1) The document describes Einstein's theory of relativity and the paradox of the twins through the example of one twin traveling in a spaceship moving at 3/5 the speed of light while the other stays on Earth. 2) It shows that due to time dilation, the traveling twin ages slower than the twin on Earth, with their clocks running at different rates. 3) After 10 years on Earth, the traveling twin has only aged to 8 years due to time passing more slowly aboard the faster moving spaceship.

2. Exam 1-slide 1-30 plus anything on space -
time

3. Dr Rob’s Story –a tale of “the paradox of the
twins”

4. a power point by Dr Rob
•

5. One day-I decided to study “what did Einstein
see . . .

6. He sees- I four 4D

7. What is 4D? And can I understand this stuff?
• How To See Spacetime Stretch - LIGO | Video
space-time
• ˈspās ˈˌtīm/
nounPhysics
noun: space-time; noun: spacetime
• the concepts of time and three-dimensional space regarded as fused
in a four-dimensional continuum

8. 4D is called “space time”
• In physics, spacetime is any mathematical model that fuses the three
dimensions of space and the one dimension of time into a single
4-dimensional continuum. Spacetime diagrams are useful in
visualizing and understanding relativistic effects such as how different
observers perceive where and when events occur.

9. famous Michelson–Morley interferometer
experiment-you do this latter (
• Einstein's theory was framed in terms of kinematics and showed how
quantification of distances and times varied for measurements made
in different reference frames.
• His theory was a breakthrough advance over Lorentz's 1904 theory of
electromagnetic phenomena and Poincaré's electrodynamic theory.
Although these theories included equations identical to those that
Einstein introduced (i.e. the Lorentz transformation), they were
essentially ad hoc models proposed to explain the results of various
experiments—including the famous Michelson–Morley
interferometer experiment—that were extremely difficult to fit into
existing paradigms

10. So I need to Add of the dimension of time

11. Like a straight line (does not change In x)

12. Limitations on space-time

13. I can not go faster than the speed of light
-or backward in time

14. If I travel faster than the speed of light – I can
time travel-nothing can go fast the light

15. And I can not go back in time

16. Let’s draw the speed of light at 45 degrees

17. Begin the Paradox of the twins

18. Time at 1 sec and distance at 3,000,000 KM
• So-we can not have anything faster than the speed of light
• And we can not go backwards-so –there are limits

19. Also-if I am in London-can I be in new York-I 5
minutes time . . The answer is no

20. Lets –use the chart to examine the “twin
Paradox

21. • Imagine that there are a set of twins-born on Earth at the same time

22. • And imagine a spaceship-traveling at speed V
• V= 3/5 C –where C is the speed of light
• And-also imagine the space ship has a hook on it

23. V= 3/5 C –where C is the speed of light

24. So- so far – I got a spaceship-with a hook
and I am going to pick up 1 twin-take her 5 years
away-like 3 light years

25. At a distance to 3 light years-the twin meets
another space craft-going the other way!!!

26. The space craft –brings the twin back at the
same speed as the first

27. As the craft returns back- I release the twin I
took away

28. • And we would like them – to celebrate there 10th birthdays together
• But-that will not be the case

29. The first twin starts at the year “naught”

30. The first twin- does not move along the space
axis –at all

31. One twin moves straight up –the graph –for a
period of 10 years

32. • Twin #1 –stays on earth-the straight line
• Twin #2-travels 3/5C (diagonal one-and diagonal 2)

33. A moving clock-appears to run slow

34. What is an Einstein clock
• You can invent a very simple clock
• Tha can consist of 2 mirrors

35. • What you do is you simply bounce light up and down
• Between the 2 mirrors-every time the light beam hits a mirror
• It essentially causes a counter to move on
• And –since the distance is known
• And since the speed of light is invariant
• Denote the time required to go from the top mirror to the bottom
mirror is :
• T = d/c

36. • And this is sense becomes “our time measure”
“one movement from the bottom mirror to the top mirror
• Constitute one time interval which equal d/c

37. • Suppose this apparatus is in a rocket-going past the earth
• So here is an observe –and an Observer on the Earth

38. And here is the rocket--And here is the mirror
in the rocket

40. The rocket is moving at velocity V-
approaching the speed of light-3/5C

41. • So what happens- the light leaves the bottom mirror, on its way up to
the top mirror-but before it gets there
The top mirror is now here-and then it is reflected – so that the time it
gets here- the rocket has moved

42. And here is the mirror in the rocket

43. So – to the observer on the Earth

44. • What the light beam has appeared to do. . . .
• Is not go up and down

45. • So how can the 2 observers –one on Earth and one inside the rocket
• Compare these two events
• Well –the person inside the rocket says ‘the light went up in a straight
line”
• The Person on the Earth says “No-it did not – I went diagonally”

