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}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 487 "The Galilean transformations dominated physics until t he beginning of the 20th century. They were based on the universality \+ of time and simple additivity of velocities between inertial (non-acce lerated) reference frames. Albert Einstein proposed to substitute the \+ following two principles for this transformation in order to resolve t he difficulties with understanding the Michelson-Morley experiment, wh ich demonstrated that the speed of light was the same in inertial refe rence frames:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "1. The laws of physics are the same for all observers sit uated in inertial frames;" }}{PARA 0 "" 0 "" {TEXT -1 84 "2. The speed of light in vacuum is the same for all observers and in all direction s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 296 "Th e first step to understand the implications of these principles is tha t the idea of simultaneity cannot be an absolute one: what appears sim ultaneous to observer O1, will not appear to be simultaneous for obser ver O2 moving with respect to O1. In fact, even the order of events ca n be reversed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 356 10 "The basics" }}{PARA 0 "" 0 "" {TEXT -1 134 "We begin with a discussion of two phenomena that arise a s a result of the stated two principles: time dilation and length cont raction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 13 "Time dilation" }}{PARA 0 "" 0 "" {TEXT -1 293 "The implications ar e that we need to understand how two observers that have their own clo cks can transform the times between their reference frames. For this p urpose one introduces a so-called light clock: a flashlamp sends of a \+ light pulse, which travels to a mirror that is a known distance " } {TEXT 257 1 "L" }{TEXT -1 13 " away in the " }{TEXT 258 1 "y" }{TEXT -1 286 "-direction. Upon return the pulse is recorded by a photocell, \+ and another pulse is triggered to leave the flashlamp. We assume that \+ there is no delay between registering at the photocell and emitting th e next pulse from the flashlamp. The clock sends pulses at a known tim e inerval of " }{XPPEDIT 18 0 "Delta*t[0] = 2*L/c;" "6#/*&%&DeltaG\" \"\"&%\"tG6#\"\"!F&*(\"\"#F&%\"LGF&%\"cG!\"\"" }{TEXT -1 169 " for an \+ observer O1 stationary with respect to the clock. The subscript makes \+ it clear that it is the time observed by someone with zero speed with \+ respect to the clock." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 180 "Now consider an observer O2 moving with a known veloc ity with respect to O1, such that the direction of travel is perpendic ular to the light path, i.e., he moves in the horizontal " }{TEXT 259 1 "x" }{TEXT -1 114 "-direction. He has such a clock in his moving fra me. Suppose a light pulse is bouncing off the mirror just at the " } {TEXT 260 1 "x" }{TEXT -1 74 "-position of O1. This means that the fla shlamp released the pulse at some " }{TEXT 261 2 "x-" }{TEXT -1 71 "di stance prior to passage of O1, and the photocell will register at an \+ " }{TEXT 262 1 "x" }{TEXT -1 47 "-distance of equal magnitude after pa ssing O1. " }}{PARA 0 "" 0 "" {TEXT -1 194 "What does O2 notice? O2 no tices a flash released from the moving clock prior to passage, and ano ther flash after the passage. This means that the light has travelled \+ a longer distance, namely 2 " }{TEXT 263 1 "D" }{TEXT -1 71 ". The tim e interval for an observer in a moving reference frame (speed " } {TEXT 264 1 "V" }{TEXT -1 43 ") with respect to the clock is, therefor e, " }{XPPEDIT 18 0 "Delta*t[V] = 2*D/c;" "6#/*&%&DeltaG\"\"\"&%\"tG6# %\"VGF&*(\"\"#F&%\"DGF&%\"cG!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 105 "The task is now to find the relationship between the tim es. The Pythagorean theorem allows one to relate " }{TEXT 267 1 "D" } {TEXT -1 5 " and " }{TEXT 266 1 "L" }{TEXT -1 93 ". The constancy of t he speed of light then implies a relationship between the time interva ls." }}{PARA 0 "" 0 "" {TEXT -1 19 "The distance along " }{TEXT 268 1 "x" }{TEXT -1 95 " that the moving clock travels between emitting the \+ flash and passing O2, as measured by O2 is " }{XPPEDIT 18 0 "V*Delta*t [V]/2;" "6#**%\"VG\"\"\"%&DeltaGF%&%\"tG6#F$F%\"\"#!\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "W e begin with the time intervals between flashes for the stationary and moving observers (O1 and O2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq1:=Dt[0]=2*L/c;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq2:=Dt[V]=2*D/c; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "Now we use the Pythagorean theorem with the statement about O2's measurement of the distance between flashes occuring:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "eq3:=D^2=L^2+(V*Dt[V]/2)^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "We square the time interval observed by O2" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq2p:=lhs(eq2)^2=rhs(eq2)^2; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "to be able to substitute the \+ unknown distance that the light pulse must have travelled (D^2):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "eq4:=subs(D^2=solve(eq3,D^2) ,eq2p);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Now we eliminate " } {TEXT 269 1 "L" }{TEXT -1 41 " by introducing the proper time interval :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq5:=subs(L=solve(eq1, L),eq4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "and solve for the dil ated time interval." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "sol: =solve(eq5,Dt[V]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "We are only interested in the positive root." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq6:=Dt[V]=sol[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "This result is commonly expressed by cancelling the c and expr essing the speed between the frames as a fraction of the speed of ligh t." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "eq6p:=Dt[V]=Dt[0]/sqr t(1-(V/c)^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "This looks clums y, and thus we often use a shorthand notation:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 56 "eq6pp:=Dt[V]=simplify(subs(V=beta*c,rhs(eq6)), symbolic);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 261 "The well-known exp erimental test of this relationship is the extension of the half-life \+ of muons produced in the upper atmosphere by cosmic radiation (reporte d first in 1941 by Rossi and Hall). The average speed of these muons p roduced typically at a height of " }{TEXT 270 1 "h" }{TEXT -1 454 "=10 km is 0.994 c. A muon produced in a laboratory at rest has a half-lif e of 1.5 microseconds. If the muons produced at 10 km height and comin g down with nearly the speed of light (299 792 km/s) were to decay at \+ this rate, their population would be cut in half about every 0.45 km. \+ Due to this exponential attenuation (executing 20 times a halving of t he current sample is calculated below) a very tiny fraction of the pro duced muons would be observed (" }{XPPEDIT 18 0 "10^(-6);" "6#)\"#5,$ \"\"'!\"\"" }{TEXT -1 2 ")." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "d_h:=1.5*10^(-6)*300000;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "0.5^(20);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Instead we have the half-life as observed by us (who are moving with " }{TEXT 271 1 " V" }{TEXT -1 39 "=0.994 c with respect to the muons) of:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "t_h:=1.5*10^(-6)/sqrt(1-0.994^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "The dilation factor is large in deed:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "1./sqrt(1-0.994^2) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "t_h*300000;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "The travel time to reach the surf ace of the Earth from a height of 10 km is a modest multiple of the ha lf-life" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "t_tr:=(10/300000 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "t_tr/t_h;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "This means that less than a quarter of t he muons produced at 10 km height do reach the surface of the Earth (a nd more of we stand on top of a high mountain)." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "We can visualize the time dilation factor by a simp le graph as a function of the speed, expressed as a fraction of the sp eed of light (" }{TEXT 19 4 "beta" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot(1/sqrt(1-beta^2),beta=0..1);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 182 "The graph shows that time dilatio n becomes appreciable for particle speeds exceeding a quarter of the s peed of light, and that it is unbounded for particles whose speed appr oaches c." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 272 18 "Length Contraction" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 321 "Now that we underst and that simultaneity of events is non-universal phenomenon, and that \+ the rate at which clocks tick depends on the velocity of the observer \+ with respect to the clock, we have to assess the question of measuring the length of an object. In Newtonian mechanics the length of an obje ct is determined by a " }{TEXT 273 12 "simultaneous" }{TEXT -1 146 " m easurement of the position of its endpoints. We know that simultaneity is not universal, i.e., there will be no universal length for any obj ect!