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Concepts

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:

PâÉá(E/c,

The total energy and momentum of a number of particles is just the sum of their individual 4-momenta. If
the total 4-momenta before a collision or decay is P_{i} and the total 4-momenta after is P_{f} the
conservation of energy and momentum are both expressed in the equation P_{i} = P_{f}. Given the definition
of the inner product from properties in dynamics, it is easy to see that:

P^{2}âÉáP.P = E^{2}/c^{2} - |

This is the most important relationship in the section.

Examples

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 .

We want to find θ in terms of E and m. We can write down the 4-momenta of the two particles.
The moving particle has P_{1} = (E/c, p, 0, 0) and the stationary particle P_{2} = (mc, 0, 0, 0), where p = . The 4-mometa after the collision are: P_{1}' = (E'/c, p'cosθ, p'sinθ, 0)
and P_{2}' = (E'/c, p'cosθ, - p'sinθ, 0), where p' = . We know from the
symmetry of the situation that the energy and momentum of the two particles must be equal after the
collision. Conserving energy gives E' = . Conserving momentum (only the x-
direction is significant since the y components cancel) gives: p'cosθ = p/2. Thus:

P_{1}' = ,,, 0

But we can take the inner product of this with itself and set it equal to m^{2}c^{2}:

m^{2}c^{2}

=

- (1 + tan^{2}θ)

âá’4m^{2}c^{4}

=

(E + mc^{2})^{2} -

âá’E^{2} + m^{2}c^{4} +2Emc^{2} -4m^{2}c^{4}

=

âá’cos^{2}θ

=

=

Which is the result desired.

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 P_{1} = (E_{1}/c, 0, p_{1}, 0) and P_{2} = (E_{2}/c, p_{2}cosθ, - p_{2}sinθ, 0); if the created particles have mass m, then, p_{1} = and
p_{2} = . 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 = P_{1} + P_{2} which implies P_{2} = P - P_{1}. Taking inner products we have:

We have made good use of the fact that the inner product of any 4-momenta with itself is just m^{2}c^{2}.
To get E_{2} we apply conservation of energy to deduce that E_{1} + E_{2} = Eâá’E_{2} = E - E_{1} = . Solving the problem in this way gets rid of the messiness of P_{2}.