Post on 15-Jan-2016
transcript
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!