Physics 2D Lecture Slides Oct 6 Vivek Sharma UCSD Physics New - - PowerPoint PPT Presentation

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

' 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

' ' ' ') ' ' ( v t x x vt t c y z x y 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

2 ' 2 ' 2

divide by dt on (1 ) There is a change in velocity in the direction to S-S' motion ' , ' ' ( H ) ' S ! R ( )

x y y y

u u dy dy dy v dt dt dx dy c u u v dy dt dx c v c γ γ γ = = = ⊥ − − = = −

Velocity Transformation Perpendicular to S-S’ motion

' 2

Similarly Z component of Ant' s velocity transforms (1 ) as

z z x

u u v c u γ = −

Inverse Lorentz Velocity Transformation

' x ' ' ' 2 ' 2 ' 2

Inverse Velocity Transform: (1 u ) 1 1 ( )

y y z x z x x x

u v vu u u v c u v c u c u u γ γ = + = + + = +

As usual, replace

v ⇒ - v

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

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

• bserver see as velocity
• f ball ?

Place S’ frame on biker Biker sees ball speed

uX’ =0.7c

Speed of ball relative to stationary observer

uX ?

Can you be seen to be born before your mother?

A frame of Ref where sequence of events is REVERSED ?!!

S S’

1 1 ' ' 1 1

( , ) ( , ) x t x t

u

2 2 ' ' 2 2

( , ) ( , ) x t x t

I t a k e

• f

f f r

• m

S D I arrive in SF

' ' 2 1 2

' For what value of v can ' v x t t t t c t γ ⎡ ∆ ⎤ ⎛ ⎞ ∆ = − = ∆ −⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ ∆ <

I Cant ‘be seen to arrive in SF before I take off from SD

S S’

1 1 ' ' 1 1

( , ) ( , ) x t x t

u

2 2 ' ' 2 2

( , ) ( , ) x t x t

' 2 2 2 ' 2 1 2

' ' For what value of v v can : Not al lowe u 1 < ' d c v x t t c c v c u v x v c t c v x t t t t c t γ ⎡ ∆ ⎤ ⎛ ⎞ ∆ = − = ∆ −⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ ∆ < ∆ ∆ ∆ = ∆ ⇒ < ⇒ > ⇒ < ⇒ ∆ >

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 =

• 1

2 Before v1’=0 v2’ 2 1 After V’ S’ S 1 2 Before v v 2 1 After V=0 P = mv –mv = 0 P = 0

' ' ' 1 2 ' ' 1 2 1 2 2 1 1 2 2 2 2 2 ' 2

' ' before after

2 0, , ' 2 1 1 1 1 1 2 2 ' ,

p p

after before

mv p mv m v v v v v V v v v V v v v v v V v v c p v v c c mV mv c c − − − − = = = = = − = + = − − − − + = ≠ + = = −

( ) ( ) ( )

2 3/2 2 2 2 2 2 2 2 3/ 2 2 2 2 3/ 2

1 ( / ) 1 2 ( )( 1 ( / ) : Relativist ) 2 1 ( / ) 1 ( / ) 1 ic 1 ( ) ( / / ) mu p mu u c d du d use dt dt du m mu u du F c dt u c u c mc mu mu du F dt c u dp d mu F dt dt u du c c dt m F c u γ = = − = ⎡ ⎤ − − ⎢ ⎥ = + × ⎢ ⎥ − − ⎣ ⎦ ⎡ ⎤ − + ⎢ ⎥ = ⎢ ⎥ − ⎛ ⎞ ⎜ ⎟ = = ⎜ ⎟ − ⎝ ⎠ ⎡ ⎣ ⎦ ⎤ ⎢ ⎥ = ⎢ ⎥ − ⎣ ⎦

• 3/2

2

Force Since Acceleration a = Note: As / 1, a 0 !!!! Its harder to accelerate when yo , F u get close to speed of l a= 1 ( / h ) ig m t u c du dt u c ⎡ ⎤ − ⎣ ⎦ ⇒ → →

• Relativistic

Momentum Force And Acceleration

Reason why you cant quite get up to the speed of light !

PEP PEP-

• II accelerator schematic and tunnel view

II accelerator schematic and tunnel view

A Linear Particle Accelerator

3/ 2 2

eE a= 1 ( / ) m u c ⎡ ⎤ − ⎣ ⎦

Accelerating Electrons Thru RF Cavities

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