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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 18 "Special Relativity" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 98 "We begin \+
at the level of Modern Physics extensions to first-year Physics texts \+
(around page 1000)." }}{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$F%\"\"#!\"\"" }{TEXT -1 2 
" ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "W
e begin with the time intervals between flashes for the stationary and
 moving observers (O1 and O2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 17 "eq1:=Dt[0]=2*L/c;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/&
%#DtG6#\"\"!,$*&%\"LG\"\"\"%\"cG!\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 18 "eq2:=Dt[V]=2*D/c; " }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%$eq2G/&%#DtG6#%\"VG,$*&%\"DG\"\"\"%\"cG!\"\"\"\"#" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 118 "Now we use the Pythagorean theorem with \+
the statement about O2's measurement of the distance between flashes o
ccuring:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "eq3:=D^2=L^2+(V
*Dt[V]/2)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/*$)%\"DG\"\"#
\"\"\",&*$)%\"LGF)F*\"\"\"*&)%\"VGF)F*)&%#DtG6#F2F)F*#F/\"\"%" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "We square the time interval observ
ed by O2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq2p:=lhs(eq2)^
2=rhs(eq2)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eq2pG/*$)&%#DtG6#%
\"VG\"\"#\"\"\",$*&*$)%\"DGF,F-F-*$)%\"cG\"\"#F-!\"\"\"\"%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 93 "to be able to substitute the unknown dist
ance 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);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq4G/*$)&%#DtG6#%\"VG\"\"#\"\"\",$*
&,&*$)%\"LGF,F-\"\"\"*&)F+F,F-F'F-#F4\"\"%F-*$)%\"cG\"\"#F-!\"\"F8" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Now we eliminate " }{TEXT 269 1 "
L" }{TEXT -1 41 " by introducing the proper time interval:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq5:=subs(L=solve(eq1,L),eq4);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq5G/*$)&%#DtG6#%\"VG\"\"#\"\"\",$*
&,&*&)&F)6#\"\"!F,F-)%\"cGF,F-#\"\"\"\"\"%*&)F+F,F-F'F-F8F-*$)F7\"\"#F
-!\"\"F:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "and solve for the dil
ated time interval." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "sol:
=solve(eq5,Dt[V]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6$*&*&%\"
cG\"\"\"&%#DtG6#\"\"!F)\"\"\"*$-%%sqrtG6#,&*$)F(\"\"#F.F)*$)%\"VGF6F.!
\"\"F.!\"\",$F&F:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "We are only \+
interested in the positive root." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "eq6:=Dt[V]=sol[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%$eq6G/&%#DtG6#%\"VG*&*&%\"cG\"\"\"&F'6#\"\"!F-\"\"\"*$-%%sqrtG6#,&*
$)F,\"\"#F1F-*$)F)F9F1!\"\"F1!\"\"" }}}{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);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eq6pG/&%#DtG6#%\"V
G*&&F'6#\"\"!\"\"\"*$-%%sqrtG6#,&\"\"\"F4*&*$)F)\"\"#F.F.*$)%\"cG\"\"#
F.!\"\"!\"\"F.F=" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "This looks cl
umsy, and thus we often use a shorthand notation:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 56 "eq6pp:=Dt[V]=simplify(subs(V=beta*c,rhs(eq6)
),symbolic);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&eq6ppG/&%#DtG6#%\"V
G*&&F'6#\"\"!\"\"\"*$-%%sqrtG6#,&\"\"\"F4*$)%%betaG\"\"#F.!\"\"F.!\"\"
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 261 "The well-known experimental \+
test of this relationship is the extension of the half-life of muons p
roduced in the upper atmosphere by cosmic radiation (reported first in
 1941 by Rossi and Hall). The average speed of these muons produced ty
pically at a height of " }{TEXT 270 1 "h" }{TEXT -1 454 "=10 km is 0.9
94 c. A muon produced in a laboratory at rest has a half-life of 1.5 m
icroseconds. If the muons produced at 10 km height and coming down wit
h nearly the speed of light (299 792 km/s) were to decay at this rate,
 their population would be cut in half about every 0.45 km. Due to thi
s exponential attenuation (executing 20 times a halving of the current
 sample is calculated below) a very tiny fraction of the produced muon
s would be observed (" }{XPPEDIT 18 0 "10^(-6);" "6#)\"#5,$\"\"'!\"\"
" }{TEXT -1 2 ")." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "d_h:=1
.5*10^(-6)*300000;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$d_hG$\"+++++X
!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "0.5^(20);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+kJuO&*!#;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 69 "Instead we have the half-life as observed by us (who are \+
moving with " }{TEXT 271 1 "V" }{TEXT -1 39 "=0.994 c with respect to \+
the muons) of:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "t_h:=1.5*
10^(-6)/sqrt(1-0.994^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$t_hG$\"
+')\\Or8!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "The dilation facto
r is large indeed:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "1./sq
rt(1-0.994^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+VKVU\"*!\"*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "t_h*300000;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"+e\\49T!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 110 "The travel time to reach the surface of the Earth from a heigh
t of 10 km is a modest multiple of the half-life" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 18 "t_tr:=(10/300000);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%t_trG#\"\"\"\"&++$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "t_tr/t_h;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+_#o1V
#!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "This means that less th
an a quarter of the muons produced at 10 km height do reach the surfac
e of the Earth (and more of we stand on top of a high mountain)." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "We can visualize the time dilatio
n factor by a simple graph as a function of the speed, expressed as a \+
fraction of the speed of light (" }{TEXT 19 4 "beta" }{TEXT -1 2 "):" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot(1/sqrt(1-beta^2),bet
a=0..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 694 225 225 {PLOTDATA 2 "6%-%'C
URVESG6$7U7$\"\"!$\"\"\"F(7$$\"1+++;arz@!#<$\"1Z*[EkP-+\"!#:7$$\"1+++X
TFwSF.$\"1u_KP=$3+\"F17$$\"1+++\"z_\"4iF.$\"1?m!3FL>+\"F17$$\"1+++S&ph
N)F.$\"1.%\\!p'4N+\"F17$$\"1+++*=)H\\5!#;$\"1H/N95b05F17$$\"1+++[!3uC
\"FD$\"1#*)=N7sy+\"F17$$\"1+++J$RDX\"FD$\"1Dl4w#>2,\"F17$$\"1+++)R'ok;
FD$\"12w'))p]T,\"F17$$\"1+++1J:w=FD$\"1#>\"oT%y!=5F17$$\"1+++3En$4#FD$
\"1,*42DlE-\"F17$$\"1+++/RE&G#FD$\"1JL/o:=F5F17$$\"1+++D.&4]#FD$\"1IO1
Q<#G.\"F17$$\"1+++vB_<FFD$\"1B@xbV5R5F17$$\"1+++v'Hi#HFD$\"1ekz,exX5F1
7$$\"1+++(*ev:JFD$\"1M8$3N'Q_5F17$$\"1+++347TLFD$\"1qOc=/(41\"F17$$\"1
+++LY.KNFD$\"1rw3?R*)o5F17$$\"1+++\"o7Tv$FD$\"1\\6(z_8*y5F17$$\"1+++$Q
*o]RFD$\"1OdX2@b)3\"F17$$\"1+++\"=lj;%FD$\"1]cyv9-+6F17$$\"1+++V&R<P%F
D$\"1kn6<3)=6\"F17$$\"1+++Xh-'e%FD$\"1%yI6C8`7\"F17$$\"1+++R\"3Gy%FD$
\"1S-TdPoQ6F17$$\"1+++.T1&*\\FD$\"1KulZ4Ka6F17$$\"1+++(RQb@&FD$\"1bv/&
*G.s6F17$$\"1,++=>Y2aFD$\"1;u(*)*zz)=\"F17$$\"1+++yXu9cFD$\"1!>:p(yY37
F17$$\"1+++\\y))GeFD$\"1*eV)[0pI7F17$$\"1+++i_QQgFD$\"1#p[nCPXD\"F17$$
\"1+++!y%3TiFD$\"1+&**=Kc)z7F17$$\"1+++O![hY'FD$\"1B)=q%4$4J\"F17$$\"1
+++#Qx$omFD$\"1zNbEj\">M\"F17$$\"1+++u.I%)oFD$\"1p_W)fM(y8F17$$\"1****
*ppe*zqFD$\"1@gHAi*fT\"F17$$\"1+++C\\'QH(FD$\"1h:!ejq<Y\"F17$$\"1+++8S
8&\\(FD$\"1nOmI()f5:F17$$\"1+++0#=bq(FD$\"1))))z*GH*o:F17$$\"1*****p?2
7\"zFD$\"1U:n%4&*[j\"F17$$\"1,++IXaE\")FD$\"1.sB+R,;<F17$$\"1+++l*RRL)
FD$\"1&oHRXn$4=F17$$\"1+++`<.Y&)FD$\"1UYr'GPd#>F17$$\"1+++8tOc()FD$\"1
'z/Un?02#F17$$\"1+++\\Qk\\*)FD$\"1ukex4YTAF17$$\"1+++p0;r\"*FD$\"1'ybQ
,`'3DF17$$\"1+++oTAq#*FD$\"12Vt\"eCmm#F17$$\"1*****\\w(Gp$*FD$\"1$Q&4E
Q1hGF17$$\"1+++B-\"\\Z*FD$\"1M^Jin7FJF17$$\"1+++!oK0e*FD$\"1()yE%*HI*[
$F17$$\"1+++<5s#y*FD$\"1z%f&yFMB[F17$%%FAILGF^[l-%'COLOURG6&%$RGBG$\"#
5!\"\"F(F(-%+AXESLABELSG6$Q%beta6\"%!G-%%VIEWG6$;F(F)%(DEFAULTG" 1 2 
0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 182 "The graph shows that time dilation becomes apprecia
ble for particle speeds exceeding a quarter of the speed of light, and
 that it is unbounded for particles whose speed approaches 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 understand that simulta
neity of events is non-universal phenomenon, and that the rate at whic
h 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 a
n object. In Newtonian mechanics the length of an object is determined
 by a " }{TEXT 273 12 "simultaneous" }{TEXT -1 146 " measurement of th
e position of its endpoints. We know that simultaneity is not universa
l, i.e., there will be no universal length for any object!" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 "The thought expe
riment to investigate the phenomenon involves two observers O1 and O2 \+
carrying rods of equal length " }{TEXT 274 1 "L" }{TEXT -1 104 " (as m
easured in their respective reference frame). Suppose 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 O1. If the rods ar
e 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. Th
e assumption of contraction or expansion in these circumstances lead t
o a contradiction with the first principle of relativity: the relative
 motion cannot lead to an observation of expansion in one frame vs con
traction in the other: O1 and O2 have agreed that while at rest with r
espect to each other rods R1 and R2 are of equal length. Now suppose t
he O2 moves by with velocity " }{TEXT 279 2 "V " }{TEXT -1 32 "and bot
h 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 contra
cted length on his rod. Now observer O1 sees from the mark that rod R2
 is longer, since from his point of view R1 is at rest and unaffected.
 Thus, we have supposed that O2 observes a contraction, while O1 obser
ves an expansion. Yet physics has to be symmetrical with respect to th
e interchange of O1 and O2, as there is nothing absolute about one or \+
the other. The contradiction is only avoided if no contraction or expa
nsion is observed under these circumstances." }}{PARA 0 "" 0 "" {TEXT 
-1 68 "Mathematically this is formulated in the following way: if O1 u
ses [" }{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 velocity " }{TEXT 292 1 "V" }{TEXT 
-1 2 "=[" }{TEXT 291 1 "V" }{TEXT -1 79 ",0,0] the coordinates orthogo
nal to the direction of the velocity transform 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 th
e 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. O
bservers O1 and O2 measure the length of the rod to be " }{TEXT 294 1 
"L" }{TEXT -1 94 "0 (the proper length in the frame where the rod is s
tationary). Now O2 takes a rod held along " }{TEXT 297 1 "x" }{TEXT 
-1 5 " (or " }{TEXT 296 1 "x" }{TEXT -1 58 "', as these directions coi
ncide) 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 p
ass between the passage of the beginning and the end of the rod (suppo
se the light clock has a start-stop trigger, and counts many pulses du
ring the passage so that the result 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 "The length of the moving rod is then to b
e calculated from" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq1:=L
[V]=V*Dt[0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/&%\"LG6#%\"VG*
&F)\"\"\"&%#DtG6#\"\"!F+" }}}{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 "(remember that the time dilation is ind
ependent of the direction of the velocity, it depends on " }{TEXT 300 
1 "V" }{TEXT -1 36 "^2). The observed time is therefore " }{TEXT 19 5 
"Dt[V]" }{TEXT -1 60 ".From this perspective 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];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$eq2G/&%\"LG6#\"\"!*&%\"VG\"\"\"&%#DtG6#F+F," }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "Observer O2 now converts the the \+
time measured from O1's clock to his proper time. We lift from the pre
vious section:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "eq6p:=Dt[
V]=Dt[0]/sqrt(1-(beta)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eq6pG
/&%#DtG6#%\"VG*&&F'6#\"\"!\"\"\"*$-%%sqrtG6#,&\"\"\"F4*$)%%betaG\"\"#F
.!\"\"F.!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "and substitute i
nto " }{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);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%$eq3G/&%\"LG6#\"\"!*&*&%\"VG\"\"\"&%#DtGF(F-\"\"\"*$-%%sqrtG6#,&F-F-
*$)%%betaG\"\"#F0!\"\"F0!\"\"" }}}{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);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$eq4G/&%\"LG6#%\"VG*&&F'6#\"\"!\"\"\"-%%sqrtG6#,&F.F.
