Physics 2D Lecture Slides Oct 6 Vivek Sharma UCSD Physics
New Rules of Coordinate Transformation Needed • The Galilean/Newtonian rules of transformation could not handles frames of refs or objects traveling fast – V ≈ C (like v = 0.1 c or 0.8c or 1.0c) • Einstein’s postulates led to – Destruction of concept of simultaneity ( ∆ t ≠ ∆ t’ ) – Moving clocks run slower – Moving rods shrink • Lets formalize this in terms of general rules of coordinate transformation : Lorentz Transformation – Recall the Galilean transform • X’ = (x-vt) • T’ = T – These rules that work ok for mule carts now must be modified for rocket ships with V ≈ C
Discovering The Correct Transformation Rule = − → = − Need to figure out x ' x vt guess x ' G ( x v t ) functional form of G ! = + → = + x x ' vt ' guess x G ( ' x vt ' ) G must be dimensionless G does not depend on x,y,z,t But G depends on v/c Do a Thought Experiment: Rocket Motion along x axis G is symmetric As v/c → 0 , G → 1 Rocket in S’ (x’,y’,z’,t’) frame moving with velocity v w.r.t observer on frame S (x,y,z,t) Flashbulb mounted on rocket emits pulse of light at the instant origins of S,S’ coincide That instant corresponds to t = t’ = 0 . Light travels as a spherical wave, origin is at O,O’ Speed of light is c for both observers Examine a point P (at distance r from O and r’ from O’ ) on the Spherical Wavefront The distance to point P from O : r = ct Clearly t and t’ must be different The distance to point P from O : r’ = ct’ t ≠ t’
Discovering Lorentz Transfromation for (x,y,z,t) Motion is along x-x’ axis, so y, z unchanged y’=y, z’ = z Examine points x or x’ where spherical wave crosses the horizontal axes: x = r , x’ =r’ = = + x ct ( ' G x vt ') = = x ' ct ' ( - G x vt ) , = γ − = γ + x ' ( x vt ) , x ( ' x vt ') G ⇒ = γ γ − + x ( ( x vt ) vt ') ⇒ = t ' ( - x vt ) c ∴ − γ + γ = γ 2 2 x x vt vt ' ∴ = = + x ct G ( ct ' vt ') ⎡ ⎤ γ γ ⎡ ⎤ 2 2 x x v t x x ∴ = − + = γ − + ' t t ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ 2 v γ γ γ γ 2 v v v v v ∴ = − + − ⎣ ⎦ 2 ⎣ ⎦ ct G ( ct vt ) v t t ⎢ ⎥ c ⎣ ⎦ 2 ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ x 1 1 v ∴ = γ + − − = −⎜ ' t t 1 , since 1 ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎟ ⇒ = − 2 2 2 2 c G c [ v ] γ 2 γ 2 v ⎝ c ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 1 ⎡ ⎤ = γ 2 or G = ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ x v vx ⇒ = γ + −⎜ − = γ −⎜ t ' ⎢ t [1 1 ⎥ t − ⎟ ⎟ 2 ⎢ ⎥ 1 ( / ) v c 2 v ⎝ c ⎠ ⎝ c ⎠ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ = γ − ' x ( x vt ) ∴
Lorentz Transformation Between Ref Frames Inverse Lorentz Transformation Lorentz Transformation = γ − = γ + x ' ( x v t ) x ( x ' vt ') = = 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 Transform for Pair of Events S S’ ruler x X ’ x 2 x 1 Can understand Simultaneity, Length contraction & Time dilation formulae from this Time dilation: Bulb in S frame turned on at t 1 & off at t 2 : What ∆ t’ did S’ measure ? two events occur at same place in S frame => ∆ x = 0 ∆ t’ = γ ∆ t ( ∆ t = proper time) Length Contraction: Ruler measured in S between x 1 & x 2 : What ∆ x’ did S’ measure ? two ends measured at same time in S’ frame => ∆ t’ = 0 ∆ x = γ ( ∆ x’ + 0 ) => ∆ x’ = ∆ x / γ ( ∆ x = proper length)
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 = γ − = γ − ' 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)
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 dy ' γ − 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 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 = u x ' x vu + 1 x ' 2 c As usual, ' u = y u replace y v v ⇒ - v γ + ' (1 ) u x 2 c ' u = u z z v γ + ' ( 1 u ) x 2 c
Does Lorentz Transform “work” ? Two rockets travel in opposite directions An observer on earth (S) measures speeds = 0.75c And 0.85c for A & B respectively What does A measure as B’s speed? 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
Can you be seen to be born before your mother? A frame of Ref where sequence of events is REVERSED ?!! u D S S 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 ⎠ ⎣ ⎦ ∆ < For what value of v can 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 t ' 0 ∆ ∆ = 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 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 = = = = = = − ' ' v 1 0, v 2 , V ' v 1 v v 2 v v V v 2 v − − − 1 1 1 1 + 1 1 2 2 2 c c c 2 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
� � mu � = = γ p mu − 2 1 ( / ) u c � � ⎛ ⎞ � dp d mu d du d = = ⎜ ⎟ = F use ⎜ ⎟ dt dt dt dt du − 2 1 ( / ) u c ⎝ ⎠ ⎡ ⎤ Relativistic − − m mu 1 2 u du ⎢ ⎥ = + × F ( )( ) ( ) Momentum ⎢ ⎥ 3/2 2 2 c dt − 2 − 2 1 ( / ) u c 1 ( / ) u c ⎣ ⎦ Force ⎡ ⎤ − + 2 2 2 mc mu mu du ⎢ ⎥ = ⎢ And F ( ) ⎥ 3/ 2 dt − 2 2 c 1 ( u / ) c ⎣ ⎦ Acceleration ⎡ ⎤ m du ⎢ ⎥ = ⎢ F : Relativist ic Force ( ) ⎥ 3/ 2 dt − 2 1 ( u / c ) ⎣ ⎦ � � du Since Acceleration a = , dt � � F 3/2 ⇒ ⎡ − ⎤ 2 a= 1 ( / u c ) Reason why you cant quite ⎣ ⎦ m get up to the speed of light ! � → → Note: As / u c 1, a 0 !!!! Its harder to accelerate when yo u get close to speed of l ig h t
A Linear Particle Accelerator � � eE 3/ 2 ⎡ ⎤ − 2 a= 1 ( / ) u c ⎣ ⎦ m PEP- PEP -II accelerator schematic and tunnel view II accelerator schematic and tunnel view
Accelerating Electrons Thru RF Cavities
Fitting a 5m pole in a 4m barnhouse Student with pole runs with v=(3/5) c farmboy sees pole contraction factor − = 2 1 (3 c /5 ) c 4/5 says pole just fits i n the barn fully! V = (3/5)c Stud ent with pole runs with v=(3/5) c Student sees barn contraction factor − = 2 1 (3 /5 ) c c 4/5 2D Student says barn is only 3.2m long , to o short farmboy to contain entire 5m pole ! Farmboy says “You can do it” Student says “Dude, you are nuts” Is there a contradiction ? Is Relativity wrong? Homework: You figure out who is right, if any and why. Hint: Think in terms of observing three events
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