[jimmahoney]

It's not entirely coincidence.

Four dimensional space-time has a metric analogous to but not quite the same as euclidean space: the time part has the opposite sign from the space parts.

  ds^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2)

ds is an "invariant proper time" which has the same value in all frames of reference. Check any special relativity textbook for the details.

Just as you can start with distance and then build up to momentum and energy in classical physics, in relativistic mechanics you can start with this metric and build up vectors in 4-space for velocity and energy/momentum. (Turns out that the time part is an energy while the space parts are momentum.) The upshot is that

  (m_0 c^2)^2 = E^2 - (p c)^2

which is the frame-invariant length of the energy-momentum 4-vector. (I think that m is better written as m_0, the rest mass, since the "m" notation sometimes means relativistic mass, which is different.)