2. 1 Recap: Relativistic definition of linear momentum and moving mass We have studied two concepts in earlier lecture: mass and momentum in SR Consider a mass moving with speed v in a rest frame Classically, p = mv, m = mass is constant and not changing with its state of motion Relativistically, the mass of a moving object changes as its speed changes: m =  m 0. m is called the relativistic mass m 0 = rest mass = the mass measured in a frame where the object is at rest. It’s value is a constant Relativistically, p = mv =  m 0 v v M I see M is at rest. Its mass is m 0, momentum, p’ = 0 O’ O I see the mass of M as m = m 0  ;momentum of M as p = mv=m 0  v

3. 2 Example The rest mass of an electron is m 0 = 9.11 x 10 -31 kg. p = m 0  u Compare it with that calculated with classical definition. If it moves with u = 0.75 c, what is its relativistic momentum? m0m0

4. 3 Solution The Lorentz factor is  = [1-(u/c) 2 ] -1/2 = [1-(0.75c/c) 2 ] -1/2 =1.51 Hence the relativistic momentum is simply p =  x m 0 x u =  x m 0 x 0.75c = 1.51 x 9.11 x 10 -31 kg x 0.75 x 3 x 10 8 m/s = 3.1 x 10 -22 kg m/s = Ns In comparison, classical momentum gives p classical = m 0 x 0.75c = 2.5 x 10 -22 Ns – about 34% lesser than the relativistic value p classical /p SR =  = [1-(u/c) 2 ] -1/2 1 when u<<c p classical /p SR =  = [1-(u/c) 2 ] -1/2 >>1 when u –>c

5. 4

6. 5 Energy in SR Recall the law of conservation of mechanical energy you have studied in classical mechanics: Work done by external force on an object (W) = the change in kinetic energy of the object, (  )

7. 6 K1K1 F K2K2 F s  = K 2 - K 1 W = F s Conservation of mechanical energy: W =  The total mechanical energy of the object, E = K + U. Ignoring potential energy, E of the object is solely in the form of kinetic energy. If K 1 = 0, then E = K 2. But in general, U also needs to be taken into account for E.

8. 7 Force, work and kinetic energy When a force is acting on an object, O, at rest with rest mass m 0, it will get accelerated (say from rest) to some speed (say u) and increase in kinetic energy from 0 to K K as a function of u can be derived from first principle based on the definition of: Force,F = dp/dt, work done, W = F dx, and conservation of mechanical energy,  K = W

9. 8 In classical mechanics, mechanical energy (kinetic + potential) of an object is closely related to its momentum and mass Since in SR we have redefined the classical mass and momentum to that of relativistic version m class (cosnt) –> m SR = m 0  p class = m class u –> p SR = (m 0  )u We will like to derive K in SR in the following slides E.g, in classical mechanics, K = p 2 /2m = 2 mu 2 /2. However, this relationship has to be supplanted by the relativistic version K = mu 2 /2 –> K = E – m 0 c 2 = mc 2 - m 0 c 2 we must also modify the relation btw work and energy so that the law conservation of energy is consistent with SR

10. 9 Derivation of relativistic kinetic energy Force = rate change of momentum Chain rule in calculus where, by definition, is the velocity of the object

11. 10 Explicitly, p =  m 0 u, Hence, dp/du = d/du(  m 0 u) = m 0 [u (d  du) +   m 0 [  + (u 2 /c 2 )    m 0  u 2  c 2  -3/2 in which we have inserted the relation integrate

12. 11 E = mc 2 is the total relativistic energy of an moving object The relativisitic kinetic energy of an object of rest mass m 0 travelling at speed u E 0 = m 0 c 2 is called the rest energy of the object. Its value is a constant for a given object Any object has non-zero rest mass contains energy as per E 0 = m 0 c 2 One can imagine that masses are ‘energies frozen in the form of masses’ as per E 0 = m 0 c 2

13. 12 Or in other words, the total relativistic energy of a moving object is the sum of its rest energy and its relativistic kinetic energy The mass of an moving object m is larger than its rest mass m 0 due to the contribution from its relativistic kinetic energy – this is a pure relativistic effect not possible in classical mechanics E = mc 2 relates the mass of an object to the total energy released when the object is converted into pure energy

14. 13 Example, 10 kg of mass, if converted into pure energy, it will be equivalent to E = mc 2 = 10 x (3 x10 8 ) 2 J = 9 x10 17 J – equivalent to a few tons of TNT explosive

15. 14 Due to mass-energy equivalence, sometimes we express the mass of an object in unit of energy Example Electron has rest mass m 0 = 9.1 x 10 -31 kg The rest mass of the electron can be expressed as energy equivalent, via m 0 c 2 = 9.1 x 10 -31 kg x (3 x 10 8 m/s) 2 = 8.19 x 10 -14 J = 8.19 x 10 -14 x (1.6x10 -19 ) -1 eV = 511.88 x 10 3 eV = 0.511 MeV

16. 15 Continue exploring the mass- energy momentum formula… In terms of relativistic momentum, the relativistic total energy can be expressed as followed Conservation of energy-momentum

17. 16 Relativistic invariance Note that, in general, E and p are frame- dependent (i.e they takes on different value in different reference frame) but the quantity is an invariant – it’s the same in value ( ) in all reference frames. We call such quantity a `relativistic invariant’