Relativistic Dynamics

Transcription

1 PH0008 Quantum Mechanics and Special Relativity Lecture (Special Relativity) v3 Relativistic Dynamics Collision, Mass depends on velocity, energy-momentum invariant, Compton Effect Prof Department of Physics Brown University Main source at Brown Course Pulisher ackground material may also e availale at Gaitskell

2 Section: Special Relativity Week 4 Homework (due for M 3/8) o Hand in now Reading (Prepare for 3/8) o SpecRel (also y French) Ch6 RelativisticKinematics Lecture 0 (M 3/8) o Relativistic Dynamics From Classical -> Relativistic What Conservation Laws can we rely on? Expressions for m, E and p Collisions Reading (Prepare for 4/ after recess) o SpecRel Revise Ch2-6 (look at Ch also) o QuantMech Ch,2 & 3 Homework #8 (M 4/) o (see we Assignments ) Lecture (W 3/20) o Relativistic Dynamics Collisions Newton s 2nd Law Compton Effect Lecture 2 (F 3/22) o Relativistic Dynamics Summary Review Prolem Set

3 4/8 (Monday) 8.30 am Start Exam II Timing

4 Homework Please pick up your HW #-4 from outside my office B&H 56 o Definitely there

5 Question Section

6 Question SpecRel L-Q Which relativistic expression correctly gives the total mass of moving particle, velocity (g) rest mass m 0? o() m m 0 o(2) m gm 0 o(3) m gm 0 o(4) m g 2 m 0

7 Question SpecRel L2-Q2 How ig was the ice sheet that just fell off Antarctic Ice Cap? o() Block Island o(2) Manhattan Island o(3) Rhode Island o(4) Who cares? Photo AFP / NY Times

8 Classical -> Relativistic Dynamics - page Conservation of Energy & Mass are comined Conservation of Momentum - still good 0 g o ut mass term is no longer invariant p gm 0 c (cp) 2 g 2 2 (m 0 g 2 ( g 2 -)(m 0 g 2 (m 0 -(m 0 E 2 -(m 0 - fi 2 g 2-2 g 2 0 ( - 2 ) 2 Consider p for << p gm 0 c Ê g ª + 2 ˆ Á Ë Ê ª + 2 ˆ Á m 0 c Ë ª cm ª m 0 v +... which is the classical form m g m 0 E g m 0 p gm 0 c E 2 (pc) 2 + (m 0 KE E - m 0 ( g -)m 0 Consider KE for << KE E - m 0 ( g -)m 0 g ( - 2 ) 2 Ê ª + 2 ˆ Á m 0 Ë ª 2 2 m ª 2 m 0v which is the classical form Ê ª + 2 ˆ Á Ë

9 Collision treated in Relativistic Dynamics

10 Elastic Collision: (Extract m(v)) Centre of Mass Frame S B A Particle A Stationary Frame S B y We will show that the masses of particles (used in momentum expression mv) are a function of the particle velocities i.e. pm(v)v Masses transform etween frames in a way that is similar to the transforms for lengths and times y xx (Assume S is moving at in S.) Usual velocity transforms apply ( x + ) ( + x ) x - x - x x y y g ( - x ) ( y g) y + x A

11 Elastic Collision: (Extract m(v)) (2) Centre of Mass Frame S B A Particle A Stationary Frame S B A (Assume S is moving at in S.) Usual velocity transforms apply x - x - x x y ( x + ) ( + x ) y y y g ( - x ) ( y g) y + x In S (Centre of Mass Frame) collision is symmetric r r A - B fi xa - xb ( -) ya - yb m A xx Conservation of Momentum in S (Elastic Collision) Dp ya + Dp yb 0 Dp ya 2 ya c Dp yb 2 yb c So ya - yb Use Velocity Transformation m ya g A -m yb g + xa B + m ya g A - m ya g 2 B xb