*$)%%betaG\"\"#\"\"\"!\"\"F7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "T
he length of a rod (length " }{TEXT 19 4 "L[0]" }{TEXT -1 66 " for a s
tationary 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 leng
th contraction was hypothesized independently in the 1890ies by FitzGe
rald 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);" }}{PARA 13 "" 1 "" {GLPLOT2D 692 218 218 {PLOTDATA 2 
"6%-%'CURVESG6$7Y7$\"\"!$\"\"\"F(7$$\"1nmm;arz@!#<$\"1)=:\"QTi(***!#;7
$$\"1LL$e9ui2%F.$\"1wg_S&)o\"***F17$$\"1nmm\"z_\"4iF.$\"1'\\s^f/2)**F1
7$$\"1mmmT&phN)F.$\"1@\"fk0E]'**F17$$\"1LLe*=)H\\5F1$\"1\\N+$H'zW**F17
$$\"1nm\"z/3uC\"F1$\"1m4$yi$*=#**F17$$\"1++DJ$RDX\"F1$\"1Fzm\"3WR*)*F1
7$$\"1nm\"zR'ok;F1$\"1ba\"Q\\n/')*F17$$\"1++D1J:w=F1$\"1Mm5ZeUA)*F17$$
\"1LLL3En$4#F1$\"1/Ayu2Py(*F17$$\"1nm;/RE&G#F1$\"1JBk#=x`t*F17$$\"1+++
D.&4]#F1$\"1q9\"H%H@#o*F17$$\"1+++vB_<FF1$\"1_^A([sOi*F17$$\"1+++v'Hi#
HF1$\"1;#e7,zAc*F17$$\"1nm\"z*ev:JF1$\"1A\\;*p8A]*F17$$\"1LLL347TLF1$
\"1X!GHZL`U*F17$$\"1LLLLY.KNF1$\"1iSu>`Yb$*F17$$\"1++D\"o7Tv$F1$\"1uaj
&Q$eo#*F17$$\"1LLL$Q*o]RF1$\"1`?'yv9l=*F17$$\"1++D\"=lj;%F1$\"1lxU3;t!
4*F17$$\"1++vV&R<P%F1$\"134%pppP**)F17$$\"1LL$e9Ege%F1$\"1EXu&e9k)))F1
7$$\"1LLeR\"3Gy%F1$\"1'))H4fp?y)F17$$\"1nm;/T1&*\\F1$\"1m$zP!>5j')F17$
$\"1mm\"zRQb@&F1$\"11&=5t$=K&)F17$$\"1***\\(=>Y2aF1$\"1e&yZ.e=T)F17$$
\"1mm;zXu9cF1$\"1/X@$oS\\F)F17$$\"1+++]y))GeF1$\"1\"yShX>b7)F17$$\"1**
**\\i_QQgF1$\"1nMlUm1rzF17$$\"1***\\7y%3TiF1$\"1Y$z*QqP8yF17$$\"1****
\\P![hY'F1$\"1XhIao;GwF17$$\"1LLL$Qx$omF1$\"1#p\\+YH?X(F17$$\"1+++v.I%
)oF1$\"18IrHv-`sF17$$\"1mm\"zpe*zqF1$\"1ynnKd;iqF17$$\"1+++D\\'QH(F1$
\"1\"QTZ_=5%oF17$$\"1KLe9S8&\\(F1$\"1&Rp()p\"*)>mF17$$\"1***\\i?=bq(F1
$\"1<HmUNxtjF17$$\"1LLL3s?6zF1$\"1phP//g;hF17$$\"1++DJXaE\")F1$\"19R=@
'eu#eF17$$\"1nmmm*RRL)F1$\"1qw!3M$zEbF17$$\"1mm;a<.Y&)F1$\"1gcj7j\"G>&
F17$$\"1LLe9tOc()F1$\"1nBuRIqH[F17$$\"1+++]Qk\\*)F1$\"1@I:/fPhWF17$$\"
1LL$3dg6<*F1$\"1v)*4GH?')RF17$$\"1mmmmxGp$*F1$\"1\"RL6R._\\$F17$$\"1++
D\"oK0e*F1$\"1=Cis=!f'GF17$$\"1+++]oi\"o*F1$\"1UJ1O#=K]#F17$$\"1++v=5s
#y*F1$\"1vO>U4Dt?F17$$\"1+D1k2/P)*F1$\"1MW(eL^zz\"F17$$\"1+]P40O\"*)*F
1$\"1H.f%oH+Z\"F17$$\"1^7.#Q?&=**F1$\"1#>z^*=&RF\"F17$$\"1+voa-oX**F1$
\"1P%\\8H')3/\"F17$$\"1\\PMF,%G(**F1$\"1&=hJ*p=ltF.7$F)F(-%'COLOURG6&%
$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q%beta6\"%!G-%%VIEWG6$;F(F)%(DEFAUL
TG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 190 "As with time dilation we observe that th
e effect becomes noticable for speeds exceeding a fifth to a quarter o
f the speed of light, and becomes really strong for highly relativisti
c objects." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 897 "The length contraction can be used to explain the extend
ed half-life of the muons from another perspective. Suppose an observe
r travels with the muons produced in the upper atmosphere (10 km heigh
t), i.e., with their speed (about 0.994 c; this known speed is the res
ult of energy balancing in a particle physics reaction). From this obs
erver's point of view the muons have a half-life as observed in a labo
ratory (where they are essentially at rest), namely 1.5 microseconds. \+
From the graph above one can infer that the vertical flight path of 10
 km to be followed by the muons is contracted substantially, namely to
 about 1/9 th of the proper length. It becomes clear that there is no \+
contradiction but complete consistency between the observations 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 this is essentially
 " }{TEXT 19 6 "L[V]/c" }{TEXT -1 89 " or 3.7 microseconds. Therefore,
 a sizeable fraction of the muons makes it to the ground." }}}{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 dilated time an
d contracted length. As stated before, we need to know how the coordin
ates transform from one reference frame to another so that we can form
ulate the laws of physics for observers O1 and O2, and make 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/her proper time w
hen looking at the laws of physics." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT 355 30 "Length/Position transformation" }}
{PARA 0 "" 0 "" {TEXT -1 118 "To derive the length transformation cons
ider the following: O1 is at rest, and observes O2 to move with a velo
city 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 " i
s at rest in the moving frame O2, i.e., it moves with velocity " }
{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 knows th
at 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 t
he 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 " stand
s for " }{TEXT 322 1 "x" }{TEXT -1 3 "'):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 26 "eq1:=x0=xp*sqrt(1-beta^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$eq1G/%#x0G*&%#xpG\"\"\"-%%sqrtG6#,&F)F)*$)%%betaG\"
\"#\"\"\"!\"\"F2" }}}{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 change with time, i.e., " }{TEXT 19 2 "xp" }
{TEXT -1 95 " is constant. Given that the particle moves with O2, obse
rver O1 can determine the location of " }{TEXT 324 1 "P" }{TEXT -1 13 
" at any time " }{TEXT 325 1 "t" }{TEXT -1 54 " by adding the displace
ment of the origin of O2, i.e.," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "eq2:=x=rhs(eq1)+V*t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2
G/%\"xG,&*&%#xpG\"\"\"-%%sqrtG6#,&F*F**$)%%betaG\"\"#\"\"\"!\"\"F3F**&
%\"VGF*%\"tGF*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "The transform
ation 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 "eq
3:=xp=solve(eq2,xp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/%#xpG*
&,&%\"xG\"\"\"*&%\"VGF*%\"tGF*!\"\"\"\"\"*$-%%sqrtG6#,&F*F**$)%%betaG
\"\"#F/F.F/!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Note that " }
{TEXT 19 3 "eq2" }{TEXT -1 132 " is not the transformation from O2 to \+
O1, as it mixes on the RHS the primed position (O2 variable) with the \+
unprimed time (O1 time)." }}{PARA 0 "" 0 "" {TEXT -1 31 "The transform
ation obtained in " }{TEXT 19 3 "eq3" }{TEXT -1 61 " reduces to the Ga
lilei transformation 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);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+)%%betaG,&%\"xG\"
\"\"*&%\"VGF'%\"tGF'!\"\"\"\"!,&F&#F'\"\"#F(#F+F/\"\"#-%\"OG6#F'\"\"%
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 220 "This is an important result,
 as the Newtonian physics relies on this transformation, and represent
s 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 mo
ving frame O2 is obtained from a linear combination of position and ti
me in the stationary frame. The important question is: how does it loo
k 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);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%)WRONGeq4G/%\"xG,&*&%#xpG\"\"\"-%%sqrtG6#,&F*F**$)%%betaG\"\"#\"\"
\"!\"\"F3F**&*&%\"VGF*%#tpGF*F3*$-F,6#F.F3!\"\"F*" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 84 "The repeated reasoning to obtain the correct inver
se transformation 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 with velocity  -V t'. He realizes that he is observing the \+
coordinate " }{TEXT 336 1 "x" }{TEXT -1 39 " in contracted form and th
en concludes:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq4:=x*sqr
t(1-beta^2)=xp+V*tp;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq4G/*&%\"x
G\"\"\"-%%sqrtG6#,&F(F(*$)%%betaG\"\"#\"\"\"!\"\"F1,&%#xpGF(*&%\"VGF(%
#tpGF(F(" }}}{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);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq4
G/%\"xG*&,&%#xpG\"\"\"*&%\"VGF*%#tpGF*F*\"\"\"*$-%%sqrtG6#,&F*F**$)%%b
etaG\"\"#F.!\"\"F.!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "This \+
result teaches us that we have to be careful in special relativity. Bl
ind substitution of equations will not lead to the correct results." }
}{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 proc
eed 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 combining the two
 transformation obtained above. If we eliminate " }{TEXT 342 1 "x" }
{TEXT -1 15 "' by combining " }{TEXT 19 3 "eq4" }{TEXT -1 5 " and " }
{TEXT 19 3 "eq3" }{TEXT -1 38 ", we arrive at the following equation:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq5:=rhs(eq3)=solve(eq4
,xp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq5G/*&,&%\"xG\"\"\"*&%\"V
GF)%\"tGF)!\"\"\"\"\"*$-%%sqrtG6#,&F)F)*$)%%betaG\"\"#F.F-F.!\"\",&*&F
(F)-F16#F3F.F)*&F+F.%#tpGF)F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 30 "eq6:=tp=expand(solve(eq5,tp));" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%$eq6G/%#tpG,&*&%\"tG\"\"\"*$-%%sqrtG6#,&\"\"\"F0*$)%%betaG\"\"#F*!
\"\"F*!\"\"F0*&*&%\"xGF0F2F*F**&%\"VG\"\"\"-F-6#F/F*F6F5" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 16 "We observe that " }{TEXT 343 1 "t" }
{TEXT -1 85 "', i.e., the proper time in the boosted 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 stationary 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 time becomes universal (a
s assumed by Galilei and Newton):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "taylor(rhs(eq6),beta=0,3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#+)%%betaG%\"tG\"\"!,&F%#\"\"\"\"\"#*&%\"xG\"\"\"%\"VG!
\"\"!\"\"\"\"#-%\"OG6#F)\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "
How can we obtain the inverse transformation?" }}{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 t
he boosted to the stationary frame and vice versa." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 20 "eq7:=x=solve(eq3,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$eq7G/%\"xG,&*&%#xpG\"\"\"-%%sqrtG6#,&F*F**$)%%betaG
\"\"#\"\"\"!\"\"F3F**&%\"VGF*%\"tGF*F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "eq8:=rhs(eq7)=rhs(eq4);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$eq8G/,&*&%#xpG\"\"\"-%%sqrtG6#,&F)F)*$)%%betaG\"\"#\"\"\"!\"
\"F2F)*&%\"VGF)%\"tGF)F)*&,&F(F)*&F5F2%#tpGF)F)F2*$-F+6#F-F2!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq9:=t=expand(solve(eq8,t));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq9G/%\"tG,&*&*&%#xpG\"\"\")%%b
etaG\"\"#\"\"\"F/*&%\"VG\"\"\"-%%sqrtG6#,&F+F+*$F,F/!\"\"F/!\"\"F+*&%#
tpGF/*$-F46#F6F/F9F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Carry ou
t the Taylor expansion of this result to verify the Galilean limit as \+
done before 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 comple
te the section with a graphical representation of our results. To unde
rstand special relativity one introduces a so-called space-time diagra
m. This is a two-dimensional graph for particles moving in one spatial
 dimension. To have consistent axes, one plots c" }{TEXT 357 1 "t" }
{TEXT -1 320 " on the vertical axis, and the particle position on the \+
horizontal 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 i
s given by 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
 slopes less than one are allowed (unless we were to consider hypothet
ical 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*c
t: x5:=ct: x6:=-ct:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "Tr:
=[[x1,ct],[x2,ct],[x3,ct],[x4,ct],[x5,ct],[x6,ct]]: coltable:=[red,blu
e,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\");" }}
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,FG7$$!1+]7G$[8j$F,FL7$$!1m;z%*frhTF,FQ7$$!1**\\ilFQ!p%F,FV7$$!1ML$3_
\"=M_F,Fen7$$!1nmTg(fJr&F,Fjn7$$!1++]7eP_iF,F_o7$$!1++]Pf!Qz'F,Fdo7$$!
1++](=ubJ(F,Fio7$$!1n;zW(*Q*y(F,F^p7$$!1LL$3F-GN)F,Fcp7$$!1LLL$e'3I))F
,Fhp7$$!1**\\7.<G&Q*F,F]q7$$!1KLLeMsw)*F,Fbq7$$!1+DJ&H\"fT5FCFgq7$$!1+
v$f)[$H4\"FCF\\r7$$!1L$ek`1l9\"FCFar7$$!1Le*[.-d>\"FCFfr7$$!1n;/Egw[7F
CF[s7$$!1m\"z%*f%)QI\"FCF`s7$$!1+voza'=N\"FCFes7$$!1n;zWho.9FCFjs7$$!1
++]i>Ad9FCF_t7$$!1+]i:jf4:FCFdt7$$!1+DJ&>r-c\"FCFit7$$!1+]P4q`;;FCF^u7
$$!1LL$eM%4n;FCFcu7$$!1++v$4v5s\"FCFhu7$$!1n\"zWn*)*p<FCF]v7$$!1++DJiY
B=FCFbv7$$!1Lek.Nyt=FCFgv7$$!1+Dc^&zj#>FCF\\w7$$!1LL3-=!y(>FCFaw7$$!1+
D\"G8O;.#FCFfw7$$!1nmm\"*\\[$3#FCF[x7$$!1n;aQz]O@FCF`x7$$!1LekG=4*=#FC
Fex7$$!1++]i4TPAFCFjx7$$!1L$3F9!z#H#FCF_y7$$!1nmmT>KUBFCFdy7$$!1+DJqJ8
&R#FCFiy7$$!1+voa-oXCFCF^z7$$!1+++++++DFCFcz-Ffz6&FhzF(FizF(-F$6$7SF'7
$$\"1LLLL3VfVFb[lF-7$$\"1nmm\"H[D:)Fb[lF27$$\"1LLLe0$=C\"F,F77$$\"1LLL
3RBr;F,F<7$$\"1nm;zjf)4#F,FA7$$\"1LL$e4;[\\#F,FG7$$\"1++]i'y]!HF,FL7$$
\"1LL$ezs$HLF,FQ7$$\"1++]7iI_PF,FV7$$\"1nmm;_M(=%F,Fen7$$\"1MLL3y_qXF,
Fjn7$$\"1+++]1!>+&F,F_o7$$\"1+++]Z/NaF,Fdo7$$\"1+++]$fC&eF,Fio7$$\"1LL
$ez6:B'F,F^p7$$\"1mmm;=C#o'F,Fcp7$$\"1mmmm#pS1(F,Fhp7$$\"1++]i`A3vF,F]
q7$$\"1nmmm(y8!zF,Fbq7$$\"1-+]i.tK$)F,Fgq7$$\"1++](3zMu)F,F\\r7$$\"1nm
m\"H_?<*F,Far7$$\"1nm;zihl&*F,Ffr7$$\"1LLL3#G,***F,F[s7$$\"1LLezw5V5FC
F`s7$$\"1++v$Q#\\\"3\"FCFes7$$\"1LL$e\"*[H7\"FCFjs7$$\"1+++qvxl6FCF_t7
$$\"1++]_qn27FCFdt7$$\"1++Dcp@[7FCFit7$$\"1++]2'HKH\"FCF^u7$$\"1nmmwan
L8FCFcu7$$\"1+++v+'oP\"FCFhu7$$\"1LLeR<*fT\"FCF]v7$$\"1+++&)Hxe9FCFbv7
$$\"1mm\"H!o-*\\\"FCFgv7$$\"1++DTO5T:FCF\\w7$$\"1nmmT9C#e\"FCFaw7$$\"1
++D1*3`i\"FCFfw7$$\"1LLL$*zym;FCF[x7$$\"1LL$3N1#4<FCF`x7$$\"1nm\"HYt7v
\"FCFex7$$\"1+++q(G**y\"FCFjx7$$\"1nm;9@BM=FCF_y7$$\"1LLL`v&Q(=FCFdy7$
$\"1++DOl5;>FCFiy7$$\"1++v.Uac>FCF^z7$$\"1+++++++?FCFcz-Ffz6&FhzF(F(F(
-F$6$7SF'7$F-F-7$F2F27$F7F77$F<F<7$FAFA7$FGFG7$FLFL7$FQFQ7$FVFV7$FenFe
n7$FjnFjn7$F_oF_o7$FdoFdo7$FioFio7$F^pF^p7$FcpFcp7$FhpFhp7$F]qF]q7$Fbq
Fbq7$FgqFgq7$F\\rF\\r7$FarFar7$FfrFfr7$F[sF[s7$F`sF`s7$FesFes7$FjsFjs7
$F_tF_t7$FdtFdt7$FitFit7$F^uF^u7$FcuFcu7$FhuFhu7$F]vF]v7$FbvFbv7$FgvFg
v7$F\\wF\\w7$FawFaw7$FfwFfw7$F[xF[x7$F`xF`x7$FexFex7$FjxFjx7$F_yF_y7$F
dyFdy7$FiyFiy7$F^zF^z7$FczFcz-Ffz6&FhzFizF(Fiz-F$6$7SF'7$$!1nmm;arz@F,
F-7$$!1LL$e9ui2%F,F27$$!1nmm\"z_\"4iF,F77$$!1mmmT&phN)F,F<7$$!1LLe*=)H
\\5FCFA7$$!1nm\"z/3uC\"FCFG7$$!1++DJ$RDX\"FCFL7$$!1nm\"zR'ok;FCFQ7$$!1
++D1J:w=FCFV7$$!1LLL3En$4#FCFen7$$!1nm;/RE&G#FCFjn7$$!1+++D.&4]#FCF_o7
$$!1+++vB_<FFCFdo7$$!1+++v'Hi#HFCFio7$$!1nm\"z*ev:JFCF^p7$$!1LLL347TLF
CFcp7$$!1LLLLY.KNFCFhp7$$!1++D\"o7Tv$FCF]q7$$!1LLL$Q*o]RFCFbq7$$!1++D
\"=lj;%FCFgq7$$!1++vV&R<P%FCF\\r7$$!1LL$e9Ege%FCFar7$$!1LLeR\"3Gy%FCFf
r7$$!1nm;/T1&*\\FCF[s7$$!1mm\"zRQb@&FCF`s7$$!1***\\(=>Y2aFCFes7$$!1mm;
zXu9cFCFjs7$$!1+++]y))GeFCF_t7$$!1****\\i_QQgFCFdt7$$!1***\\7y%3TiFCFi
t7$$!1****\\P![hY'FCF^u7$$!1LLL$Qx$omFCFcu7$$!1+++v.I%)oFCFhu7$$!1mm\"
zpe*zqFCF]v7$$!1+++D\\'QH(FCFbv7$$!1KLe9S8&\\(FCFgv7$$!1***\\i?=bq(FCF
\\w7$$!1LLL3s?6zFCFaw7$$!1++DJXaE\")FCFfw7$$!1nmmm*RRL)FCF[x7$$!1mm;a<
.Y&)FCF`x7$$!1LLe9tOc()FCFex7$$!1+++]Qk\\*)FCFjx7$$!1LL$3dg6<*FCF_y7$$
!1mmmmxGp$*FCFdy7$$!1++D\"oK0e*FCFiy7$$!1++v=5s#y*FCF^z7$$!\"\"F(FczF_
jm-%&TITLEG6#Q3spacetime:~1+1~dim6\"-%+AXESLABELSG6$%\"xG%#ctG-%(SCALI
NGG6#%,CONSTRAINEDG-%%VIEWG6$%(DEFAULTGFedn" 1 2 0 1 0 2 9 1 4 1 
1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
460 "For accelerating particles moving with variable speeds the trajec
tories become curves, and the slope at any point along the trajectory \+
allows one to deduce the instantaneous speed of the particle. The diag
ram shows only forward propagation in time, but it can be extended to \+
negative times. The lines displayed using the magenta color represent \+
photon trajectories (the muons discussed above would practically follo
w 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 plane), the limit impos
ed by all allowed photon trajectories becomes a surface (a hypersurfac
e in three spatial dimensions) called the lightcone. A particle trajec
tory is called a worldline when one refers to the spacetime diagram. T
he particle's worldline has to remain inside the lightcone 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 particle)." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 361 11 "Exercise 1:" }
}{PARA 0 "" 0 "" {TEXT -1 74 "Extend the spacetime diagram to two spat
ial dimensions. Use the help page " }{TEXT 19 11 "?spacecurve" }{TEXT 
-1 154 " to find out how to generate some worldlines. Find worldlines \+
that trace out the lightcone in 2+1 dimensions. Draw world lines for a
ccelerating particles." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 194 "We can use the spacetime diagram to illu
strate the concept of simultaneity in special relativity. Imagine a st
ationary reference frame that we consider to be the platform of a trai
n station. At " }{TEXT 362 1 "t" }{TEXT -1 30 "=0 a rocket train movin
g with " }{TEXT 363 1 "V" }{TEXT -1 171 "=c/2 passes through the stati
on, and at the same time light flashes are emitted from light clocks l
ocated at opposite ends of the platform (the clocks are synchronized a
t " }{TEXT 365 1 "t" }{TEXT -1 1 "=" }{TEXT 364 1 "t" }{TEXT -1 78 "'=
0). In the stationary reference frame the flashes correspond to the po
ints [" }{TEXT 366 1 "A" }{TEXT -1 9 ",0] and [" }{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 ext
ends symmetrically about " }{TEXT 368 1 "x" }{TEXT -1 3 "=0." }}{PARA 
0 "" 0 "" {TEXT -1 164 "It is straightforward to figure out to which s
pacetime points the two events which are simultaneous 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;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/%#xp
G*&,&%\"xG\"\"\"*&%\"VGF(%\"tGF(!\"\"\"\"\"*$-%%sqrtG6#,&F(F(*$)%%beta
G\"\"#F-F,F-!\"\"/%#tpG,&*&F+F-*$-F06#F2F-F7F(*&*&F'F(F4F-F-*&F*\"\"\"
-F06#F2F-F7F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs(beta=
V/c,V=c/2,t=0,x=A,eq3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#xpG,$*(%
\"AG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F*6#\"\"%F-#F(F," }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 34 "subs(beta=V/c,V=c/2,t=0,x=-A,eq3);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/%#xpG,$*(%\"AG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F
*6#\"\"%F-#!\"\"F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "More import
antly we wish to know the times:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "subs(beta=V/c,V=c/2,t=0,x=A,eq6);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#/%#tpG,$*&*(-%%sqrtG6#\"\"$\"\"\"-F)6#\"\"%F,%\"AG\"
\"\"F,%\"cG!\"\"#!\"\"\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
34 "subs(beta=V/c,V=c/2,t=0,x=-A,eq6);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/%#tpG,$*&*(-%%sqrtG6#\"\"$\"\"\"-F)6#\"\"%F,%\"AG\"\"\"F,%\"cG!
\"\"#F1\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "Clearly the two \+
light flashes are not simultaneous from O2's point of view. O2 observe
s the clock at the entrance to the station ([" }{TEXT 373 1 "B" }
{TEXT -1 7 ", 0]=[-" }{TEXT 372 1 "A" }{TEXT -1 191 ", 0]) flashing mu
ch later (as it flashes at the time when O2 passes the center of the p
latform and has to catch up with him; meanwhile O2 races towards the c
lock at the end of the platform ([" }{TEXT 374 1 "A" }{TEXT -1 154 ", \+
0]), and encounters the pulse from the clock located there at an earli
er time; in fact, a time that predates the passage of the center of th
e 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 Llewellyn: " }{TEXT 378 14 
"Modern Physics" }{TEXT -1 25 ", 3rd ed., Freeman 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 the moving reference fr
ame. 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 po
int [" }{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 define a map that carr
ies out the transformation for the given choice of " }{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;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$/%#xpG*&,&%\"xG\"\"\"*&%\"VGF(%\"t
GF(!\"\"\"\"\"*$-%%sqrtG6#,&F(F(*$)%%betaG\"\"#F-F,F-!\"\"/%#tpG,&*&F+
F-*$-F06#F2F-F7F(*&*&F'F(F4F-F-*&F*\"\"\"-F06#F2F-F7F," }}}{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);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%#LTGR6$%\"xG%\"tG6\"6$%)operatorG%&arrowGF)7$,$*(,
&9$\"\"\"9%#!\"\"\"\"#F2-%%sqrtG6#\"\"$\"\"\"-F86#\"\"%F;#F2F:,&*(F7F;
F<F;F3F2F?*(F1F2F7F;F<F;#F5\"\"'F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "LT(A,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,$*(%\"A
G\"\"\"-%%sqrtG6#\"\"$\"\"\"-F)6#\"\"%F,#F'F+,$F%#!\"\"\"\"'" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(1,0));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#7$$\"+R0qa6!\"*$!+%p-Nx&!#5" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(0,1));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$$!+%p-Nx&!#5$\"+R0qa6!\"*" }}}{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 ") points to (" }
{TEXT 387 1 "x" }{TEXT -1 3 "', " }{TEXT 386 1 "t" }{TEXT -1 27 "') po
ints by graphing them:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "P
tO11:=[seq([i*0.5,0],i=-6..6)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&
PtO11G7/7$$!#I!\"\"\"\"!7$$!#DF)F*7$$!#?F)F*7$$!#:F)F*7$$!#5F)F*7$$!\"
&F)F*7$F*F*7$$\"\"&F)F*7$$\"#5F)F*7$$\"#:F)F*7$$\"#?F)F*7$$\"#DF)F*7$$
\"#IF)F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "PtO21:=[seq(map
(evalf,LT(op(PtO11[i]))),i=1..nops(PtO11))];" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%&PtO21G7/7$$!+;;5kM!\"*$\"+330K<F)7$$!+Z8v')GF)$\"+tc
PV9F)7$$!+x5S4BF)$\"+R0qa6F)7$$!+330K<F)$\"+SSDg')!#57$$!+R0qa6F)$\"+%
p-Nx&F;7$$!+%p-Nx&F;$\"+Z8v')GF;7$\"\"!FG7$F?$F.F;7$F4FB7$F*$!+SSDg')F
;7$$\"+x5S4BF)F=7$$FEF)$!+tcPV9F)7$$\"+;;5kMF)F7" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 55 "PO11:=plot(PtO11,style=point,symbol=diamond,c
olor=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "PO21:=plot(Pt
O21,style=point,symbol=diamond,color=blue):" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 32 "PtO12:=[seq([0,i*0.5],i=-6..6)];" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%&PtO12G7/7$\"\"!$!#I!\"\"7$F'$!#DF*7$F'$!#?F*7$F'$
!#:F*7$F'$!#5F*7$F'$!\"&F*7$F'F'7$F'$\"\"&F*7$F'$\"#5F*7$F'$\"#:F*7$F'
$\"#?F*7$F'$\"#DF*7$F'$\"#IF*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 59 "PtO22:=[seq(map(evalf,LT(op(PtO12[i]))),i=1..nops(PtO12))];" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&PtO22G7/7$$\"+330K<!\"*$!+;;5kMF)7$
$\"+tcPV9F)$!+Z8v')GF)7$$\"+R0qa6F)$!+x5S4BF)7$$\"+SSDg')!#5$!+330K<F)
7$$\"+%p-Nx&F9$!+R0qa6F)7$$\"+Z8v')GF9$!+%p-Nx&F97$\"\"!FG7$$F0F9F=7$F
DF27$$!+SSDg')F9F'7$F?$\"+x5S4BF)7$$!+tcPV9F)$FCF)7$F:$\"+;;5kMF)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "PO12:=plot(PtO12,style=point
,symbol=cross,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
54 "PO22:=plot(PtO22,style=point,symbol=cross,color=blue):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "display([PO11,PO21,PO12,PO22],axes
=boxed,scaling=constrained,labels=[x,ct],title=\"spacetime 1+1 dim\");
" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6+-%'CURVESG6&
7/7$$!\"$\"\"!F*7$$!1+++++++D!#:F*7$$!\"#F*F*7$$!1+++++++:F.F*7$$!\"\"
F*F*7$$!1+++++++]!#;F*7$F*F*7$$\"1+++++++]F;F*7$$\"\"\"F*F*7$$\"1+++++
++:F.F*7$$\"\"#F*F*7$$\"1+++++++DF.F*7$$\"\"$F*F*-%'COLOURG6&%$RGBG$\"
*++++\"!\")F*F*-%'SYMBOLG6#%(DIAMONDG-%&STYLEG6#%&POINTG-F$6&7/7$$!1++
+;;5kMF.$\"1+++330K<F.7$$!1+++Z8v')GF.$\"1+++tcPV9F.7$$!1+++x5S4BF.$\"
1+++R0qa6F.7$$!1+++330K<F.$\"1,++SSDg')F;7$$!1+++R0qa6F.$\"1+++%p-Nx&F
;7$$!1+++%p-Nx&F;$\"1+++Z8v')GF;F<7$Fbp$FboF;7$FhoFep7$F^o$!1,++SSDg')
F;7$$\"1+++x5S4BF.F`p7$$FhpF.$!1+++tcPV9F.7$$\"1+++;;5kMF.F[p-FP6&FRF*
F*FSFVFZ-F$6&7/7$F*F(7$F*F,7$F*F07$F*F37$F*F67$F*F9F<7$F*F>7$F*FA7$F*F
D7$F*FG7$F*FJ7$F*FMFO-FW6#%&CROSSGFZ-F$6&7/7$F^oF\\o7$FcoFao7$FhoFfo7$
F]pF[p7$FbpF`p7$FgpFepF<7$FjpFbp7$FepFho7$F]qF^o7$F`pF`q7$FdqFcq7$F[pF
gqFiqFjrFZ-%&TITLEG6#Q2spacetime~1+1~dim6\"-%+AXESLABELSG6$%\"xG%#ctG-
%*AXESSTYLEG6#%$BOXG-%(SCALINGG6#%,CONSTRAINEDG-%%VIEWG6$%(DEFAULTGFau
" 1 2 0 1 0 2 9 1 2 1 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 183 "We have constructed a graph that displays how \+
sequences of events along the position axis and along the time axis re
spectively are mapped under the transformation law for the case of " }
{TEXT 390 1 "V" }{TEXT -1 10 "=c/2. The " }{TEXT 420 1 "x" }{TEXT -1 
51 "-axis is apparently mapped onto a line with slope -" }{TEXT 422 1 
"V" }{TEXT -1 12 "/2, and the " }{TEXT 421 1 "t" }{TEXT -1 32 "-axis i
nto a line with slope -1/" }{TEXT 423 1 "V" }{TEXT -1 165 ". We can ch
eck out this observation by calculating the parametric representation \+
of the two lines in question. We begin with the general case (mapping \+
of any point [" }{TEXT 425 1 "x" }{TEXT -1 2 ", " }{TEXT 424 1 "t" }
{TEXT -1 35 "]), and then specify our two lines." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 15 "evalf(LT(x,t));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#7$,&%\"xG$\"+R0qa6!\"*%\"tG$!+$p-Nx&!#5,&F)F&F%$!+%p-Nx&F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(x,0));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#7$,$%\"xG$\"+R0qa6!\"*,$F%$!+%p-Nx&!#5" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "slope:=%[2]/%[1];" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%&slopeG$!+********\\!#5" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(LT(0,t));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$,$%\"tG$!+%p-Nx&!#5,$F%$\"+R0qa6!\"*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "slope:=%[2]/%[1];" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%&slopeG$!+++++?!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 97 "An important observation is that the transformation repre
sents 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 t
ransformation 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 preserved (this is demonstrated by the stretch of the axes). These
 circumstances will require some further explorations in transformatio
n theory (beyond the well-known rotations in " }{TEXT 432 1 "n" }
{TEXT -1 20 "-dimensional space)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 117 "It is worth to examine the transformed e
vents in the spacetime diagram in some detail: the points along the or
iginal " }{TEXT 464 1 "x" }{TEXT -1 46 "-axis represent a set of simul
taneous events (" }{TEXT 465 1 "t" }{TEXT -1 132 "=0) in the stationar
y reference frame. These get transformed into a set of points that for
m a line along which the transformed time " }{TEXT 466 1 "t" }{TEXT 
-1 154 "' varies from point to point (blue diamonds), i.e., they are n
ot simultaneous in the boosted frame. This demonstrates that this line
 cannot represent the " }{TEXT 467 1 "x" }{TEXT -1 28 "' axis in the b
oosted frame." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 427 11 "Exercise 2:" }}
{PARA 0 "" 0 "" {TEXT -1 63 "Repeat the above steps for different choi
ces of boost velocity " }{TEXT 426 1 "V" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "The diagram \+
produced above is not what is usually shown, i.e., it does not show th
e (" }{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 transformed the points along the " }{TEXT 439 1 "x" }{TEXT -1 
27 "-axis (which correspond to " }{TEXT 440 1 "t" }{TEXT -1 30 "=0) an
d 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 t
ransformation law. We notice from the points used for graphing the tra
nsformed axes that they do not correspond to the axes which defined th
e (" }{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 "')-axi
s should have the property that " }{TEXT 456 1 "t" }{TEXT -1 53 "'=0 f
or all points along 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 now 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 "' a
xis 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;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6$/%#xpG*&,&%\"xG\"\"\"*&%\"VGF(%\"tGF(!\"\"\"\"\"*$-%%sq
rtG6#,&F(F(*$)%%betaG\"\"#F-F,F-!\"\"/%#tpG,&*&F+F-*$-F06#F2F-F7F(*&*&
F'F(F4F-F-*&F*\"\"\"-F06#F2F-F7F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
33 "We find 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 t
wo cases." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "PO11:=plot(PtO
11,style=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);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&tpeq0G,$%\"xG#\"\"\"\"\"#" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "PO21:=plot(tpeq0,x=-4..4,col
or=red,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "PO1
2:=plot(PtO12,style=point,symbol=cross,color=blue):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 50 "xpeq0:=solve(subs(beta=V/c,V=c/2,c=1,rhs(
eq3)),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&xpeq0G,$%\"xG\"\"#" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "PO22:=plot(xpeq0,x=-4..4,co
lor=blue,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "
display([PO11,PO12,PO21,PO22],axes=boxed,scaling=constrained,labels=[x
,ct],title=\"spacetime 1+1 dim\",view=[-3.5..3.5,-3.5..3.5]);" }}
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1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 
"To illustrate further 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 refernc
e frame for " }{TEXT 461 1 "V" }{TEXT -1 40 "=c/2) we can draw the lin
es 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 "Pt
peq1:=plot(tpeq1,x=-4..4,color=red,thickness=1):" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&tpeq1G,$*(,&*(%\"xG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F,6#
\"\"%F/F*\"\"'F*F*F+F/F0F/#F*\"#C" }}}{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,colo
r=red,thickness=1):" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&tpeq2G,$*(,&
*(%\"xG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F,6#\"\"%F/F*\"#7F*F*F+F/F0F/#F*\"#
C" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "tpeq3:=solve(subs(beta
=V/c,V=c/2,c=1,rhs(eq6)=3),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "P
tpeq3:=plot(tpeq3,x=-4..4,color=red,thickness=1):" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&tpeq3G,$*(,&*(%\"xG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F,6#
\"\"%F/F*\"#=F*F*F+F/F0F/#F*\"#C" }}}{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(xpeq1,x=-4..4,colo
r=blue,thickness=1):" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&xpeq1G,$*(,
&*(%\"xG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F,6#\"\"%F/F*!\"$F*F*F+F/F0F/#F*\"
\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "xpeq2:=solve(subs(be
ta=V/c,V=c/2,c=1,rhs(eq3)=2),t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 
"Pxpeq2:=plot(xpeq2,x=-4..4,color=blue,thickness=1):" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%&xpeq2G,$*(,&*(%\"xG\"\"\"-%%sqrtG6#\"\"$\"\"\"-F,
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{MPLTEXT 1 0 52 "xpeq3:=solve(subs(beta=V/c,V=c/2,c=1,rhs(eq3)=3),t);
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 421 "Now we have a coordinate grid in \+
the boosted frame. It can be used to understand the passage of events \+
(such as the problem of light pulses flashing simultaneously at the tw
o ends of a platform, and how this is perceived by an observer on the \+
rocket train). Points along the red lines are simultaeous in the boost
ed frame. They are clearly not simultaneous for the stationary observe
r, since they are not parallel to the " }{TEXT 470 1 "x" }{TEXT -1 6 "
-axis." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
199 "The previously discussed phenomena of time dilation and length co
ntraction can be easily demonstrated on the diagram. Let us construct \+
the worldline for a light flash emitted in the boosted frame at " }
{TEXT 472 1 "x" }{TEXT -1 52 "'=0, and then reflected back by a mirror
 located at " }{TEXT 471 1 "x" }{TEXT -1 178 "'=1 m. The worldline for
 such a light flash is represented by a straight line starting at the \+
origin with a slope of 1. It reaches the mirror at the intercept of th
e grid lines c" }{TEXT 474 1 "t" }{TEXT -1 8 "'=1 and " }{TEXT 473 1 "
x" }{TEXT -1 51 "'=1. We find this point in the stationary frame as:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eq3,eq6;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6$/%#xpG*&,&%\"xG\"\"\"*&%\"VGF(%\"tGF(!\"\"\"\"\"*$-%
%sqrtG6#,&F(F(*$)%%betaG\"\"#F-F,F-!\"\"/%#tpG,&*&F+F-*$-F06#F2F-F7F(*
&*&F'F(F4F-F-*&F*\"\"\"-F06#F2F-F7F," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 92 "sol:=solve(\{subs(beta=V/c,V=c/2,c=1,rhs(eq3)=1),subs
(beta=V/c,V=c/2,c=1,rhs(eq6)=1)\},\{x,t\});" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$solG<$/%\"xG*$-%%sqrtG6#\"\"$\"\"\"/%\"tGF(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "eq6;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%#tpG,&*&%\"tG\"\"\"*$-%%sqrtG6#,&\"\"\"F.*$)%%betaG\"
\"#F(!\"\"F(!\"\"F.*&*&%\"xGF.F0F(F(*&%\"VG\"\"\"-F+6#F-F(F4F3" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "xM1:=subs(sol,x); tM1:=subs(
sol,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$xM1G*$-%%sqrtG6#\"\"$\"
\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$tM1G*$-%%sqrtG6#\"\"$\"\"\"
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "Pflash1:=plot([[0,0],[x
M1,tM1]],style=line,color=green,thickness=3):" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 100 "Now we consider the reflected flash: it is going back
wards, i.e., has to arrive at a final location " }{TEXT 475 1 "x" }
{TEXT -1 6 "' =0, " }{TEXT 476 1 "t" }{TEXT -1 173 "'=2 (in the moving
 reference frame we ignore the motion of the reference frame and use t
he constancy of the speed of light c). In the stationary frame this lo
oks as follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "sol:=sol
ve(\{subs(beta=V/c,V=c/2,c=1,rhs(eq3)=0),subs(beta=V/c,V=c/2,c=1,rhs(e
q6)=2)\},\{x,t\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<$/%\"xG,
$*$-%%sqrtG6#\"\"$\"\"\"#\"\"#F-/%\"tG,$F)#\"\"%F-" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 35 "xM2:=subs(sol,x); tM2:=subs(sol,t);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$xM2G,$*$-%%sqrtG6#\"\"$\"\"\"#\"\"#
F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$tM2G,$*$-%%sqrtG6#\"\"$\"\"\"
#\"\"%F*" }}}{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]);" }}
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G\"eF07$F_q$!1`x8D8_daF07$Fdq$!1AW!=g)*H9&F07$Fiq$!1`x8Dt\"zz%F07$F^r$
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[E#31#Fdo7$F[w$\"1nd&>3G6G&Fdo7$F`w$\"1SCi[^FZ')Fdo7$Few$\"1xb>=*HQ>\"
F07$Fjw$\"1WA'[jp$Q:F07$F_x$\"16*G:L--(=F07$Fdx$\"16*G:>\\&4AF07$Fix$
\"1yb>)3'3YDF07$F^y$\"1XA'[aG`&GF07$Fcy$\"1yb>)Hb(4KF07$Fhy$\"16*G:\")
en_$F07$F]z$\"1XA'[n]Z'QF07$Fbz$\"1WA'[,_#)=%F07$Fgz$\"1XA'[Q)*e`%F0Fj
dlFb^m-F$6%7S7$F($!1j1FU_h>8F][o7$F.$!1'*Rg&zRZG\"F][o7$F4$!1It$f&[Ra7
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7$FH$!1'*Rga*H+7\"F][o7$FM$!1j1F\\*3s3\"F][o7$FR$!1'*Rg=aE`5F][o7$FW$!
