PHYS 3446 – Lecture #4Scattering Cross SectionLab Frame and Center of Mass FrameLab Frame and CM FrameSlide 5Now some simple arithmeticSlide 7Relationship of variables in Lab and CMScattering angles in Lab and CMDifferential cross sections in Lab and CMSome Quantities in Special RelativityRelativistic VariablesSlide 13Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt1PHYS 3446 – Lecture #4Monday, Sept. 8, 2006Dr. Andrew Brandt1. Lab Frame and Center of Mass Frame2. Relativistic Variables***HW2 due Weds***Radiation trainingMonday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt2Scattering Cross Section•For a central potential, measuring the yield as a function of , or differential cross section, is equivalent to measuring the entire effect of the scattering •So what is the physical meaning of the differential cross section?This is equivalent to the probability of certain process in a specific kinematic phase space•Cross sections are measured in the unit of barns:-24 21 barn = 10 cmMonday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt3Lab Frame and Center of Mass Frame•So far, we have neglected the motion of target nuclei in Rutherford Scattering•In reality, they recoil as a result of scattering •This complication can best be handled using the Center of Mass frame under a central potential•This description is also useful for scattering experiments with two beams of particles (moving target)Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt4Lab Frame and CM Frame•The equations of motion can be written1 1m r =r&&2 2m r =r&&i� =rwhereSince the potential depends only on relative separation of the particles, we redefine new variables, relative coordinates & coordinate of CM1 2r r r= -r r rCMR =rand1 1 2 21 2m r m rm m++r r( )1 1 2V r r- � -rr r( )2 1 2V r r- � -rr rˆiirr�+�ˆii irqq�+�ˆsinii i irfq f��1, 2i =2rrMonday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt5Lab Frame and CM Frame•From the equations in previous slides( )1 2 CMm m R+ =r&&and•What do we learn from this exercise?•For a central potential, the motion of the two particles can be decoupled when re-written in terms of –a relative coordinate–The coordinate of center of mass1 21 2m mrm m�+r&&Thusˆconstant CMR R=r&Reduced Massrm =r&&( )V r- � =rr( )ˆV rrr�-�rCMMR =r&&0Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt6Now some simple arithmetic•From the equations of motion, we obtain1 1 2 2m r m r- =r r&& &&1 1 2 21 2CMm r m rRm m+=+r r&& &&r&&( )1 2 1 2 m m r m r+ = �r r&& &&( )( )21 2 21 2mm m r m rm m- + =+r r&& &&•Since the momentum of the system is conserved:•Rearranging the terms, we obtain( )1 1 2 1m r m r r- - =r r r&& && &&( )1 2 1 2m m r m r- +r r&& &&( ) ( )1 21 2ˆ ˆ ˆ2r r V r r V rr r r� �� � �=- - =-� �� � �� �r r0 = �1 1 m r =r&&2 2m r- =r&&( )2 1m r r- -r r&& &&( )211 2 mr rm m=+r r&& &&( )1 21 2m mrm m=+r&&( )ˆr V rr�-�r( )1 21 22m mrm m+r&& �Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt7Lab Frame and CM Frame•The CM is moving at a constant velocity in the lab frame independent of the form of the central potential•The motion is that of a fictitious particle with mass (the reduced mass) and coordinate r.•In the frame where the CM is stationary, the dynamics becomes equivalent to that of a single particle of mass scattering off of a fixed scattering center.•Frequently we define the Center of Mass frame as the frame where the sum of the momenta of all the interacting particles is 0.Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt8Relationship of variables in Lab and CM•The speed of CM is•Speeds of the particles in CM frame are•The momenta of the two particles are equal and opposite!!CM CMv R= =&1v =%CM2v =%and1 CMv v- =2 11 2m vm m+CMv =1 11 2m vm m+1 11 2m vm m+1v%1vCMVCMqLabqMonday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt9• CM represents the change in the direction of the relative position vector r as a result of the collision in CM frame•Thus, it must be identical to the scattering angle for a particle with reduced mass, .•Z components of the velocities of particle with m1 in lab and CM are:•The perpendicular components of the velocities are: •Thus, the angles are related (for elastic scattering only) as:Scattering angles in Lab and CMcosLab CMv vq - =sinLabv q =1sintancosCMLabCM CMv vqqq= =+%1 2sincosCMCMm mqq=+sincosCMCMqq z+1cosCMv q%1sinCMv q%Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt10Differential cross sections in Lab and CM•The particles that scatter in lab at an angle Lab into solid angle dLab scatter at CM into solid angle dCM in CM.•Since is invariant, dLab = dCM.–Why?– is perpendicular to the direction of boost, thus is invariant.•Thus, the differential cross section becomes:( )sinLab Lab LabLabdddsq q q =W( )LabLabddsq =WreorganizeUsing Eq. 1.53( )LabLabddsq =W( )sinCM CM CMCMdddsq q qW( )CMCMddsqW( )( )coscosCMLabddqq( )CMCMddsqW( )3/ 221 2 cos1 cosCMCMz q zz q+ ++quiz!Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt11Some Quantities in Special Relativity•Fractional velocity •Lorentz factor•Relative momentum and the total energy of the particle moving at a velocity is•Square of four momentum P=(E,pc), rest mass EP =rE =v cb =rrv cb=rr211gb=-Mvg =rM cg br2 T +Re2stE =2 2 2 4P c M c+ =2Mcg2 2 2 2 2P Mc E p c= = -Monday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt12Relativistic Variables•Velocity of CM in the scattering of two particles with rest mass m1 and m2 is: •If m1 is the mass of the projectile and m2 is that of the target, for a fixed target we obtainCMvc=rCMb =r12 2 2 4 21 1 2PcP c m c m c+ +rCMb =r( )1 21 2P P cE E++r r121 2PcE m c=+rMonday, Sept. 8, 2008 PHYS 3446, Fall 2008Andrew Brandt13Relativistic Variables•At very low energies where m1c2>>P1c, the velocity reduces to:•At very high energies where m1c2<<P1c and m2c2<<P1c , the velocity can be written as:CMb =rCM CMb b= =r22 21 21 111m c m cPc Pc�� �+ +� �� �22 11 1112m c m cP P� �- -� �� �1 12 21 2m v cm c m c=+r( )1 11 2m vm m
View Full Document
Unlocking...