Physics 2D Lecture Slides Lecture 6: Jan 13th 2004 Vivek Sharma - PDF document

Quiz 1 is This Friday Bring Blue Book, check calculator battery Physics 2D Lecture Slides Lecture 6: Jan 13th 2004 Vivek Sharma UCSD Physics

Lorentz Transformation Between Ref Frames Inverse Lorentz Transformation Lorentz Transformation = γ − = γ + x ( x ' vt ') x ' ( x v t ) = = y ' y y y ' = = z ' z z z ' ⎛ ⎞ ⎛ ⎞ v x v x ' = γ − = γ + ⎜ ⎟ ⎜ ⎟ 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 = = 2 1 In S' frame, u S and S’ are measuring − x' ' ' ' t t dt ant’s speed u along x, y, z 2 1 axes v = γ − = γ − ' dx ( dx v dt ) , dt ' ( dt dx ) 2 c − S’ dx vdt = S v u , divide by dt x' v − dt dx u 2 c − u v = x u x' v u − x 1 2 c = − For v << c, u u v x' x (Galilean Trans. Resto red)

Velocity Transformation Perpendicular to S-S’ motion v = = γ − dy ' dy , dt ' ( dt dx ) Similarly 2 c dy ' dy Z component of = = ' u y v dt ' γ − Ant' s velocity ( dt dx ) 2 c transforms as divide by dt on R H S u u = ' = z u y ' u z v y v γ − γ − (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 = x ' u x vu + x ' 1 2 c As usual, ' u = y u replace y v v ⇒ - v γ + ' (1 u ) x 2 c ' u = z u z v γ + ' ( 1 u ) x 2 c

Does Lorentz Transform “work” For Topgun ? Two rockets A &B travel in y’ S’ opposite directions -0.85c B y S 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 Transform 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

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’ D S m I arrive in SF o r f f f o e k a t I ( x t , ) ( , ) x t 2 2 1 1 ' ' ( x t , ) ' ' ( , ) x t 2 2 1 1 γ ⎡ ∆ ⎤ ⎛ ⎞ v x ∆ = − = ∆ −⎜ ' ' ⎟ t ' t t ⎢ t ⎥ 2 1 ⎝ ⎠ 2 ⎣ ⎦ c ⇒ ∆ < Reversing sequence of even ts t ' 0

I Cant ‘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 γ ⎡ ∆ ⎤ ⎛ ⎞ v x ∆ = − = ∆ −⎜ ' ' ⎟ t ' t t ⎢ t ⎥ 2 1 ⎝ ⎠ ⎣ 2 ⎦ c ∆ < For what value of v can ' 0 t ∆ ∆ = v x v x v u ∆ < ⇒ ∆ < ⇒ t ' 0 t 1 < ∆ 2 2 2 c c t c v c ⇒ > ⇒ > v c : Not al lowe d c u 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 v v v 2 v V v = = = = = = − ' ' 1 2 v 0, v , V ' v 1 2 v v v v 2 V v v − − − + 1 1 1 1 1 1 2 2 2 2 c c c c − 2 mv = + = = = − ' ' ' ' p mv m v , p 2 mV ' 2 mv before 1 2 after 2 v + 1 S’ 2 c ' ≠ ' p p before after v 1 ’=0 1 2 V’ 1 2 v 2 ’ Before After

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

Relativistic Force & Acceleration � � ⎛ ⎞ � � dp d mu d du d = = ⎜ ⎟ = F use � � ⎜ ⎟ mu − dt dt 2 dt dt du ⎝ 1 ( / ) u c ⎠ = = γ p mu ⎡ ⎤ − 2 − − 1 ( / ) u c m mu 1 2 u du ⎢ ⎥ = + × F ( )( ) ( ) ⎢ ⎥ 3/2 − 2 − 2 2 c dt 2 1 ( / ) u c 1 ( u c / ) ⎣ ⎦ ⎡ ⎤ − + Relativistic 2 2 2 mc mu mu du ⎢ ⎥ = ⎢ F ( ) ⎥ 3/ 2 − dt 2 2 c 1 ( u c / ) Force ⎣ ⎦ ⎡ ⎤ And m du ⎢ ⎥ = ⎢ F : Relativistic For ce ( ) ⎥ 3/2 − dt Acceleration 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 ⎛ ⎞ ⎛ ⎞ � 2 2 du F u qE u = = − − ⎜ ⎟ ⎜ ⎟ 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

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 II accelerator schematic and tunnel view PEP