6.4 — Momentum, Energy and E = mc^2

Newton's momentum, \vec{p} = m\vec{v}, does not survive relativity. Not as an approximation that gets slightly worse at high speed — as a conservation law that flatly fails.

Here is the problem in one sentence. Take a collision that conserves momentum in one frame, transform every velocity into another frame using the velocity addition rule of Chapter 6.2, and add the momenta again: they do not balance. Since postulate 1 says the laws of physics must be the same in every frame, a conservation law that holds only in one frame is not a law at all.

Either momentum conservation is abandoned, or momentum is redefined. Conservation laws are far too valuable to give up — they come from symmetries of nature, as Noether's theorem shows — so momentum gets redefined.

The relativistic momentum

The fix is to use proper time in the definition rather than coordinate time.

Newton wrote \vec{p} = m\,d\vec{x}/dt, where t is the time in whatever frame you happen to be using. That is the problem: t is frame-dependent, so the definition is different for every observer.

Proper time \tau — the time on a clock carried along with the particle — is the same number for everybody. Use it:

\vec{p} = m\frac{d\vec{x}}{d\tau}

Since dt = \gamma\,d\tau (Chapter 6.3's time dilation), the chain rule gives:

\vec{p} = m\frac{d\vec{x}}{dt}\cdot\frac{dt}{d\tau} = \gamma m\vec{v}

\boxed{\vec{p} = \gamma m\vec{v} = \frac{m\vec{v}}{\sqrt{1-v^2/c^2}}}

With this definition, total momentum is conserved in every inertial frame. That is the whole justification, and it is enough: the definition is chosen because it makes the conservation law frame-independent, and it reduces to m\vec{v} at low speed since \gamma \to 1.

As v \to c, \gamma \to \infty, so p \to \infty. A particle can be given unlimited momentum while its speed creeps ever closer to c without reaching it. That is the mechanism by which the speed limit is enforced — not a rule forbidding faster travel, but the fact that pushing harder buys momentum instead of speed.

The "relativistic mass" trap

Older books write p = m_{\text{rel}}v with m_{\text{rel}} = \gamma m, and say mass increases with speed. This book does not, and it is worth saying why.

It gives wrong answers as soon as you push it. The force law becomes direction-dependent: pushing a fast particle sideways and pushing it forwards require different amounts of force for the same acceleration, so "mass" would need two different values at once. It also invites the idea that a fast object's gravity increases or that it might collapse into a black hole, which is false.

Mass is an invariant — the same number in every frame, a property of the object like its charge. What grows with speed is momentum and energy. Modern practice, and this book's, is: m always means rest mass, and the \gamma is written explicitly where it belongs.

Kinetic energy, derived

Energy is the work done, and work is force times distance, with force being the rate of change of momentum (Chapter 1.6):

KE = \int_0^x F\,dx = \int_0^x \frac{dp}{dt}dx

Change variables using dx = v\,dt, so \frac{dp}{dt}dx = v\,dp:

KE = \int_0^p v\,dp

Integrate by parts, \int v\,dp = pv - \int p\,dv:

KE = \gamma mv\cdot v - \int_0^v \gamma mv\,dv = \gamma mv^2 - m\int_0^v\frac{v\,dv}{\sqrt{1-v^2/c^2}}

Do that integral. Substitute u = 1 - v^2/c^2, so du = -2v\,dv/c^2 and v\,dv = -c^2du/2:

\int\frac{v\,dv}{\sqrt{1-v^2/c^2}} = -\frac{c^2}{2}\int u^{-1/2}du = -\frac{c^2}{2}\cdot 2u^{1/2} = -c^2\sqrt{1-v^2/c^2}

Evaluate from 0 to v:

\left[-c^2\sqrt{1-v^2/c^2}\right]_0^v = -c^2\sqrt{1-v^2/c^2} + c^2 = c^2\left(1-\frac{1}{\gamma}\right)

So:

KE = \gamma mv^2 - mc^2\left(1-\frac{1}{\gamma}\right) = \gamma mv^2 - mc^2 + \frac{mc^2}{\gamma}

