PHYS 1140 lecture 6. During lab this week (Tuesday to next Monday):

Transcription

1 Deadlines coming up PHYS 1140 lecture 6 During lab this week (Tuesday to next Monday): Turn in Pre-lab 3 at the START of your lab session Expt. 3: take data, start analysis during your lab session. Next week: Turn in Homework set 4 by the end of Monday (last one). Expt. 3: complete your report and turn in at end of the third weekday after your lab session. Read guide for Expt. E1 which you ll all do the following week. No more lectures after today s Check the course home page for pre-lab and lab assignments. 1

2 Today s topics Review from last week: standard deviation, SDOM, Gaussian distribution Error for counting experiments General review (Correlated measurements) 2

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4 Error on the mean (repeated measurements) If we have N measurements of a quantity x with mean x standard deviation σx then the standard deviation of the mean, SDOM (or standard error of the mean) is given by σ x = σ x N 4

5 Concept ques/on 6.1 A series of measurements of a quan/ty is taken with apparatus A and then again with apparatus B. The distribu/on of A and B results are shown. Which series of measurements has the smaller standard devia/on σ? A) A B) B C) both A and B have the same standard devia/on. D) impossible to tell from the informa/on given.

6 Concept ques/on 6.2 A series of measurements of a quan/ty is taken with apparatus A and then again with apparatus B. The distribu/on of A and B results are shown. Which series of measurements has the smaller standard devia/on of the mean σ mean? A) A B) B C) impossible to tell from the informa/on given.

7 P(x) Interpreting a result in standard form We quote x = X ± σ (mean and SDOM) We may plot it as a point with error bars The significance is measured in sigmas, with probabilities below G X,σ (x) = 1 2πσ e 1 2 (x X) 2 σ % % 95% (x-x)/! 7

8 P(x) Concept ques/on % % 99.7% (x-x)/!

9 g [m/s 2 ] A case study: measuring g Measurements of g From M1 reports of one lab section in PHYS Lab M1 (g) Expected value Measured points g =4π 2 L T Measurement No. 9

10 Standard deviation in a counting experiment Suppose we have events (e.g., radioactive decays) occurring randomly in time at constant rate R N = R t (N decays occur over time t) In a given trial N will differ from R t (random fluctuation). The standard deviation of the fluctuations is, from probability theory, σ = N Therefore the fractional error on N is δn / N = σ / N = N / N = 1 / N We measure R by counting over (exact) time t: R = N / t ± σ / t = N / t ± N / t, δr / R = δn / N = 1 / N That is, by counting over a longer time we increase N and thereby reduce the error on R. We usually call this random uncertainty the statistical error. 10

11 Concept question 6.5 The rate R of decay of a radioactive source is about 1000 per minute, from a rough estimate. If we seek a measurement of R precise within 0.5% we need to accumulate counts for A) 1 minute B) 200 minutes C) 12 seconds D) about a month E) 40 minutes We want δr / R = 1/ N to be 1/200 N = 200 N = = 40,000 t = N / R = 40,000 / 1000 = 40 11

12 Error propagation: general rule Our master equation for error propagation is Given w = f(x, y, z,... ), δw = f x 2 (δx) 2 + f y 2 (δy) 2 + f z 2 (δz)

13 Error on a sum or difference For the case z = x + y or z = x - y, the partial derivatives are just = 1, whence We add the errors in quadrature. δz = δx 2 + δy 2 13

14 And for the error on a product or quotient For, z = x y and z = x / y the error propagation formula becomes simply δz z = δx x We add the fractional errors in quadrature. 2 + δy y 2 Finally, for f(x) = ax n, dx δf f = n δx x The fractional error of a power n of x is n times the fractional error of x. 14

15 Concept ques/on 6.6 A = 1.6 ± 0.2, B = 53.2 ± 1.0 P(A,B) = A B =? Which term is going to dominate the error in P? A) A B) B

16 Concept ques/on 6.7 A = 1.6 ± 0.2, B = 53.2 ± 1.0 P(A,B) = A + B =? Which term is going to dominate the error in P? A) A B) B

17 Recipe (repeated measurements) We quote our result x = x ± σ x x = N i=1 x i N average or mean σ x = N i=1 (x i x) 2 N 1 standard deviation (error on a single measurement) σ x = σ x N SDOM, standard error of the mean 17

18 (Positively) correlated errors A mysterious, generous banker takes a big pile of pennies, cuts it exactly in two, and gives one pile to Frank, one to Jill. He tells them the piles are exactly, to the penny, the same, but he doesn t tell them how many there are. Frank does a rough count of his pile, gets F = 200 ± 30. Now he knows that Jill s share J = 200 ± 30! Note that δj and δf are perfectly correlated. If rough count of F is too high, then rough count of J is too high, also! 18

19 Concept question 6.8 In the above scenario, J = F, F = 200 ± 30, difference D = J - F, we can say that D ± δd is D = 0 A) 0 ± 0 pennies B) 0 ± 30 pennies C) 0 ± 42 pennies D) 0 ± 60 pennies E) some other answer 19

20 Concept question 6.9 In the same scenario, J = F, F = 200 ± 30, sum S = J + F, we can say that S ± δs is S = 2 F A) 400 ± 0 pennies B) 400 ± 30 pennies C) 400 ± 42 pennies D) 400 ± 60 pennies E) some other answer 20

21 Negatively (or anti-) correlated errors Now our banker counts out exactly 1000 pennies, not one more or one less. He offers some to Frank, who takes a big scoop of them. He gives the rest to Jill, telling her there were originally exactly 1000 pennies. Jill does a rough count of her pennies, and estimates J = 200 ± 30 No dummy, Jill, she uses the exactly 1000 info and quickly figures out Frank must have F = 800 ± 30. In this case δj and δf are perfectly anti-correlated. If rough count of J is too high, then rough count of F is too low! 21

22 Concept question 6.10 In the above scenario, J + F = 1000, J = 200 ± 30, difference D = F - J, we can say that D ± δd is A) 600 ± 0 pennies B) 600 ± 30 pennies C) 600 ± 42 pennies D) 600 ± 60 pennies E) some other answer 22

23 Concept question 6.11 In the same scenario, J + F = 1000, J = 200 ± 30, sum S = J + F, we can say that S ± δs is A) 1000 ± 0 pennies B) 1000 ± 30 pennies C) 1000 ± 42 pennies D) 1000 ± 60 pennies E) some other answer 23

24 Uncorrelated errors Our banker has yet another trick. This time he takes a huge sack of pennies, thousands and thousands, and offers them to Frank. Frank politely takes only a handful. He does a rough count and gets F = 300 ± 30 Then Jill gets a turn. She dips into the huge sack with both hands, pulls up a quantity of pennies. She does a rough count of her pile and gets J = 700 ± 30 This time, δf and δj are nothing to do with each other. They are statistically independent, or uncorrelated. This is the situation that we usually assume. 24

25 Concept question 6.12 In this last scenario, F = 300 ± 30, J = 700 ± 30, difference D = J - F, we can say that D ± δd is A) 400 ± 0 pennies B) 400 ± 30 pennies C) 400 ± 42 pennies D) 400 ± 60 pennies E) some other answer 25

26 Concept question 6.13 We start with a number x ± δx. We define: f(x) = x + x What is δf? A) Well, this is just like that subrule, f(x) = ax, with a =2, and so I think δf = 2δx B) No, I think it s that other subrule, for addition, so that since f(x) = x + x, I think δf = ( (δx) 2 + (δx) 2 ) 1/2 = (2) 1/2 δx = 1.4 δx 26