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 " The thought experiment to investigate the phenomenon involves two obse rvers O1 and O2 carrying rods of equal length " }{TEXT 274 1 "L" } {TEXT -1 104 " (as measured in their respective reference frame). Supp ose O1 is stationary and O2 moves with velocity " }{TEXT 276 1 "V" } {TEXT -1 2 "=[" }{TEXT 275 1 "V" }{TEXT -1 54 ",0,0] with respect to O 1. If the rods are held in the " }{TEXT 278 1 "y" }{TEXT -1 4 " or " } {TEXT 277 1 "z" }{TEXT -1 435 " directions the length measurements are not affected. The assumption of contraction or expansion in these cir cumstances lead to a contradiction with the first principle of relativ ity: the relative motion cannot lead to an observation of expansion in one frame vs contraction in the other: O1 and O2 have agreed that whi le at rest with respect to each other rods R1 and R2 are of equal leng th. Now suppose the O2 moves by with velocity " }{TEXT 279 2 "V " } {TEXT -1 32 "and both hold their rods in the " }{TEXT 280 1 "y" } {TEXT -1 545 " direction. Suppose O2 observes rod R1 to be contracted, and marks off the contracted length on his rod. Now observer O1 sees \+ from the mark that rod R2 is longer, since from his point of view R1 i s at rest and unaffected. Thus, we have supposed that O2 observes a co ntraction, while O1 observes an expansion. Yet physics has to be symme trical with respect to the interchange of O1 and O2, as there is nothi ng absolute about one or the other. The contradiction is only avoided \+ if no contraction or expansion is observed under these circumstances. " }}{PARA 0 "" 0 "" {TEXT -1 68 "Mathematically this is formulated in \+ the following way: if O1 uses [" }{TEXT 284 1 "x" }{TEXT -1 2 ", " } {TEXT 285 1 "y" }{TEXT -1 2 ", " }{TEXT 286 1 "z" }{TEXT -1 16 "], and O2 uses [" }{TEXT 283 1 "x" }{TEXT -1 3 "', " }{TEXT 282 1 "y" } {TEXT -1 3 "', " }{TEXT 281 1 "z" }{TEXT -1 96 "'] as their Cartesian \+ coordinates, then we found that for the given choice of relative veloc ity " }{TEXT 292 1 "V" }{TEXT -1 2 "=[" }{TEXT 291 1 "V" }{TEXT -1 79 ",0,0] the coordinates orthogonal to the direction of the velocity tra nsform as " }{TEXT 290 1 "y" }{TEXT -1 4 "' = " }{TEXT 289 1 "y" } {TEXT -1 5 " and " }{TEXT 288 1 "z" }{TEXT -1 4 "' = " }{TEXT 287 1 "z " }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "Now we are ready for the interesting case: O1 and O2 hold their rods in the " }{TEXT 293 1 "x" }{TEXT -1 108 "-direction, i.e., the direction of relative motion. Observers O1 and O2 measure the len gth of the rod to be " }{TEXT 294 1 "L" }{TEXT -1 94 "0 (the proper le ngth in the frame where the rod is stationary). Now O2 takes a rod hel d along " }{TEXT 297 1 "x" }{TEXT -1 5 " (or " }{TEXT 296 1 "x" } {TEXT -1 58 "', as these directions coincide) and moves with velocity \+ [" }{TEXT 295 1 "V" }{TEXT -1 285 ",0,0] past O1. O1 has a light clock and counts the number of pulses that pass between the passage of the \+ beginning and the end of the rod (suppose the light clock has a start- stop trigger, and counts many pulses during the passage so that the re sult is accurate. O1 measures a time of " }{TEXT 19 5 "Dt[0]" }{TEXT -1 28 " for the passage of the rod." }}{PARA 0 "" 0 "" {TEXT -1 58 "Th e length of the moving rod is then to be calculated from" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq1:=L[V]=V*Dt[0];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 141 "Now suppose that the moving observer O2 \+ can read the clock pulses in O1's reference frame. The clock is moving with respect to O2 with speed " }{TEXT 298 2 "V " }{TEXT -1 96 "(reme mber that the time dilation is independent of the direction of the vel ocity, it depends on " }{TEXT 300 1 "V" }{TEXT -1 36 "^2). The observe d time is therefore " }{TEXT 19 5 "Dt[V]" }{TEXT -1 60 ".From this per spective the rod is stationary and has length " }{TEXT 299 1 "L" } {TEXT -1 2 "0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq2:=L[0] =V*Dt[V];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "Observer O2 now con verts the the time measured from O1's clock to his proper time. We lif t from the previous section:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "eq6p:=Dt[V]=Dt[0]/sqrt(1-(beta)^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "and substitute into " }{TEXT 19 3 "eq2" }{TEXT -1 14 " to eliminate " }{TEXT 19 5 "Dt[V]" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 39 "eq3:=subs(Dt[V]=solve(eq6p,Dt[V]),eq2);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Now we eliminate " }{TEXT 19 5 "Dt [0]" }{TEXT -1 6 " from " }{TEXT 19 3 "eq1" }{TEXT -1 77 ", and arrive at the desired result which relates the two length measurements:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eq4:=subs(Dt[0]=solve(eq3,Dt [0]),eq1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The length of a rod (length " }{TEXT 19 4 "L[0]" }{TEXT -1 66 " for a stationary observer ) is measured to be of a shorter length " }{TEXT 19 4 "L[V]" }{TEXT -1 44 " when observed by someone moving with speed " }{TEXT 301 1 "V" }{TEXT -1 126 " in the direction of the rod. This length contraction w as hypothesized independently in the 1890ies by FitzGerald and Lorentz ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "The \+ length contraction factor is illustrated in the graph below:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot(sqrt(1-beta^2),beta=0.. 1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 190 "As with time dilation we \+ observe that the effect becomes noticable for speeds exceeding a fifth to a quarter of the speed of light, and becomes really strong for hig hly relativistic objects." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 897 "The length contraction can be used to ex plain the extended half-life of the muons from another perspective. Su ppose an observer travels with the muons produced in the upper atmosph ere (10 km height), i.e., with their speed (about 0.994 c; this known \+ speed is the result of energy balancing in a particle physics reaction ). From this observer's point of view the muons have a half-life as ob served in a laboratory (where they are essentially at rest), namely 1. 5 microseconds. From the graph above one can infer that the vertical f light path of 10 km to be followed by the muons is contracted substant ially, namely to about 1/9 th of the proper length. It becomes clear t hat there is no contradiction but complete consistency between the obs ervations in the two reference frames: from the muon's perspective it \+ is about 1.1 km that have to be traversed. The time it takes to do thi s is essentially " }{TEXT 19 6 "L[V]/c" }{TEXT -1 89 " or 3.7 microsec onds. Therefore, a sizeable fraction of the muons makes it to the grou nd." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 256 "" 0 "" {TEXT 302 22 "Lorentz transformation" }}{PARA 0 "" 0 "" {TEXT -1 364 "We are now ready to formalize the observations of d ilated time and contracted length. As stated before, we need to know h ow the coordinates transform from one reference frame to another so th at we can formulate the laws of physics for observers O1 and O2, and m ake sure that they agree. We have learned that time is not universal, \+ i.e., that we need to transform [" }{TEXT 310 1 "x" }{TEXT -1 2 ", " } {TEXT 309 1 "y" }{TEXT -1 2 ", " }{TEXT 308 1 "z" }{TEXT -1 2 ", " } {TEXT 307 1 "t" }{TEXT -1 8 "] into [" }{TEXT 306 1 "x" }{TEXT -1 3 "' , " }{TEXT 305 1 "y" }{TEXT -1 3 "', " }{TEXT 304 1 "z" }{TEXT -1 3 "' , " }{TEXT 303 1 "t" }{TEXT -1 85 "'] so that O1 and O2 can use his/he r proper time when looking at the laws of physics." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 355 30 "Length/Position transfor mation" }}{PARA 0 "" 0 "" {TEXT -1 118 "To derive the length transform ation consider the following: O1 is at rest, and observes O2 to move w ith a velocity of " }{TEXT 312 1 "V" }{TEXT -1 4 " = [" }{TEXT 311 1 " V" }{TEXT -1 28 ", 0, 0]. Suppose a particle " }{TEXT 313 1 "P" } {TEXT -1 65 " is at rest in the moving frame O2, i.e., it moves with v elocity " }{TEXT 314 1 "V" }{TEXT -1 28 " as observed by O1. At time \+ " }{TEXT 317 1 "t" }{TEXT -1 24 "=0 O1 measures particle " }{TEXT 316 1 "P" }{TEXT -1 10 " to be at " }{TEXT 315 1 "x" }{TEXT -1 42 ", and k nows that this is a measurement of " }{TEXT 19 6 "x[V], " }{TEXT -1 66 "i.e., a contracted length. If O2 were to measure the the position \+ " }{TEXT 318 1 "x" }{TEXT -1 72 "' in the O2 frame, the proper length \+ would be obtained for the position " }{TEXT 19 9 "xp = x[0]" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 98 "Therefore, O1 can state on the basis of the length contraction result obtained above that at time " }{TEXT 319 1 "t" }{TEXT -1 26 "=0 the following is true (" }{TEXT 19 2 "x0" }{TEXT -1 12 " stands for " }{TEXT 321 1 "x" }{TEXT -1 4 " at \+ " }{TEXT 320 1 "t" }{TEXT -1 8 "=0, and " }{TEXT 19 2 "xp" }{TEXT -1 12 " stands for " }{TEXT 322 1 "x" }{TEXT -1 3 "'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eq1:=x0=xp*sqrt(1-beta^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Note that the position of the particle " }{TEXT 323 1 "P" }{TEXT -1 60 " in the O2 reference frame does not cha nge with time, i.e., " }{TEXT 19 2 "xp" }{TEXT -1 95 " is constant. Gi ven that the particle moves with O2, observer O1 can determine the loc ation of " }{TEXT 324 1 "P" }{TEXT -1 13 " at any time " }{TEXT 325 1 "t" }{TEXT -1 54 " by adding the displacement of the origin of O2, i.e .," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq2:=x=rhs(eq1)+V*t; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "The transformation from " } {TEXT 328 1 "x" }{TEXT -1 4 " to " }{TEXT 327 1 "x" }{TEXT -1 43 "' is obtained by solving this equation for " }{TEXT 326 1 "x" }{TEXT -1 2 "':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq3:=xp=solve(eq2,xp );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Note that " }{TEXT 19 3 "eq 2" }{TEXT -1 132 " is not the transformation from O2 to O1, as it mixe s on the RHS the primed position (O2 variable) with the unprimed time \+ (O1 time)." }}{PARA 0 "" 0 "" {TEXT -1 31 "The transformation obtained in " }{TEXT 19 3 "eq3" }{TEXT -1 61 " reduces to the Galilei transfor mation in the limit of small " }{TEXT 19 8 "beta=V/c" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "taylor(rhs(eq3),beta=0,3) ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 220 "This is an important result , as the Newtonian physics relies on this transformation, and represen ts a very good approximation to the correct physics when the particles are moving at small fractions of the speed of light." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 282 "An important property of the position transformation is the fact that the position in the m oving frame O2 is obtained from a linear combination of position and t ime in the stationary frame. The important question is: how does it lo ok the other way around, i.e., how would we obtain " }{TEXT 331 1 "x" }{TEXT -1 6 " from " }{TEXT 330 1 "x" }{TEXT -1 6 "' and " }{TEXT 329 1 "t" }{TEXT -1 2 "'?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "How can we obtain the inverse transformation for the p osition?" }}{PARA 0 "" 0 "" {TEXT -1 118 "One needs to repeat the reas oning carried out here from the other perspective. It is incorrect to \+ substitute the time " }{TEXT 333 1 "t" }{TEXT -1 5 " for " }{TEXT 332 1 "t" }{TEXT -1 5 "' in " }{TEXT 19 4 "eq2 " }{TEXT 2 63 "under the as sumption that they are related by a time dilation (" }{TEXT 335 1 "t" }{TEXT 2 42 " is not a time interval observed by O2 as " }{TEXT 334 1 "t" }{TEXT 2 3 "'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "WRON Geq4:=subs(t=tp/sqrt(1-beta^2),eq2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "The repeated reasoning to obtain the correct inverse transforma tion goes as follows:" }}{PARA 0 "" 0 "" {TEXT -1 36 "The particle is \+ located at position " }{TEXT 19 2 "xp" }{TEXT -1 15 " for all times " }{TEXT 19 2 "tp" }{TEXT -1 106 ". O2 observes the origin O1 to move wi th velocity -V t'. He realizes that he is observing the coordinate " }{TEXT 336 1 "x" }{TEXT -1 39 " in contracted form and then concludes: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq4:=x*sqrt(1-beta^2)=x p+V*tp;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The displacement " } {TEXT 337 1 "x" }{TEXT -1 55 " grows with time as O2 zooms by O1, whic h explains why " }{TEXT 339 3 "V t" }{TEXT -1 14 "' is added to " } {TEXT 338 1 "x" }{TEXT -1 2 "'." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq4:=x=solve(eq4,x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 " This result teaches us that we have to be careful in special relativit y. Blind substitution of equations will not lead to the correct result s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 354 19 "Time transformation" }}{PARA 0 "" 0 "" {TEXT -1 127 "Now that \+ we know how the position transforms between the two reference frames, \+ we proceed with the correct time transformation." }}{PARA 0 "" 0 "" {TEXT -1 36 "We can achieve the goal of relating " }{TEXT 341 1 "t" } {TEXT -1 5 " and " }{TEXT 340 1 "t" }{TEXT -1 77 "' by simply combinin g the two transformation obtained above. If we eliminate " }{TEXT 342 1 "x" }{TEXT -1 15 "' by combining " }{TEXT 19 3 "eq4" }{TEXT -1 5 " a nd " }{TEXT 19 3 "eq3" }{TEXT -1 38 ", we arrive at the following equa tion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq5:=rhs(eq3)=solv e(eq4,xp);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq6:=tp=expan d(solve(eq5,tp));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "We observe t hat " }{TEXT 343 1 "t" }{TEXT -1 85 "', i.e., the proper time in the b oosted frame is obtained by a linear combination of " }{TEXT 345 1 "x " }{TEXT -1 5 " and " }{TEXT 344 1 "t" }{TEXT -1 25 " in the stationar y frame." }}{PARA 0 "" 0 "" {TEXT -1 49 "We can check that in the non- relativistic limit (" }{TEXT 19 10 "beta=V/c=0" }{TEXT -1 64 ") the ti me becomes universal (as assumed by Galilei and Newton):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "taylor(rhs(eq6),beta=0,3);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "How can we obtain the inverse tran sformation?" }}{PARA 0 "" 0 "" {TEXT -1 54 "Following the same ideas, \+ we should seek to eliminate " }{TEXT 346 1 "x" }{TEXT -1 90 " from the transformations that go from the boosted to the stationary frame and \+ vice versa." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq7:=x=solve (eq3,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "eq8:=rhs(eq7)=r hs(eq4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq9:=t=expand(s olve(eq8,t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Carry out the T aylor expansion of this result to verify the Galilean limit as done be fore for the inverse transformation. The definition of " }{TEXT 19 8 " beta=V/c" }{TEXT -1 21 " should be remebered." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 267 "We complete the \+ section with a graphical representation of our results. To understand \+ special relativity one introduces a so-called space-time diagram. This is a two-dimensional graph for particles moving in one spatial dimens ion. To have consistent axes, one plots c" }{TEXT 357 1 "t" }{TEXT -1 320 " on the vertical axis, and the particle position on the horizonta l axis. Particles moving with constant speeds follow straight lines in this diagram. It is convenient to express their speeds as a fraction \+ of the speed of light: the slope of the line for a particle is given b y the inverse of the speed ratio, i.e., by c/" }{TEXT 358 1 "V" } {TEXT -1 246 ". We start the particles at the origin. Obviously no slo pes less than one are allowed (unless we were to consider hypothetical tachyons). We draw space trajectories for particles moving with c/2, \+ c/3, -c/4 and c/5 in the form of parametric plots:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "x1:=1/2*ct: x2:=1/3*ct: x3:=-1/4*ct: x4:=1/5*ct: x5:= ct: x6:=-ct:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "Tr:=[[x1,c t],[x2,ct],[x3,ct],[x4,ct],[x5,ct],[x6,ct]]: coltable:=[red,blue,green ,black,magenta,magenta]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Ptr:=i->plot([op(Tr[i]),ct=0..1],color=coltable[i]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "display([seq(Ptr(i),i=1..6)],labels =[x,ct],scaling=constrained,title=\"spacetime: 1+1 dim\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 460 "For accelerating particles moving with v ariable speeds the trajectories become curves, and the slope at any po int along the trajectory allows one to deduce the instantaneous speed \+ of the particle. The diagram shows only forward propagation in time, b ut it can be extended to negative times. The lines displayed using the magenta color represent photon trajectories (the muons discussed abov e would practically follow these) moving to the left and to the right. " }}{PARA 0 "" 0 "" {TEXT -1 437 "If more space dimensions were added \+ to the diagram (e.g., one more, i.e., the y-axis coming out of the pla ne), the limit imposed by all allowed photon trajectories becomes a su rface (a hypersurface in three spatial dimensions) called the lightcon e. A particle trajectory is called a worldline when one refers to the \+ spacetime diagram. The particle's worldline has to remain inside the l ightcone centered at the origin (we assume that at " }{TEXT 360 1 "t" }{TEXT -1 11 "=0 we have " }{TEXT 359 1 "x" }{TEXT -1 21 "=0 for the p article)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 361 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 74 "Extend the spaceti me diagram to two spatial dimensions. Use the help page " }{TEXT 19 11 "?spacecurve" }{TEXT -1 154 " to find out how to generate some worl dlines. Find worldlines that trace out the lightcone in 2+1 dimensions . Draw world lines for accelerating particles." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 194 "We can use the s pacetime diagram to illustrate the concept of simultaneity in special \+ relativity. Imagine a stationary reference frame that we consider to b e the platform of a train station. At " }{TEXT 362 1 "t" }{TEXT -1 30 "=0 a rocket train moving with " }{TEXT 363 1 "V" }{TEXT -1 171 "=c/2 \+ passes through the station, and at the same time light flashes are emi tted from light clocks located at opposite ends of the platform (the c locks are synchronized at " }{TEXT 365 1 "t" }{TEXT -1 1 "=" }{TEXT 364 1 "t" }{TEXT -1 78 "'=0). In the stationary reference frame the fl ashes correspond to the points [" }{TEXT 366 1 "A" }{TEXT -1 9 ",0] an d [" }{TEXT 367 1 "B" }{TEXT -1 40 ",0] in spacetime (called events), \+ where " }{TEXT 369 1 "B" }{TEXT -1 2 "=-" }{TEXT 370 1 "A" }{TEXT -1 45 " if the platform extends symmetrically about " }{TEXT 368 1 "x" } {TEXT -1 3 "=0." }}{PARA 0 "" 0 "" {TEXT -1 164 "It is straightforward to figure out to which spacetime points the two events which are simu ltaneous to observer O1 (located in the middle of the platform, i.e., \+ at " }{TEXT 371 1 "x" }{TEXT -1 100 "=0) are transformed for observer \+ O2 travelling with the rocket train. The transformation is given as" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eq3,eq6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs(beta=V/c,V=c/2,t=0,x=A,eq3);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(beta=V/c,V=c/2,t=0,x=-A ,eq3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "More importantly we wis h to know the times:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "sub