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Lorentz Transformations

Lorentz Transformations. (px,py,px,E) are components of a 4-vector which has same Lorentz transformation px’ = g (px + uE/c 2 ) u = velocity of transform py’ = py between frames is in

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Lorentz Transformations

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  1. Lorentz Transformations (px,py,px,E) are components of a 4-vector which has same Lorentz transformation px’ = g (px + uE/c2) u = velocity of transform py’ = py between frames is in pz’ = pz x-direction. If do px’ -> px E’ = g (E + upx) + -> - Use common sense also let c = 1 Frame 1 Frame 2 (cm) Before and after scatter P460 - Relativity2

  2. center-of-momentum frame Sp = 0. Some quantities are invariant when going from one frame to another: py and pz are “transverse” momentum Mtotal = Invariant mass of system dervived from Etotal and Ptotal as if just one particle How to get to CM system? Think as if 1 particle Etotal = E1 + E2 Pxtotal = px1 + px2 (etc) Mtotal2 = Etotal2 - Ptotal2 gcm = Etotal/Ptotal and bcm = Ptotal/Etotal 1 2 at rest p1 = p 2 1 2 Lab CM P460 - Relativity2

  3. Particle production convert kinetic energy into mass - new particles assume 2 particles 1 and 2 both mass = m Lab or fixed target Etotal = E1 + E2 = E1 + m2 Ptotal = p1 --> Mtotal2 = Etotal2 - Ptotal2 Mtotal = (E12+2E1*m + m*m - p12).5 Mtotal= (2m*m + 2E1*m)0.5 ~ (2E1*m)0.5 CM: E1 = E2 Etotal = E1+E2 and Ptotal = 0 Mtotal = 2E1 1 2 at rest p1 = p 2 1 2 Lab CM P460 - Relativity2

  4. p + p --> p + p + p + pbar what is the minimum energy to make a proton-antiproton pair? • In all frames Mtotal (invariant mass) at threshold is equal to 4*mp (think of cm frame, all at rest) Lab Mtotal = (E12+2E1*m + m*m - p12).5 Mtotal= (2m*m + 2E1*m)0.5 = 4m E1 = (16*mp*mp - 2mp*mp)/2mp = 7mp CM: Mtotal = 2E1 = 4mp or E1 and E2 each =2mp 1 2 at rest p1 = p 2 1 2 Lab CM P460 - Relativity2

  5. Transform examples • Trivial: at rest E = m p=0. “boost” velocity = v E’ = g(E + bp) = gm p’ = g(p + bE) = gbm • moving with velocity v = p/E and then boost velocity = u (letting c=1) E’ = g(E + pu) p’ = g(p + Eu) calculate v’ = p’/E’ = (p +Eu)/(E+pu) = (p/E + u)/(1+up/E) = (v+u)/(1+vu) • “prove” velocity addition formula P460 - Relativity2

  6. A p=1 GeV proton hits an electron at rest. What is the maximum pt and E of the electron after the reaction? • Elastic collision. In cm frame, the energy and momentum before/after collision are the same. Direction changes. 90 deg = max pt 180 deg = max energy • bcm = Ptot/Etot = Pp/(Ep + me) • Pcm = gcmbcm*me (transform electron to cm) • Ecme = gcm*me (“easy” as at rest in lab) • pt max = Pcm as elastic scatter same pt in lab • Emax = gcm(Ecm + bcmPcm) 180 deg scatter = gcm(gcm*me + gbb*me) = g*g*me(1 + b*b) P460 - Relativity2

  7. p=1 GeV proton (or electron) hits a stationary electron (or proton) mp = .94 GeV me = .5 MeV • incoming target bcm gcm Ptmax Emax • p e .7 1.5 .4 MeV 1.7 MeV • p p .4 1.2 .4 GeV 1.4 GeV • e p .5 1.2 .5 GeV 1.7 GeV • e e .9995 30 15 MeV 1 GeV • Emax is maximum energy transferred to stationary particle. Ptmax is maximum momentum of (either) outgoing particle transverse to beam. Ptmax gives you the maximum scattering angle • a proton can’t transfer much energy to the electron as need to conserve E and P. An electron scattering off another electron can’t have much Pt as need to conserve E and P. P460 - Relativity2

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