Now simplify. Note that \frac{1}{\gamma} = \sqrt{1-v^2/c^2}, so \frac{mc^2}{\gamma} = \gamma mc^2\left(\frac{1}{\gamma^2}\right) = \gamma mc^2(1-v^2/c^2) = \gamma mc^2 - \gamma mv^2:

KE = \gamma mv^2 - mc^2 + \gamma mc^2 - \gamma mv^2

The \gamma mv^2 terms cancel:

\boxed{KE = \gamma mc^2 - mc^2 = (\gamma-1)mc^2}

Checking against Newton

At low speed, expand \gamma using the binomial series, (1-x)^{-1/2} \approx 1 + \frac{x}{2} + \frac{3x^2}{8} with x = v^2/c^2:

\gamma \approx 1 + \frac{v^2}{2c^2} + \frac{3v^4}{8c^4}

KE \approx mc^2\left(\frac{v^2}{2c^2}+\frac{3v^4}{8c^4}\right) = \frac{1}{2}mv^2 + \frac{3mv^4}{8c^2}

The first term is exactly Newton's kinetic energy. The second is the leading correction, and at 10 km/s it is a fraction 3v^2/4c^2 = 8\times10^{-10} of the first — which is why nobody found it by weighing cannonballs.

E = mc^2

Look again at the kinetic energy:

KE = \gamma mc^2 - mc^2

The first term grows with speed; the second does not depend on v at all. Einstein's step was to read the two terms as total energy minus a constant:

E_{\text{total}} = \gamma mc^2, \qquad E_{\text{rest}} = mc^2

\boxed{E = mc^2}

A stationary object has energy mc^2 simply for existing. Not stored chemically, not thermally, not from motion — energy that is there because the object has mass.

Why is this not just a bookkeeping choice? One could argue that adding a constant to energy changes nothing, since only differences matter. Einstein's insight was that the constant is not constant: if a body emits energy, its mass decreases.

His own derivation, in a three-page follow-up paper in September 1905, runs like this. A body at rest emits two equal light pulses in opposite directions, total energy E. By symmetry it stays at rest, so no momentum was transferred. Now analyse the same emission from a frame moving at v. The Doppler shifts (Chapter 6.3) are unequal, so the two pulses carry unequal momenta and the net momentum carried away is not zero — and to conserve momentum the body's own momentum must change. Its velocity has not changed, since it was symmetric in its own frame. So its mass must have. Working the algebra through gives \Delta m = E/c^2.

\boxed{\Delta m = \frac{\Delta E}{c^2}}

Mass and energy are the same thing in different units, and c^2 is merely the conversion factor — a large one, 9\times10^{16} J per kilogram, which is why the mass change is invisible in everyday energy exchanges.

Worked check. Burn a litre of petrol, releasing about 34 MJ:

\Delta m = \frac{3.4\times10^{7}}{9\times10^{16}} = 3.8\times10^{-10}\ \text{kg}

Under half a microgram out of about 750 g of fuel — one part in two billion. Every chemical reaction changes mass, and no chemical balance has ever been able to see it.

The energy–momentum relation

Two expressions, E = \gamma mc^2 and p = \gamma mv. Eliminate v between them.

E^2 = \gamma^2m^2c^4, \qquad p^2c^2 = \gamma^2m^2v^2c^2

E^2 - p^2c^2 = \gamma^2m^2c^4\left(1-\frac{v^2}{c^2}\right) = \gamma^2m^2c^4\cdot\frac{1}{\gamma^2} = m^2c^4

\boxed{E^2 = (pc)^2 + (mc^2)^2}

This is the most useful equation in particle physics, and its structure is worth reading. It is Pythagoras: a right triangle with pc and mc^2 as legs and E as the hypotenuse. And the combination E^2 - p^2c^2 is the same for every observer, even though E and p separately are not — it is an invariant, like the interval of Chapter 6.5.

