In this setion we will turn to a discussion of some interesting aspects of Special Relativity, concerning how
particle and objects gain motion, and how they interact. In this section we will arrive at an expression that looks
something like the definition of momentum, and seems to be a conserved
quantity under the new rules of Special Relativity. With this in mind consider the following setup.

As shown in , two particles have equal and opposite small speeds in the x- direction and equal
and opposite large speeds in the y-direction. The particles collide and bounce off each other as shown. Each time
one of the particles crosses one of the dotted vertical lines its clock 'ticks.' How does this look in the frame
moving in the y-direction with the same velocity as particle A? This is shown also in . Here
it is clear that the collision causes the particles to swap x-velocities. This implies that the momentum in the
x-direction of each of the particles must be the same. We know this because if particle A had p_{x} (momentum in
the x-direction) greater than particle B, the total p_{x} would not be conserved. This may seem somewhat strange
since we have not defined momentum yet, but we know from classical mechanics that the direction of the momentum
depends on the direction of the velocity and that the magnitude is proportional to the mass and the velocity. Since
the particles are identical (they have the same mass and x-velocity), if momentum is to be conserved both particles
should have the same magnitude for their x-momenta.

If the y-velocity is much greater than the x-velocity, then particle A is essentially at rest with respect to
particle B in A's frame. Time
dilation
tells us that particle B's clock must be
running slow by a factor . Particle B's clock ticks once for every vertical line crossed
(independent of frame), so particle B must be moving more slowly than A in the x-direction by a factor . Thus the magnitudes of the x-velocities of the particles are not the same. This means that the
Newtonian p_{x} = mv_{x} is not a conserved quantity because the momentum of particle B would be smaller than the
momentum of particle A by the factor 1/γ since | v_{x}| is larger for particle A. We have shown that if
momentum is to be conserved, the momenta of A and B better be the same. However, the solution to the difficulty is
not so hard: we define momentum as:

p_{x} = γmv_{x} =

A is at rest in the y-direction so γ_{A} = 1, and mv_{x} = γmv_{x}. For B however,
this we have exactly taken care of the problem: the factor by which particle B's speed was smaller is canceled out by
the γ so particle B also has momentum p_{x} = = mv_{x}.

In three dimensions the equation for relativistic momentum becomes:

We have not shown here that γmv is conserved--this is the job of experiments. What we have
done is to provide some motivation for the equation for relativistic momentum by showing that
γm (or some constant multiple of it) is the only vector of this form which has any chance of
being conserved in a collision (for instance, γ^{2}m we now know, is certainly not
conserved).

Relativistic Energy

To develop a concept of relativistic energy we will again consider a scenario and show that a particular
expression is conserved. This expression we just happen to give the label 'energy.'

In this system two identical particle of mass m both have speed u and head directly towards one
another. They collide and stick together to form a mass M which is at rest. Now consider the system
from the point of view of a frame moving to the left with speed u. The mass on the right is at rest in this
frame, M moves to the right with speed u, and the velocity addition formula tells us that the
left mass moves to the right with speed v = . The γ factor associated
with v is γ_{v} = = = . In this frame conservation of
momentum gives:

γ_{v}mv + 0 = γMuâá’m = âá’M =

Surprisingly, M is not equal to 2m, but is larger by a factor γ. However, in the limit u < < c,
M 2m as expected from the correspondence
principle.

Let us now state the expression for relativistic energy and check whether it is conserved:

EâÉáγmc^{2}

If γmc^{2} is conserved then:

γ_{v}mc^{2} +1×mc^{2}

=

γ_{u}Mc^{2}âá’m + m

=

âá’

=

This last equality is clearly true. Thus we have found a quantity that looks a little bit like classical energy
and is conserved in collisions. What happens in the limit v < < c? We can use the binomial series
expansion to expand (1 - v^{2}/c^{2})^{-1/2} as follows:

EâÉáγmc^{2}

=

1 - v^{2}/c^{2})^{-1/2}

=

mc^{2}1 + + +

=

mc^{2} + mv^{2} +

The higher order terms can be neglected for v < < c. First note that for v = 0 the second (and all higher)
terms are zero so we have the famous E = mc^{2} for a particle at rest. Second, mc^{2} is just a constant
so conservation of energy reduces to the conservation of mv^{2}/2 in this limit. Moreover, the
reduction of E = γmc^{2} to the Newtonian form in this limit justifies our choice of
γmc^{2} rather that say, 5γmc^{8} as our expression for energy.