The Twin Paradox
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Tyler Stelzer
Bob Coulson
Berit Rollay
A.J. Schmucker
Scott McKinney
"When you sit with a nice girl for two hours, it
seems like two minutes. When you sit on a hot
stove for two minutes, it seems like two hours,
that's relativity.“
-Albert Einstein
The Twin Paradox
"If I had my life to live over again, I'd be a plumber.“
-Albert Einstein
Overview
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Events and Cooridinatizations; The concept of Spacetime
Lorentz Coordinatizations; Lorentz Postulates
Minkowski Space
LorentzTransformations
Moving Reference Frames
– Time Dilation
– Length Contraction
• Lorentz-Einstien Transformations
– Boosts
• The Twins Paradox
Events and Coordinizations
The Concept of Spacetime
Events: What are they?
• An “event” is a definite happening or
occurrence at a definite place and time.
• Examples:
– A bomb explodes
– Emission of a photon, (particle of light), by an
atom; similar to switching a light on and off.
What is Spacetime?
• Let E be the set of all events as previously
defined. This set is call Spacetime.
• Let e represent a particular event, then
“e  E” means e is an event.
Modeling Spacetime
• To model spacetime we use R4. The idea
behind this is that each event e  E is
assigned a coordinate
(Te , Xe , Ye , Ze )  R4
• R4 is an ordered four-tuple
– Example: (x,y,z,w) has 4 coordinates
What Are Te , Xe , Ye , Ze ?
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•
•
•
Te Xe Ye Ze -
The time coordinate of the event.
The x position coordinate of the event.
The y position coordinate of the event.
The z position coordinate of the event.
What does this mean?
• e
(Te , Xe ,Ye , Ze ) is assumed to be a
bijection. This means that:
– It is a one to one and onto mapping
– Two different events need to occur at either different
places or different times.
• Such assignment is called a “Coordinatization of
Spacetime”. This can also be called a
“Coordinatization of E”.
Lorentz Coordinizations &
Postulates
Vector position functions and
Worldlines
• In Newtonian physics/calculus, moving
particles are described by functions
t
r(t)
• r(t) = ( x(t) , y(t) , z(t) )
This curve gives the
“history” of the
particle.
Vector position functions and
Worldlines cont…
• View this from R4 perspective
t
( t, r(t) )
• In the above ‘t’ represents time and ‘r(t)’
represents the position.
• This can be thought of as a “curve in R4”,
called the Worldline of the particle.
What is time?
•
There are 2 types of time
1. Physical Clock Time
2. Coordinate Time: the time furnished by the
coordinatization model:
e
( Te , Xe , Ye , Ze )
1st Lorentz Postulate
– For stationary events, Physical Clock Time and
Coordinate Time should agree
– That is, we assume that stationary standard
clocks measure coordinate time.
2nd Lorentz Postulate
• The velocity of light called c = 1.
• Light always moves in straight lines with
unit velocity in a vacuum.
• T|
(T, vT + r0), time and spatial position
• Note: Think of the light pulse as a moving
particle.
Minkowski Space
Geometry of Spacetime
Minkowski Space
(Geometry of Spacetime)
• The symmetric, non-degenerate bilinear form of
the inner product has the properties
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<x,y>=<y,x>
<x1 + x2, y> = <x1, y> + <x2, y>
<cx,y> = c<x,y>
The inner product does not have to be positive
definite, which means the product of it with itself
could be negative.
• Non-degenerate meaning only the zero vector is
orthogonal to all other vectors
• Spacetime has it’s own geometry described by the
Minkowski Inner Product.
Minkowski Inner Product
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Defined on R4:
u = (u0,u1,u2,u3)
v = (v0,v1,v2,v3)
<u,v>:=u0v0- u1v1- u2v2- u3v3
<•,•> also called the Lorentz Metric, the
Minkowski metric, and the Metric Tensor
• M = R4 with Minkowski Inner Product
• “•” represents the usual inner product (dot
product) in R3
• In this case you have an inner product that allows
negative length.
How is the Minkowski Inner
Product Related to the Euclidean
Inner Product?
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The Euclidean Inner Product:
r = (r1, r2, r3)
s = (s1, s2, s3)
r•s =(r1 s1 + r2 s2 + r3 s3)
Note: R4 = R1 x R3
The Minkowski Inner Product
u = (u0,(u1,u2,u3))
v = (v0,(v1,v2,v3))
<u,v>:=u0v0- (u1,u2,u3)•(v1,v2,v3)
Strange Things Can Happen In
Minkowski Space
• Such as:
• Vectors can have “negative lengths”
• Non-Zero vectors can have zero length.
• A vector v ε M is called:
– “Time Like” if <v,v> > 0
– “Null” if <v,v>
=0
(Some of these are Non-Zero Vectors with zero length.)
– “Space Like” if <v,v> < 0
(These are the negative length vectors)
Minkowski Space serves as a mathematical model of spacetime
once a Lorentz coordinization is specified. Consider an
idealized infinite pulse of light.
Consider the Problem of
Describing Light
• We think of a moving light pulse as a moving particle
emitted via a flash in spacetime. The path of this particle
is referred to as it’s worldline.
• By the second Lorentz Postulate, the worldline is given by:
T |(T, vT + r0)
• Recall: v • v = 1 ( v ε R3) r0 ε R3
• (T, vT + r0) = (T, vT) + (0 , r0) = T(1, v) + (0 , r0)
• Note: T(1, v) is a null vector because
< (1, v) , (1, v)> = 1- v • v = 1-1=0
(This is an example of a Non-Zero Vector with zero length.)
• a:=(1,v)
• b:=(0, r0)
• So, the worldline of a light pulse will be of the form
T|aT+b ε M with <a,a> = 0 (These are called null lines.)
Light Cones
• Suppose b ε M
• The light cone at b:={p ε M |<P-b,P-b> = 0}
• This is the union of all null lines passing through
b.
• The forward light cone at b = {P=(P0,P1,P2,P3) ε M
| P ε light cone at b, P0-b0>0}
• The backward light cone at b = {P=(P0,P1,P2,P3) ε
M | P ε light cone at b, P0-b0 <0}
Moving Reference Frames
Moving Reference Frames
• Recall the idea of a “coordinatization”
e  E, e  (te, xe, ye, ze)
Z
“e”
C


