Energy, momentum, and mass
A helium-4 nucleus is made of two protons and two neutrons. Weigh it, and it is lighter than those four particles weighed separately, by about three-quarters of a percent. Where did the missing mass go, and what does it have to do with the energy the Sun pours out?
The previous essay showed that time and length depend on the frame. Newton's momentum and kinetic energy are built from them, and they fail near the speed of light. This essay finds the replacements. They contain $E = mc^2$, which says that a body's mass measures the energy it has when at rest, and a relation between energy, momentum and mass that holds in every frame. That relation is what the last optional essay uses to make quantum mechanics relativistic.
Momentum and energy, rebuilt
In Newton's mechanics momentum is $mv$, and total momentum is conserved in collisions. Apply that rule in one frame, transform to another with the Lorentz transformation, and conservation fails: $mv$ is conserved in one frame but not another, which the principle of relativity forbids. The quantity that is conserved in every frame is
$$p = \gamma m v .$$Here $m$ is the ordinary mass, the one measured when the body is at rest. It is the same in every frame. (Older books, Feynman's among them, called $\gamma m$ the mass of a moving body and said that mass grows with speed. Most physicists no longer use that convention: the factor $\gamma$ belongs with the velocity, not with the mass.) At low speed $\gamma \approx 1$ and Newton's $mv$ returns; near $c$ the momentum grows without limit, which is why no amount of pushing brings a massive body to the speed of light.
The matching energy, the one conserved together with $p$, is
$$E = \gamma m c^2 .$$At low speed, with the approximation from the reminders, $E \approx mc^2 + \tfrac12 mv^2$: Newton's kinetic energy plus a constant. In Newton's mechanics a constant added to energy would be meaningless. In relativity it is not. A body at rest has energy
$$E_0 = mc^2 ,$$its rest energy, and the kinetic energy is what $\gamma$ adds on top, $(\gamma - 1)mc^2$.
The relation that holds in every frame
Energy and momentum both change from frame to frame, but one combination does not, in the same way that the spacetime interval did not. Squaring $E = \gamma mc^2$ and $p = \gamma mv$ and subtracting,
$$E^2 - p^2c^2 = \gamma^2 m^2c^4\left(1 - \frac{v^2}{c^2}\right) = m^2c^4 ,$$so
$$E^2 = p^2c^2 + m^2c^4 .$$Every observer, however they are moving, finds the same $m$ from a body's energy and momentum. That is what mass is, in relativity: the frame-independent part of energy and momentum. The relation also covers things with no mass. Setting $m = 0$ gives $E = pc$, the photon's relation used in the photons essay, where it was stated without proof. A massless particle cannot be at rest and always moves at $c$.
Energy has inertia
A thought experiment shows what $E_0 = mc^2$ means. Two identical lumps of putty, each of mass 1 kg, fly at each other at $0.6c$ ($\gamma = 1.25$) and stick. Momentum before is zero, so the merged lump is at rest. Energy is conserved, so its rest energy equals the total energy before: $2\times1.25\,mc^2 = 2.5$ kg$\times c^2$. The merged lump has a mass of 2.5 kg, not 2. In general, a system's mass measures its total energy in the frame where its total momentum is zero. The kinetic energy has not vanished into heat that weighs nothing; the heat is still there, and it has mass.
So there is no such thing, in relativity, as a collision that loses energy without changing mass. Any energy a body contains, heat, the motion of its parts, the energy of the forces binding it, contributes $E/c^2$ to its mass. Heating a kilogram of water by 100 degrees adds $4\times10^{5}$ J and so $5\times10^{-12}$ kg: real, but far below what any balance can detect. Only in nuclear physics, where binding energies are millions of times larger than chemical ones, does the change become easy to weigh.
Worked example: the helium nucleus
Masses are quoted in atomic mass units, u, with $1\ \text{u} = 931.5\ \text{MeV}/c^2$. Using whole atoms, so that the electrons balance on both sides:
$$\begin{aligned} 2\,m(^1\text{H}) + 2\,m_n &= 2(1.007825)\\[2pt] &\quad + 2(1.008665)\\[4pt] &= 4.032980\ \text{u},\\[4pt] m(^4\text{He}) &= 4.002603\ \text{u}. \end{aligned}$$The helium atom is lighter by $0.030377$ u, which is $0.030377\times931.5 = 28.3$ MeV: 0.75 percent of the mass. That is the binding energy of the helium nucleus. To pull it apart into two protons and two neutrons you must supply 28.3 MeV, and the pieces then weigh exactly that much more. Assembled, the nucleus has given that energy away, and its mass has fallen by $E/c^2$. The missing mass left as energy, carried off by radiation and moving particles.
