From Special Relativity to Feynman Diagrams.pdf

211 energy and mass we have seen that the concept of

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2.1.1 Energy and Mass We have seen that the concept of force as an action at a distance on a given particle looses its meaning in a relativistic theory. However nothing prevents us from defining the force acting on a particle as the time derivative of its relativistic momentum: F = d p dt . (2.15) Recalling the definition of p , namely, p = m (v) v , we find: F = d dt ( m (v) v ) = dm (v) dt v + m (v) d v dt , Note that F is in general no longer proportional to the acceleration a = d v dt . Writing v = v u , where u is the unit vector in the direction of motion, we obtain a = d v dt = d v dt u + v 2 ρ n , where, as is well known, the unit vector n is normal to u and oriented towards the concavity of the trajectory, ρ being the radius of curvature.
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2.1 Relativistic Energy and Momentum 45 Computing the time derivative of the relativistic mass we find dm (v) dt = m d dt 1 1 v 2 c 2 = m 1 v 2 c 2 3 / 2 v c 2 d v dt = m (v) c 2 v 1 v 2 c 2 d v dt , so that: F = m (v) 1 1 v 2 c 2 v 2 c 2 d v dt + d v dt u + m (v) v 2 ρ n = m (v) 1 v 2 c 2 d v dt u + m (v) v 2 ρ n . Note that F is not proportional to a . We are now ready to determine the relativistic expression for the kinetic energy of a particle by computing the work done by the total force F acting on it. For an infinitesimal displacement d x = v dt along the trajectory, the work reads: dW = F · d x = F · v dt = m (v) 1 v 2 c 2 v d v dt dt = 1 2 m (v) 1 v 2 c 2 d (v 2 ). Integrating along the trajectory (and changing the integration variable into x = 1 v 2 c 2 ), we easily find: W = F · d x = 1 2 m 1 v 2 c 2 3 / 2 d v 2 = − c 2 2 m dx x 3 / 2 = mc 2 1 x 1 / 2 = m 1 v 2 c 2 c 2 + const = m (v) c 2 + const . If we define the kinetic energy, as in the classical case, to be zero when the particle is at rest, then the constant is determined to be m ( 0 ) c 2 , so that, the kinetic energy E k acquired by the particle will be given by: E k (v) = m (v) c 2 mc 2 , (2.16) where, from now on, m = m ( 0 ) will always denote the rest mass. Note that in the non-relativistic limit v 2 / c 2 1 , we retrieve the Newtonian result: E k (v) = mc 2 1 v 2 c 2 mc 2 mc 2 1 + 1 2 v 2 c 2 mc 2 = 1 2 m v 2 . (2.17) where we have neglected terms of order O (v 4 / c 4 ).
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46 2 Relativistic Dynamics Let us define the total energy of a body as: E = m (v) c 2 , (2.18) The kinetic energy is then expressed, in ( 2.16 ), as the difference between the total energy and the rest energy , which is the amount of energy a mass possesses when it is at rest: E rest = E (v = 0 ) = mc 2 . (2.19) To motivate the definition of the total energy of a particle given in ( 2.18 ), we prove that the total energy in a collision process, defined as the sum of the total energies of each colliding particle, is always conserved. This immediately follows from the conservation of the total (relativistic) mass, which we have shown to be a necessary requirement for the conservation law of momentum to be covariant. Indeed by multiplying both sides of m 1 (v 1 ) + m 2 (v 2 ) + · · · + m k (v k ) = cost , by c 2 and using the definition ( 2.18 ), we find E 1 (v 1 ) + E 2 (v 2 ) + · · · + E k (v k ) = cost .
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