ū(e )(1 γ 5 )γ α v( ν e ) v( ν e )γ β (1 + γ 5 )u(e ) tr (1 γ 5 )γ α ( p ν m ν )γ β (1 + γ 5 )( p e + m e ).

Similar documents
23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

1 The muon decay in the Fermi theory

CHAPTER 7: SECOND-ORDER CIRCUITS

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Two Coupled Oscillators / Normal Modes

Discussion Session 2 Constant Acceleration/Relative Motion Week 03

Lecture 4 Kinetics of a particle Part 3: Impulse and Momentum

Final Spring 2007

Math 2214 Solution Test 1 B Spring 2016

4.5 Constant Acceleration

ψ(t) = V x (0)V x (t)

Physics 240: Worksheet 16 Name

Math 10B: Mock Mid II. April 13, 2016

MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.

Math 2214 Solution Test 1B Fall 2017

2. VECTORS. R Vectors are denoted by bold-face characters such as R, V, etc. The magnitude of a vector, such as R, is denoted as R, R, V

Theory of! Partial Differential Equations!

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle

Theory of! Partial Differential Equations-I!

Homework Solution Set # 3. Thursday, September 22, Textbook: Claude Cohen Tannoudji, Bernard Diu and Franck Lalo, Second Volume Complement G X

Rectilinear Kinematics

Scattering and Decays from Fermi s Golden Rule including all the s and c s

Roller-Coaster Coordinate System

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x

You have met function of a single variable f(x), and calculated the properties of these curves such as

INTRODUCTION TO INERTIAL CONFINEMENT FUSION

Let us start with a two dimensional case. We consider a vector ( x,

And the solution to the PDE problem must be of the form Π 1

THE BERNOULLI NUMBERS. t k. = lim. = lim = 1, d t B 1 = lim. 1+e t te t = lim t 0 (e t 1) 2. = lim = 1 2.

TP B.2 Rolling resistance, spin resistance, and "ball turn"

Math 2214 Solution Test 1A Spring 2016

and v y . The changes occur, respectively, because of the acceleration components a x and a y

3.1.3 INTRODUCTION TO DYNAMIC OPTIMIZATION: DISCRETE TIME PROBLEMS. A. The Hamiltonian and First-Order Conditions in a Finite Time Horizon

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Question 1: Question 2: Topology Exercise Sheet 3

Nature Neuroscience: doi: /nn Supplementary Figure 1. Spike-count autocorrelations in time.

EE202 Circuit Theory II

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

Solution: b All the terms must have the dimension of acceleration. We see that, indeed, each term has the units of acceleration

!!"#"$%&#'()!"#&'(*%)+,&',-)./0)1-*23)

Traveling Waves. Chapter Introduction

Chapter 6. Laplace Transforms

Non-uniform circular motion *

SMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.

5.2 GRAPHICAL VELOCITY ANALYSIS Polygon Method

BU Macro BU Macro Fall 2008, Lecture 4

KINEMATICS IN ONE DIMENSION

2 Some Property of Exponential Map of Matrix

Essential Microeconomics : OPTIMAL CONTROL 1. Consider the following class of optimization problems

CONTROL SYSTEMS. Chapter 10 : State Space Response

Solutions from Chapter 9.1 and 9.2

23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes

Introduction to SLE Lecture Notes

y = (y 1)*(y 3) t

10. State Space Methods

Math 333 Problem Set #2 Solution 14 February 2003

Solutions of Sample Problems for Third In-Class Exam Math 246, Spring 2011, Professor David Levermore

Pion Lifetime. A. George January 18, 2012

Laplace Transform. Inverse Laplace Transform. e st f(t)dt. (2)

Laplace transfom: t-translation rule , Haynes Miller and Jeremy Orloff

Lecture #8 Redfield theory of NMR relaxation

ES.1803 Topic 22 Notes Jeremy Orloff

8. Basic RL and RC Circuits

20. Applications of the Genetic-Drift Model

EECE 301 Signals & Systems Prof. Mark Fowler

Problem Set #1. i z. the complex propagation constant. For the characteristic impedance:

3, so θ = arccos

SOLUTIONS TO ECE 3084

Math 116 Second Midterm March 21, 2016

Linear Surface Gravity Waves 3., Dispersion, Group Velocity, and Energy Propagation

Problem Set If all directed edges in a network have distinct capacities, then there is a unique maximum flow.

KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

Acoustical Interference Phenomena

Y 0.4Y 0.45Y Y to a proper ARMA specification.

ME 391 Mechanical Engineering Analysis

Solutions to Assignment 1

Second Order Linear Differential Equations

Nuclear Decay kinetics : Transient and Secular Equilibrium. What can we say about the plot to the right?

Algorithmic Discrete Mathematics 6. Exercise Sheet

EE100 Lab 3 Experiment Guide: RC Circuits

y z P 3 P T P1 P 2. Werner Purgathofer. b a

CHAPTER. Forced Equations and Systems { } ( ) ( ) 8.1 The Laplace Transform and Its Inverse. Transforms from the Definition.

Today in Physics 218: radiation reaction

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

Chapter 7: Inverse-Response Systems

Second Law. first draft 9/23/04, second Sept Oct 2005 minor changes 2006, used spell check, expanded example

4.6 One Dimensional Kinematics and Integration

DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING EXAMINATIONS 2008

ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004

( ) = b n ( t) n " (2.111) or a system with many states to be considered, solving these equations isn t. = k U I ( t,t 0 )! ( t 0 ) (2.

ψ ( t) = c n ( t) t n ( )ψ( ) t ku t,t 0 ψ I V kn

AP Calculus BC Chapter 10 Part 1 AP Exam Problems

Exponential Sawtooth

k 1 k 2 x (1) x 2 = k 1 x 1 = k 2 k 1 +k 2 x (2) x k series x (3) k 2 x 2 = k 1 k 2 = k 1+k 2 = 1 k k 2 k series

where the coordinate X (t) describes the system motion. X has its origin at the system static equilibrium position (SEP).

V L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode.

Transcription:

PHY 396 K. Soluion for problem e #. Problem (a: A poin of noaion: In he oluion o problem, he indice µ, e, ν ν µ, and ν ν e denoe he paricle. For he Lorenz indice, I hall ue α, β, γ, δ, σ, and ρ, bu never µ or ν. Thu, p µα denoe he α componen of he muon 4 momenum, ec., ec. Le ar wih he muon decay ampliude M(µ e ν µ ν e = G F ū(ν µ ( γ 5 γ α u(µ ū(e ( γ 5 γ α v( ν e. ( I complex conjugae can be wrien a M = G F ū(µ γ β ( + γ 5 u(ν µ v( ν e γ β ( + γ 5 u(e, (S. where he ( γ 5 facor become ( + γ 5 becaue γ 5 γ 0 (γ 5 γ 0 = γ 5. Conequenly, M = G F ū(ν µ ( γ 5 γ α u(µ ū(µ γ β ( + γ 5 u(ν µ (S. ū(e ( γ 5 γ α v( ν e v( ν e γ β ( + γ 5 u(e and hence all M = 4 G F (( r γ 5 γ α ( p µ + M µ γ β ( + γ 5 ( p ν + m ν ( r ( γ 5 γ α ( p ν m ν γ β ( + γ 5 ( p e + m e. (S.3 Having derived eq. (S.3, we now need o evaluae he race. For he fir race, we eliminae erm conaining odd number of γ ρ marice and wrie ( r ( γ 5 γ α ( p µ + M µ γ β ( + γ 5 ( p ν + m ν = = r (( γ 5 γ α p µ γ β ( + γ 5 p ν + r (( γ 5 γ α M µ γ β ( + γ 5 m ν

( ( = r ( γ 5 γ α p µ γ β p ν ( γ 5 + M µ m ν r ( γ 5 γ α γ β ( + γ 5 = r (( γ 5 γ α p µ γ β p ν + M µ m e r (( + γ 5 ( γ 5 γ α γ β = r (( γ 5 γ α p µ γ β p ν + 0 = r (γ α p µ γ β p ν r (γ 5 γ α p µ γ β p ν = 8 p α µp β ν + p β µp α ν g αβ (p µ p ν + 8iǫ αγβδ p µγ p νδ. (S.4 Similarly, he econd race evaluae o r ( ( γ 5 γ α ( p e + m e γ β ( + γ 5 ( p ν m ν = (S.5 = 8 (p eα p νβ + p eβ p να g αβ (p e p ν + 8iǫ αρβσ p ρ νp σ e. I remain o ubiue he race formulæ (S.4 and (S.5 back ino eq. (S.3 and conrac he Lorenz indice. Thu, ( M = 6G F p α µ pβ ν + pβ µ pα ν gαβ (p µ p ν + iǫ αγβδ p µγ p νδ all ( p eα p νβ + p eβ p να g αβ (p e p ν + iǫ αρβσ p ρ νp σ e uing ymmery/aniymmery of facor under α β ( = 6G F p α µ pβ ν + pβ µ pα ν gαβ (p µ p ν p eα p νβ + p eβ p να g αβ (p e p ν ǫ αγβδ p µγ p νδ ǫ αρβσ p ρ νp σ e ( = 6G F (p µ p e (p ν p ν + (p µ p ν(p ν p e (p µ p ν (p e p ν (p µ p ν (p e p ν + 4(p µ p ν (p e p ν + (p µ p ν (p ν p e (p µ p e (p ν p ν Q.E.D. = 64G F (p µ p ν (p ν p e. (S.6

