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High School Mathematics at the Research Frontier. Don Lincoln Fermilab. http://www-d0.fnal.gov/~lucifer/PowerPoint/HSMath.ppt. What is Particle Physics?. High Energy Particle Physics is a study of the smallest pieces of matter.

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High School Mathematics at the Research Frontier

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## High School Mathematics at the Research Frontier

Don Lincoln

Fermilab

http://www-d0.fnal.gov/~lucifer/PowerPoint/HSMath.ppt

### What is Particle Physics?

High Energy Particle Physics is a study of the smallest pieces of matter.

It investigates (among other things) the nature of the universe immediately after the Big Bang.

It also explores physics at temperatures not common for the past 15 billion years (or so).

It’s a lot of fun.

Fermilab

4x10-12

seconds

Stars form

(1 billion years)

Now

(15 billion years)

Atoms form

(300,000 years)

Nuclei form

(180 seconds)

Nucleons form

(10-10 seconds)

??? (Before that)

### DØ Detector: Run II

• Weighs 5000 tons

• Can inspect 3,000,000 collisions/second

• Will record 50 collisions/second

• Records approximately 10,000,000 bytes/second

• Will record 1015(1,000,000,000,000,000) bytes in the next run (1 PetaByte).

30’

30’

50’

### Remarkable Photos

In this collision, a top and

anti-top quark were created,

helping establish their existence

This collision is the most violent

ever recorded. It required that

particles hit within 10-19 m or

1/10,000 the size of a proton

### How Do You Measure Energy?

• Go to Walmart and buy an energy detector?

• Ask the guy sitting the next seat over and hope the teacher doesn’t notice?

• Ignore the problem and spend the day on the beach?

• Design and build your equipment and calibrate it yourself.

150 lbs

?? Volts

Volts are a unit of electricity

Car battery = 12 Volts

Walkman battery = 1.5 Volts

### Build an Electronic Scale

Make a line, solve slope and intercept

y = m x + b

Voltage = (0.05) weight + 3

Implies

Weight = 20 (Voltage – 3)

This implies that you can know

the voltage for any weight.

For instance, a weight of 60 lbs

will give a voltage of 6 V.

Now you have a calibrated scale.

(Or do you?)

### Issues with calibrating.

All four of these functions go through the two calibration points. Yet all give very different predictions for a weight of 60 lbs.

What can we do to resolve this?

Easy

Hard

### Solution: Pick Two Points

Better, but still poor, representation of data

### Why don’t all the data lie on a line?

• Error associated with each calibration point.

• Must account for that in data analysis.

• How do we determine errors?

• What if some points have larger errors than others? How do we deal with this?

### First Retake Calibration Data

• Remeasure the 120 lb point

• Note that the data doesn’t always repeat.

• You get voltages near the 9 Volt ideal, but with substantial variation.

• From this, estimate the error.

### Data

While the data clusters around 9 volts, it has a range. How we estimate the error is somewhat technical, but we can say

9  1 Volts

### Redo for All Calibration Points

Both lines go through the data.

How to pick the best one?

### State the Problem

• How to use mathematical techniques to determine which line is best?

• How to estimate the amount of variability allowed in the found slope and intercept that will also allow for a reasonable fit?

• Answer will be m Dm and b Db

Looks Intimidating!

### The Problem

• Given a set of five data points, denoted (xi,yi,si) [i.e. weight, voltage, uncertainty in voltage]

• Also given a fit function f(xi) = m xi + b

• Define

### Forget the math, what does it mean?

Each term in the sum is simply the separation between the data and fit in units of error bars. In this case, the separation is about 3.

f(xi)

yi - f(xi)

si

yi

xi

### More Translation

So

Means

Since f(xi) = m xi + b, find m and b that minimizes the c2.

Calculus

### Approach

Find m and b that minimizes c2

Back to algebra

Note the common term (-2). Factor it out.

