Relativity That ClicksEpisode 4 / E=mc² ── Mass Is the Energy of Rest

Just as spacetime had an invariant separation s², energy and momentum have the very same "minus" invariant

E=mc² ── Mass Is the Energy of Rest Energy and momentum, too, have an invariant joined by the same "minus" as the spacetime interval: (mc²)² = E² − (pc)².
For an object at rest, E = mc² ── mass turns out to be another name for the energy a resting object holds.

Tools you'll need: the invariant s² from Episode 3, the γ from Episode 2 This episode's invariant: (mc²)² = E² − (pc)²

Let's take the world's most famous formula, $E=mc^2$, and instead of memorizing it, derive it. In Episode 3 we saw that spacetime has an invariant $s^2=(ct)^2-x^2$ that is the same for every observer. It turns out energy $E$ and momentum $p$ also come as a pair, like time and space, and each observer measures different values for them. And exactly the same "minus" combination as in spacetime ── $(mc^2)^2=E^2-(pc)^2$ ── is the one thing that comes out the same invariant for everyone. The identity of that invariant is mass. Read this formula for an object at rest (momentum $p=0$) and you get $E=mc^2$ ── mass is nothing but the energy a resting object carries within it. This episode simply translates the geometry of Episode 3 into the language of energy, and out drop both $E=mc^2$ and the answer to "why is the speed of light a wall?" (the homework from Episode 1), all at once.

01Were mass and energy really separate things?

In Newton's world, mass and energy were unrelated, distinct things. Mass was the amount of matter; energy was the vigor of motion or heat. Each conserved on its own. Relativity tears down this wall ── mass and energy are two faces of the same thing. The bridge that connects them is the "minus" invariant structure we met in Episode 3.

02Energy and momentum have a spacetime interval too

Just as a position in spacetime was a pair $(ct,\,x)$, a particle's motion can be written as a pair $(E,\,pc)$ (energy and momentum). Change observer, and this $E$ and $pc$ shuffle around separately, according to the $\gamma$ of Episode 2 ── exactly as time and space did. And if you combine them with the same "minus," an invariant appears.

This episode's invariant ── the "spacetime interval" of energy and momentum

$$(mc^2)^2=E^2-(pc)^2\qquad(\text{the same for every observer})$$

The look of it is a dead ringer for $s^2=(ct)^2-x^2$ from Episode 3. Just substitute $ct\to E$ and $x\to pc$. Change observers and both $E$ and $pc$ move, but this combination ── $mc^2$ ── alone stays invariant. That invariant is the true identity of the mass $m$.

03Come to rest, and E = mc²

Let's read this invariant in the simplest case ── an object at rest (momentum $p=0$).

Try it ── set p=0 $$(mc^2)^2=E^2-0\quad\Rightarrow\quad E=mc^2$$

It's at rest, its kinetic energy ought to be zero ── and yet its energy is not zero. A leftover of $mc^2$ remains. This is the rest energy ── it means that mass itself is a colossal lump of energy. Because $c^2$ is enormous ($9\times10^{16}$), a tiny amount of mass becomes a preposterous amount of energy. One gram of mass is $9\times10^{13}$ J ≈ about 20 kilotons, comparable to the Hiroshima bomb.

For an object in motion, $E=\gamma mc^2$ (the $\gamma$ of Episode 2). Made dimensionless, $E/mc^2=\gamma$ ── energy is $\gamma$ times the rest energy. At low speed $\gamma\approx1+\tfrac12\beta^2$, so ──

At low speed, Newton's kinetic energy shows its face $$E=\gamma mc^2\approx mc^2+\tfrac12 mv^2$$

$mc^2$ (rest energy) + $\tfrac12mv^2$ (the familiar kinetic energy). Newton's $\tfrac12mv^2$ was just the change left over after subtracting the rest part from the relativistic energy. Here too, "Newton is the β→0 limit."

04Let's play with it ── the energy–momentum hyperbola

The figure below is the twin of the spacetime diagram from Episode 3. The horizontal axis is momentum $pc$, the vertical axis energy $E$ (measured in units of the rest energy $mc^2$). Raise $\beta$ with the slider and that particle's $(pc,\,E)$ ── again slides along the invariant hyperbola $E^2-(pc)^2=(mc^2)^2$. Both $pc$ and $E$ grow, yet $mc^2$ (the bottom tip of the hyperbola) never budges.

