1. 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/

2. 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:
x = xproper/
or L = Lproper/
Length Contraction (or Lorentz Contraction): The length of an object measured in a reference frame that is moving with respect to the object is less than the length measured in a frame which is at rest with respect to the object (the proper length, LP.)
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!

3. Length Contraction: L’ = LP / 
x x2
Measuring the length of a moving object (in the direction of its motion) is tricky, because you have to measure the front and back of it simultaneously. [This was not an issue for perpendicular directions.]
How does length contraction follow from Lorentz Transformation?
x’ =  (x – vt), t’ =  (t – vx/c2),
  1/(1 – v2/c2)1/2
Suppose the unprimed frame is the proper frame, in which the object is at rest. Then you can measure its ends at any times (not necessarily the same times): Lp = x2-x1. To measure L’ = x2’ – x1’ in the frame it which its moving, however, you need to do the measurements such that t1’ = t2’:
(t2 –vx2/c2) = (t1 –vx1/c2)
t2 – t1 = v/c2 (x2-x1)
x2’ – x1’ =  [x2-x1 – v(t2-t1)]
x2’ – x1’ =  [(x2-x1) (1 – v2/c2)]
But (1 – v2/c2) = 1/2
x2’ – x1’ = (x2-x1) / 
Length Contraction: L’ = LP / 

4.  = 17 s =  proper =  (2.2 s)   = 17/2.2 = 7.7
Last Class we did this time dilation problem: 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  4.8 km; i.e.  > 16 s.
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
 v/c = 0.992
That was from the point of view of the observer on the ground. How does the muon ( or someone riding on its back) view the situation. In the muon’s rest frame, its lifetime is 2.2 s – how can it reach the ground?
The muon observes the ground rushing toward it, and the top of the atmosphere rushing away from it, at v = 0.992c, so from his point of view, the separation of the ground and top of the atmosphere is L’/ = 4.8 km/7.7 = 623 m. Therefore, the time for the ground to reach the muon after it is created at the top of the atmosphere is
623m/0.992c  2.2 s – its lifetime
So from the earth’s point of view, this is a time dilation problem. From the muon’s point of view, it is a length contraction problem!

5. Problem: The rest length of an alien space cruiser is 1500 m and it has a clock which
ticks with a period of 0.27 s. The cruiser passes planet Tralfamadore at a speed of x 108 m/s.
What are the length of the cruiser and period of the clock as measured by a resident of Tralfamadore?

6. Problem: The rest length of an alien space cruiser is 1500 m and it has a clock which ticks with a period of 0.27 s. The cruiser passes the planet Tralfamadore at a speed of 2.5 x 108 m/s.
What are the length of the cruiser and period of the clock as measured by a resident of Tralfamadore?
 = 1/[1 – (v/c)2]1/2 = 1 / [1 – (2.5/3)2]1/2 = 1/.3051/2 = 1.81
[Note that for this problem, the space cruiser observer measures both the proper length of his ship and the proper time of the clock!]
The proper length = 1500 m, so the length measured on Tralfamadore is
L(traf) = Lp/ = 1500m/ 1.81 = 829 m (length contraction)
The proper time period = 0.27s, so the time period measured in Tralfamadore is
T(traf) = Tp = (1.81)(0.27s) = 0.49 s (time dilation)
[As a result, an alien on the cruiser will measure less time for a trip than a Tralfamadorean at home on his planet: i.e. if the clock ticks N times during the trip, the cruiser observer measures the trip to last N*0.27s while the Tralfamadorean measures it to last N*0.49s.]

7. The most famous “paradox” about length contraction is the “pole-in-the-barn” paradox:
Suppose Jack is running along at high speed carrying a long pole toward the open front door of a barn, which also has an open back door. Jill is sitting in the barn. Suppose that the proper length of the pole is larger than the proper length of the barn, but Jack’s speed is so great that Jill measures the Lorentz contracted length of the pole to be shorter than the length of the barn so that the pole can fit in the barn. When the pole is inside, she quickly (and simultaneously) closes both doors (but then quickly opens them again so that the moving pole doesn’t smash into the back door).
front door
back door
But from Jack’s point of view, his pole did not contract but the barn did, so the pole could not fit in the barn! How could Jill have closed the doors around the pole?
Answer: In Jack’s frame the doors were not closed and opened simultaneously!

8. It is convenient to plot what is happening in “space-time” graphs:
conventional to put x on horizontal axis and time on vertical axis; time is multiplied by c so both axes have same units (e.g. meters).
Object traveling with |v|<c from + x toward –x
ct (m)
Light ray traveling to +x direction from the origin
Light ray traveling to –x from the origin
Object traveling with v< c in + x direction from origin
45o
45o
x (m)
[Any object traveling with |v|<c will have |slope| > 1.]

