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Theory of Relativity
     Albert
     Einstein




     Physics 100
      Chapt 18
watching a light flash go by
                      v


                            c



                                    2k
The man on earth sees c =       √   κ
  (& agrees with Maxwell)
watching a light flash go by
                         v


                                c




If the man on the rocket sees c-v,
    he disagrees with Maxwell
Do Maxwell’s Eqns only work in
    one reference frame?



 If so, this would be the rest frame
      of the luminiferous Aether.
If so, the speed of light should
        change throughout the year




                                 upstream,
downstream,                     light moves
 light moves                       slower
    faster     “Aether wind”
Michelson-Morley




No aether wind detected: 1907 Nobel Prize
Einstein’s hypotheses:

  1. The laws of nature are equally valid
  in every inertial reference frame.
                                  Including
                                Maxwell’s eqns

2. The speed of light in empty space is
   same for all inertial observers, regard-
   less of their velocity or the velocity of
   the source of light.
All observers see light flashes go
      by them with the same speed

                                v
No matter how fast
the guy on the rocket
is moving!!
                                    c




        Both guys see the light flash
          travel with velocity = c
Even when the light flash is
traveling in an opposite direction

                         v


                        c




 Both guys see the light flash
 travel past with velocity = c
Gunfight viewed by observer at rest
                   He sees both shots
                  fired simultaneously




       Bang
         !              Bang
                          !
Viewed by a moving observer
Viewed by a moving observer
              He sees cowboy shoot
             1st & cowgirl shoot later




 Bang
   !            Bang
                  !
Viewed by an observer in the
     opposite direction
Viewed by a moving observer
                    He sees cowgirl shoot
                   1st & cowboy shoot later




                             Bang
            Bang               !
              !
Time depends of state of motion
       of the observer!!


Events   that occur simultaneously
 according to one observer can occur
 at different times for other
 observers
Light clock
Seen from the ground
Events

y                      (x2,t2)
        (x1,t1)
          x               x




         x1              x2      x
    t
Prior to Einstein, everyone agreed
    the distance between events depends
    Same events, different observers
     upon the observer, but not the time.
      y’     y’
y                               (x2,t2)
                (x1,t1)
                x                    x

              (x1’,t1’)           (x2’,t2’)
         t’    t’
               x1’   x1’ dist’      x2’
                                 x’    x’

                x1                   x2       x
     t                    dist
Time is the 4th dimension

      Einstein discovered that there is no
      “absolute” time, it too depends upon
      the state of motion of the observer


                             Einstein
             Newton
                             Space-Time
completely   Space
 different     &            2 different aspects
 concepts    Time            of the same thing
How are the times seen
    by 2 different
  observers related?

    We can figure this out with
      simple HS-level math
       ( + a little effort)
Catch ball on a rocket ship
       Event 2: girl catches the ball




                                  w
                             v=     =4m/s
                                  t
              w=4m

                      t=1s




      Event 1: boy throws the ball
Seen from earth
                               V0=3m/s
           V0=3m/s
 Location of the 2
events is different
                                                  = 5m
 Elapsed time is                          m
                                              2
                                              )
                                       (4
    the same                   m
                                   )2 +                  w=4m

                           √ (3
 The ball appears     d=
                                     v0t=3m
 to travel faster

                                                                d
             t=1s                                          v=     = 5m/s
                                                                t
Flash a light on a rocket ship
        Event 2: light flash reaches the girl




                                   w
                              c=
                                   t0
                  w
                        t0




        Event 1: boy flashes the light
Seen from earth
                              V
               V
Speed has to
Be the same
                                   2
                                2+
                                   w
                                )
Dist is longer             √(
                             vt              w
                      d=

Time must be                      vt
    longer
                                            d
                                              =√ (vt) +w
                                                     2   2
                                       c=
         t=?                                t        t
How is t related to t0?
t= time on Earth clock                   t0 = time on moving clock
                                                    w
      c =√    (vt)2+w2                        c =
                                                    t0
                  t
     ct =   √ (vt)2+w2                        ct0 = w


     (ct)2 = (vt)2+w2

   (ct)2 = (vt)2+(ct0)2     (ct)2-(vt)2= (ct0)2          (c2-v2)t2= c2t02

                  c2                               1
           t = 2
              2
                        t 02        t2 =              t2 2
                c – v 2                        1 – v /c 0
                                                     2



                        1
           t=            t0
                    √1 – v2/c2          t = γ t0
this is called γ
Properties of γ =             1
                           √1 – v2/c2
Suppose v = 0.01c        (i.e. 1% of c)

             1                      1
  γ = √1 – (0.01c)2/c2    =
                            √1 – (0.01)2c2/c2

           1                1           1
  γ = √1 – (0.01)2   =             =
                       √1 – 0.0001   √0.9999


             γ = 1.00005
Properties 1
                            of
           γ =             √1 – v(cont’d)
                                 2
                                  /c2
Suppose v = 0.1c        (i.e. 10% of c)

             1                     1
  γ = √1 – (0.1c)2/c2     =
                            √1 – (0.1)2c2/c2

           1                  1               1
  γ = √1 – (0.1)2   =                 =
                          √1 – 0.01         √0.99


             γ = 1.005
Let’s make a chart
   v           γ =1/√(1-v2/c2)
0.01 c            1.00005
 0.1 c             1.005
Other values of                 1
  γ =                         √1 – v2/c2

Suppose v = 0.5c        (i.e. 50% of c)

             1                     1
  γ = √1 – (0.5c)2/c2     =
                            √1 – (0.5)2c2/c2

           1               1                1
  γ = √1 – (0.5)2   =             =
                      √1 – (0.25)         √0.75


             γ = 1.15
Enter into chart
   v              γ =1/√(1-v2/c2)
0.01 c               1.00005
 0.1 c                1.005
 0.5c                  1.15
Other values of                 1
  γ =                         √1 – v2/c2

Suppose v = 0.6c        (i.e. 60% of c)

            1                      1
  γ =√1 – (0.6c)2/c2      =
                            √1 – (0.6)2c2/c2

           1                  1             1
  γ = √1 – (0.6)2      =             =
                         √1 – (0.36)      √0.64


             γ = 1.25
Back to the chart
   v              γ =1/√(1-v2/c2)
0.01 c               1.00005
 0.1 c                1.005
 0.5c                  1.15
 0.6c                  1.25
Other values of                 1
  γ =                         √1 – v2/c2

Suppose v = 0.8c        (i.e. 80% of c)

             1                     1
  γ = √1 – (0.8c)2/c2     =
                            √1 – (0.8)2c2/c2

           1               1                1
  γ = √1 – (0.8)2   =             =
                      √1 – (0.64)         √0.36


             γ = 1.67
Enter into the chart
   v           γ =1/√(1-v2/c2)
0.01 c            1.00005
 0.1 c             1.005
 0.5c               1.15
 0.6c               1.25
 0.8c               1.67
Other values of                 1
  γ =                         √1 – v2/c2

Suppose v = 0.9c        (i.e.90% of c)

             1                     1
  γ = √1 – (0.9c)2/c2     =
                            √1 – (0.9)2c2/c2

           1                  1             1
  γ = √1 – (0.9)2   =                 =
                          √1 – 0.81       √0.19


             γ = 2.29
update chart
   v            γ =1/√(1-v2/c2)
0.01 c             1.00005
 0.1 c              1.005
 0.5c                1.15
 0.6c                1.25
 0.8c                1.67
 0.9c                2.29
Other values of                 1
  γ =                         √1 – v2/c2

Suppose v = 0.99c        (i.e.99% of c)

            1                       1
  γ =√1 – (0.99c)2/c2     =
                            √1 – (0.99)2c2/c2

           1                 1              1
  γ = √1 – (0.99)2   =                =
                         √1 – 0.98        √0.02


             γ = 7.07
Enter into chart
   v              γ =1/√(1-v2/c2)
0.01 c               1.00005
 0.1 c                1.005
 0.5c                  1.15
 0.6c                  1.25
 0.8c                  1.67
 0.9c                  2.29
0.99c                  7.07
Other values of                   1
  γ =                           √1 – v2/c2

Suppose v = c

             1                       1
  γ =   √1 – (c)2/c2       =
                                 √1 – c2/c2

          1                     1             1
  γ =                  =                =
        √1 – 12                √0             0


             γ =   ∞           Infinity!!!
update chart
   v            γ =1/√(1-v2/c2)
0.01 c             1.00005
 0.1 c              1.005
 0.5c                1.15
 0.6c                1.25
 0.8c                1.67
 0.9c                2.29
0.99c                7.07
1.00c                 ∞
Other values of                 1
  γ =                         √1 – v2/c2

Suppose v = 1.1c

             1                     1
  γ = √1 – (1.1c)2/c2     =
                            √1 – (1.1)2c2/c2

           1                 1              1
  γ = √1 – (1.1)2   =                =
                          √1-1.21        √ -0.21

             γ =    ???      Imaginary number!!!
Complete the chart
      v          γ =1/√(1-v2/c2)
   0.01 c           1.00005
    0.1 c            1.005
    0.5c              1.15
    0.6c              1.25
    0.8c              1.67
    0.9c              2.29
   0.99c              7.07
   1.00c               ∞
Larger than c   Imaginary number
Plot results:
                         ∞




                             Never-never land
        1
γ =
    √1 – v2/c2


                         x


                         x
                     x
             x   x
                         v=c
Moving clocks run slower
                                v

                  t0


       t=           1 t
                          0
                √1 – v2/c2

  t      t = γ t0
            γ   >1           t > t0
Length contraction
                    v



                   L0

                  time=t
                  L0 = vt
                                    te r!
                               or
                            Sh
man on   Time = t0 =t/γ
rocket   Length = vt0 =vt/γ =L0/γ
Moving objects appear shorter
                    Length measured when
                       object is at rest
         L = L0/γ
                    γ   >1  L < L0




   V=0.9999c
    V=0.86c
    V=0.1c
    V=0.99c
Length contraction
mass:                                 change in v
                       F=m0a = m0      time
                                          t0
               a               time=t0
          m0
                                             Ft0
                           change in v =
                                              m0
                                      Ft0
                          m0 =
                                  change in v

                                                   mass
         Ft              γ Ft0
m=                 =                 = γ m0     increases!!
     change in v       change in v
                                          m = γ m0
                   t=γt0
                                        by a factor   γ
Relativistic mass increase
    m0 = mass of an object when it
         is at rest  “rest mass”

mass of a moving    γ
object increases        as vc, m∞

   m = γ m0             as an object moves
                         faster, it gets
                         harder & harder
                          to accelerate
  by the γ factor

                                         v=c
summary
• Moving clocks run slow           γ
                             o f
                       o r
                  c t
• Moving objects appear shorter
                 f a
             a
           y
• Moving object’s mass increases
          B
Plot results:
                         ∞




                             Never-never land
        1
γ =
    √1 – v2/c2


                         x


                         x
                     x
             x   x
                         v=c
α-
  Twin paradox               centauri
                                         rs
                                   y   ea
                              ht
 Twin brother             lig
   & sister         4.3


                     She will travel to
                    α -centauri (a near-
                    by star on a special
He will stay home   rocket ship v = 0.9c
& study Phys 100
Light year
distance light travels in 1 year

   dist = v x time = c yr

  1cyr = 3x108m/s x 3.2x107 s
         = 9.6 x 1015 m

     We will just use cyr units
     & not worry about meters
Time on the boy’s clock

                                   r
                                cy
            0.9c       =4
                          .   3
          v=         d0

                  0.9c
                v=
            According to the boy
            & his clock on Earth:
                 d0     4.3 cyr = 4.8 yrs
        tout =       = 0.9c
                  v
                  d0    4.3 cyr = 4.8 yrs
         tback =     = 0.9c
                  v
          ttotal = tout+tback   = 9.6yrs
What does the boy see on her
           clock?
                                     yr
                0.9c          4.   3c
              v=           d=

