Muon induced neutrons in LVD - University of Minnesota€¦ · LS = (4.2 ± 1.3) x 10-4 n/(g/cm2) MC Y Fe = 1.29 x 10-3 n/(g/cm2) Y LS = 2.44 x 10-4 n/(g/cm2) The distribution of

Muon induced neutrons in LVD

Rino Persiani

University & INFN Bologna

on behalf of the LVD collaboration

AARM, Minneapolis, June 23th 2012

Large Volume Detector

R. Persiani, AARM, Minneapolis, June 2012

LVD

LNGS

2

It is located in the hall A of the Gran Sasso

National Laboratories (LNGS) at a depth

of about 3600 m w.e.

The mean muon energy is about 280 GeV.

It is made by 1000 t of liquid scintillator

and by other 1000 t of iron for the

structure.

LVD is highly modular: 105 independent

modules

1 3 2


The LVD detector

3

• 8 counters in one module

• 3 identical towers in the detector

• 35 active modules in a tower

External dimensions: 13 x 23 x 10 m3


The detector basic elements

4

External dimensions: 1.5 x 1 x 1 m3

Scint. composition: CnH2n+2

<n>=9.6 +1 g/l PPO + 0.03 g/l POPOP

Scint. density: ~ 0.8 g/cm3

Attenuation lenght: > 15m @ l=420 nm

Flash point at: ~39oC

PMT: FEU-49B

Photocathode diameter: 15 cm

Quantum efficiency: 10-15%

R. Persiani, AARM, Minneapolis, June 2012 5

LVD Physics

● LVD is mainly dedicated to the study of neutrinos from

gravitational core collapse in the galaxy (SNEWS)

● Neutrino Physics

➢ Neutrinos in time correlation with GRB, solar flares

and gravitational waves

➢ Monitoring the CNGS beam

➢ Neutrino velocity

● Physics of the cosmic rays

➢ Muon annual modulation

➢ Muon coinciding with other experiment in the

laboratories

➢ Muon vertical intensity

➢ Muon-induced neutrons


The inverse β-decay

6

Neutrinos from supernovae are detected by the inverse β-decay.

The electronic antineutrinos interact with a proton in the liquid

scintillator with the following:

This interaction has a characteristic signature made by a prompt

signal due to the positron and a second delayed signal due to the

neutron capture on proton which gives a 2.2 MeV gamma.

νe + p n + e+

n + p d + γ

HET LET


The trigger

7

• The electronics is designed to detect both the product of the Inverse Beta Decay, keeping the signal to noise ratio to an acceptable level.

• The trigger has a HIGH threshold at

~4 MeV.

• When the trigger is satisfied, the threshold is LOWered to ~1 MeV, for about 1 ms, in all the counters close to the ‘high threshold’ one, in order to detect the gamma from the neutron capture.


Muons in LVD

8

In LVD a muon is defined requiring the

time coincidence (within 250 ns) of at

least two HET signals in two different

scintillation counters.

The mean energy released by a muon

in each counter is 185 MeV.


Muon-induced neutrons in LVD

9

After each muon detected in LVD we look for the signal of a neutron

capture in those counters where the threshold is lowered to 1 MeV,

in a time window <1 ms after the muon pulse

m

n


What do we want to measure?

10

LVD is well suited to detect both muons and

neutrons.

With LVD is possible to measure the number

of muon-induced neutrons in liquid

scintillator for muons with a mean energy of

280 GeV.

A full MC simulation can sort out the

discrepancy between LVD and other

Experimental value

LVD


What do we want to measure?

LVD is well suited to detect both muons and

neutrons.

With LVD is possible to measure the number

of muon-induced neutrons in liquid

scintillator for muons with a mean energy of

280 GeV.

A full MC simulation can sort out the

discrepancy between LVD and other

Experimental value


MC simulation

Monte Carlo simulation in geant4 (v 9.3)

with the geometrical description of the

whole LVD liquid scintillator and iron

structure, the hall A at the LNGS and the

rock and concrete around it.

QGSP_BIC_HP physics list


Primary particle the muon intensity is approximately 1.1 μ/(m2 h)

Energy spectrum and angular direction are obtained with MUSIC and MUSUN


Neutron Yield

where: Nn = number of detected neutrons Nμ = number of muons Sm = fraction of active internal counters fi = fraction of detected neutrons produced in the i-th material εi = detection efficiency λi = muon mean path length i-th material

where i = LS (liquid scintillator), Fe (iron)

DATA

MC


Neutron estimation

Background

neutrons

The background rate at 1 MeV is very large in LVD: O(200 Hz). The number of neutron captures can be obtained fitting the time distribution of the LET signals, relative to the muon crossing time: The distribution has 2 components: Flat, due to uncorrelated background Exponential distribution due to neutron captures

Very important a good estimation of t !

t1 t2


τ measured with a 252Cf source

16

Neutron source outside the counter:

Neutron source in the center of the counter: t = (202 ± 2) ms

in agreement with the MC and the theoretical value (203 ms)

Neutrons uniformly generated in the whole counter, t = (182 ± 1) ms (from MC)


τ from the data

τDATA = (240 ± 10) μs

τMC = (150 ± 2) μs

τDATA = (140 ± 27) μs

τMC = (140 ± 4) μs


Test with LED

The effect observed in tank crossed by muons can be explained taking into account a derivative circuite between the PMTs and discriminator After a big energy deposition a small signal could be not detected

With a LED inside a counter we simulated the light emitted by the muon and we measure the time distribution of the events in LET


Data sample and muon selection

We analyzed a data set with the detector in its final configuration (3 towers),

from January 2005 till December 2008; the total livetime is 34659 hours.

The mean energy released by a muon in each counter is 170 MeV and the average nb. of

crossed counters is 4-5.

We select only muons that gives (within 250 ns) at least 2 distinct HET signals in 2 different

scintillation counters (one of which has to be internal), with an energy deposited > 10 MeV.

Total number of muon events (single, multiple, cascades): 8.3 M


Counter and event selection

Internal counters, not affected by hardware problems:

• Bad ADC and TDC • Time distribution in the gate not flat • Gate width shorter than 400 ms

253 internal counters satisfy the requirements (over 400). Sm = 253/400

Low energy signals: • Energy in [1, 5] MeV • Time difference from the muon pulse in [20, 400] ms • In counters NOT crossed by the muon


Muon mean path length

Lμ(Fe) = 41.1 cm

From the MC simulation we get the muon path length in iron and in liquid scintillator

for all the selected muons.

RMS = 16.4 cm

σLμ

= 5.5 x 10-3 cm

Lμ(LS) = 463 cm

RMS = 153 cm

σLμ

= 5.1 x 10-2 cm


Fraction of detected neutrons

We estimate the fraction of detected neutrons that were produced in iron or in

liquid scintillator:

We select the muon sample

We apply the event and counter

For each LET signal we look for a correspondence with a captured neutron

somewhere in te detector

Once we find it, we get the material were it was originated

NLET

74063

Ndet

(roccia) 2148 (3%)

Ndet

(Fe) 62571 (84.5%)

Ndet

(LS) 9092(12.3%)


Detection efficiency

The detection efficiency is evaluated as the number of neutrons produced in a

specific material that are detected somewhere in LVD.

materiale Nn,det Nn,prod ε (%)

ferro 62571 3731732 1.68

scintillatore 9092 772928 1.18


Neutron yield in LVD

Nn = 68300 ± 6.1% ± 30% neutroni

DATI

YFe

= (2.3 ± 0.7) x 10-3 n/(g/cm2)

YLS

= (4.2 ± 1.3) x 10-4 n/(g/cm2)

MC

YFe

= 1.29 x 10-3 n/(g/cm2)

YLS

= 2.44 x 10-4 n/(g/cm2)

The distribution of the time delay between the muon and the neutron signals in 4 years of LVD data is shown in figure 4. The fit is performed between 20 and 400 µs, where the τ value has been fixed by Monte Carlo simulation to a value of 143

We have all the ingredients to measure the neutron yield:


Neutron yield in a homogeneous material

μ

MC

Yh,Fe

= 1.19 x 10-3 n/(g/cm2)

Yh,LS

= 1.95 x 10-4 n/(g/cm2)

Muon with LNGS energy spectrum Fired through a homogeneous block of material We cut the edges of the block to avid any kind of edge effects We evaluated the neutron yield

26

Comparison block/LVD

Neutron yield in liquid scintillator (x10-4 n/(g/cm2))

MC DATI

blocco 1.95 X

LVD 2.44 4.2

Neutron yield in iron (x10-3 n/(g/cm2))

MC DATI

blocco 1.19 X

LVD 1.29 2.3

(2.1)

(3.4)

ΔLS

= 19.0%

ΔFe

= 8.5%

Summary of the results from data and from the MC simulation for homogeneous and not homogeneous detector

The ratio between the values in the MC columns estimate the difference in the neutron yield for both cases:


Possible explanation

We introduced two possible explanation that can explain the difference:

LVD is not homogeneous

Geometrical effects


Neutron yield event by event

Liquid scintillator

Iron


Neutron yield in liquid scintillator

Other points show the results from experiments at: (A) 20 m w.e., (B) 25 m w.e., (C) 32 m w.e., (D) 316 m w.e., (E) 570 m w.e., (F) 2700 m w.e., (G) 3000 m w.e., and (H) 5200 m w.e..


Conclusions

We studied the neutrons produced by cosmic muons in the LVD detector during 4 years (2005-2008), with a total number of 8.3 millions detected muons.

A full MonteCarlo simulation has been developed with Geant4.

We measured the neutron yield in LVD

• in LS: (4.2 ± 1.3) 10-4 • in iron: (2.3 ± 0.7) 10-3 (first measurement ever)

If we want to compare our result with homogeneous detectors we have to scale the neutron yield

• in LS: (3.4 ± 1.0) 10-4 • In iron: (2.1 ± 0.6) 10-3

... still work to do: neutron energy spectrum, distance from the muon track, ...


backup


1.Ricavo i coseni direttori del muone dalla

conoscenza degli angoli e

2.Genero un punto random su un cerchio con

centro nel punto centrale dell’apparato e diametro

uguale alla massima dimensione del rivelatore

3.Ruoto il cerchio in modo tale che sia

perpendicolare alla direzione del muone

Per trovare il punto di generazione del

muone, devo trovare il punto d’intersezione

tra la superficie esterna del World Volume e

la retta passante per il punto random preso

all’interno del cerchio e parallela alla

direzione del muone

Documents

Muon induced neutrons in LVD - University of Minnesota€¦ · LS = (4.2 ± 1.3) x 10-4 n/(g/cm2) MC Y Fe = 1.29 x 10-3 n/(g/cm2) Y LS = 2.44 x 10-4 n/(g/cm2) The distribution of