Momenergy and E = mc²

Warm-up

Three old friends you are about to need again: the stretch factor, a speed added to a speed, and the minus sign.

The puzzle

Here is a plan for reaching the stars faster than light. Build a good rocket and point it. Light the engine and never switch it off. Newton would say a steady push gives a steady gain, so sooner or later you must pass 1, the speed of light. Chapter 13 already showed you the catch: each burn adds the same rapidity, and the speed only creeps toward 1. That settles the speed. It opens a new question.

A real rocket muddies the question, because it throws its fuel out the back and gets lighter as it goes. So let’s make the push clean. Give a small probe a perfect mirror for a sail, and shine a giant laser on it from home. The laser never switches off, and the probe throws nothing away.

The laser never stops pouring light onto the sail, and the probe’s speed never reaches 1. So where does the endless push go?

Experiment

Below is a new kind of diagram. Momentum runs across and energy runs up, on the same scale, so the gold light lines sit at 45° as always. Each moving object is an arrow. The tip of each arrow is tied to a teal curve called its mass hyperbola. Drag a tip and that object speeds up or slows down, but the tip can never leave its curve.

There are two 1 kg balls of soft clay, A and B, heading for each other. Their arrows are joined head to tail. The bold teal arrow, from the origin (where the axes cross, bottom middle) to the last tip, is the pair taken together: the system. Its length has a name, the system mass, and the lab keeps that number hidden until you press Stick.

One mark to know before you start: ≡ marks a number every frame agrees on. In this lab, every teal number marked ≡ is a mass.

Electrons can be pushed much harder than clay. In the early 1960s William Bertozzi did the rocket experiment for real: he gave electrons more and more energy and timed how fast they went. Newton and Einstein disagree about what he should find. Here are both predictions, with his data.

What just happened?

It does sound like it. But in this course, mass is the number every frame agrees on, and A’s mass never moved off 1. Sagredo’s idea has a name and a history. We’ll meet it in a Myth box below.

The arrow every frame agrees on

Why multiply by γ? The same crash, reckoned in a rocket frame

Newton’s momentum is mass times velocity, \(mv\), and Newton says the total never changes in a collision. In Lin’s lab the clay crash obeys him: \(+0.6\) and \(-0.6\) make 0 before, and the lump sits still after. Zero equals zero. Fine.

Now let Rae ride alongside ball B. Her rocket drifts left at 0.6, so for her B is parked and A comes rushing in. After the crash, in her frame, the lump drifts right at 0.6. Let’s keep her books, one line at a time.

1. Ball A’s speed in Rae’s frame is \(0.6 \oplus 0.6 = 15/17\), from the warm-up (\(\oplus\) is Chapter 13’s velocity addition). Its stretch factor: \(1 - (15/17)^2 = 64/289\), and \(\sqrt{64/289} = 8/17\), so \(\g = 17/8\). Velocities add the Chapter 13 way, never by plain addition.
2. Newton’s ledger. Before: \(1 \times \tfrac{15}{17} + 1 \times 0 = 0.882\). After: a 2 kg lump at 0.6 carries \(2 \times 0.6 = 1.2\). The books don’t balance. A 2.5 kg lump would carry 1.5, which is further off. Same crash, same laws, different frame: if a law fails here, it isn’t a law.
3. A better guess. Newton’s momentum is mass times distance over time, \(m\,\dx/\dt\). Keep the distance \(\dx\): it is how far the ball gets in this frame, on this frame’s rulers. Change the time. Frames disagree about \(\dt\), so use the time every frame agrees on, the ball’s own wristwatch time \(\Delta\tau\). That gives \(p = m\,\dx/\Delta\tau\). Since \(\dt = \g\,\Delta\tau\) (Chapter 8), this is \(p = \g m v\). At everyday speeds \(\Delta\tau\) and \(\dt\) are almost equal, so this is still Newton’s \(mv\) there. Mass times a step along the ball’s worldline, per tick of its own wristwatch: that idea becomes the momenergy arrow below.
4. Rae’s new ledger. Before: A carries \(\tfrac{17}{8} \times \tfrac{15}{17} = \tfrac{15}{8} = 1.875\). B is parked, so it carries 0. After: we don’t know the lump’s mass yet, so call it \(M\). At 0.6, \(\g = \tfrac54\), so the lump carries \(\tfrac54 \times M \times 0.6 = 0.75\,M\). We pick the \(M\) that balances the books: \(0.75\,M = 1.875\), so \(M = 2.5\). That is the same 2.5 kg the collider showed in Lin’s lab. Two frames, one lump, one mass.
5. Now the test: energy, which we did not tune. Do the same to time that step 3 did to distance: \(m\,\dt/\Delta\tau = \g m\). That is the energy, \(E = \g m\). Before: \(\tfrac{17}{8} + 1 = \tfrac{25}{8}\). After: \(\tfrac54 \times 2.5 = \tfrac{25}{8}\). It balances. We chose \(M\) to fix the momentum. Nobody touched the energy, and it balances anyway. That is the sign the fix is right.

Newton’s momentum balances in one frame and breaks in the next. Galileo promised in Chapter 2 that the laws are the same in every free-float frame, so Newton’s rule can’t be the real one. Multiplying by \(\g\) repairs the books in every frame at once. Don’t take my word for it: test it below.

The momenergy arrow

So the right momentum is \(p = \g m v\) and the right energy is \(E = \g m\). Taylor and Wheeler gave the pair one name: momenergy. Put it on a diagram, energy up and momentum across, and you get the arrows from the lab. Take an object of mass 4 at speed 0.6. Then \(\g = 5/4\), so \(E = 5\) and \(p = 3\).

