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Special Theory of Relativity • Up to ~1895, used simple Galilean Transformations x’ = x - vt t’ = t • But observed that the speed of light, c, is always measured to travel at the same speed even if seen from different, moving frames • c = 3 x 108 m/s is finite and is the fastest speed at which information/energy/particles can travel • Einstein postulated that the laws of physics are the same in all inertial frames. With c=constant he “derived” Lorentz Transformations light sees vl =c u moving frame also sees vl = c NOT vl = c-u P460 - Relativity 1 “Derive” Lorentz Transform Bounce light off a mirror. Observe in 2 frames: A velocity=0 with respect to light source A’ velocity = v observe speed of light = c in both frames c = distance/time = 2L / t A c2 = 4(L2 + (x’/2)2) / t’2 A’ Assume linear transform (guess) x’ = G(x + vt) let x = 0 t’ = G(t + Bx) so x’=Gvt t’=Gt some algebra (ct’)2 = 4L2 + (Gvt)2 and L = ct /2 and t’=Gt gives (Gt)2 = t2(1 + (Gv/c)2) or G2 = 1/(1 - v2/c2) mirror L A P460 - Relativity A’ x’ 2 Lorentz Transformations Define b = u/c and g = 1/sqrt(1 - b2) x’ = g (x + ut) u = velocity of transform y’ = y between frames is in z’ = z x-direction. If do x’ x then t’ = g (t + bx/c) “+” “-”. Use common sense can differentiate these to get velocity transforms vx’ = (vx - u) / (1 -u vx/ c2) vy’ = vy / g / (1 - u vx / c2) vz’ = vz / g / (1 - u vx / c2) usually for v < 0.1c non-relativistic (non-Newtonian) expressions are OK. Note that 3D space point is now 4D space-time point (x,y,z,t) P460 - Relativity 3 Time Dilation • Saw that t’ = g t. The “clock” runs slower for an observer not in the “rest” frame • muons in atmosphere. Lifetime = t =2.2 x 10-6 sec ct = 0.66 km decay path = bgtc b g average in lab lifetime decay path .1 1.005 2.2 ms 0.07 km .5 1.15 2.5 ms 0.4 km .9 2.29 5.0 ms 1.4 km .99 7.09 16 ms 4.6 km .999 22.4 49 ms 15 km P460 - Relativity 4 Time Dilation • Short-lived particles like tau and B. Lifetime = 10-12 sec ct = 0.03 mm • time dilation gives longer path lengths • measure “second” vertex, determine “proper time” in rest frame If measure L=1.25 mm and v = .995c L t(proper)=L/vg = .4 ps Twin Paradox. If travel to distant planet at v~c then age less on spaceship then in “lab” frame P460 - Relativity 5 Adding velocities • Rocket A has v = 0.8c with respect to DS9. Rocket B have u = 0.9c with respect to Rocket A. What is velocity of B with respect to DS9? DS9 A B V’ =(v-u)/(1-vu/c2) v’ = (.9+.8)/(1+.9*.8)= .988c Notes: use common sense on +/if v = c and u = c v’ = (c+c)/2= c P460 - Relativity 6 Adding velocities • Rocket A has v = 0.826c with respect to DS9. Rocket B have u = 0.635c with respect to DS9. What is velocity of A as observed from B? DS9 A Think of O as DS9 and O’ as rocket B B 2) V’ =(v-u)/(1-vu/c v’ = (.826-.635)/(1-.826*.635)= .4c If did B from A get -.4c (.4+.635)/(1+.4*.635)=1.04/1.25 = .826 P460 - Relativity 7 Relativistic Kinematics • E2 = (pc)2 + (mc2)2 E = total energy m= mass and p=momentum • natural units E in eV, p in eV/c, m in eV/c2 c = 1 effectively. E2=p2+m2 • kinetic energy K = T = E - m @ 1/2 mv2 if v << c • Can show: b = p/E and g = E/m p = bgm if m.ne.0 or p=E if m=0 (many massless particles, photon, gluon and (almost massless) neutrinos) • relativistic mass m = gm0 a BAD concept P460 - Relativity 8 “Derive” Kinematics • dE = -Fdx = -dp/dt*dx = -vdp = vd(gmv) • assume p=gmv (need relativistic for p) • d(gmv) = mgdv + mvdg = mdvg3 dv dE = m vg dv = m v(1 v 3 v = final 2 v =0 2 3 / 2 /c ) dv = gmc2 - mc2 =Total Energy - “rest” energy = Kinetic Energy P460 - Relativity 9 Lorentz Transformations P and E are components of a 4 vector and so transform by same Lorentz Transformation Define b = u/c and g = 1/sqrt(1 - b2) px’ = g (px + uE) u = velocity of transform py’ = py between frames is in pz’ = pz x-direction. If do x’ x then E’ = g (E + bpx/c) “+” “-”. common sense Examples with particle with mass=m at rest “transformed” to moving with velocity v P460 - Relativity 10 • What are the momentum, kinetic, and total energies of a proton with v=.86c? · v = .86c g = 1/(1.86*.86)0.5 = 1.96 • E = g m = 1.96*938 MeV/c2 = 1840 MeV • T = E - mc2 = 1840 - 938 = 900 MeV • p = bE = gbm = .86*1840 MeV = 1580 MeV/c or p = (E2 - m2)0.5 Note units: MeV, MeV/c and MeV/c2. Usually never have to use c = 300,000 km/s in calculation P460 - Relativity 11 • Accelerate electron to 0.99c and then to 0.999c. How much energy is added at each step? · v = .99c g = 7.1 v = .999c g = 22.4 • E = g m = 7.1*0.511 MeV = 3.6 Mev • = 22.4*0.511 MeV = 11.4 MeV • step 1 adds 3.1 MeV and step 2 adds 7.8 MeV even though velocity change in step 2 is only 0.9% P460 - Relativity 12 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 - Relativity 13 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 derived from E(total) and P(total) as if just one particle How to get to C.M. system? Think as if 1 particle E(total) = E1 + E2 Px(total) = px1 + px2 (etc) M(total)2 = E(total)2 - P(total)2 g(c.m.) = E(total)/M(total) and b(c.m.) = P(total)/E(total) 1 2 at rest p1 = p2 1 2 Lab C.M. P460 - Relativity 14 Particle production convert kinetic energy into mass - new particles assume 2 particles 1 and 2 both mass = m Lab or fixed target E(total) = E1 + E2 = E1 + m2 P(total) = p1 M(total)2 = E(total)2 - P(total)2 M(total) = (E12+2E1*m + m*m - p12).5 M(total)= (2m*m + 2E1*m)0.5 ~ (2E1*m)0.5 CM: E1 = E2 E(total) = E1+E2 and P(total) = 0 M(total) = 2E1 1 2 at rest p1 = p 2 1 2 Lab CM P460 - Relativity 15 p + p p + p + p + pbar what is the minimum energy to make a protonantiproton pair? • In all frames M(total) (invariant mass) at threshold is equal to 4*mp (think of cm frame, all at rest) Lab M(total) = (E12+2E1*m + m*m - p12).5 M(total)= (2m*m + 2E1*m)0.5 = 4m E1 = (16*mp*mp - 2mp*mp)/2mp = 7mp CM: M(total) = 2E1 = 4mp or E1 and E2 each =2mp at threshold all at rest in c.m. after reaction 1 2 at rest p1 = p 2 1 2 Lab CM P460 - Relativity 16 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 - Relativity 17 • 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) = gcm(gcm*me + gbb*me) = g*g*me(1 + b*b) P460 - Relativity 180 deg scatter 18 • 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 - Relativity 19