1. Lorentz Transformation
x’ =  (x – vt)
y’ = y
z’ = z
t’ =  (t – vx/c2)
where   1/(1 - v2/c2)1/2
Problem: A rocket is traveling in the positive x-direction away from earth at speed 0.3c; it leaves earth (x = x’= 0) at t = t’= 0. An observer on earth measures the rocket to be at x1 at time t1=x1/(0.3c). What are the corresponding positions and times in the rocket’s frame?
 = 1/( )1/2 = 1/(1-0.09)1/2 = 1/0.911/2 = 1.048
x1’ = [x1- (0.3c)(x1/0.3c)] = (obviously!)
t1’ = [t1 – (0.3c)x1/c2] = t1 [ ] = 0.95t1
t1’ < t1 : Less time passed in the “moving frame” (i.e. the frame which moved with
the event so “x’ did not change in this frame”) than in the “stationary frame” (i.e.
the frame in which x is observed to change).

2. Jill is primed and Jack unprimedv
event that is stationary in Jill’s frame
Time Dilation:
Suppose Jack (S) and Jill (S’) are in different frames, moving at speed v with respect to each other. Jill uses her clock to measure the time interval between two events: tJill = tJill(2) - tJill(1) that she observes to occur at the same place: xJill = xJill(2) - xJill(1) = 0. Note that she has no difficulty in measuring the times of these events, since she can put a clock right next to the event, which is stationary in her frame.
Jack sees the event (and Jill’s clock) as moving, so sees the two events as taking place at different values of x: xJack(2)  xJack(1).
tJack = tJack(2) – tJack(1) = [tJill(2) – tJill(1) + v(xJill(2)-xJill(1))/c2] = tJill
Jill is primed and Jack unprimed

3. Jill’s time interval is called the proper time interval:
tJack = tJill
Jill’s time interval is called the proper time interval:
The proper time interval is the one measured by an observer who sees the events as occurring at the same location.
Then for any other frame (e.g. Jack’s) moving at speed v with respect to the proper time frame, t = tproper.
That is: the time interval between two events will be less in the reference frame moving with the events (the proper time frame) than in any other frame, which observes the events to be at different locations.

4. What is going on? Einstein demonstrated this with the following “Gedanken” (thought) experiment:
Jill (O’) is riding in a train moving at speed v with respect to Jack (O). She measures the time it takes for a pulse of light to go from her hand to a mirror on the ceiling and back to her hand. Since she hasn’t moved her hand, she measures the proper time: tproper = 2d/c.
Jack sees the light traveling along the diagonals, so a longer distance: [d2 + (vtJack/2)2]1/2. Since both observers measure the same speed of light, Jack measures a time interval tJack = 2 [d2 + (vtJack/2)2]1/2 /c

5. tproper = 2d/c
tJack = 2 [d2 + (vtJack/2)2]1/2 / c
(tJack)2 = 4 [d2 + (vtJack/2)2]/c2
(1-v2/c2) (tJack)2 = 4d2/c2 = (tproper)2
tJack = tproper / (1-v2/c2)1/2
tJack =  tproper
Note that we have shown this for Jack’s “light clocks”, but by the relativity postulate, it must be true for any “clock” (e.g. pendulum, quartz crystal, heart beat, earth rotation, radioactive decay, ….) in Jack’s reference frame. If not, Jack could compare this other clock to the light clock to determine that he was moving, in contradiction to the relativity postulate.
light clock
wall clock
heart rate
metabolism
radioactive decay
Any Other Clock

6. tother = tproper
 = 1/(1-v2/c2)1/2  1  A clock that is moving so that it can measure the interval between two events at the same location (the proper time clock) always runs more slowly (measures less time) than other clocks, for which the events occur at different locations: Time Dilation
[Colloquially: a moving clock runs more slowly than a stationary clock]

7. Time Dilation: t =  tproper
But   1 / (1-v2/c2)1/2
is very close to 1 unless
v  c, so not observed in “everyday” events.
v/c 
0.0001
1 + 5 x 10-9
0.001
1 + 5 x 10-7
0.01
1 + 5 x 10-5
0.1
1.005
0.2
1.021
0.3
1.048
0.5
1.155
0.75
1.512
0.9
2.29
0.99
7.09
0.999
22.4
0.9999
70.7

8. v/c 
0.0001
1 + 5 x 10-9
0.001
1 + 5 x 10-7
0.01
1 + 5 x 10-5
0.1
1.005
0.2
1.021
0.3
1.048
0.5
1.155
0.75
1.512
0.9
2.29
0.99
7.09
0.999
22.4
0.9999
70.7

9. Time Dilation: t =  tproper has been confirmed many times with extreme precision for a large range of speeds. (For example, the precision of GPS devices utilize time dilation.) One of the first examples was the observation that the lifetime of moving muons was much longer than the lifetime of stationary muons:
In the lab, stationary muons have a “proper” lifetime proper = 2.2 s. Yet muons created when cosmic ray particles hit the top of our atmosphere can make it all the way to ground: since v < c,  > L/c, where L  the thickness of the atmosphere: L  4.8 km; i.e.  > 16 s.
Problem: How fast must a muon be traveling (with respect to an observer on earth) to have a lifetime of 17 s?
 = 17 s =  proper =  (2.2 s)   = 17/2.2 = 7.7
1 / (1-v2/c2)1/2 = 7.7
(1-v2/c2)1/2 = 1/7.7 = 0.129
1-v2/c2 = =
v2/c2 = 0.983
v/c = 0.992

10. Problem: Jack and Jill are twins
Problem: Jack and Jill are twins. While Jack stays at home on earth, Jill takes a trip on a spacecraft that travels at 0.95c [no such spacecraft exists yet!] to a planet 40 light years away, and then immediately turns around and comes back to earth at 0.95c. How much do Jack and Jill each age during this trip?

