The Twin Paradox • Tyler Stelzer • Bob Coulson • Berit Rollay • A.J. Schmucker • Scott McKinney
Dec 27, 2015
"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
• 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: 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 ?
• Te - The time coordinate of the event.
• Xe - The x position coordinate of the event.
• Ye - The y position coordinate of the event.
• Ze - 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”.
Vector position functions andWorldlines
• 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 andWorldlines 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 time1. 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)
• The symmetric, non-degenerate bilinear form of the inner product has the properties
• <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• 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?• 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
• Recall the idea of a “coordinatization”
e E, e (te, xe, ye, ze)
“e”
O
C
B
A
Z
Y
X
The Idea:
i) By trig, determine “spatial coordinates” (xe, ye, ze)
ii) Assuming:
c = speed of light,
Rate X Time = Distance,
c
zyx - TT
2e
2e
2e
e
Time at which the light pulse reaches O
O
C
B
A
Z
Y
X
“e”
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:
O
Z
Y
X
O’
Z’
Y’
X’
“e”
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
duration of time between emission and return
1 unit of time =O
Z
Y
X
O’
Z’
Y’
X’
“e”
Time Dilation• 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
Mirror
L
Spaceship moving at a velocity vO’’s
perspective
Let t’ = time of ½ pendulum
L = ct’
Distance = (Rate)(Time)
O
Z
Y
X
O’
Z’
Y’
X’
“e”
Now Consider O’s Perspective
Light Source
Mirror
L
Mirror has moved since the ship has moved
vt
ct
Let t be the time of the ½ pendulum of O’’s clock as observed by O.
Observe:
(ct)2 = L2 + (vt)2
O
Z
Y
X
O’
Z’
Y’
X’
“e”
• Using some simple algebra, we can solve the previous equation for t.
22
22
22
22
2222
22222
22222
'
)(
vc
ctt
vc
Lt
vc
Lt
Lvct
Ltvtc
tvLtc
Note: From O’’s perspective L = ct’
If we let c = 1
)1
1)( : Note(
')(1
'
2
2
vv
tvtv
tt
This is the relativistic time dilation factor
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:
c = speed of light (1)
(c)(time) = 2R, c = 1,
So R = time / 2 or R = t / 2
Light source
Mirror
O
Z
Y
X
O’
Z’
Y’
X’
“e”
Now suppose the “rod” is situated in the direction of the moving frame.
Let R’ = length as measured by O’’slight scale. (R’ = t’ / 2). From O’s perspective, the rodis in motion, as O’’s lightscale functions.
R’
moving at velocity v relative to O
O
Z
Y
X
O’
Z’
Y’
X’
“e”
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
R’
moving at velocity v relative to O
O
Z
Y
X
O’
Z’
Y’
X’
“e”
Now consider t1 and t2:
Vt1 = distance the spaceship (and hence rod) travels between initial flash and mirror interception.
R vt1
ct1Light source
v
Rt
Rvt
Rvtt
vtRt
vtRct
1
)1(
1
1
11
11
11
O
Z
Y
X
O’
Z’
Y’
X’
“e”
R
Rct2 vt2
Mirror moves vt2
v
Rt
vttR
vtctR
12
22
22
Using some simple algebra, we solve for t2
Now we can revisit the total time equation t = t1 + t2
vvRt
v
R
v
Rt
ttt
1
1
1
111
21
2
2
1
2
1
2
v
Rt
vRt
O
Z
Y
X
O’
Z’
Y’
X’
“e”
Lorentz Fitzgerald Contraction
So: 21
2 and
2
''
v
Rt
tR
21
1)( ')( : Recall
vvtvt
')(1
1
1
'
1
)'2(1
1
1
2
2
2
22
22
RRv
v
v
R
v
R
Rvv
R
RvR
Rv
R
)('
'1 2
Lorentz Fitzgerald Contraction (cont.)
'1
')(
1
2 RvR
Rv
R
Lorentz Fitzgerald Contraction
')(
)('
tvt
RvR
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
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).
O
Z
Y
X
O’
Z’
Y’
X’
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)
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)”.
O
O’
E
VT
x - VT
Apply Length Contraction
• O understands that O’ will measure this as the longer distance
where
Therefore,
R
))((' vtxvx
RvR )('
Measurements of Time Lengths
• 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)
O
O’
x
x - VT*
E
VT
VT*
(in relation to “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:
The equation becomes:
')( tvt
')( tvt
)1/()(1' :So
1/')1/()( :Then
1/1)( :Since And
')()1/()(
2
2
2
vtxvt
vtvtx
vv
tvvtx
o
o
o
))((
1/)(
1/)]())(1[(
11/)(]1/)(1[
1/)()]1/()(1[
: for t' Solve
'))(()1/()(1:Substitute
'' : And ))(( x':RecallThen
2
2
22
2
vxtv
vvxt
vvtxtxv
vvvtxvtxv
vvtxvtxv
tvtxvvtxv
txtvtxv o
Solving for the Time Transformation…
Boost in the x – Direction
• Putting these elements together gives:
zz
yy
vtxvx
xvtvt
'
'
))(('
))(('
Which is called a boost in the x – direction.
Is This Boost a Lorentz Transformation?
)()(
)()()()(
))(())((
))(())((,
,
222
2222
vttxxvxxttv
vttxxvvxxttv
zzyy
vxtvvxtv
vxtvvxtvLbLa
zzyyxxttba
babababa
babababa
baba
bbaa
bbaa
bababba a
• 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.
baba
baba
babababa
xxtt
vxxvttv
vxxxxvttttv
v
vvvv
)]1()1()[(1/(1
))](1/(1[
:becomesequation the),(for ngSubstituti
)1/(1)( : So ,1/1)( :Recall
222
222
222
This supports the assumption that Lorentz transformations of Lorentz coordinatizations give Lorentz coordinatizations.
Continued…
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
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.
Scott McKinney