PHYSICS 2150EXPERIMENTAL MODERN

PHYSICS

Lecture 6Chi Square Test, Poisson Distribution

LAB REPORT FORMAT• Experiment title

•Objective : Several sentence description of scientific goal (not “to learn about….”)

• Idea/Technique: One paragraph or two to give background and physics to be tested. What is being measured and how?

•Apparatus: Explanation of equipment. Drawing diagrams strongly encouraged! (Do not copy the manual!)

OBJECTIVE (EXAMPLE)

•When hydrogen gas is placed in an electrical discharge, the exited atoms may radiate photons at specific wavelength. Hydrogen gas has four wavelengths it may emit in the visible. These are the first four emission lines in the Balmer series. Our objective in this experiment is to find the wavelength of these emission lines.

• Planck‘s constant reflects the size of discrete energy levels in quantum mechanics. In performing this experiment, we sought to accurately determine Planck‘s constant by examining the quantized emission spectrum of the Hydrogen atom.

YOUR LAB REPORT: FORMAT V

• Conclusions:

• Short discussion summarizing results (give the final result again here)

• Further comment on uncertainties: explain basis of assigning them, possible hidden errors, etc.

• Compare results to accepted value: what is level of agreement based on your error estimate?

• Stick to scientific conclusions! No opinions, no personal comments.

CONCLUSIONS (EXAMPLE)• .... Our calculated value for Planck‘s constant is around a factor

of 3 off from the known value. That is not as close as we would like to get but we are in the same power for the exponent which is not bad.

• ....The discrepancies from the accepted value of the de Broglie wavelength at these voltages are 0.0139𝜎, 0.459𝜎, 1.46𝜎 respectively. The error in the first set of measurements is considerably lower than the error in the two final sets. This could be due to our changed method of measurement taking throughout the lab....

COMPARE RESULTS TO ACCEPTED VALUE

•Measured Value:

• Accepted Value:

•Discrepancy from Accepted Value:

e

m= (1.95 ± 0.02(stat.) ± 0.06(sys.)) · 1011 C/kg

e

m= (1.758820088 ± 0.000000039) · 1011 C/kg

�e

m= (1.95� 1.76) · 1011 C/kg = 0.19 · 1011 C/kg

e/m example of lecture 2:

COMPARE RESULTS TO ACCEPTED VALUE• Significance of Discrepancy:

• Assess the probability for your result being off by 2.8 sigma:

� em

�total=

1.95�0.022 + 0.062

= 2.8

-3 -2 -1 0 1 2 3

µµ�� µ+� µ+2.8�µ�2.8�

• Probability for x inside 2.8𝜎:

• Probability for x outside 2.8𝜎:

erf�

2.8��

�= 0.9949

1� erf�

2.8��

�= 0.0051

LINEAR FITS: EXAMPLE

2 4 6 8 10

1

2

3

4

5

6

7

2 4 6 8 10

1

2

3

4

5

6

7

2 4 6 8 10

2

4

6

8

2 4 6 8 10

2

4

6

8

2 4 6 8 10

2

4

6

8

2 4 6 8 10

2

4

6

8

y = (0.4± 0.1)x + (1.1± 0.6)

y = (0.47 ± 0.07) · x + (0.6 ± 0.4)y = (0.53 ± 0.06) · x + (0.3 ± 0.3)

„True Model“ ?

y = 0.5 · x + 0.5

�y = 1.0 �y = 1.5

�y = 2.0

HOW GOOD IS THE FIT?

•Often want to know how good a fit is.

•Minimising told you what the “best fit” to your function was

• The value of that minimum can tell you how well data actually fit your function

�2

�2 = ��2�

(f(xi|p1, . . . , pN )� yi)2

CHI SQUARED TEST• test is a particular type of goodness-of-fit test

• Generally for N measurements of :

where is the expected value and error on .

•Need to know and degree of freedom (dof) : N (number of measurements) minus the number of fitting parameters (e.g. 2 for line, slope and intercept)

•Now should be about 1.

