This section is really an extension of 4-vectors which introduced the energy-momentum 4-vector. Here we see how the concept of a 4-vector, in particular the fact that the inner product is invariant between frames, can be applied to solve problems involving collisions and decays. Many such particle-particle collisions occur on the atomic or sub-atomic level; such small particles require little (by macroscopic standards) energy to accelerate them to speeds near the speed of light. Thus, Special Relativity is necessary for describing many of these interactions.
Recall that the energy-momentum 4-vector or 4-momentum is given by:
|P2âÉáP.P = E2/c2 - ||
Now let us tackle an example of first a collision problem and then a decay problem. Consider a particle with energy E and mass m. This particle moves towards another identical particle at rest. The particles collide elastically and both scatter at an angle θ with respect to the incident direction. This is illustrated in .
|P1' = ,,, 0|
|m2c2||=||- (1 + tan2θ)|
|âá4m2c4||=||(E + mc2)2 -|
|âáE2 + m2c4 +2Emc2 -4m2c4||=|
Decay problems can be solved in a similar manner; that is, by conserving energy and momentum. The situation in which a particle of mass M and energy E decays into two identical particles is also shown in . As shown, one particle heads off in the y-direction, and the other at an angle θ. Our problem is to calculate the energies of these particles resulting from the decay. Again, we begin by writing down the 4-momenta before and after the collision. Before the decay P = (E/c,, 0, 0) and after P1 = (E1/c, 0, p1, 0) and P2 = (E2/c, p2cosθ, - p2sinθ, 0); if the created particles have mass m, then, p1 = and p2 = . This problem becomes quite algebraically messy if we proceed in the same manner as we did above, conserving energy and momentum. Instead let us exploit the invariance of the inner product to solve the problem. Conservation of energy and momentum tells us that P = P1 + P2 which implies P2 = P - P1. Taking inner products we have:
|(P - P1).(P - P1) = P2.P2|
|âáP2 -2P.P1 + P12 = P22|
|âáM2c2 -2EE1/c2 + m2c2 = m2c2|
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