Wolfram Alpha:

```Energy and Momentum of a Particle
---------------------------------

The four-vector formalism is both powerful and elegant and is used
to derive the important equation that relates the physical quantities
energy, mass and momentum in Special Relativity.

The 4-velocity is the rate of change of both time and space coordinates
with respect to the proper time of the object. The 4-velocity is a
tangent vector to the world line.  Thus,

4-vector of velocity, uμ = dxμ/dτ

Time component:

x0 = ct = cγτ

dx0/dτ = cγ

Spacial components:

dτ2 = dt2 - dx2/c2

Rewrite

dτ = √{(dt2 - dx2/c2)/dt2}dt

= √(1 - v2/c2)dt

dt/dτ = 1/√(1 - v2/c2)

dx/dτ = (dx/dt)(dt/dτ) = vx/√(1 - v2/c2) same for y and z

The complete 4-vector is written notationally in terms of the time
and space components as:

dxμ/dτ = {c/√(1 - v2/c2), vx/√(1 - v2/c2)

Alternatively, and generalizing to all directions

uμ = {γc, γv}

Likewise we can identify the 4-momentum as

pμ = {γmc, γmv}

Note: The relativistic mass is defined as γm where m is the invariant/rest/proper
mass. Some physicists reject this as not really meaningful, however.

Einstein recognized p0 when multiplied by c as the KE.  Thus

E = mc2/√(1 - v2/c2)

For small v/c expand as binomial.

mc2(1 - v2/c2)-1/2 = (1 + v2/2c2)mc2

E = mc2 + mv2/2

Which is agreement with Newton (rest energy + KE).

Relationship between Energy and Momentum
----------------------------------------

E2 = m2c4/(1 - v2/c2)

Put c = 1 to simplify math.

E2 = m2/(1 - v2)

p2 = m2v2/(1 - v2)

E2 - p2 = m2

Add back in c to make it dimensionally correct.

E2 - p2c2 = m2c4

where m (the rest or invariant mass) is the mass of a body that is isolated (free) and
at rest relative to the observer.

When p = 0, E = mc2

For massless particle m = 0

E = √p2c2 = pc```