Lecture 23 HW Particle Detectors - Michigan State UniversityParticle Detectors. April 9, 2007 Carl...

PHY492: Nuclear & Particle Physics

Lecture 23

HWParticle Detectors

April 9, 2007 Carl Bromberg - Prof. of Physics 2

13.1

xmin

=mω 2

λ⎛

⎝⎜⎞

⎠⎟

1

2

xmin2 =

mω 2

λ; x

min3 =

mω 2

λ⎛

⎝⎜⎞

⎠⎟

3

2

; xmin4 =

m2ω 4

λ 2

V xmin

+ x, y( ) = −1

2mω 2 x

min+ x( )2

+ y2⎡⎣⎢

⎤⎦⎥+λ4

xmin

+ x( )2+ y2⎡

⎣⎢⎤⎦⎥

2

= −1

2mω 2 x

min2 + 2x

minx + x2 + y2⎡⎣ ⎤⎦ +

λ4

xmin4 + 4x

min3 x + 6x

min2 x2 + 4x

minx3 + x4

+ 2 xmin2 + 2x

minx + x2( ) y2 + y4

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −1

2

m2ω 4

λ− mω 2 mω 2

λ⎛

⎝⎜⎞

⎠⎟

1

2

x −1

2mω 2x2 +…

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

+1

4

m2ω 4

λ+ mω 2 mω 2

λ⎛

⎝⎜⎞

⎠⎟

1

2

x +3

2mω 2x2 +…

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= −1

4

m2ω 4

λ+ mω 2x2 +… Q.E.D.

Homework


Homework

δEδt ; τ

Γ=

6.6 ×10−22 MeV ⋅ s1.5×103 MeV

= 4.4 × 0−25s

ti= 1 fm/c = 3.3×10−24s

τti

=4.4 ×10−25s

3.3×10−24s= 1.3×10−1

t →W + b

EW+ E

b= m

t; E

W2 = m

t2 − 2m

tE

b+ E

b2

EW2 − m

W2 = p2 = E

b2 − m

b2; E

W2 = E

b2 − m

b2 + m

W2

E

b= m

t2 + m

b2 − m

W2 2m

t( ) = 69 GeV; pb= E

b2 − m

b2( )1

2 = 69 GeV/c

Ea= x

a

s

2; p

a= x

a

s

2c; E

b= x

b

s

2; p

b= −x

b

s

2c

s = Ea+ E

b( )2− p

ac + p

bc( )2

= xa2 + x

b2( ) s

4+ x

ax

b

s

2− x

a2 + x

b2( ) s

4+ x

ax

b

s

2= x

ax

bs

.350( )2

= xax

b4( ); x

ax

b= 0.03; x

a x

b= .03 = 0.18 xa

xb= 0.025

14.1 a) Top lifetime b) No time to interact

c) Top quark decayEnergy Cons.:

Momentum Cons.:

d) Parton-parton collision

top mass = 0.175 TeV sTeV =(2 TeV)2 = 4 TeV2 sLHC =(14 TeV)2 = 200 TeV2


Homework


Homework

⎧

⎨⎪⎪

⎩⎪⎪ K

0d

µ+

µ−

s

Z0

d

s

⎛⎝⎜

⎞⎠⎟=

cosθC

− sinθC

sinθC

cosθC

⎛

⎝⎜

⎞

⎠⎟

′d

′s⎛⎝⎜

⎞⎠⎟

d = ′d cosθC− ′s sinθ

C

s = ′d sinθC+ ′s cosθ

C

K0 = d ,s( )

d ,s Z = ′d , ′d Z cosθC

sinθC− ′s , ′s Z cosθ

Csinθ

C

+ ′d , ′s Z cos2θC+ ′s , ′d Z sin2θ

C

′d , ′s Z and ′s , ′d Z are both ZERO

14.5 Quark mixing matrix 2nd order weak decays (u/c)

Not seen: 1st order weak decayflavor changing neutral current

Quark content

′d , ′d Z = ′s , ′s Z ≠ 0

d ,s Z = ′d , ′d Z cosθ

Csinθ

C− ′s , ′s Z cosθ

Csinθ

C= 0

Weak interaction acts on mixed states

Z couples only to the same weak flavor quarks

But 4 quark mixing matrix gives this cancellation


HomeworkDIS

Q = ′k −k( ),ν⎡⎣ ⎤⎦; P = 0,m

pc2⎡⎣ ⎤⎦

Q ⋅ P = mpc2ν (invariant)

xP +

Q( )2= 0 = x2 P2 + 2x

P ⋅

Q + Q2

x =Q2

2P ⋅

Q=

Q2

2mpνc2

for Q2 >> x2 P2 = x2mp2c4

14.7 a) 4-vector dot product (lab frame)

b) parton absorbs Q but remains massless

Q2

c4

2m

pν / c2

x = 1

x = 0.5

x = 0.1

W = 3m

p

W = 5m

p

W = m

p

(GeV2 / c4)

20 15 10 50

5

10

15

20

i)

ii)

GeV2

c4

⎛⎝⎜

⎞⎠⎟

W 2 = m

p2 +

2mpc2ν

c4−

Q2

c4

c) from 14.6

d) shaded regions

i) ν = E − ′E = 9 GeV; Q2 = 4E ′E sin2 θ / 2( )2m

pν = 17 GeV2 / c2; Q2 = 10 GeV2

ii) ν = E − ′E = 6 GeV; Q2 = 4E ′E sin2 θ / 2( )2m

pν = 11.3 GeV2 / c2; Q2 = 10.7 GeV2

e) two points on plot

infinitemomentumframe


• Particles are detected by making them ionize atoms!• Detecting charged particles

– The electric field of a moving charged particle can ionize theatoms of the material in which it moves.

– Ionization electrons are small and low mass, and can be collectedby an electric field. Positive ions are big and heavy, and sluggish.

– A charged particle accelerated by a magnetic or electric fieldradiates photons that can ionize atoms and release electrons

• Detecting neutral particles– Interact the neutral particle with matter and in the process

release ionization electrons.– Sometimes you must completely destroy the neutral particle, but

its energy has been used to create ionization.• Must study Ionization to understand detectors

How to detect particles


Ionization

• Ionization potential minimum energy to ionize (outer e shell)– Hydrogen 13.5 eV– Helium 25 eV– Lithium 5 eV– Neon 22 eV

• Average ionization potential, includes inner shells– reaction dependent– for charged particles (e.g., electrons)

• ~16Z0.9 (eV) for Z>1• Low Z Nobel Gases (He, Ne, Ar, a little higher)


Photon induced ionization• Photon (<20 eV) induced ionization

– Only valence electrons (a few)– Non-penetrating– Gases & surfaces

• high temperature– thermionic emission

• high electric fields• ultraviolet light

– photoelectric effect– ozone

• photo-cathodes (Cs)• silicon photodiodes

• X-ray (<1 MeV) induced– All electrons (Z)– Penetrating– Gases & solid interiors

E = hν =

hc

λ=

1200 eV ⋅ nm( )λ

• Useful conversion

λ − wavelength, ν − frequency

hc = 2π c( ) = 2π 200 MeV ⋅ fm( )= 1.2 ×103 eV ⋅ nm( )

• Photon energies


Particle Physics Booklet• Particle Data Group - http://pdg.lbl.gov/• Products

– Order booklet http://pdg.lbl.gov/receive_our_products.htmlhttp://pdg.lbl.gov/2005/reviews/contents_sports.html#expmethetc

http://teachers.web.cern.ch/teachers/archiv/HST2002/

• Detector Lectures for Students/Teachers


Charged particle induced ionization

• nion is particle velocity and charge dependent (Bethe and Bloch)

S(T ) = −

dT

dx= n

ionI

T : kinetic energy of moving particle

nion

: number of electron-ion pairs unit path length

I : average energy electron-ion pair

• Moving particle, mass M, ionizes atoms in medium

S(T ) =4πQ2e2nZ

meβ 2c2

ln2m

ec2γ 2β 2

I

⎛

⎝⎜

⎞

⎠⎟ − β 2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

γ =E

Mc2; β =

pc

E; γβ =

p

Mc

S(T ) ∝

1

β 2c2=

1

v2 S minimizes, γβ 3

β = v/c = < 0.8heavy ionization

S ∝ lnγ

when β ~ 0.95minimum ionization

ultra relativisticrise of ionization

Stopping power


Ionization in gases (relevant to all particles)

Normalized Ionization vs γβIonization vs momentumfor various particle masses

γβ =

pc

Mc2

Argon Gas


Saturation of ionization in solids• Relativistic rise of ionization is due to electric field

concentration perpendicular to direction of motion

• Solids polarize and shieldfar electrons from field

• Atoms along the line of motionsee a stronger field as v -> c.

• Effect is largest in large Zgases, e.g., Xe

effect saturates at only afew % above the minimum


Minimum ionization in thin solids

• Units for energy loss– Z/A ~ 0.4 at large A, energy loss proportional to density S~ρ,– Divided by the density -> value nearly independent of material.

• (dE/dx) min tabularized for various materials in MeV/(g/cm2)

– Polystyrene scintillator: 1.95

– Iron (steel) : 1.45

S(T ) = −

dT

dx∝ nZ = ρA

0

Z

A

Z : atomic number of medium

n : number of atoms/unit volume

n =ρA

0

A; A : atomic number of medium

ρiron

= 7.87 g/cm3

−dT

dxmin

= 1.45 MeV

g/cm2

⎛⎝⎜

⎞⎠⎟ρ

iron= 11.4 MeV/cm

ρscintillator

= 1.03 g/cm3

−dT

dxmin

= 1.95 MeV

g/cm2

⎛⎝⎜

⎞⎠⎟ρ

Scintillator= 2.0 MeV/cm

Relativistic muon loses ~2 MeV/cm in plastic, ~11.4 MeV/cm in Iron

Lecture 23 HW Particle Detectors - Michigan State UniversityParticle Detectors. April 9, 2007 Carl...

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