46. The Person on the Earth says “No-it did not – I
went diagonally”

47. So how can the 2 observers –one on Earth and
one inside the rocket
Compare these two events

48. Well –the person inside the rocket says ‘the light
went up in a straight line”

49. The Person on the Earth says “No-it did not – I
went diagonally”

50. • Now –we can do some geometry

51. • And lets create a right angled triangle

52. Which we are going to call t-prime

53. • T- prime ( or t’) is the time is which the observer on the Earth
• Would decide that the rocket took to get to Here to Here

54. • Well –now we can do some basic pythagorous
• We have for equation 1:

55. • If it takes the rocket to get from here to here-than it also take the like
to travel path d the same

56. • And that means d’ =c t’ (d-prime is equal to c times t-prime

57. • After constructing the Pythagorean theorem-just divide everything by
C2 (c-squared)

58. • Then take the square root of both sides . . .. .

59. This term here –is the lorenzt transform

60. • Essentiallly- it means that the time in one frame of reference is not
necessarily the time in another frame of reference

61. • So –before we move on-let’s remind our selves what this has shown

62. • T= the time interval of the person in the space rocket

63. • For the light to travel from here to here

64. • T-prime – it the time it takes for the observer the believe the light has
travel from here to here

65. • This is distance d

66. • And that is the distance d-that we had before

67. • This is going to be distance “d prime” or d’

68. • T- prime is bigger that T-so Therefore for the Earth observer-things
seem to be moving slower than they should

69. • Each time interval-is now longer than it should be

70. • Here is the time interval for the person in the space craft

71. • Here is the time interval for the Person on Earth

72. • And if you have longer time intervals-time itself- will go more slowly

73. This is why a moving clock moves more slowly

74. • In our example- v = 3/5C- where C = 3,000,000 Km/s

75. • What then is the relationship between the person in the space craft
(t) and the time interval of the person on Earth (t’)

76. • The answer is that t equals t’ times the square root term

77. • So let’s plug in some numbers

78. • And t= 4/5t’—the time in the space craft if 4/5ths
• So this suggests the time measured by the person in the space craft
• Is 4/5 of the person on the Earth

80. • We can also say that if the person on the Earth-looking at the clock at
the space craft
• The guy on the Earth – if he has his own watch-he will see that 5
seconds has passed

81. • But-when he looks at the clock in the space craft-he will see that only
4 seconds has passed

82. • Similarily- if 5 years passed on Earth – only 4 years would of appeared
to pass on the space craft

83. • The “moving clock” on the space craft is going slower

84. • This applies both ways- if I have 2 people with different frames of
reference- and the relative speed between them in V

85. • Or –when this observer looks at this persons clock- it will appear to
be running slow

86. • When the stationary person looks at the moving persons clock-that
too will appear to be running slow

87. • So now lets go back to our diagram

88. • Lets draw it a little more carefully-lets run it up to 10 years-the x axis
is distance in light years-number one twin stays on the Earth for 10
years

89. • The Earth is not moving-and there is simply a steady movement
through time-and 10 years latter – the twin is here

90. • The other twin was picked up by a rocket travelling at 3/5th the speed
of light

91. • And that is this line of rocket coming back the other way

92. • At this point –the rocket drops the second twin off

93. • And-let us suppose there is some communication between the two
twins-and the communication travels at the speed of light-however
there is no instantenous communication- the communications- travel
at the fastests at the speed of light

94. • So lets consider the person on the Earth receiving signals from the
person in the space craft- and the person on the Earth Is 2 years old

95. • That signal-must of come from this point here-and remember-this is
the 45 degree angle-that represents the speed of light

96. • So the twin on Earth (*at point number 2)-will see the twin in the
space craft-fortunately-he is standing next to a clock-and what time
will he show on this clock?

97. • Well we look across from this chart-and say that the time on Earth
will be 1.25

98. • At the time on the space craft will be 4/5th of that-so we have to
multiply this by 4/5th’s-to get the time on the clock from the space
craft-which will be one year

99. • So what we are saying is that when this value is 2 year-they will see a
picture of there twin-showing only 1 year

100. • Now what happens- when the twin Is 4 years old-well –the twin will
have received light which has emulated here-what is this point in
Earth Time-In Earth Time- it is 2.5 years

101. • But we have to multiply this by 4/5th

102. • And that gives 2 years

103. • So now what we are saying that-the twin on the Earth is 4 years old-
He or She will see a picture of there other twin at 2 years old

104. • Now lets move forward again-when this twin on Earth is 6 years old
• On Earth-he will be 3.75 years

105. • But-we have to multiply that by 4/5ths

106. • And you get 3 years

107. • What happens when the twin on Earth is 8 years old-they are
receiving a signal which was sent-which was 5 years Earth time

108. • Then Multiply this by 4/5ths and get 4

110. • What is the distance in the demominator
• Well it is the velocity of the rocket

111. • averAnd since the speed of time is invariant