Move terms to LHS

Factor out m and b terms

Rewrite as separate sums

### Approach #2

Now distribute the terms

Approach #3

Substitution

Note the common term in the denominator

Notice that this is simply two equations with two unknowns. Very similar to

You know how to solve this

ohmigod….

yougottabekiddingme

So each number

### Approach #4

Inserting and evaluating, we get

m = 0.068781, b = 0.161967

2nd and 5th terms give biggest contribution to c2 = 2.587

Best

### Doesn’t always mean good

A new hypothetical set of data with the best line (as determined by the same c2 method) overlaid

### Goodness of Fit

Our old buddy, in which the data and the fit seem to agree

### New Important Concept

• If you have 2 data points and a polynomial of order 1 (line, parameters m & b), then your line will exactly go through your data

• If you have 3 data points and a polynomial of order 2 (parabola, parameters A, B & C), then your curve will exactly go through your data

• To actually test your fit, you need more data than the curve can naturally accommodate.

• This is the so-called degrees of freedom.

### Degrees of Freedom (dof )

• The dof of any problem is defined to be the number of data points minus the number of parameters.

• In our case,

• dof = 5 – 2 = 3

• Need to define the c2/dof

c2/dof = 22.52/(5-2) = 7.51

### Goodness of Fit

c2/dof near 1 means the fit is good.

Too small  errors were over estimated

Can calculate probability that data is represented by the given fit. In this case:

Top: < 0.1%

Bottom: 68%

In the interests of time, we will skip how to do this.

c2/dof = 2.587/(5-2) = 0.862

Recall that we found

m = 0.068781, b = 0.161967

What about uncertainty and significant figures?

If we take the derived value for one variable (say m), we can derive the c2 function for the other variable (b).

The error in b is indicated by the spot at which the c2 is changed by 1.

So  0.35

### Uncertainty in m and b #1

Recall that we found

m = 0.068781, b = 0.161967

What about uncertainty and significant figures?

If we take the derived value for one variable (say b), we can derive the c2 function for the other variable (m).

### Uncertainty in m and b #2

The error in m is indicated by the spot at which the c2 is changed by 1.

So  0.003

So now we know a lot of the story

m = 0.068781  0.003

b = 0.161967  0.35

So we see that significant figures are an issue.

Finally we can see

### Uncertainty in m and b #3

Voltage = (0.069  0.003) × Weight + (0.16  0.35)

Final complication: When we evaluated the error for m and b, we treated the other variable as constant. As we know, this wasn’t correct.

c2min + 1

c2min + 2

c2min + 3

b

Best b & m

More complicated, but shows that uncertainty in one variable also affects the uncertainty seen in another variable.

m

### Error Ellipse

Increase intercept, keep slope the same

Increase intercept, keep slope the same

To remain ‘good’, if you increase the intercept, you must decrease the slope

Decrease slope, keep intercept the same

Similarly, if you decrease the slope, you must increase the intercept

From both physical principles and strict mathematics, you can see that if you make a mistake estimating one parameter, the other must move to compensate. In this case, they areanti-correlated(i.e. if b, then m and if b, then m.)

b

m

### Error Ellipse

Best b & m

new b

within errors

bbest

When one has an m below mbest, the range of preferred b’s tends to be above bbest.

mbest

new m

within errors

Data and error analysis is crucial, whether you work in a high school lab…

### Back to Physics

Or the Frontier!!!!

### References

• P. Bevington and D. Robinson, Data Reduction and Error Analysis for the Physical Sciences, 2nd Edition, McGraw-Hill, Inc. New York, 1992.

• J. Taylor, An Introduction to Error Analysis, Oxford University Press, 1982.

• Rotated ellipses

• http://www.mecca.org/~halfacre/MATH/rotation.htm

http://www-d0.fnal.gov/~lucifer/PowerPoint/HSMath.ppt

http://worldscientific.com/books/physics/5430.html