Lorentz transformation Transformation of space and time t = g (t' + x' v/c²) t' = g (t – x v/c²) x = g (x' + vt') x' = g (x – vt) y = y' y' = y z = z' z' = z Four-vector ct x y z g g v/c 0 0 gv/c g 0 0 0 0 1 0 0 0 0 1 ct' x ' y ' z ' X = = = X' Lv X = Lv X' Spacetime interval: For two events e0 = (t0, x0, y0, z0) = (t0', x0', y0', z0') and e1 = (t1, x1, y1, z1) = (t1', x1', y1', z1') we find that the spacetime interval s2 = (cDt)2 – (Dx2 + Dy2 + Dz2) = (cDt')2 – (Dx'2 + Dy'2 + Dz'2) is Lorentz-invariant (Dt = t1 – t0 ; Dx = x1 – x0 ; …) Relativistic spacetime diagram x ct ct' x' Light 1 Unit hyperbola:

Velocity transformation ux = dx/dt dx = g (dx' + vdt') dt = g (dt' + dx' v/c²)  dx/dt = (dx'+ vdt')/ (dt'+ dx' v/c²) = (dx'/dt'+ v)/ (1 + dx'/dt' v/c²) dy/dt = dy' / g (dt'+ dx' v/c²) = dy'/dt' / ( 1 + dx'/dt' v/c²)  ux = (ux'+ v)/(1 + ux'v/c²) reverse transformation: ux' = (ux – v)/(1 – uxv/c²) uy = uy'/ g (1 + ux'v/c²) replace v by –v: uy' = uy/ g (1 – uxv/c²) uz = uz'/ g (1 + ux'v/c²) uz' = uz/ g (1 – uxv/c²) special case u = ux: (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' t = g (t' + x' v/c²) t' = g (t – x v/c²) x = g (x' + vt') x' = g (x – vt) y = y' y' = y z = z' z' = z ct x y z g g v/c 0 0 gv/c g 0 0 0 0 1 0 0 0 0 1 ct' x' y' z' X = = = X' Lv

Four-Velocity Velocity transformation: ux = (ux'+ v)/(1 + ux'v/c²) reverse transformation: ux' = (ux – v)/(1 – uxv/c²) uy = uy'/ g (1 + ux'v/c²) replace v by –v: uy' = uy/ g (1 – uxv/c²) uz = uz'/ g (1 + ux'v/c²) uz' = uz/ g (1 – uxv/c²) U (τ is proper time of the object) X X X U U' Coordinate transformation: X = Lv X'  d X = Lv d X' τ is Lorentz-invariant  U = Lv U'

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

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 = 0 t = t1 t = 2t1 t" = t2' t" = 2t2' Time passing on Earth while spaceship turns around t'= t1'=t2' t1 : x1 = xplanet ;x1' = 0  t1 = 25 y t1' :x1 = xplanet ;x1' = 0  t1' = 15 y Earth planet spaceship t' = 0 t2' : x2 = xplanet ;x2' = 0  t2' = 15 y t2 : x2 = 0 ; t2' = 15 y  t2 = 9 y t = t2 t = g (t' + x' v/c²) t' = g (t – x v/c²) x = g (x' + vt' ) x' = g (x – vt)

The Twin Paradox and Doppler Shift Earth planet

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

Cosmic Microwave Background Planck (2009) WMAP (2001) COBE (1989) Cosmic Microwave Background (Nobel prize in physics 2006: George Smoot, John Mather) COBE (COsmic Background Explorer) satellite 2.721 K 2.729 K 0 K 4 K T = 0.02 mK (Dipole subtracted)