# Higgs Without Molasses

Although almost all ordinary mass effectively arises from the kinetic and binding energy of quarks and gluons bound to protons and neutrons in atomic nuclei, the Higgs mechanism does endow some particles like quarks and weakons with intrinsic masses. Here I gently introduce the Higgs mechanism without using loose analogies like vacuum field viscosity.

STATIONARY ACTION

For simple systems, the Lagrangian is the difference between the kinetic and potential energies,

$$L = T - V.$$

Demand that the action

$$S = \int dt \,L$$

be stationary and integrate by parts

$$0 = \delta S = \int dt \left(\frac{\partial L}{\partial x} \delta x + \frac{\partial L}{\partial \dot x} \delta \dot x \right) = \int dt \left(\frac{\partial L}{\partial x}- \frac{d}{d t}\frac{\partial L}{\partial \dot x} \right)\delta x,$$

for all variations $\delta x$, to find the Euler-Lagrange motion equation

$$\frac{\partial L}{\partial x} = \frac{d}{d t}\frac{\partial L}{\partial \dot x}.$$

CLASSIC OSCILLATOR

For the simple harmonic oscillator, the Lagrangian

$$L[x,\dot x] = \frac{1}{2} m \dot x^2 - \frac{1}{2} k x^2$$

implies the Euler-Lagrange equation

$$-kx = m \ddot x,$$

which recovers familiar laws of Hooke and Newton.

SPINLESS PARTICLE

In 1+1 dimensional spacetime, represent a spinless particle by the quantum of the field $\phi[t,x]$. In units where $c = 1$ and $\hbar = 1$, the Lagrangian

$$L = \int dx\, \mathcal L = \int dx \left( \mathcal{T} - \mathcal{V} \right)$$

and the Lagrangian density

$$\mathcal{L}[\phi, \partial_t \phi, \partial_x \phi] = \frac{1}{2} \left(\partial_t \phi \right)^2 -\frac{1}{2} \left( \partial_x \phi \right)^2 -\frac{1}{2} m^2 \phi^2,$$

where the signs of the kinetic energy terms reflect the signs of the spacetime interval $d\tau^2 = dt^2 - dx^2$. The corresponding Euler-Lagrange equation

$$\frac{\partial \mathcal{L}}{\partial \phi} = \frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\partial (\partial_t\phi)} + \frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\partial (\partial_x\phi)}$$

reduces to the Klein-Gordon equation

$$-m^2 \phi = \partial_t^2 \phi - \partial_x^2 \phi$$

or

$$\left(\partial_t^2 - \partial_x^2 + m^2 \right) \phi = 0.$$

Assuming wave-particle duality, seek plane wave solutions

$$\phi = e^{i(-Et+ px)} = e^{i(-Et+ px)/\hbar}= e^{i(-\omega t+ kx)}$$

to find

$$\left( (-iE)^2 - (i p)^2 + m^2 \right) \phi = 0$$

or

$$E^2 = p^2 + m^2,$$

where $m$ is the particle mass (and $E = m = m c^2$ for $p = 0$).

Nonzero mass determines the simple harmonic oscillator potential energy $\mathcal V = m^2 \phi^2 / 2 > 0$ and the curvature of the corresponding upright parabola. Zero mass collapses the Klein-Gordon equation to the electromagnetic wave equation.

SYMMETRY BREAKING

Consider the massless Lagrangian density

$$\mathcal{L}= \frac{1}{2} \left(\partial_t \phi \right)^\dagger \left(\partial_t \phi \right) -\frac{1}{2} \left( \partial_x \phi \right)^\dagger \left( \partial_x \phi \right) + \frac{a}{2} \phi^\dagger \phi - \frac{b}{4} \left( \phi^\dagger \phi \right)^2,$$

where $\phi = \phi_m e^{\phi_a} = \phi_r + i \phi_i$, the adjoint $\phi^\dagger = \phi^*$ reduces to complex conjugation, and the parameters $a,b > 0$. For all complex arguments $\phi_a$, the potential energy density

$$\mathcal{V} = - \frac{a}{2} \phi_m^2 + \frac{b}{4} \phi_m^4$$

depends only on the complex magnitude $\phi_m$. The vanishing derivative

$$0 = \frac{d \mathcal{V}}{d \phi_m} = - a \phi_m + b \phi_m^3= \phi_m(- a + b \phi_m^2)$$

implies local maximum and minimum at $\phi_m = 0$ and $\phi_m = \sqrt{a/b} = \phi_0$, as in the figure, where the circular symmetry reflects the invariance of the Lagrangian under global phase (or gauge) transformations $\phi \rightarrow \phi \, e^{i \Lambda} = \phi_m e^{i( \phi_a + \Lambda )}$.

If the potential is a sombrero, the circular brim are ground or vacua states of nonzero field

$$\phi_r^2 + \phi_i^2 = \phi_0^2.$$

Break the circular symmetry by fixing $\phi_r = \phi_0$ and $\phi_i = 0$, but allow the field to oscillate radially and shift circularly

$$\phi[t, x] = \phi_0 + \phi_1[t, x] + i \phi_2[t, x] .$$

The Lagrangian density becomes

$$\begin{array}{c}\displaystyle\mathcal{L}= \frac{1}{2} \left(\partial_t \phi_1 \right)^2 -\frac{1}{2} \left( \partial_x \phi_1 \right)^2 - a \phi_1^2 \\ \displaystyle+ \frac{1}{2} \left(\partial_t \phi_2 \right)^2 -\frac{1}{2} \left( \partial_x \phi_2 \right)^2 \\ \displaystyle- \sqrt{a b}\,\phi_1 \left(\phi_1 ^2+\phi_2 ^2\right) -\frac{b}{4} \left(\phi_1 ^2+\phi_2^2\right)^2\\ \displaystyle+ \frac{a^2}{4 b}\end{array}$$

or

$$\begin{array}{c}\displaystyle\mathcal{L}= \frac{1}{2} \left(\partial_t \phi_1 \right)^2 -\frac{1}{2} \left( \partial_x \phi_1 \right)^2 - m_1^2 \phi_1^2 \\ \displaystyle+ \frac{1}{2} \left(\partial_t \phi_2 \right)^2 -\frac{1}{2} \left( \partial_x \phi_2 \right)^2 - m_2^2 \phi_2^2 \\ \displaystyle+\text{interaction} + \text{constant},\end{array}$$

where $m_1 = \sqrt{a} > 0$ and $m_2 = 0$.

SUMMARY

Starting with a massless field $\phi$ with nonzero vacuum states, symmetry breaking creates a field $\phi_1$, corresponding to parabolic radial motion and a massive quantum $m_1 > 0$, and a field $\phi_2$, corresponding to constant circular motion and a massless quantum $m_2 = 0$ (a Goldstone boson). This is the Higgs mechanism of mass endowment. 