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 px (momentum in the x-direction) greater than particle B, the total px 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 px = mvx 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 | vx| 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:
|px = γmvx =|
A is at rest in the y-direction so γA = 1, and mvx = γmvx. 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 px = = mvx.
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, γ2m we now know, is certainly not conserved).
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 γv = = = . In this frame conservation of momentum gives:
|γvmv + 0 = γMuâám = âáM =|
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:
If γmc2 is conserved then:
|γvmc2 +1×mc2||=||γuMc2âám + m|
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 - v2/c2)-1/2 as follows:
|EâÉáγmc2||=||1 - v2/c2)-1/2|
|=||mc21 + + +|
|=||mc2 + mv2 +|
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 = mc2 for a particle at rest. Second, mc2 is just a constant so conservation of energy reduces to the conservation of mv2/2 in this limit. Moreover, the reduction of E = γmc2 to the Newtonian form in this limit justifies our choice of γmc2 rather that say, 5γmc8 as our expression for energy.