Formula sheetThe transformation formula below assume that the primed frame is moving with velocity vˆx withrespect to the unprimed frameLorentz transformationct0x0y0z0=γ −βγ 0 0−βγ γ 0 00 0 1 00 0 0 1ctxyzβ =vcγ =1p1 − β2Velocity additionu0x=ux− v1 − uxv/c2u0y=uyγ (1 − uxv/c2)u0z=uzγ (1 − uxv/c2)Time dilation and length contractiont0is the proper time. l0is the proper length∆t = γ∆t0∆l = l0/γRelativistic Doppler shiftfD= f0s1 + β1 − βRelativistic momentum and energyp = γm0v E = γm0c2E2− p2c2= m2c44Quiz 11. Lorentz invariance Using Lorentz transformation formulae, show that if two events A and Bare observed in frame S as (tA, xA, yA, zA) and (tB, xB, yB, zB) and in frame S0as (t0A, x0A, y0A, z0A)and (t0B, x0B, y0B, z0B), the following quantity:∆S2= c2∆t2− ∆x2− ∆y2− ∆z2is the same in both frames. (4 pt.)What is the sign of ∆S2if the two events are separated by a time-like, space-like, and light-likeintervals, respectively? (1 pt.)2. Muon decay In a laboratory experiment a muon is observed to travel 800 m before disintegrating.The (rest frame) lifetime of muon is 2 × 10−6s. What was the speed of the muon? (5 pt.)3. Hot pursuit A run away criminal is driving his spaceship at3c4(with respect to earth). A policespaceship can only run onc2(with respect to earth). So the police fired a bullet with muzzelvelocityc3, aimed at the criminal. Can the bullet (eventually) reach the criminal? (2 pt.)What are the speed of the police and the bullet as measured by the criminal?(3 pt.)4. Relativistic dynamics A neutral pion of (rest) mass m and (relativistic) momentum p =34mcdecays into two photons (particles of light). One of the photons is emitted in the same directionas the original pion, and the other in the opposite direction. Find the (relativistic) energy of eachphoton. (5 pt.)5Quiz 1 solution1. Lorentz invariance From the Lorentz transformation we have:c t0A= γ (c tA− βxA) , x0B= γ (−βctA+ cxA)c t0B= γ (c tB− βxB) , x0A= γ (−βctB+ cxB)c2(t0B− t0A)2− (x0B− x0A)2= γ2[c (tB− tA) − β (xB− xA)]2− γ2[−βc (tB− tA) + (xB− xA)]2c2∆t02− ∆x02= γ2[c∆t − β∆x]2− γ2[−βc∆t + ∆x]2= γ2c2∆t2−2c∆tβ∆x + β2∆x2 − γ2β2c2∆t2−2βc∆t∆x + ∆x2 = c2γ21 − β2∆t2− γ21 − β2∆x2= c2∆t2− ∆x2It is clear that ∆y0= ∆y and ∆z0= ∆z, so:∆S02≡ c2∆t02− ∆x02− ∆y02− ∆z02= c2∆t2− ∆x2− ∆y2− ∆z2≡ ∆SSpace-like interval is when the spacial separation is more than the temporal separation and∆S < 0; Time-like interval ∆S > 0. Light-like ∆S = 0.2. Muon decay Suppose the muon is moving at speed u, it lived longer in the lab frame than itsrest frame due to time dilation:τlab= γ (u) τ0=1q1 −u2c2τ0The speed is also the distance it travels (l = 800m) before disintegrating divided by the life timein the lab frame:u = l/τlab=lq1 −u2c2τ0Solve for u:τ20u2= l21 −u2c2→ u2=c21 +c2τ20l2→ u =cq1 +c2τ20l2=c1.25= 2.4 × 108m/s3. Hot pursuit• The earth frame moves with velocity −c/2 with respect to the police spaceship frame. Thevelocity of the bullet as seen by the earth frame is:ubulletearth=ubulletpolice− uearthpolice1 −ubulletpoliceuearthpolicec2=c/3 − (−c/2)1 − (c/3)(−c/2)/c2=57c ≈ 0.71c <34cSo the bullet will not catch up to the criminal6• The police has velocity c/2 in the earth frame. In the criminal’s frame, it is:upolicecriminal=c/2 − 3c/41 − (c/2)(3c/4)/c2= −0.4c• The bullet has velocity 0.71c in the earth frame (from first part of this problem). In thecriminal’s frame, it is:ubulletcriminal=0.71c − 3c/41 − (0.71c)(3/c4)/c2≈ −0.086c4. Relativistic dynamics It was mentioned in class that since light travels at speed of light andphoton has zero mass. Therefore Ephoton= pphotonc. Suppose the energy of two photons are E1and E2, their momentums are p1= E1/c and p2= E2/c, respectively. Conservation of energyand momentum gives:Eπ=pm2πc4+ p2πc2=q1 + (3/4)2mπc2=54mc2= E1+ E2pπ=34mc = E1/c − E2/cSovle to get E1= mc2and E2=