Time dilation, length contraction, simultaneity Special Relativity Time dilation, length contraction, simultaneity

Electron-electron Interaction Center-of mass rest frame S' - - Coulombic repulsion only Electron acceleration is a'

Electron-electron Interaction Frame S in which electrons move v - v - Coulombic repulsion plus Lorentz attraction Electron acceleration is a = a' (1– b2) (b  v/c)

c in Inertial Frames To simplify we used 1/c2 = m0e0. Is that valid in all inertial frames? Is that true for all directions (isotropy)? If you move against the light, is it faster? if you move with the light, is it slower? Is c isotropic only in one absolute rest frame (the ether frame)?

Einstein’s Postulate The speed of light is isotropically c in all inertal frames.

Time Dilation

Times in Different Frames Transverse light flash in a moving car mirror v source/ detector Flash emitted, reflected, received Reference frames: S for track, S' for car

Detection time in S' Light signal travels distance 2h Detection time t' = 2h/c

Detection time in S h Light signal travels distance 2h vertically and vt horizontally Total distance is 2d h d vt Solve for detection time t, find t/t'

Time Dilation If two events at the same place in frame S' are separated by time t', in frame S they are separated by time t = t' (1 – b2)–1/2

Length Contraction

Lengths in Different Frames Longitudinal light flash in a moving car mirror v source/ detector l0 Flash emitted, reflected, received Reference frames: S for track, S' for car

Back to our Electrons a = a' (1 – b2)1/2 - - What does time dilation tell us?

Doppler Shift Of light

Moving Source Equations of Motion: Source xS = –vt 1 2 3 D Equations of Motion: Source xS = –vt ith wavefront xi = –v (i/fs) + c (t – i/fs) 15

Classical Formula 1 fD = fS 1 + b But recall Ts = gTs' So fs = fs'/g 16

Doppler Shift of Light Relativistic Doppler shift: 1 + b fD = fS' where b = vS/c Detector frame of reference (vD = 0) 17

Redshift z Dl l' l – l' l' l l' – 1 z = = = = c/fD c/fS' – 1 fD – 1

Relativistic Mechanics Momentum and energy

Momentum p = gmv

Total Energy E = gmc2

Rest Energy E = gmc2 If v = 0 then g = 1, E = mc2 Rest energy is mc2 Kinetic energy is (g–1)mc2

Correspondence At small b: Momentum gmv  mv Energy gmc2 = (1–b2)–1/2 mc2 Binomal approximation (1+x)n  1+nx for small x So (1–b2)–1/2  1 + (–1/2)(–b2) = 1 + b2/2 gmc2  mc2 + 1/2 mv2 Is this true? Let’s check:

Convenient Formula E2 = (mc2)2 + (pc)2 Derivation: show R side = (gmc2)2

A massless photon p = h/l E = hf = hc/l h = 6.62610–36 J·s (Planck constant) gmc2 incalculable: g =  and m = 0 But E2 = (mc2)2 + (pc)2 works: E2 = 0 + (hc/l)2 = (hf)2 E = hf