Time Dilation
A clock moves from event A to event B at constant velocity.
How much time  passes on the clock?
s   x    y    z   c  t 
2
2
2
2
2
2
t   
•If we were moving with the clock, we would say
the clock doesn’t move, and the time was 
x  y  z   0
s  s  c    x    y    z   c  t 
2
2
2
2 2
2
•Note that the ordinary 3D distance is given by
2
2
2
2
2

x


y


z

d
     
c   d  c  t 
2 2
2
2
2
2
 2  t 2  d 2 c2  t 2 1  d 2 t 2 c 2   t 2 1  v 2 c 2   t 2  2
t  
Moving clocks run slowly: TIME DILATION
2
Sample problem
A neutron at rest lasts an average of 886 s. How fast must it
move to make it here from the Sun, a distance of d = 1.501011 m?
d  vt   v 
v
d 2 1  v 2 c 2   v 2 2
1  v2 c2
2
2
2
2

v


d
c
d  v  v d c


2
2 2
2
2
2
1.50 10 m 
d

v  2
2
2 
2
11
8
 d c
889
s

1.50

10
m
3.0

10
m/s 

 

11
2
2
2

1.5 10


m
11
v
2
1.04 10 s
6
2
2
,
v  1.47  10 m/s.
8

2
Lorentz Contraction
A ruler of length Lp moves past us at
velocity v. How long does it look to us?
•The length Lp is the length measured by
ct
an observer moving with the ruler
•The two ends of the ruler are at x’ = 0
and x’ = Lp.
xL  0    xL  vt 
•Solve for xL
xR  Lp    xR  vt 
and xR.
ct’
x’
xL  vt
x
xR  Lp   vt
•The difference is the length
L  xR  xL  Lp   vt  vt  L p 
Moving objects
look short
L  Lp 
Contraction only in
direction of motion
Galileo almost got it right
For ordinary objects, the velocity v is much smaller
than the speed of light c

1
1 v c
2
2
1
x  x  vt
t   t  vx c 2  t
Time dilation:
vc
1
y  y
z  z
2

t    t  vx c 
Galilean
boost
t    
Lorentz contraction:
x    x  vt 
L  Lp   Lp

Reverse Lorentz Boosts
Solve these for (x,y,z,t) in terms of (x’,y’,z’,t’)
x ' vt     x  v 2 x c 2 
substitute 1
add
x    x  vt  add
y  y
z  z
2

t    t  vx c  Multiply by v
vt     vt  v 2 x c 2 
Similar computation for t
Same formula except:
•Primed  Unprimed
•Velocity v  - v
Dotted red line – easily
derived from other formulas
1
1  v2 c2
 2  1  v2 c2
x ' vt    x  2
x    x ' vt 
y  y
z  z
t    t   vx c 2 
If I observe someone moving with velocity v,
she observes me moving with velocity –v
Subtraction of velocity
t     t  vx c 2    1  vu x c 2  t ,
Suppose an object is x  u x t
moving with velocity: y  u t
y
t  t   1  vu x c 2 
u   ux , u y , uz 
z  uzt
ux  v
t
What does a moving observer see? x    x  vt     ux  v  t 
2
1

vu
c
x
u yt
y  y  u y t 
2
y’
 1  vux c
ux  v
u x 
,
2
y
1  ux v c

u
t
z
z  z  uz t 
uy
 1  vu x c 2 
u y 
,
2
 1  ux v c


Dashed Red Line –
Know how to use it
x
x’


