Physics 2D Lecture Slides Jan 13 Vivek Sharma UCSD Physics - PowerPoint PPT Presentation

Physics 2D Lecture Slides Jan 13 Vivek Sharma UCSD Physics

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

Fitting a 5m pole in a 4m barnhouse? Answer : Simultaneity ! Event A : arrival of right end of pole at left end of barn: (t =0, t'=0) is reference ' L = proper length of pole in S' 0 V = (3/5)c ' l = length of barn in S frame < L 0 0 ' − 2 In S: leng th of pol e L =L 1 ( / v c ) 0 The t imes in two frames are related: l ' l = = − = − ' 2 2 t 0 1 ( / ) v c t 1 ( / ) v c B BC 2D Student v v farmboy ' L t l ' 1 = = = ' t 0 BC C − 2 v v 1 ( / ) v c − 2 1 ( v c / ) ⇒ Time gap in S' by which events B and C A: Arrival of right end of pole at left end of barn fail to be simult aneou s B: Arrival of left end of pole at left end of barn C: Arrival of right end of pole at right end of barn Let S = Barn frame, S' = student f rame Farmboy sees two events as simultaneous 2D student can not agree Fitting of the pole in barn is relative !

Farmboy Vs 2D Student Pole and barn are in relative motion u such that lorentz contracted length of pole = Proper length of barn In rest frame of pole, Event B precedes 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 ' 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 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)

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