Physics 2D Lecture Slides Lecture 7 : Jan 12th 2005 Vivek Sharma - PDF document

Physics 2D Lecture Slides Lecture 7 : Jan 12th 2005 Vivek Sharma UCSD Physics First Quiz This Friday ! • Bring a Blue Book, calculator; check battery – Make sure you remember the code number for this couse given to you (record it some place safe!) • No “cheat Sheet” please, I will give you equations and constants that I think you need • When you come for the quiz, pl. occupy seats in the front first. • Pl. observe one seat distance in the back rows (there is plenty of space) • Academic Honesty is for you to observe and for me to enforce: – Be a good citizen, in this course and forever ! 1

Lorentz Transformation Between Ref Frames Inverse Lorentz Transformation Lorentz Transformation = � � = � + x ' ( x v t ) x ( x ' vt ') = = y ' y y y ' = = z ' z z z ' � � � � v x v x ' = � � = � + t ' t t t ' � � � � 2 2 � c � � c � As v → 0 , Galilean Transformation is recovered, as per requirement Notice : SPACE and TIME Coordinates mixed up !!! Lorentz Velocity Transformation Rule ' � ' ' x x dx = = In S' frame, u 2 1 S and S’ are measuring x' � ' ' ' t t dt ant’s speed u along x, y, z 2 1 axes v ' = � � = � � dx ( dx v d t ) , dt ' ( dt dx ) 2 c � dx vdt S’ = S u , divide by dt' v x' v � dt dx u 2 c � u v = u x x' v u � 1 x 2 c = � For v << c, u u v x' x (Gali lean Trans. Restor ed) 2

Velocity Transformation Perpendicular to S-S’ motion v = = � � dy ' dy , dt ' ( dt dx ) Similarly 2 c dy ' dy Z component of = = ' u y v dy ' � � Ant' s velocity ( dt dx ) 2 c transforms as divide by dt on R H S u u = ' u z = y ' u z v y v � � � � (1 c u ) (1 u ) x x 2 2 c There is a change in velocity in the � direction to S-S' motion ! Inverse Lorentz Velocity Transformation Inverse Velocity Transform: + u v = u x ' x vu + 1 x ' 2 c As usual, ' u = y u replace y v v ⇒ - v � + ' (1 u ) x 2 c ' u = u z z v � + ' ( 1 u ) x 2 c 3

Does Lorentz Transform “work” For Topgun ? Two rockets A &B travel in y’ S’ opposite directions -0.85c S B y An observer on earth (S) 0.75c measures speeds = 0.75c A And 0.85c for A & B respectively x’ O’ What does A measure as x B’s speed? O (Earth guy) Place an imaginary S’ frame on Rocket A ⇒ v = 0.75c relative to Earth Observer S Consistent with Special Theory of Relativity Example of Inverse Velocity Transformation Biker moves with speed = 0.8c past stationary observer Throws a ball forward with speed = 0.7c What does stationary observer see as velocity of ball ? Speed of ball relative to Place S’ frame on biker stationary observer Biker sees ball speed u X ? u X’ =0.7c 4

Velocity Transformation Perpendicular to S-S’ motion 2 bike gang leaders racing at relativistic speeds along perpendicular paths How fast does BETA recede over right shoulder of ALPHA as seen by ALPHA x Policeman measures : ALPHA : u x = 0.75 c u y = 0 BETA : u x = 0 u y = � 0.90 c y To get BETA’s speed of recession as seen by ALPHA, associate S’ frame to be ALPHA’s make Policeman’s frame to be S (measuring u x and u y ) and calculate u’ x and u’ y Velocity Transformation Perpendicular to S-S’ motion Make Alpha the S’ frame at rest, Policeman is frame S and we know what he (S) measures as the x,y components of BETA’s velocity in his frame. S ' = u x � V 0 � 0.75 c = = � 0.75 c V u x 1 � 0.0 c � 0.75 c 1 � u x V c 2 c 2 � � 2 0.75 c � 0.90 c � 1 � � � � � u y ' = c = = � 0.595 c u y � � 1 � 0 � 1 � u x V � � � c 2 � 5

Hollywood Yarns Of Time Travel ! Terminator : Can you be seen to be born before your mother? A frame of Ref where sequence of events is REVERSED ?!! u S S’ I take off from SD I arrive in SF ( x t , ) ( , ) x t 2 2 1 1 ' ' ( x t , ) ' ' ( , ) x t 2 2 1 1 � � � � � � v x � = ' � ' = � �� t ' t t t � � � 2 1 2 � c � � � � � < Reversing sequence of even ts t ' 0 6

I Can’t be seen to arrive in SF before I take off from SD u S S’ ( x , t ) ( x t , ) 2 2 1 1 ' ' ( x , t ) ' ' ( x t , ) 2 2 1 1 � � � v � x � ' � t 1 ' = � � t � � t ' = t 2 � � � � � � c 2 � � For what value of v can � t ' < 0 � t ' < 0 � � t < v � x c 2 � 1 < v � x = v u c 2 � t c 2 � v c > c u � v > c : Not allowed !! Relativistic Momentum and Revised Newton’s Laws � � Need to generalize the laws of Mechanics & Newton to confirm to Lorentz Transform = and the Special theory of relativity: Example : p mu Watching an Inelastic Collision between two putty balls S P = mv –mv = 0 P = 0 Before V=0 1 2 v v 1 2 After ' = v 1 � v ' = v 2 � v = � 2 v , V ' = V � v = 0, v 2 = � v v 1 1 � v 1 v 1 � v 1 v 1 + v 2 1 � Vv c 2 c 2 c 2 c 2 ' = � 2 mv ' + mv 2 = mv 1 = 2 mV ' = � 2 mv p before ' , p after ' 1 + v 2 c 2 S’ � p before ' ' � p after � Need to re-examine definition of relativistic momemtum v 1 ’=0 1 2 V’ 1 2 v 2 ’ Before After 7

Definition (without proof) of Relativistic Momentum � � mu � With the new definition relativistic = = � p mu momentum is conserved in all � 2 1 ( / ) u c frames of references : Do the exercise New Concepts Rest mass = mass of object measured In a frame of ref. where object is at rest 1 � = � 2 1 ( / ) u c u is velocity of the object NOT of a referen ce frame ! Nature of Relativistic Momentum � � mu � m = = � p mu u � 2 1 ( / ) u c With the new definition of Relativistic momentum Momentum is conserved in all frames of references 8

Relativistic Force & Acceleration � � � � � dp d mu d du d � � � = = = F use � mu � � � dt dt dt dt du � 2 1 ( / ) u c � � = = � p mu � � � 2 1 ( / ) u c � � m mu 1 2 u du � � = + � F ( )( ) ( ) � 3/2 2 � 2 c dt � 2 � 1 ( / ) u c 1 ( u c / ) 2 � � � � Relativistic 2 � 2 + 2 mc mu mu du � � = � F ( ) 3/2 � dt 2 � 2 Force c 1 ( u c / ) � � � � And m du � � = � F : Relativistic For ce ( ) � 3/2 Acceleration dt � 2 1 ( / ) u c � � � � du Since A ccel e r a tion a = , [rate of change of v elocity ] dt � � F 3/2 � � � 2 � a = 1 ( / ) u c � � Reason why you cant m � � � quite get up to the speed Note: As / u c 1, a 0 !!!! of light no matter how It s harder to accelerate when you get hard you try! closer to s peed of l ight A Linear Particle Accelerator - Parallel Plates + F q E= V/d F= eE -E d V Charged particle q moves in straight line Under force, work is done � � on the particle, it gains in a uniform electric field E with speed u � � Kinetic energy accelarates under f orce F=qE New Unit of Energy � � � 3/2 3/2 � � � � � du F u 2 qE u 2 = = � � a 1 = 1 � � � � 1 eV = 1.6x10 -19 Joules 2 2 dt m c m c � � � � 1 MeV = 1.6x10 -13 Joules 1 GeV = 1.6x10-10 Joules larger the potential difference V a cross plates, larger the force on particle 9

Your Television (the CRT type) is a Small Particle Accelerator ! Linear Particle Accelerator : 50 GigaVolts Accelating Potential � � eE 3/ 2 � � � 2 a= 1 ( / ) u c � � m PEP-II accelerator schematic and tunnel view PEP-II accelerator schematic and tunnel view 10

Hollywood Yarns ! 11