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Here w e consider the equations of motion of classical particles, in particul ar, the question of how to make them consistent with the new transform ation laws (Lorentz rather than Galilei transformation between referen ce frames). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "We begin by rederiving the Lorentz transformation in a slightly more formal way. (1)" }}{PARA 0 "" 0 "" {TEXT -1 82 "Then we introduc e acceptable definitions for relativistic momentum and energy. (2)" }} {PARA 0 "" 0 "" {TEXT -1 95 "Newton's equation with relativistic momen tum is illustrated on the example of an oscillator (3)" }}{PARA 0 "" 0 "" {TEXT -1 67 "How are energy and momentum transformed between iner tial frames (4)" }}{PARA 0 "" 0 "" {TEXT -1 122 "Relativistic billiard balls: transforming energy, considering momentum and energy conservat ion in elastic collisions (5,6)" }}{PARA 0 "" 0 "" {TEXT -1 64 "Inelas tic collisions: conversion of kinetic energy into mass (7)" }}{PARA 0 "" 0 "" {TEXT -1 57 "The so-called invariant mass of a system of parti cles (8)" }}{PARA 0 "" 0 "" {TEXT -1 52 "The special case of massless \+ particles (photons) (9)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{SECT 0 {PARA 257 "" 0 "" {TEXT -1 66 "1) Lorentz transformation \+ for position, velocity, and acceleration" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "Let us consider again two reference frames with a boost w ith speed " }{TEXT 258 1 "V" }{TEXT -1 24 " between them along the " } {TEXT 257 1 "x" }{TEXT -1 97 "-axis. This choice can always be made. L et us further assume that a linear transformation mixing " }{TEXT 260 1 "x" }{TEXT -1 5 " and " }{TEXT 259 1 "t" }{TEXT -1 53 " with an as y et undetermined dimensionless parameter " }{TEXT 19 5 "gamma" }{TEXT -1 12 " determines " }{TEXT 19 2 "xp" }{TEXT -1 3 " = " }{TEXT 261 1 " x" }{TEXT -1 37 "', the position in the boosted frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "restart; unprotect(gamma); gamma:=' gamma': assume(V-c); # assumption helps in sqrt's" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L1:=xp=gamma*(x-V*t);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#L1G/%#xpG*&%&gammaG\"\"\",&%\"xGF)*&%#V|irGF) %\"tGF)!\"\"F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "This is a gener alization of the Galilei transformation: for " }{TEXT 262 1 "V" } {TEXT -1 17 "/c small we need " }{TEXT 19 7 "gamma=1" }{TEXT -1 199 " \+ to reach this nonrelativistic limit. The inverse transformation has to look practically the same, if we wish the laws of physics to look the same in each reference frame. Of course the sign of the (" }{TEXT 264 3 "V t" }{TEXT -1 27 ")-term better be different:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "L1i:=x=gamma*(xp+V*tp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L1iG/%\"xG*&%&gammaG\"\"\",&%#xpGF)*&%#V| irGF)%#tpGF)F)F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The two remai ning coordinates are unchanged under the transformation (" }{TEXT 19 4 "yp=y" }{TEXT -1 2 ", " }{TEXT 19 4 "zp=z" }{TEXT -1 39 ") due to th e choice of boost along the " }{TEXT 263 1 "x" }{TEXT -1 35 "-axis. Ac cording to Galilei we had " }{TEXT 19 4 "tp=t" }{TEXT -1 75 ". That th is has changed with the generalization, and is only recovered for " } {TEXT 19 7 "gamma=1" }{TEXT -1 75 " can be seen from the combination o f the two transformations. We can solve " }{TEXT 19 3 "L1i" }{TEXT -1 5 " for " }{TEXT 19 2 "xp" }{TEXT -1 33 " and eliminate the variable f rom " }{TEXT 19 2 "L1" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "L1ip:=xp=solve(L1i,xp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L1ipG/%#xpG*&,&%\"xG\"\"\"*(%&gammaGF*%#V|irGF*%#tpGF*!\"\"\" \"\"F,!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "L1p:=subs(xp =rhs(L1ip),L1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L1pG/*&,&%\"xG\" \"\"*(%&gammaGF)%#V|irGF)%#tpGF)!\"\"\"\"\"F+!\"\"*&F+F/,&F(F)*&F,F/% \"tGF)F.F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L1p:=tp=expan d(solve(L1p,tp));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L1pG/%#tpG,(*& %\"xG\"\"\"*&%&gammaG\"\"\"%#V|irG\"\"\"!\"\"\"\"\"*&*&F,F1F)F1F*F.F0! \"\"*&F,F*%\"tGF1F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "We see cle arly how the Galilei concept of time is generalized for non-zero " } {XPPEDIT 18 0 "gamma;" "6#%&gammaG" }{TEXT -1 8 "-values." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "Now we determine \+ " }{TEXT 19 5 "gamma" }{TEXT -1 180 " from the relativity postulates t hat the speed of light be the same in each reference frame, and that t he laws of physics be the same in each inertial reference frame. Consi der at " }{TEXT 19 1 "t" }{TEXT -1 3 " = " }{TEXT 19 2 "tp" }{TEXT -1 34 " = 0 a light flash originating at " }{TEXT 19 1 "x" }{TEXT -1 3 " \+ = " }{TEXT 19 2 "xp" }{TEXT -1 141 " = 0. The wavefront of a spherical wave for observers in the unprimed and primed reference frames is spe cified by the light front conditions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "L2:=x^2+y^2+z^2=c^2*t^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G/,(*$)%\"xG\"\"#\"\"\"\"\"\"*$)%\"yGF*F+F,*$)%\"z GF*F+F,*&)%#c|irGF*F+)%\"tGF*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L2p:=xp^2+yp^2+zp^2=c^2*tp^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L2pG/,(*$)%#xpG\"\"#\"\"\"\"\"\"*$)%#ypGF*F+F,*$)%#zpGF*F+F,*&)% #c|irGF*F+)%#tpGF*F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "The two s tatements have to transform into each other. We have the freedom of " }{TEXT 19 2 "L1" }{TEXT -1 5 " and " }{TEXT 19 3 "L1p" }{TEXT -1 96 " \+ to accomplish this. First we replace the unchanged coordinates in the \+ primed coordinate system." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "L2p:=subs(yp=y,zp=z,L2p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L2 pG/,(*$)%#xpG\"\"#\"\"\"\"\"\"*$)%\"yGF*F+F,*$)%\"zGF*F+F,*&)%#c|irGF* F+)%#tpGF*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L2p1:=subs( tp=rhs(L1p),L2p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L2p1G/,(*$)%#x pG\"\"#\"\"\"\"\"\"*$)%\"yGF*F+F,*$)%\"zGF*F+F,*&)%#c|irGF*F+),(*&%\"x GF+*&%&gammaG\"\"\"%#V|irG\"\"\"!\"\"F,*&*&F;F,F9F,F+F=F?!\"\"*&F;F+% \"tGF,F,F*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L2p2:=subs( xp=rhs(L1),L2p1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L2p2G/,(*&)%&g ammaG\"\"#\"\"\"),&%\"xG\"\"\"*&%#V|irGF/%\"tGF/!\"\"F*F+F/*$)%\"yGF*F +F/*$)%\"zGF*F+F/*&)%#c|irGF*F+),(*&F.F+*&F)\"\"\"F1\"\"\"!\"\"F/*&*&F )F/F.F/F+F1FCF3*&F)F+F2F+F/F*F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Now we compare coefficients between similar terms in " }{TEXT 19 4 "L2p2" }{TEXT -1 5 " and " }{TEXT 19 2 "L2" }{TEXT -1 1 "." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "L3:=L2-L2p2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L3G/,&*$)%\"xG\"\"#\"\"\"\"\"\"*&)%&gammaGF*F+) ,&F)F,*&%#V|irGF,%\"tGF,!\"\"F*F+F5,&*&)%#c|irGF*F+)F4F*F+F,*&F8F+),(* &F)F+*&F/\"\"\"F3\"\"\"!\"\"F,*&*&F/F,F)F,F+F3FBF5*&F/F+F4F+F,F*F+F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L3:=expand(lhs(L3)-rhs(L3 ))=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L3G/,8*$)%\"xG\"\"#\"\"\" \"\"\"*&)%&gammaGF*F+F(F+!\"\"**F.F+F)F,%#V|irGF,%\"tGF,F**(F.F+)F2F*F +)F3F*F+F0*&)%#c|irGF*F+F6F+F0*&*&F8F+F(F+F+*&)F/\"\"#F+)F2\"\"#F+!\" \"F,*&*&F8F+F(F+F+*$)F2\"\"#F+FA!\"#*&*(F8F+F)F+F3F+F+F2FAF**&*(F8F+F. F+F(F+F+*$)F2\"\"#F+FAF,*&**F8F+F.F+F)F+F3F+F+F2FAFG*(F8F+F.F+F6F+F,\" \"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "L3:=collect(L3,[x,y, z,t]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L3G/,(*&,,\"\"\"F)*$)%&ga mmaG\"\"#\"\"\"!\"\"*&*&)%#c|irGF-F.F+F.F.*$)%#V|irG\"\"#F.!\"\"F)*&*$ F2F.F.*&)F,\"\"#F.)F6\"\"#F.F8F)*&*$F2F.F.*$)F6\"\"#F.F8!\"#F))%\"xGF- F.F)*(,(*&*&F2F.F+F.F.F6F8FE*&F+F.F6F)F-*&*$F2F.F.F6F8F-F)%\"tGF)FGF)F )*&,(*$F2F.F/*&F2F.F+F.F)*&F+F.)F6F-F.F/F))FOF-F.F)\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "Obviously, there are no " }{TEXT 265 1 "y " }{TEXT -1 5 "- or " }{TEXT 266 1 "z" }{TEXT -1 150 "-dependent terms left after the subtraction. Our task looks difficult: can we satisfy \+ three conditions at the same time with a single choice of gamma?" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "L3p1:=coeff(L3,x^2);" }} {PARA 8 "" 1 "" {TEXT -1 30 "Error, unable to compute coeff" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L3p1:=op(1,op(1,L3));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p1G*&,,\"\"\"F'*$)%&gammaG\"\"#\" \"\"!\"\"*&*&)%#c|irGF+F,F)F,F,*$)%#V|irG\"\"#F,!\"\"F'*&*$F0F,F,*&)F* \"\"#F,)F4\"\"#F,F6F'*&*$F0F,F,*$)F4\"\"#F,F6!\"#F')%\"xGF+F," }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "L3p1:=coeff(L3p1,x^2)=0;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p1G/,,\"\"\"F'*$)%&gammaG\"\"#\" \"\"!\"\"*&*&)%#c|irGF+F,F)F,F,*$)%#V|irG\"\"#F,!\"\"F'*&*$F0F,F,*&)F* \"\"#F,)F4\"\"#F,F6F'*&*$F0F,F,*$)F4\"\"#F,F6!\"#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "We can't use the symbol gamma to solve L 3p1 for it, as Maple still thinks that gamma is the Euler-Mascheroni c onstant. We need a change of name:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "L3p1:=subs(gamma=eta,L3p1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p1G/,,\"\"\"F'*$)%$etaG\"\"#\"\"\"!\"\"*&*&)%#c|ir GF+F,F)F,F,*$)%#V|irG\"\"#F,!\"\"F'*&*$F0F,F,*&)F*\"\"#F,)F4\"\"#F,F6F '*&*$F0F,F,*$)F4\"\"#F,F6!\"#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(L3p1,eta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6& \"\"\"!\"\"*&%#c|irG\"\"\"*$-%%sqrtG6#,&*$)%#V|irG\"\"#F'F$*$)F&F0F'F# F'!\"\",$F%F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "simplify(1 /sqrt(1-V^2/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%#c|irG\"\"\"* $-%%sqrtG6#,&*$)%#V|irG\"\"#F%!\"\"*$)F$F.F%\"\"\"F%!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The latter is identical to the fourth sol ution found above. Note that " }{TEXT 19 7 "gamma=1" }{TEXT -1 50 " is also a solution. Now try the other conditions:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 21 "L3p2:=op(2,op(1,L3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p2G*(,(*&*&)%#c|irG\"\"#\"\"\")%&gammaGF+F,F,%#V|i rG!\"\"!\"#*&F-F,F/\"\"\"F+*&*$F)F,F,F/F0F+F3%\"tGF3%\"xGF3" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "L3p2:=L3p2/(t*x)=0;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p2G/,(*&*&)%#c|irG\"\"#\"\"\")%&g ammaGF+F,F,%#V|irG!\"\"!\"#*&F-F,F/\"\"\"F+*&*$F)F,F,F/F0F+\"\"!" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "L3p2:=subs(gamma=eta,L3p2); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(L3p2,eta);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%%L3p2G/,(*&*&)%#c|irG\"\"#\"\"\")%$etaGF+F,F,% #V|irG!\"\"!\"#*&F-F,F/\"\"\"F+*&*$F)F,F,F/F0F+\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$*&%#c|irG\"\"\"*$-%%sqrtG6#,&*$)%#V|irG\"\"#F%!\"\"* $)F$F.F%\"\"\"F%!\"\",$F#F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "We are getting some good news: " }{TEXT 19 7 "gamma=1" }{TEXT -1 104 " i s excluded now, and we still have the same candidate(s) for gamma in t he running to satisfy all terms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L3p3:=op(3,op(1,L3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p3G*&,(*$)%#c|irG\"\"#\"\"\"!\"\"*&F(F+)%&gammaGF*F+\"\"\"* &F.F+)%#V|irGF*F+F,F0)%\"tGF*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "L3p3:=coeff(L3p3,t^2)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% %L3p3G/,(*$)%#c|irG\"\"#\"\"\"!\"\"*&F(F+)%&gammaGF*F+\"\"\"*&F.F+)%#V |irGF*F+F,\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "L3p3:=su bs(gamma=eta,L3p3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(L3p3,e ta);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L3p3G/,(*&)%$etaG\"\"#\"\" \")%#V|irGF*F+!\"\"*&)%#c|irGF*F+F(F+\"\"\"*$F0F+F.\"\"!" }}{PARA 11 " " 1 "" {XPPMATH 20 "6$*&%#c|irG\"\"\"*$-%%sqrtG6#,&*$)%#V|irG\"\"#F%! \"\"*$)F$F.F%\"\"\"F%!\"\",$F#F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "The magic has happened: all three conditions are satisfied by one constraint on gamma. We know how to pick out the choice for gamma tha t makes more sense: the sign of " }{TEXT 19 7 "c^2-V^2" }{TEXT -1 81 " should be positive for a real-valued answer. The Lorentz transformati on becomes:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "L1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#xpG*&%&gammaG\"\"\",&%\"xGF'*&%#V|irGF'% \"tGF'!\"\"F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "L1p;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%#tpG,(*&%\"xG\"\"\"*&%&gammaG\"\"\"% #V|irG\"\"\"!\"\"\"\"\"*&*&F*F/F'F/F(F,F.!\"\"*&F*F(%\"tGF/F/" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 44 "simplify(subs(gamma=1/sqrt(1-V^2/c^2),L1p));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%#tpG,$*&,&*&%\"xG\"\"\"%#V|irGF*F**& )%#c|irG\"\"#\"\"\"%\"tGF*!\"\"F0*&F.\"\"\"-%%sqrtG6#,&*$)F+F/F0F2*$F- F0F*F0!\"\"F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "One can convince oneself that this is equivalent to the compact form:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "L1p:=tp=gamma*(t-V*x/c^2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L1pG/%#tpG*&%&gammaG\"\"\",&%\"tGF) *&*&%#V|irGF)%\"xGF)\"\"\"*$)%#c|irG\"\"#F0!\"\"!\"\"F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The inverse transformations are given as: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "L1i;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#/%\"xG*&%&gammaG\"\"\",&%#xpGF'*&%#V|irGF'%#tpGF'F'F' " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "and an equation for " }{TEXT 267 1 "t" }{TEXT -1 48 " that has to be the symmetric equivalent of L1 p:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L1pi:=t=gamma*(tp+V*x p/c^2);;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%L1piG/%\"tG*&%&gammaG\" \"\",&%#tpGF)*&*&%#V|irGF)%#xpGF)\"\"\"*$)%#c|irG\"\"#F0!\"\"F)F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "These are supplemented by the stat ements (" }{TEXT 19 6 "y = yp" }{TEXT -1 6 ", and " }{TEXT 19 6 "z = z p" }{TEXT -1 18 ", and vice versa)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 63 "We can use the transformations to re-ite rate previous findings." }}{PARA 0 "" 0 "" {TEXT 268 33 "Transformatio n of time intervals:" }{TEXT -1 160 " take two events (x_a, t_0) and ( x_b, t_1), and translate them into the frame of a moving observer. In \+ particular look at the time interval in the primed frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "TI:=subs(x=x_b,t=t_1,rhs(L1p))-subs (x=x_a,t=t_0,rhs(L1p));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#TIG,&*&% &gammaG\"\"\",&%$t_1GF(*&*&%#V|irGF(%$x_bGF(\"\"\"*$)%#c|irG\"\"#F/!\" \"!\"\"F(F(*&F'F/,&%$t_0GF(*&*&F-F/%$x_aGF(F/*$)F2\"\"#F/F4F5F(F5" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "expand(TI);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&%&gammaG\"\"\"%$t_1GF&F&*&*(F%\"\"\"%#V|irGF&% $x_bGF&F**$)%#c|irG\"\"#F*!\"\"!\"\"*&F%F*%$t_0GF&F2*&*(F%F*F+F*%$x_aG F&F**$)F/\"\"#F*F1F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Not only the temporal separation is important, but also the spatial separation of the clocks that measured " }{TEXT 269 1 "t" }{TEXT -1 19 " in the \+ rest frame." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 270 23 "Velocity transformation" }}{PARA 0 "" 0 "" {TEXT -1 99 "The velocity transformation was also derived in the previous works heet by forming differentials of " }{TEXT 19 2 "L1" }{TEXT -1 5 " and \+ " }{TEXT 19 3 "L1p" }{TEXT -1 35 " and the corresponding derivatives. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Lv1:=up_x=(u_x-V)/(1-V* u_x/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Lv1G/%%up_xG*&,&%$u_xG \"\"\"%#V|irG!\"\"\"\"\",&F*F**&*&F+F*F)F*F-*$)%#c|irG\"\"#F-!\"\"F,F5 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Lv2:=up_y=u_y/(gamma*(1 -V*u_x/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Lv2G/%%up_yG*&%$u_ yG\"\"\"*&%&gammaG\"\"\",&\"\"\"F.*&*&%#V|irGF.%$u_xGF.F)*$)%#c|irG\" \"#F)!\"\"!\"\"\"\"\"F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " Lv3:=up_z=u_z/(gamma*(1-V*u_x/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%$Lv3G/%%up_zG*&%$u_zG\"\"\"*&%&gammaG\"\"\",&\"\"\"F.*&*&%#V|irGF. %$u_xGF.F)*$)%#c|irG\"\"#F)!\"\"!\"\"\"\"\"F7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "The inverse transformation is given by symmetry (keep \+ in mind the sign reversal on " }{TEXT 271 1 "V" }{TEXT -1 5 ") by:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Lvi1:=u_x=(up_x+V)/(1+V*up_ x/c^2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Lvi2:=u_y=up_y/(gamma*(1 +V*up_x/c^2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Lvi3:=u_z=up_z/(g amma*(1+V*up_x/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Lvi1G/%$u_ xG*&,&%%up_xG\"\"\"%#V|irGF*\"\"\",&F*F**&*&F+F*F)F*F,*$)%#c|irG\"\"#F ,!\"\"F*F4" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Lvi2G/%$u_yG*&%%up_yG \"\"\"*&%&gammaG\"\"\",&\"\"\"F.*&*&%#V|irGF.%%up_xGF.F)*$)%#c|irG\"\" #F)!\"\"F.\"\"\"F7" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Lvi3G/%$u_zG* &%%up_zG\"\"\"*&%&gammaG\"\"\",&\"\"\"F.*&*&%#V|irGF.%%up_xGF.F)*$)%#c |irG\"\"#F)!\"\"F.\"\"\"F7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "To understand relativistic dynamics we have to consider now the accelera tion. In Galilean reference frames " }}{PARA 0 "" 0 "" {TEXT -1 25 "We have to differentiate " }{TEXT 19 4 "up_x" }{TEXT -1 17 " with respec t to " }{TEXT 19 2 "tp" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Lv1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%%up_xG*&,&%$ u_xG\"\"\"%#V|irG!\"\"\"\"\",&F(F(*&*&F)F(F'F(F+*$)%#c|irG\"\"#F+!\"\" F*F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subs(tp=dtp,t=dt,x= dx,L1p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$dtpG*&%&gammaG\"\"\",&% #dtGF'*&*&%#V|irGF'%#dxGF'\"\"\"*$)%#c|irG\"\"#F.!\"\"!\"\"F'" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "First we need to form the differen tials in " }{TEXT 19 3 "Lv1" }{TEXT -1 72 ". The appearance of u_x in \+ the denominator complicates matters a little." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "What happens when we form the derivative? " }{TEXT 272 1 "V" }{TEXT -1 53 " is a consta nt, i.e., in the numerator we obtain the " }{TEXT 19 12 "diff(u_x, t) " }{TEXT -1 25 " derivative; there is no " }{TEXT 273 1 "x" }{TEXT -1 174 "-dependence, and so apart from the gamma factor we expect no surp rises from taking the derivative. However, the denominator also gets d ifferentiated with respect to time, as " }{TEXT 19 3 "u_x" }{TEXT -1 209 " appears there as well. First we form the differential by taking \+ a time derivative. For this purpose we have to introduce an explicit t ime dependence, and then remove it to recover the variables used norma lly." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "subs(u_x=u_x(t),rhs (Lv1));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "expand(diff(%,t));" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "subs((c^2-V^2)=c^2/gamma^2,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&-%$u_xG6#%\"tG\"\"\"%#V|irG!\"\"\"\"\",&F)F)*&*&F*F )F%F)F,*$)%#c|irG\"\"#F,!\"\"F+F4" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#, (*&-%%diffG6$-%$u_xG6#%\"tGF+\"\"\",&\"\"\"F.*&*&%#V|irGF.F(F.F,*$)%#c |irG\"\"#F,!\"\"!\"\"F6F.*&*(F1F,F%F.F(F,F,*&)F-\"\"#F,)F4\"\"#F,F6F.* &*&)F1\"\"#F,F%F,F,*&)F-\"\"#F,)F4\"\"#F,F6F7" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*(-%%diffG6$-%$u_xG6#%\"tGF+\"\"\")%#c|irG\"\"#\"\"\" ,&*$)%#V|irGF/F0!\"\"*$F-F0F,F,F0*$),&F6F5*&F4F,F(F,F,\"\"#F0!\"\"" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&-%%diffG6$-%$u_xG6#%\"tGF+\"\"\")% #c|irG\"\"%\"\"\"F0*&)%&gammaG\"\"#F0),&*$)F.\"\"#F0!\"\"*&%#V|irGF,F( F,F,\"\"#F0!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "dup_x:= subs(diff(u_x(t),t)=du_x,u_x(t)=u_x,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&dup_xG*&*&%%du_xG\"\"\")%#c|irG\"\"%\"\"\"F,*&)%&gammaG\"\"#F ,),&*$)F*\"\"#F,!\"\"*&%#V|irGF(%$u_xGF(F(\"\"#F,!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Now we form the difference quotient:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "dup_x/expand(rhs(subs(tp=dtp ,t=dt,x=dx,L1p)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&%%du_xG\"\" \")%#c|irG\"\"%\"\"\"F**()%&gammaG\"\"#F*),&*$)F(\"\"#F*!\"\"*&%#V|irG F&%$u_xGF&F&\"\"#F*,&*&F-F&%#dtGF&F&*&*(F-F*F6F*%#dxGF&F**$)F(\"\"#F*! \"\"F4\"\"\"FB" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "and remove the dx in the denominator by the trick used in the derivation of the velo city transformation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "sub s(dx=u_x*dt,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&%%du_xG\"\"\")% #c|irG\"\"%\"\"\"F**()%&gammaG\"\"#F*),&*$)F(\"\"#F*!\"\"*&%#V|irGF&%$ u_xGF&F&\"\"#F*,&*&F-F&%#dtGF&F&*&**F-F*F6F*F7F*F;F*F**$)F(\"\"#F*!\" \"F4\"\"\"FA" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Now we can introd uce the acceleration in the rest frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "simplify(subs(du_x=a_x*dt,%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&%$a_xG\"\"\")%#c|irG\"\"'\"\"\"F**&)%&gammaG\"\"$F* ),&*$)F(\"\"#F*F&*&%#V|irGF&%$u_xGF&!\"\"\"\"$F*!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "Apart from the c still present in the n umerator, we generated the textbook result for the acceleration in the moving reference frame, " }{TEXT 19 4 "ap_x" }{TEXT -1 147 ". Accordi ng to Galilei the acceleration is the same in two inertial reference f rames. According to Lorentz it is not, unless if we take the limit (" }{TEXT 19 7 "gamma=1" }{TEXT -1 5 " and " }{TEXT 274 1 "V" }{TEXT -1 81 "/c approximately 0). We cannot expect Newton's law to survive in t he simple form " }{TEXT 277 1 "F" }{TEXT -1 3 " = " }{TEXT 275 1 "m" } {TEXT 276 1 "a" }{TEXT -1 26 " in the relativistic case." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{SECT 0 {PARA 3 "" 0 "" {TEXT 454 43 "2) Relativistic momentum and ki netic energy" }}{EXCHG {PARA 0 "" 0 "" {TEXT 278 21 "Relativistic mome ntum" }}{PARA 0 "" 0 "" {TEXT -1 271 "We should first look at relativi stic momentum to understand the situation when no forces act on a part icle. Under these circumstances momentum is conserved, and we can ask \+ how the nonrelativistic definition of momentum should be extended to m aintain the conservation law." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 52 "Consider two observers with identical bal ls of mass " }{TEXT 279 1 "m" }{TEXT -1 82 " in the respective rest fr ame. The observers move with a boost velocity along the " }{TEXT 280 1 "x" }{TEXT -1 50 "-axis. They throw their balls in their respective \+ " }{TEXT 284 1 "y" }{TEXT -1 125 "-directions (one up and one down), s uch that the balls collide and bounce elastically, i.e., reverse their velocities in the " }{TEXT 281 1 "y" }{TEXT -1 111 "-direction. From \+ the velocity transformation discussed before we know that even the vel ocity components in the " }{TEXT 282 1 "y" }{TEXT -1 90 "-direction wi ll transform between the two frames as a result of the relative boost \+ in the " }{TEXT 283 1 "x" }{TEXT -1 122 "-direction. What will either \+ observer find when comparing the magnitudes of the v_y velocity compon ents for the two balls?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Suppose observer A is at rest. Initially, A's ball h as no " }{TEXT 285 1 "x" }{TEXT -1 45 "-component and moves with speed u0 along the " }{TEXT 286 1 "y" }{TEXT -1 32 "-axis in the positive d irection." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vA_y:=u0;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vA_yG%#u0G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "vA_x:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vA _xG\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "In B's rest frame (de noted by primed coordinates) ball B runs downward along the " }{TEXT 289 1 "y" }{TEXT -1 6 "-axis:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "vB_yp:=-u0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&vB_ypG,$%#u0G! \"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vB_xp:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&vB_xpG\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "In A's reference frame ball B has an " }{TEXT 288 1 "x" } {TEXT -1 8 "- and a " }{TEXT 287 1 "y" }{TEXT -1 11 "-component:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "vB_x:=V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vB_xG%#V|irG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% &gammaG*&\"\"\"F&*$-%%sqrtG6#,&\"\"\"F,*$)%%betaG\"\"#F&!\"\"F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "vB_y:=vB_yp/gamma/(1+V*vB _xp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vB_yG,$*&%#u0G\"\"\"-%%sqr tG6#,&F(F(*$)%%betaG\"\"#\"\"\"!\"\"F1F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 204 "Here we used the velocity transformation law from the pr imed (moving) to the unprimed (rest) frame, which simplifies for the c ase of an object moving with the boosted frame in the direction of the boost (" }{TEXT 19 7 "vB_xp=0" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 184 "The result is that ball \+ B from A's perspective moves with reduced speed, which is a consequenc e of the time dilation factor associated with the motion of B's refere nce frame along the " }{TEXT 290 1 "x" }{TEXT -1 71 "-axis. It takes m ore time for B to travel a certain distance along the " }{TEXT 291 1 " y" }{TEXT -1 146 "-direction in the unprimed frame, than the time it t akes to travel the same distance in B's rest frame (primed). In A's re st frame (unprimed) the " }{TEXT 292 1 "y" }{TEXT -1 285 "-component o f total momentum as defined according to Newton is not zero before the collision! Thus, the sum of mass times velocity for all particles is \+ not conserved in A's rest frame. The same holds in B's frame. The clas sical momentum is only conserved in the non-relativistic limit." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 326 "How can \+ we save the conservation law? The velocities were transformed accordin g to the Lorentz transformation, so they should be correct. The only o ther freedom in converting the momentum to a relativistically acceptab le quantity is to tamper with the mass. Let us call the speed of the b all from the opposite reference frame " }{TEXT 293 1 "u" }{TEXT -1 45 " to distinguish it from the own ball's speed " }{TEXT 294 1 "u" } {TEXT -1 59 "0, and to account for the motion along the boost directio n:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u:=sqrt(V^2+vB_y^2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"uG*$-%%sqrtG6#,(*$)%#V|irG\"\" #\"\"\"\"\"\"*$)%#u0GF-F.F/*&F1F.)%%betaGF-F.!\"\"F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "We formulate the momentum conservation using a mass that depends on the speed, while using velocity reversal upon im pact (the signs are equal as " }{TEXT 19 4 "vB_y" }{TEXT -1 34 " carri es a minus sign relative to " }{TEXT 303 1 "u" }{TEXT -1 3 "0):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "MC:=m(u0)*u0+m(u)*vB_y=-m(u0 )*u0-m(u)*vB_y;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#MCG/,&*&-%\"mG6# %#u0G\"\"\"F+F,F,*(-F)6#*$-%%sqrtG6#,(*$)%#V|irG\"\"#\"\"\"F,*$)F+F8F9 F,*&F;F9)%%betaGF8F9!\"\"F9F,F+F9-F26#,&F,F,*$F=F9F?F9F?,&F'F?F-F," }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "MC:=simplify(lhs(MC)-rhs(MC ))=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#MCG/,&*&-%\"mG6#%#u0G\"\" \"F+F,\"\"#*(-F)6#*$-%%sqrtG6#,(*$)%#V|irGF-\"\"\"F,*$)F+F-F9F,*&F;F9) %%betaGF-F9!\"\"F9F,F+F9-F36#,&F,F,*$F=F9F?F9!\"#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "solve(MC,m(u));" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#*&-%\"mG6#%#u0G\"\"\"*$-%%sqrtG6#,&\"\"\"F.*$)%%betaG \"\"#F(!\"\"F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "abs( u0/vB_y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&\"\"\"F$*$-%%sqrtG6#-%$ absG6#,&!\"\"\"\"\"*$)%%betaG\"\"#F$F.F$!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "limit(u,u0=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#*$-%%sqrtG6#*$)%#V|irG\"\"#\"\"\"F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "For small speeds of the balls in their respective rest fr ames " }{TEXT 295 1 "u" }{TEXT -1 48 " is given by the boost speed and the mass ratio " }{TEXT 302 1 "m" }{TEXT -1 1 "(" }{TEXT 301 1 "u" } {TEXT -1 2 ")/" }{TEXT 300 1 "m" }{TEXT -1 1 "(" }{TEXT 299 1 "u" } {TEXT -1 68 "0) just describes the increase in mass from the mass at r est (speed " }{TEXT 298 1 "u" }{TEXT -1 15 "0) to the mass " }{TEXT 297 1 "m" }{TEXT -1 1 "(" }{TEXT 296 1 "V" }{TEXT -1 265 "). This rati o is just given by the gamma factor. A stationary observer measures th e mass of a moving object to be heavier than the rest mass. This cause s the relativistic momentum of a material object to increase faster th an just proportional to the particle speed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 271 "This mass increase over the re st mass causes a problem when we attempt to accelerate particles: once particles reach a significant fraction of the speed of light c most o f the additional energy deposited goes into increasing the mass rather than speeding up the particle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 304 12 "Newton's law" }}{PARA 0 "" 0 "" {TEXT -1 84 "We can salvage Newton's law by simply demanding that rela tivistic momentum be used. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 305 99 "The rate of change of relativistic momentum of a particle is proportional to the applied net force." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 168 "This will have to be \+ verified by introducing energy: kinetic energy is defined by the work \+ done by a net force while accelerating a particle from rest to some ve locity." }}{PARA 0 "" 0 "" {TEXT -1 18 "In one dimension (" }{TEXT 306 1 "m" }{TEXT -1 24 " denotes the rest mass):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "Tkin:=Int(F,x=x0..x1)=Int(Diff(gamma*m*u,t),x=x0..x1);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%TkinG/-%$IntG6$%\"FG/%\"xG;%#x0G%#x 1G-F'6$-%%DiffG6$*&*&%\"mG\"\"\"%\"uGF7\"\"\"*$-%%sqrtG6#,&F7F7*$)%%be taG\"\"#F9!\"\"F9!\"\"%\"tGF*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 " The gamma (beta) factor refers to the actual speed " }{TEXT 311 1 "u" }{TEXT -1 21 " as attained at time " }{TEXT 310 1 "t" }{TEXT -1 14 ", \+ or position " }{TEXT 309 1 "x" }{TEXT -1 28 ". The integration paramet er " }{TEXT 308 1 "x" }{TEXT -1 30 " should be converted to speed " } {TEXT 307 1 "u" }{TEXT -1 7 " using " }{TEXT 19 17 "u(t)=diff(x(t),t) " }{TEXT -1 115 ". This results in the total differential D[m*gamma*u] to be integrated. The differential needs to be converted to d" } {TEXT 312 1 "u" }{TEXT -1 57 " which can be calculated by differentiat ing (chain rule):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "gamma: ='gamma':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "subs(diff(u(t) ,t)=du,u(t)=u,simplify(diff(m*u(t)/sqrt(1-(u(t)/c)^2),t)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*(%\"mG\"\"\"%#duGF&)%#c|irG\"\"$\"\"\"F+* $),&*$)F)\"\"#F+F&*$)%\"uGF1F+!\"\"#\"\"$F1F+!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "To carry out the integration we need to remove th e du, and multiply with the " }{TEXT 315 1 "u" }{TEXT -1 71 "-factor w hich arose during the shift of the integration variable from d" } {TEXT 314 1 "x" }{TEXT -1 5 " to d" }{TEXT 313 1 "u" }{TEXT -1 1 "." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "int(%*u/du,u=0..U);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&*()%#c|irG\"\"#\"\"\"%\"mG\"\"\",&F& F**$-%%sqrtG6#,&*$F%F(F**$)%\"UGF'F(!\"\"F(F5F*F(*$-F.6#F0F(!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Tkin:=expand(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%TkinG,&*&*&%\"mG\"\"\")%#c|irG\"\"$\"\"\" F-*$-%%sqrtG6#,&*$)F+\"\"#F-F)*$)%\"UGF5F-!\"\"F-!\"\"F)*&F4F-F(F-F9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs((c^2-U^2)=c^2/gamma^ 2,Tkin);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&*&%\"mG\"\"\")%#c|irG \"\"$\"\"\"F+*$-%%sqrtG6#*&*$)F)\"\"#F+F+*$)%&gammaG\"\"#F+!\"\"F+F8F' *&F2F+F&F+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify (%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*()%#c|irG\"\"#\"\"\"%&gamma G\"\"\"%\"mGF*F**&F%F(F+F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Tkin:=sort(%,[gamma,m,c]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%%TkinG,&*(%&gammaG\"\"\"%\"mGF()%#c|irG\"\"#\"\"\"F(*&F)F-F*F-!\"\" " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 345 "The relativistic kinetic ene rgy is made up of a term that contains the energy of the rest mass, an d then the rest mass energy is subtracted. It is of interest to Taylor expand this expression in order to see how the nonrelativistic kineti c energy is recovered. We have to substitute the gamma factor again, a nd note that we used the final speed " }{TEXT 317 1 "U" }{TEXT -1 47 " as an integration parameter. We return to use " }{TEXT 316 1 "V" } {TEXT -1 18 " in the form beta=" }{TEXT 319 1 "V" }{TEXT -1 3 "/c." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(gamma=1/sqrt(1-beta^2) ,Tkin);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&*&%\"mG\"\"\")%#c|irG\" \"#\"\"\"F+*$-%%sqrtG6#,&F'F'*$)%%betaGF*F+!\"\"F+!\"\"F'*&F&F+F(F+F4 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "taylor(%,beta=0,5);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#+)%%betaG,$*&%\"mG\"\"\")%#c|irG\"\"# \"\"\"#F(F+\"\"#,$F&#\"\"$\"\")\"\"%-%\"OG6#F(\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(beta=V/c,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&%\"mG\"\"\")%#V|irG\"\"#\"\"\"#F&F)*&*&F%F*)F(\"\"% F*F**$)%#c|irG\"\"#F*!\"\"#\"\"$\"\")-%\"OG6#*&*$)F(\"\"'F*F**$)F2\"\" 'F*F4F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "We have arrived at th e nonrelativistic result, and determined the lowest-order relativistic correction to the kinetic energy of a massive particle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "TkNRC:=convert(%,polynom);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&TkNRCG,&*&%\"mG\"\"\")%#V|irG\"\"# \"\"\"#F(F+*&*&F'F,)F*\"\"%F,F,*$)%#c|irG\"\"#F,!\"\"#\"\"$\"\")" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "TkNR:=1/2*m*V^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%TkNRG,$*&%\"mG\"\"\")%#V|irG\"\"#\"\"\"#F (F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "It is worthwhile to look \+ at a graphical presentation of these results. We pick units where c is one, and the electron mass " }{TEXT 318 1 "m" }{TEXT -1 227 "=1, and \+ graph the kinetic energy of an electron as a function of speed. To con vert to SI, or particle physics units, we note that the mass of the el ectron is 0.511 MeV; this corresponds to one energy unit on our vertic al scale." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "P1:=plot(subs( m=1,c=1,TkNR),V=0..1,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "P2:=plot(subs(m=1,c=1,TkNRC),V=0..1,color=maroon):" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "P3:=plot(subs(gamma=1/sqrt (1-V^2/c^2),m=1,c=1,Tkin),V=0..1,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "plots[display](P1,P2,P3,title=\"Kinetic energy \+ as a function of speed: relativstic and nonrelativistic\",view=[0..1,0 ..3]);" }}{PARA 13 "" 1 "" {GLPLOT2D 693 299 299 {PLOTDATA 2 "6(-%'CUR VESG6$7S7$\"\"!F(7$$\"1nmm;arz@!#<$\"1qr#)['zbP#!#>7$$\"1LL$e9ui2%F,$ \"1OY*fX0!3$)F/7$$\"1nmm\"z_\"4iF,$\"1)48&>*yw#>!#=7$$\"1mmmT&phN)F,$ \"1&)QXq%y7\\$F:7$$\"1LLe*=)H\\5!#;$\"1.VJ`M80bF:7$$\"1nm\"z/3uC\"FC$ \"1Wj.!>M,y(F:7$$\"1++DJ$RDX\"FC$\"1)49WDN\\0\"F,7$$\"1nm\"zR'ok;FC$\" 1$Q/#F,7$$\"1nm;/RE&G#FC$\"1QN%eb:7h#F,7$$\"1+++D.&4]#FC$\"1-)eSEwt7$ F,7$$\"1+++vB_>L0c&*F,7$$\"1LL$e9Ege%FC$\"1ZL^!z\"e^5FC7$$\"1LLeR\"3Gy%FC$\"1HK+ &oiP9\"FC7$$\"1nm;/T1&*\\FC$\"1Y2aFC$\"1rl8?A.i9FC7$$\"1mm;zXu9cFC$\"1tSYMyEw:FC7$$ \"1+++]y))GeFC$\"1\")QRymz)p\"FC7$$\"1****\\i_QQgFC$\"1f)=*G[5B=FC7$$ \"1***\\7y%3TiFC$\"1FC7$$\"1****\\P![hY'FC$\"1aK9ANb!4#FC7$$\" 1LLL$Qx$omFC$\"1vvKYGOBAFC7$$\"1+++v.I%)oFC$\"1d7m#ez'pBFC7$$\"1mm\"zp e*zqFC$\"1$H5#e2H1DFC7$$\"1+++D\\'QH(FC$\"1is?xK-gEFC7$$\"1KLe9S8&\\(F C$\"1.#G[p^)3GFC7$$\"1***\\i?=bq(FC$\"15DMT0voHFC7$$\"1LLL3s?6zFC$\"1k #fY(*f$HJFC7$$\"1++DJXaE\")FC$\"1k%>4IO?I$FC7$$\"1nmmm*RRL)FC$\"1+-Sox ssMFC7$$\"1mm;a<.Y&)FC$\"1\\7;PHt^OFC7$$\"1LLe9tOc()FC$\"1n^RF%)pLQFC7 $$\"1+++]Qk\\*)FC$\"1T@4_i![+%FC7$$\"1LL$3dg6<*FC$\"1*R+3J4b?%FC7$$\"1 mmmmxGp$*FC$\"1#[IFmx\"*Q%FC7$$\"1++D\"oK0e*FC$\"1m\\#GKI$*e%FC7$$\"1+ +v=5s#y*FC$\"1_Y`E:3&y%FC7$$\"\"\"F($\"1+++++++]FC-%'COLOURG6&%$RGBGF( F($\"*++++\"!\")-F$6$7SF'7$F*$\"1etibhUwBF/7$F1$\"1hJ>**)e$=$)F/7$F6$ \"1,#Q&QGDL>F:7$F<$\"1o7\\3?c4NF:7$FA$\"1/D#[>$f]bF:7$FG$\"1.)e?#*H4(y F:7$FL$\"1%=z-eG;2\"F,7$FQ$\"18Q78$)Q99F,7$FV$\"1ulnJzV1=F,7$Fen$\"1Tj TlyyjAF,7$Fjn$\"1F(HgD#\\8FF,7$F_o$\"1yP^)\\$3uKF,7$Fdo$\"1l>Nx#yp*QF, 7$Fio$\"1cdOOrOcXF,7$F^p$\"1Pmi,CQ2_F,7$Fcp$\"1v)y2/\\)[gF,7$Fhp$\"1\" 4>_bb7#oF,7$F]q$\"1MP)4r;:z(F,7$Fbq$\"1Qu@\"H.vr)F,7$Fgq$\"1WC&f)HD4)* F,7$F\\r$\"1$GyWc#e#4\"FC7$Far$\"1NU$)HaX<7FC7$Ffr$\"1*4Wpk\"**R8FC7$F [s$\"1X))GfT)4[\"FC7$F`s$\"1aF>#fpvj\"FC7$Fes$\"1#>W?&Hm#y\"FC7$Fjs$\" 1#)*et(3'*[>FC7$F_t$\"1$>s:?$oJ@FC7$Fdt$\"1[H]6:m@BFC7$Fit$\"1=*p0o.l^ #FC7$F^u$\"1*3&[Ec6YFFC7$Fcu$\"1`?H)>k['HFC7$Fhu$\"1,NL\"y')>@$FC7$F]v $\"1TQ\"Hw9&[MFC7$Fbv$\"1K**\\_=Q@PFC7$Fgv$\"1$4f8())H#*RFC7$F\\w$\"1G 2&y]s2H%FC7$Faw$\"1z\"4B0%H)f%FC7$Ffw$\"1[z!3'GbP\\FC7$F[x$\"1vGT\"\\. sd5kFC7$F _y$\"1FZG&fb%eoFC7$Fdy$\"1;B)fu4*ysFC7$Fiy$\"1s]iWKi[xFC7$F^z$\"1J@()3 Bj>#)FC7$Fcz$\"1++++++]()FC-Fhz6&Fjz$\")viobF][l$\")!\\DP\"F][l$\")%yg >%F][l-F$6$7UF'7$$\"1+++;arz@F,$\"1'*)p%*[EkP#F/7$$\"1+++XTFwSF,$\"1)4 QFDt$=$)F/7$$\"1+++\"z_\"4iF,$\"1C/i13FL>F:7$$\"1+++S&phN)F,$\"1>MS\\! p'4NF:7$$\"1+++*=)H\\5FC$\"1t$H/N95b&F:7$$\"1+++[!3uC\"FC$\"10@*)=N7sy F:7$$\"1+++J$RDX\"FC$\"1!\\_'4w#>2\"F,7$$\"1+++)R'ok;FC$\"1B2w'))p]T\" F,7$$\"1+++1J:w=FC$\"15#>\"oT%y!=F,7$$\"1+++3En$4#FC$\"1A,*42DlE#F,7$$ \"1+++/RE&G#FC$\"1sIL/o:=FF,7$F_o$\"1?IO1Q<#G$F,7$Fdo$\"1MB@xbV5RF,7$F io$\"1zdkz,exXF,7$$\"1+++(*ev:JFC$\"1\"QLJ3N'Q_F,7$$\"1+++347TLFC$\"1p pOc=/(4'F,7$$\"1+++LY.KNFC$\"1\"3n(3?R*)oF,7$$\"1+++\"o7Tv$FC$\"1[\\6( z_8*yF,7$$\"1+++$Q*o]RFC$\"1hNdX2@b))F,7$$\"1+++\"=lj;%FC$\"1(\\cyv9-+ \"FC7$$\"1+++V&RY2aFC$\"1gTx*)*zz)=FC7$$\"1++ +yXu9cFC$\"1+>:p(yY3#FC7$$\"1+++\\y))GeFC$\"1')eV)[0pI#FC7$$\"1+++i_QQ gFC$\"1Ap[nCPXDFC7$$\"1+++!y%3TiFC$\"1.]**=Kc)z#FC7$$\"1+++O![hY'FC$\" 1J#)=q%4$4JFC7$$\"1+++#Qx$omFC$\"1#zNbEj\">MFC7$$\"1+++u.I%)oFC$\"1&p_ W)fM(y$FC7$$\"1*****ppe*zqFC$\"19-'HAi*fTFC7$$\"1+++C\\'QH(FC$\"15c,ej q " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 453 28 "3) Relativistic trajectories" }} {EXCHG {PARA 0 "" 0 "" {TEXT 320 69 "Relativistic trajectories: the re lativistic harmonic oscillator (RHO)" }}{PARA 0 "" 0 "" {TEXT -1 448 " Let us solve Newton's equation in the relativistic form to get some fe eling for the effect of the increased mass. We consider a one-dimensio nal harmonic oscillator, e.g., for an electron attached to a spring (f or quite some time approximations were made in atomic physics based on oscillator strengths). We will compare the analytically known solutio ns for the nonrelativistic case with numerical solutions based on the \+ relativistic kinetic energy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "The units are chosen as before (" }{TEXT 321 1 "m" }{TEXT -1 342 "=c=1) and the spring constant, and initial displace ment is chosen to be large to make the particle acquire a substantial \+ amount of kinetic energy. We need to understand the length scale in or der to decide what we mean by this choice of parameters. Alternatively , we can just experiment around and observe the speed reached by the p article at " }{TEXT 322 1 "x" }{TEXT -1 3 "=0." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "m:=1; c:=1; k:=1; x0:=2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG \"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x0G\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "For a nonrelativistic harmonic oscillator we have the solution:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "xoft:=x0*cos(sqrt(k/m)*t); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%xoftG,$-%$cosG6#%\"tG\"\"#" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "voft:=diff(xoft,t);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%voftG,$-%$sinG6#%\"tG!\"#" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The origin is reached at " }{TEXT 323 2 "t " }{TEXT -1 143 "= Pi/2, and the velocity exceeds the speed o f light by a factor of 2. What can we find out about the relativistic \+ motion in the same force law?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 61 "We solve the Newton equation in a way tha t uses both Maple's " }{TEXT 19 6 "dsolve" }{TEXT -1 104 " integrator. For the determination of the velocity from the momentum we have the n onlinear relationship:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "e qU:=U=P/m*sqrt(1-U^2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(eqU ,U);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqUG/%\"UG*&%\"PG\"\"\"-%%s qrtG6#,&F)F)*$)F&\"\"#\"\"\"!\"\"F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #*&%\"PG\"\"\"*$-%%sqrtG6#,&*$)F$\"\"#F%\"\"\"F.F.F%!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "We use this expression for the velocity \+ in terms of the momentum for the integration, and also for display pur poses." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "NE1:=diff(p(t),t) =-k*x(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NE1G/-%%diffG6$-%\"pG6 #%\"tGF,,$-%\"xGF+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 " NE2:=diff(x(t),t)=p(t)/sqrt(p(t)^2+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NE2G/-%%diffG6$-%\"xG6#%\"tGF,*&-%\"pGF+\"\"\"*$-%%sqrtG6#,&* $)F.\"\"#F0\"\"\"F9F9F0!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "IC:=p(0)=0,x(0)=x0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6$/ -%\"pG6#\"\"!F*/-%\"xGF)\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "dt:=0.05; Nt:=trunc(4*Pi/dt);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%#dtG$\"\"&!\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NtG\"$^#" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "sol:=dsolve(\{NE1,NE2,IC\}, \{x(t),p(t)\},numeric,output=listprocedure);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG7%/%\"tGR6#F'6\"6#%aoCopyright~(c)~1993~by~the~U niversity~of~Waterloo.~All~rights~reserved.GF*9$F*F*F*/-%\"xGF)RF)6&%( rkf45_sG%)outpointG%#r1G%#r2GF+F*C&>8%-%&evalfG6#F-@$52-%$absG6#,$F9! \"\"-FA6#,&&%,loc_controlG6#\"\"#\"\"\"F9FD4-%'memberG6$&FI6#\"\"'<*$F L\"\"!FDFL!\"#FK$FDFV$FWFV$FKFVC%>FI-%%copyG6#=F*6#;FL\"#LE\\[lBFLFK\" #IFV\"#FV\"\"%$FL!\")F]oFV\"#?FZ\" \"&FgoFSFL\"\"($FL!\"*\"#@FV\"\")\"&++$\"#AFV\"\"*\"%+5\"#BFV\"#5FV\"# CFV\"#6FV\"#DFV\"#7FV\"#EFV\"#8FV\"#FFV\"#9FV\"#GFV\"#:FV\"#HFV\"#;FV> %'loc_y0G-Fhn6#=F*6#;FLFKE\\[l#FLFZFKFV>%'loc_y1G-Fhn6#=F*FgqE\\[l!@$0 F@FVC$>&FI6#FcoF9@%1%'DigitsG-%'evalhfG6#FhrC$>8$-%*traperrorG6#-Fjr6# -%=dsolve/numeric_solnall_rkf45G6,%&loc_FG-%$varG6#FI-Fis6#Fcq-Fis6#F[ r-Fis6#%'loc_F1G-Fis6#%'loc_F2G-Fis6#%'loc_F3G-Fis6#%'loc_F4G-Fis6#%'l oc_F5G-Fis6#%)loc_workG@$/F^s%*lasterrorGC%>8&-%+searchtextG6$.Fjr-%(c onvertG6$-%#opG6$FL7#F^s%%nameG>8'-Fhu6$.%)hardwareGF[v@%50FfuFV0FdvFV -Fes6,FgsFIFcqF[rFatFdtFgtFjtF]uF`u-%&ERRORG6#F^sF]w&Fcq6#FLF*6%FIFcqF [rF*/-%\"pGF)RF)F2F+F*C&>F9F:@$F>C%>FI-Fhn6#=F*F[oE\\[lBFLFKF_oFVF`oFV FKFVFaoFVFboFVFcoFVFdoFVFeoFVFfoFgoF]oFVFioFZFjoFgoFSFLF[pF\\pF^pFVF_p F`pFapFVFbpFcpFdpFVFepFVFfpFVFgpFVFhpFVFipFVFjpFVF[qFVF\\qFVF]qFVF^qFV F_qFVF`qFVFaqFV>Fcq-Fhn6#=F*FgqE\\[l#FLFZFKFV>F[r-Fhn6#=F*FgqE\\[l!@$F arC$>FdrF9@%FgrC$>F^sF_s@$FbuC%>FfuFgu>FdvFev@%FjvF]wF_wF]w&FcqFJF*Fdw F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "xt:=subs(sol,x(t)): p t:=subs(sol,p(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "xt(0. 05),pt(0.05);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"1I'*4Mn](*>!#:$!1f Vbx+%e***!#<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 161 "xL:=[[0,x0 ]]: vL:=[[0,0]]: for it from 1 to Nt do: ti:=it*dt; pti:=pt(ti); xti:= xt(ti); vti:=pti/sqrt(1+pti^2); vL:=[op(vL),[ti,vti]]: xL:=[op(xL),[ti ,xti]]: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "plot([xL,vL] ,color=[red,blue],title=cat(\"RHO for x0= \",convert(x0,string)));" }} {PARA 13 "" 1 "" {GLPLOT2D 691 299 299 {PLOTDATA 2 "6'-%'CURVESG6$7hz7 $\"\"!$\"\"#F(7$$\"1+++++++]!#<$\"1I'*4Mn](*>!#:7$$\"1+++++++5!#;$\"1B dmFg5!*>F17$$\"1+++++++:F5$\"1!evo+B!y>F17$$\"1+++++++?F5$\"1sa(H^'fh> F17$$\"1+++++++DF5$\"1fH/2mBT>F17$$\"1+++++++IF5$\"1DO?@FQ<>F17$$\"1++ +++++NF5$\"1>#3<\"*p/*=F17$$\"1+++++++SF5$\"1PT&)[g!4'=F17$$\"1+++++++ XF5$\"1wD;U#f!H=F17$$F-F5$\"1RZo'3`_z\"F17$$\"1+++++++bF5$\"1C&pr.n(f< F17$$\"1+++++++gF5$\"1Aw@u'RGs\"F17$$\"1+++++++lF5$\"1ts6)GtYo\"F17$$ \"1+++++++qF5$\"1:9w)zQak\"F17$$\"1+++++++vF5$\"1&G]kY!G0;F17$$\"1++++ +++!)F5$\"1+N]L,Kk:F17$$\"1+++++++&)F5$\"1v'Ql!4mA:F17$$\"1+++++++!*F5 $\"1`,'=G!R![\"F17$$\"1+++++++&*F5$\"1>NChFeP9F17$$\"\"\"F($\"1[vtA?I% R\"F17$$\"1++++++]5F1$\"1eFo8Fg]8F17$$\"1+++++++6F1$\"1Qv_G>`18F17$$\" 1++++++]6F1$\"1jUYE/8i7F17$$\"1+++++++7F1$\"1j#ocjLu@\"F17$$\"1++++++] 7F1$\"1b%3J[sC<\"F17$$\"1+++++++8F1$\"1-x(>3us7\"F17$$\"1++++++]8F1$\" 1AX*GIi=3\"F17$$\"1+++++++9F1$\"1m47]#ei.\"F17$$\"1++++++]9F1$\"1Z*y,g 1[!**F57$$F:F1$\"1y&)*oNiaW*F57$$\"1++++++]:F1$\"1[YI\"H*p%)*)F57$$\"1 +++++++;F1$\"1uk^+=lA&)F57$$\"1++++++];F1$\"1q$)o)=T%f!)F57$$\"1++++++ +)Qm'F57$$\"1++++++]=F1$\"1s&*3gd-(>'F57$$\"1+++++++>F1$\"1(ot7Mp %HdF57$$\"1++++++]>F1$\"1EZqbWGh_F57$F)$\"1*>#p&*y`#z%F57$$\"1++++++]? 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The particle re aches approximately " }{TEXT 19 8 "beta=0.9" }{TEXT -1 543 ", and does not exceed the speed of light. The shape of the velocity as a functio n of time reminds one of the time dependence of the angle in the mathe matical pendulum (not linearized!) for large initial angles, i.e., for cases when the pendulum starts near the top. Comparison with the nonr elativistic solution shows that the relativistic harmonic oscillator i s no longer isochronous (as the pendulum): the period depends on the a mount of energy. Let us veify the change in the period by a solution f or a smaller initial stretch of the spring:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x0:=0.2; IC:=p(0)=0,x(0)=x0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x0G$\"\"#!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #ICG6$/-%\"pG6#\"\"!F*/-%\"xGF)$\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 29 "dt:=0.05; Nt:=trunc(4*Pi/dt);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dtG$\"\"&!\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%# NtG\"$^#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "sol:=dsolve(\{N E1,NE2,IC\},\{x(t),p(t)\},numeric,output=listprocedure);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%$solG7%/%\"tGR6#F'6\"6#%aoCopyright~(c)~1993~b y~the~University~of~Waterloo.~All~rights~reserved.GF*9$F*F*F*/-%\"xGF) RF)6&%(rkf45_sG%)outpointG%#r1G%#r2GF+F*C&>8%-%&evalfG6#F-@$52-%$absG6 #,$F9!\"\"-FA6#,&&%,loc_controlG6#\"\"#\"\"\"F9FD4-%'memberG6$&FI6#\" \"'<*$FL\"\"!FDFL!\"#FK$FDFV$FWFV$FKFVC%>FI-%%copyG6#=F*6#;FL\"#LE\\[l 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"This solution is qualitatively co mparable to the nonrelativistic case (harmonic oscillation with period depending only on " }{TEXT 325 1 "k" }{TEXT -1 1 "/" }{TEXT 324 1 "m " }{TEXT -1 63 ", no dependence on the intial stretch), even though a \+ value of " }{TEXT 19 8 "beta=0.2" }{TEXT -1 51 " is reached at the pas sage of the force-free point " }{TEXT 326 1 "x" }{TEXT -1 304 " = 0. T he comparison of the qualitatively different plots helps to understand why the highly relativistic harmonic oscillator can no longer be isoc hronous: the slowdown of the mass due to its increase with velocity (e nergy is being dumped into increasing the relativistic mass), and the \+ avoidance of the " }{TEXT 328 1 "v" }{TEXT -1 127 "=1 limit the mass i s delayed and cannot execute oscillations on the same timescale as bef ore, i.e., on a timescale just set by " }{TEXT 19 9 "sqrt(k/m)" } {TEXT -1 27 ". At the equilibrium point " }{TEXT 329 1 "x" }{TEXT -1 530 "=0 all energy resides in the form of kinetic energy. By increasin g the initial stretch of the spring (i.e., x0) we can control the amou nt of kinetic energy of the mass while the spring is neither stretched nor compressed. It is important to realize that we can make the parti cle hardly go faster when doing so. The energy is dumped into the incr ease of the relativistic mass. Now that we appreciate this phenomenon, it is clear that the period of the RHO has to go down, as the mass in creases substantially during the passages of " }{TEXT 330 1 "x" } {TEXT -1 226 "=0. Note that the term RHO is a misnomer: the generaliza tion of the HO to the relativistic case leads to anharmonic motion whe n the speed of the mass becomes a significant fraction of c in the vic inity of the force-free point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 327 9 "Exercise:" }}{PARA 0 "" 0 "" {TEXT -1 19 "Find a start value " }{TEXT 19 2 "x0" }{TEXT -1 159 " for which the p eriod begins to change appreciably from the nonrelativistic harmonic o scillator case. Which maximum speed does the particle reach in this ca se?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 256 "" 0 "" {TEXT -1 60 "4) Transfo rming energy and momentum between reference frames" }}{EXCHG {PARA 0 " " 0 "" {TEXT 331 17 "Conservation laws" }}{PARA 0 "" 0 "" {TEXT -1 531 "An important aspect when studying dynamics is the investigation o f conservation laws, which are obeyed by the solutions to the equation s of motion. Conservation laws allow one to solve problems in a low nu mber of dimensions, and to impose constraints on solutions in three di mensions without looking at the details of the interactions. Examples \+ are provided by 'billiard-ball-type' scattering. In special relativity the question of which are the conserved quantities is tied closely to the properties of the Lorentz transformation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 332 45 "Lorentz transformation o f energy and momentum" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "Suppose we consider a particle moving with velocity ve ctor " }{TEXT 333 1 "u" }{TEXT -1 92 " in a reference frame S, and obs erve it also in S' which is boosted compared to S along the " }{TEXT 334 1 "x" }{TEXT -1 26 "-axis with boost velocity " }{TEXT 335 1 "V" } {TEXT -1 85 ". We can write down the unprimed and primed energies and \+ momentum components. We use " }{TEXT 19 1 "g" }{TEXT -1 11 " to denote " }{TEXT 19 5 "gamma" }{TEXT -1 101 ", and we do not assign it by the next statement, we write a (useless) equation to indicate its value: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "restart; assume(V- c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g=1/sqrt(1-u^2/c^2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"gG*&\"\"\"F&*$-%%sqrtG6#,&\"\" \"F,*&*$)%\"uG\"\"#F&F&*$)%#c|irG\"\"#F&!\"\"!\"\"F&F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E:=g*m*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG*(%\"gG\"\"\"%\"mGF')%#c|irG\"\"#\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "p[x]:=g*m*u[x];" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 15 "p[y]:=g*m*u[y];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "p[z]:=g*m*u[z];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>& %\"pG6#%\"xG*(%\"gG\"\"\"%\"mGF*&%\"uGF&F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"pG6#%\"yG*(%\"gG\"\"\"%\"mGF*&%\"uGF&F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"pG6#%\"zG*(%\"gG\"\"\"%\"mGF*&%\"uGF&F* " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "In S' gamma has a different v alue:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "gp=1/sqrt(1-up^2/c ^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#gpG*&\"\"\"F&*$-%%sqrtG6#,& \"\"\"F,*&*$)%#upG\"\"#F&F&*$)%#c|irG\"\"#F&!\"\"!\"\"F&F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Ep:=gp*m*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EpG*(%#gpG\"\"\"%\"mGF')%#c|irG\"\"#\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "pp[x] := gp*m*up[x]; pp[y] : = gp*m*up[y]; pp[z] := gp*m*up[z];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >&%#ppG6#%\"xG*(%#gpG\"\"\"%\"mGF*&%#upGF&F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%#ppG6#%\"yG*(%#gpG\"\"\"%\"mGF*&%#upGF&F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%#ppG6#%\"zG*(%#gpG\"\"\"%\"mGF*&%#upGF&F* " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "To transform these four quan tities we first express the primed gamma in terms of quantities measur ed in S. Of course, one could also proceed the other way around." }} {PARA 0 "" 0 "" {TEXT -1 26 "We need the components of " }{TEXT 336 1 "u" }{TEXT -1 32 "' in terms of the components of " }{TEXT 337 1 "u" } {TEXT -1 127 ", as given by the LT. We use the previous notation for s ubscripts rather than making use of Maple's notation for table entries :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Lv1:=up_x=(u_x-V)/(1-V *u_x/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Lv1G/%%up_xG*&,&%$u_x G\"\"\"%#V|irG!\"\"\"\"\",&F*F**&*&F+F*F)F*F-*$)%#c|irG\"\"#F-!\"\"F,F 5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Lv2:=up_y=u_y/(1/sqrt( 1-V^2/c^2)*(1-V*u_x/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Lv2G/ %%up_yG*&*&%$u_yG\"\"\"-%%sqrtG6#,&F*F**&*$)%#V|irG\"\"#\"\"\"F4*$)%#c |irG\"\"#F4!\"\"!\"\"F4F4,&F*F**&*&F2F*%$u_xGF*F4*$)F7\"\"#F4F9F:F9" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Lv3:=up_z=u_z/(1/sqrt(1-V^ 2/c^2)*(1-V*u_x/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$Lv3G/%%up _zG*&*&%$u_zG\"\"\"-%%sqrtG6#,&F*F**&*$)%#V|irG\"\"#\"\"\"F4*$)%#c|irG \"\"#F4!\"\"!\"\"F4F4,&F*F**&*&F2F*%$u_xGF*F4*$)F7\"\"#F4F9F:F9" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "up_sq:=simplify(rhs(Lv1)^2+r hs(Lv2)^2+rhs(Lv3)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&up_sqG*&* &)%#c|irG\"\"#\"\"\",0*&F'F*)%$u_xGF)F*\"\"\"*(F'F*%#V|irGF/F.F/!\"#*& F'F*)F1F)F*F/*&)%$u_yGF)F*F'F*F/*&F6F*F4F*!\"\"*&)%$u_zGF)F*F'F*F/*&F; F*F4F*F9F/F**$),&*$F'F*F/*&F1F*F.F*F9\"\"#F*!\"\"" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 38 "algsubs(u_x^2+u_y^2+u_z^2=u_sq,up_sq);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&)%#c|irG\"\"#\"\"\",,*&)%#V|irGF'F ()%$u_xGF'F(\"\"\"*(F%F(F,F/F.F/!\"#*&F+F(%%u_sqGF/!\"\"*&F%F(F+F(F/*& F%F(F3F(F/F/F(*$),&*$F%F(F/*&F,F(F.F(F4\"\"#F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "The expression for " }{TEXT 19 2 "gp" }{TEXT -1 25 " simplifies considerably:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "simplify(1/sqrt(1-up_sq/c^2),symbolic);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#*&,&*$)%#c|irG\"\"#\"\"\"\"\"\"*&%#V|irGF*%$u_xG F*!\"\"F)*$-%%sqrtG6#,$*&,&F%F**$)F,F(F)F.F*,*F%F.*$)%$u_yGF(F)F**$)F- F(F)F**$)%$u_zGF(F)F*F*F.F)!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "We recognize in the square root in the denominator a product: one \+ part belongs to " }{TEXT 19 7 "1/gamma" }{TEXT -1 18 " and the other t o " }{TEXT 19 12 "(1-u_sq/c^2)" }{TEXT 2 11 ", i.e., to " }{TEXT 19 3 "1/g" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "gp_S:=algsubs((u_x^ 2+u_y^2+u_z^2)=u_sq,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%gp_SG*&, &*$)%#c|irG\"\"#\"\"\"\"\"\"*&%#V|irGF,%$u_xGF,!\"\"F+*$-%%sqrtG6#*&,& F'F0*$)F.F*F+F,F,,&F'F0%%u_sqGF,F,F+!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Therefore the kinetic energy of the particle in S' transf orms into two contributions in S:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Ep=gamma*(E-V*p_x);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/*(%#gpG\"\"\"%\"mGF&)%#c|irG\"\"#\"\"\"*&%&gammaGF&,&*(%\"gGF&F'F+F (F+F&*&%#V|irGF&%$p_xGF&!\"\"F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "To verify by comparison with our result which up to " }{TEXT 19 5 "m*c^2" }{TEXT -1 16 " was the energy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(p_x=p[x],%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# /*(%#gpG\"\"\"%\"mGF&)%#c|irG\"\"#\"\"\"*&%&gammaGF&,&*(%\"gGF&F'F+F(F +F&**%#V|irGF&F0F+F'F+&%\"uG6#%\"xGF&!\"\"F&" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 71 "Analogous derivations follow for the components of the \+ momentum vector." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "subs(gp =gp_S,up[x]=rhs(Lv1),pp[x]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*(,& *$)%#c|irG\"\"#\"\"\"\"\"\"*&%#V|irGF+%$u_xGF+!\"\"F+%\"mGF+,&F.F+F-F/ F+F**&-%%sqrtG6#*&,&F&F/*$)F-F)F*F+F+,&F&F/%%u_sqGF+F+F*,&F+F+*&*&F-F* F.F*F**$)F(\"\"#F*!\"\"F/\"\"\"FB" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&* (%\"mG\"\"\",&%$u_xG!\"\"%#V|irGF'F')%#c|irG\"\"#\"\"\"F/*&-%%sqrtG6#, &*$F,F/F'*$)F+F.F/F*F/-F26#,&F5F'%%u_sqGF*F/!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "pp[x]=gamma*(p[x]-g*m*V);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*(%#gpG\"\"\"%\"mGF&&%#upG6#%\"xGF&*&%&gammaGF&,& *(%\"gGF&F'\"\"\"&%\"uGF*F&F&*(F0F1F'F1%#V|irGF&!\"\"F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 338 1 "y" }{TEXT -1 11 "-comp onent:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "subs(gp=gp_S,up[y ]=rhs(Lv2),pp[y]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&**,&*$)%#c|irG \"\"#\"\"\"\"\"\"*&%#V|irGF+%$u_xGF+!\"\"F+%\"mGF+%$u_yGF+-%%sqrtG6#,& F+F+*&*$)F-F)F*F**$)F(\"\"#F*!\"\"F/F*F**&-F36#*&,&F&F/*$F8F*F+F+,&F&F /%%u_sqGF+F+F*,&F+F+*&*&F-F*F.F*F**$)F(\"\"#F*F " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*(%\"mG\"\"\"%$u_yGF&%#c|irGF&\"\"\"*$-%%sqrtG6#,&*$) F(\"\"#F)F&%%u_sqG!\"\"F)!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "pp[y]=p[y];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*(%#gpG\"\"\"% \"mGF&&%#upG6#%\"yGF&*(%\"gGF&F'\"\"\"&%\"uGF*F&" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 42 "The same result follows obviously for the " } {TEXT 339 1 "z" }{TEXT -1 381 "-component. It is important to compare \+ the transformation of the momentum vector to the transformation of the position vector. For the components perpendicular to the direction of the boost we find that they are unchanged in both cases. For the comp onent along the boost direction a small manipulation reveals an intere sting fact. One can introduce the energy into this expression:" } {TEXT 340 0 "" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "-m*gamma*V*g,E;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,$**%\"mG\"\" \"%&gammaGF&%#V|irGF&%\"gGF&!\"\"*(F)\"\"\"F%F,)%#c|irG\"\"#F," }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "E:='E':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "algsubs(g*m=E/c^2,-m*gamma*V*g);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*(%&gammaG\"\"\"%#V|irGF'%\"EGF'\"\"\"*$)%# c|irG\"\"#F*!\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "- gamma*V*E/(c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*(%&gammaG\"\" \"%#V|irGF'%\"EGF'\"\"\"*$)%#c|irG\"\"#F*!\"\"!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Therefore we can write the momentum transformat ion as" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "pp[x]=gamma*(p[x] -V*E/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*(%#gpG\"\"\"%\"mGF&&%# upG6#%\"xGF&*&%&gammaGF&,&*(%\"gGF&F'\"\"\"&%\"uGF*F&F&*&*&%#V|irGF&% \"EGF&F1*$)%#c|irG\"\"#F1!\"\"!\"\"F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 215 "The Lorentz transformation for the momentum vector compo nents is now strikingly similar to the transformation for the position vector components, except that the role of time in the transformation has been taken by " }{TEXT 19 5 "E/c^2" }{TEXT -1 94 " ! Now we shoul d be curious how the energy transformation compares to the time transf ormation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Ep:='Ep':" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Ep=gamma*(E-V*p_x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%#EpG*&%&gammaG\"\"\",&%\"EGF'*&%#V|i rGF'%$p_xGF'!\"\"F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "tp=g amma*(t-V*x/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#tpG*&%&gammaG \"\"\",&%\"tGF'*&*&%#V|irGF'%\"xGF'\"\"\"*$)%#c|irG\"\"#F.!\"\"!\"\"F' " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "We observe that the role of \+ " }{TEXT 341 1 "x" }{TEXT -1 43 " in the time transformation is taken \+ uo by " }{TEXT 19 7 "p_x*c^2" }{TEXT -1 191 ". Thus, it is fair to say that in the same way that position and time transform as one entity u nder Lorentz transformations, momentum and energy transform in an anal ogous way. The factors of " }{TEXT 19 3 "c^2" }{TEXT -1 82 " are neede d to make matters correct from the point of view of physical dimension s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 129 "It is important to illustrate the derived results with some examples. (w e pick example 2-3 from the book of Tipler and Llewellyn)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 344 10 "Example 1:" }} {PARA 0 "" 0 "" {TEXT -1 25 "A micrometeroite of mass " }{TEXT 19 7 "1 0^(-9)" }{TEXT -1 75 " kg moves past Earth at speed 0.01 c. An observe r in a spaceship moving at " }{TEXT 343 1 "V" }{TEXT -1 154 " = 0.5 c \+ in the same direction as the meteorite passes by the Earth. What value s of energy and momentum will the moving observer detect for the meteo rite?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "A nswer:" }}{PARA 0 "" 0 "" {TEXT -1 247 "The total energy of the meteor ite can be calculated in the Earth's frame by the non-relativistic app roximation, or by the full relativisitic expression. We ignore the pot ential energy in the gravitational field, and use SI (MKSA) units thro ughout." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "m:=10^(-9):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "c:=3*10^8:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "u:=0.01*c:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E_rel:=m*1/sqrt(1-u^2/c^2)*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_relG$\"+O+X+!*!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "E_nr:=m*c^2+m/2*u^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_nrG$\"+++X+!*!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "p_rel:=m*1/sqrt(1-u^2/c^2)*u;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %&p_relG$\"+7+:+I!#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "p_n r:=m*u;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%p_nrG$\"+++++I!#7" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 274 "Note that despite of the fact tha t the nonrelativistic approximation is highly accurate to describe the meteorite in the Earth's reference frame, the transformation to the s paceship frame should be carried out by a Lorentz transformation, sinc e the spaceship moves at 0.5 c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "V:=0.5*c;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"VG$\"+++++:!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "unprotect(gamma);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "gamma:=1/sqrt(1-V^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&gammaG$\"+Q0qa6!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "The kinetic energy of the meteorite in the spaceship fram e:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Ep:=gamma*(E_rel-V*p_ rel);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EpG$\"+Qg3M5!\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "The energy is 15 % larger in the \+ spaceship frame. The effect is mostly due to the gamma-factor in this \+ case." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Ep/E_rel;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Oq#*[6!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pp:=gamma*(p_rel-V*E_rel/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ppG$!+oY\\(p\"!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "pp/p_rel;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+OE.ec! \")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "Obviously the second term made the big contribution here. The momentum is in the opposite direc tion, and it is huge (Tipler and Llewellyn have a buboe in their solut ion: wrong numbers and a wrong conclusion by two orders of magnitude!) ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 455 74 "5) Relativi stic billiard balls: calculating the energy in different frames" }} {EXCHG {PARA 0 "" 0 "" {TEXT 342 48 "Energy conservation: relativistic billiard balls" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 377 "Now that we have some familiarity as to how to transform the kinetic energy between inertial frames, as well as with the momen tum conservation law (which we used to find an appropriate expression \+ for relativistic momentum, we carry out an independent check: we consi der the total energy of a system of two colliding classical particles, and investigate whether it is conserved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 144 "First we return to the problem of elastic scattering of two objects of equal mass, which we considered \+ to define momentum in special relativity." }}{PARA 0 "" 0 "" {TEXT -1 79 "We simply repeat the steps that led to the dependence of the mass \+ on velocity. " }}{PARA 0 "" 0 "" {TEXT -1 85 "We had formulated the ve locities in the frame A where particle A has only a velocity " }{TEXT 345 1 "y" }{TEXT -1 64 "-component, and frame B in which particle B ha s only a velocity " }{TEXT 346 1 "y" }{TEXT -1 80 "-component, which o pposes A's motion. The catch is that frame B moves along the " }{TEXT 347 1 "x" }{TEXT -1 369 "-axis with respect to A with a relativistic s peed. This requires the use of the Lorentz transformation when calcula ting the momentum balance or the total energy. The balls are thrown su ch that they collide elastically. Due to the transformation of the vel ocity components nonrelativistic linear momentum is not conserved (whe n one transforms them according to Lorentz)." }}{PARA 0 "" 0 "" {TEXT -1 68 "We should calculate the total energy before and after the colli sion." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "restart; unprotect (gamma); assume(V-c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vA_y:=u0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "vA_x:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vA_yG%#u0G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vA_xG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "vB_y p:=-u0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vB_xp:=0;" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 8 "vB_x:=V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%& vB_ypG,$%#u0G!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&vB_xpG\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vB_xG%#V|irG" }}}{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 30 "vB_y:= vB_yp/gamma/(1+V*vB_xp);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vB_yG,$ *&%#u0G\"\"\"-%%sqrtG6#,&F(F(*$)%%betaG\"\"#\"\"\"!\"\"F1F2" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u:=sqrt(V^2+vB_y^2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"uG*$-%%sqrtG6#,(*$)%#V|irG\"\"#\" \"\"\"\"\"*$)%#u0GF-F.F/*&F1F.)%%betaGF-F.!\"\"F." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 43 "MC:=m(u0)*u0+m(u)*vB_y=-m(u0)*u0-m(u)*vB_y: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "MC:=simplify(lhs(MC)-rhs(MC))=0 ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#MCG/,&*&-%\"mG6#%#u0G\"\"\"F+F ,\"\"#*(-F)6#*$-%%sqrtG6#,(*$)%#V|irGF-\"\"\"F,*$)F+F-F9F,*&F;F9)%%bet aGF-F9!\"\"F9F,F+F9-F36#,&F,F,*$F=F9F?F9!\"#\"\"!" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 294 "We have the velocity components of particle B in \+ A's frame. How do we obtain the energy? We derived an expression for t he relativistic energy using the definition of relativistic momentum, \+ and by postulating Newton's law, and integrating free-particle motion \+ to arrive at the kinetic energy as " }{TEXT 19 15 "(gamma-1)*m*c^2" } {TEXT -1 8 ", where " }{TEXT 348 1 "m" }{TEXT -1 95 " is the rest mass , and the gamma-factor corrects this mass as a function of the particl e speed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "We should obtain then for the total energy (simply ommit the subtr action of 1 in the round bracket:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "E_A:=m*c^2/sqrt(1-u0^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_AG*&*&%\"mG\"\"\")%#c|irG\"\"#\"\"\"F,*$-%%sqrtG6# ,&F(F(*&*$)%#u0GF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "E_B:=simplify(m*c^2/sqrt(1-subs(beta=V/c,u)^2 /c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_BG*&*&%\"mG\"\"\")%#c| irG\"\"%\"\"\"F,*&-%%sqrtG6#,&*$)F*\"\"#F,F(*$)%#V|irGF4F,!\"\"F,-F/6# ,&F2F(*$)%#u0GF4F,F8F,!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "E_totA:=E_A+E_B;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'E_totAG,&*& *&%\"mG\"\"\")%#c|irG\"\"#\"\"\"F-*$-%%sqrtG6#,&F)F)*&*$)%#u0GF,F-F-*$ )F+\"\"#F-!\"\"!\"\"F-F:F)*&*&F(F-)F+\"\"%F-F-*&-F06#,&*$F*F-F)*$)%#V| irGF,F-F;F--F06#,&FDF)*$F5F-F;F-F:F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "To calculate matters from B's perspective one obtains the firs t term for B, and then hopefully for the second term the calculation g ives the same answer after transforming A's velocity." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "vA_xp:=-V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&vA_xpG,$%#V|irG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "vA_yp:=vA_y/gamma/(1-V*vA_x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&vA_ypG*&%#u0G\"\"\"-%%sqrtG6#,&F'F'*$)%%betaG\"\"#\" \"\"!\"\"F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "w:=sqrt((-V) ^2+vA_yp^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"wG*$-%%sqrtG6#,(*$ )%#V|irG\"\"#\"\"\"\"\"\"*$)%#u0GF-F.F/*&F1F.)%%betaGF-F.!\"\"F." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "E_Ap:=simplify(m*c^2/sqrt(1- subs(beta=-V/c,w)^2/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_ApG *&*&%\"mG\"\"\")%#c|irG\"\"%\"\"\"F,*&-%%sqrtG6#,&*$)F*\"\"#F,F(*$)%#V |irGF4F,!\"\"F,-F/6#,&F2F(*$)%#u0GF4F,F8F,!\"\"" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "E_Bp:=m*c^2/sqrt(1-u0^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_BpG*&*&%\"mG\"\"\")%#c|irG\"\"#\"\"\"F,*$-%%s qrtG6#,&F(F(*&*$)%#u0GF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "The two particles have swapped their rol es: this is consistent with the principle of relativity that the physi cs is the same in all intertial frames. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "E_totA-E_Ap-E_Bp;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# \"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "The dependence on " } {TEXT 349 1 "u" }{TEXT -1 105 "0 is quadratic, and therefore independe nt of the sign. The answer has to be the same after the collision." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "Suppose w e wish to calculate the energy in a frame that moves with respect to A and B along the " }{TEXT 354 1 "x" }{TEXT -1 55 "-axis. Given that B \+ moves with respect to A with speed " }{TEXT 356 1 "V" }{TEXT -1 65 ", \+ we can specify the frame C by stating that it moves with speed " } {TEXT 355 1 "W" }{TEXT -1 125 " with respect to A. To transform the en ergy of particle A is analogous to the the calculation in frame B, one just exchanges " }{TEXT 358 1 "V" }{TEXT -1 5 " for " }{TEXT 357 1 "W " }{TEXT -1 178 " in the gamma factor, etc.. The calculation of B's en ergy in frame C is trickier, as we do not know the relative speed betw een B and C! We should not take the difference between " }{TEXT 360 1 "V" }{TEXT -1 5 " and " }{TEXT 359 1 "W" }{TEXT -1 282 ", as we have r ejected the additivity of velocities in special relativity. According \+ to the velocity transformation we can calculate the energy of B in fra me C via the rest frame of A. We use the gamma factor for the rest fra me of A, and use the known velocity of particle B (namely " }{TEXT 361 1 "V" }{TEXT -1 28 ") in the transformation law." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "Let us write a procedu re that calculates the required steps. We assume that A and B are not \+ moving in the " }{TEXT 362 1 "x" }{TEXT -1 143 "-direction in their re spective frame. The energy is calculated in frame C for which we have \+ a gamma factor that relates it to the rest frame A." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "Eboost:=proc(W,V,vA_y,vB_yp) local gamma A,vA_xpp,vA_ypp,vB_xpp,vB_ypp,w_A,w_B,E_App,E_Bpp,vA_x,vB_xp,vB_x,vB_y ; vA_x:=0; vB_xp:=0; vB_x:=V; vB_y:=vB_yp*sqrt(1-V^2/c^2)/(1+V*vB_xp); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "gammaA:=1/sqrt(1-W^2/c^2); vA_y pp:=vA_y/gammaA/(1-W*vA_x/c^2); vB_ypp:=vB_y/gammaA/(1-W*vB_x/c^2);" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "vA_xpp:=-W; vB_xpp:=(vB_x-W)/(1-W* vB_x/c^2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "w_A:=sqrt(vA_xpp^2+vA _ypp^2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "w_B:=sqrt(vB_xpp^2+vB_y pp^2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E_App:=simplify(m*c^2/sqr t(1-w_A^2/c^2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E_Bpp:=simplify (m*c^2/sqrt(1-w_B^2/c^2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "E_App +E_Bpp; end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Eboost(0,c/2,c/10,-c/10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(%\"mG\"\"\")%#c|irG\"\"#\"\"\"-%%sqrtG6# \"#6F*#\"#5\"#L*(F%F*F'F*-F,6#F1F*#\"#?\"#**" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(subs(m=1,c=1,%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+@bbl@!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(u0=0.1,c=1,V=0.5,m=1,E_totA);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+@bbl@!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Eboo st(c/2,c/2,c/10,-c/10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(%\"mG\" \"\")%#c|irG\"\"#\"\"\"-%%sqrtG6#\"#6F*#\"#5\"#L*(F%F*F'F*-F,6#F1F*#\" #?\"#**" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(subs(m=1,c =1,%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+@bbl@!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "subs(u0=0.1,c=1,V=0.5,m=1,E_Ap+E_Bp );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+@bbl@!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Eboost(c/4,c/2,c/10,-c/10);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,&*(%\"mG\"\"\")%#c|irG\"\"#\"\"\"-%%sqrtG6#\"$l \"F*#\"\")\"#***(F%F*F'F*-F,6#\"#bF*#\"#9F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(subs(m=1,c=1,%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+LRv'3#!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot(evalf(subs(m=1,c=1,Eboost(W,c/2,c/10,-c/10))),W=-0.9..0.9); " }}{PARA 13 "" 1 "" {GLPLOT2D 696 299 299 {PLOTDATA 2 "6%-%'CURVESG6$ 7en7$$!1+++++++!*!#;$\"1TGX(o4i;'!#:7$$!1,+D1G\">!*)F*$\"1/B\"\\%Q7()e F-7$$!1++]7c#Q!))F*$\"1OZ1A!zKk&F-7$$!1++v=%Qdq)F*$\"1!om6)H'yU&F-7$$! 1+++D7l2')F*$\"1*)Hu@bxN_F-7$$!1+]PR4'pV)F*$\"1W))[W3+Y\\F-7$$!1,+v`1F m#)F*$\"1QJ5O6^+ZF-7$$!1,]iv:Ju!)F*$\"1u')f%HA[Y%F-7$$!1++](\\_B)yF*$ \"1TVcbVMiUF-7$$!1+++!*47*o(F*$\"1O5(44Q\\3%F-7$$!1++]#[*)e\\(F*$\"1!y 6g`O)GRF-7$$!1++veKE6rF*$\"1B%o=**zqm$F-7$$!1++v8bmanF*$\"1PXlxAJoMF-7 $$!1++v.#HaQ'F*$\"1+/T*pv`H$F-7$$!1++v$[kN+'F*$\"1+5&o,[M9$F-7$$!1,+v3 W#Hi&F*$\"1c(fEg?L,$F-7$$!1+++0$*QJ_F*$\"1W^E')o<(*GF-7$$!1++]s\\_')[F *$\"1+R&R4hq!GF-7$$!1+++:%*G)\\%F*$\"1CI2Jw(pr#F-7$$!1+++D(f%3TF*$\"1( z\"3MkwOEF-7$$!1+++&e'yKPF*$\"19G*o\"GvnDF-7$$!1++v$QR;R$F*$\"1J'f%)H/ 7^#F-7$$!1,++lB)f)HF*$\"1%)3b\\(>1X#F-7$$!1+++gwLUEF*$\"1mPu'pKVS#F-7$ $!1,+vtrfUAF*$\"1T6p\")[obBF-7$$!1,++5\"f())=F*$\"1;mUvN&oJ#F-7$$!1++v tEa+:F*$\"1#R^EoA%yAF-7$$!1++D@)o38\"F*$\"1(H$R_.eXAF-7$$!1*****\\PH:X (!#<$\"1\"4D9`^\\@#F-7$$!1.+]([`%4RFbt$\"1CB3DY\"**=#F-7$$!1K3++Dh%))) !#>$\"1b!HQ\">2m@F-7$$\"1()**\\i6pzQFbt$\"17.vZzlW@F-7$$\"1')**\\PXJMt Fbt$\"1o+]%oI(G@F-7$$\"1****\\U-a16F*$\"1(R>;oJV6#F-7$$\"1+++I\")*>\\ \"F*$\"1.Jz5V^-@F-7$$\"1++]sM4p=F*$\"1%z)Rn>+%4#F-7$$\"1***\\ig_RB#F*$ \"1*f&*Gc9()3#F-7$$\"1****\\nk1REF*$\"1O6M*[(Q'3#F-7$$\"1)******GzI+$F *$\"1wxe\"zgw3#F-7$$\"1+++v1u\"R$F*$\"1Uuxq.!G4#F-7$$\"1***\\ilDRu$F*$ \"1\\8LB^7,@F-7$$\"1,++lo&*GTF*$\"1SgGe4k9@F-7$$\"1***\\iAT7\\%F*$\"1Y JW\")*y?8#F-7$$\"1++DrF$*p[F*$\"1#Gp+o(*e:#F-7$$\"1)****\\(H4,gF*$\"1:$ptm. >F#F-7$$\"1++]dr&GQ'F*$\"1*HZgj:?L#F-7$$\"1,+Dm6YhnF*$\"1*37tf]nS#F-7$ $\"1******H*e$4rF*$\"1()4N0!HF\\#F-7$$\"1****\\F!*33vF*$\"1+)>E4()*=EF -7$$\"1******zzrkyF*$\"1&=^g$pOnFF-7$$\"1,+DE)e\\C)F*$\"1c*='\\S+x$F--% 'COLOURG6&%$RGBG$\"#5!\"\"\"\"!Fi]l-%+AXESLABELSG6$Q\"W6\"%!G-%%VIEWG6 $;$!\"*Fh]l$\"\"*Fh]l%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "It appe ars as if the total energy is minimized in a reference frame that move s with half the speed (" }{TEXT 364 1 "W" }{TEXT -1 1 "=" }{TEXT 363 1 "V" }{TEXT -1 352 "/2). This would be the result from the nonrelativ istic kinetic energy expression (+ rest energy for two massive particl es), where the sum of the squared velocities is minimized when the dif ference is split in half. A closer inspection performed with calculus \+ (performed below) shows that the minimum is not exactly at the half-po int. For larger boosts " }{TEXT 365 1 "V" }{TEXT -1 76 " the curve dev iates more dramatically from the nonrelativistic expectation. " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "solve(diff(Eboost(W,c/2,c/10 ,-c/10),W)=0,W);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&\"\"#\"\"\"*$- %%sqrtG6#\"\"$\"\"\"!\"\"F&%#c|irGF&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%#c|irG $\"*#>\\zE!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 "In the nonrela tivitic approximation we would calculate: rest energy + nonrelativisti c kinetic energy. We add the statements to verify that the total energ y is a frame-dependent quantity in the nonrelativistic case." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "E_nrA:=evalf(subs(u0=0.1,m=1 ,c=1,V=0.5,m*c^2+m/2*u0^2+m*c^2+m/2*(V^2+u0^2)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_nrAG$\"++++N@!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "E_nrpp:=evalf(subs(u0=0.1,m=1,c=1,V=0.5,m*c^2+m/2*(u0 ^2+(V/2)^2)+m*c^2+m/2*((V/2)^2+u0^2)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'E_nrppG$\"+++]s?!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "The increase in the total energy as the boost velocity approaches c o r -c is obviously associated with the fact that the two particles in A and B acquire mass. This also explains the asymmetry in the curve sho wn above. For negative " }{TEXT 368 1 "W" }{TEXT -1 42 " the increase \+ is larger than for positive " }{TEXT 367 1 "W" }{TEXT -1 98 " of equal magnitude due to the set-up of mass A being stationary, and mass B mo ving with c/2. For " }{TEXT 369 1 "W" }{TEXT -1 172 ">c/2 a substantia l mass increase for particle B occurs only for large boosts. In the op posite direction the converse is true, but note that nothing too drama tic happens at " }{TEXT 370 1 "W" }{TEXT -1 186 "=-c/2, where one migh t argue naively that the relativistic mass for particle B should hit t he roof. The relative speed between C and B reaches the speed of light in this regime only for " }{TEXT 371 1 "W" }{TEXT -1 4 "=-c." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "We should be able to verify the result calculated in the double-primed frame (p rocedure " }{TEXT 19 6 "Eboost" }{TEXT -1 102 ") by carrying out a Lor entz transformation of the total energy (e.g., from A's rest frame, i. e., from " }{TEXT 19 6 "E_totA" }{TEXT -1 24 "). We boost from A with \+ " }{TEXT 352 1 "W" }{TEXT -1 58 ", and realize that we need to know th e momentum along the " }{TEXT 353 1 "x" }{TEXT -1 209 "-axis in frame \+ A to carry out the boost. This momentum was calculated before when the momentum balance was formulated, but we need the explicit expression \+ for the mass as a function of speed for the particles." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "u;" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#*$-%%sqrtG6#,(*$)%#V|irG\"\"#\"\"\"\"\"\"*$)%#u0GF+F,F-*&F/F,)%%beta GF+F,!\"\"F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "px_totA:=m* 0/sqrt(1-u0^2/c^2)+subs(beta=V/c,m*V/sqrt(1-u^2/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(px_totAG*&*&%\"mG\"\"\"%#V|irGF(\"\"\"*$-%%sqrt G6#,&F(F(*&,(*$)F)\"\"#F*F(*$)%#u0GF4F*F(*&*&F6F*F3F*F**$)%#c|irG\"\"# F*!\"\"!\"\"F**$)F<\"\"#F*F>F?F*F>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "gammapp:=1/sqrt(1-W^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(gammappG*&\"\"\"F&*$-%%sqrtG6#,&\"\"\"F,*&*$)%\"WG\" \"#F&F&*$)%#c|irG\"\"#F&!\"\"!\"\"F&F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "EboostLT:=unapply(simplify(gammapp*(E_totA-W*(px_totA ))),W,V,u0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)EboostLTGR6%%\"WG%# V|irG%#u0G6\"6$%)operatorG%&arrowGF**&*()%#c|irG\"\"$\"\"\"%\"mG\"\"\" ,**$F0F3!\"\"*&F1F5)9%\"\"#F3F5*&)F1FF3F5*$F:F3F8F3 F8*(9$F5F;F5F?F3F5F5F3*(-F@6#,&FCF5*$)9&F " 0 "" {MPLTEXT 1 0 30 "evalf(EboostLT(c/3,c/2,c/10));" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#,$*&*&%\"mG\"\"\",&*$)%#c|irG\"\"$\"\"\"$!+++++v!#5*&)F+\"\"#F--%%sq rtG6#*$F2F-F-$!+m$yo@(F0F'F-F+!\"\"$!+5\"Q8U\"!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "evalf(Eboost(c/3,c/2,c/10,-c/10));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"mG\"\"\")%#c|irG\"\"#\"\"\"$\"+ 2gw\"4#!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "expand(simpl ify(EboostLT(c/3,c/2,c/10)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&**% \"mG\"\"\"-%%sqrtG6#\"#6\"\"\")%#c|irG\"\"#F+-F(6#F.F+#\"\"&\"#A*,F%F+ F'F+F,F+F/F+-F(6#\"\"$F+#\"#D\"$)>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "simplify(Eboost(c/3,c/2,c/10,-c/10));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(%\"mG\"\"\")%#c|irG\"\"#\"\"\"-%%sqrtG6#\"#AF* #\"\"&F.**F%F*F'F*F+F*-F,6#\"\"$F*#\"#D\"$)>" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 927 "The comparison verifies that the two calculations are \+ consistent: in EboostLT we have Lorentz transformed the total energy f or particles A and B as calculated in frame A to the reference frame C . In Eboost we have used Lorentz-transformed velocity components for p article B in A's restframe, and transformed the velocities to frame C \+ to directly evaluate the relativistic mass increase as a function of t he speed of particles A and B as seen in frame C. Note how we avoided \+ the issue of the unknown relative velocity between frames B and C. We \+ calculated the velocity of particle B as viewed by an observer in C by first transforming it into A (known relative velocity V), and then fr om A to C (known relative velocity W). The simple elastic billiard bal l scattering example was not interesting from the energy conservation \+ point of view, apart from the fact that energy conservation was ensure d by the quadratic dependence on " }{TEXT 366 1 "u" }{TEXT -1 2 "0." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "More int eresting cases involve unequal masses, and also inelastic scattering. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 260 "" 0 "" {TEXT -1 58 "6) Billiar d balls with unequal masses - elastic collisions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 176 "A non-trivial generalization would be to consider t he case of different masses. Suppose particle B has twice the rest mas s of A. Then we expect A to bounce off with more than -" }{TEXT 351 1 "u" }{TEXT -1 24 "0, and B with less than " }{TEXT 350 1 "u" }{TEXT -1 332 "0. Based on the equal-mass elastic scattering we figured out t he dependence of relativistic energy on speed and rest mass of a parti cle. We should be able to formulate what happens. Let us simplify matt ers by going to one dimension only, and formulating momentum and energ y conservation. Let us work in the rest frame of particle A." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "restart; assume(V-c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v_A:=0;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%$v_AG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "E_A:=m_A*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_AG*&%$m_AG \"\"\")%#c|irG\"\"#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v_B:=V;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$v_BG%#V|irG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "unprotect(gamma); gamma:=1/sqrt(1-V ^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&gammaG*&\"\"\"F&*$-%%sq rtG6#,&\"\"\"F,*&*$)%#V|irG\"\"#F&F&*$)%#c|irG\"\"#F&!\"\"!\"\"F&F6" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "E_B:=gamma*m_B*c^2;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_BG*&*&%$m_BG\"\"\")%#c|irG\"\"#\" \"\"F,*$-%%sqrtG6#,&F(F(*&*$)%#V|irGF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "p_A:=m_A*v_A;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$p_AG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "p_B:=gamma*m_B*v_B;" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%$p_BG*&*&%$m_BG\"\"\"%#V|irGF(\"\"\"*$-%%sqrtG6#,&F(F(*&*$)F)\"\"# F*F**$)%#c|irG\"\"#F*!\"\"!\"\"F*F8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Let us denote the velocities after the collision as primed. We can write for momentum and energy conservation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "p_Ap:=m_A*v_Ap/sqrt(1-v_Ap^2/c^2); p_Bp:=m_B* v_Bp/sqrt(1-v_Bp^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%p_ApG*& *&%$m_AG\"\"\"%%v_ApGF(\"\"\"*$-%%sqrtG6#,&F(F(*&*$)F)\"\"#F*F**$)%#c| irG\"\"#F*!\"\"!\"\"F*F8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%p_BpG*& *&%$m_BG\"\"\"%%v_BpGF(\"\"\"*$-%%sqrtG6#,&F(F(*&*$)F)\"\"#F*F**$)%#c| irG\"\"#F*!\"\"!\"\"F*F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "E_Ap:=1/sqrt(1-v_Ap^2/c^2)*m_A*c^2; E_Bp:=1/sqrt(1-v_Bp^2/c^2)*m_B*c^ 2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_ApG*&*&%$m_AG\"\"\")%#c|irG \"\"#\"\"\"F,*$-%%sqrtG6#,&F(F(*&*$)%%v_ApGF+F,F,*$)F*\"\"#F,!\"\"!\" \"F,F9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_BpG*&*&%$m_BG\"\"\")%#c |irG\"\"#\"\"\"F,*$-%%sqrtG6#,&F(F(*&*$)%%v_BpGF+F,F,*$)F*\"\"#F,!\"\" !\"\"F,F9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "MC:=p_A+p_B=p_ Ap+p_Bp;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#MCG/*&*&%$m_BG\"\"\"%#V |irGF)\"\"\"*$-%%sqrtG6#,&F)F)*&*$)F*\"\"#F+F+*$)%#c|irG\"\"#F+!\"\"! \"\"F+F9,&*&*&%$m_AGF)%%v_ApGF)F+*$-F.6#,&F)F)*&*$)F?F4F+F+*$)F7\"\"#F +F9F:F+F9F)*&*&F(F+%%v_BpGF)F+*$-F.6#,&F)F)*&*$)FLF4F+F+*$)F7\"\"#F+F9 F:F+F9F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "EC:=E_A+E_B=E_A p+E_Bp;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ECG/,&*&%$m_AG\"\"\")%#c |irG\"\"#\"\"\"F)*&*&%$m_BGF)F*F-F-*$-%%sqrtG6#,&F)F)*&*$)%#V|irGF,F-F -*$)F+\"\"#F-!\"\"!\"\"F-F=F),&*&*&F(F-F*F-F-*$-F36#,&F)F)*&*$)%%v_ApG F,F-F-*$)F+\"\"#F-F=F>F-F=F)*&*&F0F-F*F-F-*$-F36#,&F)F)*&*$)%%v_BpGF,F -F-*$)F+\"\"#F-F=F>F-F=F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sol:=solve(\{MC,EC\},\{v_Ap,v_Bp\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "So much for M aple magic..." }}{PARA 0 "" 0 "" {TEXT -1 106 "Nevertheless, we should not despair. When we substitute numbers for the masses Maple does the work for us." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "sol:=solve (\{subs(m_A=1,m_B=1,V=c/2,MC),subs(m_A=1,m_B=1,V=c/2,EC)\},\{v_Ap,v_Bp \});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6$<$/%%v_BpG,$%#c|irG# \"\"\"\"\"#/%%v_ApG\"\"!<$/F(F0/F/F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "sol:=solve(\{subs(m_A=1,m_B=2,V=c/2,MC),subs(m_A=1,m_ B=2,V=c/2,EC)\},\{v_Ap,v_Bp\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ solG6$<$/%%v_BpG,$%#c|irG#\"\"\"\"\"#/%%v_ApG\"\"!<$/F(,&*&-%%sqrtG6# \"\"$\"\"\"F*F,#!\"$\"#8F*#\"#:\"#E/F/,&F4#!#O\"$P$F*#\"$s#FD" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Note that there are always two so lutions: one represents the initial condition, the other the final out come." }}{PARA 0 "" 0 "" {TEXT -1 63 "We investigate various situation s: a light particle with speed " }{TEXT 372 1 "V" }{TEXT -1 20 " hits \+ a heavier one:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "sol:=solv e(\{subs(m_A=10,m_B=1,MC),subs(m_A=10,m_B=1,EC)\},\{v_Ap,v_Bp\});" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6$<$/%%v_ApG,$*&*(%#c|irG\"\"\" %#V|irG\"\"\",(*$)F,\"\"#F-\"#5*$)F.F3F-!#5*&F,F/-%%sqrtG6#,&F1F/F5!\" \"F-F/F/F-*&-F:6#F " 0 "" {MPLTEXT 1 0 86 "vA:=subs(sol[1],v_Ap): # we may need to pick by ha nd which one is the desired solution" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "vB:=subs(sol[1],v_Bp): # and use sol[1] or sol[2] cor respondingly." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "plot([subs (c=1,vA),subs(c=1,vB),-V],V=0..1,color=[red,blue,green],view=[0..1,-1. .1]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6'-%'CURV ESG6$7U7$\"\"!F(7$$\"1+++;arz@!#<$\"1H7^N&fR'R!#=7$$\"1+++XTFwSF,$\"1k #*zI(4pT(F/7$$\"1+++\"z_\"4iF,$\"13Q$3\\%)38\"F,7$$\"1+++S&phN)F,$\"1g a$z*41C:F,7$$\"1+++*=)H\\5!#;$\"1$o%>i8E<>F,7$$\"1+++[!3uC\"FB$\"1@\"> P4ORG#F,7$$\"1+++J$RDX\"FB$\"1#z?EOR\"G/.kF,7$$\"1+++LY.KNFB$\"1 !\\;<))eO\"oF,7$$\"1+++\"o7Tv$FB$\"1Y'RVm\"3-tF,7$$\"1+++$Q*o]RFB$\"1v UQf\\0XxF,7$$\"1+++\"=lj;%FB$\"1^XhV@vV#)F,7$$\"1+++V&RY2aFB$\"1m\"3'HPHX6FB7$$\"1+++yXu9cFB$\"1^KSmNJ17FB7$$\"1+++ \\y))GeFB$\"1>0f'z+BF\"FB7$$\"1+++i_QQgFB$\"1i4xN'>,M\"FB7$$\"1+++!y%3 TiFB$\"1#zry%)3#49FB7$$\"1+++O![hY'FB$\"1gJ>n4yc;FB7$$\"1*****ppe*zqFB$ \"1BO+q1!Hu\"FB7$$\"1+++C\\'QH(FB$\"1k\\IExZW=FB7$$\"1+++8S8&\\(FB$\"1 tPz'[.%[>FB7$$\"1+++0#=bq(FB$\"1dEl%\\.u1#FB7$$\"1*****p?27\"zFB$\"1lM X!*4<'>#FB7$$\"1,++IXaE\")FB$\"1TSaI@HZBFB7$$\"1+++l*RRL)FB$\"1MsN\\Kz 7DFB7$$\"1+++`<.Y&)FB$\"1SVWFTS3FFB7$$\"1+++8tOc()FB$\"1#)yY()RxPHFB7$ $\"1+++\\Qk\\*)FB$\"1z!*\\R$FB7$$\"1+++p0;r\"*FB$\"1D'y***zBeNFB7 $$\"1+++oTAq#*FB$\"1FU&p&[**fPFB7$$\"1*****\\w(Gp$*FB$\"1J0A/y,&*RFB7$ $\"1+++B-\"\\Z*FB$\"1C>5FKP&H%FB7$$\"1+++!oK0e*FB$\"1Dog.-kpYFB7$$\"1+ ++<5s#y*FB$\"1V/tFZj!y&FB7$%%FAILGF\\[l-%'COLOURG6&%$RGBG$\"*++++\"!\" )F(F(-F$6$7SF'7$$\"1nmm;arz@F,$!1Hl_gNZ$y\"F,7$$\"1LL$e9ui2%F,$!146%Rj \"fNLF,7$$\"1nmm\"z_\"4iF,$!1)ysBoO=3&F,7$$\"1mmmT&phN)F,$!1GF)=a?3%oF ,7$$\"1LLe*=)H\\5FB$!1Sc2kx+$f)F,7$$\"1nm\"z/3uC\"FB$!1(\\Nd!f#>-\"FB7 $$\"1++DJ$RDX\"FB$!1f'3cDG0>\"FB7$$\"1nm\"zR'ok;FB$!1%[#f`Y;l8FB7$$\"1 ++D1J:w=FB$!1ldNVij6 CFB7$$\"1nm\"z*ev:JFB$!1L&p2y./d#FB7$$\"1LLL347TLFB$!1DdWM#f)fFFB7$$\" 1LLLLY.KNFB$!1Z84JFB7$$\"1LLL$Q*o ]RFB$!1e?O?tVwKFB7$$\"1++D\"=lj;%FB$!1eV%GFe3Y$FB7$$\"1++vV&R4AzTFB7$$\"1mm\"zRQb@&FB$!1^9a/FOtVFB7$$ \"1***\\(=>Y2aFB$!1W'p3mcNa%FB7$$\"1mm;zXu9cFB$!1C&)Q?NrGZFB7$$\"1+++] y))GeFB$!1V0\\$)fd@\\FB7$$\"1****\\i_QQgFB$!1wUq57$>6&FB7$$\"1***\\7y% 3TiFB$!1F@Ha'=yH&FB7$$\"1****\\P![hY'FB$!1!*HK(4bj]&FB7$$\"1LLL$Qx$omF B$!1lzgug#ep&FB7$$\"1+++v.I%)oFB$!1TMbf&>0!fFB7$$\"1mm\"zpe*zqFB$!15dH +'Q$)3'FB7$$\"1+++D\\'QH(FB$!13pa@aZ'H'FB7$$\"1KLe9S8&\\(FB$!1kq`e-F& \\'FB7$$\"1***\\i?=bq(FB$!1*o$RRd[1nFB7$$\"1LLL3s?6zFB$!1Le<6Ly;pFB7$$ \"1++DJXaE\")FB$!1e%>l+D:9(FB7$$\"1nmmm*RRL)FB$!1G&)oiZ3jtFB7$$\"1mm;a <.Y&)FB$!1Q&\\z_Zdf(FB7$$\"1LLe9tOc()FB$!1cTFe-xLyFB7$$\"1+++]Qk\\*)FB $!1^!)>!)>!)*FB-F^[l6&F`[lF(F(Fa[l-F$6$7SF'7$Fh[l$!1 nmm;arz@F,7$F]\\l$!1LL$e9ui2%F,7$Fb\\l$!1nmm\"z_\"4iF,7$Fg\\l$!1mmmT&p hN)F,7$F\\]l$!1LLe*=)H\\5FB7$Fa]l$!1nm\"z/3uC\"FB7$Ff]l$!1++DJ$RDX\"FB 7$F[^l$!1nm\"zR'ok;FB7$F`^l$!1++D1J:w=FB7$Fe^l$!1LLL3En$4#FB7$Fj^l$!1n m;/RE&G#FB7$F^o$!1+++D.&4]#FB7$Fco$!1+++vB_Y2aFB7$Fdcl$!1mm;zXu9cFB7$Ficl$!1+++]y))Ge FB7$F^dl$!1****\\i_QQgFB7$Fcdl$!1***\\7y%3TiFB7$Fhdl$!1****\\P![hY'FB7 $F]el$!1LLL$Qx$omFB7$Fbel$!1+++v.I%)oFB7$Fgel$!1mm\"zpe*zqFB7$F\\fl$!1 +++D\\'QH(FB7$Fafl$!1KLe9S8&\\(FB7$Fffl$!1***\\i?=bq(FB7$F[gl$!1LLL3s? 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Note, however, that in the highly relativistic li mit a crossover occurs!" }}{PARA 0 "" 0 "" {TEXT -1 227 "The green lin e shows what would happen if one assumed naively that in the limit of \+ the hit particle being very massive no energy transfer were to occur, \+ and the impinging particle would rebound with its original speed rever sed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 91 "plot([subs(c=1,vA), subs(c=1,-vB)],V=0..1,color=[red,blue],view=[0..1,0..1],numpoints=1500 );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG 6$7hhp7$\"\"!F(7$$\"1*****\\\\U(zp!#>$\"1Wy!fOW!p7F,7$$\"1+++7zF08!#=$ \"1McwrVBtBF,7$$\"1+++'Ra#))>F2$\"1SvSqV,:OF,7$$\"1+++#4edn#F2$\"1^Qkd -.l[F,7$$\"1+++.U**fLF2$\"1g>\")[664hF,7$$\"1+++m'oV*RF2$\"1+`1'>SDE(F ,7$$\"1+++ymA^YF2$\"1>h!zUdoX)F,7$$\"1+++t,bI`F2$\"1/7o5B.#p*F,7$$\"1+ ++q]p2gF2$\"1#*y<*yDB4\"F27$$\"1+++G=A/nF2$\"1ItquQ(*=7F27$$\"1+++vHs< tF2$\"1uOi$)o_I8F27$$\"1,++g*z$3!)F2$\"1-5)3w5hX\"F27$$\"1*****z%G(=q) 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the massive hit particle A picks up more energy t han the impinging light particle B retains in its rebound at an impact velocity of " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "solve(sub s(v_Ap=vA,m_A=10,c=1,E_Ap-m_A*c^2)=subs(v_Bp=vB,m_B=1,c=1,E_Bp-m_B*c^2 ),V);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%,$*$-%%sqrtG6#\"#T\"\"\"#\"\" *\"#h,$F$#!\"*F,\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "eva lf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%$\"+![KsW*!#5$!+![KsW*F%\"\" !" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "These results are clearly in compatible with our intuition based on the nonrelativistic limit." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 373 9 "Exercise: " }}{PARA 0 "" 0 "" {TEXT -1 127 "Explore how these crossover points m ove as the mass ratio is increased, i.e., as the hit particle is chose n to be more massive." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "The other limit of a massive particle im pinging on a light particle is 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G-%%VIEWG6$;F_`hlF]`hlF[d[o" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "simplify(lim it(vA,V=1)),simplify(limit(vB,V=1));" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6$\"\"\"#\"#**\"$,\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "It appear s as if there were fewer surprises in this case...?" }}{PARA 0 "" 0 " " {TEXT -1 265 "The hit particle is 10 times lighter in this example, \+ and one can see how its speed approaches c when the incident massive p article speed exceeds 0.8 c. For the kinetic energies this can imply a crossover where the hit particle acquires a big mass due to its speed ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "plot([subs(v_Ap=vA,m_ A=1,c=1,E_Ap-m_A*c^2),subs(v_Bp=vB,m_B=10,c=1,E_Bp-m_B*c^2)],V=0..1,co lor=[red,blue],numpoints=1500,view=[0..1,0..15]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6$7jhp7$$\"1*****\\\\U( zp!#>$\"1*ey+6oB0)!#A7$$\"1+++\"3E;+\"!#=$\"1T/GLRFe;!#@7$$\"1+++7zF08 F1$\"1+b(Q\"37;GF47$$\"1+++'Ra#))>F1$\"1c4A\"4lT`'F47$$\"1+++#4edn#F1$ \"1'3^/=EM=\"!#?7$$\"1+++.U**fLF1$\"1/))ptC1m=FD7$$\"1+++m'oV*RF1$\"1A c)zaDsj#FD7$$\"1+++ymA^YF1$\"1Cm!ezBfd$FD7$$\"1+++t,bI`F1$\"1EWYCyx'p% FD7$$\"1+++q]p2gF1$\"1[Y-([$)e'fFD7$$\"1+++G=A/nF1$\"1tnC;[\\HuFD7$$\" 1+++vHs()p5!#<$\"18klNF*7$$\"1+++='>:`\"Fip$\"1YdU0 /yxQF*7$$\"1+++Qo[*f\"Fip$\"1:C#*GQoHUF*7$$\"1+++yc3q;Fip$\"12J\\$*\\S 6YF*7$$\"1+++v@aJFip$\"1'eO\\'Gz\"= 'F*7$$\"1+++]%z%)*>Fip$\"1=f+\\/%Rg'F*7$$\"1+++Pxaq?Fip$\"1KQj?g,*3(F* 7$$\"1+++*H/`8#Fip$\"1=4'[ld&RvF*7$$\"1+++]dW/AFip$\"1kV+Jy%f.)F*7$$\" 1+++#=)4nAFip$\"1@XnteV*\\)F*7$$\"1+++RQfNBFip$\"1%yw&>4/@!*F*7$$\"1++ +8H/+CFip$\"1I)3yRNg_*F*7$$\"1+++m2TnCFip$\"144x$[io+\"F17$$\"1+++d^FL DFip$\"1ErEUeOh5F17$$\"1+++p!HAg#Fip$\"1l%\\t\"y'*>6F17$$\"1+++$))R'oE Fip$\"1YROF***y<\"F17$$\"1+++JXbOFFip$\"1tVDywlQ7F17$$\"1+++7o!R!GFip$ \"1\"fYjTC/I\"F17$$\"1+++jlzlGFip$\"1LG$pY4&e8F17$$\"1+++-#Hn$HFip$\"1 bvv@ekE9F17$$\"1++++A<+IFip$\"1'>DXg.!*[\"F17$$\"1+++oc\"y1$Fip$\"1(yX U$H'pb\"F17$$\"1+++*4fD8$Fip$\"1&eU1wKMi\"F17$$\"1+++e$[W?$Fip$\"1g0=/ 8())p\"F17$$\"1+++t\"*pmKFip$\"1w.,jNelF17$$\"1+++Dl=sMFip$\"1STL(yY\\ *>F17$$\"1+++8A'H`$Fip$\"1=+\"f;wa1#F17$$\"1+++G'y@g$Fip$\"1[r([O)HZ@F 17$$\"1+++o#Q!pOFip$\"1SL!)R\"\\yA#F17$$\"1+++IU&et$Fip$\"1aMruX$)4BF1 7$$\"1+++)QCC!QFip$\"1\\`V.u*HR#F17$$\"1+++quPmQFip$\"1r&p+f\"GuCF17$$ \"1+++WC^NRFip$\"1t.?\\`ojDF17$$\"1+++)3.9+%Fip$\"1OCRdnP]EF17$$\"1+++ )))\\22%Fip$\"1*p!)GBzJu#F17$$\"1+++Ta_LTFip$\"1/TBy7dGGF17$$\"1+++zT! 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\\ffl7$Fg\\gl$\"1%\\,EzF'z6F\\ffl7$F\\]gl$\"1?7K_9I)=\"F\\ffl7$Fa]gl$ \"1Ih*>-By>\"F\\ffl7$Ff]gl$\"1g?h>:!z?\"F\\ffl7$F[^gl$\"1[#[eFOo@\"F\\ ffl7$F`^gl$\"1uQ;U/mE7F\\ffl7$Fe^gl$\"1u\"*>pj+P7F\\ffl7$Fj^gl$\"1+g@( *zKZ7F\\ffl7$F__gl$\"1-H^b)4vD\"F\\ffl7$Fd_gl$\"1pD\"p@[!p7F\\ffl7$Fi_ gl$\"12*4([Hjz7F\\ffl7$F^`gl$\"1'o9^\"*p6H\"F\\ffl7$Fc`gl$\"1P!*[+?%=I \"F\\ffl7$Fh`gl$\"12P!=lbPJ\"F\\ffl7$F]agl$\"1$oP+R4_K\"F\\ffl7$Fbagl$ \"1c63_[WP8F\\ffl7$Fgagl$\"1+!30ox'\\8F\\ffl7$F\\bgl$\"1bVZ!*Gyi8F\\ff l7$Fabgl$\"1OQS+fqv8F\\ffl7$Ffbgl$\"1*=mxlS#*Q\"F\\ffl7$F[cgl$\"1\\()) )\\j*HS\"F\\ffl7$F`cgl$\"1@b-X=%fT\"F\\ffl7$Fecgl$\"1'pG5Pb6V\"F\\ffl7 $Fjcgl$\"1=!Q04>^W\"F\\ffl7$F_dgl$\"146eDkRg9F\\ffl7$Fddgl$\"1]&o*)>8a Z\"F\\ffl7$Fidgl$\"10p(f&Hc#\\\"F\\ffl7$F^egl$\"1)z#o)oQy]\"F\\ffl7$Fc egl$\"1LrA]8oD:F\\ffl7$Fhegl$\"1h*\\'ecVU:F\\ffl7$F]fgl$\"1`mg=;Lh:F\\ ffl7$Fbfgl$\"1rM;?49y:F\\ffl7$Fgfgl$\"1**z8p!*)yf\"F\\ffl7$F\\ggl$\"1A ,W0Ih<;F\\ffl7$Faggl$\"1ddW'[,!Q;F\\ffl7$Ffggl$\"1K***ywK!f;F\\ffl7$F[ hgl$\"1*y1h;d*z;F\\ffl7$F`hgl$\"1PIo\\iU.F\\ffl7$F][hl$\"1`(yrQ-h& >F\\ffl7$Fb[hl$\"1ti?3:\"[*>F\\ffl7$Fg[hl$\"1*ez,+B?.#F\\ffl7$F\\\\hl$ \"1KF#4%=Zs?F\\ffl7$Fa\\hl$\"1r[!*R\\T;@F\\ffl7$Ff\\hl$\"1mRN$H]'f@F\\ ffl7$F[]hl$\"1t.J4T=4AF\\ffl7$F`]hl$\"1jA)GQ@CE#F\\ffl7$Fe]hl$\"1S%*)[ ;(pCBF\\ffl7$Fj]hl$\"1LY2%*y8&Q#F\\ffl7$F_^hl$\"1p6))GDuiCF\\ffl7$Fd^h l$\"1I1z)=n=a#F\\ffl7$Fi^hl$\"1$)f8N6?JEF\\ffl7$F^_hl$\"1(z\"yx(>Ru#F \\ffl7$Fi_hl$\"1p[1$QoD)GF\\ffl7$Fc`hl$\"1N#)HsN([/$F\\ffl7$$\"1+++\" \\b**)**Fhjm$\"1)Hl6jfY:$F\\ffl7$Fh`hl$\"1e&G)[=M$H$F\\ffl7$$\"1+++*oy \\***Fhjm$\"1W7iVwq!Q$F\\ffl7$$\"1******eCl'***Fhjm$\"1l>&>A7$*[$F\\ff l7$$\"1+++X$*[(***Fhjm$\"1T\"*[q^ycNF\\ffl7$$\"1+++IiK)***Fhjm$\"1Db(f >H'ROF\\ffl7$$\"1+++:J;****Fhjm$\"1;cJ\\sW_PF\\ffl7$$\"\"\"Feahl$\"1,+ ++++]SF\\ffl-F_ahl6&FaahlFeahlFeahlFbahl-%+AXESLABELSG6$Q#V|ir6\"%!G-% %VIEWG6$;FeahlFi\\jm;Feahl$\"#:Feahl" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "fs olve(subs(v_Ap=vA,m_A=1,c=1,E_Ap-m_A*c^2)=subs(v_Bp=vB,m_B=10,c=1,E_Bp -m_B*c^2),V=0.5..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+\"[KsW*!#5 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "We recognize from the compar ison of the numerical results that the crossover in the energies occur s for the same incident speed in the cases of corresponding mass ratio s for " }{TEXT 19 7 "m_A/m_B" }{TEXT -1 5 " and " }{TEXT 19 7 "m_B/m_A " }{TEXT -1 378 " respectively. Crossover in the present case of a hea vy particle hitting a lighter one, which results in the heavy particle continuing its forward travel while the light particle zooms off, mea ns that at this point the light particle has acquired more kinetic ene rgy than the heavy one. These results refer to the rest frame of the p article initially at rest (laboratory frame)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 258 "" 0 "" {TEXT -1 53 "7) Inelastic collisions in 1d: energy -mass conversion" }}{EXCHG {PARA 0 "" 0 "" {TEXT 374 19 "Inelastic col lision" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 606 "We now proceed with an example of an inelastic collision. The ine lasticity will play a different role than in nonrelativistic mechanics . Inelasticities in billiard ball scattering are associated with 'stic kiness', a part of the mechanical energy is given up to heat. We say t hat it is dissipated and rely only on momentum conservation to calcula te the dynamics of the one-dimensional motion. Complete inelasticity m eans that the two masses are stuck together, and proceed with a common velocity. The energy balance can be used to calculate the amount of m echanical energy that was dissipated in the system." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 510 "In the relativistic co ncept we do not calculate the mechanical energy only. The rest energy \+ is included automatically in the energy expression. We know from chemi cal and nuclear reactions that the energy released from these is comin g from the mass-to-energy conversion. The final products have a smalle r mass than the initial ingredients to the reaction. In chemistry the \+ mass differences which are converted into energy are small (but measur able); in nuclear physics they are considerably larger per reaction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 804 "Thus, \+ we will take the point of view that if we collide two objects in a com pletely inelastic way, such that they stick together, then the energy \+ removed from the mechanical energy balance is not necessarily dissipat ed, but rather stored in the bound object. We can think of having comp ressed a spring to form the bond, and that at some point later (perhap s only induced by some reaction), the composite system can decay, and \+ release the stored energy of the compressed spring. According to the e nergy-mass equivalence the rest mass of the composite object should ac count for the stored energy. We can use the energy conservation law to calculate this change in mass. Consistency of these ideas with specia l relativity will be demonstrated by performing the calculation in two different reference frames." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 375 20 "Center of mass frame" }}{PARA 0 "" 0 "" {TEXT -1 70 "In the so-called center-of-mass frame the identical particles o f mass " }{TEXT 376 1 "m" }{TEXT -1 28 " approach with equal speeds " }{TEXT 377 1 "u" }{TEXT -1 383 "0 from left and right. After the colli sion they stick together at remain at rest in this frame (it is also c alled the zero-momentum frame as the total momentum vector vanishes at all times in this frame). If one looks at a worldline diiagram in thi s frame it looks like a Mercedes star (or a deformation thereof): for \+ negative times the particles come in on worldlines with slopes 1/" } {TEXT 379 1 "u" }{TEXT -1 9 "0 and -1/" }{TEXT 378 1 "u" }{TEXT -1 118 "0 from left and right respectively. These worldlines merge at the origin (assuming that the collision takes place at c" }{TEXT 380 2 "t " }{TEXT -1 66 "= 0 in this frame, and emerge as a vertical worldline along the (c" }{TEXT 381 1 "t" }{TEXT -1 43 ")-axis for the newly for med object of mass " }{TEXT 382 1 "M" }{TEXT -1 28 ". The mass M is no t simply 2" }{TEXT 383 1 "m" }{TEXT -1 77 ", but has to be calculated \+ to conserve energy. Let us denote this frame as S." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 384 16 "Laboratory frame" }} {PARA 0 "" 0 "" {TEXT -1 55 "In the laboratory frame one of the two or iginal masses " }{TEXT 385 1 "m" }{TEXT -1 74 " sits at rest (in the o rigin), while the other approaches with some speed " }{TEXT 386 1 "u" }{TEXT -1 35 "' from the left. The initial speed " }{TEXT 387 1 "u" } {TEXT -1 13 "' would be 2 " }{TEXT 388 1 "u" }{TEXT -1 195 "0 in the n onrelativistic limit, but we anticipate a complication, since we canno t simply add velocities naively. The laboratory frame S' moves with re spect to the center-of-mass frame with speed " }{TEXT 389 1 "u" } {TEXT -1 52 "0 to the right, otherwise the mass that moves with -" } {TEXT 390 1 "u" }{TEXT -1 200 "0 in S would not be at rest (it represe nts the target in the laboratory, while the other mass is a particle i n the accelerator beam). Now that we have specified the boost velocity between S and S' as " }{TEXT 392 2 "V " }{TEXT -1 1 "=" }{TEXT 393 2 " u" }{TEXT -1 124 "0, we have to be careful when determining the velo city of the impinging mass in S', i.e., we cannot simply say that it i s 2 " }{TEXT 391 1 "u" }{TEXT -1 150 "0. We will determine it from con sistency with energy conservation or from the Lorentz transformation. \+ After the sticky collision the composite object " }{TEXT 394 1 "M" } {TEXT -1 51 " dashes off to the right with some speed less than " } {TEXT 395 1 "u" }{TEXT -1 79 "', as the impinging mass conserves momen tum, i.e., the momentum that propelled " }{TEXT 396 1 "m" }{TEXT -1 12 " with speed " }{TEXT 397 1 "u" }{TEXT -1 27 "' is the same that pr opels " }{TEXT 399 1 "M" }{TEXT -1 12 " with speed " }{TEXT 398 1 "U" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 28 "Can we figure out the \+ speed " }{TEXT 400 1 "U" }{TEXT -1 93 " without a calculation? Apparen tly yes, as we know the boost velocity between S and S' to be " } {TEXT 402 1 "V" }{TEXT -1 2 " =" }{TEXT 401 2 " u" }{TEXT -1 7 "0, and " }{TEXT 403 1 "M" }{TEXT -1 29 " is at rest in S. Therefore, " } {TEXT 406 1 "M" }{TEXT -1 24 " has to move in S' with " }{TEXT 404 1 " U" }{TEXT -1 1 "=" }{TEXT 405 1 "u" }{TEXT -1 84 "0 after the collisio n. The worldlines in S have to look as follows: the target mass " } {TEXT 408 1 "m" }{TEXT -1 25 " sits at the origin for c" }{TEXT 407 1 "t" }{TEXT -1 36 "' < 0, while the beam particle mass " }{TEXT 410 1 " m" }{TEXT -1 25 " comes in with a slope 1/" }{TEXT 409 1 "u" }{TEXT -1 17 "' from the left. " }}{PARA 0 "" 0 "" {TEXT -1 15 "At the origin c" }{TEXT 411 1 "t" }{TEXT -1 7 "' = 0, " }{TEXT 412 1 "x" }{TEXT -1 16 "' = 0 they form " }{TEXT 414 1 "M" }{TEXT -1 31 " which moves off \+ for positive c" }{TEXT 413 1 "t" }{TEXT -1 20 "' with a slope of 1/" } {TEXT 416 1 "U" }{TEXT -1 10 ", i.e., 1/" }{TEXT 415 1 "u" }{TEXT -1 2 "0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 " Let us figure out the incident particle speed " }{TEXT 417 1 "u" } {TEXT -1 127 "' from momentum and energy conservation in S'. Before th e collision in the lab frame all momentum resides in the beam particle ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "restart; assume(u0-c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pb_p:=m/sqrt(1-u_ p^2/c^2)*u_p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%pb_pG*&*&%\"mG\"\" \"%$u_pGF(\"\"\"*$-%%sqrtG6#,&F(F(*&*$)F)\"\"#F*F**$)%#c|irG\"\"#F*!\" \"!\"\"F*F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Eb_p:=m*c^2+ m/sqrt(1-u_p^2/c^2)*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%Eb_pG,& *&%\"mG\"\"\")%#c|irG\"\"#\"\"\"F(*&*&F'F,F)F,F,*$-%%sqrtG6#,&F(F(*&*$ )%$u_pGF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F;F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "After the collision we have" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "pa_p:=M/sqrt(1-u0^2/c^2)*u0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%pa_pG*&*&%\"MG\"\"\"%$u0|irGF(\"\"\"*$-%%sqrtG6#,&F( F(*&*$)F)\"\"#F*F**$)%#c|irG\"\"#F*!\"\"!\"\"F*F8" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 29 "Ea_p:=M/sqrt(1-u0^2/c^2)*c^2;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%%Ea_pG*&*&%\"MG\"\"\")%#c|irG\"\"#\"\"\"F,*$-% %sqrtG6#,&F(F(*&*$)%$u0|irGF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F9" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol:=solve(\{Eb_p=Ea_p,pb_p= pa_p\},\{M,u_p\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<$/%\"MG, $*&*(%\"mG\"\"\"%#c|irGF,-%%sqrtG6#,&*$)F-\"\"#\"\"\"F,*$)%$u0|irGF4F5 !\"\"F5F5,&F2F9F6F,!\"\"!\"#/%$u_pG,$*&*&F3F5F8F,F5,&F6F,F2F,F;F4" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "To look at the solution in detail \+ we extract:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "M_sp:=simpli fy(subs(sol,M));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%M_spG,$*&*(%\"m G\"\"\"%#c|irGF)-%%sqrtG6#,&*$)F*\"\"#\"\"\"F)*$)%$u0|irGF1F2!\"\"F2F2 ,&F/F6F3F)!\"\"!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "This is un derstood better by cancelling a factor of c in numerator and denominat or." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "u_sp:=simplify(subs( sol,u_p));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%u_spG,$*&*&)%#c|irG\" \"#\"\"\"%$u0|irG\"\"\"F+,&*$)F,F*F+F-*$F(F+F-!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "taylor(u_sp,u0=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++%$u0|irG\"\"#\"\"\",$*&\"\"\"F)*$)%#c|irG\"\"#F) !\"\"!\"#\"\"$,$*&F)F)*$)F,\"\"%F)F.F%\"\"&-%\"OG6#\"\"\"\"\"(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "We recover the expected nonrelati vistic result in the limit of c being infinite (or u0/c being small). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "Now l et us perform the calculation in S, the center-of-mass, or zero-moment um frame: " }}{PARA 0 "" 0 "" {TEXT -1 136 "The momentum is zero befor e and after the collision, which we do not formulate as an equation, b ut simply draw the conclusion that mass " }{TEXT 418 1 "M" }{TEXT -1 23 " moves with zero speed." }}{PARA 0 "" 0 "" {TEXT -1 34 "The energy conservation law reads:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "M:='M': Eb:=2*m/sqrt(1-u0^2/c^2)*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EbG,$*&*&%\"mG\"\"\")%#c|irG\"\"#\"\"\"F-*$-%%sqrtG6#,&F)F)*& *$)%$u0|irGF,F-F-*$)F+\"\"#F-!\"\"!\"\"F-F:F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "Ea:=M*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #EaG*&%\"MG\"\"\")%#c|irG\"\"#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "M:=simplify(solve(Ea=Eb,M));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG,$*&*&%#c|irG\"\"\"%\"mGF)\"\"\"*$-%%sqrtG6#,&*$) F(\"\"#F+F)*$)%$u0|irGF3F+!\"\"F+!\"\"F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "We can verify that the energy expressions obtained in ei ther frame are compatible with each other. Let us start in the CM fram e and transform the energy to the lab frame: we carry out the calculat ion for the energy after the collision, " }{TEXT 19 2 "Ea" }{TEXT -1 53 ", first, and later perform the other calculation for " }{TEXT 19 2 "Eb" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 423 27 "Energy after the collision:" }}{PARA 0 "" 0 "" {TEXT -1 108 "To Lorentz transform to the lab frame we need to calculate (th e object is not moving, i.e., has momentum 0):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 36 "E_lab:=1/sqrt(1-u0^2/c^2)*(Ea+u0*0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_labG,$*&*&%\"mG\"\"\")%#c|irG\"\"$\"\" \"F-*&-%%sqrtG6#,&F)F)*&*$)%$u0|irG\"\"#F-F-*$)F+\"\"#F-!\"\"!\"\"F--F 06#,&*$)F+F7F-F)*$F5F-F " 0 "" {MPLTEXT 1 0 23 "E_lab:=simplify(E_lab);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%& E_labG,$*&*&)%#c|irG\"\"%\"\"\"%\"mG\"\"\"F+,&*$)F)\"\"#F+!\"\"*$)%$u0 |irGF1F+F-!\"\"!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simp lify(Ea_p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&)%#c|irG\"\"%\"\" \"%\"mG\"\"\"F),&*$)F'\"\"#F)!\"\"*$)%$u0|irGF/F)F+!\"\"!\"#" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "Can we transform the other way to o? Suppose we start with the lab frame result, we have the single mass M moving with " }{TEXT 420 1 "u" }{TEXT -1 2 "0." }}{PARA 0 "" 0 "" {TEXT -1 53 "We are transforming to a frame that is boosted with -" } {TEXT 419 1 "u" }{TEXT -1 36 "0 compared to this frame. Therefore," }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "E_cm:=simplify(1/sqrt(1-u0^ 2/c^2)*(E_lab-u0*M/sqrt(1-u0^2/c^2)*u0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_cmG,$*&*&%\"mG\"\"\")%#c|irG\"\"$\"\"\"F-*$-%%sqrt G6#,&*$)F+\"\"#F-F)*$)%$u0|irGF5F-!\"\"F-!\"\"F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "Ea;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&% \"mG\"\"\")%#c|irG\"\"$\"\"\"F+*$-%%sqrtG6#,&*$)F)\"\"#F+F'*$)%$u0|irG F3F+!\"\"F+!\"\"F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "This was s imple. Now transform the energy before the collision. Let us go from t he center of mass to the lab frame, as we know the velocities there." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 424 28 "Energy before the collision:" }}{PARA 0 "" 0 "" {TEXT -1 30 "The lab frame i s boosted with " }{TEXT 422 1 "V" }{TEXT -1 3 " = " }{TEXT 421 1 "u" } {TEXT -1 31 "0 with respect to the CM frame." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "E_lab:=1/sqrt(1-u0^2/c^2)*((m*c^2/sqrt(1-u0^2/c^ 2)+u0*u0)+(m*c^2/sqrt(1-u0^2/c^2)-u0*u0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_labG,$*&*&%\"mG\"\"\")%#c|irG\"\"#\"\"\"F-,&F)F)*& *$)%$u0|irGF,F-F-*$)F+\"\"#F-!\"\"!\"\"F6F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "Eb_p;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&%\"mG \"\"\")%#c|irG\"\"#\"\"\"F&*&*&F%F*F'F*F**$-%%sqrtG6#,&F&F&*&*$)%$u_pG F)F*F**$)F(\"\"#F*!\"\"!\"\"F*F9F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "simplify(subs(u_p=2*u0/(1+u0^2/c^2),Eb_p));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&)%#c|irG\"\"%\"\"\"%\"mG\"\"\"F),&*$)F '\"\"#F)!\"\"*$)%$u0|irGF/F)F+!\"\"!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(E_lab);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#, $*&*&)%#c|irG\"\"%\"\"\"%\"mG\"\"\"F),&*$)F'\"\"#F)!\"\"*$)%$u0|irGF/F )F+!\"\"!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 260 "We notice that t he results agree when we insert the speed of the beam particle (which \+ we determined from momentum conservation). We could have used the Lore ntz transformation of the energy from the CM (zero-momentum) frame to \+ determine the beam particle speed:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve(Eb_p=E_lab,u_p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,$*&*&)%#c|irG\"\"#\"\"\"%$u0|irG\"\"\"F),&*$)F*F(F)F+*$F&F)F+! \"\"F(,$F$!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "For completene ss one more check: the transformation from the lab frame to the CM fra me. We begin with a definition of the beam particle speed " }{TEXT 19 3 "u_p" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "u _p:=2*u0/(1+u0^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$u_pG,$*&% $u0|irG\"\"\",&\"\"\"F**&*$)F'\"\"#F(F(*$)%#c|irG\"\"#F(!\"\"F*F3F." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "E_cm:=1/sqrt(1-u0^2/c^2)*( (m*c^2+u0*0)+(m*c^2/sqrt(1-u_p^2/c^2)-u0*m/sqrt(1-u_p^2/c^2)*u_p));" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_cmG*&,(*&%\"mG\"\"\")%#c|irG\"\" #\"\"\"F)*&*&F(F-F*F-F-*$-%%sqrtG6#,&F)F)*&*$)%$u0|irGF,F-F-*&),&F)F)* &*$F7F-F-*$)F+\"\"#F-!\"\"F)\"\"#F-)F+\"\"#F-FA!\"%F-FAF)*&*&F7F-F(F-F -*&-F26#F4F-F;\"\"\"FA!\"#F-*$-F26#,&F)F)F " 0 "" {MPLTEXT 1 0 21 "E_cm:=simplify(E_cm);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E_cmG,$*&*&%\"mG\"\"\")%#c|irG\"\"$\"\"\"F-*$-%%sqrt G6#,&*$)F+\"\"#F-F)*$)%$u0|irGF5F-!\"\"F-!\"\"F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "simplify(Eb);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&%\"mG\"\"\")%#c|irG\"\"$\"\"\"F+*$-%%sqrtG6#,&*$)F)\"\"#F+ F'*$)%$u0|irGF3F+!\"\"F+!\"\"F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "We summarize the progress made in this section:" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 202 "1) we understand now h ow to transform energy between inertial frames (such as lab vs cm fram e) for systems of particles (demonstrated mostly in one dimension, but the generalization is straightforward);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "2) we considered elastic and (com pletely) inelastic billiard ball scattering. For completely inelastic \+ collisions in the center of mass frame we demonstrated how all kinetic energy was converted into an additional mass for a new particle of ma ss " }{TEXT 425 1 "M" }{TEXT -1 706 " in this zero-momentum frame. We \+ can think of the inelasticity being effected by a virtual spring that \+ was loaded and locked during the collision process; the converted ener gy that resulted in a mass increase can become available when the two \+ particles break apart One concrete example of such a process would be \+ the alpha-decay, or fissioning of nuclei, where after a tunneling proc ess, or by collisional fragmentation the two daughter nuclei are pushe d apart from each other by the Coulomb repulsion. The daughter product s appear asymptotically will well-defined kinetic energies. As long as the spring is loaded (the system sticks together) the stored energy i s noticeable via the increased mass (over 2" }{TEXT 426 1 "m" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 333 "Usually bound systems in ph ysics represent situations where the composite system has a lower ener gy (mass) than the sum of the masses of the constituents. This situati on makes the composite system stable against break-up. The gain in ene rgy by forming the composite system is the driving process for the com pound system to be formed." }}{PARA 0 "" 0 "" {TEXT -1 104 "This impli es that the idea of a rest mass for a composite system of fundamental \+ particles may be tricky." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 259 "" 0 "" {TEXT -1 42 "8) Invariant mass for systems of particles" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 427 14 "Invariant mass" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 203 "We noticed tha t the components of the momentum vector and energy transformed under L Ts in analogy to the transformation of position and time. For the LT o f a spacetime interval an invariant can be found:" }}{PARA 0 "" 0 "" {TEXT 19 38 "Ds^2 = (c*Dt)^2 - (Dx^2 + Dy^2 + Dz^2)" }}{PARA 0 "" 0 " " {TEXT -1 363 "(the mathematical reason for this quantity to be conse rved is that it forms the inner product of a vector with itself in the four-dimensional spacetime. The LTs represent rotations in this space (that is why there is the peculiar sign chosen between the space and \+ time contributions to ds), and leave the length of four-vectors invari ant under the transformation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 135 "It is of interest to consider the invari ant quantity associated with the four-vector made up of the energy and the momentum components." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "restart; unprotect(gamma); assume(c >0,u0-c,m>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "gamm a:=1/sqrt(1-u^2/c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&gammaG*&\" \"\"F&*$-%%sqrtG6#,&\"\"\"F,*&*$)%\"uG\"\"#F&F&*$)%#c|irG\"\"#F&!\"\"! \"\"F&F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "px:=m*gamma*ux; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#pxG*&*&%#m|irG\"\"\"%#uxGF(\"\" \"*$-%%sqrtG6#,&F(F(*&*$)%\"uG\"\"#F*F**$)%#c|irG\"\"#F*!\"\"!\"\"F*F9 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "py:=m*gamma*uy: pz:=m*g amma*uz:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "E:=gamma*m*c^2; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG*&*&%#m|irG\"\"\")%#c|irG\" \"#\"\"\"F,*$-%%sqrtG6#,&F(F(*&*$)%\"uGF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F 9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "E^2-c^2*(px^2+py^2+pz^ 2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&*&)%#m|irG\"\"#\"\"\")%#c|i rG\"\"%F)F),&\"\"\"F.*&*$)%\"uGF(F)F)*$)F+\"\"#F)!\"\"!\"\"F6F.*&)F+F( F),(*&*&F&F))%#uxGF(F)F)F-F6F.*&*&F&F))%#uyGF(F)F)F-F6F.*&*&F&F))%#uzG F(F)F)F-F6F.F.F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "algsubs (ux^2+uy^2+uz^2=u^2,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&*()%#c| irG\"\"%\"\"\")%#m|irG\"\"#F))%\"uGF,F)F),&*$)F'F,F)\"\"\"*$F-F)!\"\"! \"\"F4*&*&F*F)F&F)F),&F2F2*&*$F-F)F)*$)F'\"\"#F)F5F4F5F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&)%#c|irG\"\"%\"\"\")%#m|irG\"\"#F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "The invariant magnitude (under LT) of the fourv ector made up of " }{TEXT 429 1 "E" }{TEXT -1 5 " and " }{TEXT 428 1 " p" }{TEXT -1 32 " is the rest energy of the mass " }{TEXT 430 1 "m" } {TEXT -1 262 ". We can state that any inertial-frame observer will mea sure the same value for the rest energy, and therefore the mass of an \+ isolated system (since c is universal). There is a special reference f rame where the particle is at rest, namely the frame specified by " } {TEXT 431 1 "p" }{TEXT -1 58 "=0. In this frame the total energy equal s the rest energy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 432 15 "kinematic state" }{TEXT -1 74 " of a \+ system of particles is specified by the invariant spacetime interval" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 19 38 "Ds^2 = (c*Dt)^2 - (Dx^2 \+ + Dy^2 + Dz^2)" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 433 13 "dyn amic state" }{TEXT -1 150 " of a relativistic system of particles can \+ be described by the following expression for the invariant total rest \+ mass in terms of energy and momentum:" }}{PARA 0 "" 0 "" {TEXT 19 25 " (m*c^2)^2 = E^2 - (p*c)^2" }}{PARA 0 "" 0 "" {TEXT -1 240 "This relati on allows to deduce the rest mass from an energy and momentum measurem ent. This is important, as one does not have to be in the rest frame o f a particle to determine its rest mass - which is often encountered i n particle physics." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 272 "Even if there are no interactions between the \+ particles a system of moving particles does experience a peculiarity. \+ The total energy and momentum of a system of particles was used to det ermine the invariant rest energy. We will now show that this invariant rest energy is " }{TEXT 434 3 "not" }{TEXT -1 58 " the sum of the res t energies of the individual particles." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 175 "A simple example of two freely movi ng particles suffices to make the point. We return to a set-up discuss ed in the context of inelastic collisions. Two particles of rest mass \+ " }{TEXT 437 1 "m" }{TEXT -1 18 " with velocities +" }{TEXT 436 1 "u" }{TEXT -1 7 "0 and -" }{TEXT 435 1 "u" }{TEXT -1 168 "0 approach each \+ other (CM frame). We can consider the invariant mass calculation also \+ in different frames, such as the lab frame, where one of the particles is at rest." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "For each particle we can calculate its energy" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "E:=sqrt(m^2*c^4+p^2*c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG*$-%%sqrtG6#,&*&)%#c|irG\"\"%\"\"\")%#m|irG \"\"#F.\"\"\"*&)%\"pGF1F.)F,F1F.F2F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "The total momentum in the CM frame is zero (it is also called t he zero-momentum frame). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "P_tot:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&P_totG\"\"!" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_tot:=2*E;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_totG,$*$-%%sqrtG6#,&*&)%#c|irG\"\"%\"\"\")%#m |irG\"\"#F/\"\"\"*&)%\"pGF2F/)F-F2F/F3F/F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "The invariant mass of the system of two free particles is now calculated to be:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "M :=simplify(sqrt((E_tot/c^2)^2-(P_tot/c)^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG,$*&*$-%%sqrtG6#,&*&)%#c|irG\"\"#\"\"\")%#m|irGF/ F0\"\"\"*$)%\"pGF/F0F3F0F0F.!\"\"F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 638 "We notice that the rest mass of the system is greater than twi ce the rest mass of the particles! It does not make sense to associate the excess in the invariant rest mass of the system over twice the re st masses of the constituents with the motion of individual particles \+ (even though the origin of this excess does lie in the non-vanishing m omentum vectors). To illustrate this, we calculate the same invariant \+ mass by transforming to another reference frame, and calculating the i nvariant mass of the entire system there (to check whether it really i s a property of the system, i.e., independent of the reference frame o f the observer)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 234 "Before we carry out this calculation we express the mome ntum of the particles in terms of their speeds. Given that the square \+ of the momentum is required we do not need to distinguish the signs fo r the left-and right-moving particles:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "p:=m/sqrt(1-u0^2/c^2)*u0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG*&*&%#m|irG\"\"\"%$u0|irGF(\"\"\"*$-%%sqrtG6#,&F( F(*&*$)F)\"\"#F*F**$)%#c|irG\"\"#F*!\"\"!\"\"F*F8" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 22 "simplify(eval(E_tot));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&)%#c|irG\"\"$\"\"\"%#m|irG\"\"\"F)*$-%%sqrtG6#,&* $)F'\"\"#F)F+*$)%$u0|irGF3F)!\"\"F)!\"\"F3" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "M_cm:=simplify(eval(M));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%M_cmG,$*&*&%#c|irG\"\"\"%#m|irGF)\"\"\"*$-%%sqrtG6#, &*$)F(\"\"#F+F)*$)%$u0|irGF3F+!\"\"F+!\"\"F3" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 65 "We evaluate the expressions to make the use of the expr ession of " }{TEXT 443 1 "p" }{TEXT -1 22 " in terms of the mass " } {TEXT 442 1 "m" }{TEXT -1 11 " and speed " }{TEXT 441 1 "u" }{TEXT -1 2 "0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Suppose we pick a frame that moves with " }{TEXT 438 1 "V " }{TEXT -1 3 " = " }{TEXT 439 1 "u" }{TEXT -1 166 "0 with respect to \+ the center of mass, i.e., one of the masses is at rest, and the other \+ (e.g., a beam particle) moves with an increased velocity vector (not q uite - 2" }{TEXT 440 1 "u" }{TEXT -1 68 "0 according to a calculation \+ done above, where the correct value of " }{TEXT 19 3 "v_p" }{TEXT -1 17 " was determined)." }}{PARA 0 "" 0 "" {TEXT -1 47 "Now one particle is at rest and one moves with:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "u_p:=simplify(-2*u0/(1+u0^2/c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$u_pG,$*&*&%$u0|irG\"\"\")%#c|irG\"\"#\"\"\"F-,&*$F*F -F)*$)F(F,F-F)!\"\"!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "We cou ld repeat the calculation of the invariant mass from scratch:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "E_tot:=m*c^2+simplify(m/sqrt (1-u_p^2/c^2)*c^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&E_totG,&*&)% #c|irG\"\"#\"\"\"%#m|irG\"\"\"F,*&*(F+F*,&*$F'F*F,*$)%$u0|irGF)F*F,F,F 'F*F*,&F0F,F1!\"\"!\"\"F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "P_tot:=simplify(m*0-m/sqrt(1-u_p^2/c^2)*u_p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&P_totG,$*&*(%#m|irG\"\"\"%$u0|irGF))%#c|irG\"\"#\"\" \"F.,&*$F+F.F)*$)F*F-F.!\"\"!\"\"F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "M_lab:=simplify(sqrt((E_tot/c^2)^2-(P_tot/c)^2));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&M_labG,$*&*&%#c|irG\"\"\"%#m|irGF) \"\"\"*$-%%sqrtG6#,&*$)F(\"\"#F+F)*$)%$u0|irGF3F+!\"\"F+!\"\"F3" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "M_cm;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&%#c|irG\"\"\"%#m|irGF'\"\"\"*$-%%sqrtG6#,&*$)F&\" \"#F)F'*$)%$u0|irGF1F)!\"\"F)!\"\"F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "It is impressive that Maple 5 does recognize these two express ions to be the same once the restricition -c < " }{TEXT 444 1 "u" } {TEXT -1 24 "0 < c is imposed in the " }{TEXT 19 6 "assume" }{TEXT -1 26 " statement (invoked after " }{TEXT 19 7 "restart" }{TEXT -1 35 " a t the beginning of this section)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 198 "It is important to recognize that the sa me invariant mass was calculated for the system in both frames. This i s remarkable as the relativistic mass calculation is different in the \+ CM and Lab frames:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "RM_cm:=2*m/sqrt(1-u0^2/c^2);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%&RM_cmG,$*&%#m|irG\"\"\"*$-%%sqrtG6#,&\"\"\"F. *&*$)%$u0|irG\"\"#F(F(*$)%#c|irG\"\"#F(!\"\"!\"\"F(F8F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "RM_lab:=simplify(m+m/sqrt(1-u_p^2/c ^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'RM_labG,$*&*&)%#c|irG\"\"# \"\"\"%#m|irG\"\"\"F+,&*$F(F+F-*$)%$u0|irGF*F+!\"\"!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "plot([subs(c=1,m=1,RM_cm),subs(c=1, m=1,RM_lab)],u0=0..1,color=[red,blue],view=[0..1,0..10]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6$7U7$\"\"!$\" \"#F(7$$\"1+++;arz@!#<$\"1%*yH&Gv/+#!#:7$$\"1+++XTFwSF.$\"1[0luOm,?F17 $$\"1+++\"z_\"4iF.$\"1TKhTl'Q+#F17$$\"1+++S&phN)F.$\"12))4Q$>q+#F17$$ \"1+++*=)H\\5!#;$\"1f3qG?56?F17$$\"1+++[!3uC\"FD$\"1%yPqCWd,#F17$$\"1+ ++J$RDX\"FD$\"1]I>_&Q9-#F17$$\"1+++)R'ok;FD$\"19_t(R,$G?F17$$\"1+++1J: w=FD$\"1%QiL)o:O?F17$$\"1+++3En$4#FD$\"1-)>9]I`/#F17$$\"1+++/RE&G#FD$ \"1hm3OJOa?F17$$\"1+++D.&4]#FD$\"1gs7wMkl?F17$$\"1+++vB_7#F17$$\"1+++LY.KNFD$\"1U`Y2aFD$\"1K[&z*ffx BF17$$\"1+++yXu9cFD$\"1!QIQvNpT#F17$$\"1+++\\y))GeFD$\"1xro(4\"QhCF17$ $\"1+++i_QQgFD$\"1%Q(\\$\\u!4DF17$$\"1+++!y%3TiFD$\"1,!*zVErfDF17$$\"1 +++O![hY'FD$\"1Yw.%*='=i#F17$$\"1+++#Qx$omFD$\"1er5`E$Qo#F17$$\"1+++u. I%)oFD$\"1R0*o>puv#F17$$\"1*****ppe*zqFD$\"1V?fWC*>$GF17$$\"1+++C\\'QH (FD$\"1AJgr7aBHF17$$\"1+++8S8&\\(FD$\"1MtKhu>@IF17$$\"1+++0#=bq(FD$\"1 wxfz&ey8$F17$$\"1*****p?27\"zFD$\"1%3V$*=!zpKF17$$\"1,++IXaE\")FD$\"10 WZ+y-KMF17$$\"1+++l*RRL)FD$\"1q$fy!\\t=OF17$$\"1+++`<.Y&)FD$\"1$GHMdu9 &QF17$$\"1+++8tOc()FD$\"1#f4%[8/TTF17$$\"1+++\\Qk\\*)FD$\"1ZH#H[%F1 7$$\"1+++p0;r\"*FD$\"1r:rFgI<]F17$$\"1+++oTAq#*FD$\"18'oM;\\KL&F17$$\" 1*****\\w(Gp$*FD$\"1n2>_w7AdF17$$\"1+++B-\"\\Z*FD$\"1o-jCNDaiF17$$\"1+ ++!oK0e*FD$\"1td`))fgypF17$$\"1+++<5s#y*FD$\"1f*=rb&oY'*F17$%%FAILGF^[ l-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F(-F$6$7WF'7$F,$\"1XQa$o]4+#F17$F3$ \"1nx?L(GL+#F17$F8$\"10&)He0u2?F17$F=$\"1(G^+V)o\"4#F17$F[o$\"17fXZR?5@F17$F`o$\"1bNB&[TM8#F17$F eo$\"1*>pVsv%f@F17$Fjo$\"1\\gCbSH(=#F17$F_p$\"1#=KG1M]@#F17$Fdp$\"1PV8 ekJ^AF17$Fip$\"1w2![Uo]G#F17$F^q$\"10$y;!)3\"GBF17$Fcq$\"1ZEH\"Q\"*)pB F17$Fhq$\"1x#zV\\%4?CF17$F]r$\"1'[WH!zbsCF17$Fbr$\"1T&G8yfE`#F17$Fgr$ \"10)z()R,Kf#F17$F\\s$\"1/:S+P\"\\m#F17$Fas$\"10r99AKZFF17$Ffs$\"1;auk 8[EGF17$F[t$\"19=2>#*y?HF17$F`t$\"1ne-a%)>HIF17$Fet$\"1x&fpSFx9$F17$Fj t$\"1NAT4W1wKF17$F_u$\"1OLq&*)zqV$F17$Fdu$\"1/@%\\&)y9g$F17$Fiu$\"1x+! f===!QF17$F^v$\"12K6.145SF17$Fcv$\"1+MQ$yYNF%F17$Fhv$\"1k`;]q\"Qc%F17$ F]w$\"1&oxFByI#\\F17$Fbw$\"1T@;TRwX`F17$Fgw$\"1\"f9;T2%*)eF17$F\\x$\"1 !)*zm;@wa'F17$Fax$\"1Qh)[>GpT(F17$Ffx$\"1m*=Ys6Td)F17$F[y$\"1#e3GYH[+ \"!#97$F`y$\"1Vby))zme7Fjcl7$Fey$\"1L>)>Lx@U\"Fjcl7$Fjy$\"16$)RVs8P;Fj cl7$$\"1+++&**)4A%*FD$\"1F)Rb\"z)=y\"Fjcl7$F_z$\"11v^dVyb>Fjcl7$$\"1++ +_9sF&*FD$\"1B:a^Pgo@Fjcl7$Fdz$\"1o8;xq/NCFjcl7$Fiz$\"1wj*=6FHl%FjclF] [l-F`[l6&Fb[lF(F(Fc[l-%+AXESLABELSG6$Q$u0|ir6\"%!G-%%VIEWG6$;F($\"\"\" F(;F($\"#5F(" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "Clearly the invariant mass is not equal to the sum of the two relativistic masses of the particles! It \+ has to be calculated according to the prescription for the dynamic sta te of the relativistic system." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 445 9 "Exercise:" }}{PARA 0 "" 0 "" {TEXT -1 268 "Verify the invariant mass \+ of the two-particle system by carrying out the Lorentz transformation \+ of the energy from the CM to the Lab frame, and using the dynamic stat e of the system. Then illustrate the value of the invariant mass by co nsidering the following values: " }{TEXT 446 1 "m" }{TEXT -1 9 " = 4 k g, " }{TEXT 447 1 "u" }{TEXT -1 10 "0 = 0.6 c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 261 "" 0 "" {TEXT -1 35 "9) Massless particles: photons,...?" }}{EXCHG {PARA 0 "" 0 "" {TEXT 448 18 "Massless particles" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 339 "It is possible to consider the dynamic state in the limit of zero rest mass. In princip le, there could also be particles with negative rest mass (tachyons), \+ but so far they have eluded observation if they indeed exist. For part icles without a rest mass we find from the dynamic state the following relationship between energy and momentum:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "E:=subs(m=0,sqrt(m^2*c^4+p^2*c^2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG*$-%%sqrtG6#*&)%\"pG\"\"#\"\"\")%\"cGF,F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "We can determine the speed of the partic le from the following observation. For massive particles the speed is \+ related to energy and momentum as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "p_m:=m/sqrt(1-u^2/c^2)*u;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$p_mG*&*&%\"mG\"\"\"%\"uGF(\"\"\"*$-%%sqrtG6#,&F(F(*& *$)F)\"\"#F*F**$)%\"cG\"\"#F*!\"\"!\"\"F*F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "E_m:=m/sqrt(1-u^2/c^2)*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_mG*&*&%\"mG\"\"\")%\"cG\"\"#\"\"\"F,*$-%%sqrtG6#,& F(F(*&*$)%\"uGF+F,F,*$)F*\"\"#F,!\"\"!\"\"F,F9" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 12 "p_m*c^2/E_m;" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#%\"uG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "This relationship is \+ valid irrespective of the mass of the particle, and presumably carries over to the massless case:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "simplify(p*c^2/E,symbolic);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%\" cG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Therefore a massless partic le can only move with one speed, namely the speed of light." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "What is the pho ton's momentum? We should get " }{TEXT 450 2 "p " }{TEXT -1 2 "= " } {TEXT 449 1 "E" }{TEXT -1 61 "/c according to our derivation above. In sertion of the speed " }{TEXT 452 1 "u" }{TEXT -1 14 " = c and mass " }{TEXT 451 1 "m" }{TEXT -1 111 " = 0 into the momentum expression resu lts in an undefined answer (0/0). The appropriate step is to replace t he " }{TEXT 19 17 "m/sqrt(1-u^2/c^2)" }{TEXT -1 15 " expression by " } {TEXT 19 5 "E/c^2" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 378 "The only particles known to be definitel y massless are photons. Neutrinos are candidates for massless particle s, although there is some evidence for a lower bound on the mass. Grav itons, the postulated qunatum carriers of gravity also should be massl ess (for photons and gravitons the masslessness is connected with the \+ inverse squared force laws that appear in these theories)." }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "19 0 0" 0 } {VIEWOPTS 1 1 0 1 1 1803 }