1. Relativistic Momentum p  mu/[(1 – (u/c) 2 ] 1/2 =  mu   1     

2. In Newtonian mechanics, the work done on an object = change in its kinetic energy (K). Let’s keep that as the definition of kinetic energy in relativity. dW = F  dx = dK  K =  dp/dt  dx For simplicity, let’s stick to one-dimension:  K =  dp/dt  dx = m  (d/dt {u / [ 1 – (u/c) 2 ] 1/2 } dx  K = m  {1/ [ 1 – (u/c) 2 ] 1/2 + (u/c) 2 / [ 1 – (u/c) 2 ] 3/2 } (du/dt) dx  K = m  (du/dt) dx / [ 1 – (u/c) 2 ] 3/2 Now, setting dx = u dt  (du/dt) dx = (du/dt) u dt = u du  K = m  u du / [ 1 – (u/c) 2 ] 3/2 = m/2  d(u 2 ) / [ 1 – (u/c) 2 ] 3/2 = {mc 2 /[1-(u/c) 2 ] 1/2 } Let’s assume that the particle starts from rest (u initial = 0) and its final velocity = u, so that K initial = 0 and K final =  K = K K = mc 2 {1/ [ 1 – (u/c) 2 ] 1/2 – 1} K = (  - 1)mc 2 u initial u final

3. K = mc 2 {1/ [ 1 – (u/c) 2 ] 1/2 – 1} K = (  - 1)mc 2 Check that this gives the expected Newtonian result when u << c: [1 – (u/c) 2 ] -1/2  1 + ½ (u/c) 2  K  [1 + ½ (u/c) 2 -1]mc 2 K  ½ mu 2  classical limit K   as u  c

4. K = mc 2 {1/ [ 1 – (u/c) 2 ] 1/2 – 1} K = (  - 1)mc 2 Problem: Consider a collision between an electron and positron in a collider. In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. What are the kinetic energies of the electron and positron in the lab frame and in the electron’s frame? (m e = m p = 9.11 x 10 -31 kg)

5. K = mc 2 {1/ [ 1 – (u/c) 2 ] 1/2 – 1} K = (  - 1)mc 2 Problem: Consider a collision between an electron and positron in a collider. In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. What are the kinetic energies of the electron and positron in the lab frame and in the electron’s frame? (m e = m p = 9.11 x 10 -31 kg) Lab frame: Each particle has  = 10.01, so each has kinetic energy K = (9.01) (9.11 x 10 -31 kg) (3 x 10 8 m/s) 2 K = 7.39 x 10 -13 J Electron’s frame:  e = 1, so K e = 0.  p = 196, so K p = (195) (9.11 x 10 -31 kg) (3 x 10 8 m/s) 2 K p = 1.60 x 10 -11 J [Note: K p (electron’s frame)/K p (lab frame)  22 ! (In classical physics, it would only increase x 4.)]

6. Lab frame: Each particle has K = 7.39 x 10 -13 J Electron’s frame: K p = 1.60 x 10 -11 J For subatomic particles, it is customary to express energies in eV (electron volts). [or keV, MeV, GeV, …] 1 eV = 1.6 x 10 -19 J (= kinetic energy of an electron after being accelerated through 1 volt) So Lab frame: K = (7.39 x 10 -13 J) (1 eV/1.6 x 10 -19 J) = 4.6 x 10 6 eV = 4.6 MeV Electron’s frame: K p = (1.60 x 10 -11 J) (1 eV/1.6 x 10 -19 J) =10 8 eV = 100 MeV = 0.10 GeV

7. p =  mu K = (  - 1)mc 2 Problem: The kinetic energy of the innermost electron (m e = 9.11 x 10 -31 kg) in a uranium atom is K  116 keV. What is its speed and momentum? K = (116 x 10 3 eV) (1.6 x 10 -19 J/eV) = 1.86 x 10 -14 J  = 1 + K/mc 2  = 1 + (1.86 x 10 -14 J)/[(9.11 x 10 -31 kg) (3 x 10 8 m/s) 2 ]  = 1.23

8. p =  mu K = (  - 1)mc 2 Problem: The kinetic energy of the innermost electron (m e = 9.11 x 10 -31 kg) in a uranium atom is K  116 keV. What is its speed and momentum? K = (116 x 10 3 eV) (1.6 x 10 -19 J/eV) = 1.86 x 10 -14 J   = 1 + (1.86 x 10 -14 J)/[(9.11 x 10 -31 kg) (3 x 10 8 m/s) 2 ]  = 1.23 1/ [1 – (u/c) 2 ] 1/2 = 1.23 1 – (u/c) 2 = 0.66 u/c = 0.58 p = (1.23) (9.11 x 10 -31 kg) (0.58 x 3 x 10 8 m/s) p = 1.95 x 10 -22 kg  m/s

9. K = (  - 1)mc 2 Einstein associated mc 2 with the energy the object has when its at rest: Rest Energy: E R  mc 2 That is, mass (inertia) is itself a form of energy. [This is sometimes called the “equivalence of mass and energy”.] For objects that contain other particles (e.g. protons, people, stars, …), E R (and therefore the object’s mass) includes contributions from the kinetic energies of these particles (k j ) and potential energies of interactions between these particles (v ij ), as well as the (constant) rest masses (m j ) of these particles: E R = mc 2 =  k j +  v ij +  m j c 2. [Notes: 1) E R does not include the potential energy of the object in an external force, such as gravity: that is the subject of General Relativity. 2) If the k j ’s and/or v ij ’s change, E R and m change --- i.e. the rest mass is not necessarily constant for an object with constituents.]

10. E R = mc 2 =  k j +  v ij +  m j c 2 For ordinary objects, the magnitude of the kinetic and interaction energies of the constituent particles is much less than the total mc 2 ; i.e. m   m j and is not noticeably changed by typical experiments. (i.e. m and E R are  constants). Example 1: The heat energy (  E) needed to change the temperature of 1g of water from 0 o C to 100 o C (  E goes into the kinetic and interaction energies of the water molecules) is  E = 419 J. Therefore  E/c 2 = 419 J / (3 x 10 8 m/s) 2 = 4.7 x 10 -15 kg = 4.7 x 10 -12 g. So the relative increase in mass due to heating up the water =  m/m = 4.7 x 10 -12.

11. Kinetic Energy: K = (  - 1)mc 2 Rest Energy (includes internal kinetic and potential energy): E R  mc 2 For an object moving in an inertial frame (i.e. no external force or potential energy), Total energy : E = K + E R E =  mc 2 In any process, the total energy is conserved!

12. E R = mc 2, E =  mc 2, K = (  -1) mc 2 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 + … Energy must be conserved: If the rest energy of the A’s is smaller than that of the B’s: m(A 1 ) + m(A 2 ) + m(A 3 ) + … < m(B 1 ) + m(B 2 ) + m(B 3 ) + … then the reaction is endothermic: extra energy must be applied to the A’s, e.g. through the incoming kinetic energy. However, if the rest energy of the A’s is larger than that of the B’s: m(A 1 ) + m(A 2 ) + m(A 3 ) + … > m(B 1 ) + m(B 2 ) + m(B 3 ) + … then the reaction is exothermic, and energy will be released, through kinetic energy of the B’s.

13. E R = mc 2, E =  mc 2, K = (  -1) mc 2 In classical physics, u<<c    1 + ½ mu 2, so K  ½ mu 2. Also, typical constituent kinetic and interaction energies << E R [E R = mc 2 =  k j +  v ij +  m j c 2 ], so changes in E R and total mass due to changes in these are extremely small and not measurable. [i.e. m is due overwhelmingly to the constituent rest masses so  constant.] Example 2: Consider the interaction of hydrogen and oxygen to make m = 1 kg of water. It is a very exothermic reaction that releases  E = 13.4 J. Therefore, the rest mass of the water is slightly less than that of the oxygen and hydrogen that formed it:  m = 13.4 J/(3 x 10 8 m/s) 2  m = 1.5 x 10 -16 kg ! For practical, every day experiences, including chemical reactions, we can assume that total mass is conserved, so including mc 2 in the energy just shifts the “zero” of the energy.

14. Example 3: It takes 13.5 eV to ionize a hydrogen atom: H  proton + electron. What is the relative difference in mass between a hydrogen atom and a separated proton and electron? (The mass of a hydrogen atom  1.67 x 10 -27 kg.) Because the ionization reaction is endothermic, the mass of the hydrogen atom must be less than the separated proton and electron.  m =  E/c 2 = [(13.5 eV) (1.6 x 10 -19 J)/eV] / (3 x 10 8 m/s) 2 = 2.4 x 10 -35 kg  m / m = (2.4 x 10 -35 ) / (1.67 x 10 -27 ) = 1.4 x 10 -8.

15. E R = mc 2 =  k j +  v ij +  m j c 2 For practical, every day experiences, including chemical reactions, we can assume that total mass is  conserved, so including mc 2 in the energy just shifts the “zero” of the energy. But: Total energy is always exactly conserved! In nuclear reactions, the change in  m j can result in significant changes in K: A 1 + A 2 + A 3 + ….  B 1 + B 2 + B 3 + … Energy that is released in exothermic reactions goes into the kinetic energy of the B’s, and this can be a significant fraction of the total mass (x c 2 ). In a reactor, the reaction is slow enough that these fast moving B’s are confined and hit lots of other “stuff”; the energy is used to increase the temperature of the stuff, drive a turbine, etc. In a bomb, the reaction is fast and the fast moving particles are not confined.

16. For atoms and subatomic particles, masses are often given in atomic mass units : 12 u = mass of carbon atom. 1 u = 1.66 x 10 -27 kg (1 u)c 2 = (1.66 x 10 -27 kg)(3 x 10 8 m/s) 2 = 1.49 x 10 -10 J (1 u)c 2 = (1.49 x 10 -10 J) (1 eV/1.6 x 10 -19 J) = 9.31 x 10 8 eV (1 u)c 2 = 931 MeV

17. Consider the fusion reaction (one we are trying to use to generate energy): 2 D 1 + 3 T 1  4 He 2 + n, where 2 D 1 is a deuterium atom, with m(D) = 2.0141 u, 3 T 1 is a tritium atom, with m(T) = 3.0161 u, 4 He 2 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  m / m  0.0189 u / 5.0 u = 0.38% Therefore, for every kilogram of “fuel” used, one produces: E = (0.0038) (1 kg) (3 x 10 8 m/s) 2 = 3.4 x 10 14 J of energy. This is the 3.5 day output of the Big Sandy power plant, for which they burn  6 x 10 7 kg of coal!.

18. Consider the following fission reaction, in which a uranium atom captures a neutron, causing it to divide: 235 U 92 + n  144 Ba 56 + 90 Kr 36 + 2n m(U) = 235.044 u m(Ba) = 143.923 u m(Kr) = 89.92 u m(n) = 1.009 u  m = 0.192 u = 179 MeV/c 2 (i.e.  m/m = 0.192/236 = 0.08%) i.e. for each uranium atom, 179 MeV are released. Since more neutrons (which carry most of the kinetic energy) are released than used, this can start a chain reaction (bomb). In a reactor, half the neutrons are captured by other atoms so that fission proceeds slowly.  m/m(U)  0.192/235 = 0.082%. Therefore, fission of 1 kg of uranium produces E = (0.00082) (1 kg) (3x10 8 m/s) 2 = 7.4 x 10 13 J (This is the Big Sandy power plant production in 19 hr, for which they burn 1.3 x 10 7 kg of coal.)

19. Problem: A nuclear power plant produces power at an average rate of P = 1100 MW. After 3 years, its fuel cells must be replaced. How much has the mass of the fuel cells decreased? --------------------------------------  m =  E/c 2  E = P  t P = 1100 MW = 1.1 x 10 9 J/s.  t = (3 years) (365 days/year) (24 hours/day) (3600 s/hr) = 9.46 x 10 7 s  E = (1.1 x 10 9 J/s) (9.46 x 10 7 s) = 1.04 x 10 17 J  m = (1.04 x 10 17 J) / (3 x 10 8 m/s) 2 = 1.16 kg [Note: Big Sandy has a maximum power output of 1100 MW. If it ran at that level for 3 years, it would have to burn 1.8 x 10 10 kg  4 x 10 6 tons of coal.] (These examples are not intended to argue for/against nuclear power, but to illustrate how much energy can be obtained from small losses in mass.)