Solutions to someexercises

Luca BaldiniINFN–Pisa and University of Pisa

luca.baldini@pi.infn.it

Fermi Summer School 2012Lewes, June 2012

Fermi LAT cheat-sheet

Quantity Value

LAT side 1.50 mLAT total thickness on axis 10.1 X0

TKR layer side 36 cmSpacing between two consecutive TKR layers 3.5 cmTKR total depth on axis 1.5 X0

TKR strip pitch 228 µm

CAL log dimensions (length × width × depth) 32.6 × 2.7 × 2.0 cm3

CAL total depth on axis 8.6 X0

CsI critical energy Ec 11.2 MeVEarth radius 6400 kmEarth magnetic field at the equator 0.25 gauss

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 2 / 12

Energy reconstruction

Exercise: What is the fraction of energy escaping out the back of the CAL fora 500 GeV photon on-axis?

]0

Depth in CsI [X0 5 10 15 20 25

]0

[GeV

/XdtdE

0

20

40

60

80

100

0 9.60 X100 GeV photon

0.10.25 0.5

0.750.9

0 10.63 X280 GeV photon

0.1

0.25 0.5

0.75

0.9

0 11.21 X500 GeV photon

0.1

0.250.5

0.75

0.9

0 11.61 X750 GeV photon

0.1

0.250.5

0.75

0.9

0 11.90 X1000 GeV photon

0.1

0.250.5

0.75

0.9

The longitudinal shower profile can be parametrized with a gamma function:

dE

dt=

E0b

Γ(a)(bt)(a−1)e(−bt) (1)

(t = x/X0, b ≈ 0.5 and a is given by tmax = (a− 1)/b = ln(E/Ec) + 0.5, withEc ≈ 11.2 MeV in CsI).

The fraction of energy escaping for a 500 GeV photon on-axis is of the order of70%.

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 3 / 12

High-energy effective area

Exercise: Estimate the high-energy on-axis effective area of the LAT.

The geometric area of the LAT is approximately:

Ageo ≈ 1.5× 1.5 = 2.25 m2 (2)

Not all the area is active, though (look at the gaps between CAL modules inone of the event displays to get an idea). We can give a rough estimate of theactive (CAL) surface based on the length Llog of a CAL log:

Aactive ≈ 16× L2log = 16× 0.326× 0.326 = 1.70 m2 (3)

(16, here, is the number of tower modules in the LAT). The probability ofconverting in the TKR (which is 1.5 X0 thick on axis) is:

εconv ≈ 1− e−79×1.5 = 0.69 (4)

We’ll assume 100% triggering and filtering efficiency (εtrg = εobf = 1) and a70% selection efficiency in the background rejection stage (εsel = 0.7):

Aeff = Aactive × εconv × εtrg × εobf × εsel = 0.82 m2 (5)

Compare this figure with the P7SOURCE on-axis effective area. Why does thisdiscussion only apply to high energies?

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 4 / 12

High-energy effective area

Energy [MeV]210 310 410 510

]2 [m

eff

On-

axis

A

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

P7SOURCE_V6 TotalP7SOURCE_V6 FrontP7SOURCE_V6 Back

(a)

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 5 / 12

High-energy acceptance (1/2)

Exercise: Estimate the high-energy acceptance of the LAT.

The maximum off-axis angle a particle can trigger the LAT is fixed by thethree-in-a-row trigger condition and depends on the spacing ∆z ≈ 3.5 cmbetween two consecutive TKR layers and the side s ≈ 36 cm of a TKR Si plane:

θmax =2∆z

s≈ 79 cos θmax ≈ 0.2 (6)

(it is not a coincidence that IRFs are tabulated up to cos θ = 0.2). Above thatangle the effective area is zero. We can’t calculate analytically the effectivearea as a function of θ, but we can connect the dots, as shown in figure:

θcos0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

eff

/On-

axis

Aef

fA

0

0.2

0.4

0.6

0.8

1

Continues on next slide. . .Luca Baldini (INFN and UniPi) Fermi Summer School 2012 6 / 12

High-energy acceptance (2/2)

At this point (assuming Aeff does not depend on φ) the acceptance integral istrivial:

A =

∫Aeff (θ, φ) dΩ = 2π

∫ 1

0.2

Aeff (θ) d(cos θ) =

= 2πAeff (θ = 0)(1− 0.2)

2≈ 2.08 m2 sr (7)

Compare this figure with the P7SOURCE acceptance (again: this is only relevantat high energy).The FoV is easy to calculate:

FoV =A

Aeff (θ = 0)≈ 2.54 sr (8)

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 7 / 12

High-energy acceptance/FoV

Energy [MeV]210 310 410 510

sr]

2A

ccep

tanc

e [m

0

0.5

1

1.5

2

2.5 P7SOURCE_V6 TotalP7SOURCE_V6 FrontP7SOURCE_V6 Back

(a)

Energy [MeV]210 310 410 510

FO

V [s

r]

0

0.5

1

1.5

2

2.5

3

3.5

P7SOURCE_V6 FrontP7SOURCE_V6 BackP7SOURCE_V6 Combined

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 8 / 12

High-energy PSF (1/3)

Exercise: Estimate the asymptotic high-energy PSF for front- andback-converting events. Why are they different?

(The following estimation is so rough that I’ll happily give you a factor of 2/3at each step. The main point is really to understand the orders of magnitude.)

The high-energy PSF is dictated by the strip pitch in the TKR (p = 228 µm)and the spacing ∆z ≈ 3.5 cm between two consecutive TKR layers. Thereadout being digital, the hit resolution is

σx = σy =p√12≈ 66 µm (9)

If we had to measure a slope (i.e. in the x–z or y–z plane) based on twomeasurements only, the error on theta, on axis, would be:

σθxz = σθyz =

√2p√

12∆z. (10)

For the space angle we need an extra√

2:

σθ =2p√12∆z

(11)

Continues on next slide. . .Luca Baldini (INFN and UniPi) Fermi Summer School 2012 9 / 12

High-energy PSF (2/3)

Though the track reconstruction is much more complicated than this, we shallassume that the error on the direction scales with the number of measurementss like 1/

√n as in a least squares fit. For the front (back) section we can have

anything between 7 and 18 (3 and 6) point, therefore we’ll assume:

〈n〉Front =18 + 7

2= 12.5 〈n〉Back =

3 + 6

2= 4.5 (12)

Putting all together:

σFrontθ =

2p√12〈n〉Front∆z

≈ 0.06

σBackθ =

2p√12〈n〉Back∆z

≈ 0.10 (13)

Compare this with the Monte Carlo PSF. Though the hit resolution is the same,the PSF for the front section is better because we have more measurements.

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 10 / 12

High-energy PSF (3/3)

Exercise: Estimate the rollover energy of the transition between the tworegimes.

(This is even more crude than the previous two slides.)

We shall take the average angular errors for front and back we just calculated:

σθ =σFrontθ + σBack

θ

2≈ 0.08 (14)

An electron (positron) traverses 〈t〉 = 1.5/2 ≈ 0.75 X0 on average. Thecorresponding multiple scattering angle is

∆θms ≈√

20.0136

E [GeV]

√〈t〉 rad (15)

therefore the transition between the two regimes happens at

σθ ≈ ∆θms (16)

or

E ≈0.0136

√2〈t〉

σθ≈ 12 GeV (17)

(you might want to multiply by 2 since this is the electron or the positron, butat this point a factor of 2 is noise).

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 11 / 12

High-energy PSF

Energy [MeV]210 310 410 510

]°C

onta

inm

ent a

ngle

[

-110

1

10 P7SOURCE_V6 Back 68%P7SOURCE_V6 Combined 68%P7SOURCE_V6 Front 68%P6_V3_DIFFUSE Back 68%P6_V3_DIFFUSE Combined 68%P6_V3_DIFFUSE Front 68%

(a)

Luca Baldini (INFN and UniPi) Fermi Summer School 2012 12 / 12