9. Let’s use the numbers in the text’s example: v = 0.75 c, so  = 1.51
proper lengths: Lp(barn) = 10m, Lp(pole) = 15m
Then from Jill’s point of view (at rest in the barn), the pole has length L’(pole) = 15m/ = 9.9 m, and should just fit in the barn.
Jill’s space-time graph of what’s happening:
Front and back of pole each traveling at v; they are 9.9 meters apart in Jill’s view.
Close and open doors now
Doors 10 m apart at rest in Jill’s frame, so vertical.
Leading edge of pole at front door at t = 0

10. v = 0.75 c, so  = 1.51
proper lengths: Lp(barn) = 10m, Lp(pole) = 15m
Then from Jack’s point of view (i.e. the pole’s frame), the barn has length L’(barn) = 10m/ = 6.6 m, so the pole cannot fit in the barn.
Jack’s space-time graph of what’s happening:
Front and back of pole 10 meters apart in Jack’s view and not moving, so vertical lines.
Doors 6.6 m apart at rest in Jack’s frame and moving toward -x at speed v.
Leading edge of pole at front door at t = 0

11. Jill’s view: Doors close and open at same time
Jack’s view: Rear door closes and opens before front door closes and opens

12. Problem: A spacecraft (S) leaves earth (E) with a constant velocity = 0.6c, such that the origins of E and S coincide at tE = tS = 0. At t1S = 333 s, the spacecraft fires a rocket (R) at a target in front of it, and the rocket hits the target at t2S = 383 s and x2S = 1 x 1010 m.
At what time does the rocket hit the target as measured on earth?
Note: Neither S nor E measures the proper time or proper length for the rocket’s trip, so to transform from S to E must use Lorentz transformation.
Note that E is traveling at velocity v = -0.6c with respect to S:
  = 1 / (1 – v2/c2)1/2
 = 1/(1-0.36)1/2 = 1/0.641/2 = 1/0.8
 = 1.25
t2E =  (t2S - vx2S/c2)
t2E = 1.25 [383 - (-0.6 x 1 x 1010 )/3x108] s = 1.25 x 403 s
t2E = 504 s E S 20

13. x’ =  (x – vt)
y’ = y
z’ = z
t’ =  (t – vx/c2)
Problem: Observer S’ is traveling at speed 0.8c in the +x direction with respect to S. S measures two supernova to occur at x1 = 30 ly, t1 = 20 y and x2 = 40 ly, t2 = 45 y. Where and when does S’ observe them?
= 1 / [( 1 – (v/c)2]1/2 = 1 / [ ]1/2 = 1/0.361/2 = 1/0.6 = 1.67
x1’ = 1.67 [(30 – (0.8) (20)] ly = ly [1 year = 1 light year / c]
t1’ = 1.67 [20 – (0.8)(30)] y = y
x2’ = 1.67 [40 – (0.8)(45)] ly = 6.67 ly
t2’ = 1.67 [ 45 – (0.8)(40)] y = y
[The minus times for t1’ means that in the S’ frame, the supernova occurred before S and S’ passed each other at x = x’= 0 and t = t’=0 (so S’ could have at that instant warned S to watch for it).]

14. An interesting way to think about the Lorentz Transformation
Consider x’2 + y’2 + z’2 – c2t’2 :
x’2 + y’2 + z’2 – c2t’2 = y2 + z2 + 2 [(x2-2xvt+v2t2) - (c2t2-2vxt + v2x2/c2)]
= y2 + z2 + 2[x2 (1-v2/c2) – c2t2 (1 – v2/c2)]
but (1 –v2/c2) = 1/2
x’2 + y’2 + z’2 – c2t’2 = x2 + y2 + z2 – c2t in all inertial frames
i.e. frames moving at constant velocity

15. x2 + y2 + z2 – c2t2 has same value in all inertial frames:
it is “invariant” under a Lorentz Transformation.
Analogous to a rotation in x-y-z space: the length of any vector
doesn’t change if you rotate the axes: x’2 + y’2 + z’2 = x2 + y2 + z2
e.g. in 2D: x’ = x cos + y sin
y’ = -x sin + y cos
r2 = x2 + y2 = x’2 + y’2
The analog to the length of a vector in x-y-z space is an interval in space-time.
An event is a point in space-time, (x,y,z,t). The “length of the interval” between this event and the origin (0,0,0,0) = [x2 + y2 + z2 – c2t2 ]1/2. This interval is invariant – i.e. the same in all frames moving at constant velocity.
So a Lorentz transformation is a “rotation in space-time”. [Unlike ordinary rotation, (ct)2 has a minus sign.]

16. Detection of Gravity Waves at LIGO
(Laser Interferometer Gravity Wave Observatory)
General Relativity: A gravitational field is manifest as a “warping” of space-time.
Because gravity is so weak, these distortions are only noticeable around very massive objects.

17. If a mass moves, the gravitational field doesn’t change instantaneously but propagates away from the mass at the speed of light. Therefore an oscillating (or revolving) mass will be a source of gravitational waves: oscillating changes in distances, i.e. the lengths of objects.
These propagate as “quadrupole” waves:
Because gravity is so weak, these distortions of space (L/L) are tiny. The best chance to measure a small L is to make L large.

18. 2 Michelson Interferometers
x/ = /2
Mirrors tilted so light bounce ~ 250 times in each arm before recombining: x ~ 2 x 250 x L

20. The bandwidth (frequency range) of the interferometer is  10 Hz – 7 kHz (i.e. cannot observe orbiting planets in the solar system).
10 Hz:  = 30,000 km
7 kHz:  = 43 km
To be sure not measuring noise, a signal has to be observed in both interferometers (with appropriate time delay).
The first event was reported in February 2016 with L/L  10-21
(i.e. L  x 4 km  proton radius/200

21. Prediction: Calculated merger of two black holes (masses = 36 and 29 solar masses), spiraling into each other and merging.
Signal lasts ~ 0.1 sec, frequency chirps from ~ 30 Hz to 300 Hz
3 more black hole mergers have been observed since then (the last with a 3rd interferometer in Italy).

22. August 2017: Observed event modeled as > 10 second spiraling merger of two neutron-stars (1-2 solar masses). Resulting “kilonova” explosion (gamma ray burst) observed with gamma ray telescopes 1.7 seconds later, and “afterglow” observed with electromagnetic radiation (visible light, infrared, radio waves) for days to months.