Introduction Radiative habhas can be used as one of the calibrations of the aar electromagnetic calorimeter (EM). Radiative habha events (e ; e +! e ;

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1 aar Note # 52 September 28, 2 Kinematic Fit for the EM Radiative habha alibration J. M. auer University of Mississippi bstract For the radiative habha calibration of aar's electromagnetic calorimeter, the measured energy of a photon cluster is being compared with the energy obtained via a kinematic t involving other quantities from that event. The details of the tting algorithm are described in this note, together with its derivation and checks that ensure that the tting routine is working properly.

2 Introduction Radiative habhas can be used as one of the calibrations of the aar electromagnetic calorimeter (EM). Radiative habha events (e ; e +! e ; e + ) deposit photons over a large energy range everywhere in the calorimeter. If the momenta of the incoming and outgoing electrons and positrons, as well as the photon's angular position are known, the photon energy can be obtained via a kinematic t. This t results in an absolute measurement of the photon energy which then can be compared to the measured photon energy to obtain calibration constants. The radiative habha module is part of aar's Online Prompt Reconstruction (OPR) executable. Initial cuts select good electrons, positrons, and photons. Then all possible combinations of triplets (one electron, one positron, one photon) are formed. Each triplet is sent to the tting routine to calculate its 2 est, the \estimated 2 ". The triplet with the lowest 2 est is then submitted to the full kinematic t which returns, among other quantities, the tted photon energy E f and the error matrix of the tted quantities. The ratio E meas =E f is later used to calibrate the calorimeter. Note that no information on the measured photon energy E meas goes into the kinematic t or 2 est. This note is the complete documentation on the algorithm for tting the radiative habha events for the purpose of calibrating the calorimeter. It describes the whole tting procedure: the quantities for the kinematic t and 2 est the derivation and formulas for 2 est the derivation and algorithm for the kinematic t tests to check the quality of the kinematic t. The note details all formulas which go into the computer program so that the program can be checked directly against this document. The derivations contain more details than needed to understand the concept, but the details help to derive, check and recheck all necessary formulas. ctual results of the tting procedure using real data are not included in this note to keep it a pure code documentation. 2 Dening the quantities and constraints 2. Measured quantities From the experiment come the following measurements, which shall form the 4-dimensional vector y: P ix; y P iy; y 2 P iz; y 3 msrd momentum in x, y, and z of incoming e ; P ix+ y 4 P iy+ y 5 P iz+ y 6 msrd momentum in x, y, and z of incoming e + P ox; y 7 P oy; y 8 P oz; y 9 msrd momentum in x, y, and z of outgoing e ; P ox+ y P oy+ y P oz+ y 2 msrd momentum in x, y, and z of outgoing e + o y 3 o y 4 measured and of the photon The momenta of the incoming electron and positron and their errors are changing run-by-run. The errors of the incoming leptons are given as covariance matrices: V i; = V ixx; V ixy; V ixz; V ixy; V iyy; V iyz; V ixz; V iyz; V izz; V i+ = The errors on P i; and P i+ are assumed to be independent. V ixx+ V ixy+ V ixz+ V ixy+ V iyy+ V iyz+ V ixz+ V iyz+ V izz+

3 The errors of P o; =(P ox; P oy; P oz;) and P o+ =(P ox+ P oy+ P oz+ ) are also assumed to be independent from each other.they are given in two 3 3 error matrices: V o; = V oxx; V oxy; V oxz; V oxy; V oyy; V oyz; V o+ = V oxx+ V oxy+ V oxz+ V oxy+ V oyy+ V oyz+ V oxz; V oyz; V ozz; V oxz+ V oyz+ V ozz+ The errors on o and o appear in the current analysis without --correlations since they were found to be negligibly small, but we still use this 2 2 sub-set of the larger 4 4 error matrix of the Emcluster:! V V o = o V o V o V o ll the errors can be combined in one 4 4 error matrix V all. Its format is like this: V all = V i; V i+ V o; V o+ V o = 2

4 2.2 Quantities for the kinematic t The kinematic t determines the following numbers: f ix; f f iy; f 2 f iz; f 3 x, y, andz momentum of incoming e ; f ix+ f 4 f iy+ f 5 f iz+ f 6 x, y, andz momentum of incoming e + f ox; f 7 f oy; f 8 f oz; f 9 x, y, andz momentum of outgoing e ; f ox+ f f oy+ f f oz+ f 2 x, y, andz momentum of outgoing e + f f 3 f f 4 E f h and, and energy of the photon 2 3 four Lagrange multipliers for momentum and energy conservation constraints 4 The variables f to f 4 have corresponding measurements. The variable h, the photon energy, is called a \hidden variable". The vector shall be dened as a 9-element composite of f (4 elements), h ( element), and (4 elements). 2.3 onstraints We have four constraint equations that have to be satised in the kinematic t: Here we use,e.g., p ix; + p ix+ ; p x; ; p x+ ; E f sin f cos f = momentum in x p iy; + p iy+ ; p y; ; p y+ ; E f sin f sin f = momentum in y p iz; + p iz+ ; p z; ; p z+ ; E f cos f = momentum in z E i; + E i+ ; E ; ; E + ; E f = energy E i; = q p 2 ix; + p2 iy; + p2 iz; + m2 e q f 2 + f f m2 e 3 The estimated 2 : 2 est This function is calculated for any given electron-positron-gamma triplet to determine which triplet should be used for the kinematic t. t the end of this subsection, we willhave a complete analytical formula for calculating 2 est. The formula is based on the dierence between the initial and nal momentum, P P i ; P o. The initial momentum P i is the sum of the momenta of the incoming electron and positron as dened earlier: P i; and P i+. The measured momenta of the outgoing electron, positron are given by P ; and P +. For the outgoing photon, we only have its angles E = E i; + E i+ ; E ; ; E + we may substitute the unknown photon energy E P (E i; + E i+ ; E ; ; E + ) sin cos sin sin cos 3 and. Using the energy constraint with measured values, and we obtain: E n x n y n z E n

5 Of course, n is the normal vector, the direction of the photon. alculating the dierence to form vector P is easy: P P x P y P z = P i; + P i+ ; P o; ; P o+ ; P o In the ideal world, this vector would be exactly zero. For its error matrix V p, we convert V all, the error matrix of y, via a transformation matrix T into V p : V p = T t V all T For the transformation matrix T we have to calculate expressions like j = x y z: (E n j ) = P ix; P n j i; P ix; The transformation matrix is a 3 4 matrix: T = P ix; P iy; P iz; P ix+ P iy+ P iz+ ; ; ; P ix; P iy; P iz; P ix+ P iy+ P iz+ ; ; ; P ix; P iy; P iz; P ix+ P iy+ P iz+ ; ; ; = ; P ix; E i; n x ; P iy; E i; n x ; P iz; E i; n x ; P ix+ E i+ n x ; P iy+ E i+ n x ; P ix; E i; n y ; P iy; E i; n y ; P iz; E i; n y ; P ix+ E i+ n y ; P iy+ E i+ n y ; P iz+ n x ; P iz+ n y E i+ E i+ ;+ P x; E n P x; x ; E n y ; P y; E n x ;+ P y; ; E n y ; P z; E n x ; Px. We note that for Pix; ; P ix; E i; n z ; P iy; E i; n z ; P iz; E i; n z ; P ix+ E i+ n z ; P iy+ E i+ n z ; P iz+ E i+ n z P x; E n z ; P y; E n z ; P z; E n y ;+ P z; ; E n z ; ;+ P x+ P x+ P x+ n x n y n z E + E + E + P y+ n x ;+ P y+ P y+ n y n z E + E + E + P z+ P z+ n x n y ;+ P z+ n z E + E + E + ;E cos cos ;E cos sin E sin E sin sin ;E sin cos 4

6 Now wehave V ; p = T t V ; T, and hence we may calculate 2 est: all 2 est = P t V ; p P What is the meaning of this 2? We can say that the 4 input variables are used to measure P,and 2 est tells us the deviation of the measured P from the expected P,which is zero. 4 The kinematic t For the derivation of the kinematic t algorithm, we follow the description of Louis Lyons, page 5, 52 []. 4. The 2 -Function The real 2 -function can be written down in the following way: 2 = (f ; m) t V ; all (f ; m) + [p xi; + p xi+ ; p xo; ; p xo+ ; E sin cos ] + 2 [p yi; + p yi+ ; p yo; ; p yo+ ; E sin sin ] + 3 [p zi; + p zi+ ; p zo; ; p zo+ ; E cos ] + 4 [E i; + E i+ ; E o; ; E o+ ; E ] The constraint equations are here included via Lagrange multipliers. To minimize this 2, we could use a standard package like MINUIT, but standard packages are always slower than specially adapted code. Since the 2 -minimization is being done millions of times, it pays o to write special code for the minimization. In addition, MINUIT is not supported in aar's Online Prompt Reconstruction. 4.2 Derivation of kinematic t algorithm t the minimum of 2, its rst derivatives are to be zero. Lyons uses for this the following equations: = for i =to4 h The three equation sets can be written as: = here h = E = 5 k = here = 6 etc. 2 G (f ; m)+d t = E t = = The factor 2 in front ofv ; is missing in Lyons' book []. We could easily remove this factor from our formulas all by re-dening the Lagrange multipliers in the 2 -function with a factor 2. This would not change the t result or errors, as long as the subsequent calculations were carried out consistently. 5

7 where G is the 4 4 inverse error matrix of the measurements which we also call V ; all. = ::: = 4 = 5 D = = ::: = 4 and E = = 5 3 = ::: 3 = 4 4 = ::: 4 = 4 3 = 5 4 = 5 We now expand the constraint equations around f and h, and we obtain for the four equations k with k =to4: k () k + 4X i= () (f i ; f () i f i )+ () h (h ; h() )= () We may rewrite this into: 4X i= () f i (f i ; m i )+ () h (h ; h() )=; () k + 4X i= () f i (f () i ; m () i ) Now we collect everything, use the denitions for M, Y,andZ, M = 2G Dt E t D E Y = f ; m h ; h Z = ;R with (f h ) R = () ; D(f ; m) = 2 (f h ) 3 (f h ) ; D(f ; m) 4 (f h ) and we see: MY = Z This is the equation we have to solve. Since the constraint equations = contain non-linear functions like sin, Eq. () is only an approximation, and we have to iterate as described in the next section. 4.3 Recipe for the kinematic t algorithm The matrix M and the vector Z are functions of the measurements and their error matrices as well as of the parameters. The vector Y is, as mentioned above, Y = f ; m h ; h and can be calculated with: Y = M ; Z 6

8 Here is the iteration: Initially, we will use for the t quantities f = m, i.e. the measured quantities. For h = h, we calculate the photon energy via simple energy conservation. These together with the measured quantities allow us to calculate M and Z. We multiply the inverse of M with Z and obtain Y. This result will then give us a better set of f and h, which we again use to calculate M and Z, and then a better Y. nd we continue until our constraint equations are suciently fullled and the quantities f and h are stable. It might happen that the iteration does not converge at the minimum, but wanders o into unphysical numbers. In that case, it would be good to have a certain boundary box around the point. If the step would make the point lie outside the box, then the program would change the step so that the point would be back inside. It might be good to implement this, although the radiative habha tting does not seem to need this part of the algorithm. 4.4 Details of matrices and vectors used in the kinematic t We dene the following variables: E i = 8 >< >: E i; for i =,2,3 (p ix;, p iy;, p iz;) E i+ for i =4,5,6 (p ix+, p iy+, p iz+ ) E ; for i =7,8,9 (p x;, p y;, p z;) E + for i =,, 2 (p x+, p y+, p z+ ) s i = ( for i 6 (pix;, p iy;, p iz;, p ix+, p iy+, p iz+ ) ; for i>6 (p x;, p y;, p z;, p x+, p y+, p z+ ) For 4 4 matrix D we need the following expressions: Row j = to 3, columns i = to 2: j = ( Row j =, column i =3: Row j =, column i =4: Row j = 2, column i =3: s i if i = j or i = j +3 or i = j +6or i = j +9 else 3 = ;E f cos f cos f = ; 5 cos 3 cos 4 4 = E f sin f sin f = 5 sin 3 sin 4 3 = ;E f cos f sin f = ; 5 cos 3 sin 4 7

9 Row j = 2, column i =4: Row j = 3, column i =3: 4 = ;E f sin f cos f = ; 5 sin 3 cos = E f sin f = 5 sin 3 Row j = 3, column i =4: 3 4 = Row j =4,columnsi = to 2: Row j = 4, column i = 3 and 4: 4 = s i i E i 4 3 = 4 4 = The 4 matrix E is: E = ; sin f cos f ; sin f sin f ; cos f ; = ; sin 3 cos 4 ; sin 3 sin 4 ; cos 3 ; 4.5 The error matrix of the t The second partial derivatives of 2 appear in the error matrix of the t parameters: H = 2! ; j So in our case, H is a 9 9 matrix. The detailed expressions for the second derivatives of 2 will be given in the following section. 4.6 Tests for goodness of t fter completing the iteration on the kinematic t, one wants to make sure that all quantities are indeed correct. esides the obvious tests that the constraint equations are satised, one can check that indeed a minimum was reached. For this, one may wiggle each nal value to 4 and recalculate 2. In our case we have in the 2 -function the terms with the Lagrange multipliers. Just recalculating the 2 function will not lead to correct results, since the found vector is a minimum only when 8

