Nothing you have touched is elementary#

Pick up your coffee mug. Its mass is almost entirely protons and neutrons, and a proton is not a fundamental particle. It is three quarks held together by gluons, and the quarks’ own masses account for only about one percent of its mass. The rest is the energy of the quarks and gluons moving and interacting inside it, so the proton’s mass comes almost entirely from the strong force that holds it together. Protons and neutrons then bind into nuclei, nuclei grab electrons to make atoms, atoms make molecules, and molecules make mugs.

So the visible Universe is a ladder of bound states. The first rungs, protons and light nuclei, were climbed within the first few minutes after the Big Bang. Atoms formed a few hundred thousand years later, and heavier nuclei were built inside stars.

Now look at how we usually talk about dark matter. We picture a single elementary particle, a WIMP or an axion, drifting through the galaxy and occasionally leaving a trace in a detector. That picture might be right. But if the dark sector has forces of its own, I find it hard to see why its particles would politely stay apart forever when ours did not. Dark matter could have climbed a ladder too. That possibility goes by the name composite dark matter, and once you take it seriously, almost everything we expect about dark matter changes.

The first rung: a dark hydrogen atom#

The simplest thing a dark sector can build is a copy of hydrogen. Give it a heavy dark proton, a light dark electron and a dark photon that pulls them together with coupling $\alpha'$. How big is the atom, and how tightly is it bound? You don’t need to solve the Schrödinger equation to find out. The uncertainty principle is enough.

Confine the dark electron to a region of size $r$. Its momentum is then at least of order $p \sim 1/r$ (in natural units, $\hbar = c = 1$), so it carries kinetic energy $p^2/2\mu' \sim 1/(2\mu' r^2)$, where $\mu'$ is the reduced mass. The attraction contributes $-\alpha'/r$. The total energy is

Squeezing the electron closer lowers the potential energy but raises the kinetic energy, so there is a best size in between. Setting $dE/dr = 0$ gives

and putting this back into $E(r)$ gives

Both results agree exactly with the hydrogen ground state, which is a bit of luck for such a rough argument, but the scalings are what matter. A heavier dark electron or a stronger dark force makes a smaller, more tightly bound atom. For our own hydrogen, $\alpha = 1/137$ and $\mu \approx 511$ keV give the familiar $13.6$ eV and $0.53$ Å.

This “atomic dark matter” ¹ already behaves differently from a single particle. In the early Universe it is a hot plasma of dark ions and dark photons, which rings with sound waves the same way our own plasma did before the CMB was released. Later the atoms bump into each other with roughly their geometric size as a cross section, so they are self-interacting. They can also cool by radiating dark photons, which lets them sink and clump more than ordinary cold dark matter would. The shapes of real galaxies limit how much of this is allowed.

Copying QCD instead of QED gives dark protons, dark pions and dark glueballs, and people have built a whole zoo of models this way. I’ll skip that zoo here, because the more interesting question for me is what happens if the binding never stops.

Why our nuclei stop growing, and dark ones might not#

In our Universe, Big Bang nucleosynthesis turns protons and neutrons into deuterium, helium and a pinch of lithium, and then stops. Two things in particular stop it. There are no stable nuclei with five or eight nucleons, so the chain of fusions hits a gap. And protons repel each other electrically, so as the Universe cools the Coulomb barrier shuts fusion off. Even inside stars, where fusion goes further, it ends at iron. Past iron the electrical repulsion between all those protons grows faster than the strong force that holds them, so fusing heavier nuclei no longer releases energy.

You can see where iron comes from with a few lines of the liquid-drop model, which treats a nucleus as a drop of incompressible nuclear fluid. A nucleus with $A$ nucleons and $Z$ protons has binding energy

The first term says each nucleon is bound to its neighbors, so binding grows with volume. The second corrects for nucleons on the surface, which have fewer neighbors. The third is the electrical repulsion of $Z$ protons packed into a sphere of radius $\propto A^{1/3}$. Take $Z \approx A/2$ and divide by $A$ to get the binding per nucleon,

The surface penalty shrinks as nuclei grow, while the Coulomb penalty grows, so there is a sweet spot. Setting $d(B/A)/dA = 0$,

With one common set of fitted values, $a_S \approx 17.8$ MeV and $a_C \approx 0.71$ MeV, this gives $A_\ast \approx 50$. The real peak is at iron and nickel, $A \approx 56$ to $62$, which is remarkably close for something this crude. Most of the gap comes from setting $Z = A/2$. Heavier nuclei carry extra neutrons, which lowers their Coulomb cost, and adding that effect back moves the peak to about $A = 60$.

Now do the same for a dark sector whose nucleons attract through a short-range force and feel no long-range repulsion. Set $a_C = 0$, and the binding per nucleon becomes

which only ever increases with $A$, approaching $a_V$ from below. There is no sweet spot. Every time two dark nuclei merge into a bigger one energy is released, so nothing tells them to stop.

When people followed this through the early Universe, the answer was startling. Dark nucleosynthesis can build nuclei with a hundred million constituents or more ². These are not particles in any everyday sense. They are tiny chunks of dark nuclear matter.

The same thing happens in an even simpler model, which is the one I work on. Take a dark fermion and a light scalar field that it couples to. Exchanging the scalar pulls every fermion towards every other one, and the fermions collect into large bound states, usually called nuggets ³. Once a nugget is bigger than the range of the force, each fermion only feels its neighbors, and the nugget behaves like a drop of nuclear matter. Its density stops changing as it grows. These nuggets can also be bound remarkably tightly. The binding energy per fermion can be an order-one fraction of the fermion’s own mass ⁴. A sizeable piece of the dark matter’s mass simply turns into energy when the nugget forms.

Keep that in mind, because the obvious follow-up question is where all that energy goes. That is what Part II is about.

An aside on boiling water#

There is one more way to make giant composites, and it is older than all of the above. In 1984 Witten asked what would happen if the transition from a plasma of quarks and gluons to protons and neutrons in the early Universe was first order ⁵.

A first-order transition is the boiling-water kind. When water boils, it doesn’t smoothly thin out into steam. Bubbles of steam appear inside the liquid, both phases sit side by side, and the bubbles grow until the liquid is gone. Witten imagined the same thing happening with bubbles of ordinary hadronic matter growing inside a sea of free quarks. As the bubbles expand, the quarks get squeezed into shrinking pockets between them. If those pockets are stable, they survive as macroscopic quark nuggets, and they could be the dark matter.

It is a beautiful idea, and it turns out not to be how our Universe works. Lattice QCD shows that, at the low densities of the early Universe, the quark–hadron change is a smooth crossover with no bubbles at all ⁶. A dark sector, though, is free to have a properly first-order transition, and then the bubbles come back. Dark quark nuggets made this way can range from about $10^{-7}$ grams to $10^{23}$ grams ⁷. The colliding bubbles would also shake the plasma hard enough to leave gravitational waves behind.

How do you catch something that heavy?#

Here is the part I find most fun, because composites turn the usual detection problem upside down.

We know roughly how much dark matter is around us, $\rho_{\rm DM} \approx 0.3$ GeV per cubic centimeter, moving at $v \approx 300$ km/s. If it comes in lumps of mass $M$, the number density of lumps is $\rho_{\rm DM}/M$, and the number crossing a unit area per unit time is

Let’s put the numbers in, one conversion at a time. The mass flux is

A square meter is $10^4$ cm² and a year is about $3\times10^7$ s, so per square meter per year this becomes

Dividing by $M$, one lump crosses each square meter per year when $M \sim 3\times10^{18}$ GeV. Since $1$ GeV is $1.8\times10^{-24}$ g, that is about five micrograms. This is a few times below the Planck mass of about 22 micrograms, and around the mass of a fine grain of sand. Make the lumps much heavier and a detector a meter across will wait years to see even one, however strongly they interact. So the Planck mass is roughly where meter-sized underground detectors run out.

The flip side is that heavy composites can interact very strongly, and the first reason is coherence. Suppose a nucleus scatters off a composite made of $A$ constituents at positions $\mathbf{r}_j$, with momentum transfer $\mathbf{q}$. Each constituent contributes an amplitude $f$ with a phase $e^{i\mathbf{q}\cdot\mathbf{r}_j}$ that depends on where it sits, and quantum mechanics says we add amplitudes, not probabilities,

If the nucleus hits gently, so that $q$ is small compared to the inverse size $1/R$ of the composite, every phase is close to one and the sum is simply $A$. The cross section, which goes as $|\mathcal{M}|^2$, is then

where $\sigma_1 \propto |f|^2$ is the cross section for a single constituent. For a harder hit, $qR \gtrsim 1$, the phases point in all directions, they cancel, and the enhancement goes away ⁸. This is the same physics that makes ordinary WIMP–nucleus scattering grow as the square of the nucleus’s mass number.

There is a catch. At fixed dark matter density, packing constituents into composites of $A$ means there are $A$ times fewer things to scatter off. The event rate is number density times cross section,

so overall the rate grows as $A$, not $A^2$. That is still a big win. As long as the hits stay gentle enough for the enhancement to hold, a composite of a billion dark fermions is a billion times easier to see than the same fermions spread out on their own.

The second reason is that a large cross section means the composite does not scatter once. It scatters again and again as it passes through the detector, roughly $N \sim n\,\sigma L$ times for target density $n$ and path length $L$, leaving a line of nuclear recoils. How straight is that line? In an elastic collision between a heavy mass $M$ and a nucleus of mass $m_N$ at rest, the largest fraction of the kinetic energy that can be handed over is

For a xenon nucleus, $m_N \approx 120$ GeV, and a composite of $10^{18}$ GeV, that is about $5\times10^{-16}$ per collision. Even after a billion collisions the composite has lost less than a millionth of its energy. It plows through in an essentially straight line, which looks nothing like background. There is also a nice subtlety about the kilometers of rock above an underground lab. You might worry that a strongly interacting composite would be slowed to a stop before it ever reaches the detector. But it can only hit as many nuclei as are actually in its path, so beyond a certain cross section adding more interaction strength doesn’t slow it any further. Together with dedicated searches for multiple scatters, this means xenon and bubble-chamber experiments such as LZ, XENON, PandaX and PICO could find dark matter up to the Planck mass and beyond ⁹.

What I take away from this#

The thing I like most about composite dark matter is that it separates two questions we usually treat as one. How weakly does a dark matter particle interact? and How hard is dark matter to detect? sound like the same question, but they aren’t. Each dark fermion might be hopelessly weakly coupled to us, and the nugget it lives in could still light up a detector.

There is a second, quieter consequence. Building a nugget releases a lot of energy, and in a dark sector that does not talk to us, that energy has nowhere to go except into the dark sector itself. In Part II I follow that energy, and show how it could leave a fingerprint on the cosmic microwave background.

If you want to read more#

• J. M. Cline, Dark Atoms and Composite Dark Matter (Les Houches lectures), arXiv:2108.10314. The friendliest place to start.
• J. Bramante, Very Heavy and Composite Dark Matter: Theory and Experimental Searches, Ann. Rev. Nucl. Part. Sci. 76 (2026) 319, arXiv:2602.23708. The most up-to-date overview, especially of how to search for heavy composites.
• G. D. Kribs and E. T. Neil, Review of strongly-coupled composite dark matter models and lattice simulations, arXiv:1604.04627. The dark-QCD zoo I skipped.