A
X
Y
O
B
The Idea:
i)
By trig, determine “spatial coordinates”
(xe, ye, ze)
ii) Assuming:
c = speed of light,
Rate X Time = Distance,
x e  ye  ze
Te  T c
2
2
Z
“e”
C


2
Time at which the light
pulse reaches O
A
X
O
B
Y
Interesting Math Problem
• Suppose there is a 2nd coordinate system,
moving at a constant velocity v, in the
direction of the x-axis. Suppose O, O’ both
employ the same procedure for
coordinatizing E:
(T, X, Y, Z) (Stationary Frame)
(T’, X’, Y’, Z’) (Moving Frame)
How are these two frames
related?
• First, let’s look at a picture:
Z
Z’
“e”
Y
O
O’
Y’
X
X’
Assume constant
velocity c = 1
Solution:
• Assume O, O’ have standard clocks.
e.g. Einstein – Langevin clock
(light pendelum)
• A rigid rod (or tube) of length L
Light
source
Mirror
Z
Z’
“e”
1 unit of time =
duration of time between
emission and return
Y
O
O’
X
X’
Y’
Time Dilation
Z
• How does O regard O’’s clock?
Think of O’’s clock as sitting in a
moving vehicle
(e.g. a train or spaceship)
Light
Source
L
O’’s
perspective
Mirror
Let t’ = time of ½ pendulum
Z’
“e”
Y
O
O’
X
X’
Spaceship
moving at a
velocity v
L = ct’
Distance = (Rate)(Time)
Y’
Now Consider O’s Perspective
Mirror has moved
since the ship has
moved
Light
Source
Z
Z
’
“e”
ct
L
O’
vt
Y
O
Mirror
X
X’
Let t be the time of the
½ pendulum of O’’s
clock as observed by O.
Observe:
(ct)2 = L2 + (vt)2
Y’
• Using some simple algebra, we can solve
the previous equation for t.
c 2t 2  L2  v 2t 2
c 2t 2  v 2t 2  L2
t 2 (c 2  v 2 )  L2
2
L
t  2
c  v2
L
t
2
2
c v
ct '
t
c2  v2
2
If we let c = 1
t
t'
1 v2
t   (v)t '
( Note :  (v) 
Note: From O’’s
perspective L = ct’
1
1 v
This is the relativistic
time dilation factor
2
)
Light Scales
• Consider a “rod” of length R. Since we
assume (Rate)(Time) = Distance, the
length R may be measured by light rays as
follows:
Light
source
c = speed of light (1)
(c)(time) = 2R, c = 1,
So R = time / 2 or R = t / 2
Mirror
Z
Z
’
“e”
Y
O
O’
X
X’
Y’
Now suppose the “rod” is situated in the
direction of the moving frame.
R’
moving at velocity
v relative to O
Let R’ = length as measured by O’’s
light scale. (R’ = t’ / 2).
From O’s perspective, the rod
is in motion, as O’’s light
scale functions.
Z
Z
’
“e”
Y
O
O’
X
X’
Y’
moving at velocity
v relative to O
R’
Let:
t1 = time on O’s clock until the flash
reaches the mirror.
t2 = the time the flash takes between the
mirror and returning to the light source.
So:
The total time (on O’s) clock is
t = t1 + t2
“e”
O’
X
X’
Z
Z
’
Y
O
Y’
Now consider t1 and t2:
Vt1 = distance the spaceship (and hence rod)
travels between initial flash and mirror
interception.
ct1
Light
source
R
ct1  R  vt1
t1  R  vt1
vt1
Z
Z
’
“e”
t1  vt1  R
t1 (1  v )  R
R
t1 
1 v
Y
O
O’
X
X’
Y’
R
Mirror moves vt2
ct2
vt2
R
R  t 2  vt2
Using some simple
algebra, we solve for t2
Now we can revisit the total
time equation t = t1 + t2
t  t1  t 2
R
R
t

