research32 min studyLesson 703 in Physics

Yang–Mills classical equations and conserved currents

Derive covariant field equations and test a homogeneous nonlinear solution ansatz without promoting it to quantum dynamics.

Before you begin

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The question for today

What you’ll understand

Distinguish a covariantly conserved color current from componentwise ordinary conservation.
Reduce a declared classical ansatz to a nonlinear ordinary differential equation.
Check energy density and units in the reduced dynamics.

The Yang-Mills action turns connection geometry into nonlinear field dynamics. Covariant differentiation enters both the curvature and its equation of motion, reflecting the fact that gauge fields themselves carry the non-Abelian interaction. A matter source must fit the resulting covariant conservation law. These classical equations are useful in their own regime, but their solutions do not automatically describe the quantum vacuum, prove confinement, or include fluctuations generated by a path integral.

Vary the canonically normalized action

Use Minkowski metric with signature plus, minus, minus, minus and the curvature convention already fixed. The pure gauge Lagrangian is minus one quarter of the color-component field-strength contraction. A connection variation changes curvature by the antisymmetrized covariant derivative of that variation. Integrating by parts with appropriate boundary conditions produces a covariant divergence of F, rather than an ordinary divergence acting independently on every color.

If gauge-invariant matter is included, define the source current with the sign and coupling that make the field equation D F equal J; equivalently it is minus the functional derivative of the matter action with respect to the connection. This definition avoids mixing a source convention with a differently normalized gauge field. In four dimensions current has inverse-length-cubed units. The source-free equation follows by setting J to zero, not by merely choosing time-independent potentials.

\mathcal L_{\rm YM}=-\tfrac14F^a_{\mu\nu}F^{a\mu\nu},\qquad D_\mu F^{\mu\nu}=J^\nu,\qquad J^{a\nu}=-\frac{\delta S_{\rm matter}}{\delta A_\nu^a}

The current definition includes the relevant coupling factors and assumes a compatible gauge-invariant matter action. Boundary variations vanish in the action principle; indices use the stated Minkowski metric.↔ Scroll wide equations horizontally.

Apply the Bianchi identity and source consistency

The Jacobi identity for covariant derivatives gives the cyclic Bianchi identity. It is a geometric identity obeyed by any smooth connection, whereas the covariant divergence equation is dynamical. Applying another covariant derivative to the equation of motion gives covariant conservation of the source. The antisymmetry of F and the curvature commutator make the would-be obstruction cancel. An arbitrarily prescribed color current need not satisfy that requirement.

Covariant conservation permits exchange among color components through the connection. It therefore does not generally imply that each component has zero ordinary divergence. One may move the commutator term to define a gauge-dependent combined current with an ordinary continuity relation, but that rearrangement does not create separately measurable local color charges. Boundary conditions and allowed gauge transformations matter when defining global charges. Energy-momentum conservation follows from spacetime symmetry with its own gauge-invariant stress tensor.

D_\lambda F_{\mu\nu}+D_\mu F_{\nu\lambda}+D_\nu F_{\lambda\mu}=0,\qquad D_\nu J^\nu=0

The first identity is classical and off shell; the current condition follows on shell for a consistent source. Covariant conservation is different from ordinary componentwise conservation.↔ Scroll wide equations horizontally.

A homogeneous nonlinear subsector

In temporal gauge choose two spatial SU(2) potentials of equal amplitude along generators one and two, with all others zero. Assume spatial homogeneity in flat space, interpreted through energy density or a periodic box. The Gauss constraint is satisfied because each potential commutes with its own time derivative. Their commutator produces a third-color magnetic field. Unlike a single-color homogeneous free field, this ansatz experiences a restoring interaction even without spatial gradients.

Write the amplitude as a fixed inverse-length scale M times a dimensionless q, and time as dimensionless tau equal to M times physical time. Substituting into the action gives a quartic potential and a cubic restoring term. The conserved energy density contains two electric contributions and one magnetic contribution. Its amplitude-dependent oscillation differs from a linear harmonic oscillator. This finite ansatz is an exact classical reduction under its assumptions, not a solver for arbitrary gauge fields or their quantum state.

A_0=0,\ A_1=Mq(\tau)T^1,\ A_2=Mq(\tau)T^2,\ \tau=Mt,\qquad q''+g^2q^3=0,\quad \frac{\mathcal E}{M^4}=(q')^2+\tfrac12g^2q^4

Primes are derivatives with respect to dimensionless tau. The reduction assumes spatially uniform pure SU(2) fields and the stated canonical normalization; M has inverse-length units.↔ Scroll wide equations horizontally.

Work it through

Initial acceleration of a classical color configuration

In the homogeneous ansatz take $g=0.5,q(0)=2,q'(0)=0$ . Find initial dimensionless acceleration, magnetic curvature coefficient, and conserved energy density. Is this turning-point configuration static?

The reduced equation gives $q''(0)=-(0.5)^2(2)^3=-2$ . A zero initial velocity therefore does not imply zero acceleration or a static solution.
The magnetic coefficient is $F_{12}^3=gM^2q^2=0.5(4)M^2=2M^2$ . Electric fields vanish initially because the amplitude derivative is zero.
Energy density is $\mathcal E/M^4=0+\tfrac12(0.5)^2(2)^4=2$ . Initially all this energy is magnetic; subsequent classical evolution exchanges magnetic and electric contributions while preserving the total.

The result

The initial acceleration is minus two, curvature coefficient two M-squared, and energy density two M-to-the-fourth; this turning point evolves rather than remaining static.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect the amplitude scaling behind a quartic energy

Reset and set coefficient one, position one; predict quadratic output one. Compare a fresh Reset run at coefficient one, position two; predict output four. Squaring those two outputs gives the quartic scaling one versus sixteen used in the magnetic energy. The lab plots only a quadratic and does not integrate the nonlinear Yang-Mills amplitude equation or impose Gauss constraints.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA quartic magnetic energy scales differently from a harmonic potential; this algebraic comparison does not simulate its dynamics.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The field equation and Bianchi consistency give D-nu J-nu equal to zero. The connection mixes color components, so ordinary componentwise conservation is a stronger and generally incorrect replacement.

Your turn

Try it yourself

For the same homogeneous reduction take $g=1,q(0)=1,q'(0)=1$ . Find initial acceleration and energy density. Assuming the trajectory reaches a positive turning point with zero velocity, find its amplitude.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the conserved quartic energy to evaluate a turning point, rather than treating the motion as harmonic.

Compare with the solution

The acceleration is $q''(0)=-1$ . The dimensionless energy density is $1^2+\tfrac12(1)^4=1.5$ . At a turning point it equals $\tfrac12q_{\rm turn}^4$ , so $q_{\rm turn}^4=3$ and $q_{\rm turn}=3^{1/4}\approx1.316$ . The calculation conserves energy density in the spatially homogeneous classical subsector and does not assert a quantum excitation energy.

Keep the idea

Take it with you

Yang-Mills dynamics uses covariant derivatives and includes field self-interaction.
Bianchi identities constrain consistent sources without replacing equations of motion.
A controlled classical ansatz remains distinct from full quantum gauge dynamics.

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