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The relativistic kinetic energy of the particle is given by: ( ) \u239f \u239f \u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = 1 1 1 2 2 icrelativist cv mcK A spreadsheet program to graph Krelativistic and Kclassical is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form A8 A7+0.05 v/c + 0.05 B7 0.5*$B$3*A7^2 2 2 1 mv C7 $B$3*(1/((1\u2212A7^2)^0.5)\u22121) ( ) \u239f \u239f \u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 1 1 1 2 2 cv mc A B C 1 2 3 mc2= 100 MeV 4 5 v/c 2 2 1 mv ( ) 2 21 1 mc cv \u239f \u239f \u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 6 Kclassical Krelativistic 7 0.00 0.00 0.00 8 0.05 0.13 0.13 9 0.10 0.50 0.50 10 0.15 1.13 1.14 23 0.80 32.00 66.67 24 0.85 36.13 89.83 25 0.90 40.50 129.42 26 0.95 45.13 220.26 The solid curve is the graph of the relativistic kinetic energy. The relativistic formula, represented by the continuous curve, begins to deviate from the classical curve around v/c \u2248 0.4. Special Relativity 1075 0 50 100 150 200 250 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 v /c E (M eV ) Kclassical Krelativistic 45 \u2022\u2022 [SSM] (a) Show that the speed v of a particle of mass m and total energy E is given by ( ) 212 22 1 \u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u2212= E mc c v and that when E is much greater than mc2, this can be approximated by ( )2 2221 Emccv \u2212\u2248 . Find the speed of an electron with kinetic energy of (b) 0.510 MeV and (c) 10.0 MeV. Picture the Problem We can solve the equation for the relativistic energy of a particle to obtain the first result and then use the binomial expansion subject to E >> mc2 to obtain the second result. In parts (b) and (c) we can use the first expression obtained in (a), with E = E0 + K, to find the speeds of electrons with the given kinetic energies. See Table 39-1 for the rest energy of an electron. (a) The relativistic energy of a particle is given by Equation R-15: 2 2 2 1 c v mcE \u2212 = Solving for v/c yields: ( ) 21 2 22 1 \u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 \u2212= E mc c v (1) Expand the radical expression binomially to obtain: ( ) ( ) sorder term-higher 2 111 2 22 2 22 +\u2212=\u2212= E mc E mc c v Chapter R 1076 Because the higher-order terms are much smaller than the 2nd-degree term when E >> mc2: ( ) 2 22 2 1 E mc c v \u2212\u2248 (b) Solve equation (1) for v: ( ) 2 22 1 E mccv \u2212= Because E = E0 + K: ( ) 2 0 2 0 2 0 1 111 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + \u2212=+\u2212= E K c KE Ecv For an electron whose kinetic energy is 0.510 MeV: ( ) c cv 866.0 MeV0.511 MeV510.01 11MeV510.0 2 = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + \u2212= (c) For an electron whose kinetic energy is 10.0 MeV: ( ) c cv 999.0 MeV0.511 MeV0.101 11MeV0.10 2 = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + \u2212= 46 \u2022\u2022 Use the binomial expansion and Equation R-17 to show that when pc << mc2, the total energy is given approximately by E \u2248 mc2 + p2/(2m). Picture the Problem We can solve Equation R-17 for E and factor ( )22mc from under the resulting radical to obtain an expression to which we can apply the binomial expansion to write the radical expression as a power series. Finally, we can invoke the condition that pc << mc2 to complete the argument that the total energy is given approximately by ( )mpmcE 222 +\u2248 . From Equation R-17 we have: ( )2222 mccpE += Factor ( )22mc under the radical and simplify to obtain: ( ) ( ) ( ) 2 2 2 22 22 2 22 22 22 111 mc pmc mc cpmc mc cpmcE +=+=\u23a5\u23a5\u23a6 \u23a4 \u23a2\u23a2\u23a3 \u23a1 += Special Relativity 1077 Expand the radical factor binomially to obtain: sorder term-higher 2 111 22 221 22 2 ++=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + cm p cm p For pc << mc2: 22 221 22 2 2 111 cm p cm p +\u2248\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + Substituting for 21 22 2 1 \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b + cm p and simplifying yields: m pmc cm pmcE 22 11 2 2 22 2 2 +=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b += 47 \u2022\u2022 Derive the equation E2 = p2c2 + m2c4 (Equation