Four-Momentum: Where E=mc^2 Comes From

Building the four-momentum

In Newtonian physics momentum is mass times velocity. The relativistic upgrade is forced on us: multiply the four-velocity by the particle's rest mass m (an invariant scalar) and you get a genuine four-vector, the four-momentum P^\mu=mU^\mu. Writing out its components reveals something remarkable in the time slot.

P^{\mu} = m\,U^{\mu} = \big(\gamma m c,\ \gamma m \vec v\big) = \Big(\tfrac{E}{c},\ \vec p\Big)

The four-momentum. The spatial part is the relativistic momentum \vec p=\gamma m\vec v; the time part, times c, is the relativistic energy E=\gamma mc^2.

Reading energy off the time component of momentum is the deep unification: in relativity, energy and momentum are the time and space parts of a single object, exactly as t and \vec x are parts of x^\mu. This is why energy and momentum conservation are not two laws but one — the conservation of a single four-vector.

The energy-momentum relation

Now take the invariant length of the four-momentum. Since P=mU and U\cdot U=c^2, we get P\cdot P=m^2c^2 instantly — but computing it from the components gives the physics:

P\cdot P = \Big(\tfrac{E}{c}\Big)^2 - |\vec p|^2 = m^2 c^2

The invariant magnitude of four-momentum equals (mc)^2 in every frame. Rearranging gives the energy-momentum relation.

E^2 = (pc)^2 + (mc^2)^2

The master equation of relativistic dynamics — a right triangle with legs pc and mc^2 and hypotenuse E.

Two limits live at the triangle's corners. Set p=0: the particle is at rest and E=mc^2, the rest energy — mass-energy equivalence, the statement that mass is a form of energy. Set m=0: a massless particle like the photon has E=pc and must move at exactly c. And for slow massive particles, expanding \gamma recovers Newton with a constant offset:

E = \gamma m c^2 \approx mc^2 + \tfrac12 m v^2 + \tfrac{3}{8}\frac{mv^4}{c^2} + \cdots

The low-speed expansion. The rest energy mc^2 is the constant Newton never saw; the second term is the familiar kinetic energy \tfrac12 mv^2.

Conservation and a worked collision

The law of relativistic dynamics is stark: in any interaction, the total four-momentum is conserved, \sum P^\mu_{\text{in}}=\sum P^\mu_{\text{out}}. Its time component is energy conservation; its space components are momentum conservation. Because m^2c^2=P\cdot P, the invariant mass of a system of particles is generally not the sum of the individual masses — kinetic energy contributes to it. Let us see that shockingly directly.

Worked example. Two identical lumps of clay, each of rest mass m, fly toward each other head-on, each with speed v (Lorentz factor \gamma) in the lab, and collide and stick. Find the rest mass M of the single lump they form.

Set up four-momenta. Clay A: P_A=(\gamma mc,\ +\gamma m v). Clay B, moving oppositely: P_B=(\gamma mc,\ -\gamma m v). This is the lab = center-of-momentum frame, since total momentum is zero.
Add them (conservation). P_{\text{tot}}=P_A+P_B=(2\gamma mc,\ 0). The spatial momenta cancel; the energies add.
The final lump is at rest (zero total momentum), so P_{\text{final}}=(Mc,\ 0). Matching time components: Mc=2\gamma mc.
Solve. M=2\gamma m > 2m. The composite is heavier than the two originals by 2(\gamma-1)m — exactly the kinetic energy that got converted to rest mass, \Delta M\,c^2 = 2(\gamma-1)mc^2 = the total kinetic energy lost.