The Twin Paradox

1. The Twin Paradox • Tyler Stelzer • Bob Coulson • Berit Rollay • A.J. Schmucker • Scott McKinney

2. "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

3. 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

4. Events and CoordinizationsThe Concept of Spacetime

5. 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.

6. 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.

7. 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

8. 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.

9. 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”.

10. Lorentz Coordinizations & Postulates

11. 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. Z r(t) X Y

12. 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.

13. What is time? • There are 2 types of time • Physical Clock Time • Coordinate Time: the time furnished by the coordinatization model: e ( Te , Xe , Ye , Ze )

14. 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.

15. 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.

16. Minkowski Space Geometry of Spacetime

17. 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.

18. 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.

19. 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)

20. 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.

21. 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.)

22. 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}

23. Moving Reference Frames

24. Moving Reference Frames • Recall the idea of a “coordinatization” e  E, e  (te, xe, ye, ze) Z “e” C  Y B O  A X

25. Z C  Y O B  A X The Idea: • By trig, determine “spatial coordinates” (xe, ye, ze) • Assuming: c = speed of light, Rate X Time = Distance, “e” Time at which the light pulse reaches O

26. 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)

27. Z Z’ “e” Y O Y’ O’ X X’ How are these two frames related? • First, let’s look at a picture: Assume constant velocity c = 1

28. Z Light source Mirror Z’ “e” Y O Y’ O’ X X’ Solution: • Assume O, O’ have standard clocks. e.g. Einstein – Langevin clock (light pendelum) • A rigid rod (or tube) of length L 1 unit of time = duration of time between emission and return

29. Light Source Z Z’ “e” L Y O Y’ Mirror O’ X X’ 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) Spaceship moving at a velocity v O’’s perspective L = ct’ Let t’ = time of ½ pendulum Distance = (Rate)(Time)

30. Z Z’ “e” Y O Y’ O’ X X’ Now Consider O’s Perspective Mirror has moved since the ship has moved Light Source ct L Mirror vt Let t be the time of the ½ pendulum of O’’s clock as observed by O. Observe: (ct)2 = L2 + (vt)2

31. If we let c = 1 • Using some simple algebra, we can solve the previous equation for t. Note: From O’’s perspective L = ct’ This is the relativistic time dilation factor

32. Z Z’ “e” Y O Y’ O’ X X’ 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

33. moving at velocity v relative to O R’ Z Z’ “e” Y O Y’ O’ X X’ Now suppose the “rod” is situated in the direction of the moving frame. 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.

34. Z Z’ “e” Y O Y’ O’ X X’ 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

35. Z Z’ “e” Y O Y’ O’ X X’ Now consider t1 and t2: Vt1 = distance the spaceship (and hence rod) travels between initial flash and mirror interception. ct1 Light source R vt1

36. Z Z’ “e” Y O Y’ O’ X X’ R Mirror moves vt2 ct2 vt2 R Using some simple algebra, we solve for t2 Now we can revisit the total time equation t = t1 + t2

37. Lorentz Fitzgerald Contraction So:

38. Lorentz Fitzgerald Contraction (cont.) Lorentz Fitzgerald Contraction

39. Hendrick A. Lorentz 6/18/1853 – 2/4/1928 Lorentz Transformations

40. 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.

41. 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.

42. 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”

43. 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

44. Lorentz – Einstein Transformations (Boosts)

45. Z Z’ Y O Y’ O’ X X’ 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).

46. O VT O’ x - VT E 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)”.

47. Apply Length Contraction • O understands that O’ will measure this as the longer distance where Therefore,

48. (in relation to “O”) O VT O’ VT* x - VT* E x 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)

49. 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)

50. Apply Time Dilation When o adjusts for time dilation where: The equation becomes: