Lecture 3: Tools of the Trade

• High Energy Units

• 4-Vectors

• Cross-Sections

• Mean Free Path

Section 1.5, Appendix A, Appendix B

Useful Sections in Martin & Shaw:

''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

ℏ = 6.58 10-22 MeV s

m E

L 1/E

p E

t 1/E

High Energy Units (''natural" units)

E mc2 1 eV/c2 = 1.782x10-36 kg

E = ℏc/ƛ ℏc/L ℏc = 197 MeV fm (1 fm = 10-15m)

Epc 1 eV/c = 5.346 10-28 kg m/s

E=ℏ ℏ/t

Don’t worry the combination will be unique !

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

The use of natural units is not always strictly followedand sometimes is even done in a mixed manner !

However

Example: Convert the following to MKS units of acceleration:

2.18x1034 GeV

want to multiply by units LT2E1 ℏ E T c L T1

ℏacb Ea Ta LbTb

comparing a = 1 & b = 1

factor = cℏ

3x108m/s6.58x105 GeV s= = 4.5x1032 m/s2/GeV

(2.18x1034 GeV)(4.5x1032 m/s2/GeV) = 9.8 m/s2 = g

= Ea Ta-b Lb

Steve’s Tips for Becoming a Particle Physicist

1) Be Lazy2) Start Lying

There are different formulations and differentsign conventions used for this... so beware!!

But (luckily) the basics are pretty straight-forward:

Define the 4-component vector P (E, px, p

y, p

z) = (E, p)

(''natural" units, otherwise E E/c)

Define the scalar product of 2 such beasts as

P1 P

2 = E

1E

2 - p

1 p

2

Note that this means

P P

= P2 = E2 - p2 = m2 relativistic invariant !!

4-Momenta

true for the square of any linear combination of 4-vectors!

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 .Find the total CM energy in the limit E≫m.What is this for a head-on collision?

P1

2 + P22 + 2P1 · P2 = P

CM2

ECM

2 = 2E1E

2 (1 - cos)

for E≫m, ignore m1 & m

2 and take p

1 E

1, p

2 E

2

Example:

p1 p

2

m1

2 + m2

2 + 2E1E

2 - 2p

1p

2cos = E

CM2 – 02

P1 + P2 = PT

( )2 2LAB CM

PCM

= (ECM

, 0)

= 2E1E

2 (1 + cos)

so, for a head-on collision ECM

≈ 2(E1E

2)1/2

P1 +P2 = PF

where: PF = (E

3+E

4+E

5+E

6 , p

3+p

4+p

5+p

6)

(P1 + P2)2 = P

F2

Example:

mp2 + mp

2 + 2E1E2 – 2p1·p2 = (4m

p)2 - (0)2

What is the threshold energy for antiprotonproduction in the fixed target proton-proton

collision: p + p p + p + p + p ?

1 2 3 4 5 6 (fixed)

= (4mp , 0) in CM at threshold

(in lab, p2= 0 , so E

2=m

p)

2E

1m

p + 2m

p2 = 16m

p2

E

1 = 7m

p

If target proton is moving with a kinetic energy of 30 MeV, what isthe threshold kinetic energy ?

T2 = E m

p = m

p - m

p = m

p(1)

=1.03 & = 0.253

E = ( E p ) = 1.03 (E10.253p

1)

( p1 = E

12 - m

p2 = 6.93m

p)

= 5.4mp

T1 = E m

p = 4.4m

p

Example:A high energy neutron travelling at velocity v

n

undergoes beta decay: n p + e+ e

The opening angle of the charged particle productsis measured, along with their energies. Determine the angle of the neutrino trajectory with respect to the incoming neutron direction.

Pn = P

p + P

e + P

Pn P= P

p + P

e ( )2 ( )2

Pn

2 P22(E

nE - pn

pcos) = Pp

2 Pe22(E

pE

e - p

pp

ecos)

mn

2 02E(En - p

ncos) = m

p2 m

e22(E

pE

e - p

pp

ecos)

[mp

2 me2m

n2 2(E

pE

e - p

pp

ecos)]

2pnE

cos = En / p

n

p

e

n

e

= [m

p2 m

e2m

n2 2(E

pE

e - p

pp

ecos)]

2pn(E

nEp E

e )

vn

(m/mv)

Keeping with the vector idea, we can also write the Lorentz transformation as :

E´ E =p||´ p|| () ( )( )

where = velocity of moving frame

= ( 1 - 2 ) -1/2

p|| = component of p parallel to

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 )

K ( , k ) ℏ

k||||(for photon)

RelativisticDoppler Shift

k||

More fun with 4-Vectors!

Start with X ( t, x ), P ( E, p )

Take time-derivative of X ...but which “time?” d/d

d/d X = [(dt/d d/dt) t, (dt/d d/dt) x]

Similary, d/dP = (dE/dt, f ) F

= ( , v ) V

Pions can decay via the reaction + + . Find the energy of the neutrino in the rest frame of the pion.

P = P + P

P = P - P ( & in same frame)

P2 = P

2 + P2 - 2P· P rest frame rest frame

m2 = m

2 + 0 – 2(EE – p·p)

m2 - m

2 = 2(EE –p·p)

E= (m2 - m

2 )/2m

3sheet 1

= 2Em

Derive an expression relating the emission angle of themuon or neutrino with respect to the beam in the CM tothat in the lab frame.

CM frame

pCM

lab frame

plab

Transverse momentum: plabsin = pcmsin

Longitudinal momentum: plabcos = (pcmcosEcm

divide: tan = sin[(cosEcm/pcm

= sin[(cos/ppcm/Ecm = p

(particle)

Are there any limits to the lab directions of the muon or neutrino?

To find maximum angle, set d/d = 0 cosmax = -p

Not possible for since p (i.e. no restriction), but a restriction is possible for

Thus, a solution is well defined if p, or if p

3sheet 1

A cloud of matter is ejected with high velocity from a distant galactic nucleus. The cloud moves from A to B in the diagram, emitting photons seen, over a period of several years, by an observer on Earth. In the Earth’s frame of reference, the cloud’s velocity has constant magnitude v and is at an angle with the line of sight.

A

BC

The points B, C represent the space coordinates of space-time points on the paths of photons seen at different times by the observer. Derive an expression, in terms of v and for the apparent transverse velocity of the cloud which would be deduced by the observer from the difference between the arrival times of photons with the paths shown and from the measured spatial separation of B and C.

Time for matter to get from A to B = L/v

Time for light to get from A to C = L cos/c

Distance from C to B = L sin

Apparent velocity = DCB

t

= L sin

(L/v - L cos/c)

Lv

= v sin

(1 - cos)

7sheet 1

face area A

L

Interaction Probability = Fraction of area occupied by targets = N

T / A

# Interactions = Nint

= NT N

B /A

projectedtarget area

(# targets = NT)

beam

(# in beam = NB)

Cross-Sections etc. (a quick review of useful definitions and relations)

(transition rate per unit time)

Alternate expression to eliminate A :

NT = (#/Volume) x Volume = L A

Alternate expression to connect with theory:

Rate = Nint

/t

Nint

= NB L

vB = beam velocity t = L/v

B

Nint

= NT N

B /A

W = vB / VProb/Time = Rate/(N

B N

T)

= vB N

B N

T / V= v

BN

BN

T / (LA)

Target

L

= (NB - N

TR)/(N

BL)

Example: Transmission experiment

Beam Counters N

B

Transmission Counter N

TRNint

= NB - N

TR

Nint

= NB L

= (1-fTR

)/(L)

Relate the interaction cross-section to easily measurable target properties and the fraction of the beam which survives intact after passing through the target

(Interaction Length)

Area average length travelled per interaction

= volume per interaction = V/N = 1/

P = e-x/

Mean Free Path

x

Probability to go some distance xwithout interacting:

(Poisson) P = n e- / n!

In this case n = 0 and = N = V/() = x/