1j1FX2V>5F][o7$Ffn$!1(Ht$\\iFY)*F07$F[o$!1k*Rg<I(R&*F07$F`o$!1Jmq->j%>
*F07$Ffo$!1ImqAm6[))F07$F[p$!1JmqU\\=9&)F07$F`p$!1k*Rg)H%4@)F07$Fep$!1
(Ht$p*e.&yF07$Fjp$!1)Ht$4q*[a(F07$F_q$!1JmqK@d*=(F07$Fdq$!1*Ht$4%\\](o
F07$Fiq$!1KmqK\"o*HlF07$F^r$!1Jmq_\"p8?'F07$Fcr$!1)Ht$*e5&eeF07$Fhr$!1
)Ht$*RfOa&F07$F^s$!1m*Rg&)\\S?&F07$Fds$!1m*Rg)4H^[F07$Fis$!1Kmq_L@WXF0
7$F^t$!1l*Rg4hD@%F07$Fct$!1Kmqi=$*pQF07$Fht$!1Mmq-gtMNF07$F]u$!1LmqsnT
5KF07$Fbu$!1LmqibJ]GF07$Fgu$!1*Ht$4'[n_#F07$F\\v$!1KmqA=F\"=#F07$Fav$!
1m*Rg]=#o=F07$Ffv$!1KmqU&of_\"F07$F[w$!1+LP**z$R?\"F07$F`w$!1Jj1FHBt')
Fdo7$Few$!1**Ht$*)3AQ&Fdo7$Fjw$!1Dj1F<\"o$>Fdo7$F_x$\"1W.gR_^\"Q\"Fdo7
$Fdx$\"1S.gRQ)\\x%Fdo7$Fix$\"13qE1GNS\")Fdo7$F^y$\"1oLHPxFB6F07$Fcy$\"
1,ni!\\/xZ\"F07$Fhy$\"1M+'R+3Zz\"F07$F]z$\"1nLHn)*pK@F07$Fbz$\"1nLH27?
cCF07$Fgz$\"1nLHxv%Q!GF0FjdlFb^m-F$6&7$7$F*F*7$$\"1x)ov!30K<F0Fi]p-F\\
[l6&F^[lF*F_[lF*-Fc[l6#\"\"$-%&STYLEG6#%%LINEG-F$6&7$Fh]p7$$\"1^#z$Q0q
a6F0$\"1.&en2,%4BF0-F\\[l6&F^[lF_[lF*F_[lF]^pF`^p-%&TITLEG6#Q2spacetim
e~1+1~dim6\"-%+AXESLABELSG6$%\"xG%#ctG-%(SCALINGG6#%,CONSTRAINEDG-%%VI
EWG6$;F*$\"#X!\"\"F_`p" 1 2 0 1 0 2 9 1 4 1 1.000000 45.000000 
45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 477 11 "Exercise 3:" }}
{PARA 0 "" 0 "" {TEXT -1 194 "Repeat the calculation and graph for oth
er mirror distances than 1 m. Repeat the whole section for reference f
rames with different boosts: from less relativistic to more relativist
ic situations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 478 11 "Exercise 4:" }}{PARA 0 "" 0 "" {TEXT -1 312 "In the prob
lem discussed above the worldlines were constructed for a light pulse,
 i.e., the speed in the moving frame was given as c (with reversed dir
ection for the second half). Consider another textbook problem: an imp
regnated 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 ges
tation period is known to be 21 months. At the time of birth a radiosi
gnal is sent back toward Earth to announce the birth. When will the ra
diosignal arrive?" }}{PARA 0 "" 0 "" {TEXT -1 514 "This is a two-step \+
problem: in step 1 one determines the time dilation, i.e., one determi
nes at what point in time on Earth the elephant will give birth; in st
ep 2 one calculates the distance the spaceship has travelled during th
is time; in step 3 one finds out how long it takes for the radiosignal
 to travel (the two times are 31.7 and 23.8 months respectively). Dete
rmine these times from the relevant formulae. Draw the worldlines for \+
the two processes (why does the worldline for the elephant lie along t
he (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 fr
ames implies that it is to be drawn along the " }{TEXT 485 1 "x" }
{TEXT -1 19 "'-axis (choice of c" }{TEXT 486 1 "t" }{TEXT -1 84 "'=0 t
o determine its length), 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 ", nevertheless the same time for both ends)." }}{PARA 0 "" 0 "" 
{TEXT -1 177 "Construct the relevant spacetimediagram and use the poin
t-and-click interface 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 app
earance of solid objects to a moving observer. Length contraction occu
rs only in the direction of motion. Therefore the shape of a two-dimen
sional object (e.g., a profile) is distorted." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "Suppose a square which i
s rotated by 30 degrees resides in a reference frame that moves with s
ome 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], colo
r=red):\ncr:=rotate(c,Pi/6,[1,1]):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 65 "display(cr, scaling=constrained,view=[0..5,0..5],labe
ls=[xp,yp]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&
-%)POLYGONSG6$7&7$$\"\"\"\"\"!F(7$$\"+330KF!\"*$\"+++++?F.7$$\"+330K<F
.$\"+330KPF.7$F*F,-%'COLOURG6&%$RGBG$\"*++++\"!\")F*F*-%+AXESLABELSG6$
%#xpG%#ypG-%(SCALINGG6#%,CONSTRAINEDG-%%VIEWG6$;F*$\"\"&F*FJ" 1 2 0 1 
0 2 9 1 4 1 1.000000 45.000000 45.000000 0 }}}}{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 lengt
h contraction of the " }{TEXT 491 1 "x" }{TEXT -1 24 "-coordinates, wh
ile 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 determined relative to
 the vertex [1,1]. Let us first find the other vertices in the moving \+
reference frame." }}{PARA 0 "" 0 "" {TEXT -1 84 "A rotation matrix tha
t 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)]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$R30G7$7
$,$*$-%%sqrtG6#\"\"$\"\"\"#\"\"\"\"\"##!\"\"F07$F.F'" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 139 "To transform the vertices, we have to remove t
he dispalcement of the first vertex, apply the rotation, and then add \+
the displacement again:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "
with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Pt2p:=eval
m([1,1] + R30 &* ([3,1]-[1,1]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
%Pt2pG-%'vectorG6#7$,&\"\"\"F**$-%%sqrtG6#\"\"$\"\"\"F*\"\"#" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Pt3p:=evalm([1,1] + R30 &* (
[3,3]-[1,1]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Pt3pG-%'vectorG6#
7$*$-%%sqrtG6#\"\"$\"\"\",&\"\"#\"\"\"F)F1" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 42 "Pt4p:=evalm([1,1] + R30 &* ([1,3]-[1,1]));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Pt4pG-%'vectorG6#7$\"\"!,&\"\"\"F+*
$-%%sqrtG6#\"\"$\"\"\"F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "We pi
ck a boost velocity as a fraction of the speed of light c:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "beta:=8/10;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%betaG#\"\"%\"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 77 "Let us apply a Lorentz transformation to the observer's referen
ce frame. The " }{TEXT 497 1 "x" }{TEXT -1 64 "-ccordinates are contra
cted according the beta-dependent factor." }}{PARA 0 "" 0 "" {TEXT -1 
34 "We need to unprotect the variable " }{TEXT 19 5 "gamma" }{TEXT -1 
108 " in Maple if we wish to use it for our own purposes (instead of r
eferring to the Euler-Mascheroni constant):" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "unprotect(gamma);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&gammaG#\"\"&\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "Pt1:=[1/gamma,1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%$Pt1G7$#\"\"$\"\"&\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "Pt2:=[Pt2p[1]/gamma,Pt2p[2]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%$Pt2G7$,&#\"\"$\"\"&\"\"\"*$-%%sqrtG6#F(\"\"\"F'\"\"#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Pt3:=[Pt3p[1]/gamma,Pt3p[2]];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Pt3G7$,$*$-%%sqrtG6#\"\"$\"\"\"#F+
\"\"&,&\"\"#\"\"\"F'F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "P
t4:=[Pt4p[1]/gamma,Pt4p[2]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Pt4
G7$\"\"!,&\"\"\"F(*$-%%sqrtG6#\"\"$\"\"\"F(" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 124 "display(polygon([Pt1,Pt2,Pt3,Pt4], color=red, lin
estyle=3, thickness=2),view=[0..5,0..5],scaling=constrained,labels=[x,
y]);\n" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%)POL
YGONSG6&7&7$$\"+++++g!#5$\"\"\"\"\"!7$$\"+&[I#R;!\"*$\"\"#F-7$$\"+&[I#
R5F1$\"+330KPF17$F-$\"+330KFF1-%'COLOURG6&%$RGBG$\"*++++\"!\")F-F--%*T
HICKNESSG6#F3-%*LINESTYLEG6#\"\"$-%+AXESLABELSG6$%\"xG%\"yG-%(SCALINGG
6#%,CONSTRAINEDG-%%VIEWG6$;F-$\"\"&F-FV" 1 2 0 1 0 2 9 1 4 1 1.000000 
45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "It is e
vident that the rotated square profile has turned into a parallelogram
 for the stationary observer who sees the profile whiz by." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 556 11 "Exercise 6:" }}
{PARA 0 "" 0 "" {TEXT -1 127 "Explore the appearance of the rotated sq
uare shape for different orientations. What happens as the boost veloc
ity approaches c?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 286 "The observation of three-dimensional objects requ
ires an additional consideration: the time when the light left the obs
erved object has to be taken into account. The photons which arrive si
multaneously at the moving observer's eye: the ones arriving from the \+
more deeply located parts (" }{TEXT 498 1 "z" }{TEXT -1 82 "-direction
) must have left at an earlier time when the observer was at a previou
s " }{TEXT 499 1 "x" }{TEXT -1 154 "-location. As a result the observe
r sees a tilted object (or rather sees the side face that would be hid
den to a stationary observer of the solid object)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 353 23 "V
elocity transformation" }}{PARA 0 "" 0 "" {TEXT -1 280 "An interesting
 remaining question is the transformation property of velocities. Reme
mber that our main goal is to understand the constancy of the speed of
 light, i.e., to understand how the simple additivity of velocities in
 the Galilei transformation 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 de
fine the velocities as the derivatives of proper length with respect t
o proper time. The posed question then is how to relate the two measur
ements of these quantities. This can be achieved by differentiating th
e transformations for position and time." }}{PARA 0 "" 0 "" {TEXT -1 
36 "First we summarize the main results:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "LT1:=eq3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$LT1G/%
#xpG*&,&%\"xG\"\"\"*&%\"VGF*%\"tGF*!\"\"\"\"\"*$-%%sqrtG6#,&F*F**$)%%b
etaG\"\"#F/F.F/!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "LT2:
=eq6;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$LT2G/%#tpG,&*&%\"tG\"\"\"*
$-%%sqrtG6#,&\"\"\"F0*$)%%betaG\"\"#F*!\"\"F*!\"\"F0*&*&%\"xGF0F2F*F**
&%\"VG\"\"\"-F-6#F/F*F6F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
10 "LT1p:=eq4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%LT1pG/%\"xG*&,&%#
xpG\"\"\"*&%\"VGF*%#tpGF*F*\"\"\"*$-%%sqrtG6#,&F*F**$)%%betaG\"\"#F.!
\"\"F.!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "LT2p:=eq9;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%LT2pG/%\"tG,&*&*&%#xpG\"\"\")%%be
taG\"\"#\"\"\"F/*&%\"VG\"\"\"-%%sqrtG6#,&F+F+*$F,F/!\"\"F/!\"\"F+*&%#t
pGF/*$-F46#F6F/F9F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "The diffe
rentiation is accomplished by replacing the variables 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);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%LT1dG/%$dxpG*&,&%#dxG\"\"\"*&%\"VGF*%#dtGF*!\"\"\"\"
\"*$-%%sqrtG6#,&F*F**$)%%betaG\"\"#F/F.F/!\"\"" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 33 "LT2d:=subs(tp=dtp,x=dx,t=dt,LT2);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%%LT2dG/%$dtpG,&*&%#dtG\"\"\"*$-%%sqrtG6#,&\"
\"\"F0*$)%%betaG\"\"#F*!\"\"F*!\"\"F0*&*&%#dxGF0F2F*F**&%\"VG\"\"\"-F-
6#F/F*F6F5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "We form the veloci
ty expression in the boosted frame (think of dxp and dtp as finite dif
ferences, and do not worry about mathematical rigour for a moment):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "VT1:=vp=rhs(LT1d)/rhs(LT2d
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$VT1G/%#vpG*&,&%#dxG\"\"\"*&%
\"VGF*%#dtGF*!\"\"\"\"\"*&-%%sqrtG6#,&F*F**$)%%betaG\"\"#F/F.F/,&*&F-F
/*$-F26#F4F/!\"\"F**&*&F)F*F6F/F/*&F,\"\"\"-F26#F4F/F>F.\"\"\"F>" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "To introduce the velocity in the s
tationary (unprimed, O1) frame we make the substitution:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "VT1:=expand(simplify(subs(dx=v*dt,V
T1)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$VT1G/%#vpG,&*&*&%\"VG\"\"
\"%\"vG\"\"\"F-,&F*F+*&F,F+)%%betaG\"\"#F-!\"\"!\"\"F+*&*$)F*F2F-F-F.F
4F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "taylor(rhs(VT1),beta
=0,3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+)%%betaG,&%\"vG\"\"\"%\"VG!
\"\"\"\"!,&*&*$)F&\"\"#\"\"\"F0F(!\"\"F'F&F)\"\"#-%\"OG6#F'\"\"%" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "This allows us to transform the ve
locity of a particle in the O1 frame, i.e., " }{TEXT 347 1 "v" }{TEXT 
-1 59 " to a boosted frame O2. Since the boosted 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 f
or fixed boost velocity (" }{TEXT 19 4 "beta" }{TEXT -1 2 ", " }{TEXT 
391 1 "V" }{TEXT -1 115 " determined) how the transformed velocity dep
ends 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));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%&VT1_1G/%#vpG,$*&,&%\"vG\"\"$!\"\"\"\"\"\"\"\",&!\"$F-F*F-!\"
\"F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "VT1_2:=simplify(sub
s(beta=1/2,V=c/2,c=1,VT1)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
48 "VT1_3:=simplify(subs(beta=2/3,V=c*2/3,c=1,VT1)):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 177 "PL1:=plot([rhs(VT1_1),rhs(VT1_2),rhs(VT1
_3)],v=-1..1,color=[red,blue,green],title=\"Lorentz velocity transform
ation\",labels=[v,vp],scaling=constrained,thickness=3): display(PL1);
" }}{PARA 13 "" 1 "" {GLPLOT2D 694 323 323 {PLOTDATA 2 "6)-%'CURVESG6%
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F.7$F_u$!1`ibxVxTYF.7$Fdu$!1J6JJ7d#G%F.7$Fiu$!12yn_t[qQF.7$F^v$!1LBEeD
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7$Fey$\"15*y,T3N'\\F.7$Fjy$\"1k3+[\\,3kF.7$$\"1++++PDj$*F.$\"1W/8*y8f<
(F.7$F_z$\"1F))*>u]4+)F.7$$\"1++v=5s#y*F.$\"1]ImL\"\\)e*)F.Fcz-Fgz6&Fi
zF*FjzF*F][l-%&TITLEG6#Q@Lorentz~velocity~transformation6\"-%+AXESLABE
LSG6$%\"vG%#vpG-%(SCALINGG6#%,CONSTRAINEDG-%%VIEWG6$;F(Fdz%(DEFAULTG" 
1 2 0 1 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 133 "We can observe how the transformation allows for
 a small range of boost and particle velocity values for which additiv
ity works: the " }{TEXT 393 1 "y" }{TEXT -1 86 " intercepts 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 velocities 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 where 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 transformation\",labels=[v,vp
],scaling=constrained,view=[-1..1,-1..1],thickness=2): display(PL2);" 
}}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6)-%'CURVESG6%7S
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "The diagram shows that the Galile
i transformation is only adequate for the smallest boost velocity show
n (" }{TEXT 400 1 "V" }{TEXT -1 55 "=c/3), when the particle velocity \+
is restricted to 0 < " }{TEXT 401 1 "v" }{TEXT -1 32 " < c/3 (which re
sults in -c/3 > " }{TEXT 402 1 "v" }{TEXT -1 121 "' > 0). For the case
s of larger boosts of the reference frame the Galilei transformation n
ever works (not even for small " }{TEXT 409 1 "v" }{TEXT -1 242 " !), \+
apart from the crossings which correspond to the cases where the parti
cle velocity is zero, or where the boost velocity and particle velocit
y add up to zero (these are trivial cases for which adding velocities \+
represent the only option)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 81 "On the other hand the graphs show that for part
icle velocities in the range -c < " }{TEXT 403 1 "v" }{TEXT -1 57 " < \+
c the transformed 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 "Obser
ve the behaviour of the transformation in the limit as the boost veloc
ity " }{TEXT 406 1 "V" }{TEXT -1 47 " approaches c. What result does o
ne obtain 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 consid
er the transformation of the velocity components that are orthogonal t
o the direction of the boost. They are affected due to the time transf
ormation. Let us derive the velocity transformation for the " }{TEXT 
410 1 "y" }{TEXT -1 11 "-component:" }}}{EXCHG {PARA 0 "> " 0 "" 
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 do by a substitution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L
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6#F-F(F4F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "VT2:=vyp=simp
lify(subs(dy=vy*dt,dx=vx*dt,rhs(LT3d)/rhs(LT2d)));" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$VT2G/%$vypG,$*&*(%#vyG\"\"\"%\"VGF+-%%sqrtG6#,&F+F
+*$)%%betaG\"\"#\"\"\"!\"\"F5F5,&F,F6*&%#vxGF+F2F5F+!\"\"F6" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "An analogous result is obtained fo
r the relationship between the  " }{TEXT 411 1 "z" }{TEXT -1 128 " com
ponents of the velocity vector. Physically the meaning of this result \+
is the following: a particle moves with some velocity " }{TEXT 414 2 "
v " }{TEXT -1 3 "= [" }{TEXT 413 2 "vx" }{TEXT -1 2 ", " }{TEXT 412 2 
"vy" }{TEXT -1 84 ", 0] in a stationary rest frame. An observer in a b
oosted frame with boost velocity " }{TEXT 416 1 "V" }{TEXT -1 4 " = [
" }{TEXT 415 1 "V" }{TEXT -1 35 ", 0, 0] observes a velocity vector " 
}{TEXT 419 1 "v" }{TEXT -1 5 "' = [" }{TEXT 418 2 "vx" }{TEXT -1 3 "',
 " }{TEXT 417 2 "vy" }{TEXT -1 70 "', 0] with the transformed velocity
 components given by the equations " }{TEXT 19 3 "VT1" }{TEXT -1 5 " a
nd " }{TEXT 19 3 "VT2" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 192 
"The non-vanishing component perpendicular to the boost direction depe
nds in its transformation properties on the boost velocity and the par
ticle's velocity in this direction in the rest frame." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "R1:=subs(beta=V/c,V=c/2,c=1,rhs(VT2
));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R1G,$*&*(%#vyG\"\"\"-%%sqrtG
6#\"\"$\"\"\"-F+6#\"\"%F.F.,&#!\"\"\"\"#F)%#vxG#F)F1!\"\"#F4\"\")" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "plot3d(R1,vx=-1..1,vy=-1..1,
axes=boxed,shading=zhue,style=patchnogrid,orientation=[160,40]);" }}
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "At first sight one might think tha
t there is a problem with the resulting velocity component " }{TEXT 
433 2 "vy" }{TEXT -1 123 "' exceeding the speed of light. One has to b
ear in mind, however, that the range of vy is restricted for a given c
hoice of " }{TEXT 434 2 "vx" }{TEXT -1 73 " (the magnitude of the velo
city vector is limited by the speed of light)." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 165 "vx0:=0.7; plot(subs(vx=vx0,R1),vy=-evalf(sqrt
(1-vx0^2))..evalf(sqrt(1-vx0^2)),labels=[vy,vyp],title=cat(\"transform
ation of vy to vy' for vx= \",convert(vx0,string)));" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%$vx0G$\"\"(!\"\"" }}{PARA 13 "" 1 "" {GLPLOT2D 
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[\"**4LTF^r7$$!1#z\\`Op)\\q!#>$!1)*G482(GR*F^s7$$\"1!RY[3/&yIF^r$\"1b%
o#*)\\j,TF^r7$$\"1wAfg-s>eF^r$\"1m/g?b)Qv(F^r7$$\"1_*o%G)3.y)F^r$\"1!>
6^hT)p6F*7$$\"1P&)G^'))Q=\"F*$\"18I]E-Nx:F*7$$\"1DWx136$[\"F*$\"1\"4t.
!z,w>F*7$$\"1+O#ziBEx\"F*$\"1cA&RP\\<O#F*7$$\"16]Ab#yS4#F*$\"1=g%*4%Q+
z#F*7$$\"1eu7N&>HQ#F*$\"1E3rd^([<$F*7$$\"1t+O`#>8p#F*$\"1j)3x_qde$F*7$
$\"1\"e*GN_xqHF*$\"19q)R!G5eRF*7$$\"1!>OvV%HwKF*$\"1h2!yKg^O%F*7$$\"1j
?=(>kPc$F*$\"1w[(\\tp\"[ZF*7$$\"1m\"em#HDkQF*$\"1;hhC\\_[^F*7$$\"1x-1%
eN!eTF*$\"1dmv+`%*RbF*7$$\"1W1++))flWF*$\"1QLifhs\\fF*7$$\"1IlqJ(==w%F
*$\"1-BKIQRWjF*7$$\"1:&oeRYZ1&F*$\"17nsE&)*zu'F*7$$\"1gj-mc;l`F*$\"1x#
pC<h#[rF*7$$\"1.I'\\x>7k&F*$\"1EkBu41;vF*7$$\"11Y\\O*3w&fF*$\"19zC`5gP
zF*7$$\"1]5Fu6fSiF*$\"1`:8D:j9$)F*7$$\"1q%o>E4Ba'F*$\"1$GMikCmr)F*7$$
\"1<aDq>4JoF*$\"1ugHcLQ,\"*F*7$$\"1+++H%G99(F*$\"1T#pi8f[^*F*-%'COLOUR
G6&%$RGBG$\"#5!\"\"\"\"!F`[l-%&TITLEG6#QGtransformation~of~vy~to~vy'~f
or~vx=~.76\"-%+AXESLABELSG6$%#vyG%$vypG-%%VIEWG6$;$!+H%G99(!#5$\"+H%G9
9(Fa\\l%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 
}}}}{EXCHG {PARA 0 "" 0 "" {TEXT 437 11 "Exercise 8:" }}{PARA 0 "" 0 "
" {TEXT -1 64 "Check whether the resulting components in the transform
ed 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 interval" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 321 "The length and time elements do not re
main constant under Lorentz boosts. The length interval is contracted \+
by a factor of gamma, while the time interval is dilated by the same f
actor. The question that arises is whether there is an interval involv
ing space and time that remains invariant under a Lorentz transformati
on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 138 "T
he answer is that such an interval can be found, although it looks unn
atural at first. We specify 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;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$STiG/*$)%#dsG\"\"#\"\"\",&*&)%\"cGF)F*)%#dtGF)F*\"
\"\"*$)%#dxGF)F*!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The sepa
ration " }{TEXT 19 4 "ds^2" }{TEXT -1 49 " can be positive, negative o
r zero. The interval " }{TEXT 19 2 "ds" }{TEXT -1 489 " in spacetime d
enotes the separation between two events. The claim is that it is inva
riant under Lorentz transformations; this is of interest as it will al
low to make statements that are valid for all observers. The interval \+
is called lightlike when ds=0, as it represents the position-time rela
tionship of a particle moving with c. For ds>0 the time separation dom
inates, and the interval is called timelike; for ds<0 the space separa
tion 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 fol
low timelike worldlines. We can use example 4 to illustrate the invari
ance of ds as follows: consider a freshly impregnated elephant in a sp
aceship 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 frame (d" }
{TEXT 502 1 "x" }{TEXT -1 47 "'=0); we denote the two increments in Ma
ple by " }{TEXT 19 3 "dtp" }{TEXT -1 5 " and " }{TEXT 19 3 "dxp" }
{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;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%%betaG*&%\"VG\"\"\"%\"cG!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%&gammaG*&\"\"\"F&*$-%%sqrtG6#,&\"\"\"F,*&*
$)%\"VG\"\"#F&F&*$)%\"cG\"\"#F&!\"\"!\"\"F&F6" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 61 "Given the time interval in the elephant's (spaceship) \+
frame (" }{TEXT 19 3 "dtp" }{TEXT -1 63 ") we compute the time passed \+
in the stationary observer frame (" }{TEXT 19 2 "dt" }{TEXT -1 2 "):" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "dtp:=21*c*_months;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$dtpG,$*&%\"cG\"\"\"%(_monthsGF(\"#@
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "dt:=dtp*gamma;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dtG,$*&*&%\"cG\"\"\"%(_monthsGF)\"
\"\"*$-%%sqrtG6#,&F)F)*&*$)%\"VG\"\"#F+F+*$)F(\"\"#F+!\"\"!\"\"F+F9\"#
@" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "Now we calculate the distan
ce travelled by the elephant in the observer's frame until the elephan
t gives birth:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "dx:=dt*be
ta*c;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dxG,$*&*(%\"cG\"\"\"%(_mon
thsGF)%\"VGF)\"\"\"*$-%%sqrtG6#,&F)F)*&*$)F+\"\"#F,F,*$)F(\"\"#F,!\"\"
!\"\"F,F9\"#@" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "ds2:=simpl
ify((c*dt)^2-dx^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ds2G,$*&)%\"
cG\"\"%\"\"\")%(_monthsG\"\"#F*\"$T%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 73 "Evidently this agrees with the spacetime interval in the spaces
hip frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dsp2:=(c*dtp)
^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%dsp2G,$*&)%\"cG\"\"%\"\"\")%
(_monthsG\"\"#F*\"$T%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 503 27 "Relativ
istic Doppler effect" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 383 "We assume familiarity with the Doppler effect for soun
d waves, i.e., the change in the frequency of a monochromatic sound so
urce that is moving towards or away from an observer. For light waves \+
we do expect a difference, as the light waves are not being created in
 a medium as in the case of sound. Let us consider a light source movi
ng with respect to a stationary frame with speed " }{TEXT 504 1 "V" }
{TEXT -1 171 ", and emitting monochromatic light in two directions: on
e stationary observer is behind the light source (we call that one B),
 while the other sits in its flight path (A)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "Consider a spacetime diag
ram where the two reference frames (moving and 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: conside
r d" }{TEXT 510 1 "t" }{TEXT -1 38 " to be the time interval during wh
ich " }{TEXT 511 1 "N" }{TEXT -1 49 " waves are emitted: the source mo
ves 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 wavetrain is observed to h
ave a distance (c-" }{TEXT 515 1 "V" }{TEXT -1 3 ") d" }{TEXT 516 1 "t
" }{TEXT -1 64 ", and the wavelength is determined by dividing this di
stance by " }{TEXT 517 1 "N" }{TEXT -1 98 ". The frequency follows fro
m the usual relationship with the wavelength and the propagation speed
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "dt:='dt': dx:='dx': be
ta:='beta':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "lambda:=(c*d
t-V*dt)/N;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'lambdaG*&,&*&%\"cG\"
\"\"%#dtGF)F)*&%\"VGF)F*\"\"\"!\"\"F-%\"NG!\"\"" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 12 "f:=c/lambda;" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"fG*&*&%\"cG\"\"\"%\"NG\"\"\"F(,&*&F'F*%#dtGF*F**&%\"VGF*F-F(!\"
\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "f:=simplify(subs
(V=beta*c,c/lambda),symbolic);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
fG,$*&%\"NG\"\"\"*&%#dtG\"\"\",&!\"\"\"\"\"%%betaGF.\"\"\"!\"\"F-" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "In the moving reference frame of \+
the light source the so-called proper frequency and wavelength are det
ermined as (fp=c/lambda'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "dtp:='dtp':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "fp:=N/dt
p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fpG*&%\"NG\"\"\"%$dtpG!\"\"" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "The time interval " }{TEXT 19 
3 "dtp" }{TEXT -1 20 " is the proper time " }{TEXT 19 3 "tau" }{TEXT 
-1 36 ", as all light waves are emitted at " }{TEXT 518 1 "x" }{TEXT 
-1 15 "'=0, and thus, " }{TEXT 19 5 "dxp=0" }{TEXT -1 39 " between the
 emission of the first and " }{TEXT 519 1 "N" }{TEXT -1 60 "th wave. T
herefore, dtp and dt are related by time dilation." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%&gammaG*&\"\"\"F&*$-%%sqrtG6#,&\"\"\"F,*$)%%betaG\"
\"#F&!\"\"F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "dt:=ga
mma*dtp;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dtG*&%$dtpG\"\"\"*$-%%s
qrtG6#,&\"\"\"F-*$)%%betaG\"\"#F'!\"\"F'!\"\"" }}}{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));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#fAG,$*&*&-%%sqrtG6#,&\"\"\"F,*$)%%betaG\"\"#\"\"\"!
\"\"F1%#fpGF,F1,&F2F,F/F,!\"\"F2" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 47 "plot(fA/fp,beta=0..1,title=\"blueshift factor\");" }}
{PARA 13 "" 1 "" {GLPLOT2D 694 207 207 {PLOTDATA 2 "6&-%'CURVESG6$7U7$
\"\"!$\"\"\"F(7$$\"1+++;arz@!#<$\"1,Cuw*R?-\"!#:7$$\"1+++XTFwSF.$\"1$4
Xn[G;/\"F17$$\"1+++\"z_\"4iF.$\"1n[XQ[9k5F17$$\"1+++S&phN)F.$\"1laMQYO
(3\"F17$$\"1+++*=)H\\5!#;$\"1J4=+L166F17$$\"1+++[!3uC\"FD$\"1C@M.\\fL6
F17$$\"1+++J$RDX\"FD$\"1I&4iAIv:\"F17$$\"1+++)R'ok;FD$\"1<.E')\\(H=\"F
17$$\"1+++1J:w=FD$\"1tBtWb347F17$$\"1+++3En$4#FD$\"1yj'H(yxO7F17$$\"1+
++/RE&G#FD$\"1&otSm>>E\"F17$$\"1+++D.&4]#FD$\"1g8@C`7\"H\"F17$$\"1+++v
B_<FFD$\"1%>\\%*G$[@8F17$$\"1+++v'Hi#HFD$\"1HyD?Qz^8F17$$\"1+++(*ev:JF
D$\"1QQo[UG!Q\"F17$$\"1+++347TLFD$\"1GpQjMX:9F17$$\"1+++LY.KNFD$\"1]Hc
a4VY9F17$$\"1+++\"o7Tv$FD$\"1VqnB)\\R[\"F17$$\"1+++$Q*o]RFD$\"1,D$GB0'
=:F17$$\"1+++\"=lj;%FD$\"11AH$fI$e:F17$$\"1+++V&R<P%FD$\"1:gk]h'zf\"F1
7$$\"1+++Xh-'e%FD$\"1J[kN[QT;F17$$\"1+++R\"3Gy%FD$\"1?\"z;N%H$o\"F17$$
\"1+++.T1&*\\FD$\"1!*ebg;\"4t\"F17$$\"1+++(RQb@&FD$\"19Vr^6J$y\"F17$$
\"1,++=>Y2aFD$\"19Ru*)fjJ=F17$$\"1+++yXu9cFD$\"1k[(Gs\"*p)=F17$$\"1+++
\\y))GeFD$\"1UpWni/[>F17$$\"1+++i_QQgFD$\"1G%)))o^27?F17$$\"1+++!y%3Ti
FD$\"1v)=I]D'y?F17$$\"1+++O![hY'FD$\"1h&eQI)fe@F17$$\"1+++#Qx$omFD$\"1
?igunvOAF17$$\"1+++u.I%)oFD$\"1?wh4p*yK#F17$$\"1*****ppe*zqFD$\"13C>*p
:&=CF17$$\"1+++C\\'QH(FD$\"1z@a#Rmz_#F17$$\"1+++8S8&\\(FD$\"1aGGBF\"Gk
#F17$$\"1+++0#=bq(FD$\"1$>*G51(yx#F17$$\"1*****p?27\"zFD$\"1()pB![%HGH
F17$$\"1,++IXaE\")FD$\"1j[nP-a5JF17$$\"1+++l*RRL)FD$\"1vr\\FMG<LF17$$
\"1+++`<.Y&)FD$\"1zXt'[y9d$F17$$\"1+++8tOc()FD$\"1bXXHYa$)QF17$$\"1+++
\\Qk\\*)FD$\"1/&pEs)[ZUF17$$\"1+++p0;r\"*FD$\"10N0u*y$4[F17$$\"1+++oTA
q#*FD$\"1J>>Y`kQ^F17$$\"1*****\\w(Gp$*FD$\"1St;conTbF17$$\"1+++B-\"\\Z
*FD$\"1'y-_H^+4'F17$$\"1+++!oK0e*FD$\"1<%o48TA$oF17$$\"1+++<5s#y*FD$\"
1y.wbW)=a*F17$%%FAILGF^[l-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%&TITLEG6#Q1
blueshift~factor6\"-%+AXESLABELSG6$Q%betaFj[l%!G-%%VIEWG6$;F(F)%(DEFAU
LTG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 132 "The frequency shift is quite appreciable
 even for moderate speeds. This is caused by a series expansion with a
dditive contributions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "t
aylor(fA/fp,beta=0,5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+/%%betaG\"
\"\"\"\"!F%\"\"\"#F%\"\"#\"\"#F(\"\"$#\"\"$\"\")\"\"%-%\"OG6#F%\"\"&" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 156 "In complete analogy the redshi
ft (frequency reduction compared to the proper frequency) can be calcu
lated for observer B from whom the light source recedes." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "lambda:=(c*dt+V*dt)/N;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%'lambdaG*&,&*&*&%\"cG\"\"\"%$dtpGF*\"\"\"*$-%
%sqrtG6#,&F*F**$)%%betaG\"\"#F,!\"\"F,!\"\"F**&*&%\"VGF*F+F,F,*$-F/6#F
1F,F7F*F,%\"NGF7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "f:=c/la
mbda;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&*&%\"cG\"\"\"%\"NG\"
\"\"F(,&*&*&F'F*%$dtpGF*F(*$-%%sqrtG6#,&F*F**$)%%betaG\"\"#F(!\"\"F(!
\"\"F**&*&%\"VGF*F.F(F(*$-F16#F3F(F9F*F9" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "f:=simplify(subs(V=beta*c,c/lambda),symbolic);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&*&-%%sqrtG6#,&\"\"\"F+*$)%%bet
aG\"\"#\"\"\"!\"\"F0%\"NGF+F0*&%$dtpG\"\"\",&F+F+F.F+\"\"\"!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "fB:=simplify(algsubs(N=dtp*f
p,f));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fBG*&*&-%%sqrtG6#,&\"\"\"
F+*$)%%betaG\"\"#\"\"\"!\"\"F0%#fpGF+F0,&F+F+F.F+!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot(fB/fp,beta=0..1,title=\"redshi
ft factor\");" }}{PARA 13 "" 1 "" {GLPLOT2D 694 207 207 {PLOTDATA 2 "6
&-%'CURVESG6$7U7$\"\"!$\"\"\"F(7$$\"1nmm;arz@!#<$\"1p'*[&3`Vy*!#;7$$\"
1LL$e9ui2%F.$\"1<W(*y=N+'*F17$$\"1nmm\"z_\"4iF.$\"1V3_Jq@(R*F17$$\"1mm
mT&phN)F.$\"1q!zt*pa'>*F17$$\"1LLe*=)H\\5F1$\"12%oYG(Q+!*F17$$\"1nm\"z
/3uC\"F1$\"1x7.PM\\@))F17$$\"1++DJ$RDX\"F1$\"1lXhfK3R')F17$$\"1nm\"zR'
ok;F1$\"1D9#[6kKX)F17$$\"1++D1J:w=F1$\"1,e3'Q82F)F17$$\"1LLL3En$4#F1$
\"1mBD%GEb3)F17$$\"1nm;/RE&G#F1$\"19/**>ZVCzF17$$\"1+++D.&4]#F1$\"1+!f
\">:=XxF17$$\"1+++vB_<FF1$\"1G0&4Aasc(F17$$\"1+++v'Hi#HF1$\"1l3NLyd(R(
F17$$\"1nm\"z*ev:JF1$\"1\"y_!HX)[C(F17$$\"1LLL347TLF1$\"1@!Rrtt[1(F17$
$\"1LLLLY.KNF1$\"1@1'e&)oN\"pF17$$\"1++D\"o7Tv$F1$\"1[kXABxQnF17$$\"1L
LL$Q*o]RF1$\"18'H0#)*)\\e'F17$$\"1++D\"=lj;%F1$\"11of#eBrT'F17$$\"1++v
V&R<P%F1$\"1!*[HN[&zD'F17$$\"1LL$e9Ege%F1$\"1fW_lkT#4'F17$$\"1LLeR\"3G
y%F1$\"1uU'4jJ2%fF17$$\"1nm;/T1&*\\F1$\"1S;pYBIxdF17$$\"1mm\"zRQb@&F1$
\"1f=5$QYvg&F17$$\"1***\\(=>Y2aF1$\"1,0`\"3+'faF17$$\"1mm;zXu9cF1$\"1]
Al3.W*H&F17$$\"1+++]y))GeF1$\"1b[i,$[L8&F17$$\"1****\\i_QQgF1$\"1d%)pX
K**p\\F17$$\"1***\\7y%3TiF1$\"11i\"oSr3\"[F17$$\"1****\\P![hY'F1$\"1Vm
f+fjKYF17$$\"1LLL$Qx$omF1$\"1cf$Rye2Z%F17$$\"1+++v.I%)oF1$\"13E\">(Gs&
H%F17$$\"1mm\"zpe*zqF1$\"1*ROKXnZ8%F17$$\"1+++D\\'QH(F1$\"1sTw*y[d&RF1
7$$\"1KLe9S8&\\(F1$\"1#ox!zt%Qy$F17$$\"1***\\i?=bq(F1$\"1t\"y>pz)*f$F1
7$$\"1LLL3s?6zF1$\"1MR%)*3d\\T$F17$$\"1++DJXaE\")F1$\"1k?\"oiv[@$F17$$
\"1nmmm*RRL)F1$\"1Fr(>![^9IF17$$\"1mm;a<.Y&)F1$\"15ctl3'**z#F17$$\"1LL
e9tOc()F1$\"1;D!y=n\\d#F17$$\"1+++]Qk\\*)F1$\"1v:&QKKVN#F17$$\"1LL$3dg
6<*F1$\"1Z;=M0Fz?F17$$\"1mmmmxGp$*F1$\"1;Cxdz]/=F17$$\"1++D\"oK0e*F1$
\"1\"*fWt&[OY\"F17$$\"1+++]oi\"o*F1$\"1p6qZ`&=F\"F17$$\"1++v=5s#y*F1$
\"1>!>$45,[5F17$$\"1+]P40O\"*)*F1$\"19u^-BH!R(F.7$F)F(-%'COLOURG6&%$RG
BG$\"#5!\"\"F(F(-%&TITLEG6#Q0redshift~factor6\"-%+AXESLABELSG6$Q%betaF
h[l%!G-%%VIEWG6$;F(F)%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ta
ylor(fB/fp,beta=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#+1%%betaG\"\"\"
\"\"!!\"\"\"\"\"#F%\"\"#\"\"##F'F*\"\"$#\"\"$\"\")\"\"%#!\"$F0\"\"&-%
\"OG6#F%\"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "We note that wh
ile the graphs for the two expressions look quite different, the first
 order terms in the frequency shifts are symmetrical, i.e., to first o
rder the magnitude of the frequency shift is proportional to beta for \+
either red or blueshift." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 141 "An example where redshifts and blueshifts are rel
evant to astronomers is the estimation of the orbiting motion of stars
 and distant galaxies." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 201 "Astronomers deal with redshifts all the time as the
 galaxies recede from each other according to the expansion of the uni
verse (Hubble's law). They express the redshift as a fractional freque
ncy shift:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "z:='z': eq:=z
=simplify((fp-fB)/fB);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/%\"zG
,$*&,(!\"\"\"\"\"%%betaGF**$-%%sqrtG6#,&F+F+*$)F,\"\"#\"\"\"F*F5F+F5*$
-F/6#F1F5!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "beta=so
lve(eq,beta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%%betaG*&*&%\"zG\"\"
\",&F'F(\"\"#F(F(\"\"\",(*$)F'F*F+F(F'F*F*F(!\"\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 26 "Transverse Doppler effect:" }}{PARA 0 "" 0 "" 
{TEXT -1 286 "Naively one might think that in a direction perpendicula
r to the direction of motion of the source no frequency shifts will be
 observed. This is not correct. One has to generalize the above argume
nts in one dimension for an observer who looks at waves arriving at an
 angle theta to the " }{TEXT 520 1 "x" }{TEXT -1 56 "-axis, assuming t
hat the source still moves only in the " }{TEXT 521 1 "x" }{TEXT -1 
133 "-direction. When one carries out the analysis one obtains an expr
ession in which even for theta=Pi/2 a time dilation factor survives." 
}}{PARA 0 "" 0 "" {TEXT -1 34 "The general expression is given as" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "f:=fp/gamma/(1-beta*cos(thet
a));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&*&%#fpG\"\"\"-%%sqrtG6
#,&F(F(*$)%%betaG\"\"#\"\"\"!\"\"F1F1,&F(F(*&F/F(-%$cosG6#%&thetaGF(F2
!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eval(f,theta=Pi/2)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%#fpG\"\"\"-%%sqrtG6#,&F%F%*$)%
%betaG\"\"#\"\"\"!\"\"F." }}}{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 for both the tim
e dilation and the Doppler effect. Twins A and B are separated at " }
{TEXT 523 1 "t" }{TEXT -1 3 " = " }{TEXT 524 1 "t" }{TEXT -1 80 "' = 0
 with A remaining stationary on Earth, and B travelling on a spaceship
 for " }{TEXT 525 1 "n" }{TEXT -1 104 " years with some beta-value. Th
en B turns around and comes back to visit A on Earth again after anoth
er " }{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 they repeat the exp
eriment, and agree to send light pulses (perhaps even videoimages of t
heir faces) to each other with a frequency of 1/year. They do conclude
 that their observation is consistent with special relativity." }}
{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 through 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 worldlin
es go off at c" }{TEXT 528 1 "t" }{TEXT -1 3 " = " }{TEXT 529 1 "j" }
{TEXT -1 2 ", " }{TEXT 557 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 slope -1, and reach A
 (the c" }{TEXT 530 1 "t" }{TEXT -1 28 " axis) at certain intervals." 
}}{PARA 0 "" 0 "" {TEXT -1 221 "Initially, while the twins are recedin
g from each other both have to wait long times for the transmissions t
o arrive! The frequency of the 1/year transmission clock is dramatical
ly reduced, and the situation is symmetric." }}{PARA 0 "" 0 "" {TEXT 
-1 415 "It is when B turns around that matters begin to change dramati
cally: out of a sudden B is receiving the videoimages that travelled f
rom A at a fast rate; shortly before B arrives A will see a rapid succ
ession of videopulses from B, but they will not make up entirely for t
he difference. A net difference remains between the ageing processes f
or A and B! A had to send more images than B, and therefore has lost o
ut." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 346 "T
he 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 th
eory of relativity, but the conclusions remain the same. So it seems t
hat 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 worldlines for the spacesh
ip 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) coordinates (or unprimed an
d primed ones)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "beta:='b
eta': V:='V':" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "We re-derive equ
ations 3 and 6 that were used several times before" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 26 "eq1:=x0=xB*sqrt(1-beta^2);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$eq1G/%#x0G*&%#xBG\"\"\"-%%sqrtG6#,&F)F)*$)%%bet
aG\"\"#\"\"\"!\"\"F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq2
:=xA=rhs(eq1)+V*tA;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/%#xAG,&
*&%#xBG\"\"\"-%%sqrtG6#,&F*F**$)%%betaG\"\"#\"\"\"!\"\"F3F**&%\"VGF*%#
tAGF*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq3:=xB=solve(eq
2,xB);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/%#xBG*&,&%#xAG\"\"\"
*&%\"VGF*%#tAGF*!\"\"\"\"\"*$-%%sqrtG6#,&F*F**$)%%betaG\"\"#F/F.F/!\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "eq4:=xA*sqrt(1-beta^2
)=xB+V*tB;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq4G/*&%#xAG\"\"\"-%%
sqrtG6#,&F(F(*$)%%betaG\"\"#\"\"\"!\"\"F1,&%#xBGF(*&%\"VGF(%#tBGF(F(" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "eq4:=xA=solve(eq4,xA);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq4G/%#xAG*&,&%#xBG\"\"\"*&%\"VGF*
%#tBGF*F*\"\"\"*$-%%sqrtG6#,&F*F**$)%%betaG\"\"#F.!\"\"F.!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "eq5:=rhs(eq3)=solve(eq4,xB);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq5G/*&,&%#xAG\"\"\"*&%\"VGF)%#
tAGF)!\"\"\"\"\"*$-%%sqrtG6#,&F)F)*$)%%betaG\"\"#F.F-F.!\"\",&*&F(F)-F
16#F3F.F)*&F+F.%#tBGF)F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 
"eq6:=tB=expand(solve(eq5,tB));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$
eq6G/%#tBG,&*&%#tAG\"\"\"*$-%%sqrtG6#,&\"\"\"F0*$)%%betaG\"\"#F*!\"\"F
*!\"\"F0*&*&%#xAGF0F2F*F**&%\"VG\"\"\"-F-6#F/F*F6F5" }}}{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;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(tB_turnG\"\"
$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "beta:=0.8; V:=0.8; c:=
1; gamma:=1/sqrt(1-beta^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%beta
G$\"\")!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG$\"\")!\"\"" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG\"\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&gammaG$\"+nmmm;!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "tA_turn:=tB_turn*gamma;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%(tA_turnG$\"+,+++]!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 19 "xA_turn:=V*tA_turn;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(xA_t
urnG$\"+,+++S!\"*" }}}{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 552 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]);" }}{PARA 13 "" 1 "" {GLPLOT2D 611 324 324 
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{PARA 0 "" 0 "" {TEXT -1 422 "Observe how in the first half matters ar
e symmetric: At the end of three years in B's frame the first videosig
nal 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 one year). Both A and B are surprised that their twin partner
 appears to be younger, but then they can both figure out that the sig
nal travelled for a long time..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 161 "Now what happens? B turns around the shi
p, and races back to Earth. Within a year B will to receive a bunch of
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45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 327 "We see that B tra
nsmits his videopicture in yearly intervals while receiving three imag
es per year from A after the turnaround. Shortly before B's arrival A \+
receives also a rapid succession of transmissions (three images within
 one year), but A cannot catch up, and 10 years have passed for A, whi
le only 6 years went by for B." }}{PARA 0 "" 0 "" {TEXT -1 450 "We can
 appreciate the symmetry from the point of view of Doppler shifts of t
he frequencies: both observers see the pulses redshifted initially by \+
the same rate (factor of 3 slowdown), and later blueshifted by the sam
e amount (factor of 3 increase). Nevertheless, the outcome of the agei
ng process is non-symmetric. As mentioned above, there is really no pa
radox, as the situation is not completely symmetric: observer A whose \+
worldline follows the (c" }{TEXT 554 1 "t" }{TEXT -1 150 ")-axis remai
ns in a rest frame without change, while B undergoes qualitatively dif
ferent motion. It is the turnaround that makes the whole difference." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 555 11 "Exerci
se 9:" }}{PARA 0 "" 0 "" {TEXT -1 163 "Construct the spacetime diagram
 for another situation, such as: B moves with beta=0.9 for 5 years, an
d then turns around to reach back to Earth in another 5 years." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "281" 0 }
{VIEWOPTS 1 1 0 1 1 1803 }