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Transformation of space and time t =  (t' + x' v/c²)t' =  (t – x v/c²) x =  (x' + vt')x' =  (x – vt) y = y'y' = y z = z'z' = z ct x y z  v/c 00 

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Presentation on theme: "Transformation of space and time t =  (t' + x' v/c²)t' =  (t – x v/c²) x =  (x' + vt')x' =  (x – vt) y = y'y' = y z = z'z' = z ct x y z  v/c 00 "— Presentation transcript:

1 Transformation of space and time t =  (t' + x' v/c²)t' =  (t – x v/c²) x =  (x' + vt')x' =  (x – vt) y = y'y' = y z = z'z' = z ct x y z  v/c 00  v/c  ct' x ' y ' z ' X = = = X ' Spacetime interval: For two events e 0 = (t 0, x 0, y 0, z 0 ) = (t 0 ', x 0 ', y 0 ', z 0 ') and e 1 = (t 1, x 1, y 1, z 1 ) = (t 1 ', x 1 ', y 1 ', z 1 ') we find that the spacetime interval s 2 = (c  t) 2 – (  x 2 +  y 2 +  z 2 ) = (c  t') 2 – (  x' 2 +  y' 2 +  z' 2 ) is Lorentz-invariant (  t = t 1 – t 0 ;  x = x 1 – x 0 ; …) Lv Lv X = L v X ' Lorentz transformation Four-vector x ct ct' x' Light 1 Unit hyperbola: Relativistic spacetime diagram

2 t =  (t' + x' v/c²)t' =  (t – x v/c²) x =  (x' + vt')x' =  (x – vt) y = y'y' = y z = z'z' = z ct x y z  v/c 00  v/c  ct' x' y' z' X = = = X ' Velocity transformation: u x = dx/dt dx =  (dx' + vdt') dt =  (dt' + dx' v/c²)  dx/dt = (dx'+ vdt')/ (dt'+ dx' v/c²) = (dx'/dt'+ v)/ (1 + dx'/dt' v/c²) dy/dt = dy' /  (dt'+ dx' v/c²) = dy'/dt' / ( 1 + dx'/dt' v/c²)  u x = (u x '+ v)/(1 + u x 'v/c²)reverse transformation: u x ' = (u x – v)/(1 – u x v/c²) u y = u y '/  (1 + u x 'v/c²)replace v by –v:u y ' = u y /  (1 – u x v/c²) u z = u z '/  (1 + u x 'v/c²)u z ' = u z /  (1 – u x v/c²) special case u = u x : (that is: u is parallel to the relative velocity v of the reference frames) u = (u'+ v)/(1 + uv/c²) u' = (u – v)/(1 – uv/c²)  dt' Lv Lv Velocity transformation

3 Velocity transformation: u x = (u x '+ v)/(1 + u x 'v/c²)reverse transformation: u x ' = (u x – v)/(1 – u x v/c²) u y = u y '/  (1 + u x 'v/c²)replace v by –v:u y ' = u y /  (1 – u x v/c²) u z = u z '/  (1 + u x 'v/c²)u z ' = u z /  (1 – u x v/c²) Four-Velocity U U ' U (τ is proper time of the object) X X X Coordinate transformation: X = L v X'  d X = L v d X' τ is Lorentz-invariant  U = L v U'

4 x t y Structure of spacetime Elsewhere x Past Future ct y Elsewhere ct' x'x' s² = 0 s² < 0 t > 0 t' < 0 3e8 m x ct Past Future s² > 0 ct' x'x' Light (c = Δx/ Δt) 1 s 1 m

5 The Twin Paradox A and B are twins of age 20. B decides to join an expedition to a nearby planet. The distance between Earth and the planet is d = 20 light years and the space ship used for the expedition travels at a speed of v = 4/5 c. -What time is it on Earth and on the spaceship respectively when B arrives at the planet? After a short stop at the planet the spaceship returns with the same speed. -How old are A and B at B’s return? t =  (t' + x' v/c²)t' =  (t – x v/c²) x =  (x' + vt' )x' =  (x – vt) t = 0 t = t 1 t = 2t 1 t'= t 1 '=t 2 ' t' = 0 t" = t 2 ' t" = 2t 2 ' Earth planet spaceship Time passing on Earth while spaceship turns around t 1 : x 1 = x planet ;x 1 ' = 0  t 1 = 25 y t 1 ' :x 1 = x planet ;x 1 ' = 0  t 1 ' = 15 y t 2 ' : x 2 = x planet ;x 2 ' = 0  t 2 ' = 15 y t 2 : x 2 = 0 ; t 2 ' = 15 y  t 2 = 9 y t = t 2

6 The Twin Paradox and Doppler Shift Earth planet

7 Relativistic Doppler Effect f R = f E [ (1 – v/c) / (1 + v/c) ] ½ for source moving away from receiver (or vice versa) f R = f E [ (1 + v/c) / (1 – v/c) ] ½ for source moving towards receiver (or vice versa) (f R : frequency of receiver in receiver’s frame; f E frequency of emitter in emitter’s frame) “longitudinal Doppler shift” Transversal effect due to time dilation: f R = f E  1 ct' x' Plain wave emitted source at rest in S' Plain wave emitted source at rest in S x 1 ct 0

8 0 K4 K K2.729 K  T = 0.02 mK (Dipole subtracted) Cosmic Microwave Background (Nobel prize in physics 2006: George Smoot, John Mather) COBE (COsmic Background Explorer) satellite Planck (2009) WMAP (2001) COBE (1989)


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