Sphericity

Lee PondromMay 9, 2011

References for sphericity and thrust

• Original application from Spear

G. Hanson et al., PRL 35. 1609 (1975).Useful lecture slides by Steve Mrennain a description of Pythia:

http://cepa.fnal.gov/psm/simulation/mcgen/lund/pythia_manual/pythia6.3/pythia6301/node213.html

Definitions

• Sαβ = Σi pαipβ

i/Σipi² . • 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)Σipi².• This is the form originally proposed by

Bjorken (PRD 1, 1416(1970)). We will use Mrenna’s definition.

Eigenvalues of S

• Diagonalize S• λ1 • S’ = RSR-1 = λ2 , • λ3 • and order them λ1>λ2>λ3, so λ1

• 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.

Some formulas

• The matrix S is symmetric, so we have to calculate six components: S11, S12, S13, S22, S23, and S33. The trace is an invariant, S11 + S22 + S33 = 1. The diagonalization procedure gives a cubic equation: λ3 – λ2 + q λ + r = 0, where q and r are functions of the components of S.

More formulas

• q=(S11S22 + S11S33 + S22S33 – (S13)² - (S23)² - (S12)²), and

• r=-S11S22S33 – 2S12S13S23 + (S13)²S22 + (S 23)²S11 + (S12)²S33.The cubic equation may be solved with

the substitution λ = x + 1/3. This eliminates the squared term: x3 + ax + b = 0.

Cubic equation

• x = λ – 1/3; x3 + ax + b = 0.• a = (3q -1)/3; and b = (-2 +9q

+27r)/27.• Define K = b²/4 + a3/27.• 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

Solutions to the cubic equation

• K<0 is the usual case for sphericity

• Then xn = 2 (-a/3)1/2cos((φ + 2πn)/3), for n=0,1,2.

• cosφ = (27b2/(-4a3))1/2 , + if b<0.

More about the cubic equation

• It can be written in terms of the trace and determinant of the matrix S

• λ3–Tr(S)λ2-.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 λ3=0. The other two roots are

• λ±=(1±(1-4q).5)/2, where q=-.5(Tr(S²)-Tr(S)²)

Eigenvectors

• The cosine of the polar angle of λ1 was calculated from

• Sψ = λ1ψ, with components of ψ (a,b,c) satisfying a² + b² + c² =1, and the ratios

• a/c=(S12S23-S13(S22-λ1))/denom

• b/c=(S12S13-S23(S11-λ1))/denom

• denom=(S11-λ1)(S22-λ1) – (S12)²

Transverse eigenvector

• To calculate the azimuthal angle φ the thrust was used in the transverse plane.

• Thrust = ∑i|n∙pi|/∑i|pi|, where n and p are transverse vectors, and n is determined so that Thrust is maximized. ½<Thrust<1.

Simple example

• Consider a three body decay M->3, and define x1=2E1/M, 0<x1<1. x1+x2+x3=2.

• Phase space

0 x1 1

1 x2 0

Allowed

Generate the events

• Pick x1 and x2 and check that the point is inside the allowed triangle.

• Calculate x3 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.

Analyze the events

• The three momentum vectors are coplanar, which means that r=0, and λ3=0.

• The two other roots are

• λ = (1(1-4q)1/2)/2, with λ+ = λ1.

• The direction cosines of λ1 give the thrust direction, and λ2 gives the transverse momentum in the decay plane.

The results for 1000 events generated

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).

10000 jet20 events tower eta distribution

• Left hand plot is before any cuts. Note the ring of fire.

• Right hand plot has tower ET>1 GeV and tower |η|<2.

Before cutsBefore cuts

Before cutsBefore cuts

Before cutsBefore cutstechnical computing 4/8/2011

Cal towers jet 20 φ and ET

Cal towers sphericity and λ1

λ1 η φ

Jet data from jet20 fileET and η

Jet data φ and Zvertex

Met variables

sumET and metsig

Met variables

• Look normal – no cuts applied.

Jet1 compared to λ1

Δ η Δφ

Delta R=(Δη²+Δφ²)1/2

Look at the second jet in the event

Φ resolution for jets and thrust

Transform tower η to the dijet center of mass

• Define ηcm= (ηjet1 + ηjet2)/2

• Then tower ηcm = tower η – ηcm

• Also correct tower η to the event vertex• For CHA use r=154 cm to the iron face,

and tanθ = tanθ0/(1-zvtanθ0/r)• For PHA use d = 217 cm from the origin to

the iron face, and • tanθ=tanθ0/(1-zv/d). Not much difference.

Comparison of ηcm and tower ηsphericity λ1 η distribution

Δη λ1- jet1 ΔR

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.

Check the lorentz transformation by comparing jets and towers

Definitions for the previous slide

• labeta = (jet1η +jet2η)/2 ignores jet3

• y* = .5*log((1+β*)/(1-β*))

• β* = ∑i pzi / ∑i Ei summed over all towers with ET > 1 GeV.

What about jet3 in the jet20 data?

Compare transverse energy balance, 3 jets and sum towers

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

Nothing really improves things

• About 90% of the events with jet1ET>15 GeV have a third ‘jet’, which has an average ET ≈ 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.

Lorentz transformation to the event center of mass

• Using the towers, define a total momentum vector ptot = ∑ipxi x + ∑ipyi y + ∑ipzi z, where (x,y,z) are unit vectors

• And a total energy Etot= ∑I towEi

• Then β* = ptot/Etot , and L = R-1LzR, where R is a space rotation placing the z axis along ptot , and Lz is a Lorentz transformation along the new z axis.

Total momentum in the event center of mass should vanish, and

it does.

And the other two components

So the Lorentz transformation to the event center of mass works

Event c of m and longitudinal Lorentz transformation are close

Compare to jet1 in the event

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 σinel=60 mb• <L>E32 <n> Pr(0) Pr(1) Pr(>=2) >=2/1data• .5 1.2 .3 .36 .34 .15• 2.0 4.8 .008 .039 .95 .34• <n> is much less than estimated from <L>

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

Ntowers with ET>1 GeV

Events with extra vertices

• They have lots of extra tower hits:

• 1 vertex <Ntowers> = 518,

• >=2 vertices <Ntowers> = 636.

• However, a cut on tower ET>1 GeV virtually wipes out the minbias background. 1Vertex <Ntowers>=12.3;

• >=2 vertices <Ntowers>=13.4.

• So the sphericity analysis, which requires towerET>1 GeV is not affected by extra vertices.

Δvertex gjt1bk

sphericity

Tower sum energy in cm

Thrust axis cm η