�2 =� �

Ei � yi

�i

�2

y1, . . . , yN

Ei �i yi

�2

�i

�̃2 =�2

dof

CHI SQUARED TEST

• Look up the probabilities using Appendix D in Taylor :If is high (>5%) then the fit is good

• Use of this test: determine if the fit is good

• if bad, either data is bad, or fit function is wrong

Prob(�̃2 � �̃20)

COUNTING EXPERIMENTS

• Start with a large of radioactive material with a very long halflife (compared to the experiment).

•Measure the decays with a detector at a fixed distance.

•Measure the number of decays in five 100 s intervals.

• Should the five intervals all have the same number of decays?

• Should there be a consistent trend?

COUNTING EXPERIMENTS

• Should there be a consistent trend? (in this case Should the decay rate be changing?)

•No! The long half-life assures that over the time of the experiment, the decay rate isn’t changing significantly.

• Should the five intervals all have the same number of decays?

•No! The decay is a random process.

A POISSON PROCESS• Assume a decay rate of 0.01 decays/sec.

•On average will get 1 decay in 100s.

• Sometimes we will get zero.

• Sometimes we will get two.

• Rarely will even get three or more.

•What is the probability of getting exactly n decays in 100s?

• Answer: The Poisson Distribution. Applies to any discrete process where events occur randomly at a constant rate.

THE POISSON DISTRIBUTION

•Distribution is asymmetric

•Distribution is discrete

• is a positive number

Pµ(n) =µn

n!eµ

Probability to observe n counts if the mean number of counts µ

2 4 6 8 10

5

10

15

20

Number of counts nProb

abilit

y to

obs

erve

n c

ount

s (%

)

µ = 3

µ

PROPERTIES OF THE POISSON DISTRIBUTION

• Average number of counts:

• Average count is characterising parameter of the Poisson distribution

�n� =��

n=0

nPµ(n)

=��

n=0

nµn

n!e�µ = e�µ

��

n=1

nµn

n!= µ e�µ

��

n=1

µn�1

(n� 1)!

= µ e�µ��

n=0

µn

n!= µ e�µeµ

�n� = µ

PROPERTIES OF THE POISSON DISTRIBUTION

• Standard deviation:

• Standard deviation is square root of the mean

�2 = �(n � �n�)2� = �n2 � 2n�n� + �n�2�

= �n2� � 2�n��n� + �n�2 = �n2� � �n�2

d2

dµ2

� ��

n=0

Pµ(n) = 1

�

0 = µ2 � 2µ�

Pµ(n) +�

(n2 � n)Pµ(n)

= �n2� � µ2 � µ

� =�

µ

SMALL MEAN• Consider Poisson distribution

with

• Gaussian: ,

• For small , Gaussian and Poisson is very different

• Poisson:

• Gaussian:

µ

0 2 4 6 8 10

0.1

0.2

0.3

0.4

µ = 1

� = 1µ = 1

p(0 � n � 2) =

P1(0) + P1(1) + P1(2) = 73.6%

p(0 � x � 2) = 68%

0 2 4 6 8 10

0.1

0.2

0.3

0.4

LARGE MEAN: GAUSSIAN LIMIT•Now let‘s consider a large

mean, e.g.

•Well matched by a Gaussian with and

• Poisson:

• 68% vs. 64%: close enough

µ = 7.5

0 2 4 6 8 10 12 14

0.05

0.10

0.15

µ = 7.5 � =�

7.5

p(7.5± 2.7) = P1(0) + . . . + P1(10) = 64%p(7.5± 2.7) = P1(0) + . . . + P1(10) = 64%

Often use Gaussian approximation for large , generally for

µµ > 5

0 2 4 6 8 10 12 14

0.05

0.10

0.15

GAUSS VS. POISSON

• Symmetric

• Continuous (x is real)

•Mean : is the most probable value

• Standard deviation:

