Kinetic Energy: K = (γ - 1)mc 2 Rest Energy (includes internal kinetic and potential energy): E R mc 2

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1 Kinetic Energy: K = (γ - 1)mc 2 Rest Energy (includes internal kinetic and potential energy): E R mc 2 For an object moving in an inertial frame), Total energy : E = K + E R = γmc 2 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) a) m = 2.5 x 10-6 kg E R = (2.5 x 10-6 kg) (3 x 10 8 m/s) 2 = 2.25 x J [This is the energy output of Big Sandy power plant in 3.5 minutes 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 mv 2 = 0.5 (2.5 x 10-6 kg) (20 m/s) 2 = 5 x 10-4 J So K/E R = 5 x 10-4 / 2.25 x = 2.2 x For everyday objects at ordinary speeds, relativistic corrections to energy are negligible classical mechanics works fine.

2 Kinetic Energy: K = (γ - 1)mc 2 Rest Energy (includes internal kinetic and potential energy): E R mc 2 For an object moving in an inertial frame, Total energy : E = K + E R E = γmc 2 In any process, the total energy is conserved! And: if no external forces, momentum is also conserved!

3 Consider a general reaction in which some species (A j ) react to form new species (B j ): A 1 + A 2 + A 3 +. B 1 + B 2 + B 3 + Conservation of Energy: c 2 A s (γm) = c 2 B s (γm) Conservation of Momentum: A s (γmv) = B s (γmv) [Today: Because masses will be given in atomic mass units (u), we ll use v s, not u s, for particle velocities.] (1u) c 2 = 931 MeV

4 Consider the fusion reaction: 2 D T 1 4 He 2 + n, where 2 D 1 is a deuterium atom, with m(d) = u, 3 T 1 is a tritium atom, with m(t) = u, 4 He 2 is a helium atom, with m(he) = u and n is a neutron, with m(n) = u. The incoming mass = m(d) + m(t) = u. The outgoing mass = m(he) + m(n) = u. Therefore, exothermic with m c 2 = ( u) c 2 Use (1 u)c 2 = 931 MeV m c 2 = 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).

5 m(he) = u = 3726 MeV/c 2, m(n) = u = 939 MeV/c 2, m c 2 = 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 (γ(n) 1) m(n)c 2 = 14.1 MeV γ(n) -1 = 14.1 MeV / 939 MeV = γ= [1-(v/c) γ(n) = ] -1/2 v(n) = 0.17c p(n) = γ m v = (1.015) (939 MeV/c 2 ) (0.017c) =162 MeV/c K(He) = 3.5 MeV (γ(he) -1) m He c 2 = 3.5 MeV γ(he) -1 = 3.5 MeV / 3726 MeV = 9.4 x 10-4 γ(he) = v(he) = 0.043c p(he) = γmv = ( ) (3726 MeV/c 2 ) (.043c) = 162 MeV/c

6 Problem 3: A particle of mass M 0 at rest divides into two particles, with masses m 1 = 1u and m 2 = 4u. The 1u particle has speed v 1 = 0.8c. a) What is the speed of the 4u particle? b) What is the value of M 0? c) What is the total kinetic energy of the outgoing particles? a) p initial = 0 so p 2 = -p 1 γ 2 m 2 v 2 = γ 1 m 1 v 1 4v 2 /[1-(v 2 /c) 2 ] 1/2 = v 1 /[1-(v 1 /c) 2 ] 1/2 = 0.8c / [1-0.64] 1/2 = 0.8c/0.6 v 2 /[1-(v 2 /c) 2 ] 1/2 = c/3 (3v 2 /c) 2 = 1-(v 2 /c) 2 (v 2 /c) 2 = 1/10 v 2 = c

7 Problem 3: A particle of mass M 0 at rest divides into two particles, with masses m 1 = 1u and m 2 = 4u. The 1u particle has speed v 1 = 0.8c. a) What is the speed of the 4u particle? b) What is the value of M 0? c) What is the total kinetic energy of the outgoing particles? a) v 2 = c b) Since M 0 is at rest, its energy = M 0 c 2 M 0 c 2 = γ 1 m 1 c 2 + γ 2 m 2 c 2 M 0 = 1u [1/( ) 1/2 + 4/( ) 1/2 M 0 = 1u [1/0.36 1/2 + 4/0.9 1/2 ] = ( )u = 5.89 u c) Method 1: K total = K 1 + K 2 = (γ 1-1) m 1 c 2 + (γ 2-1)m 2 c 2 = ( (1-1/0.9 1/2 ) uc 2 = ( ) x 931 MeV = 829 MeV Method 2: Since M 0 is at rest, K total = (M 0 -m 1 -m 2 ) c 2 = ( )uc 2 = 0.89 x 931 MeV = 829 MeV

8 Problem 4: A particle of mass M 0 = 15 u at rest decays into three particles. The first particle has mass m 1 = 1 u and velocity v 1 = 0.8 c i. The second particle has mass m 2 = 3 u and velocity v 2 = - v 1. What are the mass and velocity of the third particle? γ 1 = γ 2 = 1/[1 v 12 /c 2 ] 1/2 = 1/[ ] 1/2 = 1/ [1 0.64] 1/2 = 1/0.36 1/2 = 1/0.6 = 5/3 Consvtn. of Energy: M 0 c 2 = γ 1 m 1 c 2 + γ 2 m 2 c 2 + γ 3 m 3 c 2 γ 3 m 3 = [15 u 5/3 (1 u + 3 u] = (25/3) u Consvtn. of momentum: initial momentum = final momentum 0 = γ 1 m 1 v 1 + γ 2 m 2 v 2 + γ 3 m 3 v 3 0 = 5/3 (1 u) (0.8 c)i 5/3 (3 u) (0.8c) i + (25/3 u) v 3 v 3 = (3/25) (5/3) (1-3) (0.8) i = c i (= 9.6 x 10 7 m/s in +x direction) Therefore γ 3 = 1/[1-(v 3 /c) 2 ] 1/2 = 1/[ ] 1/2 = Therefore m 3 = 25/3 u / = 7.90 u (This doesn t correspond to a real particle.)

