax = {500, 500, 4000, 2000, 4000, 4000,4000, 4000};

for (int i = 0; i < NCURVE; i++)// For each curve

{

u[0] = (float)ustart[i];

ft[0] = (float)tstart[i];

nstep[0] = 0.0f;

for (int n = 1; n < stepmax[i]; n++) // Iterate for each bounce

{

. . .// THE CODE THAT DOES THE WORK GOES HERE (next slide)

}

}

}

nstep[1] = (float)n; // The iteration number. Cast to float

u[1] = Math.abs(u[0] + ft[0] - .5f); // u_n+1 = |u_n + {t_n} - 1/2|

ft[1] = ft[0] + (M/u[1]);// The time of the bounce

while (ft[1] > 1.0f)// Get its fractional part

ft[1] -= 1.0f; // could use ft[1] - (float)Math.floor((double)ft[1])

// Phase plot

pc.symbol(ft[1], u[1], colors[i]);// Plot a symbol at the new point

// Time plot

tc.curve(nstep, u,colors[i]);// Plot line between the points

// Prepare for the next iteration

u[0] = u[1];

ft[0] = ft[1];

nstep[0] = nstep[1];

// add a delay so we can watch the plot happening

try { // This is new. Some methods throw an exception if a problem

Thread.sleep(5); // In this case, sleep throws one when its over

// 5 millisecs

}

catch(InterruptedException ex) {

}

Remember that the two canvases

are Swing components and that Swingis processing many events. It is a bad idea to call methods in Swingfrom outside the EventQueue.

See next slide for the correct way.

What happens if ft, u changebefore they get plotted?

Can only pass final variables intoinvokeLater so that the things wepass in do not change

- final float u1 = u[1];
- final float ft1 = ft[1];
- final int i1 = i;
- // Phase plot
- EventQueue.invokeLater(new Runnable() {
- public void run() {
- pc.symbol(ft1, u1, colors[i1]);
- } // end of run()
- } ); //end of call
- // Time plot
- final float uf[] = {u[0], u[1]};
- final float nstepf[] = {nstep[0], nstep[1]};
- EventQueue.invokeLater(new Runnable() {
- public void run() {
- tc.curve(nstepf, uf,colors[i1]);
- }
- });

Note that we are defininga new anonymous class(we never need it's name)to spawn a Thread thatwill sit on the Swing eventqueue until Swing isready to process it.

We must implement therun() method

There is alsoEventQueue.invokeAndWait(new Runnable()…)

for when we want our call processed before continuing

- The strategy is to make a class that extends JPanel (one of the Java containers). It is a rectangle that we can draw in.
- We will add PhaseCanvas to the Swing Component palate and then we can add to to our GUI. We can do this because all Swing components are Java Beans.
- All we have to do is to override the paintComponent() method of the JPanel to make our plot.

public class PhaseCanvas extends JPanel {

public PhaseCanvas() {}// Constructor

@Override

public void paintComponent(Graphics g) {

super.paintComponent(g);

// More code to come

}

public void symbol(float x, float y, Color c ){

pts.add(new DataPoint(x, y, c));

this.repaint(); // Force JPanel to be redrawn with this point in it

}

private float M = 10.0f; // Data members of PhaseCanvas class

private Vector<DataPoint> pts = new Vector<DataPoint>();

private int npts = 0;

// Define a new internal class to hold the data of one point

private class DataPoint {

protected float x;// protected variables can be accessed

protected float y;// by classes in this package

protected Color c;

private DataPoint(float x, float y, Color c) {

this.x = x;

this.y = y;

this.c = c;

}

} // End of DataPoint class

}// End of PhaseCanvas class

symbol() is called by FermiCalc

when it has some data to plot.

It puts the data into a DataPoint

and adds it to the end of a

Vector

<DataPoint> tells the compiler

that this vector is composed

of DataPoints. If you try to add

something else, you get a

compiler error.

@Override

public void paintComponent(Graphics g) {

super.paintComponent(g);

Graphics2D g2d = (Graphics2D)g; // cast to Graphics2D which has

int width = this.getWidth(); // many drawing methods

int height = this.getHeight(); // Get the JPanel dimensions

g2d.drawRect(0,0,width-1,height-1); // Outline the panel

AffineTransform t = new AffineTransform(); // Next Slide

t.translate(0, (double) height);

t.scale(1.0, -1.0);

t.scale((double)width, (double)height/(2.0*M));

for(int i = 0; i < pts.size(); i++) { // Draw all the points

DataPoint dp = (DataPoint)pts.elementAt(i);

g2d.setColor(dp.c);

// We need to draw in pixel space, so must use the transform

Point2D.Float inPt = new Point2D.Float(dp.x, dp.y);

Point2D.Float outPt = new Point2D.Float();

t.transform(inPt, outPt);

g2d.drawRect((int)outPt.x -1, (int)outPt.y -1, 1, 1);

}

}

- FermiCalc calculated u in units of M , and {t} goes from 0 to 1. How do we plot this in pixel space?
- We use an affine transform which is some fancy matrix algebra (good reason to do more math!)
- But I can never remember how it works, so I always draw some pictures to help me break it down into easy-to-understand steps

1.0, 2M

0, 0

0.0, 2M

width, 0

PhaseCanvas

FermiCalc

?

0, height

0.0, 0.0

width,height

1.0, 0.0

1.0, 2M

0, 0

0.0, 2M

width, 0

PhaseCanvas

FermiCalc

?

0, height

0.0, 0.0

width,height

1.0, 0.0

scale(width/1.0, height/2M)

translate(0.0, height)

width, -height

width, height

0.0, -height

0.0, height

scale(1.0, -1.0)

0.0, 0.0

width, 0.0

0.0, 0.0

width, 0.0

t.translate(0, (double) height);

t.scale(1.0, -1.0);

t.scale((double)width, (double)height/(2.0*M));

We do the steps backwards

in the code

public class TimeCanvas extends JPanel {

public TimeCanvas() {

}

@Override

public void paintComponent(Graphics g) {

super.paintComponent(g);

// More code to come . . .

}

public void curve(float x[], float y[], Color c) {

linesegs.add(new Lines(x, y, c));

this.repaint(); // Repaint TimeCanvas to display new curve

}

private float M = 10.0f;

Vector<Lines> linesegs = new Vector<Lines>()

// Define a private class to store the plot vectors

private class Lines {

protected Point2D.Float[] p2d = new Point2D.Float[2];

protected Color c;

protected Lines(float x[], float y[], Color c) {

p2d[0] = new Point2D.Float(x[0], y[0]);

p2d[1] = new Point2D.Float(x[1], y[1]);

this.c = c;

}

}

}

curve() is called from FermiCalc

when new iteration s are ready

to be plotted

@Override

public void paintComponent(Graphics g) {

super.paintComponent(g);

Graphics2D g2d = (Graphics2D)g; // cast to Graphics2D

int width = this.getWidth();

int height = this.getHeight();

g2d.drawRect(0,0,width-1,height-1);// Draw a box around the plot

AffineTransform t = new AffineTransform();

t.translate(0, (double) height);

t.scale(1.0, -1.0);

t.scale((double)width/4000.0, (double)height/(2.0*M));

for(int i = 0; i < linesegs.size(); i++) {

Lines lns = (Lines)linesegs.elementAt(i);

g2d.setColor(lns.c);

// We need to draw in pixel space, so must use the transform

Point2D.Float[] inPts = lns.p2d;

Point2D.Float[] outPts = new Point2D.Float[2];

t.transform(inPts, 0, outPts, 0, 2);

// Note that we have to draw in integer coordinates (pixels)

g2d.drawLine((int)outPts[0].x, (int)outPts[0].y,

(int)outPts[1].x, (int)outPts[1].y);

}

}

- First, right-click the FermiPlot project and pick Clean and build (otherwise this will not work) - The palette components are taken from the compiled canvases (in the jar file in the dist directory)
- Click Tools, Palette, Swing/AWT Components

Add components from your project

PhaseCanvas

TimeCanvas

Labels

- In design view, Add the two canvases and all the labels.
- Make sure that the labels are pinned to their canvases, and that the canvases are pinned to the sides and bottom of the mainPanel.
- I used copy and paste to replicate the identical labels
- Save, clean and build, and run your project. Make sure that the labels and windows behave properly when you resize things.

These are the names you called

the two canvases in the GUI builder

public FermiPlotView()

{

initComponents(); // NetBeans code

// instantiate the calculation

FermiCalc fc = new FermiCalc(phaseCanvas, timeCanvas);

fc.setM(10.0f);// I left these out before, setting M=10 in the code

phaseCanvas.setM(10.0f);

timeCanvas.setM(10.0f);

disableDoubleBuffering(phaseCanvas);// So you can watch the plot

disableDoubleBuffering(timeCanvas); // Otherwise it happens off screen

// start the calculation

fc.start(); // this starts FermiCalc in its thread by calling fc.run()

}

/**

*

* @param c - the component to have double buffering disabled

*/

private void disableDoubleBuffering( Component c ) {

RepaintManager currentManager = RepaintManager.currentManager(c);

currentManager.setDoubleBufferingEnabled(false);

}

// . . . the rest of the code

This will change the title of the Application window

- Click on the FermiPlot Source Packages
- Run/Generate JavaDoc
- file:///Users/jar/Documents/TGA/Projects/FermiPlot/dist/javadoc/index.html will show you documentation for your Project
- But we did not insert many JavaDoc comments…