Sphericity
Lee Pondrom
May 9,
2011
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/pyth
pythia6301/node213.html
Definitions
• Sαβ = Σi pαipβi/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
1
Allowed
x2
0
0
x1
1
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
Before cuts
Before
BeforeBefore
cuts cuts
cuts
• Left hand plot is before any cuts. Note the ring
of fire.
• Right hand plot has tower ET>1 GeV and
Before
Before
cuts
cuts
technical
computing
4/8/2011
tower |η|<2.
Cal towers jet 20 φ and ET
Cal towers sphericity and λ1
λ1 η φ
Jet data from jet20 file
ET and η
Jet data φ and Zvertex
Met variables
sumET and metsig
Met variables
• Look normal – no cuts applied.
Jet1 compared to λ1
Δη
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 η
sphericity
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