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  MeV s m  E L  1/E p  E t  1/E High Energy Units High Energy Units (''natural" units) E  mc 2 1 eV/c 2 = 1.782x kg E = ℏ c/ ƛ  ℏ c/L ℏ c = 197 MeV fm (1 fm = m) E  pc 1 eV/c =  kg m/s E=ℏ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 followed and sometimes is even done in a mixed manner ! However Dealing with High Energy Units

Example: Convert the following to MKS units of acceleration: 2.18x10  34 GeV want to multiply by units LT  2 E  1 ℏ  E T c  L T  1 ℏacbℏacb  E a T a L b T  b comparing  a =  1 & b = 1 factor = cℏcℏ 3x10 8 m/s 6.58x10  5 GeV s = = 4.5x10 32 m/s 2 /GeV (2.18x10  34 GeV)(4.5x10 32 m/s 2 /GeV) = 9.8 m/s 2 = g = E a T a-b L b Conversion Example

Steve’s Tips for Becoming a Particle Physicist 1) Be Lazy Tips 1 2) Start Lying

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, p x, p y, p z ) = (E, p) (''natural" units, otherwise E  E/c) Define the scalar product of 2 such beasts as P 1  P 2 = E 1 E 2 - p 1  p 2 Note that this means P  P = P 2 = E 2 - p 2 = m 2  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? P P P 1 · P 2 = P CM 2  E CM 2 = 2E 1 E 2 (1 - cos  ) for E ≫ m, ignore m 1 & m 2 and take p 1  E 1, p 2  E 2 Example:    p1p1 p2p2 Example: beam collisions m m E 1 E 2 - 2p 1 p 2 cos  = E CM 2 – 0 2 P 1 + P 2 = P T ( ) 2 2 LAB CM P CM = (E CM, 0) = 2E 1 E 2 (1 + cos  ) so, for a head-on collision E CM ≈ 2(E 1 E 2 ) 1/2

P 1 +P 2 = P F where: P F = (E 3 +E 4 +E 5 +E 6, p 3 +p 4 +p 5 +p 6 ) (P 1 + P 2 ) 2 = P F 2 Example: Example: threshold production  m p 2 + m p 2 + 2E 1 E 2 – 2p 1 ·p 2 = (4m p ) 2 - (0) 2 What is the threshold energy for antiproton production in the fixed target proton-proton collision: p + p  p + p + p + p ? (fixed) = (4m p, 0) in CM at threshold (in lab, p 2 = 0, so E 2 =m p ) 2E 1 m p + 2m p 2 = 16m p 2 E 1 = 7m p If target proton is moving with a kinetic energy of 30 MeV, what is the threshold kinetic energy ? T 2 = E  m p =  m p - m p = m p (  1)   =1.03 &  = E =  ( E  p )= 1.03 (E 1  0.253p 1 ) ( p 1 =  E m p 2 = 6.93m p ) = 5.4m p T 1 = E  m p = 4.4m p

Example: Example: beta decay A high energy neutron travelling at velocity v n undergoes beta decay: n  p + e  + e 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. P n = P p + P e + P  P n  P  = P p + P e ( ) 2 P n 2  P 2    2(E n E - p n p cos  ) = P p 2  P e 2    2(E p E e - p p p e cos  ) m n 2  0    2E (E n - p n cos  ) = m p 2  m e 2    2(E p E e - p p p e cos  ) [m p 2  m e 2     m n 2   2(E p E e - p p p e cos  )] 2p n E cos  =  E n / p n p ee n e   = [m p 2  m e 2     m n 2   2(E p E e - p p p e cos  )] 2p n (E n   E p  E e )  v n (  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  Lorentz Transformations 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) Relativistic Doppler Shift  k || 

More fun with 4-Vectors! More 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/d  P = (  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) P  2 = P  2 + P 2 - 2P  · P  rest frame  rest frame m  2 = m  – 2(E  E – p  ·p ) m  2 - m  2 = 2(E  E  –p  ·p ) E  = (m  2 - m  2 )/2m  3 sheet 1 = 2E m 

Derive an expression relating the emission angle of the muon or neutrino with respect to the beam in the CM to that in the lab frame. CM frame  p CM lab frame  p lab Transverse momentum: p lab sin  = p cm sin  Longitudinal momentum: p lab cos  =  (p cm cos  E cm  divide: tan  = sin  [  (cos  E cm /p cm   = sin  [  (cos  /  p   p cm /E cm =  p (particle) Are there any limits to the lab directions of the muon or neutrino? To find maximum angle, set d  /d  = 0 cos  max = -  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 3 sheet 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 B C 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 = D CB tt = L sin  (L/v - L cos  /c) L v = v sin  (1 -  cos  )  7 sheet 1

face area A L Interaction Probability = Fraction of area occupied by targets =  N T / A # Interactions = N int =  N T N B /A projected target area  (# targets = N T ) beam (# in beam = N B ) Cross-Sections Cross-Sections etc. (a quick review of useful definitions and relations)

(transition rate per unit time) Alternate expression to eliminate A : N T = (#/Volume) x Volume =   L A Alternate expression to connect  with theory: Rate = N int /t Alternative expressions N int =  N B   L v B = beam velocity t = L/v B N int =  N T N B /A W = v B  / V Prob/Time = Rate/(N B N T ) = v B N B N T  / V= v B N B N T  / (LA)

Target L  = (N B - N TR )/(N B   L) Transmission Experiment Example: Transmission experiment Beam Counters  N B Transmission Counter  N TR N int = N B - N TR N int =  N B   L = (1-f TR )/(   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 x without interacting: (Poisson) P =  n e -  / n! In this case n = 0 and  = N = V/(  ) = x/