•Distribution describes results of measurements with accuracy

• Asymmetric

• Continuous (n is integer)

•Mean : most probable value is

• Standard deviation:

•Distribution describes results from “counting experiment”

Pµ(n) =µn

n!eµGµ,�(x) = 1�

2��e�

12 ( x�µ

� )2

-4 -2 2

0.1

0.2

0.3

0.4

2 4 6 8 10

5

10

15

20

µ

�

µ� µ

�

HOW TO USE POISSON DISTRIBUTIONS IN COUNTING• Uncertainty:

(for large , can be interpreted similarly to Gaussian )

• Statistical uncertainty on the number of counts is thus the square root of the number of counts.

• Background processes may be contributing to your rate: . Often, can measure by turning off the signal, so when you then measure , you can subtract the background to measure your signal rate.

• How do you handle the error in this situation?

� =�

µ�µ

µ = µsig + µbkg µbkg

µ

• Claim to have 160 accidents in a 10 year period at the intersection of Broadway and Tablemesa Drive.

• The City of Boulder decides to install an additional pedestrian light.

•Over the next two years they record 26 accidents at the intersection.

• The city makes the claim that the accident rate has gone down. Can they justify their claim statistically??

AN EXAMPLE

AN EXAMPLE: ANALYSIS• Rate before installation of pedestrian light:

• Rate after installation of pedestrian light:

• Change:

• Yes, the city‘s claim is backed up by statistics!

rateb =160±

�160

10= 16.0± 1.3

ratea =26±

�26

2= 13.0± 2.6

(3.0± 2.8) per year

SCIENCE EXAMPLE:RADIOISOTOPE LIFETIME

•Measure count rate (background) with no radioactive material present

• Introduce radioactive material

• Count the number of decays in a 1s period; remeasure every 30 seconds

• Subtract the background rate

• Fit the rate vs. time to an exponential lifetime

I: MEASURING THE BACKGROUND• Background rate should be constant, so each trial is

a re-measurement of the same thing

• Get the background rate by taking mean: 402/10=40.2

• Note that this is exactly equivalent to simply measuring for 10 seconds and dividing by 10 to find the rate.

• Uncertainty in the rate: (402±√402) counts/10s

• Background rate is 40.2±2.0 counts/s

TrialCounts (in 1 s)

1 442 423 394 365 346 457 498 379 3310 43

Total 402

DATA FROM RADIOACTIVE MATERIAL

• Are all the trials measuring the same rate?

• What are the uncertainties in the number of counts?

• What is the measured signal rate?

• Subtract background (40.2 counts/s)

• Statistical error remains the square root of the total number of counts

• Also there is a systematic error (not shown) due to uncertainty in the background

Seconds Counts Corrected0 2007±45 1967±4530 1464±38 1424±3860 973±31 933±3190 698±26 658±26120 526±23 486±23150 353±19 313±19180 285±17 245±17210 217±15 177±15240 150±12 110±12270 122±11 82±11300 112±11 72±11330 86±9 46±9360 78±9 38±9390 53±7 13±7420 51±7 11±7

FITTING FOR THE MEAN HALF LIFE

• Plot counts vs. time

• Fit the decay:

• Half Life:

100 200 300 400

500

1000

1500

2000

Time (seconds)

Cou

nts N(t) = N(0) · e�t/�

100 200 300 400

500

1000

1500

2000

� = (83.8± 1.1)s

WHAT YOU SHOULD KNOW FROM LECTURES

• Error propagation for an arbitrary equation of which variables’ error are known

• Understand the difference between statistical and systematic uncertainties.

• Calculate mean, standard deviation, std on mean and know what they mean.

•Determine if measurements are compatible and how to compute weighted average.

WHAT YOU SHOULD KNOW FROM LECTURES

• Perform least-squares fit to data and extract the slope and intercepts and their uncertainties.

• How to determine if two variables are correlated.

• Understand what a counting experiment is and its distribution can be different from Gaussian.