s(beta=V/c,V=c/2,t=0,x=A,eq6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(beta=V/c,V=c/2,t=0,x=-A,eq6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "Clearly the two light flashes are not simultaneous from \+ O2's point of view. O2 observes the clock at the entrance to the stati on ([" }{TEXT 373 1 "B" }{TEXT -1 7 ", 0]=[-" }{TEXT 372 1 "A" }{TEXT -1 191 ", 0]) flashing much later (as it flashes at the time when O2 p asses the center of the platform and has to catch up with him; meanwhi le O2 races towards the clock at the end of the platform ([" }{TEXT 374 1 "A" }{TEXT -1 154 ", 0]), and encounters the pulse from the cloc k located there at an earlier time; in fact, a time that predates the \+ passage of the center of the platform (" }{TEXT 377 1 "t" }{TEXT -1 1 "=" }{TEXT 376 1 "t" }{TEXT -1 25 "'=0) and the flash from [" }{TEXT 375 1 "B" }{TEXT -1 53 ", 0]. (Example from Paul Tipler and Ralph Llew ellyn: " }{TEXT 378 14 "Modern Physics" }{TEXT -1 25 ", 3rd ed., Freem an 1999)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "We can ask which curves are traced out by the transformation to t he moving reference frame. We pick the chosen velocity " }{TEXT 379 1 "V" }{TEXT -1 29 "=c/2 and draw the result for " }{TEXT 19 2 "xp" } {TEXT -1 5 " and " }{TEXT 19 2 "tp" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 12 "Each point [" }{TEXT 381 1 "x" }{TEXT -1 3 ", c" }{TEXT 380 1 "t" }{TEXT -1 28 "] is mapped to a new point [" }{TEXT 383 1 "x " }{TEXT -1 4 "', c" }{TEXT 382 1 "t" }{TEXT -1 90 "']. Let us first d efine a map that carries out the transformation for the given choice o f " }{TEXT 19 4 "beta" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 38 " We choose units sich that c=1, i.e., c" }{TEXT 385 1 "t" }{TEXT -1 3 " = " }{TEXT 384 1 "t" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eq3,eq6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "LT:=unapply([rhs(subs(beta=V/c,V=c/2,c=1,eq3)),rhs(subs(beta=V/c,V=c/ 2,c=1,eq6))],x,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "LT(A,0 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(1,0));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(0,1));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "We can illustrate the correspondence of ( " }{TEXT 389 1 "x" }{TEXT -1 2 ", " }{TEXT 388 1 "t" }{TEXT -1 13 ") p oints to (" }{TEXT 387 1 "x" }{TEXT -1 3 "', " }{TEXT 386 1 "t" } {TEXT -1 27 "') points by graphing them:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "PtO11:=[seq([i*0.5,0],i=-6..6)];" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 59 "PtO21:=[seq(map(evalf,LT(op(PtO11[i]))),i=1. .nops(PtO11))];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "PO11:=pl ot(PtO11,style=point,symbol=diamond,color=red):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 56 "PO21:=plot(PtO21,style=point,symbol=diamond,co lor=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "PtO12:=[seq([ 0,i*0.5],i=-6..6)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "PtO2 2:=[seq(map(evalf,LT(op(PtO12[i]))),i=1..nops(PtO12))];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "PO12:=plot(PtO12,style=point,symbol =cross,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "PO22: =plot(PtO22,style=point,symbol=cross,color=blue):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 102 "display([PO11,PO21,PO12,PO22],axes=boxed,sc aling=constrained,labels=[x,ct],title=\"spacetime 1+1 dim\");" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 183 "We have constructed a graph that \+ displays how sequences of events along the position axis and along the time axis respectively are mapped under the transformation law for th e case of " }{TEXT 390 1 "V" }{TEXT -1 10 "=c/2. The " }{TEXT 420 1 "x " }{TEXT -1 51 "-axis is apparently mapped onto a line with slope -" } {TEXT 422 1 "V" }{TEXT -1 12 "/2, and the " }{TEXT 421 1 "t" }{TEXT -1 32 "-axis into a line with slope -1/" }{TEXT 423 1 "V" }{TEXT -1 165 ". We can check out this observation by calculating the parametric representation of the two lines in question. We begin with the genera l case (mapping of any point [" }{TEXT 425 1 "x" }{TEXT -1 2 ", " } {TEXT 424 1 "t" }{TEXT -1 35 "]), and then specify our two lines." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(x,t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(x,0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "slope:=%[2]/%[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(0,t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "slope:=%[2]/%[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "An important observation is that the transformation represents \+ a shear: we start with orthogonal " }{TEXT 429 1 "x" }{TEXT -1 6 " and c" }{TEXT 428 1 "t" }{TEXT -1 47 " axes we wind up after the transfor mation with " }{TEXT 431 1 "x" }{TEXT -1 7 "' and c" }{TEXT 430 1 "t" }{TEXT -1 56 "' axes that are non-orthogonal, and the length interval \+ " }{TEXT 19 23 "ds=sqrt(dx^2+ c^2*dt^2)" }{TEXT -1 189 " is not preser ved (this is demonstrated by the stretch of the axes). These circumsta nces will require some further explorations in transformation theory ( beyond the well-known rotations in " }{TEXT 432 1 "n" }{TEXT -1 20 "-d imensional space)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 117 "It is worth to examine the transformed events in the spa cetime diagram in some detail: the points along the original " }{TEXT 464 1 "x" }{TEXT -1 46 "-axis represent a set of simultaneous events ( " }{TEXT 465 1 "t" }{TEXT -1 132 "=0) in the stationary reference fram e. These get transformed into a set of points that form a line along w hich the transformed time " }{TEXT 466 1 "t" }{TEXT -1 154 "' varies f rom point to point (blue diamonds), i.e., they are not simultaneous in the boosted frame. This demonstrates that this line cannot represent \+ the " }{TEXT 467 1 "x" }{TEXT -1 28 "' axis in the boosted frame." }}} {EXCHG {PARA 0 "" 0 "" {TEXT 427 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 63 "Repeat the above steps for different choices of boost vel ocity " }{TEXT 426 1 "V" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "The diagram produced above i s not what is usually shown, i.e., it does not show the (" }{TEXT 469 1 "x" }{TEXT -1 4 "', c" }{TEXT 468 1 "t" }{TEXT -1 106 "') coordinate axes (e.g., in fig. 1-22 in Tipler and Llewellyn). We have transforme d the points along the " }{TEXT 439 1 "x" }{TEXT -1 27 "-axis (which c orrespond to " }{TEXT 440 1 "t" }{TEXT -1 30 "=0) and the points along the c" }{TEXT 441 1 "t" }{TEXT -1 27 "-axis (which correspond to " } {TEXT 442 1 "x" }{TEXT -1 165 "=0) according to the transformation law . We notice from the points used for graphing the transformed axes tha t they do not correspond to the axes which defined the (" }{TEXT 455 1 "x" }{TEXT -1 4 "', c" }{TEXT 454 1 "t" }{TEXT -1 27 "') coordinate \+ system. The (" }{TEXT 457 1 "x" }{TEXT -1 38 "')-axis should have the \+ property that " }{TEXT 456 1 "t" }{TEXT -1 53 "'=0 for all points alon g it and vice versa for the (c" }{TEXT 458 1 "t" }{TEXT -1 8 "')-axis. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "We n ow calculate for the boosted frame the location of the axes which are \+ specified as follows: for the " }{TEXT 444 1 "x" }{TEXT -1 44 "' axis \+ it is the locus of all points where c" }{TEXT 443 1 "t" }{TEXT -1 14 " '=0; for the c" }{TEXT 446 1 "t" }{TEXT -1 29 "' axis it is the locus \+ where " }{TEXT 445 1 "x" }{TEXT -1 58 "' vanishes. This is consistent \+ with the definition of the " }{TEXT 448 1 "x" }{TEXT -1 6 " and c" } {TEXT 447 1 "t" }{TEXT -1 6 " axes." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eq3,eq6;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "We fi nd the axes for the case of " }{TEXT 453 1 "V" }{TEXT -1 18 "=1/2 from setting " }{TEXT 452 1 "x" }{TEXT -1 6 "' and " }{TEXT 451 1 "t" } {TEXT -1 37 "' to zero respectively, and graphing " }{TEXT 450 1 "t" } {TEXT -1 8 " versus " }{TEXT 449 1 "x" }{TEXT -1 19 " for the two case s." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "PO11:=plot(PtO11,styl e=point,symbol=diamond,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "tpeq0:=solve(subs(beta=V/c,V=c/2,c=1,rhs(eq6)),t);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "PO21:=plot(tpeq0,x=-4..4,c olor=red,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "P O12:=plot(PtO12,style=point,symbol=cross,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "xpeq0:=solve(subs(beta=V/c,V=c/2,c= 1,rhs(eq3)),t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "PO22:=pl ot(xpeq0,x=-4..4,color=blue,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 129 "display([PO11,PO12,PO21,PO22],axes=boxed,scaling=c onstrained,labels=[x,ct],title=\"spacetime 1+1 dim\",view=[-3.5..3.5,- 3.5..3.5]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "To illustrate furt her the new coordinate system (which represents the (" }{TEXT 459 1 "x " }{TEXT -1 10 "')- and (c" }{TEXT 460 1 "t" }{TEXT -1 69 "')-axes, i. e., the coordinate axes in the boosted refernce frame for " }{TEXT 461 1 "V" }{TEXT -1 40 "=c/2) we can draw the lines of constant " } {TEXT 463 1 "x" }{TEXT -1 7 "' and c" }{TEXT 462 1 "t" }{TEXT -1 53 "' , e.g., for a spacing of one unit along either axis." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "tpeq1:=solve(subs(beta=V/c,V=c/2,c=1,rhs( eq6)=1),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Ptpeq1:=plot(tpeq1,x =-4..4,color=red,thickness=1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "tpeq2:=solve(subs(beta=V/c,V=c/2,c=1,rhs(eq6)=2),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Ptpeq2:=plot(tpeq2,x=-4..4,color=red,thic kness=1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "tpeq3:=solve(s ubs(beta=V/c,V=c/2,c=1,rhs(eq6)=3),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Ptpeq3:=plot(tpeq3,x=-4..4,color=red,thickness=1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "xpeq1:=solve(subs(beta=V/c,V=c/2,c= 1,rhs(eq3)=1),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "Pxpeq1:=plot(x peq1,x=-4..4,color=blue,thickness=1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "xpeq2:=solve(subs(beta=V/c,V=c/2,c=1,rhs(eq3)=2),t); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "Pxpeq2:=plot(xpeq2,x=-4..4,colo r=blue,thickness=1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "xpe q3:=solve(subs(beta=V/c,V=c/2,c=1,rhs(eq3)=3),t);" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 51 "Pxpeq3:=plot(xpeq3,x=-4..4,color=blue,thickness=1): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 150 "display([PO21,PO22,Ptp eq1,Ptpeq2,Ptpeq3,Pxpeq1,Pxpeq2,Pxpeq3],scaling=constrained,labels=[x, ct],title=\"spacetime 1+1 dim\",view=[-4.5..4.5,-4.5..4.5]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 421 "Now we have a coordinate grid in \+ the boosted frame. It can be used to understand the passage of events \+ (such as the problem of light pulses flashing simultaneously at the tw o ends of a platform, and how this is perceived by an observer on the \+ rocket train). Points along the red lines are simultaeous in the boost ed frame. They are clearly not simultaneous for the stationary observe r, since they are not parallel to the " }{TEXT 470 1 "x" }{TEXT -1 6 " -axis." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 199 "The previously discussed phenomena of time dilation and length co ntraction can be easily demonstrated on the diagram. Let us construct \+ the worldline for a light flash emitted in the boosted frame at " } {TEXT 472 1 "x" }{TEXT -1 52 "'=0, and then reflected back by a mirror located at " }{TEXT 471 1 "x" }{TEXT -1 178 "'=1 m. The worldline for such a light flash is represented by a straight line starting at the \+ origin with a slope of 1. It reaches the mirror at the intercept of th e grid lines c" }{TEXT 474 1 "t" }{TEXT -1 8 "'=1 and " }{TEXT 473 1 " x" }{TEXT -1 51 "'=1. We find this point in the stationary frame as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eq3,eq6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "sol:=solve(\{subs(beta=V/c,V=c/2,c=1,rhs( eq3)=1),subs(beta=V/c,V=c/2,c=1,rhs(eq6)=1)\},\{x,t\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "eq6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "xM1:=subs(sol,x); tM1:=subs(sol,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "Pflash1:=plot([[0,0],[xM1,tM1]],style=lin e,color=green,thickness=3):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "N ow we consider the reflected flash: it is going backwards, i.e., has t o arrive at a final location " }{TEXT 475 1 "x" }{TEXT -1 6 "' =0, " } {TEXT 476 1 "t" }{TEXT -1 173 "'=2 (in the moving reference frame we i gnore the motion of the reference frame and use the constancy of the s peed of light c). In the stationary frame this looks as follows:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "sol:=solve(\{subs(beta=V/c,V =c/2,c=1,rhs(eq3)=0),subs(beta=V/c,V=c/2,c=1,rhs(eq6)=2)\},\{x,t\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "xM2:=subs(sol,x); tM2:=su bs(sol,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "Pflash2:=plot ([[xM1,tM1],[xM2,tM2]],style=line,color=magenta,thickness=3):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 160 "display([PO21,PO22,Ptpeq1,P tpeq2,Ptpeq3,Pxpeq1,Pxpeq2,Pxpeq3,Pflash1,Pflash2],scaling=constrained ,labels=[x,ct],title=\"spacetime 1+1 dim\",view=[0..4.5,0..4.5]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 477 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 194 "Repeat the calculation and graph for other mirror distan ces than 1 m. Repeat the whole section for reference frames with diffe rent boosts: from less relativistic to more relativistic situations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 478 11 "Exerci se 4:" }}{PARA 0 "" 0 "" {TEXT -1 312 "In the problem discussed above \+ the worldlines were constructed for a light pulse, i.e., the speed in \+ the moving frame was given as c (with reversed direction for the secon d half). Consider another textbook problem: an impregnated elephant is sent off in a spaceship at time t=t'=0 travelling at a given speed of " }{TEXT 479 1 "V" }{TEXT -1 172 "=0.75c. The gestation period is kno wn to be 21 months. At the time of birth a radiosignal is sent back to ward Earth to announce the birth. When will the radiosignal arrive?" } }{PARA 0 "" 0 "" {TEXT -1 514 "This is a two-step problem: in step 1 o ne determines the time dilation, i.e., one determines at what point in time on Earth the elephant will give birth; in step 2 one calculates \+ the distance the spaceship has travelled during this time; in step 3 o ne finds out how long it takes for the radiosignal to travel (the two \+ times are 31.7 and 23.8 months respectively). Determine these times fr om the relevant formulae. Draw the worldlines for the two processes (w hy does the worldline for the elephant lie along the (c" }{TEXT 480 1 "t" }{TEXT -1 11 "')-axis?). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 481 11 "Exercise 5:" }}{PARA 0 "" 0 "" {TEXT -1 104 "Illustrate length contraction using the spacetime diagram. Draw a meterstick at rest in a moving frame (" }{TEXT 482 2 "V " }{TEXT -1 47 "= 0.75 c, or some other choice) extending from " }}{PARA 0 "" 0 " " {TEXT 483 1 "x" }{TEXT -1 10 "1'=1 m to " }{TEXT 484 1 "x" }{TEXT -1 96 "2'=2 m. A measurement of the meterstick in both frames implies \+ that it is to be drawn along the " }{TEXT 485 1 "x" }{TEXT -1 19 "'-ax is (choice of c" }{TEXT 486 1 "t" }{TEXT -1 84 "'=0 to determine its l ength), and that one is interested in the projection onto the " } {TEXT 487 1 "x" }{TEXT -1 69 "-axis in the rest frame (or any parallel axis to use some other time " }{TEXT 488 1 "t" }{TEXT -1 44 ", nevert heless the same time for both ends)." }}{PARA 0 "" 0 "" {TEXT -1 177 " Construct the relevant spacetimediagram and use the point-and-click in terface to read off the length in the rest frame. Carry out the length contraction calculation and compare." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 489 36 "Appearance of rapidly moving objects" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 236 "An interesting problem concerns the question of appearance of sol id objects to a moving observer. Length contraction occurs only in the direction of motion. Therefore the shape of a two-dimensional object \+ (e.g., a profile) is distorted." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 114 "Suppose a square which is rotated by 30 \+ degrees resides in a reference frame that moves with some speed along \+ the " }{TEXT 490 1 "x" }{TEXT -1 11 "-direction." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "with(plottools):\nc := rectangle([1,1], [3,3], color=red):\ncr:=rotat e(c,Pi/6,[1,1]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "display (cr, scaling=constrained,view=[0..5,0..5],labels=[xp,yp]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "A stationary observer (" }{TEXT 493 1 "x " }{TEXT -1 2 ", " }{TEXT 494 1 "y" }{TEXT -1 58 ")-coordinate system \+ notices the length contraction of the " }{TEXT 491 1 "x" }{TEXT -1 24 "-coordinates, while the " }{TEXT 492 1 "y" }{TEXT -1 30 "-coordinates remain the same (" }{TEXT 496 1 "y" }{TEXT -1 1 "=" }{TEXT 495 1 "y" }{TEXT -1 158 "'). The two sides that determine the shape can be deter mined relative to the vertex [1,1]. Let us first find the other vertic es in the moving reference frame." }}{PARA 0 "" 0 "" {TEXT -1 84 "A ro tation matrix that rotates with respect to the origin by 30 degrees is given by:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "R30:=[[cos(Pi /6),-sin(Pi/6)],[sin(Pi/6),cos(Pi/6)]];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "To transform the vertices, we have to remove the dispalc ement of the first vertex, apply the rotation, and then add the displa cement again:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linal g):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Pt2p:=evalm([1,1] + \+ R30 &* ([3,1]-[1,1]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "P t3p:=evalm([1,1] + R30 &* ([3,3]-[1,1]));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 42 "Pt4p:=evalm([1,1] + R30 &* ([1,3]-[1,1]));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "We pick a boost velocity as a frac tion of the speed of light c:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "beta:=8/10;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "Let us apply a Lorentz transformation to the observer's reference frame. The " } {TEXT 497 1 "x" }{TEXT -1 64 "-ccordinates are contracted according th e beta-dependent factor." }}{PARA 0 "" 0 "" {TEXT -1 34 "We need to un protect the variable " }{TEXT 19 5 "gamma" }{TEXT -1 108 " in Maple if we wish to use it for our own purposes (instead of referring to the E uler-Mascheroni constant):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "unprotect(gamma);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ga mma:=1/sqrt(1-beta^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "P t1:=[1/gamma,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Pt2:=[P t2p[1]/gamma,Pt2p[2]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "P t3:=[Pt3p[1]/gamma,Pt3p[2]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Pt4:=[Pt4p[1]/gamma,Pt4p[2]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "display(polygon([Pt1,Pt2,Pt3,Pt4], color=red, linest yle=3, thickness=2),view=[0..5,0..5],scaling=constrained,labels=[x,y]) ;\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "It is evident that the ro tated square profile has turned into a parallelogram for the stationar y observer who sees the profile whiz by." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 535 11 "Exercise 6:" }}{PARA 0 "" 0 "" {TEXT -1 127 "Explore the appearance of the rotated square shape for d ifferent orientations. What happens as the boost velocity approaches c ?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 286 "The observation of three-dimensional objects requires an addition al consideration: the time when the light left the observed object has to be taken into account. The photons which arrive simultaneously at \+ the moving observer's eye: the ones arriving from the more deeply loca ted parts (" }{TEXT 498 1 "z" }{TEXT -1 82 "-direction) must have left at an earlier time when the observer was at a previous " }{TEXT 499 1 "x" }{TEXT -1 154 "-location. As a result the observer sees a tilted object (or rather sees the side face that would be hidden to a statio nary observer of the solid object)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 353 23 "Velocity transfo rmation" }}{PARA 0 "" 0 "" {TEXT -1 280 "An interesting remaining ques tion is the transformation property of velocities. Remember that our m ain goal is to understand the constancy of the speed of light, i.e., t o understand how the simple additivity of velocities in the Galilei tr ansformation is modified for high speeds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 337 "In the reference frames O1 and O2 (the latter denoted by primed coordinates), we can define the velocit ies as the derivatives of proper length with respect to proper time. T he posed question then is how to relate the two measurements of these \+ quantities. This can be achieved by differentiating the transformation s for position and time." }}{PARA 0 "" 0 "" {TEXT -1 36 "First we summ arize the main results:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "L T1:=eq3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "LT2:=eq6;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "LT1p:=eq4;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "LT2p:=eq9;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "The differentiation is accomplished by replacing the var iables by differentials. This we do by a substitution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "LT1d:=subs(xp=dxp,x=dx,t=dt,LT1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "LT2d:=subs(tp=dtp,x=dx,t= dt,LT2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "We form the velocity expression in the boosted frame (think of dxp and dtp as finite diffe rences, and do not worry about mathematical rigour for a moment):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "VT1:=vp=rhs(LT1d)/rhs(LT2d); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "To introduce the velocity in \+ the stationary (unprimed, O1) frame we make the substitution:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "VT1:=expand(simplify(subs(dx =v*dt,VT1)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "taylor(rhs (VT1),beta=0,3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "This allows u s to transform the velocity of a particle in the O1 frame, i.e., " } {TEXT 347 1 "v" }{TEXT -1 59 " to a boosted frame O2. Since the booste d frame moves with " }{TEXT 349 1 "V" }{TEXT -1 2 "=[" }{TEXT 348 1 "V " }{TEXT -1 54 ", 0, 0] , the Galilei transformation (obtained in the \+ " }{TEXT 19 6 "beta=0" }{TEXT -1 28 " limit) shows a velocity of " } {TEXT 352 1 "v" }{TEXT -1 4 "' = " }{TEXT 351 1 "v" }{TEXT -1 3 " - " }{TEXT 350 1 "V" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "We can display for fixed boost velocity (" }{TEXT 19 4 "beta" } {TEXT -1 2 ", " }{TEXT 391 1 "V" }{TEXT -1 115 " determined) how the t ransformed velocity depends on the velocity in the original reference \+ frame. Suppose we pick " }{TEXT 19 8 "beta=1/2" }{TEXT -1 8 ", i.e., \+ " }{TEXT 392 1 "V" }{TEXT -1 5 "=c/2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "VT1_1:=simplify(subs(beta=1/3,V=c/3,c=1,VT1));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "VT1_2:=simplify(subs(beta=1/ 2,V=c/2,c=1,VT1)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "VT1_3 :=simplify(subs(beta=2/3,V=c*2/3,c=1,VT1)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 177 "PL1:=plot([rhs(VT1_1),rhs(VT1_2),rhs(VT1_3)],v=-1 ..1,color=[red,blue,green],title=\"Lorentz velocity transformation\",l abels=[v,vp],scaling=constrained,thickness=3): display(PL1);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "We can observe how the transforma tion allows for a small range of boost and particle velocity values fo r which additivity works: the " }{TEXT 393 1 "y" }{TEXT -1 86 " interc epts allow one to read off the chosen boost velocities (the crossings \+ list the " }{TEXT 396 2 "v'" }{TEXT -1 28 " values for which result in " }{TEXT 395 1 "v" }{TEXT -1 45 "=0 in the rest frame for boost veloc ities of " }{TEXT 394 1 "V" }{TEXT -1 25 "=c/3, c/2 and 2c/3); the " } {TEXT 399 1 "x" }{TEXT -1 87 " intercepts correspond to the points whe re the boost is offset by particle motion with " }{TEXT 398 1 "v" } {TEXT -1 1 "=" }{TEXT 397 1 "V" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 46 "Let us plot for comparison the Galilei result:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 181 "PL2:=plot([v-1/3,v-1/2,v-2/ 3],v=-1..1,color=[red,blue,green],title=\"Galilei velocity transformat ion\",labels=[v,vp],scaling=constrained,view=[-1..1,-1..1],thickness=2 ): display(PL2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "display (PL1,PL2,title=\"Galilei-Lorentz comparison\",axes=boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "The diagram shows that the Galilei trans formation is only adequate for the smallest boost velocity shown (" } {TEXT 400 1 "V" }{TEXT -1 55 "=c/3), when the particle velocity is res tricted to 0 < " }{TEXT 401 1 "v" }{TEXT -1 32 " < c/3 (which results \+ in -c/3 > " }{TEXT 402 1 "v" }{TEXT -1 121 "' > 0). For the cases of l arger boosts of the reference frame the Galilei transformation never w orks (not even for small " }{TEXT 409 1 "v" }{TEXT -1 242 " !), apart \+ from the crossings which correspond to the cases where the particle ve locity is zero, or where the boost velocity and particle velocity add \+ up to zero (these are trivial cases for which adding velocities repres ent the only option)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "On the other hand the graphs show that for particle ve locities in the range -c < " }{TEXT 403 1 "v" }{TEXT -1 57 " < c the t ransformed velocities remain in the range -c < " }{TEXT 404 1 "v" } {TEXT -1 41 "' < c irrespective of the boost velocity " }{TEXT 405 1 " V" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 408 11 "Exercise 7:" }}{PARA 0 "" 0 "" {TEXT -1 79 "Observe the \+ behaviour of the transformation in the limit as the boost velocity " } {TEXT 406 1 "V" }{TEXT -1 47 " approaches c. What result does one obta in for " }{TEXT 407 1 "V" }{TEXT -1 5 " < 0?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 284 "The discussion of the \+ velocity transformation is not complete yet. We have to consider the t ransformation of the velocity components that are orthogonal to the di rection of the boost. They are affected due to the time transformation . Let us derive the velocity transformation for the " }{TEXT 410 1 "y " }{TEXT -1 11 "-component:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "LT3:=yp=y;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "LT2;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "The differentiation is accomplish ed by replacing the variables by differentials. This we do by a substi tution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "LT3d:=subs(yp=dy p,y=dy,LT3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "LT2d;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "VT2:=vyp=simplify(subs(dy=vy *dt,dx=vx*dt,rhs(LT3d)/rhs(LT2d)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "An analogous result is obtained for the relationship between th e " }{TEXT 411 1 "z" }{TEXT -1 128 " components of the velocity vecto r. Physically the meaning of this result is the following: a particle \+ moves with some velocity " }{TEXT 414 2 "v " }{TEXT -1 3 "= [" }{TEXT 413 2 "vx" }{TEXT -1 2 ", " }{TEXT 412 2 "vy" }{TEXT -1 84 ", 0] in a \+ stationary rest frame. An observer in a boosted frame with boost veloc ity " }{TEXT 416 1 "V" }{TEXT -1 4 " = [" }{TEXT 415 1 "V" }{TEXT -1 35 ", 0, 0] observes a velocity vector " }{TEXT 419 1 "v" }{TEXT -1 5 "' = [" }{TEXT 418 2 "vx" }{TEXT -1 3 "', " }{TEXT 417 2 "vy" }{TEXT -1 70 "', 0] with the transformed velocity components given by the equ ations " }{TEXT 19 3 "VT1" }{TEXT -1 5 " and " }{TEXT 19 3 "VT2" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 192 "The non-vanishing compo nent perpendicular to the boost direction depends in its transformatio n properties on the boost velocity and the particle's velocity in this direction in the rest frame." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "R1:=subs(beta=V/c,V=c/2,c=1,rhs(VT2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "plot3d(R1,vx=-1..1,vy=-1..1,axes=boxed,shading=z hue,style=patchnogrid,orientation=[160,40]);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 93 "At first sight one might think that there is a problem \+ with the resulting velocity component " }{TEXT 433 2 "vy" }{TEXT -1 123 "' exceeding the speed of light. One has to bear in mind, however, that the range of vy is restricted for a given choice of " }{TEXT 434 2 "vx" }{TEXT -1 73 " (the magnitude of the velocity vector is lim ited by the speed of light)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 165 "vx0:=0.7; plot(subs(vx=vx0,R1),vy=-evalf(sqrt(1-vx0^2))..evalf( sqrt(1-vx0^2)),labels=[vy,vyp],title=cat(\"transformation of vy to vy' for vx= \",convert(vx0,string)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 437 11 "Exercise 8:" }}{PARA 0 "" 0 "" {TEXT -1 64 "Check whether the \+ resulting components in the transformed frame " }{TEXT 435 2 "vx" } {TEXT -1 6 "' and " }{TEXT 436 2 "vy" }{TEXT -1 55 "' together result \+ in speeds limited by c. Use equation " }{TEXT 19 3 "VT1" }{TEXT -1 11 " to obtain " }{TEXT 438 2 "vx" }{TEXT -1 2 "'." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 500 22 "The spacetime in terval" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 321 "The length and time elements do not remain constant under Lorentz boosts. The length interval is contracted by a factor of gamma, while the time interval is dilated by the same factor. The question that ar ises is whether there is an interval involving space and time that rem ains invariant under a Lorentz transformation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 138 "The answer is that such \+ an interval can be found, although it looks unnatural at first. We spe cify it for a case of one spatial dimension." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "c:='c':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "STi:=ds^2=(c*dt)^2-dx^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The separation " }{TEXT 19 4 "ds^2" }{TEXT -1 49 " can be positive, n egative or zero. The interval " }{TEXT 19 2 "ds" }{TEXT -1 489 " in sp acetime denotes the separation between two events. The claim is that i t is invariant under Lorentz transformations; this is of interest as i t will allow to make statements that are valid for all observers. The \+ interval is called lightlike when ds=0, as it represents the position- time relationship of a particle moving with c. For ds>0 the time separ ation dominates, and the interval is called timelike; for ds<0 the spa ce separation dominates, and the interval is called spacelike. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 354 "Material particles have been observed only to travel with speeds smaller than \+ c. They follow timelike worldlines. We can use example 4 to illustrate the invariance of ds as follows: consider a freshly impregnated eleph ant in a spaceship travelling at some value of beta. In the elephant's reference frame birth is given after a 21-month gestation time (d" } {TEXT 501 1 "t" }{TEXT -1 46 "') with no distance traversed in this fr ame (d" }{TEXT 502 1 "x" }{TEXT -1 47 "'=0); we denote the two increme nts in Maple by " }{TEXT 19 3 "dtp" }{TEXT -1 5 " and " }{TEXT 19 3 "d xp" }{TEXT -1 49 " respectively. Thus the interval in spacetime is " } {TEXT 19 9 "dsp=c*dtp" }{TEXT -1 107 ". From the point of view of the \+ stationary observer the elephant is travelling during this time a dist ance " }{TEXT 19 2 "dx" }{TEXT -1 75 ", which is determined by the spe ed of the spaceship times the dilated time " }{TEXT 19 12 "dt=gamma*dt p" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "The \+ invariance holds for any spaceship speed, i.e., we keep it symbolic:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "beta:=V/c;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Given the time interval in the ele phant's (spaceship) frame (" }{TEXT 19 3 "dtp" }{TEXT -1 63 ") we comp ute the time passed in the stationary observer frame (" }{TEXT 19 2 "d t" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "dtp:= 21*c*_months;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "dt:=dtp*ga mma;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "Now we calculate the dis tance travelled by the elephant in the observer's frame until the elep hant gives birth:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "dx:=dt *beta*c;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "ds2:=simplify(( c*dt)^2-dx^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Evidently this \+ agrees with the spacetime interval in the spaceship frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsp2:=(c*dtp)^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 503 27 "Relativistic Doppler effect" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 383 "We assume familiarity \+ with the Doppler effect for sound waves, i.e., the change in the frequ ency of a monochromatic sound source that is moving towards or away fr om an observer. For light waves we do expect a difference, as the ligh t waves are not being created in a medium as in the case of sound. Let us consider a light source moving with respect to a stationary frame \+ with speed " }{TEXT 504 1 "V" }{TEXT -1 171 ", and emitting monochroma tic light in two directions: one stationary observer is behind the lig ht source (we call that one B), while the other sits in its flight pat h (A)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "Consider a spacetime diagram where the two reference frames (moving a nd stationary) coincide at " }{TEXT 509 2 "t " }{TEXT -1 1 "=" }{TEXT 508 2 " t" }{TEXT -1 29 "' = 0 when a finite train of " }{TEXT 507 1 " N" }{TEXT -1 21 " waves is emitted at " }{TEXT 506 2 "x " }{TEXT -1 2 "= " }{TEXT 505 1 "x" }{TEXT -1 6 "' = 0." }}{PARA 0 "" 0 "" {TEXT -1 22 "Observer A: consider d" }{TEXT 510 1 "t" }{TEXT -1 38 " to be the \+ time interval during which " }{TEXT 511 1 "N" }{TEXT -1 49 " waves are emitted: the source moves a distamce (" }{TEXT 513 1 "V" }{TEXT -1 2 " d" }{TEXT 514 1 "t" }{TEXT -1 36 "), while the first wave travels (c d" }{TEXT 512 1 "t" }{TEXT -1 75 ") in the stationary frame. The wave train is observed to have a distance (c-" }{TEXT 515 1 "V" }{TEXT -1 3 ") d" }{TEXT 516 1 "t" }{TEXT -1 64 ", and the wavelength is determi ned by dividing this distance by " }{TEXT 517 1 "N" }{TEXT -1 98 ". Th e frequency follows from the usual relationship with the wavelength an d the propagation speed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "dt:='dt': dx:='dx': beta:='beta':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "lambda:=(c*dt-V*dt)/N;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f:=c/lambda;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "f:=simplify(subs(V=beta*c,c/lambda),symbolic);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "In the moving reference frame of the light sou rce the so-called proper frequency and wavelength are determined as (f p=c/lambda'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dtp:='dtp' :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "fp:=N/dtp;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "The time interval " }{TEXT 19 3 "dtp" } {TEXT -1 20 " is the proper time " }{TEXT 19 3 "tau" }{TEXT -1 36 ", a s all light waves are emitted at " }{TEXT 518 1 "x" }{TEXT -1 15 "'=0, and thus, " }{TEXT 19 5 "dxp=0" }{TEXT -1 39 " between the emission o f the first and " }{TEXT 519 1 "N" }{TEXT -1 60 "th wave. Therefore, d tp and dt are related by time dilation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 14 "dt:=gamma*dtp;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "fp:='fp':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "fA:=simplify(algsubs(N=dtp*fp,f));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plot(fA/fp,beta=0..1,title=\"blueshift factor\");" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "The frequency shift is quite app reciable even for moderate speeds. This is caused by a series expansio n with additive contributions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "taylor(fA/fp,beta=0,5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 156 "In complete analogy the redshift (frequency reduction compared to the proper frequency) can be calculated for observer B from whom the \+ light source recedes." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "la mbda:=(c*dt+V*dt)/N;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f:= c/lambda;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "f:=simplify(su bs(V=beta*c,c/lambda),symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "fB:=simplify(algsubs(N=dtp*fp,f));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot(fB/fp,beta=0..1,title=\"redshift fac tor\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "taylor(fB/fp,bet a=0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "We note that while the \+ graphs for the two expressions look quite different, the first order t erms in the frequency shifts are symmetrical, i.e., to first order the magnitude of the frequency shift is proportional to beta for either r ed or blueshift." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "An example where redshifts and blueshifts are relevant t o astronomers is the estimation of the orbiting motion of stars and di stant galaxies." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 201 "Astronomers deal with redshifts all the time as the gala xies recede from each other according to the expansion of the universe (Hubble's law). They express the redshift as a fractional frequency s hift:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "z:='z': eq:=z=simp lify((fp-fB)/fB);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "beta=s olve(eq,beta);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Transverse Dopp ler effect:" }}{PARA 0 "" 0 "" {TEXT -1 286 "Naively one might think t hat in a direction perpendicular to the direction of motion of the sou rce no frequency shifts will be observed. This is not correct. One has to generalize the above arguments in one dimension for an observer wh o looks at waves arriving at an angle theta to the " }{TEXT 520 1 "x" }{TEXT -1 56 "-axis, assuming that the source still moves only in the \+ " }{TEXT 521 1 "x" }{TEXT -1 133 "-direction. When one carries out the analysis one obtains an expression in which even for theta=Pi/2 a tim e dilation factor survives." }}{PARA 0 "" 0 "" {TEXT -1 34 "The genera l expression is given as" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f:=fp/gamma/(1-beta*cos(theta));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eval(f,theta=Pi/2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 522 23 "Twin (or clock) paradox" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 120 "This famous paradox is a good example fo r both the time dilation and the Doppler effect. Twins A and B are sep arated at " }{TEXT 523 1 "t" }{TEXT -1 3 " = " }{TEXT 524 1 "t" } {TEXT -1 80 "' = 0 with A remaining stationary on Earth, and B travell ing on a spaceship for " }{TEXT 525 1 "n" }{TEXT -1 104 " years with s ome beta-value. Then B turns around and comes back to visit A on Earth again after another " }{TEXT 526 1 "n" }{TEXT -1 305 " years. To B's \+ dismay, A has aged more than B. To keep track as to how this happens t hey repeat the experiment, and agree to send light pulses (perhaps eve n videoimages of their faces) to each other with a frequency of 1/year . They do conclude that their observation is consistent with special r elativity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "How does one illustrate the situation with a spacetime diagram? " }}{PARA 0 "" 0 "" {TEXT -1 74 "The spaceship with B has a worldline \+ that starts at the origin with slope " }{TEXT 19 6 "1/beta" }{TEXT -1 80 " (somewhat higher than the slope of 1 for a light worldline throug h the origin)." }}{PARA 0 "" 0 "" {TEXT -1 19 "Observer A sits at " } {TEXT 527 1 "x" }{TEXT -1 68 "=0 and fires off videopulses once a year : the worldlines go off at c" }{TEXT 528 1 "t" }{TEXT -1 3 " = " } {TEXT 529 1 "j" }{TEXT -1 2 ", " }{TEXT 536 1 "j" }{TEXT -1 60 "=1,2,. .. as lines with slope 1, and intersect B's worldline." }}{PARA 0 "" 0 "" {TEXT -1 77 "B sends off videoimages which form worldlines of slo pe -1, and reach A (the c" }{TEXT 530 1 "t" }{TEXT -1 28 " axis) at ce rtain intervals." }}{PARA 0 "" 0 "" {TEXT -1 221 "Initially, while the twins are receding from each other both have to wait long times for t he transmissions to arrive! The frequency of the 1/year transmission c lock is dramatically reduced, and the situation is symmetric." }} {PARA 0 "" 0 "" {TEXT -1 415 "It is when B turns around that matters b egin to change dramatically: out of a sudden B is receiving the videoi mages that travelled from A at a fast rate; shortly before B arrives A will see a rapid succession of videopulses from B, but they will not \+ make up entirely for the difference. A net difference remains between \+ the ageing processes for A and B! A had to send more images than B, an d therefore has lost out." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 346 "The reason why B manages to beat A when it comes \+ to extend life is that B changes directions, while A remains in a rest frame. To understand the details of changing directions one needs to \+ consult the general theory of relativity, but the conclusions remain t he same. So it seems that to extend life one has to move fast, rather \+ than sit at rest!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "To generate the spacetime diagram we need to find the intercepts between the worl dlines for the spaceship and the worldlines of the light signals." }} {PARA 0 "" 0 "" {TEXT -1 147 "We have to decide for how long B travels away from the Earth. We use the (ct_A, x_A) and (ct_B, and x_B) coord inates (or unprimed and primed ones)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "beta:='beta': V:='V':" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "We re-derive equations 3 and 6 that were used several tim es before" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eq1:=x0=xB*sqr t(1-beta^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq2:=xA=rhs (eq1)+V*tA;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq3:=xB=solv e(eq2,xB);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "eq4:=xA*sqrt( 1-beta^2)=xB+V*tB;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq4:= xA=solve(eq4,xA);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq5:=r hs(eq3)=solve(eq4,xB);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "e q6:=tB=expand(solve(eq5,tB));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 " To find the turnaround point for the worldline of the traveller (B) we solve:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "tB_turn:=3;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "beta:=0.8; V:=0.8; c:=1; gam ma:=1/sqrt(1-beta^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "tA _turn:=tB_turn*gamma;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "xA _turn:=V*tA_turn;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "P1:=p lot([[0,0],[xA_turn,tA_turn],[0,2*tA_turn]],style=line,color=red,thick ness=3): P2:=plot([[-1,1],[0,0],[6,6]],style=line,color=green,linestyl e=2,thickness=2): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "P3:=p lot([seq(i+x,i=1..9)],x=0..6,color=blue,linestyle=2,thickness=2):" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "To graph the worldline of the tra veller B and the worldlines of the transmissions from A in the station ary frame was easy; we also included a broadcast at " }{TEXT 531 1 "t " }{TEXT -1 21 "=0 from both A and B." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display(P1,P2,P3,scaling=constrained,view=[-1..6,0..1 0],labels=[x,ct]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "To figure \+ out the broadcasts from the travelling spaceship we need to figure out how equal time spacings in B's frame translate into the A frame." }} {PARA 0 "" 0 "" {TEXT -1 57 "First we work on the years while B is mov ing away from A." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "tA_WL:= tA=tA_turn/xA_turn*xA;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "We wis h to intersect the worldline at times tB=1,2,3... with worldlines of p hotons moving to the left." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 1 to 3 do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sol:=s olve(\{rhs(eq6)=i,tA_WL\},\{xA,tA\});" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "tA1:=subs(sol,tA): xA1:=subs(sol,xA): signalB[i]:=[xA1+p,tA1-p,p =0..-5]: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "P4:=plot([s eq(signalB[i],i=1..3)],color=orange,linestyle=1,thickness=2):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "display(P1,P2,P3,P4,scaling= constrained,view=[-1..6,0..10],labels=[x,ct]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 422 "Observe how in the first half matters are symmetric : At the end of three years in B's frame the first videosignal arrives from A showing that A aged by one year. Similarly, A waits for three \+ years for the first videosignal to arrive from B (showing B aged by on e year). Both A and B are surprised that their twin partner appears to be younger, but then they can both figure out that the signal travell ed for a long time..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161 "Now what happens? B turns around the ship, and races \+ back to Earth. Within a year B will to receive a bunch of signals, whi le not much changes for A for a while!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "tA_WL:=tA=2*tA_turn-tA_turn/xA_turn*xA;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "Clearly B will travel back for as long as it took to get to " }{TEXT 19 18 "[xA_turn, tA_turn]" }{TEXT -1 51 ", i.e., 3 years. We use time dilation to translate " }{TEXT 19 2 "tB" } {TEXT -1 6 " into " }{TEXT 19 2 "tA" }{TEXT -1 16 ", and calculate " } {TEXT 19 2 "xA" }{TEXT -1 117 " on the worldline. Then we generate the worldline for the photon going to Earth (crossing A's worldline, i.e. , the (c" }{TEXT 532 1 "t" }{TEXT -1 8 ")-axis)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 4 to 6 do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "tA1:=tA_turn+(i-3)*gamma;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "xA1:=solve(subs(tA=tA1,tA_WL),xA): signalB[i]:=[xA1+p ,tA1-p,p=0..-5]: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "P5: =plot([seq(signalB[i],i=4..6)],color=orange,linestyle=1,thickness=2): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "display(P1,P2,P3,P4,P5, scaling=constrained,view=[-1..6,0..10],labels=[x,ct]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 327 "We see that B transmits his videopicture in yearly intervals while receiving three images per year from A afte r the turnaround. Shortly before B's arrival A receives also a rapid s uccession of transmissions (three images within one year), but A canno t catch up, and 10 years have passed for A, while only 6 years went by for B." }}{PARA 0 "" 0 "" {TEXT -1 450 "We can appreciate the symmetr y from the point of view of Doppler shifts of the frequencies: both ob servers see the pulses redshifted initially by the same rate (factor o f 3 slowdown), and later blueshifted by the same amount (factor of 3 i ncrease). Nevertheless, the outcome of the ageing process is non-symme tric. As mentioned above, there is really no paradox, as the situation is not completely symmetric: observer A whose worldline follows the ( c" }{TEXT 533 1 "t" }{TEXT -1 150 ")-axis remains in a rest frame with out change, while B undergoes qualitatively different motion. It is th e turnaround that makes the whole difference." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 534 11 "Exercise 9:" }}{PARA 0 " " 0 "" {TEXT -1 163 "Construct the spacetime diagram for another situa tion, such as: B moves with beta=0.9 for 5 years, and then turns aroun d to reach back to Earth in another 5 years." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 2 0" 50 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }