Mass Without Friction: What a Ferromagnet Teaches About the Higgs Mechanism — Epoche C1
Popular accounts of the Higgs mechanism reach for the same image: an invisible field fills space, and particles acquire mass because that field resists their motion like treacle. The image is memorable and it is wrong, and it is worth being specific about how, because each of its failures points to the feature of the real mechanism that replaces it. What the Higgs field actually does can be read off two systems that sit in laboratories rather than in accelerators — a magnet and the conduction electrons of a metal. This essay proceeds through those two in turn and then to the electroweak vacuum, where the same bookkeeping produces the masses of the weak force carriers. Three ways the treacle picture fails The first failure is dynamical. A resisting medium exerts a force that depends on velocity, so a body moving through it obeys an equation of the form $m\dot{v} = -bv$ and its speed decays; mass, by contrast, appears in Newton's law multiplying acceleration and contributes no velocity-dependent term at all. A ball in honey comes to rest. A massive particle in empty space keeps its velocity indefinitely, which is what inertia means. Whatever gives a particle mass cannot be something it ploughs through. The second failure is relativistic. A medium singles out the frame in which it is at rest, and in that frame alone the physics would look isotropic; observers moving relative to it would find directional effects. Experiment excludes this to extraordinary precision. Herrmann and colleagues (2009) compared the resonance frequencies of two orthogonal optical cavities while rotating the apparatus continuously for a year, and found no dependence of the speed of light on orientation at the level of parts in $10^{17}$. There is no ether, and the Higgs field is not one. The third failure is selective. The photon travels through exactly the same vacuum as the $W$ boson and remains exactly massless. A universal medium that drags everything cannot explain why it drags some things and not others, and it certainly cannot explain why the exemption is precisely the particle associated with the one symmetry that survives. Any correct account must produce that exemption as a consequence rather than an exception. A magnet: symmetric law, asymmetric state The first laboratory system supplies the central idea, that a state can have less symmetry than the law it obeys. The Heisenberg model describes a magnet as a lattice of atomic spins — quantum magnetic moments, each an arrow of fixed length — interacting through the energy $$H = -J \sum_{\langle ij \rangle} \mathbf{S}_i \cdot \mathbf{S}_j ,$$ where $\mathbf{S}_i$ is the spin at site $i$, the sum runs over neighbouring pairs, and $J$, the exchange coupling, is positive so that aligned neighbours have lower energy. Only relative angles appear, since a dot product is unchanged when both vectors are rotated together. The law therefore has no preferred direction whatever. Yet below the Curie temperature — $770\,^{\circ}\mathrm{C}$ for iron — the ground state is magnetised along some particular direction. Nothing selected it; a fluctuation did, and the whole sample followed because alignment lowers the energy. The symmetry has not been removed from the theory, only from the state, which is what "spontaneously broken" means, and it leaves a signature in the excitations. Tilting all the spins together costs nothing, being a symmetry operation, so a slowly varying tilt — a spin wave, or magnon — costs very little, and the cost falls to zero as the wavelength grows. For the Heisenberg ferromagnet the magnon energy is proportional to $k^2$, where $k$ is the wavenumber, so it vanishes as $k \to 0$ (Ashcroft and Mermin, 1976). Goldstone (1961) proved that this is general rather than an accident of magnets: whenever a continuous symmetry is spontaneously broken, the spectrum contains excitations whose energy tends to zero at long wavelength, one for each independent symmetry direction that the ground state fails to respect. The proof assumes a locally conserved current and interactions of short range, and both assumptions matter — as the next system shows by violating one of them. A metal: a photon that behaves as if massive The second system adds the ingredient the magnet lacks, which is electric charge, and with it a long-range force. The conduction electrons of a metal form a plasma: a gas of mobile negative charges neutralised on average by the fixed positive ions. A transverse electromagnetic wave crossing this plasma obeys $$\omega^2 = \omega_p^2 + c^2 k^2 ,$$ which follows from Maxwell's equations once the current is written in terms of the free electrons' response to the wave's electric field. Here $\omega_p$ is the plasma frequency, the natural frequency at which the electron gas oscillates bodily against the ionic background; in the Gaussian units of the condensed-matter literature it is $\omega_p^2 = 4\pi n_e e^2/m_e$, with $n_e$ the electron density and $e$ and $m_e$ the electron's charge and mass. The square root of a density divided by a mass is what one expects of a restoring force proportional to displaced charge acting on inertia, which is exactly the oscillation being described. Now set that dispersion relation beside the relativistic energy–momentum relation $E^2 = (mc^2)^2 + (pc)^2$, using $E = \hbar\omega$ and $p = \hbar k$. They are the same equation, with $\hbar\omega_p$ in the role of the rest energy. Inside the metal, the photon propagates exactly as a particle of mass $\hbar\omega_p/c^2$ would. The number is large: for $n_e = 10^{23}\,\mathrm{cm^{-3}}$, a typical value for a good metal, the formula gives $\omega_p \approx 1.8\times10^{16}\,\mathrm{rad\,s^{-1}}$ and $\hbar\omega_p \approx 12\,\mathrm{eV}$, against the $2$ to $3\,\mathrm{eV}$ of visible photons. Waves below the plasma frequency have imaginary $k$ and cannot propagate, which is the reason such a metal reflects visible light and is transparent in the far ultraviolet. Anderson (1963) a