Special Relativistic “Paradoxes”

Special Relativistic “Paradoxes” PHYS 206 – Spring 2014

Simultaneous in rest frame NOT simultaneous in moving frame (B occurs beforeA)

Distance measured in rest frame defined by lines parallel to t-axis Distance measured in moving frame defined by lines parallel to t’-axis

Lorentz Transformation Rest frame: Moving frame: Events that are seen as simultaneous in the rest frame are not simultaneous in the moving frame. 2.18 m in the moving frame appears to be 2 m as measured in the rest frame.

Dilation Formulae • All quantities measured in rest frame (t,x) • Time Dilation: • Length Contraction: Elapsed time seen on rest clock Velocity of moving clock Elapsed time seen on moving clock Velocity of moving object Apparent length of moving object Length of same object at rest

The Pole in the Barn Paradox

A person carrying a 3 m long pole runs very fast (!) at a 2 m long barn, reasoning that if his velocity is fast enough, the pole will length-contract and fit nicely inside the barn. Once inside, the doors on either end of the barn close, completely trapping the pole. As seen from the frame of reference of the barn (rest), the the pole moving at v = 0.745 looks to be 2 m long, and will fit entirely in the barn and the doors can both be closed.

BUT…..

If we look from the frame of reference of the pole, the pole is still 3 m long, but now the barn is shorter!! As seen from the frame of reference of the pole (rest), the the barn is moving at v = 0.745 and looks to be 1.33 m long!!! The pole can never fit in the barn, and the doors certainly cannot both close!

So…..How is it possible that the pole fits entirely in the barn in one frame of reference with both doors closed, while in another frame of reference it doesn’t? Answer: Lorentz transformations and breakdown of simultaneity!

t At every instant of time in the barn’s frame, the barn occupies the space between x=0 and x=2. Each line of constant t (parallel to x-axis) is a “snapshot” of all space at that instant of time. Barn at t = 2 Barn at t = 1 Barn at t = 0 x Barn at t = -1 Barn at t = -2

t World-sheet As the barn sits in its rest frame, it “smears” out an area of spacetime called its “world-sheet” as time elapses. Objects exist in any reference frame as slices of their world-sheet at constant time. Barn at t = 2 Barn at t = 1 Barn at t = 0 x Barn at t = -1 Barn at t = -2

t’ [v=0.75] t x’ [v=0.75] In the moving, the pole is between x’ = 0 and x’ = 3 m. The calibration hyperbola tells us where these are. x’ = 3 Pole x’ = 0 x

t’ t x’ Just as the barn occupies the space betweenx=0 and x=2 in its frame, the pole occupies x’=0 and x’=3 in its frame. x’ = 0 x’ = 3 x

t’ t The calibration hyperbolae tell us where to find tick-marks of t’ x’ t’ = 2 t’ = 1 t’ = 0 x t’ = -1

t’ t At every instant of time in the pole’s frame (t’), it occupies the space between x’=0 and x’=3. Each line of constant t’ (parallel to x’-axis) is a “snapshot” of all space at that instant of time. x’ Pole at t’ = 1 Pole at t’ = 0 Pole at t’ = -1 x

t’ t x’ Pole at t’ = 1 As the pole sits in its rest frame, it also “smears” out an area of spacetime called its “world-sheet”. Pole at t’ = 0 Objects exist in any reference frame as slices of their world-sheet at constant time. Pole at t’ = -1 x

t’ t x’ Overlay of the barn and pole world sheets. Note the pole fits perfectly in the barn, at least in the barn frame! x

t’ t x’ What does the pole look like in thebarnframe? Slices of pole world-sheet at constant t (barn frame). 2 m long pole, moving toward and away from the barn. t < 0 : entering the barn t = 0 : completely contained in the barn; doors can be simultaneously shut. t > 0 : exiting the barn Pole at t = 2 Pole at t = 1 Pole at t = 0 x Pole at t = -1

Barn frame Pole completely in barn. Both doors shut simultaneously. “2 m” pole v = 0.75 2 m barn

t’ t x’ How NOT to read a spacetime diagram! Pole at t’ = 0 At t = -1 s in the barn frame, the pole only exists at x=-0.5m... … and therefore each other bit of the pole exists at different points in time.... ?!?!?! Pole at t’ = -1 x

t’ t x’ What does the barn look like in the pole frame? Slices of barn world-sheet at constant t’ (pole frame). Barn at t’ = 1 Barn at t’ = 0 Barn at t’ = -1 x

t’ t x’ What does the barn look like in the pole frame? Slices of barn world-sheet at constant t’ (pole frame). Barn at t’ = 0 x The apparent length of the barn is determined by where the line of constant x = 2 m intersects the x’-axis (about x’= 1.3 m).

t’ t x’ Pole at t’ = 1 t’ = 0: back of the pole is at the barn entrance (x=0), but the front is sticking out beyond the other end. Pole at t’ = 0 Pole at t’ = -1 x At some t’ < 0, the pole has entered the barn and reaches the end, but again is not completely contained. Pole at t’ = -2.

t’ t x’ Pole at t’ = 0 Left door shuts at t’=0 x Pole at t’ = -2.26 Right door shuts at t’= -2.26

Pole frame 1.3 m barn V’ = -0.75 3 m pole Pole too long for barn. When front end reaches back of barn, back end is still outside (t’ = -2.26)

Pole frame 1.3 m barn V’ = -0.75 Right door opens, pole passes through. 3 m pole Back door closes once end back end of pole is inside (t’ = 0)

Lorentz Transformations The simultaneous events A (t=0,x=0) and B (t=0,x=2) in the rest frame and are not simultaneous in the moving frame. In fact, B occurs before A in the moving frame.

Twin Paradox

Two twins decide to test special relativity. One stays on Earth, while the other departs on a spaceship. The goal is to see whose clock has recorded less passage of time when “space twin” returns to Earth. The paradox: Both observe each other’s clock to be running slow! So, whose actually does?

t The proper time in any frame of reference is: Vectors of the same length in hyperbolic spacetime are not the same length “on paper.” Δτ = 4 Δτ = 4 The moving clocks appear to run slowly. Δτ = 4 x

t The proper time in any frame of reference is: Longest in the rest frame ( Δx= 0 ): Δt x

t The proper time in any frame of reference is: Δτ Δt x

t A photon (light signal) travels at v = c = 1, so is proper time is NULL VECTOR Δτship Δt Δτphoton = 0 x

t So, if the rest frame is the Earth-bound twin and the blue moving frame is the space-bound twin, the spacetime diagram of the round trip will be: Δt Δτphoton= 0 Δτship Δτship Δt Δτphoton= 0 The twin on Earth ages the most, because his spacetime path is the longest. x

t But wait: why can’t this scenario be reversed? The spaceship twin sees the Earth-bound twin as the one who is moving. Δτphoton= 0 Δt Δτship Δτship The difference: one accelerates, one does not! The acceleration implies the spaceship frame is notinertial, and special relativity doesn’t hold. Δt Δτphoton= 0 x

t The shortest path between two points on a spacetime diagram has the longest proper time! Δt Δτ Δτphoton= 0 Δτ Δt Δτphoton= 0 x

Lorentz Transformation Notations 1. Coordinate 2. Matrix 3. Algebraic 4. Index

Special Relativistic “Paradoxes”