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# Transformation of space and time t = g ( t' + x' v/c²) t' = g (t – x v/c²)

X = = = X '. g g v/c 0 0 g v /c g 0 0 0 0 1 0 0 0 0 1. X = L v X '. L v. ct'. Light. ct. x'. x. 1. ct x y z. ct' x ' y ' z '. Lorentz transformation. Transformation of space and time t = g ( t' + x' v/c²) t' = g (t – x v/c²)

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## Transformation of space and time t = g ( t' + x' v/c²) t' = g (t – x v/c²)

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1. X = = = X' gg v/c 0 0 gv/c g 0 0 0 0 1 0 0 0 0 1 X = LvX' Lv ct' Light ct x' x 1 ct x y z ct' x ' y ' z ' 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 • Relativistic spacetime diagram 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; …) Unit hyperbola:

2. X = = = X' gg v/c 0 0 gv/c g 0 0 0 0 1 0 0 0 0 1 Lv ct x y z ct' x' y' z' Velocity transformation 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

3. 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) X XX U (τis proper time of the object) Coordinate transformation: X= Lv X' d X= Lvd X' τ is Lorentz-invariant U= Lv U' UU'

4. ct t ct ct' Future x' y Elsewhere y x x x Past Structure of spacetime ct' s² = 0 s² > 0 Future t > 0 t' <0 x' Elsewhere s² < 0 Past

5. t = 2t1 t"= t2' Time passing on Earth while spaceship turns around t = t1 t"= t2' t= 0 spaceship planet Earth 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'= t1'=t2' t1 : x = xplanet;x' = 0  t1= 25 y t1' :x = xplanet;x' = 0  t1' = 15 y t' = 0 t2' : x = xplanet;x' = 0  t2' = 15 y t2 : x = 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)

6. The Twin Paradox and Doppler Shift planet Earth

7. ct' ct x' 1 1 x 0 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; fEfrequency of emitter in emitter’s frame) “longitudinal Doppler shift” Transversal effect due to time dilation: fR = fEg Plain wave emitted source at rest in S' Plain wave emitted source at rest in S

8. 2.721 K 0 K 2.729 K 4 K Cosmic Microwave Background (Nobel prize in physics 2006: George Smoot, John Mather) COBE (COsmic Background Explorer) satellite COBE (1989) WMAP (2001) T = 0.02 mK (Dipole subtracted) Planck (2009)

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