References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Post 28.

[Griffiths’s approach to the relativistic four-velocity is similar to that of Moore, although rather confusingly, he uses different notation (as well as keeping factors of in the equations rather than setting ). To keep the notation consistent with Griffiths, I’ll use his notation here, but anyone attempting to follow both books should beware…]

One way of defining linear momentum of an object of mass in relativity is to multiply the mass by the velocity, but since we have an ‘ordinary’ velocity and a four-velocity, we need to choose which one to use. It turns out that if we want momentum to be conserved, we need to use the four-velocity. We can see this as follows.

If we define momentum using ordinary velocity, so that

we can rework our collision problem from earlier. In an inertial frame we have a collision between a particle (mass and velocity ) and another particle (mass and velocity ). During the collision, some of the mass of gets transferred to , so that afterwards we have particles and with masses and velocities . If momentum is conserved, then

In inertial frame which moves with velocity relative to , the velocities of the particles transform using the formulas

The LHS of 2 transforms as

We can’t use 2 to convert this into the transform of the RHS (where is replaced by and by ) because of the different factors of and multiplying each term, that is, for the component for example, we can’t factor out a term from this expression to set it equal to .

If we use the four-velocity to define momentum, however, things work out properly. We define

(remember that Griffiths uses for four-velocity). To make this a four-vector, we define

Now we start with three-momentum (using the spatial components of the four-momentum) conserved in :

To convert to we can use Lorentz transformations, since the masses are scalars (constants) and the s are four-vectors. The LHS becomes

If we now apply 9 to each spatial component separately, we see that this last line is equal to

In order for this last expression to be equal to the Lorentz transform of the RHS of 9, the one remaining term containing and terms must equal its corresponding term in the transform of the RHS. That is, we must have

This is used as the motivation to define the relativistic energy as

With these definitions, we can see that four-momentum (which is 3-momentum and energy together) are conserved.