Kinetic Energy: K = ( - 1)mc2 Rest Energy (includes internal kinetic and potential energy): ER  mc2 For an object moving in an inertial frame), Total energy : E = K + ER = mc2 Problem 1: A mosquito has a mass of 2.5 mg. a) What is its rest energy? b) By what fraction does it change its energy if it is flying at 20 m/s ( 40 miles/hour) m = 2.5 x 10-6 kg ER = (2.5 x 10-6 kg) (3 x 108 m/s)2 = 2.25 x 1011 J [This is the energy output of Big Sandy power plant in 3.5 minutes  0.0024 days. Therefore, if you could completely convert the mass of  400 mosquitoes/day to energy (we can’t !!), you could match the energy output of Big Sandy!] b) Since v <<c, so can use K  ½ mv2 = 0.5 (2.5 x 10-6 kg) (20 m/s)2 = 5 x 10-4 J So K/ER = 5 x 10-4 / 2.25 x 1011 = 2.2 x 10-15 For everyday objects at ordinary speeds, relativistic corrections to energy are negligible – classical mechanics works fine.

E = mc2 In any process, the total energy is conserved! Kinetic Energy: K = ( - 1)mc2 Rest Energy (includes internal kinetic and potential energy): ER  mc2 For an object moving in an inertial frame, Total energy : E = K + ER E = mc2 In any process, the total energy is conserved! And: if no external forces, momentum is also conserved!

Consider a general reaction in which some species (Aj) react to form new species (Bj): A1 + A2 + A3 + ….  B1 + B2 + B3 + … Conservation of Energy: c2 A’s (m) = c2 B’s (m) Conservation of Momentum: A’s (mv) = B’s (mv) [Today: Because masses will be given in atomic mass units (u), we’ll use v’s, not u’s, for particle velocities.] (1u) c2 = 931 MeV

Consider the fusion reaction: 2D1 +3T1  4He2 + n, where 2D1 is a deuterium atom, with m(D) = 2.0141 u, 3T1 is a tritium atom, with m(T) = 3.0161 u, 4He2 is a helium atom, with m(He) = 4.0026 u and n is a neutron, with m(n) = 1.0087 u. The incoming mass = m(D) + m(T) = 5.0302 u. The outgoing mass = m(He) + m(n) = 5.0113 u. Therefore, exothermic with m c2= (0.0189 u) c2 Use (1 u)c2 = 931 MeV  m c2 = 17.6 MeV ------------------------------------------------------------------------------- Problem 2: Of this energy, K(He) = 3.5 MeV and K(n) = 14.1 MeV. Show that these values are consistent with conservation of momentum when the incoming deuterium and tritium have v << c (i.e. p(D)  p(T)  0).

m(He) = 4.0026 u = 3726 MeV/c2 , m(n) = 1.0087 u = 939 MeV/c2, m c2= 17.6 MeV Of this energy, K(He) = 3.5 MeV and K(n) = 14.1 MeV. ---------------------------------------------------------------------------------------------- Show that these values are consistent with conservation of momentum: Because the incoming particles are  at rest, the total momentum  0, so the neutron and helium atom should have equal and opposite momenta. K(n) = 14.1 MeV  ((n) – 1) m(n)c2 = 14.1 MeV (n) -1 = 14.1 MeV / 939 MeV = 0.0150 (n) = 1.015  v(n) = 0.17c  p(n) =  m v = (1.015) (939 MeV/c2) (0.017c) =162 MeV/c K(He) = 3.5 MeV  ((He) -1) mHec2 = 3.5 MeV (He) -1 = 3.5 MeV / 3726 MeV = 9.4 x 10-4 (He) = 1.00094  v(He) = 0.043c  p(He) = mv = (1.00094) (3726 MeV/c2) (.043c) = 162 MeV/c  = [1-(v/c)2]-1/2

Problem 3: A particle of mass M0 at rest divides into two particles, with masses m1 = 1u and m2 = 4u. The 1u particle has speed v1 = 0.8c. What is the speed of the 4u particle? What is the value of M0? What is the total kinetic energy of the outgoing particles? pinitial = 0 so p2 = -p1 2m2v2 = 1m1v1 4v2/[1-(v2/c)2]1/2 = v1/[1-(v1/c)2]1/2 = 0.8c / [1-0.64]1/2 = 0.8c/0.6 v2 /[1-(v2/c)2]1/2 = c/3 (3v2/c)2 = 1-(v2/c)2 (v2/c)2 = 1/10 v2 = 0.316 c

Problem 3: A particle of mass M0 at rest divides into two particles, with masses m1 = 1u and m2 = 4u. The 1u particle has speed v1 = 0.8c. What is the speed of the 4u particle? What is the value of M0? What is the total kinetic energy of the outgoing particles? v2 = 0.316 c Since M0 is at rest, its energy = M0c2 M0c2 = 1m1c2 + 2m2c2 M0 = 1u [1/(1-0.82)1/2 + 4/(1 – 0.3162)1/2 M0 = 1u [1/0.361/2 + 4/0.91/2] = (1.67 + 4.22)u = 5.89 u c) Method 1: Ktotal = K1 + K2 = (1-1) m1c2 + (2-1)m2c2 = (0.67 + 4 (1-1/0.91/2) uc2 = (0.67 + 0.22) x 931 MeV = 829 MeV Method 2: Since M0 is at rest, Ktotal = (M0-m1-m2) c2 = (5.89 -1 -4)uc2 = 0.89 x 931 MeV = 829 MeV

