Mass That ClicksBonus ② / Clearing up two familiar misconceptions

Neither "you get heavier the faster you move" nor "mass is conserved" is quite accurate

Mass Is Not Conserved In Episode 1 we said mass is the "invariant floor." But two misconceptions survive out in the world.
Going back to E=mc² and the Episode 1 formula, we pin down the true meaning of "invariant mass."

Tools you'll need: Episode 1's \(E^2=(mc^2)^2+(pc)^2\), Episode 3's "not the sum of the parts" What's conserved isn't mass but energy and momentum

In Episode 1, mass was "the floor of energy that remains, that you can't erase, once you strip away momentum." This "floor" has an important property ── it's the same for everyone (invariant). Yet out in the world, two misconceptions that conflict with this property stubbornly survive ── "you get heavier the faster you move" and "mass is conserved." This time we correct both, going back to E=mc² and the Episode 1 formula. Where we end up is a little surprising ── the sum of rest masses is not conserved, and you can build a massive system out of nothing but massless particles.

01Misconception 1: "you get heavier the faster you move"

Old textbooks had a phrasing: "relativistic mass \(m_{\rm rel}=\gamma m\) (larger the faster you go)." But we don't use it anymore. Go back to the Episode 1 formula and the reason is clear.

Seen through the Episode 1 formula ── what actually increases
$$E^2=(mc^2)^2+(pc)^2,\qquad E=\gamma mc^2,\quad p=\gamma mv$$

Move fast and both \(E\) and \(p\) increase. But \(m\) (the floor, the intercept) doesn't budge. What increases is energy, not mass.

The reason we dropped the phrase "relativistic mass" is that it fools you into thinking \(m\) changes with speed or direction, and it muddles the \(m\) in \(E=mc^2\). Mass is a "frame-independent property" of a thing ── speed it up, turn it around, and it doesn't change.

02Mass is "invariant" ── the same for everyone

This is the crux. The mass \(m\) is a Lorentz invariant. Viewed at rest or viewed while flying by at tremendous speed, it's the same value. This is Episode 1's "floor" not changing its height when you change frame.

Mass = the "length" of the four-momentum (invariant)
$$(mc^2)^2 = E^2-(pc)^2\qquad(\text{the same value in every frame})$$

By contrast, energy \(E\) is frame-dependent ── at rest it's \(mc^2\), viewed in motion it's \(\gamma mc^2\). The same object gives a different \(E\) to different observers, but the same \(m\) to everyone. The "you get heavier when moving" misconception mixed up this distinction. That's why the accurate term today is "invariant mass" rather than "rest mass."

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03Misconception 2: "mass is conserved" ── the sum is not conserved

The "conservation of mass" you learn in chemistry is actually an approximation. In a reaction, the total rest mass is not conserved. The most vivid example is the annihilation of an electron and a positron.

Let's try it ── mass vanishes in annihilation

electron + positron → 2 photons

$$\underbrace{0.511+0.511}_{\text{has rest mass}}\ \text{MeV}\ \longrightarrow\ \underbrace{2\ \text{photons}}_{\text{zero rest mass}}\ (\text{total energy }1.022\ \text{MeV})$$

Before the reaction there's \(1.022\) MeV worth of rest mass. After, only zero-rest-mass photons. The sum of rest masses vanished, \(1.022\to0\) ── but the energy stays \(1.022\) MeV, conserved as light. The "mass defect" of nuclear fusion (the Sun) and fission (reactors) is the same story.

What is really conserved

Conserved: total energy and total momentum (four-momentum).
Not conserved: the "sum" of rest masses (\(\sum m_i\)).

04A massive system can be built from massless parts

This is the most interesting part today. The mass of a system made of several particles (the system's invariant mass \(M\)) is not the sum of the parts' rest masses. It's determined by the total energy and total momentum.

The system's invariant mass (not the sum of the parts)
$$(Mc^2)^2=\Big(\textstyle\sum E_i\Big)^2-\Big(\textstyle\sum \vec p_i\,c\Big)^2$$

And then something astonishing happens ── even two zero-rest-mass photons can, as a system, have a mass \(M\neq0\). Two photons flying in opposite directions have their momenta cancel (\(\sum\vec p=0\)), yet the total energy remains (\(\sum E\neq0\)), so \(M\neq0\). Check it in the figure below by changing the opening angle.

Figure: the mass of the "system" formed by two photons of energy \(E\) (zero rest mass, amber). Change the opening angle θ and the system's mass \(Mc^2=2E\sin(\theta/2)\) changes. Same direction (θ=0) gives \(M=0\); opposite directions (θ=180°) gives the maximum
Move the opening angle and the system's mass changes.
photon (zero rest mass) system mass M

This is the purest form of Episode 3's "99% of the proton is the energy of the gluon field." Mass is not a "property of the parts" but a "combination of the system's energy and momentum." Episode 1's floor \((mc^2)^2=E^2-(pc)^2\) holds for the whole system, not just a single particle ── that is the true meaning of "invariant mass."

An honest line ── the precise meaning of "not conserved"

What "mass is not conserved" refers to is strictly that the "sum" of individual rest masses \(\sum m_i\) is not conserved. The total energy and total momentum of a closed, isolated system (= the system's invariant mass \(M\)) are properly conserved. Everyday "conservation of mass" is just a low-energy approximation that holds well enough in practice, because only a tiny fraction of the energy shows up as a mass difference.

Also, relativistic mass \(\gamma m\) is less "wrong" than old notation; the modern standard calls only invariant mass "mass." This time we're doing standard special relativity, unrelated to the watchword \(c\cdot t=\text{constant}\).

Practice problems (solvable with today's material)
  1. What is wrong with "you get heavier the faster you move"?
    See the answer
    What increases is the energy \(E=\gamma mc^2\) and momentum \(p=\gamma mv\), while the mass \(m\) is invariant (Lorentz invariant, Episode 1's "floor"). The speed-dependent "relativistic mass \(\gamma m\)" is old notation and isn't used anymore.
  2. In \(e^+e^-\to2\gamma\), what is conserved and what isn't?
    See the answer
    Conserved: total energy and total momentum. Not conserved: the sum of rest masses (the \(1.022\) MeV worth all moves into the energy of the zero-rest-mass photons).
  3. Why can two zero-rest-mass photons form a massive system?
    See the answer
    The system's invariant mass is \((Mc^2)^2=(\sum E)^2-(\sum \vec p\,c)^2\). Two photons flying in opposite directions have their momenta cancel (\(\sum\vec p=0\)) while the total energy remains (\(\sum E\neq0\)), so \(M\neq0\). Mass is not the sum of the parts but a combination of the system's energy and momentum.

Bonus ② summaryMass is "invariant," but its "sum" is not conserved

The mass \(m\) is Lorentz invariant (the same for everyone) ── so "you get heavier the faster you move" is a misconception; what increases is energy and momentum. Meanwhile, in a reaction the "sum" of rest masses is not conserved (annihilation, mass defect); what's conserved is total energy and total momentum. And the mass of a system is not the sum of the parts but \((Mc^2)^2=(\sum E)^2-(\sum \vec p c)^2\) ── even two massless photons can build a massive system.

Episode 1's floor \((mc^2)^2=E^2-(pc)^2\) is an "invariant length" that holds for one particle and for a whole system. That is what "invariant mass" is, the same story as Episode 3's 99%-of-the-proton. Invariant when you set it in motion, but a different mass once a reaction changes the system ── with this you can tell two easily-confused things apart.

This document is Bonus ② of the "Mass That Clicks" series, a piece for physics-loving high-schoolers and undergraduates. The following are all established content of special relativity: that relativistic mass (\(\gamma m\)) is not standardly used in modern physics, and "mass" refers to the invariant mass (a Lorentz invariant, \((mc^2)^2=E^2-(pc)^2\)); that energy \(E=\gamma mc^2\) and momentum \(p=\gamma mv\) are frame-dependent and increase with speed; that in a reaction the total rest mass is not conserved while total energy and total momentum (four-momentum) are (annihilation \(e^+e^-\to2\gamma\), the mass defect of nuclear reactions); that the invariant mass of a multi-particle system is given by \((Mc^2)^2=(\sum E_i)^2-(\sum\vec p_i c)^2\) and is not the sum of the parts' rest masses; and that even a system of massless particles can have a nonzero invariant mass. The total four-momentum of a closed, isolated system (and hence the system's invariant mass) is conserved. The figure is a schematic of a two-photon system in units of \(E=1\), with \(Mc^2=2E\sin(\theta/2)\). ── To print, use your browser's "Print" and "Save as PDF" (in the print version the slider and answers are static and hidden).

Print / PDF: Ctrl+P (Cmd+P on Mac). On screen, change the opening angle with the slider and a system's mass emerges from two massless photons. Click "See the answer" to open each solution.