For a massless particle, m = 0:

E = pc

A photon has energy and momentum despite having no mass, and its momentum is p = E/c — exactly the radiation pressure result of Chapter 4.7, now derived rather than asserted. And since m = 0 makes \gamma m indeterminate, there is no contradiction in a massless particle moving at exactly c: the formulas E = \gamma mc^2 and p = \gamma mv simply do not apply to it, while E^2 = (pc)^2 + (mc^2)^2 does.

For a slow massive particle, pc \ll mc^2, expand the square root:

E = mc^2\sqrt{1+\frac{p^2}{m^2c^2}} \approx mc^2 + \frac{p^2}{2m}

which is rest energy plus Newtonian kinetic energy. Everything is consistent.

Where the numbers get interesting: nuclear energy

Chemical reactions rearrange electrons and release a few electron-volts per atom. Nuclear reactions rearrange nucleons and release millions. The reason is entirely in the mass.

Nuclear fission

Uranium-235 absorbs a neutron and splits. A representative reaction:

^{235}\text{U} + n \to ^{141}\text{Ba} + ^{92}\text{Kr} + 3n

Masses, in atomic mass units (1 u = 1.66054\times10^{-27} kg, and 1\ \text{u}\cdot c^2 = 931.5 MeV):

Before:

235.043930 + 1.008665 = 236.052595\ \text{u}

After:

140.914411 + 91.926156 + 3(1.008665) = 235.866562\ \text{u}

Mass difference:

\Delta m = 236.052595 - 235.866562 = 0.186033\ \text{u}

Energy released:

E = 0.186033\times931.5 = 173\ \text{MeV}

The full accounting including prompt gamma rays and later decays brings it to about 200 MeV per fission.

Scale it up. One kilogram of ²³⁵U contains:

N = \frac{1000\ \text{g}}{235\ \text{g/mol}}\times6.022\times10^{23} = 2.56\times10^{24}\ \text{nuclei}

E = (2.56\times10^{24})(200\times10^{6})(1.602\times10^{-19}) = 8.2\times10^{13}\ \text{J}

82 terajoules per kilogram. Compare with coal at 3\times10^{7} J/kg — a factor of 2.7 million. One kilogram of uranium releases the energy of 2,700 tonnes of coal.

Note the fractional mass change: 0.186/236 = 0.079 %. Less than a tenth of one percent of the mass became energy, and that is enough to power a city.

Nuclear fusion

The proton–proton chain in the Sun, net:

4\,^{1}\text{H} \to ^{4}\text{He} + 2e^+ + 2\nu

Before: 4\times1.007825 = 4.031300 u. After: 4.002603 u (plus the positrons, which annihilate and whose energy stays in the star).

\Delta m = 0.028697\ \text{u} \quad\Longrightarrow\quad E = 0.028697\times931.5 = 26.7\ \text{MeV}

Fractional mass change: 0.0287/4.031 = 0.71 % — nine times better than fission per unit mass, which is why fusion is worth the enormous engineering difficulty.

The Sun's mass loss. Its output is 3.85\times10^{26} W:

\frac{dm}{dt} = \frac{P}{c^2} = \frac{3.85\times10^{26}}{9\times10^{16}} = 4.3\times10^{9}\ \text{kg/s}

The Sun converts 4.3 million tonnes of mass into energy every second. In its 4.6-billion-year life so far it has lost about 6\times10^{26} kg — which sounds enormous and is 0.03 % of its mass.

Annihilation

Matter meeting antimatter converts all the mass:

e^- + e^+ \to 2\gamma

Each electron has rest energy m_ec^2 = 0.511 MeV, so each photon carries 0.511 MeV. That specific energy is the signature that positron emission tomography (PET) scanners look for: a radioactive tracer emits positrons, each annihilates within a millimetre or two, and the pair of 0.511 MeV photons flies out back-to-back. Detecting both ends of that line locates the annihilation, and thousands of such lines reconstruct an image of where the tracer went.