The Sun. The Sun builds helium from four hydrogen nuclei, which releases 26.7 MeV per helium nucleus: a little less than 28.3 MeV, because two of the protons must first turn into the slightly heavier neutrons. That energy, apart from about 2 percent carried off by neutrinos, is what the Sun radiates. Its output, $3.83\times10^{26}$ W, corresponds to losing $3.83\times10^{26}/c^2 = 4.3\times10^{9}$ kg of mass every second, over four million tonnes. Converting one gram entirely into energy would release $9\times10^{13}$ J, about the energy of 21 kilotonnes of TNT.
Check: chemistry. Binding an electron to a proton releases 13.6 eV, a fraction $1.4\times10^{-8}$ of the atom's rest energy. The same principle holds, but the mass change is too small to weigh, which is why chemists could treat mass as exactly conserved.
Optional: why $p = \gamma mv$ and not $mv$
Consider a glancing collision seen from two frames. One ball is nearly at rest in the ground frame, the other nearly at rest in a frame moving along $x$ at $v$; each has a small sideways velocity along $y$, and the collision reverses them. Sideways distances are the same in both frames, but by time dilation each frame judges the other ball's clock slow, so it judges the other ball's sideways speed smaller by $\gamma$. If momentum were $mv$, the two sideways kicks could not cancel in both frames at once. Weighting each velocity by the $\gamma$ of the ball's own motion makes them cancel in every frame. Carrying the same argument through for the energy, so that energy and momentum transform together the way $ct$ and $x$ do, gives $E = \gamma mc^2$.
Two questions
1. A fast electron. An electron (rest energy 0.511 MeV) is given 1.0 MeV of kinetic energy. What are its total energy, its momentum (as $pc$, in MeV) and its speed?
Worked answer
Total energy $E = 0.511 + 1.0 = 1.511$ MeV. From $E^2 = p^2c^2 + m^2c^4$, $pc = \sqrt{1.511^2 - 0.511^2} = 1.422$ MeV. Speed: $v/c = pc/E = 0.941$.
The slip to avoid is using $\tfrac12 mv^2 = 1.0$ MeV, which gives $v/c = \sqrt{2\times1.0/0.511} = 1.98$: faster than light. When the kinetic energy is comparable with the rest energy, Newton's formula is not a small error but a wrong answer.
2. Is mass converted into energy? People often say that in the Sun "mass is converted into energy". Using the helium example, what is a more careful statement?
Worked answer
Energy is conserved throughout; nothing turns from one kind of stuff into another. Before fusion, part of the energy is locked up as the rest energy of the separate particles. After, the helium nucleus has less rest energy, and the difference is carried away as the kinetic energy of particles and as radiation. The helium's mass is lower because it contains less energy. Total energy is unchanged, and so is the total mass of everything, radiation included, if you count the radiation's energy divided by $c^2$.
The tempting wrong turn is to picture mass as a fuel that disappears. What $E = mc^2$ says is that mass and rest energy are the same quantity in different units; the Sun loses mass because it loses energy.
Further reading
- Einstein, Relativity: The Special and the General Theory (1916, tr. Lawson), §XV "General Results of the Theory": the argument that a body absorbing energy gains inertia in proportion.
- Feynman, Lectures on Physics, Vol. I, Ch. 15, §15-8 "Relativistic dynamics" and §15-9 "Equivalence of mass and energy", and Ch. 16 "Relativistic Energy and Momentum", §16-5 "Relativistic energy". Feynman's Chs. 15–16 use the older convention throughout, explicitly in §16-4 "Relativistic mass", which also gives the glancing-collision argument of the fold-out above.
Where this leaves us
The helium nucleus is lighter than its parts because it contains less energy: assembling it releases 28.3 MeV, and every joule released takes $1/c^2$ kilograms of mass with it. That is $E_0 = mc^2$, one consequence of rebuilding momentum and energy so that they are conserved in every frame. The other is the relation $E^2 = p^2c^2 + m^2c^4$, which picks out mass as the part of energy and momentum that every observer agrees on.
That relation is the key to the last optional essay, where Dirac asks what wave equation an electron obeys if it must respect it. Before that, there is a short excursion into gravity, where Einstein carried the idea that clocks depend on circumstances one step further. It is not needed for the Dirac essay, and readers heading straight there can skip it.