The bar on the right breaks down the energy. The lower rest energy $mc^2$ stays fixed, while the kinetic energy stacked on top swells with $\beta$. Push $\beta\to1$ and the hyperbola closes in on the 45° asymptote ($E=pc$), and the energy $E$ diverges without bound. Here is the answer to the homework we left hanging in Episode 1 ── getting matter to the speed of light would take infinite energy, so $\beta=1$ is a wall you cannot cross.

Figure: the energy–momentum hyperbola E²−(pc)²=(mc²)² (the twin of the Episode 3 spacetime diagram). Raise β and the point slides along the hyperbola; pc and E both grow, but mc² stays invariant. Right bar = the breakdown of E (rest mc² + kinetic). As β→1, E→∞ = the wall of light speed.

β = v/c 0.60

particle's (pc, E) / invariant hyperbola rest energy mc² (invariant) light E=pc (45° asymptote)

05Zero mass, and the wall of light speed

Put $m=0$ into the invariant, and everything ties together.

Zero mass ── all energy is motion, and hence c

$$m=0\ \Rightarrow\ E^2=(pc)^2\ \Rightarrow\ E=pc,\quad \beta=\frac{pc}{E}=1$$

A massless particle (a photon) has zero rest energy, and all of its energy is motion, $pc$. And $\beta=pc/E=1$ ── it always travels at the speed of light. This shakes hands completely with "Fields That Click" Episode 2 ("a massless field goes at c") and Episode 3 ("massless means infinite range"). Zero mass cannot come to rest and must go at c; having mass means you can never reach c ── mass was also "the license to be able to sit still."

And the true destructive power of $E=mc^2$ is that mass and energy can trade places. Mass is "frozen energy," and when conditions are right it melts back out. The Sun shines by converting 4 million tons of mass into light energy every second; nuclear fusion and fission release the difference in binding energy as a difference in mass; and an electron and a positron, when they meet, convert their entire mass into light (energy) and vanish (annihilation). Conversely, as we saw in "Fields That Click" Episode 6, 99% of a proton's mass is the confined energy of quarks and gluons turned into mass via $E=mc^2$ ── your body weight is mostly frozen energy. Mass and energy were two faces of one and the same thing.

◇ ◇ ◇

The honest line ── which "mass" do we mean?

The invariant $(mc^2)^2=E^2-(pc)^2$, the rest energy $E=mc^2$, $E=\gamma mc^2$ (which at low speed becomes $mc^2+\tfrac12mv^2$), $E=pc$ and $\beta=1$ for zero mass, the mass⇄energy conversion in nuclear reactions and annihilation, and the fact that most of a nucleon's mass comes from binding energy ("Fields That Click" Episode 6) ── all of these are established physics.

The $m$ here is rest mass (invariant mass) ── a quantity intrinsic to the particle, the same for all observers. The old textbooks' "relativistic mass $\gamma m$ that grows with speed" is a convention no longer used (what grows is the energy $E=\gamma mc^2$, not the mass). The $m$ in this article does not change when the object moves. Also, $E=mc^2$ is the energy of a resting object; a moving object has $E=\gamma mc^2$, and in general $E^2=(pc)^2+(mc^2)^2$ is the correct form. The "1 g ≈ 20 kt" figure is for an idealized complete conversion; real nuclear reactions release only a tiny fraction of the mass.

Practice problems (solvable with just this episode's formulas)

For a particle with $\beta=0.6$ ($\gamma=1.25$), how many times the rest energy is its energy? What fraction of the rest energy is its kinetic energy?
See the answer
$E=\gamma mc^2=1.25\,mc^2$. 1.25 times the rest energy. Kinetic energy $=E-mc^2=0.25\,mc^2$, or 25% of the rest energy.
Why can matter never reach the speed of light $\beta=1$? Answer in the language of energy.
See the answer
In $E=\gamma mc^2$, as $\beta\to1$ we have $\gamma\to\infty$, i.e. $E\to\infty$. Reaching light speed would take infinite energy, so with finite energy you can never get there. That's the true nature of the "wall" from Episode 1.
Apply the invariant to a massless photon: what is the relation between E and p? What is β?
See the answer
With $m=0$, $E^2=(pc)^2\Rightarrow E=pc$. And $\beta=pc/E=1$. Zero mass always travels at light speed (it cannot rest).
What does "most of your body weight is frozen energy" mean? Connect it to "Fields That Click" Episode 6.
See the answer
Your weight is mostly protons and neutrons, and about 99% of their mass is the binding energy of confined quarks and gluons turned into mass via $E=mc^2$. Mass is "frozen energy"; the bare mass from the Higgs is only a small part.

Episode 4 SummaryMass is rest energy ── (mc²)² = E² − (pc)²

Energy $E$ and momentum $p$, like time and space, come as a pair and shuffle around from observer to observer. But $(mc^2)^2=E^2-(pc)^2$, built with the very same "minus" as Episode 3, alone stays invariant ── and that invariant is the mass $m$. For an object at rest ($p=0$), $E=mc^2$: mass is another name for the energy a resting object holds. In motion, $E=\gamma mc^2$ ($E/mc^2=\gamma$), and at low speed $mc^2+\tfrac12mv^2$ — Newton's kinetic energy shows its face.

On the hyperbola diagram, raising $\beta$ slides the point and $pc,E$ grow, but $mc^2$ stays put; as $\beta\to1$, $E\to\infty$ ── the reason light speed is a wall (the Episode 1 homework). With zero mass, $E=pc,\ \beta=1$ (always at light speed, a handshake with "Fields That Click"). And mass and energy trade places ── the Sun, nuclear reactions, annihilation, and 99% of your body weight is frozen energy. Mass, an invariant, runs the whole show from behind the scenes of the unit-bearing E and p.

Coming next ── Episode 5 "Simultaneity is up to the observer ── but causality is invariant." So far, both time dilation and E=mc² we've accepted quietly. Next comes the heart of what is most counterintuitive in relativity ── the very concept of "simultaneous" differs from observer to observer. Two events simultaneous for one person can appear ordered for another. So do cause and effect get flipped too? No ── the sign of the Episode 3 spacetime interval $s^2$ (timelike or spacelike) keeps the order of causality alone invariant for everyone. The watchman for that is "nothing exceeds c." We'll take it all in with the light cone.

This document is Episode 4 of the "Relativity That Clicks" series, a reading piece for physics-loving high-schoolers and undergraduates. The four-momentum invariant $(mc^2)^2=E^2-(pc)^2$, the rest energy $E=mc^2$, $E=\gamma mc^2$ and its low-speed expansion $E\approx mc^2+\tfrac12mv^2$, $\beta=pc/E$, $E=pc$ and $\beta=1$ for zero mass, the fact that $\beta\to1$ implies $E\to\infty$ so a massive particle cannot reach light speed, the mass–energy conversion in nuclear reactions and annihilation, and the fact that most of a nucleon's mass comes from binding energy ── all of these are established standard physics. That the $m$ here is the invariant mass (rest mass) and that the speed-dependent "relativistic mass $\gamma m$" is a convention not used in the modern approach (what grows is energy, not mass), that $E=mc^2$ is the relation at rest while in general $E^2=(pc)^2+(mc^2)^2$ is correct, and that "1 g ≈ 20 kt" is the ideal complete-conversion value, are all noted in the "honest line" section. The figure is a schematic of the energy–momentum hyperbola $E^2-(pc)^2=(mc^2)^2$ (normalized to $mc^2$ as unit). ── To print, use your browser's "Print" and "Save as PDF" (in the print version the slider and answers are frozen and hidden). Neighboring episodes: Episode 3 The Spacetime Interval / Table of Contents / sister series Fields That Click, Episode 6.

Print / Save as PDF: ⌘+P (Ctrl+P on Windows). On screen, use the β slider to slide the point along the hyperbola E²−(pc)²=(mc²)², and E diverges as β→1. Click "See the answer" to open a solution.

E=mc² ── Mass Is the Energy of Rest Energy and momentum, too, have an invariant joined by the same "minus" as the spacetime interval: (mc²)² = E² − (pc)².For an object at rest, E = mc² ── mass turns out to be another name for the energy a resting object holds.

01Were mass and energy really separate things?

02Energy and momentum have a spacetime interval too

03Come to rest, and E = mc²

04Let's play with it ── the energy–momentum hyperbola

05Zero mass, and the wall of light speed

Episode 4 SummaryMass is rest energy ── (mc²)² = E² − (pc)²

E=mc² ── Mass Is the Energy of Rest Energy and momentum, too, have an invariant joined by the same "minus" as the spacetime interval: (mc²)² = E² − (pc)².
For an object at rest, E = mc² ── mass turns out to be another name for the energy a resting object holds.