Twin Paradox A2290-19 1Twin Paradox RevisitedRelativity and AstrophysicsLecture 19Terry HerterA2290-19 Twin Paradox 2Outline Simultaneity Again Sample Problem L-2 Twin Paradox Revisited Time dilation viewpoint Length contraction viewpoint Paradox & why it’s not! Problem L-13, page 117 (due Friday/Monday) Will hand back on Monday if you hand it in on FridayTwin Paradox A2290-19 2A2290-19 Twin Paradox 3Revisiting Simultaneity and Paradoxes We will look one more time at Simultaneity Time dilation / twin paradoxAs we showed previously If two events occur in our lab frame separated by distance x and time t, we can use the Lorentz transformation to find a “rocket”frame in which – Events occur at same location (for x < t) Events occur at same time (for x > t)A2290-19 Twin Paradox 4Timelike Intervals In lab frame we see two events (event A and event B) separated by a distance x (= xB– xA) and time t (= tB– tA)  Can we now find a rocket frame in which the events occur at the same location? Let the coordinates for event A be zero in both frames. Then we havexA= x′A= 0 xB= known x′B= 0tA= t′A= 0 tB= known t′B= unknown We want the speed vrelof the rocket such that x′B= 0. We use the Lorentz transformation equations Again vrelis the required speed of the rocket so that x′B= 0.  Note that since vrel< 1, we must have xB< t Thus t2B- x2B> 0 (the spacetime interval is timelike) BrelBBtvxxBrelBtvxBBreltxv x′B= 0Twin Paradox A2290-19 3A2290-19 Twin Paradox 5Events at the same locationx′ = 0 line, vrel= xb/tbv= 100Events occur at same locationWhen xb< tb(separation < delta time) for two events, we can always choose a frame in which the events happens at the same location.xtA2290-19 Twin Paradox 6Spacelike Intervals In lab frame we see two events (event A and event B) separated by a distance x (= xB– xA) and time t (= tB– tA)  Can we find a rocket frame in which the events are simultaneous? Let the coordinates for event A be zero in both frames. Then we havexA= x′A= 0 xB= known x′B= unknowntA= t′A= 0 tB= known t′B= 0 We want the speed vrelof the rocket such that t′B= 0. We use the Lorentz transformation equations vrelis the required speed of the rocket so that t′B= 0.  Note that since vrel< 1, we must have xB> t Thus t2B- x2B< 0 (the spacetime interval is spacelike) BrelBBxvttBrelBxvtBBrelxtv t′B= 0Twin Paradox A2290-19 4A2290-19 Twin Paradox 7Events at the same timet′ = 0 line, vrel= tb/xbv= 1Events occur at same timeWhen xb> tb(separation > delta time) for two events, we can always choose a frame in which the events happens at the same time.xtA2290-19 Twin Paradox 8Simultaneity Revisited: Sample Prob. L-2  Problem Setup Julius Caesar was murdered approximate 2000 years ago. Is there a way using the laws of physics to save his life? Let Caesar’s death be the reference event with coordinates, xo= 0, to= 0  Event A is the present time which is xA= 0 lyr, tA~ 2000 yr Simultaneous with event A, the Starship Enterprise sets off a fire cracker in the Andromeda galaxy Event B is the firecracker at xB= 2106lyr, tB= 2000 yr.  Let x′o= 0, t′o= 0 for the Enterprise (Caesar’s murder is the reference event)Twin Paradox A2290-19 5A2290-19 Twin Paradox 9Sample Problem L-2 (cont’d) How fast must the Enterprise be going in the Earth frame in order that Caesar’s murder is happening NOW, that is, t′B= 0?  In lab frame we see two events (event A and event B) separated by a distance x (= xB– xA) and time t (= tB– tA) Can we find an Enterprise frame in which the events are simultaneous? Let the coordinates for event A be zero in both frames. Then we havexA= x′A= 0 xB= 2x106yr x′B= unknowntA= t′A= 0 tB= 2x103yr t′B= 0 We want the speed vrelof the rocket such that t′B= 0. We use the Lorentz transformation equations Thus vrel= 10-3= 0.001 we have t′B= 0, that is, Caesar’s murder happens NOW in the Enterprise frame,BrelBBxvttBrelBxvtBBrelxtv t′B= 0yr 102yr 10263A2290-19 Twin Paradox 10Sample Problem L-2 (cont’d) Draw a spacetime diagram which contains: Event O (Caesar’s death), Event A (you here), event B (firecracker in Andromeda), your line of NOW simultaneity, the position of the Enterprise, the worldline of the Enterprise,and the Enterprise NOW line of simultaneity.spacetimeEarth line of simultaneityEnterprise line of simultaneityAYou are viewing thisOCaesar’s deathBEnterprise sets off firecracker in AndromedaSegment of Enterprise worldlinetB= vrelxB(t′B= 0 line)Twin Paradox A2290-19 6A2290-19 Twin Paradox 11Sample Problem L-2 (cont’d) In the Enterprise frame what are the x and t coordinates of the firecracker. Note that we choose the relative velocity of the Enterprise so that t′B= 0, which gives vrel= 10-3. Compute  and use the inverse Lorentz transformation.yr 010210210363BBrelBtxvt6633610210110210102BrelBBtvxx210121~11622relrelvvlyr 10999999.11022102101102101210166126666Coordinate is not much different in the two frames because the velocity is smallA2290-19 Twin Paradox 12Sample Problem L-2 (cont’d) Can the Enterprise firecracker explosion warn Caesar, thus changing the course of history? There exists a frame (the rest frame of the Enterprise) in which Caesar’s death and the firecracker explosion occur at the same time. In this frame a signal connecting the two events would have to travel faster than light. Enterprise can’t warn Caesar. The spatial coordinate is hugely different Events separated in this fashion are called spacelike Spacelike events cannot be causally linked, that is, they cannot have a cause and effect relationship.Twin Paradox A2290-19 7A2290-19 Twin Paradox 13Twin Paradox A starship leaves Earth (Event 1) and travel at 95% light speed, later arriving at Proxima Centauri (Event 2), which lies 4.3 light-years from Earth.1. What are the space and time separation between the two events asmeasured in the Earth frame, in years?2. What are the space and time separations in the frame of the starship? Part 1: The distance separation