Antonio Ereditato University of Bern · by Lorentz transformations and Einstein special relativity holds. 3 space coordinates and time are inter-connected. A.Ereditato SS 2017 5 Time

A.Ereditato SS 2017 1

Elementarteilchenphysik

Antonio Ereditato

University of Bern

Introduction: kinematics, decays and collisions


•  In this course we deal with elementary particles: what does it mean ? •  the electron came first (>100 years ago) and then…all the others

•  circa 1960 >100 particles ! really elementary ?

•  today: a handful elementary particles

•  tomorrow ?

0 1000 1950 1800 today 1900

Elem

enta

ry ‘

part

icle

s’ 10

100

1


Elementary= no inner structure = pointlike

The pointlike nature depends on the space resolution of the probe!

λ A λ

A

De Broglie: λ = h/p Optical resolution: Δr ~ λ/sinθ

Δr ~ λ/sinθ = h/p sinθ ∼ h/q (q is the momentum transferred to the projectile)

Example: q = 10 GeV implies a space resolution of 10-16 m (HIGH ENERGY !)

HIGH ENERGY is also needed to produce high mass (NEW) particles (E=mc2) Note: 1) the “pointlike” nature of a particle is not absolute but “operational”

2) elementary particles are described by quantum mechanics


Lorentz transformations

Elementary particles have usually high speed (~c). Galileian transformations must be replaced by Lorentz transformations and Einstein special relativity holds. 3 space coordinates and time are inter-connected.

4


Time dilatation and space contraction…

Example: the proper life time (at rest) of a 10 GeV π+ meson is 26 ns. With a speed of ~c in the laboratory it should travel, on the average, 0.8 m. Instead it travels for ~57 m!


7

!

gµ" =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 #1

!

p " p = pµg µ# p#µ#

$ = c2r p "

r p % E " E = %m

0

2c4

INTRODUCTION TO RELATIVISTIC KINEMATICS

Some times the time-like component (E) is defined with an i (imaginary unit). In

this case gµν becomes the unity matrix and we are back in an Euclide’s space.

!

p " p = pµgµ#

p#µ#

$ = c2r p "

r p + iE( ) " iE( )

7

!

gµ" =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 #1

!

p " p = pµg µ# p#µ#

$ = c2r p "

r p % E " E = %m

0

2c4


Some times the time-like component (E) is defined with an i (imaginary unit). In

this case gµν becomes the unity matrix and we are back in an Euclide’s space.

!

p " p = pµgµ#

p#µ#

$ = c2r p "

r p + iE( ) " iE( )

6


Energy and momentum are components of the same four-vector:

The module squared of the four-momentum is a scalar called invariant mass and operatively

is defined as the mass measured in the reference standing with the moving body

!

E

m0c2

= "Homework: show that

and more in detail:

!

p " #(cm0

vx

c,cm

0

vy

c,cm

0

vz

c,m

0c2)

!

p " (cpx,cpycpz,E)

!

p.p " c2# (px # px + py # py + pz # pz ) $ E # E = $m

0

2# c

4

!

E2

= p2c2

+ m0

2c4

The square of a 4-vector is defined as follows and it is a Lorenz Invariant quantity

The invariant mass

4-vector formulation


0 1c ε= = =h

International Unit System

Energy : 1 eV = 1.602·10-19 J Length : 1 fm = 10-15 m = 1 Fermi Speed of light : c = 299 792 458 ms-1 Reduced Plank Constant : h/2π = 1.055·10-34 J s

natural units:

26

15 1

24 1

1 5.610 10

1 5.068 10

1 1.519 10

kg GeV

m GeV

s GeV

−

−

= ⋅

= ⋅

= ⋅

2 -2

1

L eV

1 197.3 MeV fm1

1 fm= MeV197.3

c

σ

−

# $ # $= =% & % &= =h

24 2 2

3 2

1barn 10 cm 100 fm

1mbarn 10 barn 0.1 fm

−

−

= =

= =

Fine structure constant:

In natural units: E2 = p2 + m2

€

β =vc

=pE

γ =Em


Energy [eV]

Energy [Joule]

Speed electron [m = 511keV]

Speed proton [m = 938 MeV]

1 eV 1.6 x 10-19 J 593 km/s 0.002 c

14 km/s 0.00005 c

1 keV 1.6 x 10-16 J 18730 km/s 0.062 c

438 km/s 0.0015 c

1 MeV 1.6 x 10-13 J 282128 km/s 0.94 c

13832 km/s 0.046 c

1 GeV 1.6 x 10-10 J 299792 km/s 0.9999998 c

262338 km/s 0.88 c

1 TeV 1.6 x 10-7 J 299792 km/s 0.9999999999998 c

299792 km/s 0.9999996 c

7 TeV 1.1 x 10-6 J 299792 km/s 0.999999999999997 c

299792 km/s 0.999999991 c

Some examples

Energy of one of the proton colliding beams of the LHC at CERN


Kinematics and invariant quantities

The concept of invariant mass can be extended to the case of a system of particles. If the particles “do not interact” we have a simple expression: The square of the mass is an invariant (positive) quantity: Since s is invariant, we can compute it in any reference frame. In the CM reference frame, where the total momentum is null, we have:

Namely, the invariant mass m of a system of non interacting particles is equivalent to the total energy in the CM system.


Processes in particle physics: decays and collisions

Decay of an unstable particle into other particles: a ! b + c + …

•  The final state is defined by what is actually measured. •  We can calculate the decay rate (# of events/unit time) of the specific final state. •  Recalling the uncertainty principle ΔEΔt ≥ h/4π one can relate the width Γbc of the

specific decay channel a ! b + c to its lifetime τb: Γbc~ 1/τbc •  If there are more decay channels, one talks of partial widths: Γbc , Γde , Γbcd , …

Their sum is the total width Γ, corresponding to the total lifetime τ of the decaying state. We will see that the measurement of the width allows to infer otherwise not measurable lifetimes.

•  We define the branching ratio of a to decay into b and c as Rbc = Γbc/Γ

a

b

c

DECAY


Particle collisions (1)

a + b = c + d + e +… (general inelastic reaction)

a + b = a + b (elastic reaction) Note: 1) a particle excited state is considered as a “different” particle

2) particles are considered to be “free” in the initial and final states (“fast” interaction)

In the laboratory reference frame: s = (Ea + mb)2 – pa

2 = ma2 + mb

2 + 2mbEa and, neglecting the masses vs the energy, s ≈ 2mbEa

In the CM system (also here neglecting masses): Ea* ≈ pa* and Eb* ≈ pb* Therefore, being the momenta opposite, we have s ≈ (2E*)2

Example: fixed target accelerator collisions vs head-on collisions :


Particle collisions (2)

a + b = c + d (two-body scattering)

Note: in general, the initial and final momenta in the CM frame are different (unlike in an elastic scattering) Since s is invariant, it must be the same in the two systems as well as before and after the collision: We can define other invariant quantities made up of combinations of initial and final momenta: One can show that the following relation holds: s, t and u are called Mandelstam variables.

Note: that s ≥ 0 while t ≤ 0 and u ≤ 0


Cross section

Definition of cross section (dimensions of a surface):

Number of particle interactions per unit time per target/incident flux

Where the flux is the number of incident particles per unit time and per unit surface In practice: the cross section is the surface of the

“shadow” of the nucleus We can define the differential cross section:

Number of particle interactions per unit time per target per solid angle/incident flux

Prof. M.A. Thomson Michaelmas 2009 27

Cross section definition

• The “cross section”, !, can be thought of as the effective cross-sectional area of the target particles for the interaction to occur.