uz
uz 
.
2
 1  u x v c 
Paradox Problems
•Because relativity is so counter-intuitive, it often leads to apparent contradictions
•These apparent contradictions lead to apparent paradoxes
•These paradoxes can always be resolved by careful analysis
•Most common resolution – simultaneity is implicitly assumed
•Sometimes their resolution leads to important insights
The Twin Pair o’ Docs
A pair of identical twin PhD’s both start at Earth.
David stays on Earth, while Daniel travels to the star
Arcturus ( d = 37 light-yr) at 0.900c. When he gets
there, he returns to Earth at the same speed.
How much has each of them aged when he returns?
d 37c  yr
 41.1 yr
One way: t  
v 0.900c
  t   t 1  v 2 c 2   41.1 yr  1  0.92
 17.9 yr
Round Trip:
David = 82.2 years
Daniel = 35.8 years
Twin Pair o’ Docs = Paradox?
David
Daniel
ct
•Why is this an apparent paradox?
•According to David, Daniel is moving
•According to Daniel, David is moving
•How can we say who is really moving?
•The two cases are not interchangeable
•Daniel changed speeds, David did not
•If you want to compute using Daniel’s frame of
reference, you must also include a Lorentz
transformation at the turnaround point
Arcturus
x
Note: Twin that goes at
constant velocity is older
Pole Vaulter Paradox
A pole vaulter runs towards a barn with a
pole of length L. The barn is also length L.
Farmer says, “I’ll shut the
door when you are inside”
Because the man is running, his pole is shortened
It looks like the pole will fit in the barn
Vaulter says, “I’ll stop when
the pole hits the back”
From the viewpoint of the man, the barn is shortened,
and it will hit the back before he gets inside
Pole Vaulter Resolution
Consider the moment when the pole hits the back
of the barn from the vaulter’s perspective
x  0
t  0
x   L
t  0
x   L
t    Lv c
t 0
What does the farmer see?
•At t = -Lv/c2, the runner stops
•Far end of pole continues on until t = 0
•Pole stretches and hits back wall
t    t   vx c 2 
What are these coordinates
in the unprimed frame?
x0
2
x    x  vt  
Paradox resolved?
•What if vaulter uses a rigid pole?
•What does rigid mean?
•When you stop one end, other end stops instantaneously
•Far end can’t “know” that this end stopped instantaneously, because
influences can’t travel faster than light
•Real poles: information travels along pole at speed of sound
•Speed of sound must be slower than speed of light
In relativity, perfectly
rigid objects cannot exist
Binomial Expansion
It is often helpful to have approximate formulas
2
3
1
1
1



1

n


n
n

1


n
n

1
n

2

 


 
2 
6 
n
when is small

1

1
 1  v c
2
1 v c
2
2
 1 v c
2
2
 1  v c
2

v 2 3v 4
 1 2  4 
2c 8c

v2
v4
 1 2  4 
2c 8c
2 1/2
2 1/2
Often, your calculator will fail
where approximations will succeed
Sample Problem
Professor Carlson is sprinting at 10 m/s. He is 1.95 m tall, and his
nose is 3 cm long. How does his height and nose length change?
Contraction only in direction of motion
L p  0.03 m
No change in height
2


v
L  L p   L 1 
p
2 
2
c



v  v Lp 10 m/s   0.03 m 

L  L p  L  Lp  Lp 1  2  
2
2
8
2
c
2
c
2  3.00 10 m/s 


2
2
 1.67 1017 m
.0167 fm shorter
2
t2  2
1
t0  0

v cos 
T1     d  v cos  c
T0  d c
d  v cos 
d
•Imagine source moving at velocity v at angle 
•Sending out pulses at time interval  in its own frame
•What is time interval between signals received by us?
t  
T  T1  T0    d c  v cos c  d c   1  v cos c 
•Frequency of pulses is reciprocal of time gap between pulses
f0
1
1
f 


T  1  v cos  c   1  v cos  c 
To observer
Relativistic
Doppler t  
Shift
Relativistic Doppler Shift
2
2
f0
1 v c
f 
 f0
 1  v cos  c 
1  v cos  c
 
f  f0
 0
1 v c
1 v c
f  f0
  12 
f  f0 1  v2 c2
1 v c
1 v c
Summary: Formulas you need
s   x    y    z   c  t   c 2 2
2
2
2
2
x    x  vt 
 
y  y
z  z
t     t  vx c
2
x    x  vt  
y  y
z  z
t    t   vx c 2 

2
2
1
1 v
2
c
L  Lp 
t  
2
ux  v
u x 
,
2
1  ux v c
u y 
uy
 1  u x v c
2

,
uz
uz 
.
2
 1  u x v c 
2
1
1



1

n


n
n

1

 
 
2 
n
f0
f 
 1  v cos  c 