Lecture 3: Tools of the Trade
•
High Energy Units
•
4-Momenta
•
Cross-Sections
•
Mean Free Path
Useful Sections in Martin & Shaw:
Section 1.5, Appendix A, Appendix B
High Energy Units (''natural" units)
''c = ℏ = 1"
but really it’s just
LYING!
Say that you’re talking about mass, momentum, time and
length but actually convert everything to units of energy
using fundamental constants and pretend not to notice
1 eV/c2 = 1.782x10-36 kg
m∝E
E ⇔ mc2
L ∝ 1/E
E = ℏc/
p∝E
λ = ℏ/p = L
t ∝ 1/E
E=ℏω ⇔ ℏ/t
⇔ ℏc/L
E⇔pc
ℏc = 197 MeV fm (1 fm = 10-15m)
1 eV/c = 5.346 × 10-28 kg m/s
ℏ = 6.58 × 10-22 MeV s
However
The use of natural units is not always strictly followed
and sometimes is even done in a mixed manner !
Basically, just look at what units you have and decide
what units you would like given the nature of the
quantity being calculated. Then just figure out what
combinations of ℏ and c you need to get there
Don’t worry ⇒ the combination will be unique !
Example:
Convert the following to MKS units of acceleration:
a = 2.18x10−34 GeV
want to multiply by units
ℏ∝ET
c ∝ L T−1
LT−2E−1
ℏncm ∝ En Tn LmT−m= En Tn-m Lm
comparing ⇒ n = −1 & m = 1
c
factor =
=
ℏ
3x108m/s
6.58x10−25 GeV s
=
4.5x1032 m/s2/GeV
a = (2.18x10−34 GeV)(4.5x1032 m/s2/GeV) = 9.8 m/s2 = g
Steve’s Tips for Becoming a Particle Physicist
1) Be Lazy
2) Start Lying
4-Momenta
There are different formulations and different
sign conventions used for this... so beware!!
But (luckily) the basics are pretty straight-forward:
Define the 4-component vector P ≡ (E, px, py, pz) = (E, p)
(''natural" units, otherwise E → E/c)
Define the scalar product of 2 such beasts as
P1 • P2 = E1E2 - p1 • p2
Note that this means
P • P = P2 = E2 - p2 = m2
true for the square of
any linear combination
of 4-vectors!
⇒ relativistic invariant !!
Basic Recipe Idea:
1) Represent the reaction in terms of 4-momenta
(statement of energy and momentum conservation)
2) Algebraically manipulate 4-vectors to simplify
and insure the result will be couched in terms
of relevant angles, energies etc.
3) “Square” both sides of the equation, choosing
convenient reference frames for each side
4) Let cool for 15 minutes and serve warm with
plenty of custard and a nice, hot mug of tea
2 particle beams cross with angle θ.
Find the total CM energy in the limit E≫m.
What is this for a head-on collision?
Example:
)θ
p1
α
LAB
p2
( P1 + P2 )2 = PT2
P12 + P22 + 2P1 · P2 = PCM2
CM
PCM = (ECM, 0)
m12 + m22 + 2E1E2 - 2p1p2cosα = ECM2 – 02
for E≫m, ignore m1 & m2 and take p1 ∼ E1, p2 ∼ E2
⇒ ECM2 = 2E1E2 (1 - cosα) = 2E1E2 (1 + cosθ)
so, for a head-on collision ECM ≈ 2(E1E2)1/2
Example:
P1 +P2 = PF
What is the threshold energy for antiproton
production in the fixed target proton-proton
collision: p + p → p + p + p + p ?
1 2
3
4 5 6
(fixed)
where: PF = (E3+E4+E5+E6 , p3+p4+p5+p6) = (4m , 0) in CM at threshold
p
(P12 + P2)2 = PF2 ⇒
mp2 + mp2 + 2E1E2 – 2p1·p2 = (4mp)2 - (0)2
(in lab, p2= 0 , so E2=mp)
2E1mp + 2mp2 = 16mp2
E1 = 7mp
If target proton is moving with a
kinetic energy of 30 MeV, what is
the threshold kinetic energy ?