                        0.9c
                      v=
                  According to the boy
                  her clock runs slower
                   tout 4.8 yrs
           t =
            out
                        = 2.3 = 2.1 yrs
                    γ
                   tback 4.8 yr
            tback = γ =         = 2.1 yrs
                          2.3
            ttotal = tout+tback           = 4.2yrs
So, according to the boy:

                                    yr
                   0.9c      4.   3c
                 v=       d=

                       0.9c
                     v=


      his clock     her clock
  out: 4.8yrs        2.1yrs
 back: 4.8yrs        2.1yrs                ges
                                     She a
 total: 9.6yrs       4.2yrs             less
But, according to the
 girl, the boy’s clock is
moving &, so, it must be
                                                     0.9c
     running slower                                v=



                                   According to her, the
                                boy’s clock on Earth says:

                            tout 2.1 yrs
                    tout = γ =           = 0.9 yrs
                                   2.3
                            tback 2.1 yrs = 0.9 yrs
                     tback =     = 2.3
      .9c                     γ
   v=0
                      ttotal = tout+tback   = 1.8yrs
Her clock advances 4.2 yrs
& she sees his clock advance
        only 1.8 yrs,

  contradict ion??
A
She should think he has aged
 less than her!!
Events in the boy’s life:
                  As seen by him   As seen by her

 She leaves
                    4.8 yrs           0.9 yrs

  She arrives
 & starts turn
                    short time         ????

  Finishes turn
  & heads home
                    4.8 yrs          0.9 yrs

   She returns      9.6+ yrs       1.8 + ??? yrs
turning around as seen by her
                 According to her, these
                 2 events occur very,very
                far apart from each other



                                            He sees her
He sees her
                                            finish turning
start to turn


        Time interval between 2 events depends
        on the state of motion of the observer
Gunfight viewed by observer at rest
                   He sees both shots
                  fired simultaneously




       Bang
         !              Bang
                          !
Viewed by a moving observer
Viewed by a moving observer
              He sees cowboy shoot
             1st & cowgirl shoot later




 Bang
   !            Bang
                  !
In fact, ???? = 7.8+ years
                   as seen by him   as seen by her
   She leaves
                      4.8 yrs          0.9 yrs

    She arrives
   & starts turn
                     short time       7.8+ yrs
                                        ???

  Finishes turn
  & heads home
                      4.8 yrs          0.9 yrs

    She returns      9.6+ yrs       1.8 + ???yrs
                                      9.6+ yrs
No paradox: both twins agree



  The twin that
 “turned around”
    is younger
Ladder & Barn Door paradox
Stan & Ollie puzzle over how
to get a 2m long ladder thru
a 1m wide barn door


   ???
                               1m




                               2m


                                    ladder
Ollie remembers Phys 100 & the
        theory of relativity
Stan, pick up
the ladder &
run very fast
                 1m




                 2m            tree

                      ladder
View from Ollie’s ref. frame



                1m
 Push,
 Stan!
                2m/γ




                       V=0.9c
Ollie        Stan      (γ=2.3)
View from Stan’s ref. frame

                           But it
                  1m/    doesn’t fit,
                  γ
                            Ollie!!
V=0.9c
(γ=2.3)      2m




Ollie             Stan
If Stan pushes both ends of the
ladder simultaneously, Ollie sees the
  two ends move at different times:

 Too late
                      1m
   Too
  Stan!
   soon          nk
  Stan!      clu           clan
                                  k

                                      V=0.9c
 Ollie       Stan
                Stan                  (γ=2.3)
Fermilab proton accelerator




                V=0.9999995c

                γ =1000
          2km
Stanford electron accelerator
v=0.99999999995 c




                    3k
                      m
γ=100,000
status
Einstein’s theory of “special relativity” has
  been carefully tested in many very precise
  experiments and found to be valid.

Time is truly the 4th dimension of space &
  time.
test




γ=29.3

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Theory of relativity

  • 1. Theory of Relativity Albert Einstein Physics 100 Chapt 18
  • 2. watching a light flash go by v c 2k The man on earth sees c = √ κ (& agrees with Maxwell)
  • 3. watching a light flash go by v c If the man on the rocket sees c-v, he disagrees with Maxwell
  • 4. Do Maxwell’s Eqns only work in one reference frame? If so, this would be the rest frame of the luminiferous Aether.
  • 5. If so, the speed of light should change throughout the year upstream, downstream, light moves light moves slower faster “Aether wind”
  • 6. Michelson-Morley No aether wind detected: 1907 Nobel Prize
  • 7. Einstein’s hypotheses: 1. The laws of nature are equally valid in every inertial reference frame. Including Maxwell’s eqns 2. The speed of light in empty space is same for all inertial observers, regard- less of their velocity or the velocity of the source of light.
  • 8. All observers see light flashes go by them with the same speed v No matter how fast the guy on the rocket is moving!! c Both guys see the light flash travel with velocity = c
  • 9. Even when the light flash is traveling in an opposite direction v c Both guys see the light flash travel past with velocity = c
  • 10. Gunfight viewed by observer at rest He sees both shots fired simultaneously Bang ! Bang !
  • 11. Viewed by a moving observer
  • 12. Viewed by a moving observer He sees cowboy shoot 1st & cowgirl shoot later Bang ! Bang !
  • 13. Viewed by an observer in the opposite direction
  • 14. Viewed by a moving observer He sees cowgirl shoot 1st & cowboy shoot later Bang Bang ! !
  • 15. Time depends of state of motion of the observer!! Events that occur simultaneously according to one observer can occur at different times for other observers
  • 17. Seen from the ground
  • 18. Events y (x2,t2) (x1,t1) x x x1 x2 x t
  • 19. Prior to Einstein, everyone agreed the distance between events depends Same events, different observers upon the observer, but not the time. y’ y’ y (x2,t2) (x1,t1) x x (x1’,t1’) (x2’,t2’) t’ t’ x1’ x1’ dist’ x2’ x’ x’ x1 x2 x t dist
  • 20. Time is the 4th dimension Einstein discovered that there is no “absolute” time, it too depends upon the state of motion of the observer Einstein Newton Space-Time completely Space different & 2 different aspects concepts Time of the same thing
  • 21. How are the times seen by 2 different observers related? We can figure this out with simple HS-level math ( + a little effort)
  • 22. Catch ball on a rocket ship Event 2: girl catches the ball w v= =4m/s t w=4m t=1s Event 1: boy throws the ball
  • 23. Seen from earth V0=3m/s V0=3m/s Location of the 2 events is different = 5m Elapsed time is m 2 ) (4 the same m )2 + w=4m √ (3 The ball appears d= v0t=3m to travel faster d t=1s v= = 5m/s t
  • 24. Flash a light on a rocket ship Event 2: light flash reaches the girl w c= t0 w t0 Event 1: boy flashes the light
  • 25. Seen from earth V V Speed has to Be the same 2 2+ w ) Dist is longer √( vt w d= Time must be vt longer d =√ (vt) +w 2 2 c= t=? t t
  • 26. How is t related to t0? t= time on Earth clock t0 = time on moving clock w c =√ (vt)2+w2 c = t0 t ct = √ (vt)2+w2 ct0 = w (ct)2 = (vt)2+w2 (ct)2 = (vt)2+(ct0)2  (ct)2-(vt)2= (ct0)2  (c2-v2)t2= c2t02 c2 1  t = 2 2 t 02  t2 = t2 2 c – v 2 1 – v /c 0 2 1  t= t0 √1 – v2/c2  t = γ t0 this is called γ
  • 27. Properties of γ = 1 √1 – v2/c2 Suppose v = 0.01c (i.e. 1% of c) 1 1 γ = √1 – (0.01c)2/c2 = √1 – (0.01)2c2/c2 1 1 1 γ = √1 – (0.01)2 = = √1 – 0.0001 √0.9999 γ = 1.00005
  • 28. Properties 1 of γ = √1 – v(cont’d) 2 /c2 Suppose v = 0.1c (i.e. 10% of c) 1 1 γ = √1 – (0.1c)2/c2 = √1 – (0.1)2c2/c2 1 1 1 γ = √1 – (0.1)2 = = √1 – 0.01 √0.99 γ = 1.005
  • 29. Let’s make a chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005
  • 30. Other values of 1 γ = √1 – v2/c2 Suppose v = 0.5c (i.e. 50% of c) 1 1 γ = √1 – (0.5c)2/c2 = √1 – (0.5)2c2/c2 1 1 1 γ = √1 – (0.5)2 = = √1 – (0.25) √0.75 γ = 1.15
  • 31. Enter into chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15
  • 32. Other values of 1 γ = √1 – v2/c2 Suppose v = 0.6c (i.e. 60% of c) 1 1 γ =√1 – (0.6c)2/c2 = √1 – (0.6)2c2/c2 1 1 1 γ = √1 – (0.6)2 = = √1 – (0.36) √0.64 γ = 1.25
  • 33. Back to the chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25
  • 34. Other values of 1 γ = √1 – v2/c2 Suppose v = 0.8c (i.e. 80% of c) 1 1 γ = √1 – (0.8c)2/c2 = √1 – (0.8)2c2/c2 1 1 1 γ = √1 – (0.8)2 = = √1 – (0.64) √0.36 γ = 1.67
  • 35. Enter into the chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67
  • 36. Other values of 1 γ = √1 – v2/c2 Suppose v = 0.9c (i.e.90% of c) 1 1 γ = √1 – (0.9c)2/c2 = √1 – (0.9)2c2/c2 1 1 1 γ = √1 – (0.9)2 = = √1 – 0.81 √0.19 γ = 2.29
  • 37. update chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29
  • 38. Other values of 1 γ = √1 – v2/c2 Suppose v = 0.99c (i.e.99% of c) 1 1 γ =√1 – (0.99c)2/c2 = √1 – (0.99)2c2/c2 1 1 1 γ = √1 – (0.99)2 = = √1 – 0.98 √0.02 γ = 7.07
  • 39. Enter into chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07
  • 40. Other values of 1 γ = √1 – v2/c2 Suppose v = c 1 1 γ = √1 – (c)2/c2 = √1 – c2/c2 1 1 1 γ = = = √1 – 12 √0 0 γ = ∞ Infinity!!!
  • 41. update chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07 1.00c ∞
  • 42. Other values of 1 γ = √1 – v2/c2 Suppose v = 1.1c 1 1 γ = √1 – (1.1c)2/c2 = √1 – (1.1)2c2/c2 1 1 1 γ = √1 – (1.1)2 = = √1-1.21 √ -0.21 γ = ??? Imaginary number!!!
  • 43. Complete the chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07 1.00c ∞ Larger than c Imaginary number
  • 44. Plot results: ∞ Never-never land 1 γ = √1 – v2/c2 x x x x x v=c
  • 45. Moving clocks run slower v t0 t= 1 t 0 √1 – v2/c2 t t = γ t0 γ >1  t > t0
  • 46. Length contraction v L0 time=t L0 = vt te r! or Sh man on Time = t0 =t/γ rocket Length = vt0 =vt/γ =L0/γ
  • 47. Moving objects appear shorter Length measured when object is at rest L = L0/γ γ >1  L < L0 V=0.9999c V=0.86c V=0.1c V=0.99c
  • 49. mass: change in v F=m0a = m0 time t0 a time=t0 m0 Ft0 change in v = m0 Ft0 m0 = change in v mass Ft γ Ft0 m= = = γ m0 increases!! change in v change in v m = γ m0 t=γt0 by a factor γ
  • 50. Relativistic mass increase m0 = mass of an object when it is at rest  “rest mass” mass of a moving γ object increases as vc, m∞ m = γ m0 as an object moves faster, it gets harder & harder to accelerate by the γ factor v=c
  • 51. summary • Moving clocks run slow γ o f o r c t • Moving objects appear shorter f a a y • Moving object’s mass increases B
  • 52. Plot results: ∞ Never-never land 1 γ = √1 – v2/c2 x x x x x v=c
  • 53. α- Twin paradox centauri rs y ea ht Twin brother lig & sister 4.3 She will travel to α -centauri (a near- by star on a special He will stay home rocket ship v = 0.9c & study Phys 100
  • 54. Light year distance light travels in 1 year dist = v x time = c yr 1cyr = 3x108m/s x 3.2x107 s = 9.6 x 1015 m We will just use cyr units & not worry about meters
  • 55. Time on the boy’s clock r cy 0.9c =4 . 3 v= d0 0.9c v= According to the boy & his clock on Earth: d0 4.3 cyr = 4.8 yrs tout = = 0.9c v d0 4.3 cyr = 4.8 yrs tback = = 0.9c v ttotal = tout+tback = 9.6yrs
  • 56. What does the boy see on her clock? yr 0.9c 4. 3c v= d= 0.9c v= According to the boy her clock runs slower tout 4.8 yrs t = out = 2.3 = 2.1 yrs γ tback 4.8 yr tback = γ = = 2.1 yrs 2.3 ttotal = tout+tback = 4.2yrs
  • 57. So, according to the boy: yr 0.9c 4. 3c v= d= 0.9c v= his clock her clock out: 4.8yrs 2.1yrs back: 4.8yrs 2.1yrs ges She a total: 9.6yrs 4.2yrs less
  • 58. But, according to the girl, the boy’s clock is moving &, so, it must be 0.9c running slower v= According to her, the boy’s clock on Earth says: tout 2.1 yrs tout = γ = = 0.9 yrs 2.3 tback 2.1 yrs = 0.9 yrs tback = = 2.3 .9c γ v=0 ttotal = tout+tback = 1.8yrs
  • 59. Her clock advances 4.2 yrs & she sees his clock advance only 1.8 yrs, contradict ion?? A She should think he has aged less than her!!
  • 60. Events in the boy’s life: As seen by him As seen by her She leaves 4.8 yrs 0.9 yrs She arrives & starts turn short time ???? Finishes turn & heads home 4.8 yrs 0.9 yrs She returns 9.6+ yrs 1.8 + ??? yrs
  • 61. turning around as seen by her According to her, these 2 events occur very,very far apart from each other He sees her He sees her finish turning start to turn Time interval between 2 events depends on the state of motion of the observer
  • 62. Gunfight viewed by observer at rest He sees both shots fired simultaneously Bang ! Bang !
  • 63. Viewed by a moving observer
  • 64. Viewed by a moving observer He sees cowboy shoot 1st & cowgirl shoot later Bang ! Bang !
  • 65. In fact, ???? = 7.8+ years as seen by him as seen by her She leaves 4.8 yrs 0.9 yrs She arrives & starts turn short time 7.8+ yrs ??? Finishes turn & heads home 4.8 yrs 0.9 yrs She returns 9.6+ yrs 1.8 + ???yrs 9.6+ yrs
  • 66. No paradox: both twins agree The twin that “turned around” is younger
  • 67. Ladder & Barn Door paradox Stan & Ollie puzzle over how to get a 2m long ladder thru a 1m wide barn door ??? 1m 2m ladder
  • 68. Ollie remembers Phys 100 & the theory of relativity Stan, pick up the ladder & run very fast 1m 2m tree ladder
  • 69. View from Ollie’s ref. frame 1m Push, Stan! 2m/γ V=0.9c Ollie Stan (γ=2.3)
  • 70. View from Stan’s ref. frame But it 1m/ doesn’t fit, γ Ollie!! V=0.9c (γ=2.3) 2m Ollie Stan
  • 71. If Stan pushes both ends of the ladder simultaneously, Ollie sees the two ends move at different times: Too late 1m Too Stan! soon nk Stan! clu clan k V=0.9c Ollie Stan Stan (γ=2.3)
  • 72. Fermilab proton accelerator V=0.9999995c γ =1000 2km
  • 74. status Einstein’s theory of “special relativity” has been carefully tested in many very precise experiments and found to be valid. Time is truly the 4th dimension of space & time.