2026-10-01 · chapter Yang–Mills and the mass gapPart 2 Physical preparationthe action principle and Noether

Lagrangians and the action — motion can be written as a stationary point of "one number attached to the whole path"

Newton's equation of motion $m\ddot q = F$ is a local rule that fixes the acceleration at each moment. In this chapter we write the same motion in another form. Fix the start and the arrival, and give the whole path between them a single number $S$ (the action). The actual motion is chosen as the path for which $S$ does not change at first order when the path is shifted slightly. This is the action principle (also called the principle of least action).

There are three advantages. It does not depend on coordinates (§04). Conservation laws come out automatically from symmetries (§06). It is the language of field theory: Yang–Mills theory is defined by an action, the integral of $\operatorname{tr}F^2$ (3-03), and the path integral too sums over paths with the action in the exponent (2-07).

As a metaphor, the equation of motion is "step-by-step directions", and the action principle is the view of "comparing whole routes on a map". To be exact, for systems with conservative forces both give the same equation (§04), and the only difference is what is easy to see.

For a particle on a line, let $q(t)$ be its position at time $t$, $\dot q = dq/dt$ its velocity, $m$ its mass and $V(q)$ the potential energy. The Lagrangian $L$ is the kinetic energy minus the potential energy, a different quantity from the energy, which is their sum. The number obtained by integrating it along the path is the action.

The square brackets show that it is a map that takes a whole path and returns a number (a functional).

Example. Take $V = 0$, $m = 1$, and ends $q(0) = 0$, $q(1) = 1$. For the uniform path $q = t$, $S = 0.5$. For the detour $q = t + 0.3\sin\pi t$ joining the same ends, $S = \tfrac12\int_0^1 (1 + 0.3\pi\cos\pi t)^2\,dt = \tfrac12(1 + 0.09\pi^2/2) \approx 0.722$. Taking a detour increases the action.

An extremum of a function was a point where the first-order change under a small change of the variable is 0. For the action we move the path. Taking a function $\eta(t)$ that vanishes at the ends, $S(\varepsilon) = S[q + \varepsilon\eta]$ is an ordinary function of the real number $\varepsilon$. The action principle is the requirement "$S'(0) = 0$ for every $\eta$ with fixed ends", and a path satisfying it is called a stationary point.

Discretised, it becomes a function of finitely many variables

Dividing time into $N$ equal parts (step $h$), making the path a broken line and approximating the integral by the trapezoidal rule, the action becomes a function of many variables, the intermediate positions $q_1, \dots, q_{N-1}$ (the ends $q_0$, $q_N$ fixed).

The action principle becomes "the gradient is 0", and this is a difference version of $m\ddot q = -V'(q)$ (used again in §07).

Example: a particle on a spring

Take $V = \tfrac12 m\omega^2 q^2$, $m = 1$, time length $T = 1$, and ends $q(0) = 1$, $q(1) = -1$. The solution of the equation of motion is $q_{\mathrm{cl}}(t) = \sin\bigl(\omega(\tfrac12 - t)\bigr)/\sin(\omega/2)$ (cl stands for "classical path"). For the shift $\eta_n = \sin n\pi t$ ($n = 1, 2, 3$), the action is a quadratic in $\varepsilon$.

The absence of a first-order term is what "stationary" means. The second-order term comes from $\tfrac12\int(\dot\eta^2 - \omega^2\eta^2)\,dt$. Moving $\varepsilon$ in Figure 1 moves the action along a parabola, with slope 0 at $\varepsilon = 0$.

ComputedThe curves draw the formula above in the browser, and the formula was checked against numerical integration on this machine (Table 1). The dotted line is the classical path, and the dots are the variables when discretised with $N = 10$.

ComputedThe difference between numerical integration by Simpson's rule (20000 intervals) and the formula is at most $4\times10^{-15}$. The first-order coefficient is also at most $5\times10^{-13}$.

We compute the condition of §03 without discretising. Differentiate with respect to $\varepsilon$ and integrate the second term by parts (the boundary term vanishes because $\eta = 0$ at the ends).