11. Problem: Jack and Jill are twins
Problem: Jack and Jill are twins. While Jack stays at home on earth, Jill takes a trip on a spacecraft that travels at 0.95c [no such spacecraft exists yet!] to a planet 40 light years away, and then immediately turns around and comes back to earth at 0.95c. How much do Jack and Jill each age during this trip?
Let d = 40 light years, v = 0.95c. The distance Jill travels = 2d = 80 light years.
Therefore, as measured on earth, tearth = tJack = 2d/v = 80 light year/(0.95c)
tJack = 84.2 years
Since, in Jill’s frame, earth and the planet are moving to/from her, Jill measures the proper time: tJill = tJack / , where  = 1/[1-(v/c)2]1/2 = 1/( )1/2 = 1/.09751/2 = 3.20
tJill = 84.2 years/3.20 = 26.3 years.
Therefore, Jill will be younger than Jack when she returns!
They can compare their bodies and see the difference.
Jill
Jack

12. tJack = 84. 2 years and tJill = 26. 3 years
tJack = 84.2 years and tJill = 26.3 years. Therefore, Jill will be younger than Jack when she returns! They can compare their bodies and see the difference.
Wait a minute, cries Jill! How do you know I was moving. I think I was sitting at rest in my spacecraft while the earth (and Jack) hurtled away from me at 0.95c and then returned at 0.95c. Therefore, I should end up older than Jack!
But when they get together, they can compare bodies and see who is older: this is the so-called “twin-paradox”
In fact, Jill’s analysis is mistaken: she can determine that she was moving because she had to turn around, which means that she needed to accelerate at some point in her trip and therefore feel a force.
Jack, on the other hand, did not feel any forces during her trip.
A correct analysis of the time in Jill’s “accelerating frame” requires
general relativity, with the result that
tJack = 84.2 years and tJill = 26.3 years
(as Jack concluded in his stationary, and non-accelerating frame).

13. Problem: A GPS satellite orbits the earth with a velocity  3900 m/s
Problem: A GPS satellite orbits the earth with a velocity  3900 m/s. Suppose the satellite broadcasts a timing signal with a separation T. What is the fractional change in the period of the signal as measured on earth?

14. The satellite is in the proper frame. Tearth =  Tsatellite
Problem: A GPS satellite orbits the earth with a velocity  3900 m/s. Suppose the satellite broadcasts a radio signal with a separation T. What is the fractional change in the period of the signal as measured on earth?
The satellite is in the proper frame. Tearth =  Tsatellite
[Tearth – Tsatellite] / Tsatellite =  - 1
 = {1 – [(3900) /(3x108)]2 }-1/2 = { 1 – (1.3 x10-5)2}-1/2
  1 + ½ (1.3 x 10-5)2
  x 10-11
Therefore, [Tearth – Tsatellite] / Tsatellite  8.5 x 10-11
While this effect seems tiny, it is cumulative. In one day, if the clock were not adjusted, it would amount to a timing error T(1day) = (8.6 x 104 s)(8.5 x 10-11) = 7 s, and therefore a position error d = c T(1day) = 2 km!
[This is only the Special Relativity shift; there is also a larger shift (with opposite sign) in clock rate due to General Relativity: the net error would be 38 s/day.]

15. Problem: A rocket travels away from earth at constant speed v to planet Q.
The trip takes 100 years, as measured on earth but only 25 years as measured on the rocket. What is v?
Solution: The rocket measures the proper time: trocket = tproper = 25 y.
Therefore tearth = 100 y = trocket
 = tearth /trocket =100/25 = 4
 v/c = 0.968
Note that, as measured on earth, the distance from earth to Q must be
xearth = (0.968c) x 100 y
xearth = 96.8 lightyears
However, the distance from earth to Q as measured in the rocket must be
xrocket = (0.968c) x 25 y
xrocket = 24.2 lightyears
That is, xrocket = xearth/

16. xrocket = xearth/
Note that the earth observer is at rest with respect to Q (and the earth), so can take his/her time in measuring the distance between them; we say that the earth observer measures the “proper length”. So any other observer measures a smaller length:
Length contraction (or Lorentz contraction) : x = xproper/
Proper time: the time interval between two events measured by the observer who sees them happening at the same location.
Proper length: The length of an object (or distance between two points) measured by an observer who is at rest with respect to the object (or the two points).
In general, the proper length and proper time are not measured by the same observer!