Efficiency comparison, as a fraction of mass converted:

Process	Fraction
Chemical burning	\sim10^{-10}
Nuclear fission	0.08 %
Nuclear fusion	0.7 %
Accretion onto a black hole	up to 40 %
Matter–antimatter annihilation	100 %

The black hole entry is not a typo, and Chapter 12.3 explains it: matter spiralling into a rotating black hole radiates a large fraction of its rest energy before crossing the horizon, which makes quasars the most efficient engines in the universe.

Binding energy and where mass actually comes from

Here is the point that reverses everyone's intuition about mass.

A helium nucleus has less mass than its parts. Two protons and two neutrons weighing 2(1.007276) + 2(1.008665) = 4.031882 u combine into a helium nucleus of mass 4.001506 u. The difference, 0.030376 u or 28.3 MeV, is the binding energy — the energy released on assembly, which had to leave, taking its mass with it.

Bound systems weigh less than their parts. A hydrogen atom weighs less than a proton plus an electron by 13.6 eV worth of mass. Every stable structure in nature is a mass deficit.

Now the surprise. A proton's mass is not the sum of its quarks. A proton contains two up quarks and one down quark, with rest masses of about 2.2, 2.2 and 4.7 MeV — a total of roughly 9 MeV. The proton's mass is 938 MeV.

Over 98 % of the proton's mass is not the quarks at all. It is the energy of the gluon field binding them and the kinetic energy of the quarks confined inside, converted into mass by E = mc^2 running backwards.

And since you are made of protons and neutrons, over 98 % of your mass is field energy and confinement energy, not the mass of any constituent particle. The Higgs field (Chapter 8.4) gives the quarks and electrons their intrinsic masses, and that accounts for less than 2 % of what a bathroom scale reads. The rest is E/c^2.

Relativistic collisions

In Newtonian mechanics, mass is conserved and energy may not be (in inelastic collisions). Relativity replaces both with one statement: total energy is conserved, and total momentum is conserved, and mass is not conserved at all.

Take two identical lumps of mass m, each moving at speed v towards each other, which collide and stick.

Momentum: zero before, zero after. The combined lump is at rest.

Energy: 2\gamma mc^2 before. After, the lump is at rest, so its energy is entirely rest energy Mc^2:

M = 2\gamma m

The resulting object is more massive than the two originals combined, by 2(\gamma-1)m, which is exactly the kinetic energy divided by c^2. The kinetic energy did not disappear into heat and vanish from the mass ledger — it became mass, in the form of thermal energy of the atoms inside, which contributes to the object's rest mass.

This is completely general. A hot object weighs more than a cold one. A compressed spring weighs more than a relaxed one. A charged battery weighs more than a flat one — by about 10^{-11} kg for a phone battery, which is real and unmeasurable.

And it is how particle physics makes new particles. Collide two protons at high energy and the products can include particles far heavier than the protons. Nothing was created from nothing; kinetic energy was converted into rest mass. At the LHC, two 6.5 TeV protons collide with 13 TeV available, which is enough to make a 125 GeV Higgs boson a hundred times over — the difficulty is not energy but that it happens in only about one collision in ten billion.

Where this shows up in your life

Nuclear power supplies about 10 % of world electricity, and every joule is a mass deficit computed exactly as above.

PET scans rely on the 0.511 MeV annihilation signature.

The Sun shines by converting mass at 4.3 million tonnes per second, and everything alive is running on it.

Particle accelerators used in medicine. Radiotherapy machines accelerate electrons to about 20 MeV, where \gamma = 40, and their design is entirely relativistic.

Smoke detectors contain americium-241, whose alpha decay is a mass-deficit reaction.

And the mass on your bathroom scale is 98 % binding energy and quark confinement energy, which is as direct an application of E = mc^2 as exists.

What the next chapter fixes

Space and time have been mixed by the Lorentz transformation, and energy and momentum have been mixed the same way. That parallel is not a coincidence, and taking it seriously produces a cleaner way of thinking about all of it: instead of space and time as separate stages, one four-dimensional spacetime with a geometry of its own. Chapter 6.5 introduces the invariant interval — the quantity every observer agrees on — draws the diagrams that make paradoxes obvious rather than puzzling, and shows exactly why nothing can outrun light without breaking cause and effect.