If this is 0 for every $\eta$, the bracket is 0 at each time (the bracket is continuous, so if it is positive somewhere it is positive nearby, and taking a $\eta$ positive only nearby makes the integral positive: the fundamental lemma of the calculus of variations). This gives the Euler–Lagrange equations. With several degrees of freedom there is one for each component.

Example. If $L = \tfrac12 m\dot q^2 - V(q)$, then $m\ddot q = -V'(q)$: Newton's equation for a conservative force itself.

No choice of coordinates

This derivation does not use what $q$ represents. It may be an angle (generalised coordinates), and equations of the same form hold Known. This is because the same path gets the same action whatever the coordinates (1-05). Writing a pendulum with the angle $\theta$, $L = \tfrac12 ml^2\dot\theta^2 + mgl\cos\theta$ gives $ml^2\ddot\theta = -mgl\sin\theta$. The tension does not appear.

Adding $dG(q,t)/dt$ to $L$ does not change the equations (the action changes only by a quantity fixed by the end values) Known. The θ term of Yang–Mills theory is the four-dimensional version of this, with the form of a four-dimensional divergence (a total divergence) instead of a total time derivative. To be exact, the θ term does not change the classical equations, but it matters in quantum theory through the topology of the field (3-05).

The action principle asks only that the first-order change be 0; whether it is a minimum is decided by the second-order term. In the example of §03 it is the sign of the coefficient $((n\pi)^2 - \omega^2)/4$. If $\omega T \lt \pi$ it is positive for every $n$, and a general $\eta$, expanded in a sine series, splits into a sum over each $n$, so the classical path really is a minimum (because the action is quadratic).

When $\omega T$ exceeds $\pi$, the coefficient of $n = 1$ becomes negative, and in that direction the action decreases (a saddle point). Moving $\omega T/\pi$ either side of 1 in Figure 1 turns the parabola of $n = 1$ over. At exactly $\omega T = \pi$, moving in the direction of $n = 1$ does not change the action (it increases in the direction of $n \ge 2$), and the solutions joining the same ends form a family $\cos\pi t + c\sin\pi t$. In general, a time at which a family of nearby solutions leaving the same point at time 0 gathers again at one point is called a conjugate point of 0 Known. The conjugate points in this example are $t = \pi/\omega, 2\pi/\omega, \dots$, and $\omega T = \pi$ is the case where the final time falls exactly on the first conjugate point.

The second-order term is an eigenvalue problem for a matrix

Discretised, the second-order term is the Hessian matrix $H_{jk} = \partial^2 S_N/\partial q_j\partial q_k$. It is real symmetric, so it can be diagonalised (1-01), and a local minimum means all eigenvalues are positive. The number of negative eigenvalues is the number of directions in which the action can be reduced, called the Morse index (here a name for the discretised matrix). For the harmonic oscillator $H$ is tridiagonal (diagonal $2m/h - m\omega^2 h$, neighbours $-m/h$); its eigenvalues divided by $h$ approach $m((n\pi/T)^2 - \omega^2)$, and $\lfloor\omega T/\pi\rfloor$ of them are negative.

ComputedAll eigenvalues found with numpy's eigvalsh. With $N = 20$ the numbers are the same, and the smallest eigenvalues differ by about 0.02.

Without discretising too, the directions that reduce the action in this example are those among $\sin n\pi t$ with $n\pi \lt \omega T$, and their number equals the number of conjugate points $k\pi/\omega$ in $(0, T)$. This is an instance of the Morse index theorem (the number of negative directions of the second-order term equals the number of conjugate points in $(0, T)$, counted with multiplicity) Known.

This view is extended to infinite dimensions in 1-10. The smallest eigenvalue here (of the second-order term of the action) and the mass gap of 4-11 (the gap in the spectrum of the Hamiltonian) are different things.

If $L$ does not contain the coordinate $q$, (EL) says that $p = \partial L/\partial\dot q$ (the conjugate momentum) is conserved. $L = \tfrac12 m(\dot r^2 + r^2\dot\varphi^2) - V(r)$, a central force written in polar coordinates, does not contain $\varphi$, and the angular momentum $mr^2\dot\varphi$ is conserved. Noether's theorem is this detached from coordinates.