1 v 1 v
1 
 1
t  R


1 v 1 v 
R  ct2  vt2
R
t2 
1 v
Z
 2 
t  R
2 
1 v 
2R
t
1 v2
Z
’
“e”
Y
O
O’
X
X’
Y’
Lorentz Fitzgerald Contraction
So:
t'
2R
R'  and t 
2
1 v2

1
Recall : t   (v)t '   (v) 
2
1 v

2R
1

(2 R' )
2
1 v
1 v2
R
R'

1 v2
1 v2
1 v2
( R)  R'
2
1 v
R
 R'
1 v
R '   (v ) R
2




Lorentz Fitzgerald Contraction
(cont.)
1
R
R'
 (v )
R  1  v R'
2
Lorentz Fitzgerald
Contraction
R '   (v ) R
t   (v)t '
Lorentz Transformations
Hendrick A. Lorentz
6/18/1853 – 2/4/1928
Lorentz Coordinatiztions
• Suppose E is modeled by M = R4 by a
Lorentz coordinatization:
e|
(Te, Xe, Ye, Ze)
• Suppose (t,x,y,z) are the original Lorentz
coordinates and (t’,x’,y’,z’) are the new
Lorentz coordinates. We can regard this as
a bijection.
Which Bijections Preserve
Lorentz coordinatizations?
• Worldlines of light pulses are exactly the
null lines in M. (at + b, <a,a> = 0)
• Therefore, F must map null lines to null
lines
• (t’(at+b), x’(at+b), y’(at+b), z’(at+b))
should be a null line.
Examples of Null Line Preserving
Transformations in M
• Translations: L(v) = V + C, (V,C in M)
• Scalar Multiplications: L(v) = SV
• Metric Preserving Linear Maps:
<LU,LV> = <U,V>
(Minkowski inner product)
These are called “Lorentz Transformations”
Proofs of Null Line Preserving
Maps
• Translations :
L(at + b) = (at + b) + c = at + (b + c)
• Scalar Multiplications:
L(at + b) = S(at + b) = Sat + Sb
<Sa,Sb> = S2<a,a> = 0
• Linear Maps (Lorentz Transformations):
L(at + b) = L(at) + L(b) = tL(a) + L(b)
<L(a),L(a)> = <a,a> = 0
Lorentz – Einstein
Transformations
(Boosts)
Moving Reference Frames
• At time t = 0, O = O’
• (t’,x’,y’,z’) moves at velocity v in the x direction
relative to the stationary frame (t,x,y,z).
Z
Z’
Y
O
O’
X
X’
Y’
Suppose an event “e” occurs along the x-axis and is given coordinates
(t,x,y,z) by O and (t’,x’,y’,z’) by O’. ( y’ = y, z’ = z)
O
VT
O’
x - VT
E
Since Distance = Rate * Time, At time “t”, the x coordinate of O’
will be “vt” (from O’s perspective). Therefore the distance between
the light source and O’ will be “(x-vt)”.
Apply Length Contraction
• O understands that O’
will measure this as
the longer distance
where
R
R '   (v ) R
Therefore,
x'   (v)( x  vt)
Measurements of Time Lengths
(in relation to “O”)
O
VT
O’
VT*
x - VT*
E
x
• Let t`o = the time that e reaches O`
• Let t* = this time as measured by O’s clock
Note that t* > t, and (t* - t) = transit time between e and O` as
measured by O.
(where t is time of emission as measured by O)
Compute the Time difference
• Again, since Distance = Rate * Time,
C (t*-t) = x – vt*, where C = 1
t*-t = x – vt*
• Now solve for t* :
t* + vt* = x + t
t*(1 + v) = x + t
t* = (x + t) / (1 + v)
Apply Time Dilation
When o adjusts for time dilation where:
t   (v)t '
The equation becomes:
( x  t ) /(1  v )   (v )t ' o
And Since :  (v )  1 /
t   (v)t '
1 v
Then : ( x  t ) /(1  v )  t 'o /
So : t 'o 
2
1 v2
1  v 2 ( x  t ) /(1  v )
Solving for the Time Transformation…
Then Recall : x'   (v)( x  vt)
And : t 'o  x't
Substitute : 1  v 2 ( x  t ) /(1  v)   (v)( x  vt)  t '
Solve for t' :
 [ 1  v 2 ( x  t ) /(1  v)]  ( x  vt) / 1  v 2
 [ 1  v ( x  t ) / 1  v ]  ( x  vt) / 1  v 1  v
 [(1  v)( x  t )  ( x  vt)] / 1  v 2
 (t  vx) / 1  v 2
  (v)(t  vx)
Boost in the x – Direction
• Putting these elements together gives:
t '   (v )(t  xv)
x '   (v )( x  vt)
y'  y
z'  z
Which is called a boost in the x – direction.
Is This Boost a Lorentz
Transformation?
• To show that the boost is a Lorentz transformation
we must show that the Minkowski inner product of
the transformation is equal to the Minkowski inner
product of the original vectors.
 a, b  t a tb  xa xb  ya yb  z a zb
 La, Lb   (v )(t a  vxa ) (v )(tb  vxb )
  (v )(t a  vxa ) (v )(tb  vxb )
 y a yb  z a z b
  (v ) 2 (t a tb  xa xb v 2 )   (v ) 2 ( xa xb  t a tb v 2 )
  (v ) 2 (t a tb  xa xb v 2  xa xb  t a tb v 2 )
Continued…
Recall :  (v)  1 / 1  v 2 , So :  (v) 2  1 /(1  v 2 )
Substituti ng for  (v), the equation becomes :
[1 /(1  v 2 )](t a tb  t a tb v 2  xa xb  xa xb v 2 )
 1 /(1  v )[(t a tb (1  v )  xa xb (1  v )]
2
2
2
 t a tb  xa xb
This supports the assumption that Lorentz
transformations of Lorentz coordinatizations give
Lorentz coordinatizations.
The Twin Paradox
Scott McKinney
The Twin Paradox
(An application of the time dilation)
• The idea with these examples is that due to
Special Relativity and the idea of time
dilation, if two twins are born and one stays
on Earth while the other boards a space ship
and rockets away at a fraction of the speed
of light then the twin on board the space
ship will age more slowly than the one on
the planet based on the speed of the ship.
Scott McKinney
The Transformation and Variables
Recall :
t
t'
1 v
t   (v)t '
2
( Note :  (v) 
1
1 v
2
)
Scott McKinney
Example 1
• If the space ship’s velocity is equal to .5c, and ten
years pass on Earth, how many years would pass
on the ship?
• If 10 years passes on Earth then only 8.66 years
would have passed on the space ship.
Scott McKinney
Example 2
• Let’s say the space ship in Example 1 passes us as
its clocks read 12 noon. In our reference, how far
away will it be when its clocks read 1 pm?
• If one hour passes on the ship then 1.155 hours
passes on Earth. Since the ship is moving at .5c or
3.35*10^8 mph. The ship would be 3.867*10^8
miles away when its clocks read
1 pm. 3.35*10^8 mph*1.155= 3.867*10^8 miles
Scott McKinney
Example 3
• How fast must a space ship travel in order that its
occupants will only age 10 years while 100 years
passes on Earth?
• So if 10 years passed on the ship going .995c then
100 years would pass on Earth.
References
• Relativistic Electrodynamics and
Differential Geometry, Parrot, SpringerVerloy 1987.