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A pocketful of change…10 pennies!...are tossed up in the air. About how many heads (tails) do you expect when they land? What’s the probability of no heads at all in the batch? What’s the probability on all heads? What’s the probability of finding just one head? Relative Probability of N heads in 10 flips of a coin 0 1 2 3 4 5 6 7 8 9 10 12 dice are rolled. About how many 6s do you expect when they land? What’s the probability of no sixes at all in the batch? What’s the probability on all sixes? What’s the probability of finding just one six? Relative probability of getting N sixes in a toss of 12 dice. 0 1 2 3 4 5 6 7 8 9 10 11 12 244140625 / 2176782336 585937500 644531250 429687500 Log P 193359375 61875000 14437500 2475000 309375 27500 1650 60 1 0 1 2 3 4 5 6 7 8 9 10 11 12 The counts for RANDOM EVENTS fluctuate •Geiger-Meuller tubes clicking in response to a radioactive source •Oscilloscope “triggering” on a cosmic ray signal Cosmic rays form a steady background impinging on the earth equally from all directions measured rates NOT literally CONSTANT long term averages are just reliably consistent These rates ARE measurably affected by •Time of day •Direction of sky •Weather conditions You set up an experiment to observe some phenomena …and run that experiment for some (long) fixed time… but observe nothing: You count ZERO events. What does that mean? If you observe 1 event in 1 hour of running Can you conclude the phenomena has a ~1/hour rate of occurring? Random events arrive independently •unaffected by previous occurrences •unpredictably 0 sec time 1 A reading of could result from the lucky capture of an exceeding rare event better represented by a much lower rate (~0?). or the run period could have just missed an event (starting a moment too late or ending too soon). A count of 1 could represent a real average as low as 0 or as much as 2 1±1 A count of 2 2 ± 1? ± 2? A count of 37 37 ± at least a few? A count of 1000 1000 ± ? The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time Dt can be very small: p << 1 •cosmic rays arrive at a fairly stable, regular rate when averaged over long periods •the rate is not constant nanosec by nanosec or even second by second •this average, though, expresses the probability per unit time of a cosmic ray’s passage Example: a measured rate of 1200 Hz = 1200/sec would mean in 5 minutes we should expect to count about A. 6,000 events B. 12,000 events C. 72,000 events D. 360,000 events E. 480,000 events F. 720,000 events Example: a measured rate of 1200 Hz = 1200/sec would mean in 3 millisec we should expect to count about A. 0 events B. 1 or 2 events C. 3 or 4 events D. about 10 events E. 100s of events F. 1,000s of events Example: a measured rate of 1200 Hz = 1200/sec would mean in 100 nanosec we should expect to count about A. 0 events B. 1 or 2 events C. 3 or 4 events D. about 10 events E. 100s of events F. 1,000s of events The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time Dt can be very small: p << 1 for example (even for a fairly large surface area) 72000/min=1200/sec =1200/1000 millisec =1.2/millisec = 0.0012/msec =0.0000012/nsec The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time Dt can be very small: p << 1 The probability of NO cosmic rays passing through that area during that interval Dt is A. p D.( p - 1) B. p2 C. 2p E. ( 1 - p) The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time Dt can be very small: p << 1 If the probability of one cosmic ray passing during a particular nanosec is P(1) = p << 1 the probability of 2 passing within the same nanosec must be A. p D.( p - 1) B. p2 C. 2p E. ( 1 - p) The probability of a single COSMIC RAY passing through a small area of a detector within a small interval of time Dt is p << 1 the probability that none pass in that period is (1-p)1 While waiting N successive intervals (where the total time is t = NDt ) what is the probability that we observe exactly n events? pn n “hits” ??? × ( 1 - p )N-n ??? “misses” N-n“misses” While waiting N successive intervals (where the total time is t = NDt ) what is the probability that we observe exactly n events? P(n) N! = nn ! ( N -N n)! C n p (1-p N-n ) N-n = ??? ln (1-p) ln (1-p)N-n (N-n) ln (1-p)N-n = (N-n) ln (1-p) 2 3 4 x x x ln( 1 x ) x 2 3 4 and since p << 1 ln (1-p) - p ln (1-p)N-n = (N-n) (-p) from the basic definition of a logarithm this means ???? == ???? e-p(N-n) (1-p)N-n P(n) N! = n! ( N - n)! pn ( 1 - p )N-n P(n) N! = n! ( N - n)! pn e-p(N-n) If we have to wait a large number of intervals, N, for a relatively small number of counts,n n<<N P(n) N! = n! ( N - n)! pn e-pN P(n) N! = n! ( N - n)! -pN n p e And since N! N (N - 1) (N - 2) (N - n 1) ( N - n)! N - (n-1) N (N) (N) … (N) = Nn for n<<N P(n) N! = n! ( N - n)! -pN n p e n N n -pN P(n) = p e n! n ( Np ) -Np P(n) = e n! n ( Np Np ) -Np P(n) = e n! Hey! What does Np represent? 2 3 4 x x x ex 1 x 2! 3! 4! xn ex n! n0 n 0 n 0 m, mean = n P(n) 0 n 1 e - Np n ( Np ) n n! e - Np n ( Np) n n! n=0 term 0 e - Np n 1 n / n! = 1/(???) n ( Np ) n n! m, mean e - Np n 1 (Np ) e - Np ( Np ) n (n - 1)! n 1 (Np ) e - Np ( Np)?? (n - 1)! n 1 (Np ) e - Np ( Np ) n -1 (n - 1)! m ( N p ) let m = n-1 i.e., n(m=)! m 0 what’s this? m, mean e - Np n 1 (Np) e- Np ( Np ) n (n - 1)! m 0 ( Np) m (m)! m = (Np) e-Np eNp m = Np m = Np m e-m n P(n) = n! Poisson distribution probability of finding exactly n events within time t when the events occur randomly, but at an average rate of m (events per unit time) (4) P(n) = n! n 4 e e-4 = 0.018315639 If the average rate of some random event is p = 24/min = 24/60 sec = 0.4/sec what is the probability of recording n events in 10 seconds? P(0) = 0.018315639 P(1) = 0.073262556 P(2) = 0.146525112 P(3) = 0.195366816 P(4) = 0.195366816 P(5) = 0.156293453 P(6) = 0.104195635 P(7) = 0.059540363 Probability of Observing N Events When the Average Expected Counts Should Be 1 0.4 0.35 0.3 m=1 0.25 0.2 0.15 0.1 0.05 0 0 1 2 3 4 5 6 7 8 9 10 Probability of Observing N Events When the Average Count Expected Should Be 4 0.25 Probability 0.2 m=4 0.15 0.1 0.05 0 0 1 2 3 4 5 6 7 8 9 10 Number of Events Counted Probability of Observing N Events When the Average Count Expected Should Be 8 0.16 0.14 0.12 m=8 Probability 0.1 0.08 0.06 0.04 0.02 0 0 1 2 3 4 5 6 Number of Events Counted 7 8 9 10 Another abbreviation (notation): mean, m = x (the average x value) i.e. 1 N x n 1 x n N 3 different distributions with the same mean describe the spread in data by a calculation of the average distance each individual data point is from the overall mean N s= (xi – m)2 i=1 N-1 mean, m Recall: The standard deviation s is a measure of the mean (or average) spread of data away from its own mean. It should provide an estimate of the error on such counts. N 1 2 s ( xi - m ) N i 1 2 or s (x - m ) 2 for short 2 The standard deviation s should provide an estimate of the error in such counts n - n 2 2 n - 2n n n n - 2n n n 2 2 2 2 2 n - 2n n n - n 2 n -m 2 2 2 2 What is n2 for a Poisson distribution? m -m m -m n n e n e n! (n - 1) ! n 0 n 1 2 n 2 n first term in the series is zero e -m m n ( n 1 ) ! n 1 n factor out e-m which is independent of n What is n2 for a Poisson distribution? n e 2 -m me m n ( n 1 ) ! n 1 -m n Factor out a m like before n -1 m n ( n 1 ) ! n 1 Let j = n-1 n = j+1 me -m m ( j 1) ( j ) ! j 0 j What is n2 for a Poisson distribution? j m 2 -m n me ( j 1) ( j) ! j 0 mj -m me j j 0 ( j ) ! m ( j ) ! j 0 j This is just n me -m 2 m m 2 em again! me m e m The standard deviation s should provide an estimate of the error in such counts n - n 2 n -m 2 2 m m -m m 2 In other words s 2= m s = m 2 Assuming any measurement N usually gives a result very close to the true m the best estimate of error for the reading is N We express that statistical error in our measurement as N ± N 1000 Cosmic Ray Rate (Hz) 500 0 Time of day How many pages of text are there in the new Harry Potter and the Order of the Phoenix? 870 What’s the error on that number? A. 0 B. 1 C. 2 D. /870 29.5 E. 870/2 = 435 A punted football has a hang time of 5.2 seconds. What is the error on that number? Scintillator is sanded/polished to a final thickness of 2.50 cm. What is the error on that number? You count events during two independent runs of an experiment. Run Events Counted Error 1 64 8 2 100 10 In summarizing, what if we want to combine these results? 164 18 ? ?? But think: adding assumes each independent experiment just happened to fluctuate in the same way Fluctuations must be random…they don’t conspire together! How different is combining these two experiments from running a single, longer uninterrupted run? 1 64 8 2 100 10 164 18 ?? Run Events Counted Error 0 164 12.8 ? 8 2 + 10 2 12.8 164 12.8 What if these runs were of different lengths in time? How do you compare the rates from each? Run Elapsed Time Events Counted Error 8 1 10 minutes 64 Rate 6.4 0.8 ?? /min 10 ?? /min 5.0 0.5 2 20 minutes 100 How do errors COMPOUND? d Dd velocity = t Dt Can’t add Dd + Dt = or even d ??? t (Dd)2 + (Dt)2 • the units don’t match! • it ignores whether we’re talking about km/hr , m/sec , mi/min , ft/sec , etc That question provides a clue on how we handle these errors How do we know it scales correctly for any of those? Look at: x vt taking derivatives: divide by: Fixes units! dx tdv vdt x vt v x/t or vt dx dv dt x v t dx x dv - 2 dt t t v x/t x/t dv dx dt v x t And we certainly don’t expect Though we still shouldn’t be simply adding the random errors. two separate errors to magically cancel each other out. Whether multiplying or dividing, we add the relative errors in quadrature (taking the square root of the sum of the squares) 2 dx dv dt x v t 2 2 dv dx dt v x t 2 What about a rate - background calculation? ( R DR) - ( B DB) R - B (DR - DB) The units match nicely! but we can’t guarantee that the errors will cancel! Once again: ( DR) ( DB) 2 2 -dE/dx = (4pNoz2e4/mev2)(Z/A)[ln{2mev2/I(1-b2)}-b2] I = mean excitation (ionization) potential of atoms in target ~ Z10 GeV -dE/dx [ MeV·g-1cm2 ] 10 8 6 4 Minimum Ionizing: 3 2 1 0.01 1 – 1.5 MeV2 g/cm 0.1 1.0 10 100 Muon momentum [ GeV/c ] 1000 A typical gamma detector has a light-sensitive photomultiplier attached to a small NaI crystal. The scinitillator responds to the dE/dx of each MIP track passing through If an incoming particle initiates a shower, each track segment (averaging an interaction length) will leave behind an ionization trail with about the same energy deposition. The total signal strength Number of track segments Basically Emeasured N tracks E MIP avg Measuring energy in a calorimeter is a counting experiment governed by the statistical fluctuations expected in counting random events. Since E Ntracks and DN = N we should expect DE E and the relative error DE E E E 1 = E DE = AE a constant that characterizes the resolution of a calorimeter