For transformations $q \mapsto \Phi_s(q)$ labelled continuously by a real number $s$ and equal to the identity at $s = 0$ (translations, rotations by the angle $s$), let $K(q) = \frac{d}{ds}\Phi_s(q)\big|_{s=0}$ be the direction of motion. Noether's theorem: if $L\bigl(\Phi_s(q), \tfrac{d}{dt}\Phi_s(q)\bigr) = L(q, \dot q)$ for every path, then the following $Q$ is constant along the motion Known.

Proof. Differentiating the assumption with respect to $s = 0$ gives $\sum_i (\partial L/\partial q_i) K_i + \sum_i (\partial L/\partial\dot q_i)\,\dot K_i = 0$. Replacing $\partial L/\partial q_i$ by $\frac{d}{dt}(\partial L/\partial\dot q_i)$ using (EL), the left-hand side is the derivative of a product $dQ/dt$. ∎

The third row follows by differentiating directly, from $dE/dt = -\partial L/\partial t$ ($E = \tfrac12 m\dot q^2 + V$).

Checking with a figure: breaking the symmetry removes the conserved quantity

Give a particle in the plane ($m = 1$) the potential $V = \tfrac12 r^2 + \tfrac14 r^4(1 + \delta\cos 4\theta)$ in polar coordinates. If $\delta = 0$ it is rotationally symmetric, and $L_z$ is conserved. With $\delta \gt 0$ the rotations that leave $V$ unchanged are only multiples of 90°, and the symmetry drops to a discrete one. There are no infinitesimally small transformations for the theorem to use, so the theorem says nothing, and in fact $L_z$ is not conserved. $V$ does not depend on time, so $E$ is conserved.

ComputedInitial position $(1, 0)$, velocity $(0, 0.9)$. The orbit is computed by the browser with the velocity Verlet method ($h = 0.01$), and Table 3 gives the values from running the same procedure on this machine (1.4e−14 means $1.4\times10^{-14}$). The two bands on the right have the same vertical scale.

For $\delta = 0.3$ and $0.5$ even the sign of $L_z$ changes, and the particle reverses its direction of rotation partway. The deviation of $E$ is about $10^{-5}$, and looks flat in the figure.

The $s$ of §06 was a constant (a global symmetry). A symmetry under which the action is unchanged even when transforming by a $s(t)$ that differs from time to time is a local symmetry, and it is at the centre of gauge theory (3-01). We look at a toy with two variables, $x(t)$ and $\lambda(t)$.

For any $s(t)$, $\dot x - \lambda$ does not change. $\lambda$ appears without time derivatives, so its equation is a constraint.

The second is the derivative of the first, and only one is independent. $x$ remains as an arbitrary function that the equations do not fix. Noether's second theorem (1918) states that when the action is unchanged under transformations containing arbitrary functions, some of the Lagrange equations follow from the rest Known. To be exact, as many identities as there are arbitrary functions hold among the left-hand sides of the Lagrange equations and their derivatives. In this toy the arbitrary function is the one $s(t)$, and the identity is also one: "the left-hand side for $(x)$ is the time derivative of the left-hand side for $(\lambda)$". The $Q = m(\dot x - \lambda)$ made by (N) from a constant $s$ is, far from conserved, constrained to be 0.

In electromagnetism the scalar potential $\phi$ plays the role of $\lambda$, the vector potential $\mathbf A$ the role of $x$, and Gauss's law plays the role of the equation for $\lambda$ (2-03, 4-01). To be exact, in electromagnetism $s(t)$ becomes a function at each point $\chi(t,\mathbf x)$, $\mathbf A$ shifts by $\nabla\chi$ and $\phi$ by $-\partial\chi/\partial t$ ((G) of 2-03; the sign of the shift of $\phi$ is opposite to the $\lambda$ of the toy). The degrees of freedom the equations do not fix are a redundancy of description, and handling them is one of the big problems of quantising Yang–Mills theory (4-08).

The toy itself has no physical content (the only solution is $\lambda = \dot x$, with $x$ arbitrary).

What this chapter can say

Sources and reproduction

Next chapter: 2-02 The idea of a field (replacing $q_i$ by a field $\phi(x)$, the degrees of freedom become infinitely many).

← 1-12 Detour: what happens only in four dimensionsContents2-02 The idea of a field →