R-17) by eliminating v from Equations R-10 and R-16. Picture the Problem We can solve Equation R-16 for v and substitute in Equation R-10 to eliminate v. Simplification of the resulting expression leads to ( ) .22222 mccpE += Express the relativistic momentum of a particle using Equation R-10: 221 cv mvp \u2212= From Equation R-16 we have: E pc c v = \u21d2 E pcv 2 = Substitute for v and simplify to obtain: ( ) 2 22 2 1 c Epc E pcm p \u2212 = or 2 22 2 1 1 E cpE mc \u2212 = Multiply both sides of the equation by 2 22 1 E cpE \u2212 : 2 2 22 1 mc E cpE =\u2212 Chapter R 1078 Square both sides of the equation: ( )222 222 1 mcEcpE =\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 Solve for E2 to obtain Equation R-17: ( )22222 mccpE += 48 \u2022\u2022 The rest energy of a proton is about 938 MeV. If its kinetic energy is also 938 MeV, find (a) its momentum and (b) its speed. Picture the Problem (a) We can solve the relation for total energy, momentum, and rest energy of the proton for its momentum and evaluate this expression for K = E0. In Part (b) we can start with the expression for the relativistic energy of the proton, solve for u, and evaluate the resulting expression for K = E0. (a) Relate the total energy, momentum, and rest energy of a proton: ( )22222 mccpE += Solving for p yields: ( ) c KEK c EKE c EE p 0 2 2 0 2 0 2 0 2 2+= \u2212+=\u2212= If K = E0: c E c EE p EK 0 2 0 2 0 32 0 =+== Substitute numerical values and evaluate p: ( ) c c p EK GeV/ 62.1 MeV 9383 0 = == (b) The relativistic energy of a particle is given by Equation R-15: ( )2 2 1 cv mcE \u2212 = \u21d2 2 2 01 E Ecv \u2212= Because E = E0 + K: ( )20 2 01 KE Ecv +\u2212= Special Relativity 1079 For K = E0: ( ) cc c EE Ecv EK 866.0750.0 4 111 2 00 2 0 0 == \u2212=+\u2212== 49 \u2022\u2022 What percentage error is made in using 221 mv for the kinetic energy of a particle if its speed is (a) 0.10c and (b) 0.90c? Picture the Problem We can use the expressions for the classical and relativistic kinetic energies of a particle to obtain a general expression for the fractional error made in using 221 mv for the kinetic energy of a particle as a function of its speed. The percentage error made in using 2 2 1 mv for the kinetic energy of a particle is given by: rel classical rel classicalrel rel 1 K K K KK K K \u2212=\u2212=\u394 The relativistic kinetic energy of the particle is given by Equation R-14: ( ) ( ) 2 2 2 2 2 rel 1 1 1 1 mc cv mc cv mcK \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 = \u2212 \u2212 = Substitute for Krel in the expression for relK K\u394 and simplify to obtain: ( ) ( ) 2 2 2 2 2 2 2 1 rel 1 1 12 1 1 1 1 1 c cv v mc cv mv K K \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 \u2212= \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 \u2212=\u394 (a) Evaluate f for v = 0.10c: ( ) ( ) ( ) %75.0 1 10.01 12 10.0110.0 2 2 2 = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 \u2212= c ccf Chapter R 1080 (b) Evaluate f for v = 0.90c: ( ) ( ) ( ) %69.0 1 90.01 12 90.0190.0 2 2 2 = \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212 \u2212= c ccf General Problems 50 \u2022 A spaceship departs from Earth for the star Alpha Centauri, which is 4.0 c\u22c5y away in the reference frame of Earth. The spaceship travels at 0.75c. How long does it take to get there (a) as measured on Earth and (b) as measured by a passenger on the spaceship? Picture the Problem We can find the duration of the trip, as measured on earth, using the definition of average speed; that is, \u394t = L/v. The elapsed time measured by the passenger is the proper time and is related to \u394t through the time dilation equation (Equation R-3). (a) Relate the time \u394t for the trip as measured on earth to its length L and the speed u of the spaceship: v Lt =\u394 Substitute numerical values and evaluate \u394t: y3.5y33.575.0 y0.4 \u394 ==\u22c5= c ct (b) Use Equation R-3, the time dilation equation, to express the proper time measured by a passenger: 2 0 1\u394\u394 \u239f\u23a0 \u239e\u239c\u239d \u239b\u2212= c vtt Substitute numerical values and evaluate \u394t0: ( ) ( ) y5.375.01y33.5\u394 2 2 p =\u2212= c ct 51 \u2022 The total energy of a particle is three