1. Sphericity Lee Pondrom May 9, 2011

2. References for sphericity and thrust Original application from Spear G. Hanson et al., PRL 35. 1609 (1975). Useful lecture slides by Steve Mrenna in a description of Pythia: http://cepa.fnal.gov/psm/simulation/mcgen/lund/pythia_manual/pythia6.3/ pythia6301/node213.html

3. Definitions S αβ = Σ i p α i p β i /Σ i p i ². Where the sum is over all particles in the event, and α,β refer to the coordinate axes x,y,z. Gail Hanson uses a definition which interchanges the eigenvalues, namely: T = (1 – S)Σ i p i ². This is the form originally proposed by Bjorken (PRD 1, 1416(1970)). We will use Mrenna’s definition.

4. Eigenvalues of S Diagonalize S  λ1  S’ = RSR -1 =  λ2 ,  λ3  and order them λ1>λ2>λ3, so λ 1 is the ‘jet axis’. A two body final state would have λ1 = 1, and λ3 = λ2 = 0, which is as jetty as you can get. A spherical event would have λ1 = λ2 = λ3 = 1/3. The sphericity is defined as Sp = 3(λ2 + λ3 )/2, 0<Sp<1.

5. Some formulas The matrix S is symmetric, so we have to calculate six components: S 11, S 12, S 13, S 22, S 23, and S 33. The trace is an invariant, S 11 + S 22 + S 33 = 1. The diagonalization procedure gives a cubic equation: λ 3 – λ 2 + q λ + r = 0, where q and r are functions of the components of S.

6. More formulas q=(S 11 S 22 + S 11 S 33 + S 22 S 33 – (S 13 )² - (S 23 )² - (S 12 )²), and r=-S 11 S 22 S 33 – 2S 12 S 13 S 23 + (S 13 )²S 22 + (S 23 )²S 11 + (S 12 )²S 33. The cubic equation may be solved with the substitution λ = x + 1/3. This eliminates the squared term: x 3 + ax + b = 0.

7. Cubic equation x = λ – 1/3; x 3 + ax + b = 0. a = (3q -1)/3; and b = (-2 +9q +27r)/27. Define K = b²/4 + a 3 /27. If K>0 there are one real and two conjugate imaginary roots. If K=0 there are three real roots, at least two are equal. If K<0 there are three real unequal roots

8. Solutions to the cubic equation K<0 is the usual case for sphericity Then x n = 2 (-a/3) 1/2 cos((φ + 2πn)/3), for n=0,1,2. cosφ =  (27b 2 /(-4a 3 )) 1/2, + if b<0.

9. More about the cubic equation It can be written in terms of the trace and determinant of the matrix S λ 3 –Tr(S)λ 2 -.5(Tr(S²)–Tr(S)²)λ–det|S|=0 Here Tr(S)=1, and r=det|S|. If det|S|=0, S is singular, and one root λ 3 =0. The other two roots are λ±=(1±(1-4q).5 )/2, where q=-.5(Tr(S²)- Tr(S)²)

10. Eigenvectors The cosine of the polar angle of λ 1 was calculated from Sψ = λ 1 ψ, with components of ψ (a,b,c) satisfying a² + b² + c² =1, and the ratios a/c=(S 12 S 23 -S 13 (S 22 -λ 1 ))/denom b/c=(S 12 S 13 -S 23 (S 11 -λ 1 ))/denom denom=(S 11 -λ 1 )(S 22 -λ 1 ) – (S 12 )²

11. Transverse eigenvector To calculate the azimuthal angle φ the thrust was used in the transverse plane. Thrust = ∑ i |n∙p i |/∑ i |p i |, where n and p are transverse vectors, and n is determined so that Thrust is maximized. ½<Thrust<1.

12. Simple example Consider a three body decay M->3 , and define x 1 =2E 1 /M, 0<x 1 <1. x 1 +x 2 +x 3 =2. Phase space 0 x1 1 1 x2 0 Allowed

13. Generate the events Pick x 1 and x 2 and check that the point is inside the allowed triangle. Calculate x 3 and the angles  12  13 in the decay plane. Orient the plane at random relative to the master xyz coordinates with a cartesian rotation (α,β,  ). Calculate 9 momentum components.

14. Analyze the events The three momentum vectors are coplanar, which means that r=0, and λ 3 =0. The two other roots are λ  = (1  (1-4q) 1/2 )/2, with λ+ = λ 1. The direction cosines of λ 1 give the thrust direction, and λ 2 gives the transverse momentum in the decay plane.

15. The results for 1000 events generated

16. Next try it with jet20 data Use calorimeter towers as energy vectors Calculate S for the event, with a tower threshold of 1 GeV. Two problems: 1. cal towers are in detector coordinates (fixable). 2. Events are in the center of mass only on average (also fixable).

17. 10000 jet20 events tower eta distribution Left hand plot is before any cuts. Note the ring of fire. Right hand plot has tower E T >1 GeV and tower |η|<2. Before cuts technical computing4/8/2011

18. Cal towers jet 20 φ and E T

19. Cal towers sphericity and λ 1

20. λ 1 η φ

21. Jet data from jet20 file E T and η

22. Jet data φ and Zvertex

23. Met variables

24. sumE T and metsig

25. Met variables Look normal – no cuts applied.

26. Jet1 compared to λ 1 Δ η Δφ

27. Delta R=(Δη² + Δφ² ) 1/2

28. Look at the second jet in the event

29. Φ resolution for jets and thrust

30. Transform tower η to the dijet center of mass Define η cm = (ηjet1 + ηjet2)/2 Then tower η cm = tower η – η cm Also correct tower η to the event vertex For CHA use r=154 cm to the iron face, and tanθ = tanθ 0 /(1-z v tanθ 0 /r) For PHA use d = 217 cm from the origin to the iron face, and tanθ=tanθ 0 /(1-z v /d). Not much difference.

31. Comparison of η cm and tower η sphericity λ 1 η distribution

32. Δη λ 1 - jet1 ΔR

33. Jet triggers L1 L2 L3 ST5 (100) CL20 (50) Jet20 ST5 (100) CL40 (1) Jet50 ST10(8) CL60 (1) Jet70 ST20(1) CL90 (1) Jet100 Prescales in parenthesis, from Physics_5_05 trigger table.

34. Check the lorentz transformation by comparing jets and towers

35. Definitions for the previous slide labeta = (jet1η +jet2η)/2 ignores jet3 y* =.5*log((1+β*)/(1-β*)) β* = ∑ i p zi / ∑ i E i summed over all towers with E T > 1 GeV.

36. What about jet3 in the jet20 data?

37. Compare transverse energy balance, 3 jets and sum towers

38. Transverse energy balance is not perfect, and is about the same for towers and jets. Longitudinal tower sum energy is sharpened by the lorentz transformation

39. Nothing really improves things About 90% of the events with jet1E T >15 GeV have a third ‘jet’, which has an average E T ≈ 7 GeV, and cuts off at 3 GeV! Tower sums do not balance in the transverse plane any better than the 3 jets do. Longitudinally (η1+η2)/2 sharpens up the tower sum pz, but it is far from perfect.

40. Lorentz transformation to the event center of mass Using the towers, define a total momentum vector p tot = ∑ i p xi x + ∑ i p yi y + ∑ i p zi z, where (x,y,z) are unit vectors And a total energy E tot = ∑ I towE i Then β* = p tot /E tot, and L = R -1 L z R, where R is a space rotation placing the z axis along p tot, and L z is a Lorentz transformation along the new z axis.

41. Total momentum in the event center of mass should vanish, and it does.

42. And the other two components

43. So the Lorentz transformation to the event center of mass works

44. Event c of m and longitudinal Lorentz transformation are close

45. Compare to jet1 in the event

46. Two vertex events Analysis so far has been Jet20 triggers gjt1ah (1->4) Aug 04->Sep 05 low luminosity Now run on Jet20 in a later set of runs gjt1bk (14->17) Oct 07->Apr 08. 396 nsec bunch crossing and σ inel =60 mb E32 Pr(0) Pr(1) Pr(>=2) >=2/1data.5 1.2.3.36.34.15 2.0 4.8.008.039.95.34 is much less than estimated from

47. Tower occupancy gjt1bk – cut at ntower=560:63%1v,19%2v.

48. Ntowers with E T >1 GeV

49. Events with extra vertices They have lots of extra tower hits: 1 vertex = 518, >=2 vertices = 636. However, a cut on tower E T >1 GeV virtually wipes out the minbias background. 1Vertex =12.3; >=2 vertices =13.4. So the sphericity analysis, which requires towerE T >1 GeV is not affected by extra vertices.

50. Δvertex gjt1bk

51. sphericity

52. Tower sum energy in cm

53. Thrust axis cm η