( p1 = √E12 - mp2 = 6.93mp )
T2 = E − mp = γmp - mp = mp(γ − 1)
⇒ γ=1.03 & β = 0.253
E′ = γ ( E − β p ) = 1.03 (E1− 0.253p1) = 5.4mp
T1′ = E′ − mp = 4.4mp
A high energy neutron travelling at velocity vn
undergoes beta decay: n → p + e−+ νe
The opening angle of the charged particle products
is measured, along with their energies. Determine
the angle of the neutrino trajectory with respect to
the incoming neutron direction.
Example:
νe
θ
n
p
α
Pn = Pp + Pe + Pν
e−
(Pn − Pν)2 = (Pp + Pe)2
Pn2 + Pν2 − 2(EnEν - pnpνcosθ) = Pp2 + Pe2 + 2(EpEe - pppecosα)
mn2 + 0 − 2Eν(En - pncosθ) = mp2 + me2 + 2(EpEe - pppecosα)
[mp2 + me2 − mn2 + 2(EpEe - pppecosα)]
cosθ =
2pnEν
[mp +
2
=
me2 −
− En / pn
mn + 2(EpEe - pppecosα)]
(γm/γmv)
2
2pn(En − Ep − Ee )
− 1/vn
Keeping with the vector idea, we can also
write the Lorentz transformation as :
( )(
E´
γ
−γ β
=
p∥´
−γ β
γ
)( )
E
where β = velocity of moving frame
γ = ( 1 - β2 ) -1/2
p∥
p∥ = component of p parallel to β
Note: this transformation matrix also applies for other
types of similarly defined 4-vectors as well!
Basic 4-Vectors:
X ≡ ( t, x ), P ≡ ( E, p )
ℏ
K ≡ ( ω, k )
ω′ = γ (ω − βk∥)
Relativistic
ω′/ω = γ (1 − βk /ω) = γ (1−
(1−β )
Doppler
Shift
(for photon)
∥
∥
More fun with 4-Vectors!
Start with X ≡ ( t, x ), P ≡ ( E, p )
Take time-derivative of X ...but which “time?”
d/dτ
d/dτ X = [(dt/dτ d/dt) t, (dt/dτ d/dt) x] = ( γ, γ v ) ≡ V
Similary,
d/dτ P = (γ dE/dt, γ f ) ≡ F
Cross-Sections etc.
(a quick review of useful
definitions and relations)
beam
(# in beam = NB)
projected
target area σ
(# targets = NT)
L
face area A
Interaction Probability = Fraction of area occupied by targets
= σ NT / A
# Interactions = Nint = σ NT NB /A
Alternate expression to eliminate A :
NT = (#/Volume) x Volume = ρΤ L A
Nint = σ NT NB /A
Nint = σ NB ρΤL
Alternate expression to connect σ with theory:
vB = beam velocity
Rate = Nint/t
t = L/vB
= vBNBNTσ / (LA)
Prob/Time = Rate/(NB NT)
= vB NB NT σ / V
W = vB σ / V
(transition rate per unit time)
Example: Transmission experiment
Relate the interaction cross-section to
easily measurable target properties and
the fraction of the beam which survives intact
after passing through the target
L
Beam Counters
⇓
NB
Target
Nint = NB - NTR
Transmission
Counter
⇓
NTR
Nint = σ NB ρΤL
σ = (NB - NTR)/(NBρΤL) = (1-fTR)/(ρΤL)
Mean Free Path
(Interaction Length)
Area σ
σλ = volume per interaction
= V/N = 1/ρ
⇒
λ ⇒ average length
travelled per interaction
λ = 1/(σρ)
Probability to go some distance x
without interacting:
(Poisson) P = µn e-µ / n!
In this case n = 0 and µ = N = V/(σλ) = x/λ
⇒
P = e-x/λ
x