1. Relativistic Momentum Classical physics: Definition of momentum: p = mv Conservation of momentum:p 1 + p 2 = p 3 + p 4 Coordinate transformation (Galilei; velocity of object: u; frame: v): p' = m u'  p 1 ' + p 2 ' p1p1 p2p2 m M p3p3 p4p4 Relativistic physics: Same definition of momentum does not work due to complicated velocity transformation  Use four-velociy (has simpler transformation law!): define four-momentum: P = m U Assume four-momentum conservation in one frame: P 1 + P 2 = P 3 + P 4 Coordinate transformation (Lorentz): P ' = m U '  P 1 ' + P 2 ' U =  u cucu ( )  P = m  u cucu ( ) Space component (velocity of object: v): p =  mv This is the relativistic momentum (conserved in all inertial frames!) Define dynamic mass: m d =  m  p = m d v (equivalent to classical case) To distinguish between the masses we call m the rest mass = m(u + v)= p + mv = p 1 + p 2 + mv + Mv = p 3 + p 4 + mv + Mv= p 3 ' + p 4 ' = m L v -1 U = L v -1 m U = L v -1 P = L v -1 P 1 + L v -1 P 2 = L v -1 ( P 1 + P 2 )= L v -1 ( P 3 + P 4 )= P 3 ' + P 4 '

2. Relativistic Momentum and Energy Coordinate transformation: P = m U = m L v U ' = L v P ' Time component :  mc = ––––––– mc  1 – v 2 /c 2 ≈ mc (1 + – –– ) 1 v 2 2 c 2 = – (mc 2 + – mv 2 ) 1 c 2 Time component of four-momentum conservation (same approximation, v << c ): (mc 2 + ½ mv 1 2 ) + (Mc 2 + ½ Mv 2 2 ) = (mc 2 + ½ mv 3 2 ) + (Mc 2 + ½ Mv 4 2 ) E 1 + E 2 = E 3 + E 4 Conservation of Energy! Total energy: E =  mc 2 Rest energy: E 0 = mc 2 (Energy for v = 0) Kinetic energy: E kin = E – E 0 =  mc 2 – mc 2 = (  – 1) mc 2 Some important relations Analog to the spacetime interval we calculate the ‘magnitude’ of the four-momentum: P 2 = (E/c) 2 – p 2 = For m  0 (p 2 c 2 >> m 2 c 4 ): Velocity in terms of E and p : = ––––––– energy c (  mc) 2 – (  mv) 2 =  2 m 2 c 2 (1 – v 2 /c 2 ) = m 2 c 2  E 2 = p 2 c 2 + m 2 c 4 E = pc v = p/(  m) = pc 2 /E ; for m  0 : v = c = m d c 2  E' =  (E – v p x ); p x ' =  (p x – v/c 2 E); p y ' = p y ; p z ' = p z

3. Relativistic Momentum and Energy v p  Energy transformation and Doppler shift Energy transformation E' =  (E – v p x ) Identical to Doppler shift formulas  Energy proportional to frequency (for m = 0)  Can be used to easily deduce general formula for Doppler shift: E' =  (E – v p x )  f E = f R  (1 – v/c cos  ) pxpx p x = p cos  =  (E – v E/c) for massless objects ( p = E/c; assume p = p x ) =  (E – v p cos  ) = E ––––––– 1 – v/c  1 – v 2 /c 2 = E –––– 1 – v/c 1 + v/c  = E  (1 – v/c cos  ) (for m = 0) If p x = 0: E' =  E

4. Relativistic Momentum and Energy v p  Energy transformation and Doppler shift Energy transformation E' =  (E – v p x ) Identical to Doppler shift formulas  Energy proportional to frequency (for m = 0)  Can be used to easily deduce general formula for Doppler shift: E' =  (E – v p x )  f E = f R  (1 – v/c cos  ) pxpx p x = p cos  L M L M Equivalence of Mass and Energy: Einstein’s box Centre of mass cannot shift in isolated system  light pulse has a mass equivalent: Conservation of momentum: p = E/c = Mv Total time of process: t = L/c CM shift of box: M  x = Mvt = p L/c = E/c 2 L CM shift of light: m light  x light = m light L  m light = E/c 2 xx =  (E – v E/c) for massless objects ( p = E/c; assume p = p x ) =  (E – v p cos  ) = E ––––––– 1 – v/c  1 – v 2 /c 2 = E –––– 1 – v/c 1 + v/c  = E  (1 – v/c cos  ) (for m = 0) If p x = 0: E' =  E

5. Headlight Effect v = 0.9 c E-M Radiation emitted isotropically Rest frame of emitter Laboratory frame v = 0.99 c v = 0.9 c Galilean transf.

6. uncharged wire S S' charged wire at rest in S at rest in S' I v – Q S I S' Lorentz force (attractive) Electrostatic force (attractive) Relativistic Wire B = µI/2  d F B = – Qv×B B = µI/2  d F B = – Qv×B = 0

7. General Relativity: Spacetime curvature

9. General Relativity: Gravitational Lensing

10. General Relativity: Spacetime Diagram, Light Cones x Past Future ct y Elsewhere x'x' ct' Object …. Black hole falling into Special Relativity General Relativity