Vxvx vyvy vzvz Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for...

vx

vy

vzClassically, for free particlesE = ½ mv2 = ½ m(vx

2 + vy2 + vz

2 )

Notice for any fixed E, m this definesa sphere of velocity points all which give the same kinetic energy.

The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere).

dV = 4v2dv

Classically, for free particlesE = ½ mv2 = ½ m(vx

2 + vy2 + vz

2 )

dv mvdE mE

Ed

mv

Eddv

2

We just argued the number of accessible states (the “density of states”) is proportional to 4v2dv

mE

dE

m

ECdvvCdN

2

244 2

dEEm

CdN 2/1

2/32

4

dN

dE E1/2

T absolute zero

The Maxwell-Boltzmann distribution describes not only how rms velocity increases with T but the spread about in the distribution as well

With increasing energy (temperature)a wider range of velocities become probable.

There are MORE combinations of molecules that can display this total E.

That’s exactly what “density of states” describes.

The absolute number is immaterial.It’s the slope with changing E,

dN/dE that’s needed.

For a particle confined to a cubic box of volume L3

the wave number vector p

is quantum mechanically constrained

ii nL

2

2222

222222

22 zyx nnnmLmm

pE

The continuum of a free particle’s momentum is realized by the limit L

For a LARGE VOLUME these energy levels may be spaced very closely with many states corresponding to a small range in wave number

xx dL

dn 2

The total number of states within a momentum range

33

3

2d

LdndndnN zyx

d

Vdd

L 23

23

3

2

4

2

dV

V

N 232

4

The total number of final states per unit volume

32

23 2

4

2

4

dpp

dV

dN

or

From E=p2/2m dE=p dp/m dE = v dp

dEv

p

V

dN3

2

2

4

number of final states per unit volume whose energies lie within the range from E to E+dE

v

p

dE

dN

VE 3

2

2

41

Giving (FINALLY) the transition rate of

EiN EVEW 2||

2

f

fiN

v

pEVE 3

22

2

4||

2

Since transition rate = FLUX × cross-section

/nvi

Particle density of incoming beam

but when dealing withone-on-one collisions n=1

If final states of target and projectile have non-zero spins

2

24

2)12)(12(||

ffi

dciN pvv

ssEVE

or

rate/FLUX

Total cross sections for incident proton,antiproton, positiveand negative pions,and positive andnegative kaons onproton and neutrontargets.

Breakdown of RutherfordScattering formula

When an incident particlegets close enough to the targetPb nucleus so that they interact

through the nuclear force (inaddition to the Coulomb forcethat acts when they are furtherapart) the Rutherford formula

no longer holds.

The point at which this breakdown occurs gives a

measure of the size of the nucleus.

R.M.Eisberg and C.E.PorterRev. Mod. Phys, 33, 190 (1961)

Total cross sections for + p + p and + p o++

The total +p cross section

40Ca

12C

16O

Electron scattering off nuclei

Left: total cross sections for K+ onprotons (top)deuterons (bottom)

Above: K-p total cross sections

Direct Production in pp Interactions

Jet Production in pp and pp Interactions

Scattering Cross Sections (light projectile off heavy target) a + b c + d

)12)(12(

)12)(12(2

24

2

ba

dc

fi

f

ss

ss

vv

p

if

M

counting numberof final states

counting numberof initial states

from (E)=dN/dE“flux”

Note: If a and b are unpolarized (randomly polarized)the experiment cannot distinguish between

or separate out the contributions from different possibilities, but measures the

scattered total of all possible spin combinations:

444321 total

avg

averaging over all possible (equally probable) initial states)2)(2(

)12)(12(

ba ss

We will find that everything is in fact derivable from a comprehensive

Lagrangian (better yet a Lagrangian density, where . L3dxL

As a preview assume L describes some generic fields: fi, fj, fk

wave functions…of matter fields …or interaction fields (potentials)

Invariance principles (symmetries) will guarantee that schematically we will find 3 basic type of terms:

1 mass terms:

2 kinetic energy terms:

3 interaction terms:

mfi2 or m2fi

2

Fermion Boson

fi fi or fi fi

g fi fj fkmay also contain derivativesor >3 fields…or both

Cross sections or decay rates are theoretically computed/predictedfrom the Matrix Elements (transition probabilities)

ief dri LM4

~

pick out the relevant terms in Lcan be expanded in a series of approximations

[the coefficients of each term being powers of a coupling constant, g]

Griffiths outlines the Feynmann rules that translate L terms into M factors

mass & kinetic terms

interaction terms

propagator

vertex

g

22~

mp

i

g

g

Vertices get hooked together with propagators, with eachvertex contributing one power of coupling to the calculation

of the matrix element for the process.

g 22

2

~mp

ig

You know from Quantum Mechanics: Amplitudes are, in general,COMPLEX NUMBERS

In e+e collisions,can’t distinguish

e+

e+

e+

e+

e- e-

e-e-

All diagrams with the same initial and final states

must be added, then squared.

The cross terms introduced by squaring describe interference between the diagrams

(sometimes suppressing rates!)

+

d = [flux] × | M |2 × (E) × 4

initialstate

properties

statistical factorcounting the number

of ways final stateproduced

will be a Lorentzinvariant phase space

dp3

2E

kinematic constraints on 4-momentum

•Matrix elements get squared

•Basic vertex of any interaction introduces a coupling factor, g

•g2 is the minimum factor associated with any process…usually the expansion parameter (coefficient) of any series approximation for the matrix element

e

electromagnetic

e2 1137

gweak

g2 GF10-5GeV-2

W±

gs

strong

gs2 s 0.1

g

/1 rpif

ieV

)r(

Remember that when we are dealing with “free” (unbounded) particles

/1 rpif e

V)r(

f

rkiie

rkie

f

3* )(),( drrVkkF

V

iffi

iffi

M

3)()( drrVeqFi

rqi

q q

The matrix

element M

is the fouriertransform ofthe potential!

pi

pi q = ki kf =(pi-pf )/ħ

Proton-proton (strong interaction) cross sections

The strong force has a very short effective range (unlike the coulomb force)

If assume a simple “black disk” model with fixed geometric cross section:

2~ ppp r typical hadron sizerp~1fm = 10-13cm

226103~ cm mb30(1 barn=10-24 cm2)

This ignores the dependence on E or resonances,

but from 1 to several 1000 GeV of beam energyits approximately correct!

35-40 mb

pp collisions

pp collisions

Note:Elastic scattering1/E dependence

Letting one cat out of the bag:Protons, anti-protons, neutrons are each composed of 3 quarks

The (lighter) mesons (+, 0, -, K+, K0, K-, …) … 2 quarks

Might predict: pnppppnp ~38mb ~42mb

and

3

2pp

Kp

pp

p

±p~ 25mb

K±p~ 20mb

n p + ee

ee + Ne* Ne +

N C + e e

Pu U +

20 10

20 10

13 7

13 6

236 94

232 92

Fundamental particle decays

Nuclear decays

Some observed decays

The transition rate, W (the “Golden Rule”) of initialfinalis also invoked to understand

ab+c (+ )decays

How do you calculate an “overlap” between ???nep e |

It almost seems a self-evident statement:

Any decay that’s possible will happen!

What makes it possible?What sort of conditions must be satisfied?

initialtotal mm Total charge q conserved.

J conserved.

NdtdN teNtN )0()(

tet )(Pprobability of surviving

to time t

mean lifetime = 1/

For any free particle (separation of space-time components)

/0)0()( tiEet

Such an expression CANNOT describe an unstable particle

since2//22 )0()0()( 00 tiEtiE eet

Instead mathematically introduce the exponential factor:

2//0)0()( ttiE eet

/22 )0()( tet

2//0)0()( ttiE eet

then

a decaying probabilityof surviving Note: =ħ

/)2/(0)0()( tiEiet

Also notice: effectively introduces an imaginary part to E

/)2/(0)0()( tiEiet

Applying a Fourier transform:

0

/)()( dtetEg iEt

0

]/)(2

[

0

/)2/(/

0

0

)0(

)0()(

dte

dteEg

tEEi

tiEiiEt

)](2

[0

EEit

e

2/)()(

0

iEEEg

still complex!

What’s this represent?

E distribution ofthe unstable state

4/)(

4/)(

220

2

max

EE

E

Breit-Wigner Resonance Curve

Expect

4/)(

4/)()(*~)(

220

2

max

EEEgegE

some constant

Eo E

1.0

0.5

MAX

= FWHM

When SPIN of the resonant state is included:

4/)(

4/

)12)(12(

)12()(

220

2

max

EEss

JE

ba

130-eV neutron resonancesscattering from 59Co

Transmission

-ray yield for neutron radiative

capture

+p elastic scattering cross-section in the region of the Δ++ resonance.

The central mass is 1232 MeV with a width =120 MeV

Cross-section for the reaction

e+e anything

near the Z0 resonanceplotted against

cms energy

In general: cross sections for free body decays (not resonances)are built exactly the same way as scattering cross sections.

DECAYS (2-body example) (2-body) SCATTERING

except for how the “flux” factor has to be defined

pE

ofcons

space

phasefluxd

,42 M

pE

ofcons

space

phasefluxd

,42 M

)()2(

2)2(2)2(2

1

32144

33

33

23

322

1

ppp

E

pcd

E

pcd

m

M

)()2(

2)2(2)2()(4

432144

33

34

23

332

121

2

pppp

E

pcd

E

pcd

pEE

M

in C.O.M.

in Lab frame:

cpm 12

2

4

enforces conservationof energy/momentum

when integratingover final states

Now the relativisticinvariant phase space

of both recoilingtarget and

scattered projectile

Number scatteredper unit time = (FLUX) × N × total

)()2(2)2(2)2()(4 4321

44

33

34

23

332

121

2

ppppE

pcd

E

pcd

pEEd

M

(a rate)/cm2·sec

A concentrationfocused into a small spot and

small time interval

densityof targets size of

eachtarget

Notice: is a

function of flux!