9 Problem 5: An object with mass m 1 = 900 kg and traveling at speed v 1 = 0.85c collides with a stationary object with mass m 2 = 1400 kg and the two objects stick together. What are the speed and mass of the composite object (#3)? Energy conservation: γ 1 m 1 c 2 + m 2 c 2 = γ 3 m 3 c 2 Momentum conservation: γ 1 m 1 v 1 = γ 3 m 3 v 3 Energy: γ 1 = 1/( ) 1/2 = 1.90 γ 3 m 3 = (1.90 x ) kg = 3110 kg Momentum: γ 3 m 3 v 3 = 1.90 x 900 kg * 0.85c = 1454 kg c v 3 = (1454/3110) c = c γ 3 = 1/( ) 1/2 = 1.13 m 3 = 3110 kg/1.13 = 2750 kg

10 Problem 6: A 6.6 GeV (= 6600 MeV) proton (M P = u) collides with a proton at rest to create a new particle, X. What are the mass and energy of the new particle? stationary proton moving proton Conservation of Energy: (1 + γ P )M P c 2 = γ X M X c 2 Conservation of Momentum: γ P M P v P = γ X M X v X The rest energy of the proton E RP = x 931 MeV = 938 MeV. Therefore the moving proton has γ P = 6600/938 = Since 7.04 = γ P = 1/(1-v P2 /c 2 ) 1/2, the moving proton has speed v P /c = Conservation of Energy: γ X M X = 8.04M P = 8.10 u E X = 8.10 x 931 MeV = 7540 MeV = 7.54 GeV Conservation of Momentum: v X = 7.04 M P (0.990 c) / 8.04 M P = c Then γ X = 1/( ) 1/2 = 2.01 M X = 8.10 u / 2.01 = 4.0 u

11 For a particle with velocity v, p = γmv and E = γmc 2 E 2 p 2 c 2 = (mc) 2 (c 2 -v 2 )/ [1 v 2 /c 2 ] = (mc 2 ) 2 E 2 = p 2 c 2 + (mc 2 ) 2 This equation is true for all particles, even those with mass = 0 (e.g. photons, gravitons, gluons). For a particle with m=0: E R = mc 2 = 0, so all the energy is kinetic. K = γmc 2 0 γ = v = c (Massless particles must travel at speed c.) Then p = (γmv) = E/c. For example, a photon of frequency f and wavelength λ = c/f has energy E = hf. Therefore it has momentum p = hf/c = h/λ.

12 Problem 7: A pion (m π c 2 = 140 MeV) at rest decays to a muon (m µ c 2 = 106 MeV) and antineutrino (m ν 0). Find the energy of the antinuetrino. Since the antineutrino s mass 0, essentially all of its energy is kinetic: E ν = p ν c. Conservation of energy: Conservation of momentum: m π c 2 = γ µ m µ c 2 + p ν c γ µ m µ v µ = p ν m π c 2 = γ µ m µ c 2 (1 + v µ /c) m π / m µ = (1 + v µ /c)/ [1 (v µ /c) 2 ] 1/2 But [1 (v µ /c) 2 ] = (1 v u /c) (1 + v u /c) m π / m µ = [(1 + v µ /c)/ (1 v µ /c)] 1/2 140/106 = 1.32 = [(1+v u /c)/ (1 v u /c) ] 1/ = [(1+v u /c)/ (1 v u /c) ] v u /c = 1 + v u /c 2.74 v u /c = 0.74 v u /c = 0.27 γ µ = E ν = p ν c = m π c 2 - γ µ m µ c 2 = [140 (1.039)106] MeV E ν = 29.9 MeV

13 Recommended reading: Feynman s Lectures on Physics, Volume I, Chapters Available (for free!) online at (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 m 0. What we denote as γm 0 is often called the relativistic mass, m. Quantity Us (Serway/Jewett) Feynman m = mass m 0 = rest mass γm m = (relativistic) mass Momentum (p) γmu γm 0 u = mu kinetic energy (K) (γ-1) mc 2 (γ-1) m 0 c 2 = (m-m 0 )c 2 rest energy (E R ) mc 2 m 0 c 2 total energy (E) γmc 2 γm 0 c 2 = mc 2

Physics 202. Professor P. Q. Hung. 311B, Physics Building. Physics 202 p. 1/2

Physics 202. Professor P. Q. Hung. 311B, Physics Building. Physics 202 p. 1/2 Physics 202 p. 1/2 Physics 202 Professor P. Q. Hung 311B, Physics Building Physics 202 p. 2/2 Momentum in Special Classically, the momentum is defined as p = m v = m r t. We also learned that momentum