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0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 21 "Relativistic dynamics" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "In the wo rksheet " }{TEXT 19 14 "SpecialRel.mws" }{TEXT -1 447 " we introduced \+ the phenomena of time dilation and length contraction, and motivated t he transformation of position and time to relate two inertial frames i n order to be consistent with the two postulates of relativity. 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);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "This is a generalization of the Galilei t ransformation: 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);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The two r emaining coordinates are unchanged under the transformation (" }{TEXT 19 4 "yp=y" }{TEXT -1 2 ", " }{TEXT 19 4 "zp=z" }{TEXT -1 39 ") due to the choice of boost along the " }{TEXT 263 1 "x" }{TEXT -1 35 "-axis. According to Galilei we had " }{TEXT 19 4 "tp=t" }{TEXT -1 75 ". That this has changed with the generalization, and is only recovered for \+ " }{TEXT 19 7 "gamma=1" }{TEXT -1 75 " can be seen from the combinatio n of the two transformations. We can solve " }{TEXT 19 3 "L1i" }{TEXT -1 5 " for " }{TEXT 19 2 "xp" }{TEXT -1 33 " and eliminate the variabl e from " }{TEXT 19 2 "L1" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "L1ip:=xp=solve(L1i,xp);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "L1p:=subs(xp=rhs(L1ip),L1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L1p:=tp=expand(solve(L1p,tp));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "We see clearly 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 that the speed of light be the same in \+ each reference frame, and that the laws of physics be the same in each inertial reference frame. Consider 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 an d primed reference frames is specified by the light front conditions: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "L2:=x^2+y^2+z^2=c^2*t^2 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L2p:=xp^2+yp^2+zp^2=c^ 2*tp^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "The two statements hav e 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 coordi nate system." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "L2p:=subs(y p=y,zp=z,L2p);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L2p1:=sub s(tp=rhs(L1p),L2p);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L2p2 :=subs(xp=rhs(L1),L2p1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Now w e 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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "L3:=expand(lhs(L3)-rhs(L3))=0;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 26 "L3:=collect(L3,[x,y,z,t]);" }}}{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 le ft after the subtraction. Our task looks difficult: can we satisfy thr ee conditions at the same time with a single choice of gamma?" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "L3p1:=coeff(L3,x^2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L3p1:=op(1,op(1,L3));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "L3p1:=coeff(L3p1,x^2)=0;" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "We can't use the symbol gamma to solve L3p1 for it, as Maple still thinks that gamma is the Euler-Masc heroni constant. We need a change of name:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "L3p1:=subs(gamma=eta,L3p1);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 16 "solve(L3p1,eta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "simplify(1/sqrt(1-V^2/c^2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The latter is identical to the fourth solution found a bove. Note that " }{TEXT 19 7 "gamma=1" }{TEXT -1 50 " is also a solut ion. Now try the other conditions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L3p2:=op(2,op(1,L3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "L3p2:=L3p2/(t*x)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "L3p2:=subs(gamma=eta,L3p2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(L3p2,eta);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "We are getting some good news: " }{TEXT 19 7 "gamma=1" }{TEXT -1 104 " is excluded now, and we still have the same candidate(s) for gam ma in the running to satisfy all terms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "L3p3:=op(3,op(1,L3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "L3p3:=coeff(L3p3,t^2)=0;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "L3p3:=subs(gamma=eta,L3p3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(L3p3,eta);" }}}{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;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "L1p;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "si mplify(subs(gamma=1/sqrt(1-V^2/c^2),L1p));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "One can convince oneself that this is equivalent to the c ompact form:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "L1p:=tp=gam ma*(t-V*x/c^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The inverse tr ansformations are given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "L1i;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "and an equation for " }{TEXT 267 1 "t" }{TEXT -1 48 " that has to be the symmetric equivalen t of L1p:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L1pi:=t=gamma* (tp+V*xp/c^2);;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "These are supp lemented by the statements (" }{TEXT 19 6 "y = yp" }{TEXT -1 6 ", and \+ " }{TEXT 19 6 "z = zp" }{TEXT -1 18 ", and vice versa)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "We can use the trans formations to re-iterate previous findings." }}{PARA 0 "" 0 "" {TEXT 268 33 "Transformation of time intervals:" }{TEXT -1 160 " take two ev ents (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 prime d 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));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "expand(TI);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Not only the temporal separation is important, but also the spati al 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 pr evious worksheet by forming differentials of " }{TEXT 19 2 "L1" } {TEXT -1 5 " and " }{TEXT 19 3 "L1p" }{TEXT -1 35 " and the correspond ing derivatives." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Lv1:=up _x=(u_x-V)/(1-V*u_x/c^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Lv2:=up_y=u_y/(gamma*(1-V*u_x/c^2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Lv3:=up_z=u_z/(gamma*(1-V*u_x/c^2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "The inverse transformation is given by sy mmetry (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=(u p_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 "Lv i3:=u_z=up_z/(gamma*(1+V*up_x/c^2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "To understand relativistic dynamics we have to consider now th e acceleration. In Galilean reference frames " }}{PARA 0 "" 0 "" {TEXT -1 25 "We have to differentiate " }{TEXT 19 4 "up_x" }{TEXT -1 17 " with respect to " }{TEXT 19 2 "tp" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Lv1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subs(tp=dtp,t=dt,x=dx,L1p);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 43 "First we need to form the differentials in " }{TEXT 19 3 "Lv1" }{TEXT -1 72 ". The appearance of u_x in the denominator compl icates 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 constant, i.e., in the numerat or we obtain the " }{TEXT 19 12 "diff(u_x, t)" }{TEXT -1 25 " derivati ve; there is no " }{TEXT 273 1 "x" }{TEXT -1 174 "-dependence, and so \+ apart from the gamma factor we expect no surprises from taking the der ivative. However, the denominator also gets differentiated with respec t to time, as " }{TEXT 19 3 "u_x" }{TEXT -1 209 " appears there as wel l. First we form the differential by taking a time derivative. For thi s purpose we have to introduce an explicit time dependence, and then r emove it to recover the variables used normally." }}}{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,%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "dup_ x:=subs(diff(u_x(t),t)=du_x,u_x(t)=u_x,%);" }}}{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)) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "and remove the dx in the d enominator by the trick used in the derivation of the velocity transfo rmation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "subs(dx=u_x*dt, %);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Now we can introduce the a cceleration in the rest frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "simplify(subs(du_x=a_x*dt,%));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "Apart from the c still present in the numerator, we gen erated the textbook result for the acceleration in the moving referenc e frame, " }{TEXT 19 4 "ap_x" }{TEXT -1 147 ". According to Galilei th e acceleration is the same in two inertial reference frames. According to Lorentz it is not, unless if we take the limit (" }{TEXT 19 7 "gam ma=1" }{TEXT -1 5 " and " }{TEXT 274 1 "V" }{TEXT -1 81 "/c approximat ely 0). We cannot expect Newton's law to survive in the 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 kinetic energy" }} {EXCHG {PARA 0 "" 0 "" {TEXT 278 21 "Relativistic momentum" }}{PARA 0 "" 0 "" {TEXT -1 271 "We should first look at relativistic momentum to understand the situation when no forces act on a particle. Under thes e circumstances momentum is conserved, and we can ask how the nonrelat ivistic definition of momentum should be extended to maintain the cons ervation law." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Consider two observers with identical balls of mass " } {TEXT 279 1 "m" }{TEXT -1 82 " in the respective rest frame. The obser vers 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), such that the \+ balls collide and bounce elastically, i.e., reverse their velocities i n the " }{TEXT 281 1 "y" }{TEXT -1 111 "-direction. From the velocity \+ transformation discussed before we know that even the velocity compone nts in the " }{TEXT 282 1 "y" }{TEXT -1 90 "-direction will 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 f ind when comparing the magnitudes of the v_y velocity components for t he two balls?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "Suppose observer A is at rest. Initially, A's ball has no " }{TEXT 285 1 "x" }{TEXT -1 45 "-component and moves with speed u0 a long the " }{TEXT 286 1 "y" }{TEXT -1 32 "-axis in the positive direct ion." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vA_y:=u0;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "vA_x:=0;" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 81 "In B's rest frame (denoted 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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vB_xp:=0;" }}}{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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "vB_y:=vB_yp/gamma/(1+V*vB_xp);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 204 "Here we used the velocity transfo rmation law from the primed (moving) to the unprimed (rest) frame, whi ch simplifies for the case 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, w hich is a consequence of the time dilation factor associated with the \+ motion of B's reference frame along the " }{TEXT 290 1 "x" }{TEXT -1 71 "-axis. It takes more time for B to travel a certain distance along the " }{TEXT 291 1 "y" }{TEXT -1 146 "-direction in the unprimed fram e, than the time it takes to travel the same distance in B's rest fram e (primed). In A's rest frame (unprimed) the " }{TEXT 292 1 "y" } {TEXT -1 285 "-component of total momentum as defined according to New ton is not zero before the collision! Thus, the sum of mass times velo city for all particles is not conserved in A's rest frame. The same ho lds in B's frame. The classical 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 w ere transformed according to the Lorentz transformation, so they shoul d be correct. The only other freedom in converting the momentum to a r elativistically acceptable quantity is to tamper with the mass. Let us call the speed of the ball from the opposite reference frame " } {TEXT 293 1 "u" }{TEXT -1 45 " to distinguish it from the own ball's s peed " }{TEXT 294 1 "u" }{TEXT -1 59 "0, and to account for the motion along the boost direction:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u:=sqrt(V^2+vB_y^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "We f ormulate the momentum conservation using a mass that depends on the sp eed, while using velocity reversal upon impact (the signs are equal as " }{TEXT 19 4 "vB_y" }{TEXT -1 34 " carries 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;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "MC:=simplify(lhs(MC)-rhs(MC) )=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "solve(MC,m(u));" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "abs(u0/vB_y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "limit(u,u0=0);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 62 "For small speeds of the balls in their respective \+ rest frames " }{TEXT 295 1 "u" }{TEXT -1 48 " is given by the boost sp eed 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 rest (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 "). Thi s ratio is just given by the gamma factor. A stationary observer measu res the mass of a moving object to be heavier than the rest mass. This causes the relativistic momentum of a material object to increase fas ter than just proportional to the particle speed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 271 "This mass increase over \+ the rest mass causes a problem when we attempt to accelerate particles : once particles reach a significant fraction of the speed of light c \+ most of 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 relativistic momentum be used. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 305 99 "The rate of change of relativistic momen tum 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 th e work done by a net force while accelerating a particle from rest to \+ some velocity." }}{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..x 1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "The gamma (beta) factor re fers 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 parameter " }{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 d ifferential needs to be converted to d" }{TEXT 312 1 "u" }{TEXT -1 57 " which can be calculated by differentiating (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)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "To carry out the integration we need to remove the du, and multiply w ith the " }{TEXT 315 1 "u" }{TEXT -1 71 "-factor which arose during th e 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);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 16 "Tkin:=expand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs((c^2-U^2)=c^2/gamma^2,Tkin);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Tkin:=sort(%,[gamma,m,c]);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 345 "The relativistic kinetic energy is made up of a term t hat contains the energy of the rest mass, and then the rest mass energ y is subtracted. It is of interest to Taylor expand this expression in order to see how the nonrelativistic kinetic energy is recovered. We \+ have to substitute the gamma factor again, and note that we used the f inal speed " }{TEXT 317 1 "U" }{TEXT -1 47 " as an integration paramet er. We return to use " }{TEXT 316 1 "V" }{TEXT -1 18 " in the form bet a=" }{TEXT 319 1 "V" }{TEXT -1 3 "/c." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(gamma=1/sqrt(1-beta^2),Tkin);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "taylor(%,beta=0,5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(beta=V/c,%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "We have arrived at the nonrelativistic result, and deter mined the lowest-order relativistic correction to the kinetic energy o f a massive particle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Tk NRC:=convert(%,polynom);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "TkNR:=1/2*m*V^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "It is worth while to look at a graphical presentation of these results. We pick un its 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 convert to SI, or particle physics units, we note that th e mass of the electron is 0.511 MeV; this corresponds to one energy un it on our vertical 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=\"Ki netic energy as a function of speed: relativstic and nonrelativistic\" ,view=[0..1,0..3]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 752 "We notice from the graph that an electron moving at 90% of the speed of light h as a kinetic energy that exceeds the rest mass energy. The general con clusion here is that the first relativistic correction may be adequate for particles moving at up to beta=0.6, but that for faster massive p articles the full relativistic kinetic energy expression is required. \+ The kinetic energy of a slow particle is negligible compared to its re st mass energy: an electron in the ground state of a hydrogen atom has a typical energy of 13.6 eV. This is small compared to the 511 keV re st energy, despite of the fact that the electron moves fast (on our li fe experience scale): it moves at about 1/137 times c (1/137 is the fi ne structure constant of electromagnetism)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 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 relativis tic harmonic oscillator (RHO)" }}{PARA 0 "" 0 "" {TEXT -1 448 "Let us \+ solve Newton's equation in the relativistic form to get some feeling f or the effect of the increased mass. We consider a one-dimensional har monic oscillator, e.g., for an electron attached to a spring (for quit e some time approximations were made in atomic physics based on oscill ator strengths). We will compare the analytically known solutions for \+ the nonrelativistic case with numerical solutions based on the relativ istic 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 displacement \+ is chosen to be large to make the particle acquire a substantial amoun t of kinetic energy. We need to understand the length scale in order t o decide what we mean by this choice of parameters. Alternatively, we \+ can just experiment around and observe the speed reached by the partic le 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;" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 63 "For a nonrelativistic harmonic oscillator we have the s olution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "xoft:=x0*cos(sq rt(k/m)*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "voft:=diff(x oft,t);" }}}{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 of light by a factor of 2. What can we find out about the r elativistic 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 that uses both Maple's " }{TEXT 19 6 "dsolve" }{TEXT -1 104 " inte grator. For the determination of the velocity from the momentum we hav e the nonlinear relationship:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "eqU:=U=P/m*sqrt(1-U^2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "so lve(eqU,U);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "We use this expre ssion for the velocity in terms of the momentum for the integration, a nd also for display purposes." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "NE1:=diff(p(t),t)=-k*x(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "NE2:=diff(x(t),t)=p(t)/sqrt(p(t)^2+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "IC:=p(0)=0,x(0)=x0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "dt:=0.05; Nt:=trunc(4*Pi/dt);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "sol:=dsolve(\{NE1,NE2,IC\}, \{x(t),p(t)\},numeric,output=listprocedure);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "xt:=subs(sol,x(t)): pt:=subs(sol,p(t)):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "xt(0.05),pt(0.05);" }}} {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)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "We observe that the motion is no longer linear. T he particle reaches approximately " }{TEXT 19 8 "beta=0.9" }{TEXT -1 543 ", and does not exceed the speed of light. The shape of the veloci ty as a function of time reminds one of the time dependence of the ang le in the mathematical pendulum (not linearized!) for large initial an gles, i.e., for cases when the pendulum starts near the top. Compariso n with the nonrelativistic solution shows that the relativistic harmon ic oscillator is no longer isochronous (as the pendulum): the period d epends on the amount of energy. Let us veify the change in the period \+ by a solution for 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;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "dt:=0.05; Nt:=trunc(4*Pi/dt) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "sol:=dsolve(\{NE1,NE2, IC\},\{x(t),p(t)\},numeric,output=listprocedure);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 39 "xt:=subs(sol,x(t)): pt:=subs(sol,p(t)):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "xt(0.05),pt(0.05);" }}} {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)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "This solution is qualitatively comparable 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 passage of the \+ force-free point " }{TEXT 326 1 "x" }{TEXT -1 304 " = 0. The compariso n of the qualitatively different plots helps to understand why the hig hly relativistic harmonic oscillator can no longer be isochronous: the slowdown of the mass due to its increase with velocity (energy is bei ng dumped into increasing the relativistic mass), and the avoidance of the " }{TEXT 328 1 "v" }{TEXT -1 127 "=1 limit the mass is delayed an d cannot execute oscillations on the same timescale as before, i.e., o n 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 increasing the initial stre tch of the spring (i.e., x0) we can control the amount of kinetic ener gy of the mass while the spring is neither stretched nor compressed. I t is important to realize that we can make the particle hardly go fast er when doing so. The energy is dumped into the increase of the relati vistic mass. Now that we appreciate this phenomenon, it is clear that \+ the period of the RHO has to go down, as the mass increases substantia lly during the passages of " }{TEXT 330 1 "x" }{TEXT -1 226 "=0. Note \+ that the term RHO is a misnomer: the generalization of the HO to the r elativistic case leads to anharmonic motion when the speed of the mass becomes a significant fraction of c in the vicinity 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 period begins to change \+ appreciably from the nonrelativistic harmonic oscillator case. Which m aximum speed does the particle reach in this case?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 256 "" 0 "" {TEXT -1 60 "4) Transforming energy and momentum b etween reference frames" }}{EXCHG {PARA 0 "" 0 "" {TEXT 331 17 "Conser vation laws" }}{PARA 0 "" 0 "" {TEXT -1 531 "An important aspect when \+ studying dynamics is the investigation of conservation laws, which are obeyed by the solutions to the equations of motion. Conservation laws allow one to solve problems in a low number of dimensions, and to imp ose constraints on solutions in three dimensions without looking at th e 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 of energy and momentum" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "Suppose we cons ider a particle moving with velocity vector " }{TEXT 333 1 "u" }{TEXT -1 92 " in a reference frame S, and observe it also in S' which is boo sted compared to S along the " }{TEXT 334 1 "x" }{TEXT -1 26 "-axis wi th 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 (u seless) 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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E:=g*m*c^2;" }}}{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];" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "In S' gamma has a different value: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "gp=1/sqrt(1-up^2/c^2); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Ep:=gp*m*c^2;" }}} {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];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "To transform these four quantities we first express the primed gamma in terms of quantities measured 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 subscripts 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);" }}}{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));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Lv3:=up_z=u_z/(1/sq rt(1-V^2/c^2)*(1-V*u_x/c^2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "up_sq:=simplify(rhs(Lv1)^2+rhs(Lv2)^2+rhs(Lv3)^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "algsubs(u_x^2+u_y^2+u_z^2=u_sq,up_s q);" }}}{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),symbol ic);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "We recognize in the squar e root in the denominator a product: one part belongs to " }{TEXT 19 7 "1/gamma" }{TEXT -1 18 " and the other to " }{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,%);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Therefore the kinetic energy of t he particle in S' transforms into two contributions in S:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Ep=gamma*(E-V*p_x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "To verify by comparison with our result w hich 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],%);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "Analogous derivations follow for t he components of the momentum vector." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "subs(gp=gp_S,up[x]=rhs(Lv1),pp[x]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 25 "pp[x]=gamma*(p[x]-g*m*V);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 4 "The " }{TEXT 338 1 "y" }{TEXT -1 11 "-component:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "subs(gp=gp_S,up[y]=rhs(Lv2), pp[y]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "pp[y]=p[y];" }}}{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 comp are the transformation of the momentum vector to the transformation of the position vector. For the components perpendicular to the directio n of the boost we find that they are unchanged in both cases. For the \+ component along the boost direction a small manipulation reveals an in teresting 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;" }}}{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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "-gamma*V*E /(c^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Therefore we can write the momentum transformation as" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "pp[x]=gamma*(p[x]-V*E/c^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 215 "The Lorentz transformation for the momentum vector components \+ is now strikingly similar to the transformation for the position vecto r components, except that the role of time in the transformation has b een taken by " }{TEXT 19 5 "E/c^2" }{TEXT -1 94 " ! Now we should be c urious how the energy transformation compares to the time transformati on." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Ep:='Ep':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Ep=gamma*(E-V*p_x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "tp=gamma*(t-V*x/c^2);" }}}{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 under Lore ntz transformations, momentum and energy transform in an analogous way . The factors of " }{TEXT 19 3 "c^2" }{TEXT -1 82 " are needed to make matters correct from the point of view of physical dimensions." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 129 "It is im portant to illustrate the derived results with some examples. (we 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 "10^(-9)" }{TEXT -1 75 " kg moves past Earth at speed 0.01 c. An observer in a s paceship moving at " }{TEXT 343 1 "V" }{TEXT -1 154 " = 0.5 c in the s ame direction as the meteorite passes by the Earth. What values of ene rgy and momentum will the moving observer detect for the meteorite?" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Answer:" }}{PARA 0 "" 0 "" {TEXT -1 247 "The total energy of the meteorite can \+ be calculated in the Earth's frame by the non-relativistic approximati on, or by the full relativisitic expression. We ignore the potential e nergy in the gravitational field, and use SI (MKSA) units throughout. " }}}{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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "E_nr:=m*c^2+m/2*u^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "p_rel:=m*1/sqrt(1-u^2/c^2)*u;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "p_nr:=m*u;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 274 "Note that despite of the fact that the nonrelativistic a pproximation is highly accurate to describe the meteorite in the Earth 's reference frame, the transformation to the spaceship frame should b e carried out by a Lorentz transformation, since the spaceship moves a t 0.5 c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "V:=0.5*c;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "unprotect(gamma);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ga mma:=1/sqrt(1-V^2/c^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "The ki netic energy of the meteorite in the spaceship frame:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Ep:=gamma*(E_rel-V*p_rel);" }}} {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;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pp:=gamma*(p_rel-V*E_rel/c^2 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "pp/p_rel;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "Obviously the second term made the big c ontribution here. The momentum is in the opposite direction, and it is huge (Tipler and Llewellyn have a buboe in their solution: wrong numb ers 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) Relativistic 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 be tween inertial frames, as well as with the momentum conservation law ( which we used to find an appropriate expression for relativistic momen tum, we carry out an independent check: we consider the total energy o f a system of two colliding classical particles, and investigate wheth er it is conserved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 144 "First we return to the problem of elastic scattering o f 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 velocities in the f rame A where particle A has only a velocity " }{TEXT 345 1 "y" }{TEXT -1 64 "-component, and frame B in which particle B has only a velocity " }{TEXT 346 1 "y" }{TEXT -1 80 "-component, which opposes 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 speed. This requir es the use of the Lorentz transformation when calculating the momentum balance or the total energy. The balls are thrown such that they coll ide elastically. Due to the transformation of the velocity components \+ nonrelativistic linear momentum is not conserved (when one transforms \+ them according to Lorentz)." }}{PARA 0 "" 0 "" {TEXT -1 68 "We should \+ calculate the total energy before and after the collision." }}}{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;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 11 "vB_yp:=-u0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "vB _xp:=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "vB_x:=V;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "gamma:=1/sqrt(1-beta^2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "vB_y:=vB_yp/gamma/(1+V*vB_xp );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u:=sqrt(V^2+vB_y^2); " }}}{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:=simplif y(lhs(MC)-rhs(MC))=0;" }}}{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 the relativistic energy using the definition of relativistic momentum, and by postulating Newton's \+ law, and integrating free-particle motion to arrive at the kinetic ene rgy 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 correc ts this mass as a function of the particle speed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "We should obtain then for the total energy (simply ommit the subtraction of 1 in the round brac ket:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "E_A:=m*c^2/sqrt(1-u 0^2/c^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "E_B:=simplify( m*c^2/sqrt(1-subs(beta=V/c,u)^2/c^2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "E_totA:=E_A+E_B;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "To calculate matters from B's perspective one obtains the first t erm for B, and then hopefully for the second term the calculation give s the same answer after transforming A's velocity." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "vA_xp:=-V;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "vA_yp:=vA_y/gamma/(1-V*vA_x);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "w:=sqrt((-V)^2+vA_yp^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "E_Ap:=simplify(m*c^2/sqrt(1-subs(beta=-V/c,w) ^2/c^2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "E_Bp:=m*c^2/sq rt(1-u0^2/c^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "The two parti cles have swapped their roles: this is consistent with the principle o f relativity that the physics is the same in all intertial frames. " } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "E_totA-E_Ap-E_Bp;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "The dependence on " }{TEXT 349 1 " u" }{TEXT -1 105 "0 is quadratic, and therefore independent of the sig n. The answer has to be the same after the collision." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "Suppose we wish to cal culate 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 re spect to A with speed " }{TEXT 356 1 "V" }{TEXT -1 65 ", we can specif y the frame C by stating that it moves with speed " }{TEXT 355 1 "W" } {TEXT -1 125 " with respect to A. To transform the energy 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 energy in frame C \+ is trickier, as we do not know the relative speed between 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 rejected the addi tivity of velocities in special relativity. According to the velocity \+ transformation we can calculate the energy of B in frame C via the res t frame of A. We use the gamma factor for the rest frame 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 procedure that calculate s the required steps. We assume that A and B are not moving in the " } {TEXT 362 1 "x" }{TEXT -1 143 "-direction in their respective frame. T he energy is calculated in frame C for which we have a gamma factor th at relates it to the rest frame A." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "Eboost:=proc(W,V,vA_y,vB_yp) local gammaA,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; v B_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_ypp:=vA_y/gam maA/(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_ypp^2);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E_App:=simplify(m*c^2/sqrt(1-w_A^2/ c^2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E_Bpp:=simplify(m*c^2/sqr t(1-w_B^2/c^2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "E_App+E_Bpp; en d:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Eboost(0,c/2,c/10,-c/10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(subs(m=1,c=1,%));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "subs(u0=0.1,c=1,V=0.5,m=1,E_totA);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Eboost(c/2,c/2,c/10,-c/10); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(subs(m=1,c=1,%)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "subs(u0=0.1,c=1,V=0.5,m =1,E_Ap+E_Bp);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Eboost(c/ 4,c/2,c/10,-c/10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf (subs(m=1,c=1,%));" }}}{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);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "It appears as if the total energy is minimized in a reference frame that moves with half the speed (" } {TEXT 364 1 "W" }{TEXT -1 1 "=" }{TEXT 363 1 "V" }{TEXT -1 352 "/2). T his would be the result from the nonrelativistic kinetic energy expres sion (+ rest energy for two massive particles), where the sum of the s quared velocities is minimized when the difference is split in half. A closer inspection performed with calculus (performed below) shows tha t the minimum is not exactly at the half-point. For larger boosts " } {TEXT 365 1 "V" }{TEXT -1 76 " the curve deviates more dramatically fr om the nonrelativistic expectation. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "solve(diff(Eboost(W,c/2,c/10,-c/10),W)=0,W);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 "In the nonrelativitic approximation we would cal culate: rest energy + nonrelativistic kinetic energy. We add the state ments to verify that the total energy 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)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "E_nrpp:=eval f(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)));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "The increase in th e total energy as the boost velocity approaches c or -c is obviously a ssociated with the fact that the two particles in A and B acquire mass . This also explains the asymmetry in the curve shown above. For negat ive " }{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 t he set-up of mass A being stationary, and mass B moving with c/2. For \+ " }{TEXT 369 1 "W" }{TEXT -1 172 ">c/2 a substantial mass increase for particle B occurs only for large boosts. In the opposite direction th e converse is true, but note that nothing too dramatic happens at " } {TEXT 370 1 "W" }{TEXT -1 186 "=-c/2, where one might argue naively th at the relativistic mass for particle B should hit the roof. The relat ive speed between C and B reaches the speed of light in this regime on ly 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 (procedure " }{TEXT 19 6 "Eboost" }{TEXT -1 102 ") by carrying out a Lorentz transformation o f 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 the momentum along the " }{TEXT 353 1 "x" }{TEXT -1 209 "-axis in frame A to carry out the b oost. This momentum was calculated before when the momentum balance wa s formulated, but we need the explicit expression for the mass as a fu nction of speed for the particles." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "u;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "px_to tA:=m*0/sqrt(1-u0^2/c^2)+subs(beta=V/c,m*V/sqrt(1-u^2/c^2));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "gammapp:=1/sqrt(1-W^2/c^2); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "EboostLT:=unapply(simpl ify(gammapp*(E_totA-W*(px_totA))),W,V,u0);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "evalf(EboostLT(c/3,c/2,c/10));" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 34 "evalf(Eboost(c/3,c/2,c/10,-c/10));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "expand(simplify(EboostLT(c/3 ,c/2,c/10)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "simplify(E boost(c/3,c/2,c/10,-c/10));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 927 "T he comparison verifies that the two calculations are consistent: in Eb oostLT we have Lorentz transformed the total energy for particles A an d B as calculated in frame A to the reference frame C. In Eboost we ha ve used Lorentz-transformed velocity components for particle B in A's \+ restframe, and transformed the velocities to frame C to directly evalu ate the relativistic mass increase as a function of the speed of parti cles 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 ve locity of particle B as viewed by an observer in C by first transformi ng it into A (known relative velocity V), and then from A to C (known \+ relative velocity W). The simple elastic billiard ball scattering exam ple was not interesting from the energy conservation point of view, ap art from the fact that energy conservation was ensured by the quadrati c dependence on " }{TEXT 366 1 "u" }{TEXT -1 2 "0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "More interesting cases in volve 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) Billiard balls with unequal \+ masses - elastic collisions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 176 "A \+ non-trivial generalization would be to consider the case of different \+ masses. Suppose particle B has twice the rest mass of A. Then we expec t 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 the dependence of relati vistic energy on speed and rest mass of a particle. We should be able \+ to formulate what happens. Let us simplify matters by going to one dim ension only, and formulating momentum and energy 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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "E_A:=m_A*c^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "v_B:=V; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "unprotect(gamma); gamma :=1/sqrt(1-V^2/c^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "E_B :=gamma*m_B*c^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "p_A:=m_ A*v_A;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "p_B:=gamma*m_B*v_ B;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Let us denote the velociti es 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);" }}} {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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "MC:=p_A+p_B=p_Ap+p_Bp;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 22 "EC:=E_A+E_B=E_Ap+E_Bp;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 32 "sol:=solve(\{MC,EC\},\{v_Ap,v_Bp\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "So much for Maple magic..." }}{PARA 0 "" 0 "" {TEXT -1 106 "Nevertheless, we should not despair. When we substi tute 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\});" }}}{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\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Note that there are always two solutions: one represents the initial condition, the other the final outcome." }}{PARA 0 "" 0 " " {TEXT -1 63 "We investigate various situations: a light particle wit h speed " }{TEXT 372 1 "V" }{TEXT -1 20 " hits a heavier one:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "sol:=solve(\{subs(m_A=10,m_B =1,MC),subs(m_A=10,m_B=1,EC)\},\{v_Ap,v_Bp\});" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 86 "vA:=subs(sol[1],v_Ap): # we may need to pick b y hand 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]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "limit(vA,V=c),limi t(vB,V=c);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 289 "For small speeds w e observe the known result from nonrelativistic energy and momentum co nservation: the light impinging particle bounces back with appreciable speed and the hit particle runs off with a smaller speed. Note, howev er, that in the highly relativistic limit a crossover occurs!" }} {PARA 0 "" 0 "" {TEXT -1 227 "The green line shows what would happen i f 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 reversed." }}}{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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fsolve(subs(c=1,vA+vB),V=0.9..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "What can we say about the energies (in t he rest frame of A)? Let us graph the kinetic energy of A and B after \+ the collision:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "plot([su bs(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=0..1,color=[red,blue],numpoints=1500,view=[0..1,0..5]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "As far as kinetic energy is conce rned the crossover point is earlier: the massive hit particle A picks \+ up more energy than the impinging light particle B retains in its rebo und at an impact velocity of " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "solve(subs(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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "ev alf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "These results are clea rly incompatible with our intuition based on the nonrelativistic limit ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 373 9 "Exer cise:" }}{PARA 0 "" 0 "" {TEXT -1 127 "Explore how these crossover poi nts move as the mass ratio is increased, i.e., as the hit particle is \+ chosen 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 solved less satisfactorily by Maple." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "sol:=solve(\{subs(m_A=1,m_ B=10,c=1,MC),subs(m_A=1,m_B=10,c=1,EC)\},\{v_Ap,v_Bp\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "### WARNING: allvalues now returns a list of symbolic values instead of a sequence of lists of numeric v alues\nsol1:=allvalues(sol)[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "The solution no longer recognizes the initial condition as a solu tion to the set of equations, even though they do satisfy the equation s. A call to " }{TEXT 19 14 "allvalues(sol)" }{TEXT -1 62 " reveals a \+ set of solutions, but no simplification is obvious." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "vA:=subs(sol1,v_Ap):" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 20 "vB:=subs(sol1,v_Bp):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "plot([vA,vB,V],V=0..1,color=[red,blue,green],view=[0. .1,0..1],numpoints=1500);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "simplify(limit(vA,V=1)),simplify(limit(vB,V=1));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "It appears 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 approache s c when the incident massive particle speed exceeds 0.8 c. For the ki netic energies this can imply a crossover where the hit particle acqui res 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,color=[red,blue],numpoints=1500,v iew=[0..1,0..15]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "fsolv e(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);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "We recogniz e from the comparison of the numerical results that the crossover in t he energies occurs for the same incident speed in the cases of corresp onding mass ratios 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 pr esent case of a heavy particle hitting a lighter one, which results in the heavy particle continuing its forward travel while the light part icle zooms off, means that at this point the light particle has acquir ed more kinetic energy than the heavy one. These results refer to the \+ rest frame of the particle 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 collisio ns in 1d: energy-mass conversion" }}{EXCHG {PARA 0 "" 0 "" {TEXT 374 19 "Inelastic collision" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 606 "We now proceed with an example of an inelastic coll ision. The inelasticity will play a different role than in nonrelativi stic mechanics. Inelasticities in billiard ball scattering are associa ted with 'stickiness', a part of the mechanical energy is given up to \+ heat. We say that it is dissipated and rely only on momentum conservat ion to calculate the dynamics of the one-dimensional motion. Complete \+ inelasticity means that the two masses are stuck together, and proceed with a common velocity. The energy balance can be used to calculate t he amount of mechanical energy that was dissipated in the system." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 510 "In the r elativistic concept we do not calculate the mechanical energy only. Th e rest energy is included automatically in the energy expression. We k now from chemical and nuclear reactions that the energy released from \+ these is coming from the mass-to-energy conversion. The final products have a smaller mass than the initial ingredients to the reaction. In \+ chemistry the mass differences which are converted into energy are sma ll (but measurable); in nuclear physics they are considerably larger p er 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 ob jects in a completely inelastic way, such that they stick together, th en the energy removed from the mechanical energy balance is not necess arily dissipated, but rather stored in the bound object. We can think \+ of having compressed a spring to form the bond, and that at some point later (perhaps only induced by some reaction), the composite system c an decay, and release the stored energy of the compressed spring. Acco rding to the energy-mass equivalence the rest mass of the composite ob ject should account for the stored energy. We can use the energy conse rvation law to calculate this change in mass. Consistency of these ide as with special relativity will be demonstrated by performing the calc ulation 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 ident ical particles of 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 collision they stick together at remain at rest in this fra me (it is also called the zero-momentum frame as the total momentum ve ctor vanishes at all times in this frame). If one looks at a worldline diiagram in this frame it looks like a Mercedes star (or a deformatio n thereof): for negative times the particles come in on worldlines wit h 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 vert ical worldline along the (c" }{TEXT 381 1 "t" }{TEXT -1 43 ")-axis for the newly formed object of mass " }{TEXT 382 1 "M" }{TEXT -1 28 ". Th e mass M is not 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 "Laborato ry frame" }}{PARA 0 "" 0 "" {TEXT -1 55 "In the laboratory frame one o f the two original masses " }{TEXT 385 1 "m" }{TEXT -1 74 " sits at re st (in the origin), 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 nonrelativistic limit, but we anticipate a complicati on, since we cannot simply add velocities naively. The laboratory fram e S' moves with respect 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 r est (it represents the target in the laboratory, while the other mass \+ is a particle in 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 deter mining the velocity of the impinging mass in S', i.e., we cannot simpl y say that it is 2 " }{TEXT 391 1 "u" }{TEXT -1 150 "0. We will determ ine it from consistency with energy conservation or from the Lorentz t ransformation. 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 c onserves momentum, 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 propels " }{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 calc ulation? Apparently 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. Theref ore, " }{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 collision. The worldlines in S have to look as follows: the targe t 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 " whic h moves off for positive c" }{TEXT 413 1 "t" }{TEXT -1 20 "' with a sl ope 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 the collision in the lab frame all momentum resides in the be am particle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "restart; as sume(u0-c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pb_p:= m/sqrt(1-u_p^2/c^2)*u_p;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Eb_p:=m*c^2+m/sqrt(1-u_p^2/c^2)*c^2;" }}}{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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Ea_p:=M/sqrt(1-u0^2/c^2)*c^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol:=solve(\{Eb_p=Ea_p,pb_p=pa_p\},\{M,u_ p\});" }}}{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:=simplify(subs(sol,M));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "Thi s is understood better by cancelling a factor of c in numerator and de nominator." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "u_sp:=simplif y(subs(sol,u_p));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "taylor (u_sp,u0=0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "We recover the e xpected nonrelativistic 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 let us perform the calculation in S, the center-of-ma ss, or zero-momentum frame: " }}{PARA 0 "" 0 "" {TEXT -1 136 "The mome ntum is zero before and after the collision, which we do not formulate as an equation, but 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;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "Ea:=M*c^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "M:=simplify(solve(Ea=Eb,M));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "We can verify that the energy expression s obtained in either frame are compatible with each other. Let us star t in the CM frame and transform the energy to the lab frame: we carry \+ out the calculation for the energy after the collision, " }{TEXT 19 2 "Ea" }{TEXT -1 53 ", first, and later perform the other calculation fo r " }{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 (the 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)*(E a+u0*0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "E_lab:=simplify (E_lab);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(Ea_p); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "Can we transform the other w ay too? 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));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "Ea;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "This was simple. Now transform the energy before the collision. Let us go \+ from the center of mass to the lab frame, as we know the velocities th ere." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 424 28 " Energy before the collision:" }}{PARA 0 "" 0 "" {TEXT -1 30 "The lab f rame is 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));" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 5 "Eb_p;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "simplify(subs(u_p=2*u0/(1+u0^2/c^2),Eb_p));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(E_lab);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 260 "We notice that the results agree when we insert the s peed of the beam particle (which we determined from momentum conservat ion). We could have used the Lorentz 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);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "For completeness one more check: the transformation from the lab frame to the CM frame. 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);" }}}{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));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "This is a formidable expr ession at first. Nevertheless, it simplifies to the desired result:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "E_cm:=simplify(E_cm);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "simplify(Eb);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "We summarize the progress made in this se ction:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 202 "1) we understand now how to transform energy between inertial fra mes (such as lab vs cm frame) for systems of particles (demonstrated m ostly in one dimension, but the generalization is straightforward);" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "2) we c onsidered elastic and (completely) inelastic billiard ball scattering. For completely inelastic collisions in the center of mass frame we de monstrated how all kinetic energy was converted into an additional mas s for a new particle of mass " }{TEXT 425 1 "M" }{TEXT -1 706 " in thi s zero-momentum frame. We can think of the inelasticity being effected by a virtual spring that was loaded and locked during the collision p rocess; the converted energy that resulted in a mass increase can beco me available when the two particles break apart One concrete example o f such a process would be the alpha-decay, or fissioning of nuclei, wh ere after a tunneling process, or by collisional fragmentation the two daughter nuclei are pushed apart from each other by the Coulomb repul sion. The daughter products appear asymptotically will well-defined ki netic energies. As long as the spring is loaded (the system sticks tog ether) the stored energy is noticeable via the increased mass (over 2 " }{TEXT 426 1 "m" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 333 "U sually bound systems in physics represent