• In general this has nothing to do with the physical size of thetarget although there are exceptions, e.g. neutron absorption

!here is the projective area of nucleus

no of interactions per unit time/per target incident flux

! =

Differential Cross sectionno of particles per sec/per target into d"

incident flux =d!

d"

integrate over all other particles

#e–

e–

p

Flux = number ofincident particles/unit area/unit time

d!d$$$

or generally

with


• Consider a single particle of type a with velocity, va, traversing a region of area A containing nb particles of type b per unit volume

vaA vbIn time %t a particle of type a traverses region containingparticles of type b

A! !Interaction probability obtained from effective

cross-sectional area occupied by the particles of type b

• Interaction Probability =

nb v !Rate per particle of type a =

• Consider volume V, total reaction rate = =

example

• As anticipated: Rate = Flux x Number of targets x cross section







! =


incident flux =d!

d"


#e–

e–

p


d!d$$$

or generally

with









example








! =


incident flux =d!

d"


#e–

e–

p


d!d$$$

or generally

with









example



Transition rate (e.g. for a collision reaction)

Assume a reaction

€

a + b→ c + d where a beam of particles a with density na and velocity vi hits a target of thickness dx containing nb particles b per unit volume. c and d are the reaction products







! =


incident flux =d!

d"


#e–

e–

p


d!d$$$

or generally

with









example








! =


incident flux =d!

d"


#e–

e–

p


d!d$$$

or generally

with









example


4

We first want to calculate W, the reaction rate per (one) target particle.

Assume each particle b covers an effective area σ (=reaction cross section). If an incoming particle a arrives within this area, a reaction occurs.

Assume incoming particles are distributed uniformly within an area A. Each incoming particle then has a reaction probability equal to σ/A.

(happy with σ, but want to replace A)

Let incoming particles come with speed

The rate of incoming particles (crossing an area) is then given by

The reaction/probability/transition rate is then

iaaaaa Avn

dtdxAn

dtAxnd

dtVnd

dtdN

====)()(

)15.2 Eq.( )( prob.)reaction ((time)

incoming) of(number φσσσ===×= ia

a vnAdt

dNW

iv

aV

anaN

a

a

incoming offlux : volume:

incoming ofdensity : incoming ofnumber :

φ

[A disappears (as it should)]

The reaction rate W

4








iaaaaa Avn

dtdxAn

dtAxnd

dtVnd

dtdN

====)()(



a vnAdt

dNW

iv

aV

anaN

a

a



φ


The reaction rate W

φ =naA× t

=ρ ×VA× t

=ρ × vi × A

A= ρ × vi

The transition rate W is equal to:

And hence the cross section per target particle is equal to the reaction rate per unit incident flux:

W = φ σ4








iaaaaa Avn

dtdxAn

dtAxnd

dtVnd

dtdN

====)()(



a vnAdt

dNW

iv

aV

anaN

a

a



φ


The reaction rate W


Particle quantum state transitions

•  Particle physics: “observe” interactions between particles and decays of unstable particles, described by quantum wave functions

•  These are state transitions that one wants to calculate (predict)

•  Use first order quantum mechanics perturbation theory: Fermi Golden Rule

W is the (not Lorentz Invariant) rate of transitions from state to Mfi is the matrix element, describing the action of the “perturbation” Hamiltonian to leading order

ρ(E)f is the density of available final states dN/dEf The matrix element contains the physics of the transition (specific interaction); the phase space is a sort of “weight” including the actual kinematics of the reaction


Cross Sections and Decay Rates• In particle physics we are mainly concerned

with particle interactions and decays, i.e.transitions between states

• Calculate transition rates from Fermi’s Golden Rule

is Transition Matrix Element

is density of final states

is number of transitions per unit time from initial stateto final state – not Lorentz Invariant !

! Rates depend on MATRIX ELEMENT and DENSITY OF STATES

the ME contains the fundamental particle physics

" these are the experimental observables of particle physics

just kinematics

is the perturbingHamiltonian


The first five lectures

e– !–

e+ !"#

! Aiming towards a proper calculation of decay and scattering processesWill concentrate on: e– e–

qq

• e+e–! !+!–

• e– q ! e– q(e– q!e– q to probeproton structure)

# Need relativistic calculations of particle decay rates and cross sections:

# Need relativistic calculation of interaction Matrix Element: Interaction by particle exchange and Feynman rules

# Need relativistic treatment of spin-half particles:Dirac Equation

+ and a few mathematical tricks along, e.g. the Dirac Delta Function











just kinematics




e– !–

e+ !"#


qq

• e+e–! !+!–






W =2πhM fi

2ρ(E) f











just kinematics




e– !–

e+ !"#


qq

• e+e–! !+!–






Mfi

5

Fermi’s Golden Rule:

(Goto QM books to see derived)

fifMW ρπ 22h

=

energy eappropriat with states final ofdensity ==dEdN

fρ

∫==

=

dVxxUx

M

i

if

)()()(

Upotential todue ions wavefunctstate final and initialbetween integral overlap

state final and initialbetween element matrix

*f

rrr ψψ

(non-rel.)


Compute the cross section of the two particle collision

We consider 1 single particle in 1 dimension, confined in a segment [0, L] Impose periodicity in the segment (wave function vanishing outside the box) with number of possible states. Then:

6

dEdpL

dEdN

pLN

NpLexLx

ipx

h

h

hh

π

π

πψ

2

states possible ofnumber 2

2/ i.e. ,)( periodic Need

0 toconfineddimension 1in particle single aConsider /

=

==

=∝

≤≤

2.17] Eq.[ 22

(E)

:dimensions 3in particle Single

23

33

≈Ω⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛== ∫∫ d

dEdppLpd

dEdL

dEdN

hh ππρ

But we have 2 outgoing (final state) particles (c and d) not just one.

Do we have more phase space?

No, energy-momentum conservation fixes the other particle.

(So p here is the (absolute value of the) momentum of either c or d.)

Phase space of final states

)( 3LV =

6

dEdpL

dEdN

pLN

NpLexLx

ipx

h

h

hh

π

π

πψ

2




=

==

=∝

≤≤

2.17] Eq.[ 22

(E)


23

33

≈Ω⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛== ∫∫ d

dEdppLpd

dEdL

dEdN

hh ππρ






)( 3LV =

6

dEdpL

dEdN

pLN

NpLexLx

ipx

h

h

hh

π

π

πψ

2




=

==

=∝

≤≤

2.17] Eq.[ 22

(E)


23

33

≈Ω⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛== ∫∫ d

dEdppLpd

dEdL

dEdN

hh ππρ






)( 3LV =

6

dEdpL

dEdN

pLN

NpLexLx

ipx

h

h

hh

π

π

πψ

2




=

==

=∝

≤≤

2.17] Eq.[ 22

(E)


23

33

≈Ω⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛== ∫∫ d

dEdppLpd

dEdL

dEdN

hh ππρ






)( 3LV =

A single particle in 3D:

If we go back to our 2 particle collision reaction: , we have 2 outgoing (final state) particles (c and d) and not just one, but energy-momentum conservation fixes the kinematics of the other particle. Therefore p is the (absolute value of the) momentum of either c or d. Then, the the cross section of the process can be derived:

6

dEdpL

dEdN

pLN

NpLexLx

ipx

h

h

hh

π

π

πψ

2




=

==

=∝

≤≤

2.17] Eq.[ 22

(E)


23

33

≈Ω⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛== ∫∫ d

dEdppLpd

dEdL

dEdN

hh ππρ






)( 3LV =

€

a + b→ c + d

7

( )2.18] Eq.[

212

221

2

32

3

2

23

2

2

≈×=Ω

Ω⎟⎠⎞

⎜⎝⎛==

==

∫

ai

if

ifiaia

iaif

nL

dEdpp

vM

dd

ddEdppLM

vnvnW

vnMW

hh

hh

h

ππσ

ππσ

σρπ

NB: extra factor dropped in Perkins (will cancel out in the end)

Then continue in center of mass frame:

fdcd

f

c

f

df

f

cf

f

f

dfcfdcf

vvvEp

Ep

mp

p

mp

pdpdE

mpmpEpppp

=+=+=+

++

=

+++====

22220

22220

2

2

2

2

and rr

(relative velocity)

( ) 2.19] Eq.[ 4

1)( 2

22

42 ≈×=+→+Ω

Vvvp

Mdcbadd

fi

fif

hπσ

Putting it all together:

( )"1" =aVn

The cross section then becomes ....

(Anticipating the normalisation of the wavefunction)

) 22

rel.-non fromrelation (same22

0d

f

c

f

mp

mp

E +=

(Similar expression for relativistic case)

7

( )2.18] Eq.[

212

221

2

32

3

2

23

2

2

≈×=Ω

Ω⎟⎠⎞

⎜⎝⎛==

==

∫

ai

if

ifiaia

iaif

nL

dEdpp

vM

dd

ddEdppLM

vnvnW

vnMW

hh

hh

h

ππσ

ππσ

σρπ



fdcd

f

c

f

df

f

cf

f

f

dfcfdcf

vvvEp

Ep

mp

p

mp

pdpdE

mpmpEpppp

=+=+=+

++

=

+++====

22220

22220

2

2

2

2

and rr

(relative velocity)

( ) 2.19] Eq.[ 4

1)( 2

22

42 ≈×=+→+Ω

Vvvp

Mdcbadd

fi

fif

hπσ


( )"1" =aVn



) 22


0d

f

c

f

mp

mp

E +=



7

( )2.18] Eq.[

212

221

2

32

3

2

23

2

2

≈×=Ω

Ω⎟⎠⎞

⎜⎝⎛==

==

∫

ai

if

ifiaia

iaif

nL

dEdpp

vM

dd

ddEdppLM

vnvnW

vnMW

hh

hh

h

ππσ

ππσ

σρπ



fdcd

f

c

f

df

f

cf

f

f

dfcfdcf

vvvEp

Ep

mp

p

mp

pdpdE

mpmpEpppp

=+=+=+

++

=

+++====

22220

22220

2

2

2

2

and rr

(relative velocity)

( ) 2.19] Eq.[ 4

1)( 2

22

42 ≈×=+→+Ω

Vvvp

Mdcbadd

fi

fif

hπσ


( )"1" =aVn



) 22


0d

f

c

f

mp

mp

E +=


If we continue in the CM system: We finally obtain: This is a non-relativistic formulation that also does not take into account the spin of the particles involved in the initial and final states.

7

( )2.18] Eq.[

212

221

2

32

3

2

23

2

2

≈×=Ω

Ω⎟⎠⎞

⎜⎝⎛==

==

∫

ai

if

ifiaia

iaif

nL

dEdpp

vM

dd

ddEdppLM

vnvnW

vnMW

hh

hh

h

ππσ

ππσ

σρπ



fdcd

f

c

f

df

f

cf

f

f

dfcfdcf

vvvEp

Ep

mp

p

mp

pdpdE

mpmpEpppp

=+=+=+

++

=

+++====

22220

22220

2

2

2

2

and rr

(relative velocity)

( ) 2.19] Eq.[ 4

1)( 2

22

42 ≈×=+→+Ω

Vvvp

Mdcbadd

fi

fif

hπσ


( )"1" =aVn



) 22


0d

f

c

f

mp

mp

E +=


7

( )2.18] Eq.[

212

221

2

32

3

2

23

2

2

≈×=Ω

Ω⎟⎠⎞

⎜⎝⎛==

==

∫

ai

if

ifiaia

iaif

nL

dEdpp

vM

dd

ddEdppLM

vnvnW

vnMW

hh

hh

h

ππσ

ππσ

σρπ



fdcd

f

c

f

df

f

cf

f

f

dfcfdcf

vvvEp

Ep

mp

p

mp

pdpdE

mpmpEpppp

=+=+=+

++

=

+++====

22220

22220

2

2

2

2

and rr

(relative velocity)

( ) 2.19] Eq.[ 4

1)( 2

22

42 ≈×=+→+Ω

Vvvp

Mdcbadd

fi

fif

hπσ


( )"1" =aVn



) 22


0d

f

c

f

mp

mp

E +=


Documents

Antonio Ereditato University of Bern · by Lorentz transformations and Einstein special relativity holds. 3 space coordinates and time are inter-connected. A.Ereditato SS 2017 5 Time