Inside the Large Hadron Collider, protons are whipped up to 99.9999991\% of the speed of light. Push harder — pour in more energy — and they go faster by only the tiniest sliver; they seem to hit an invisible wall at c. If Newton were right, a constant force should keep accelerating them without limit. Something in the old bookkeeping of momentum and energy has to give. Fixing it hands us the most famous equation in all of physics, E = mc^2, essentially for free.

This page rebuilds momentum and energy so that they survive at high speed. We'll see why the schoolbook formula p = mv quietly breaks, patch it into p = \gamma m v, discover that a moving body's total energy is E = \gamma m c^2 — which is not zero even when the body sits still — and tie momentum and energy together with one clean, frame-independent relation.

Why p = mv has to fail

Momentum earns its keep by being conserved: in any collision, the total momentum before equals the total after, no matter which (non-accelerating) frame you watch from. That's the whole point of the quantity. But "the same in every frame" now means the frames are related by the Lorentz transformation, not the old Galilean one — and if you take a collision that conserves \sum m\vec v in one frame and Lorentz-transform it into another, the totals no longer match. The Newtonian definition simply isn't conserved once velocities add relativistically.

So either momentum conservation is wrong (it isn't — it's rooted in the deepest symmetry we know, the uniformity of space) or the definition is wrong. Physics keeps the law and repairs the definition. The cure is to measure the change in position not per unit of the observer's time t, but per unit of the particle's own proper time \tau — a quantity everyone agrees on. Since dt = \gamma\, d\tau (time dilation), that inserts exactly one factor of \gamma:

At everyday speeds \gamma \approx 1 and we recover p \approx mv — which is why nobody noticed for two centuries. But as v \to c, \gamma \to \infty, so the momentum of a massive particle grows without any bound. That is the invisible wall: it takes ever more momentum to coax out ever less extra speed.

Energy: work done, and a stunning leftover

Energy is defined the same way it always was — the work a force does — but now using the corrected momentum. Carefully totalling up the work needed to accelerate a particle from rest to speed v (an integral of F = dp/dt over distance) gives a strikingly simple result for its total energy:

Look at what happens when the particle is at rest, v = 0, so \gamma = 1. The energy does not vanish. It sits at

the rest energy: energy an object owns purely by virtue of having mass, even stone still. Mass is a form of energy — an intensely concentrated one, since c^2 is a huge number. One gram of anything hides about 9 \times 10^{13}\ \text{J}, the yield of roughly twenty kilotonnes of TNT. This is the energy released, in part, when nuclei split or fuse; the Sun shines by turning about four million tonnes of its mass into sunlight every second.

The kinetic energy — the energy of motion alone — is the total minus the rest energy:

Sanity check: where did \tfrac12 m v^2 go?

A new formula for kinetic energy had better reduce to the trusty \frac12 m v^2 at ordinary speeds, or it's wrong. It does — and watching it happen is satisfying. For small \beta, expand \gamma = (1 - \beta^2)^{-1/2} with the binomial series:

Therefore

The leading term is exactly \frac12 m v^2 — Newton's kinetic energy is just the first term of Einstein's. The corrections are suppressed by powers of v^2/c^2, utterly negligible for a thrown ball but dominant for a particle in an accelerator. The graph below plots the true kinetic energy against the Newtonian guess: they hug each other near \beta = 0 and part company dramatically as \beta \to 1, where the real curve rockets to infinity.

The relation that ties it all together

We now have two headline quantities, E = \gamma m c^2 and p = \gamma m v. Both contain the awkward factor \gamma and depend on v. Watch what happens when we combine them to eliminate both \gamma and v. Compute E^2 - (pc)^2:

The \gamma's cancel completely, leaving something that depends only on the mass:

This is the workhorse of particle physics. You rarely know a particle's speed, but you can measure its energy and momentum, and this relation instantly gives its mass — the same number in every lab, however the particle is moving. It's a right triangle: mc^2 and pc are the legs, E the hypotenuse.

Worked examples

Example 1 — kinetic energy of a fast proton. A proton (mc^2 = 938\ \text{MeV}) moves at \beta = 0.8, so \gamma = 1/\sqrt{1 - 0.64} = 1.667. Its total energy is E = \gamma m c^2 = 1.667 \times 938 \approx 1563\ \text{MeV}, and its kinetic energy is

Newton would have guessed \frac12 m v^2 = \frac12(938)(0.8)^2 = 300\ \text{MeV} — barely half the true value. At \beta = 0.8 the old formula is already hopeless.

Example 2 — energy from momentum and mass. An electron (mc^2 = 0.511\ \text{MeV}) has momentum pc = 2.0\ \text{MeV}. Its total energy comes straight from the energy–momentum relation:

Because pc \gg mc^2 here, the electron is "ultra-relativistic" and E \approx pc — it behaves almost like a massless particle.

Example 3 — mass into energy. When an electron meets its antiparticle, a positron, both at rest, they annihilate entirely into light. All the rest energy converts: E = 2 m c^2 = 2 \times 0.511 = 1.022\ \text{MeV}, shared between two photons of 0.511\ \text{MeV} each. Mass has become pure radiation — E = mc^2 running in reverse.

Momentum's runaway growth

The chart below drives home the same lesson for momentum. Newton predicts p/mc = \beta, a straight line — double the speed, double the momentum. The relativistic p/mc = \gamma\beta agrees at first but curls upward and blows up at \beta = 1. You can pump limitless momentum (and energy) into a particle and it still never quite reaches c.