Toy-model Subhasis Samanta Jan Kochanowski University, Institute of - - PowerPoint PPT Presentation

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Toy-model Subhasis Samanta Jan Kochanowski University, Institute of - - PowerPoint PPT Presentation

Toy-model Subhasis Samanta Jan Kochanowski University, Institute of Physics, Kielce, Poland Subhasis Samanta April 15, 2020; Kielce 1 / 7 Generating uncorrelated particles 80 Multiplicity distribution 60 Poisson distribution with mean N


slide-1
SLIDE 1

Toy-model

Subhasis Samanta

Jan Kochanowski University, Institute of Physics, Kielce, Poland

Subhasis Samanta April 15, 2020; Kielce 1 / 7

slide-2
SLIDE 2

Generating uncorrelated particles

Multiplicity distribution ⋆ Poisson distribution with mean N =30 pT, φ distributions ⋆ P(pT) = pT e−6pT ⋆ φ = Uniform(-π, +π)

mult

N 20 40 60

  • No. of events

20 40 60 80 (GeV/c)

T

p 0.5 1 1.5 2 )

T

P(p 0.005 0.01 0.015 0.02 0.025 (rad) φ 4 − 2 − 2 4 ) φ P( 0.005 0.01 0.015

Total number of events = 1000

Subhasis Samanta April 15, 2020; Kielce 2 / 7

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SLIDE 3

(px, py) distribution ⋆ px = pT cos(φ) ⋆ py = pT sin(φ)

5 −

10

4 −

10

3 −

10

(GeV/c)

x

p 2 − 1 − 1 2 (GeV/c)

y

p 2 − 1 − 1 2

Subhasis Samanta April 15, 2020; Kielce 3 / 7

slide-4
SLIDE 4

Second factorial moment

F2(M) = 1

M2

M2

i=1 ni(ni − 1)

1

M2

M2

i=1 ni

Equivalent formula: F2(M) = 2M2 N2 Npp(M) M - number of bins in px and py ni - number of particles in i-th bin N - event multiplicity Npp(M) - total number of particle pairs in M bins in an event

Npp(M) =?

Subhasis Samanta April 15, 2020; Kielce 4 / 7

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SLIDE 5

Npp in a single event

n = 34

0.01 0.02 0.03 0.04 0.05

(GeV/c)

x

p 2 − 1 − 1 2 (GeV/c)

y

p 2 − 1 − 1 2

n = 9, # pair = 36 n = 6, # pair = 15 n = 9, # pair = 36 n = 10, # pair = 45 = 132

pp

N

Subhasis Samanta April 15, 2020; Kielce 5 / 7

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SLIDE 6

Npp in a single event

n = 34

0.01 0.02 0.03 0.04 0.05

(GeV/c)

x

p 2 − 1 − 1 2 (GeV/c)

y

p 2 − 1 − 1 2

n = 9, # pair = 36 n = 6, # pair = 15 n = 9, # pair = 36 n = 10, # pair = 45 = 132

pp

N 0.01 0.02 0.03 0.04 0.05

(GeV/c)

x

p 2 − 1 − 1 2 (GeV/c)

y

p 2 − 1 − 1 2

(0,0) (0,0) (0,0) (0,0) (9,36) (6,15) (9,36) (9,36) (1,0) = 123

pp

N

M = 1, Npp = 561 M = 2, Npp = 132 M = 4, Npp = 123 ⋆ Particles in different bins are not uniformly distributed for M ≥ 4 ⋆ (Binning in φ would give better result)

Subhasis Samanta April 15, 2020; Kielce 5 / 7

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SLIDE 7

Npp distributions

(M =1)

pp

N 200 400 600 800 1000 1200

  • No. of events

20 40 60 80

hNppM1

Entries 1000 Mean 452 RMS 163.7

(M =2)

pp

N 100 200 300

  • No. of events

5 10 15

hNppM2

Entries 1000 Mean 113.6 RMS 41.99

(M =4)

pp

N 100 200 300

  • No. of events

5 10 15

hNppM4

Entries 1000 Mean 111.2 RMS 41.13

M = 1, < Npp >= 452 M = 2, < Npp >= 113 M = 4, < Npp >= 111

Subhasis Samanta April 15, 2020; Kielce 6 / 7

slide-8
SLIDE 8

< Npp >, F2 vs. M2

2

M 1 10

2

10

3

10

4

10

5

10 (M) >

pp

< N 100 200 300 400

2

M 1 10

2

10

3

10

4

10

5

10 (M)

2

F 5 10 15 20

F2(M) = 2M2 N2 Npp(M)

Subhasis Samanta April 15, 2020; Kielce 7 / 7

slide-9
SLIDE 9

Thank you

Subhasis Samanta April 15, 2020; Kielce 7 / 7