From Special Relativity to Feynman Diagrams.pdf

# This fact has no correspondence in classical

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This fact has no correspondence in classical mechanics where we know that, as opposed to the total linear momentum, which is always conserved in collision processes, the conservation of mechanical energy only holds in elastic collisions. This apparent clash between the classical and the relativistic laws of energy conservation is obviously a consequence of the fact that the rest energy can be transformed into other forms of energy, like kinetic energy, etc. We can give a clear illustration of this by considering again the collision of two particles with rest masses m 1 , m 2 and velocities v 1 e v 2 . Suppose that the collision is perfectly inelastic, so that the two particles stick together into a single one of rest mass M . It is convenient to describe the process in the center of mass frame, in which the final particle is at rest. Let us first describe the collision in the context of Newtonian mechanics. The conservation of momentum reads: p 1 + p 2 = P = 0 , or, equivalently m 1 v 1 + m 2 v 2 = 0 . Moreover conservation of the classical mass is also assumed. M = m 1 + m 2 . (2.20)

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2.1 Relativistic Energy and Momentum 47 The initial and final mechanical (kinetic) energies are however different since E i k = m 1 v 2 1 2 + m 2 v 2 2 2 , E f k = 0 . and thus E k = E f k E i k = − m 1 v 2 1 2 + m 2 v 2 2 2 = 0 . In Newtonian physics, the interpretation of this result is that the kinetic energy of the particles in the initial state has been converted into heat, increasing the thermal energy of the final body, that is the disordered kinetic energy of the constituent molecules. Let us now describe the same process from the relativistic point of view. Conservation of momentum and energy give the following two equations: m 1 γ (v 1 ) v 1 + m 2 γ (v 2 ) v 2 = 0 , m 1 (v 1 ) c 2 + m 2 (v 2 ) c 2 = M ( 0 ) c 2 , where we have set M (v = 0 ) = M ( 0 ). Using ( 2.16 ) to separate the rest masses from the (relativistic) kinetic energies, we obtain E k (v 1 ) + m 1 c 2 + E k (v 2 ) + m 2 c 2 = M ( 0 ) c 2 . (2.21) Since the kinetic energy E f k of the mass M ( 0 ) is zero, it follows 4 c 2 M ( 0 ) c 2 ( M ( 0 ) m 1 m 2 ) = − ( 0 E k (v 1 ) E k (v 1 )) = − E k . (2.22) From the above relation we recognize that the loss of kinetic energy has been transformed in an increase of the final rest mass M = M ( 0 ) ; thus M is not the sum of the rest masses of the initial particles (as it was instead assumed in the classical case, see ( 2.20 )). If we consider the inverse process in which a particle of rest mass M decays, in its rest frame, into two particles of rest masses m 1 and m 2 , we see that part of the initial rest mass is now converted into the kinetic energy of the decay products. The importance of this effect obviously depends on the size of the ratio (v 2 / c 2 ). These examples illustrate an important implication of relativistic dynamics: the rest mass m of an object can be regarded as a form of energy, the rest energy mc 2 , which can be converted into other forms of energy (kinetic etc.). Let us illustrate this property in an other example. Consider a body of mass M at some given temperature: M will be given by the sum of the relativistic masses
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• Fall '17
• Chris Odonovan

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