12 Elastic Collision: (Extract m(v)) (3) Centre of Mass Frame S B A Particle A Stationary Frame S B (Assume S is moving at in S.) Usual velocity transforms apply x - x - x x y ( x + ) ( + x ) y y y g ( - x ) ( y g) y + x xx So we have Re - express in terms of xb ( xb xb + ) ( + xb ) Given xb ( 2) xb + 2 So 2 ( - xb ) 2 g xb A If we wish to ensure Conservation of Momentum in S frame, we must conclude that apparent mass in a given frame is transformed m g m 0 where m 0 is the rest mass (in the rest frame of the particle)

13 Centre of Mass Frame S B m B Particle A Stationary Frame S B Elastic Collision: (Extract m(v)) (4) A (Assume S is moving at in S.) Usual velocity transforms apply ( x + ) ( + x ) x - x - x x y y y g ( - x ) ( y g) y + x y So we have Re - express in terms of xb ( xb xb + ) ( + xb ) Given xb ( 2) xb + 2 So 2 ( - xb ) 2 xx g xb The final "trick" is to imagine the collision in the limiting case that the y components of velocities go to zero... In this case yb, yb Æ 0 xb Æ B ya, ya Æ 0 xa Æ 0 (rest frame of A) The eqn 2 ( - xb ) 2 can e re - expressed as the more general relation m 0 2 ( - B ) 2 g xb g B m 0 A Æ 0 So if we wish to ensure Conservation of Momentum in S frame, we must conclude that apparent mass in any given frame is transformed m g m 0 where m 0 is the rest mass (in the rest frame of the particle)

14 r F dr p dt d gm r 0u dt v u r K F.d r l Ú v 0 Consider special case along x Fdx v u Ú v 0 u Ú 0 u dt d mu dx Ú u d( mu) since dx dt u 0 u Ú u mdu + udm 0 Can show (see next slide) mudu + u 2 dm dm m Ú dm m 0 m - m 0 g - m 0 since m g m 0 K E - m 0 Newton s 2nd Law & Kinetic Energy Consider KE for << KE E - m 0 ( g -)m 0 g ( - 2 ) 2 Ê ª + 2 ˆ Á m 0 Ë ª 2 2 m ª 2 m 0 v which is the classical form Ê ª + 2 ˆ Á Ë Rememer! m g m 0 E m g m 0 p mv g m 0 ( c) gm 0 c

15 Newton s 2nd Law & Kinetic Energy (2) Discussion of the following steps u Ú () u d mu 0 u Ú u mdu + udm 0 (2) mudu + u 2 dm dm m Ú dm m 0 () Differentiation, y parts gives, d( mu) mdu + udm The full (technically correct) expression is, d( mu) (mu) du + (mu) dm u m const m u const mdu + udm (2) This step is a little trickier mudu + u 2 dm dm We know that m 0 m gm 0 ( - u 2 ) 2 So m 2 - m 2 u 2 m 2 0 Form the derivative d(m 2 ) - d(m 2 u 2 ) - d(m 2 0 ) 0 d(m 2 Ê ) dm dm - (m2 u 2 ) dm + (m2 u 2 ) Á Ë m u const u m const ˆ du The last term is d(m 0 2 ) is zero as argument is a constant 2m dm - 2mu 2 dm + m 2 2udu 0 fi 2m( - u 2 )dm - 2m 2 udu 0 Divide y 2m and rearrange fi u 2 dm + mudu dm

16 Energy and Momentum Given m g m 0 E g m 0 p gm 0 c Then (cp) 2 g 2 2 (m 0 But g 2 - fi 2 g 2-2 g 2 So (cp) 2 ( g 2 -)(m 0 g 2 (m 0 -(m 0 or E 2 -(m 0 E 2 (pc) 2 + (m 0 E 2 p 2 + m 0 2 with c Lorentz Invariant E 2 - (pc) 2 (m 0 (m 0 E 2 - ( p c) 2 E 2 - ( p c) 2 E 2 - ( p c) 2... (Consider also rest frame E and p) Massless particle E 2 (pc) 2 E pc Note also that p E gm 0c g m 0 c v