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

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

4. This term here –is the lorenzt transform

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

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

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

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

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

10. • This is distance d

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

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

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

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

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

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

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

18. This is why a moving clock moves more slowly

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

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

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

22. • So let’s plug in some numbers

23. • 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

25. • 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

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

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

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

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

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

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

32. • So now lets go back to our diagram

33. • 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

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

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

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

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

38. • 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

39. • 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

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

41. • 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?

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

43. • 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

44. • 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

45. • 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

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

47. • And that gives 2 years

48. • 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

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

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

51. • And you get 3 years

52. • 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

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

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

56. • averAnd since the speed of time is invariant