10 also requiring the constraints. So one has to redo the t while forcing the selected element of to the o-minimum value. This wiggling allows us to map out the minimum, and it also tells us whether the t error returned for that parameter is reasonable. If we xe f to be t way from the real t result, then the 2 should rise by in either direction. When mapping out this rise, one will see the shape of a parabola. When the formulas are complicated and/or one is far away from the minimum, the parabola will be distorted. In our case, we can indeed calculate the t error for E f, but if this would be impossible, one can nd the t error by mapping out the minimum with the above described re-tting with xed values. The -error is then dened to be where 2 is unit above the minimum. s mentioned, this function may be distorted when far away from the minimum. complicated 2 -function might even distort the -area. In this case, one can take the minimum and two points very close to it, t a parabola through these three points, and take the sigma from that parabola as the error. The same process also works for the hidden parameter (tted photon energy), and we denitely have to re-t since the tted photon energy only appears in the constraints, where the Lagrange multipliers would inuence the outcome. Here is how we have to modify the formulas for re-tting: 4.6. Re-tting with xed E f We want to redo the t with the photon energy xed to E x = E f +. To the 2 -function, we add the term + X (E ; E x ) 2 where X is a large number compared to the original 2. If we now minimize this new 2 -function, the additional term adds a large penalty toany deviation of E from E x. Going through the derivation again, we nd the following places that have to be changed in the code: First partial derivative 2 for i =5[fori = k] has the additional term \+2X(E i ; E x )". No change to second partial derivatives. Matrix M has the additional term \+2X" at (5,5). This means that the (5,5)-element of M is no longer zero.. Vector Z has an additional term at position 5: Z = ;2X(h ; E x ) ;R These are all necessary changes. The iteration should converge again, but this time always result in E = E x for suciently large X. 9

11 4.6.2 Re-tting with xed f k Let us now wiggle one of the measurement variables to 4. When xing f k to f k = f k x, we add the term + X (f k ; f k x ) 2 to the 2 -function. gain, X is a large number compared to the original 2. The following changes have to be made in the formulas of the algorithm: The rst partial derivative = gets for i = k the additional term \+2X(f k ; f k x )". gain no change to second partial derivatives. Matrix M gets at position (k k) the additional term \+2X". Vector Z has at position k the entry \;2X(m k ; f k x )" ondence Level If all errors of the measurements are nicely described by Gaussian distributions, and if all events are what we think they are, i.e., (in our case) radiative habhas, then the 2 values of the ts should be distributed like the 2 -distribution for n = 3 (3 because out t is a 3-constraint t). Instead of looking at the 2 distributions directly, it is easier to map the 2 to a at distribution with values between and. This value is then called the condence level (.L.) of the event. If the 2 is really distributed as it should be, the condence level will have a at distribution. So we are looking for two things in the.l. distribution: () Most of the region should have a at distribution. If not, the errors used in the t might be too large or too small. If the errors are underestimated, the 2 will be larger than expected, and the condence level distribution will be tilted downward (when going from to ). Vice-versa, if the errors are overestimated, the.l. distribution will be tilted upward. More information on the validity of errors might be obtained from the \pull" distributions described later. (2) peak at zero indicates events that do not fulll the kinematics of radiative habhas at all. They will result in very large 2 (=very small.l., close to zero). These events can come from backgrounds or misidentied tracks. What can we do? We can improve our selection criteria. Or we can cut out all events belonging to that peak, taking only those events that are part of the at distribution. cut on the condence level is, of course, equivalent to a cut on The \Pull" For each measured variable, one can plot the so-called \pull" [2] or \normalized stretch values" [3] [4]: pull p = meas ; t p meas ; t The minus sign in the square root comes from the strong correlation between the measured and the tted quantity, and \still puzzles many users" [2]. If all measured errors were estimated correctly