Problem 4: A particle of mass M0 = 15 u at rest decays into three particles. The first particle has mass m1 = 1 u and velocity v1 = 0.8 c i. The second particle has mass m2 = 3 u and velocity v2 = - v1. What are the mass and velocity of the third particle? 1 = 2 = 1/[1 –v12/c2]1/2 = 1/[1 – 0.82]1/2 = 1/ [1 – 0.64]1/2 = 1/0.361/2 = 1/0.6 = 5/3 Consvtn. of Energy: M0c2 = 1m1c2 + 2m2c2 + 3m3c2 3m3 = [15 u – 5/3 (1 u + 3 u] = (25/3) u Consvtn. of momentum: initial momentum = final momentum 0 = 1m1 v1 + 2m2 v2 + 3m3 v3 0 = 5/3 (1 u) (0.8 c)i – 5/3 (3 u) (0.8c) i + (25/3 u) v3 v3 = (3/25) (5/3) (1-3) (0.8) i = + 0.32 c i (= 9.6 x 107 m/s in +x direction) Therefore 3 = 1/[1-(v3/c)2]1/2 = 1/[1-0.322]1/2 = 1.056 Therefore m3 = 25/3 u / 1.056 = 7.90 u (This doesn’t correspond to a real particle.)

Problem 5: An object with mass m1 = 900 kg and traveling at speed v1 = 0.85c collides with a stationary object with mass m2 = 1400 kg and the two objects stick together. What are the speed and mass of the composite object (#3)? Energy conservation: 1m1c2 + m2c2 = 3m3c2 Momentum conservation: 1m1v1 = 3m3v3 Energy: 1 = 1/(1-0.852)1/2 = 1.90  3m3 = (1.90 x 900 +1400) kg = 3110 kg Momentum: 3m3v3 = 1.90 x 900 kg * 0.85c = 1454 kgc  v3 = (1454/3110) c = 0.467 c  3 = 1/(1-0.4672)1/2 = 1.13 m3 = 3110 kg/1.13 = 2750 kg

Conservation of Energy: (1 + P)MPc2 = XMXc2 Problem 6: A 6.6 GeV (= 6600 MeV) proton (MP = 1.007 u) collides with a proton at rest to create a new particle, X. What are the mass and energy of the new particle? Conservation of Energy: (1 + P)MPc2 = XMXc2 Conservation of Momentum: PMP vP = XMXvX The rest energy of the proton ERP = 1.007 x 931 MeV = 938 MeV. Therefore the moving proton has P = 6600/938 = 7.04 . Since 7.04 = P = 1/(1-vP2/c2)1/2, the moving proton has speed vP/c = 0.990. Conservation of Energy: XMX = 8.04MP = 8.10 u  EX = 8.10 x 931 MeV = 7540 MeV = 7.54 GeV Conservation of Momentum: vX = 7.04 MP (0.990 c) / 8.04 MP = 0.867 c Then X = 1/(1-0.8672)1/2 = 2.01 MX = 8.10 u / 2.01 = 4.0 u stationary proton moving proton

For a particle with velocity v, p = mv and E = mc2 E2 – p2c2 = (mc)2 (c2-v2)/ [1 – v2/c2] = (mc2)2 E2 = p2c2 + (mc2)2 This equation is true for all particles, even those with mass = 0 (e.g. photons, gravitons, gluons). For a particle with m=0: ER = mc2 = 0, so all the energy is kinetic. K = mc2  0   =   v = c (Massless particles must travel at speed c.) Then p = (mv) = E/c. For example, a photon of frequency f and wavelength  = c/f has energy E = hf. Therefore it has momentum p = hf/c = h/.

Problem 7: A pion (mc2 = 140 MeV) at rest decays to a muon (mc2 = 106 MeV) and antineutrino (m  0). Find the energy of the antinuetrino. Since the antineutrino’s mass  0, essentially all of its energy is kinetic: E = pc. Conservation of energy: mc2 = mc2 + pc Conservation of momentum: mv = p mc2 = mc2 (1 + v/c) m/ m = (1 + v/c)/ [1 – (v/c)2]1/2 But [1 – (v/c)2] = (1 – vu/c) (1 + vu/c)  m/ m = [(1 + v/c)/ (1 – v/c)]1/2 140/106 = 1.32 = [(1+vu/c)/ (1 – vu/c) ]1/2 1.74 = [(1+vu/c)/ (1 – vu/c) ] 1.74 – 1.74 vu/c = 1 + vu/c 2.74 vu/c = 0.74 vu/c = 0.27  = 1.039 E = pc = mc2 - mc2 = [140 – (1.039)106] MeV E = 29.9 MeV

Recommended reading: Feynman’s Lectures on Physics, Volume I, Chapters 15-17. Available (for free !) online at http://www.feynmanlectures.caltech.edu/ (or hardcopy for ~ $40). Caution: Many older texts (e.g. Feynman) use different names and notation: What we call mass, m, is often called the “rest mass” and denoted m0. What we denote as m0 is often called the relativistic mass, m. Quantity Us (Serway/Jewett) Feynman m = mass m0 = rest mass m m = (relativistic) mass Momentum (p) mu m0u = mu kinetic energy (K) (-1) mc2 (-1) m0c2 = (m-m0)c2 rest energy (ER) mc2 m0c2 total energy (E) mc2 m0c2 = mc2