Problem (b: A explained in he Pekin & Schroeder exbook, he parial rae of a decay proce (in he re frame of he iniial paricle i given by dγ = M 0 M dp (S.7 where M i he decay ampliude, M i M averaged over he unknown iniial and ummed over he unmeaured final, and dp i he infinieimal phae pace facor for he final paricle. For hree final paricle, dp = d 3 p (π 3 (E d 3 p (π 3 (E d 3 p 3 (π 3 (E 3 (π3 δ (3 (p +p +p 3 (πδ(e +E +E 3 M 0 (S.8 where he energy-momenum conervaion law apply in he re frame, hu p + p + p 3 = p o = 0 and E + E + E 3 = E o = M 0. We ar by uing he momenum-conervaion δ funcion o eliminae eliminae he p 3 a independen variable, hu dp = d3 p d 3 p δ(e + E + E 3 E o 56π 5 E E E 3 Nex, we ue pherical coordinae for he wo remaining momena, p3= (p +p. (S.9 d 3 p = p dp d Ω, d 3 p = p dp d Ω, (S.0 and hen replace he d Ω decribing he direcion of he econd paricle momenum relaive o he fixed exernal frame wih d Ω ( = dθ in θ dφ ( decribing he ame direcion of p relaive o he frame cenered on he p. Conequenly, d Ω d Ω = d Ω d Ω ( = d Ω dφ ( dθ in θ d 3 Ω d(coθ (S. 3

and hence dp = d3 Ω 56π 5 p p dp dp d(coθ δ(e + E + E 3 E o E E E 3 p3= (p +p. (S. Nex, we ue he coine heorem which give p 3 = (p + p = p + p + p p co θ (for fixed p, p and herefore d(coθ = p 3 dp 3 p p dp = d3 Ω 56π 5 p p p 3 E E E 3 dp dp dp 3 δ(e + E + E 3 E o. (S.3 Finally, we noice ha for a relaiviic paricle of any ma, pdp = EdE and hence dp = d3 Ω 56π 5 de de de 3 δ(e + E + E 3 E o. (S.4 Subiuing hi formula ino eq. (S.7 give eq. (3 for he parial decay rae. I remain o deermine he limi of kinemaically allowed way o diribue he ne energy E o = M 0 of he proce among he hree final paricle. Such limi follow from he riangle inequaliie for he hree momena, p p + p 3, p p + p 3, p 3 p + p, (S.5 which look imple bu produce raher complicaed inequaliie for he energie E = p + m, E = p + m, and E 3 = p 3 + m3. However, when all hree final paricle are male, he kinemaic rericion become imply E E + E 3 = M 0 E (S.6 and dio for he oher wo inequaliie, or equivalenly 0 E, E, E 3 M 0, while E + E + E 3 = M 0. (5 4

Problem (c: In ligh of eq. ( and (3, he parial decay rae of he muon a re i given by dγ(µ e ν µ ν e = G F 8π 5 M µ (p µ p ν (p e p ν de e de ν de ν d 3 Ωδ(E e +E ν +E ν M µ. (S.7 Specializing o he muon frame, we have (p µ p ν = M µ E ν (S.8 while (p e p e = E e E ν p e p ν coθ eν neglecing m e, m ν, m ν = E e E ν + p e + p ν p ν = E e E ν + E e + E ν E ν (S.9 = (E e + E ν E ν uing E e + E ν = M µ E ν = M µ(m µ E ν. Hence, dγ(µ e ν µ ν e = G F 6π 5 M µe ν (M µ E ν de e de ν de ν d 3 Ωδ(E e +E ν +E ν M µ. (S.0 A hi poin we are ready o inegrae over he final-ae variable. In ligh of d 3 Ω = 8π and he kinemaic limi (5, we immediaely obain Γ(µ e ν µ ν e = G F M µ π 3 = G F M µ π 3 = G F M µ π 3 Mµ de e de ν de ν E ν (M µ E ν δ(e e + E ν + E ν M µ 0 Mµ 0 Mµ 0 de e Mµ de ν E ν(m µ E ν Mµ Ee de e E e ( M µ 3 E e. (S. In oher word, he parial muon decay rae wih repec o he final elecron energy i given 5

by dγ de e = G F M µ π 3 E e(3m µ 4E e (S. or raher { dγ G F π M 3 µ Ee (3M µ 4E e for E e < M µ, de e 0 for E e > M µ. (S.3 Graphically, dγ/de e E e M µ Noe how hi curve moohly reache i maximum a E e = M µ and hen abruply fall down o zero. I remain o calculae he oal decay rae of he muon by inegraing he parial rae (S.3 over he elecron energy. The reul i Γ o (µ eν ν = G F M5 µ 9π 3. (S.4 6

Problem : A he ree level, here are wo Feynman diagram for he Bhabha caering e + e e + e : ougoing e e + e e + + (S.5 incoming e e + e e + Conequenly, he ree-level ampliude i M ree = M + M, M = e ū(p γ µ u(p v(kγ µ v(k, M = + e v(kγµ u(p ū(p γ µ v(k, (S.6 where p and p are momena of he iniial / final elecron e while k and k are momena of he iniial / final poiron e +. Noe he ign of he wo ampliude: The channel diagram ha a direc poiron-in-poiron-ou line, hence an overall minu ign; he u channel diagram doe no have hi ign facor becaue all i fermionic line involve an elecron e on one end or he oher. Le u um over he ; pleae bear eq. ( in mind! Saring wih he fir diagram, we have M = = ū(p γ µ u(pū(pγ ν u(p v(kγ µ v(k v(k γ ν v(k r ( p + mγ µ ( p + mγ ν r ( k mγ µ ( k mγ ν r p γ µ pγ ν r kγ µ k γ ν (S.7 7

= 4 p µ p ν + p ν p µ g µν (p p 4 k µk ν + k νk µ g µν (k k = 6 (k p (k p + (k p(k p (k k(p p (k k(p p + 4(k k(p p = 3 (k p (k p + (k p(k p = 8 + u (S.7 where he la equaliy follow from he kinemaic relaion (k p = (k p = m, (p p = (k k = m, (S.8 (k p = (p k = m u u. In a imilar manner, he econd diagram yield M = v(kγ µ u(pū(pγ ν v(k ū(p γ µ v(k v(k γ ν u(p = r( k mγ µ ( p + mγ ν r ( p mγ µ ( k mγ ν r kγ µ pγ ν r p γ µ k γ ν (S.9 = 4 k µ p ν + k ν p µ g µν (k p 4 k µ p ν + k ν p µ g µν(k p = 6 (k k(p p + (k p(k p. (k p (k p (k p (k p + 4(k p (k p = 3 (k k(p p + (k p(k p = 8 + u. (S.9 8

The inerference erm i more inereing: M M = e4 ū(pγ µ u(p ū(p γ ν v(k v(k γ µ v(k v(kγ ν u(p = e4 r ( p + mγ µ ( p + mγ ν ( k mγ µ ( k mγ ν e4 r pγ µ p γ ν k γ µ kγ ν uing γ ν k γ µ kγ ν = kγ µ k = + e4 r pγ µ p kγ µ k uing γ µ p kγ µ = 4(p k = +8 e4 (p k r p k = +3 e4 (p k (p k = +8 e4 u. (S.30 Subiuing he pin um (S.7, (S.9 and (S.30 ino eq. ( and dividing by 4 (for he 4 pin ae of he wo iniial paricle, we arrive a 4 all M = e 4 + u + + u = e 4 ( + ( = e 4 4 + 4 + u 4 + + u ( u + u (S.3 where he la equaliy follow from + = 4m u u. In he cener of ma frame = 4E 4p, = p ( co θ, (S.3 u = p ( + coθ, 9

and hence 4 + 4 + u 4 = 6 + ( coθ4 + ( + co θ 4 4( coθ = (3 + co θ ( coθ. (S.33 Therefore, dσ dω cm = 64π 4 M = α 4 (3 + co θ ( coθ. (S.34 all 0

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback