The Twin Paradox

• 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

• 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 CoordinizationsThe 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 ?

• 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”.

Lorentz Coordinizations & Postulates

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

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

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 Transformations

Hendrick A. Lorentz6/18/1853 – 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).

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

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

Scott McKinney

)1

1)( : Note(

')(1

'

:Recall

2

2

vv

tvtv

tt

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

References

• Relativistic Electrodynamics and Differential Geometry, Parrot, Springer-Verloy 1987.