Now do Chapter 6’s favorite move and subtract squares: \(E^2 - p^2 = 25 - 9 = 16 = 4^2\). The mass comes back out. Try another speed. At \(v = 0.8\) the same object has \(E = 20/3\) and \(p = 16/3\), and \(400/9 - 256/9 = 144/9 = 16\) again. In general:

\[ \begin{gathered} \lab{E}^2 - \lab{p}^2 = \g^2 m^2 - \g^2 m^2 v^2 \\ = \g^2 m^2 (1 - v^2) = \inv{m^2}. \end{gathered} \]

Look at that equation next to the interval, \(\inv{\tau^2} = \lab{\dt^2} - \lab{\dx^2}\). Swap \(t\) for \(E\), \(x\) for \(p\) and \(\tau\) for \(m\), and it is the same sentence. That’s no coincidence. The momenergy arrow is the object’s mass times one step along its own worldline, so it points the way the object travels through spacetime. Different frames tilt it: E and p change, the way Δt and Δx did. Every frame agrees on its length, measured the spacetime way. That length is the mass. Mass is to momenergy what wristwatch time is to a trip.

The arrow also tells you the speed: how far it leans away from the E axis, across over up, \(v = p/E\). For anything with mass, \(|p| < E\), so \(v < 1\). That is the whole answer to the rocket puzzle, in one picture.

Push the probe with kicks from outside, each adding the same amount of energy, and the energy has no ceiling. But speed is \(p/E\), and as the arrow grows it swings toward the light line, \(p \to E\), without arriving. A push from outside changes where the tip sits on its hyperbola, never which hyperbola it is on, and on any hyperbola \(|p| < E\). Only something with zero mass rides the light line itself.

A real rocket is messier. It throws mass out the back, so as it burns its tip drops to lower and lower hyperbolas, and its energy doesn’t even grow without limit. But whatever is left still has mass, so \(|p| < E\) still holds and the speed still creeps toward 1. That is Chapter 13’s rocket stack: each stage adds the same rapidity, never the same speed.

Where did ½mv² go?

At rest, \(v = 0\) and \(\g = 1\), so \(E = m\): a resting object still carries energy, its rest energy. The extra energy that comes with motion is the kinetic energy:

\[ K = E - m = (\g - 1)\,m. \]

Newton wrote \(K = \tfrac12 m v^2\). Is he wrong? Only at high speed. In Chapter 8 you met the small-speed shortcut \(\g \approx 1 + \tfrac12 v^2\). Put it in: \(K \approx \tfrac12 v^2 \times m\). That’s Newton’s formula, hiding inside Einstein’s. Need a refresher?

At everyday speeds, then, Newton is superb. At 0.6 he isn’t: a proton at 0.6 has \(K = 0.25 \times 938.3 = 234.6\) MeV, while Newton’s formula gives only 168.9 MeV.

E = mc², or why a hot kettle weighs more

In our units energy and mass come in the same unit, and a resting object has \(E = m\). In everyday units you need a conversion factor, the speed of light squared, and the same statement reads \(E = mc^2\). That is all the famous equation says: energy in a body at rest is mass. Any energy. Not only nuclear.

Heat 1 kg of water from 20 °C to 100 °C. That takes 334.9 kJ (80 degrees at 4.186 kJ per kilogram per degree), and the water is now heavier by \(334\,900 / (8.988 \times 10^{16}) = 3.7 \times 10^{-12}\) kg. Nobody can weigh that difference. But it is exactly the clay story: jiggling molecules carry energy, and energy inside a body at rest is mass.

How much heat did the clay make? Its extra half kilogram, times \(c^2\), is \(4.49 \times 10^{16}\) J: enough to boil about 134 billion kettles. That is what two balls at 0.6 carry. Rockets that fast are not on the market.

A box of light has mass

Light is the edge case. A flash of light, a photon, has energy and momentum with \(E = |p|\), so \(E^2 - p^2 = 0\): its mass is zero, and its arrow lies right on the gold 45° line. Now take two photons of energy 1 flying in opposite directions. Their arrows, head to tail, go up-right then up-left and end at \(E = 2\), \(p = 0\). The system mass is \(\sqrt{4 - 0} = 2\). Two massless things make a system of mass 2.

So momenergy gives us two rules that work in every frame: total momenergy is conserved, and every object’s mass is the length of its arrow. Next, Nature gets a vote.

Ask Nature

Four experiments, from a tabletop electron race to the discovery of antimatter, all reading the same arrow.

Practice

Three tiers. Fractions, square roots and powers of ten like 3.7e-12 are all fine as answers. The core moves are always the same: find each arrow, add the arrows, read off the length.

◇ Back of the envelope

◆ Pencil and paper

★ Head-scratcher

For quick reference, every formula in this chapter is on the formula card.

Build it

Four small functions, and the collider above starts running on your physics. Once momenergy passes, the lab draws every arrow from your numbers; once invariantMass passes, the teal system mass is yours too.

function total(...items) {
  let sum = 0;
  for (const x of items) sum += x.price;
  return { sum, count: items.length };
}
total({ price: 3 }, { price: 4 });   // { sum: 7, count: 2 }

Checkpoint

Five graded items and one to write. Feedback waits until you submit. Items marked “Key idea” must all be right to pass.

Explain it to Sagredo

Going further

Reading: Taylor and Wheeler, Spacetime Physics (2nd ed., 1992), chapters 7 and 8, where the word “momenergy” comes from; free at eftaylor.com. The Feynman Lectures on Physics, Vol. I, chapters 15 and 16, free at feynmanlectures.caltech.edu. Feynman uses an older bookkeeping in which his \(m\) changes with speed; read his \(m\) as our \(E\).