situations where the composi te system has a lower energy (mass) than the sum of the masses of the \+ constituents. This situation makes the composite system stable against break-up. The gain in energy by forming the composite system is the d riving process for the compound system to be formed." }}{PARA 0 "" 0 " " {TEXT -1 104 "This implies that the idea of a rest mass for a compos ite 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 part icles" }}{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 that the components of the momentum vector and energy tran sformed under LTs in analogy to the transformation of position and tim e. For the LT of 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 quant ity to be conserved is that it forms the inner product of a vector wit h itself in the four-dimensional spacetime. The LTs represent rotation s 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 invariant under the transformation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "It is of interest to consider \+ the invariant 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 "gamma:=1/sqrt(1-u^2/c^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "px:=m*gamma*ux;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " py:=m*gamma*uy: pz:=m*gamma*uz:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "E:=gamma*m*c^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 " E^2-c^2*(px^2+py^2+pz^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "algsubs(ux^2+uy^2+uz^2=u^2,%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 " The invariant magnitude (under LT) of the fourvector 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 ca n state that any inertial-frame observer will measure the same value f or the rest energy, and therefore the mass of an isolated system (sinc e c is universal). There is a special reference frame where the partic le is at rest, namely the frame specified by " }{TEXT 431 1 "p" } {TEXT -1 58 "=0. In this frame the total energy equals 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 partic les 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 "dynamic 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 relation allows to \+ deduce the rest mass from an energy and momentum measurement. This is \+ important, as one does not have to be in the rest frame of a particle \+ to determine its rest mass - which is often encountered in particle ph ysics." }}{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 determine the invariant rest energy. We will now show that this invariant rest ener gy is " }{TEXT 434 3 "not" }{TEXT -1 58 " the sum of the rest energies of the individual particles." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 175 "A simple example of two freely moving pa rticles suffices to make the point. We return to a set-up discussed 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 o ther (CM frame). We can consider the invariant mass calculation also i n 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);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "The total momentum in the CM frame is zer o (it is also called the zero-momentum frame). " }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 9 "P_tot:=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_tot:=2*E;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "T he invariant mass of the system of two free particles is now calculate d to be:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "M:=simplify(sqr t((E_tot/c^2)^2-(P_tot/c)^2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 638 "We notice that the rest mass of the system is greater than twice \+ the rest mass of the particles! It does not make sense to associate th e excess in the invariant rest mass of the system over twice the rest \+ masses of the constituents with the motion of individual particles (ev en though the origin of this excess does lie in the non-vanishing mome ntum vectors). To illustrate this, we calculate the same invariant mas s by transforming to another reference frame, and calculating the inva riant mass of the entire system there (to check whether it really is a property of the system, i.e., independent of the reference frame of t he 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;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 22 "simplify(eval(E_tot));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "M_cm:=simplify(eval(M));" }}}{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));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "We could repeat the calculation of the invariant mass fro m 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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "P_tot:=simplify(m*0-m/sqrt(1-u_p^2/c^2)*u_p);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "M_lab:=simplify(sqrt((E_tot/ c^2)^2-(P_tot/c)^2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "M_c m;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "It is impressive that Mapl e 5 does recognize these two expressions to be the same once the restr icition -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 " at the beginning of this section). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 198 "It i s important to recognize that the same invariant mass was calculated f or the system in both frames. This is remarkable as the relativistic m ass 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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "R M_lab:=simplify(m+m/sqrt(1-u_p^2/c^2));" }}}{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]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "Clearly the invariant mass is not equal to the sum of th e two relativistic masses of the particles! It has to be calculated ac cording to the prescription for the dynamic state 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 sys tem by carrying out the Lorentz transformation of the energy from the \+ CM to the Lab frame, and using the dynamic state of the system. Then i llustrate the value of the invariant mass by considering the following values: " }{TEXT 446 1 "m" }{TEXT -1 9 " = 4 kg, " }{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 dynam ic state in the limit of zero rest mass. In principle, there could als o be particles with negative rest mass (tachyons), but so far they hav e eluded observation if they indeed exist. For particles without a res t mass we find from the dynamic state the following relationship betwe en energy and momentum:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r estart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "E:=subs(m=0,sqrt (m^2*c^4+p^2*c^2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "We can de termine the speed of the particle 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;" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "E_m:=m/sqrt(1-u^2/c^2)*c^2 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "p_m*c^2/E_m;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "This relationship is valid irresp ective 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);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Therefore \+ a massless particle 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 photon's momentum? We should get " }{TEXT 450 2 "p " } {TEXT -1 2 "= " }{TEXT 449 1 "E" }{TEXT -1 61 "/c according to our der ivation above. Insertion of the speed " }{TEXT 452 1 "u" }{TEXT -1 14 " = c and mass " }{TEXT 451 1 "m" }{TEXT -1 111 " = 0 into the momentu m expression results in an undefined answer (0/0). The appropriate ste p is to replace the " }{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 definitely massless are photons. Neutrinos are candidates for m assless particles, although there is some evidence for a lower bound o n the mass. Gravitons, the postulated qunatum carriers of gravity also should be massless (for photons and gravitons the masslessness is con nected with the inverse squared force laws that appear in these theori es)." }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "10 21 \+ 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }