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

2

farmboy sees pole contraction factor 1 (3 Student with pole runs /5 ) 4/5 says pole just fits i with v=(3/5) n the barn fully! c c c − =

2D Student farmboy

2

Student sees barn contraction factor 1 (3 /5 ) 4/5 says barn is only 3.2m long Stud , to ent with pole runs

• short

to contain entire 5m pole ! with v=(3/5) c c c − =

Farmboy says “You can do it” Student says “Dude, you are nuts” V = (3/5)c 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?

' ' 2 '

L = proper length of pole in S' Event A : arrival of right end of pole at = length of barn in S < L left end of barn: (t =0, t'=0) is reference In frame L =L 1 S: leng ( / th of pol ) The t e imes in l v c −

' 2 2 B ' ' C 2 2

' t 1 ( / ) 1 ( / ) ' 1 t 1 ( / ) two frames are related: Time gap in S' by which events B and C fail to be simult 1 ( aneou / s )

BC BC

l l v c t v c v v L t l v v v c v c = = − = − = = ⇒ = − − 2D Student farmboy V = (3/5)c

Answer : Simultaneity !

Farmboy sees two events as simultaneous 2D student can not agree Fitting of the pole in barn is relative ! A: Arrival of right end of pole at left end of barn B: Arrival of left end of pole at left end of barn C: Arrival of right end of pole at right end of barn

S = Barn frame, S' = student f Let rame

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

' guess ' ( ) ( ' ' ' ' guess ) x x v G x x vt x x vt t x vt G x = − → = = + + → = −

Need to figure out functional form of G ! G must be dimensionless G does not depend on x,y,z,t But G depends on v/c 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’ Do a Thought Experiment: Rocket Motion along x axis Speed of light is c for both

• bservers

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 The distance to point P from O : r’ = ct’ Clearly t and t’ must be different

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’

2 2 2 2 2 2 2

' ' ( - ) , ' ( - ) ( ) [ ] 1

• r

= 1 ( / ) ( ' ') ( ' ') '

( )

x ct G x vt G t x vt c v ct G ct t c c x ct G x vt x ct G G c v ct vt G v vt v c t

x x vt

γ

γ

= = ⇒ = ∴   ∴ = − + −     ⇒ = − = = + = = − + = ∴

= −

2 2 2 2 2 2 2 2 2

1 since 1 , ( ' ') ( ( ) ') ' 1 ' 1 , ' ( ) ' ' [1 x x vt x x vt vt x x vt x x vt vt x x t x x t t v v v v x t t v x v v v c v t v c t γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ γ = + ⇒ = − + ∴     ∴ = − + = − +            − = −           ∴ = +  −  = − − + =          ⇒ = −   +  

2

1 vx t c γ      −      − =         

Lorentz Transformation Between Ref Frames

2

' ' ' ' ( ) y y z z v t t x t c x v x γ γ =   = −    = −  =

Lorentz Transformation

2

' ' ' ) ' ' ( y y z v t x x v t c t x z γ γ   = +    = = +  =

Inverse Lorentz Transformation As v→0 , Galilean Transformation is recovered, as per requirement

Notice : SPACE and TIME Coordinates mixed up !!!

Lorentz Transform for Pair of Events

Can understand Simultaneity, Length contraction & Time dilation formulae from this Time dilation: Bulb in S frame turned on at t1 & off at t2 : What ∆t’ did S’ measure ? two events occur at same place in S frame => ∆x = 0

∆t’ = γ ∆t (∆t = proper time)

S

x

S’

X’

Length Contraction: Ruler measured in S between x1 & x2 : 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)

x1 x2 ruler

Lorentz Velocity Transformation Rule

' ' ' 2 1 x' ' ' ' 2 1 x' 2 x' 2 x' 2 '

In S' frame, u , u , u 1 For v << c, u (Gali divide by dt' ' lean Trans. Restor ( ) ( ed) )

x x x

x x dx t t dt dx vdt v dt dx c v dt dt dx dx v dx c u u u d v v t c v γ γ − = = − − = − − = = = = − − − −

S S’ v

u

S and S’ are measuring ant’s speed u along x, y, z axes

Does Lorentz Transform “work” ?

Two rockets travel in

• pposite 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