12 and the conditions for the t were satised (e.g., the event was really a radiative habha event), then the pull quantity will be distributed like a Gaussian centered at with =. If an error is for example overestimated, the pull quantity will have a more narrow distribution. In this case, the condence level should also be aected, displaying a tilt in its distribution. To check whether a systematic increase or decrease of one or more errors would improve the pull and/or the condence level distributions, one can redo the whole analysis with increased or decreased errors. Perhaps one can nd a set of corrections that create nice pull distributions and a nice condence level distribution. If the errors are really not correct, one should talk with the colleagues who are responsible for the errors. However, abnormal pull quantities might not be always created by incorrect errors. Systematically shifted measurements could also cause such symptoms Function First Derivatives For this set of equations, we will use the following notation: L i = 8 >< >: for i =,4,7, (p ix;, p ix+, p x;, p x+ ) 2 for i =2,5,8, (p iy;, p iy+, p y;, p y+ ) 3 for i =3,6,9,2 (p iz;, p iz+, p z;, p z+ ) Now we calculate the rst partial derivatives of the 2 -function, i.e., the 9 equations =. For i =to2: For i = 3: For i = 4: = 2 = 2 X4 j= X4 j= = 2 =2 X4 j= V ; all ij (f j ; m j )+s i L i + s i 4 f i E i V ; all ij (f j ; m j ) ; E cos cos ; 2 E cos sin + 3 E sin V ; all ij ( j ; m j ) ; 6 5 cos 3 cos 4 ; 7 5 cos 3 sin sin 3 = 2 X4 j= X4 j= V ; all ij (f j ; m j )+ E sin sin ; 2 E sin cos V ; all ij ( j ; m j )+ 6 5 sin 3 sin 4 ; 7 5 sin 3 cos 4 For i = 5: = ; sin cos ; 2 sin sin ; 3 cos ; 4 = ; 6 sin 3 cos 4 ; 7 sin 3 sin 4 ; 8 cos 3 ; 9

13 For i = 6: = p xi; + p xi+ ; p x; ; p x+ ; E sin cos = + 4 ; 7 ; ; 5 sin 3 cos 4 For i = 7: = p yi; + p yi+ ; p y; ; p y+ ; E sin sin = ; 8 ; ; 5 sin 3 sin 4 For i = 8: = p zi; + p zi+ ; p z; ; p z+ ; E cos = ; 9 ; 2 ; 5 cos 3 For i = 9: = E i; + E i+ ; E ; ; E + ; E = E i; + E i+ ; E ; ; E + ; Function Second Derivatives For i =to2 and j = to 2: = 2 V ; j + all s Ei;fi ij i 2 =E i 4 E 2 i i = 2 V ; all ij ; s i 4 fi fj E 3 i = 2 V ; all ij = 2 V ; all ij ; s i 9 2 i ;E 2 i E 3 i = 2 V ; all ij ; s i 9 i j E 3 i else if i = j if E i = E j by denition For i =to2 and j =3to4: For i =to2 and j = 5: =2V ; all ij = For i =to2 and j =6to8: = s i if L i = L j by denition = else 2

14 For i =to2 and j = 9: For i = 3 and j =3: = s i f i E i = s i i E i = 2 V ; j + all ij E sin cos + 2 E sin sin + 3 E cos i For i = 3 and j =4: For i = 3 and j =5: = 2 V ; all ij sin 3 cos sin 3 sin cos 3 = 2 V ; j + all ij E cos sin ; 2 E cos cos i = 2 V ; all ij cos 3 sin 4 ; 7 5 cos 3 cos 4 = ; cos cos ; 2 cos sin + 3 sin = ; 6 cos 3 cos 4 ; 7 cos 3 sin sin 3 For i = 3 and j =6: = ;E cos cos = ; 5 cos 3 cos 4 For i = 3 and j =7: = ;E cos sin = ; 5 cos 3 sin 4 For i = 3 and j =8: For i = 3 and j =9: = E sin = 5 sin 3 = For i = 4 and j =4: = 2 V ; j + all ij E sin cos + 2 E sin sin i = 2 V ; all ij sin 3 cos sin 3 sin 4 3

15 For i = 4 and j =5: = sin sin ; 2 sin cos = 6 sin 3 sin 4 ; 7 sin 3 cos 4 For i = 4 and j =6: = E sin sin = 5 sin 3 sin 4 For i = 4 and j =7: For i = 4 and j = 8 and 9: For i = 5 and j =5: j i = ;E sin cos = ; 5 sin 3 cos 4 = = For i = 5 and j =6: For i = 5 and j =7: For i = 5 and j =8: For i = 5 and j =9: = ; sin cos = ; sin 3 cos 4 = ; sin sin = ; sin 3 sin 4 = ; cos = ; cos 3 = ; For i =6to9andj =6to9: = 4

16 References [] Lyons, L., Statistics for nuclear and particle physicists (ambridge Univ. Press, 986). [2] Eadie, W. T. et al., Statistical methods in experimental physics (North-Holland, msterdam, 986). [3] V. lobel, Least squares methods, p. I 27, in ock, R. K. et al. (eds.), Formulae and methods in experimental data evaluation with emphasis on high energy physics. (European Phys. Soc., 984) [4] Roe,. P., Probability and statistics in experimental physics, (Springer-Verlag, 992). cknowledgments I thank ill Dunwoodie (SL) for his patience with my numerous questions. His help and expertise were invaluable in understanding all issues on the estimated 2 and the kinematic t. 5