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

16. Light clock

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

48. Length contraction

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

73. Stanford electron accelerator
v=0.99999999995 c
3k
m
γ=100,000

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.

75. test
γ=29.3