research32 min studyLesson 705 in Physics

Covariant-gauge propagators and gauge parameters

Invert the free Euclidean gauge-field kernel and isolate the gauge-dependent longitudinal sector.

Before you begin

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The question for today

What you’ll understand

Construct longitudinal and transverse momentum projectors.
Invert the free gauge-fixed quadratic operator with its gauge parameter.
Test a conserved probe contraction without assuming every off-shell quantity is physical.

A gauge-field quadratic action has null directions before gauge fixing, so its naive inverse does not exist. Covariant gauge fixing supplies a controlled longitudinal term and makes perturbative propagators possible. The resulting propagator depends on a gauge parameter because it describes a redundant field, not directly an observable particle probability. Projectors make this dependence transparent and provide a simple diagnostic: a transverse linearized probe cannot couple to the extra longitudinal piece.

Choose a signature before inverting

For this lesson use Euclidean momentum with positive $k^2$ away from zero. The positive gauge-fixing term is the squared ordinary divergence divided by twice xi, with xi positive before taking limiting gauges. Expanding about the zero connection gives the usual quadratic gauge action; g appears in interactions rather than this free kernel. This Euclidean setup avoids confusing a matrix inverse with Minkowski causal prescriptions, which require a separate analytic continuation.

Define the longitudinal projector along k and the transverse projector onto its orthogonal complement. They are idempotent, mutually annihilating, and sum to identity. In four dimensions their ranks are one and three for nonzero Euclidean k. This decomposition concerns off-shell momentum components; it must not be read as three physical polarizations of a real massless gauge boson. Physical asymptotic polarization counting uses constraints and an on-shell Lorentzian analysis.

P^L_{\mu\nu}=\frac{k_\mu k_\nu}{k^2},\qquad P^T_{\mu\nu}=\delta_{\mu\nu}-P^L_{\mu\nu},\qquad (P^{L,T})^2=P^{L,T},\quad P^LP^T=0

The projectors are dimensionless and require nonzero Euclidean momentum. This lesson does not assign them a Minkowski pole prescription or identify their ranks with physical polarization counts.↔ Scroll wide equations horizontally.

Invert each projector coefficient

The free quadratic operator multiplies the transverse projector by k squared and the longitudinal projector by k squared divided by xi. Its inverse therefore divides the transverse sector by k squared and multiplies the longitudinal sector by xi over k squared. The color factor is a Kronecker delta at this free level. Multiplying the kernel by this propagator recovers identity independently in both sectors.

Feynman gauge sets xi one and gives a particularly simple diagonal tensor. Landau gauge is obtained as the xi-to-zero limit and suppresses the longitudinal propagator. At strictly zero xi the unconstrained quadratic form is not inverted in the same naive way; it is understood through the limit or the transverse constrained measure. Zero momentum and residual gauge modes also need their own treatment, matching the qualifications already encountered in the Faddeev-Popov determinant.

K=k^2P^T+\frac{k^2}{\xi}P^L,\qquad \mathcal D^{ab}_{\mu\nu}(k)=\frac{\delta^{ab}}{k^2}\left(P^T_{\mu\nu}+\xi P^L_{\mu\nu}\right)

This is the tree-level Euclidean propagator for canonically normalized gauge fields and k squared greater than zero. The kernel has inverse-length-squared units and the propagator has length-squared units.↔ Scroll wide equations horizontally.

Locate gauge dependence before calculating a diagram

Contracting the propagator with k leaves a longitudinal term proportional to xi. A dimensionless transverse probe vector J instead projects entirely into the transverse sector, so its quadratic contraction is independent of xi at this free level. This is a useful algebraic test of a conserved linearized source. A general non-Abelian matter amplitude requires the complete gauge-consistent set of vertices, external-state conditions, and loop contributions.

Individual propagators, off-shell Green functions, and selected diagrams can depend on the gauge parameter. Observable independence is a statement about a consistently defined physical quantity, proved under the relevant BRST and renormalization assumptions. It does not permit setting xi differently in separate pieces without recomputing their matching contributions. Nor does a smooth Euclidean propagator by itself establish a positive physical spectral density in a strongly coupled theory.

k_\mu\mathcal D^{ab}_{\mu\nu}=\delta^{ab}\xi\frac{k_\nu}{k^2},\qquad k\cdot J=0\ \Longrightarrow\ J_\mu\mathcal D^{aa}_{\mu\nu}J_\nu=\frac{J^2}{k^2}\quad\text{for one fixed color a}

The final expression has no sum over a and uses a real dimensionless probe J. It is a tree-level tensor identity, not a claim that a colored probe alone defines a complete gauge-invariant observable.↔ Scroll wide equations horizontally.

Work it through

Invert a diagonal momentum example

Take Euclidean $k=(2M,0,0,0)$ and gauge parameter $\xi=2$ . For one fixed color find the propagator diagonal and the contraction with dimensionless transverse probe $J=(0,3,0,0)$ . What changes when xi becomes one?

Here $k^2=4M^2$ and the longitudinal projector has diagonal $(1,0,0,0)$ . The other three directions form the transverse projector.
The propagator diagonal is $(2,1,1,1)/(4M^2)$ , namely $(0.5,0.25,0.25,0.25)M^{-2}$ . Changing xi to one changes only its first entry to $0.25M^{-2}$ .
The probe has zero first component, so its contraction is $3^2(0.25)M^{-2}=2.25M^{-2}$ in both gauges. Its transverse character is checked by $k\cdot J=0$ before using any physical interpretation.

The result

The free longitudinal entry changes with xi, while the specified transverse contraction remains 2.25 inverse-M-squared.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Transverse and longitudinal projections

Treat the horizontal axis as a two-dimensional momentum direction. Reset and set angle ninety degrees, magnitude two; predict horizontal projection zero. Compare a fresh Reset run at angle zero, magnitude two; predict horizontal projection two. This Euclidean geometric analogy distinguishes transverse from longitudinal components but does not construct a four-dimensional gauge propagator or prove gauge independence of an interacting amplitude.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 projection test is useful algebra, while physical gauge independence requires the complete observable and its assumptions.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The inverse kernel has coefficients one over k squared in the transverse sector and xi over k squared in the longitudinal sector. Its off-shell gauge dependence is not itself an observable dependence.

Your turn

Try it yourself

Let $k=(M,M,0,0)$ and compare dimensionless probes $J=(1,-1,0,0)$ and $L=(1,1,0,0)$ . Find each quadratic propagator contraction for arbitrary positive xi and identify the transverse probe.

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

Give me a hint

Both probes have squared length two, but only one is orthogonal to the momentum.

Compare with the solution

The momentum squared is $2M^2$ . J is orthogonal to k, so $J\mathcal DJ=J^2/k^2=M^{-2}$ . L is parallel to k and therefore $L\mathcal DL=\xi L^2/k^2=\xi M^{-2}$ . At xi one the numerical values coincide, but their different dependence on xi reveals distinct tensor sectors. The calculation is an off-shell free-field test and excludes zero-momentum complications.

Keep the idea

Take it with you

Signature and zero-mode conventions belong in a propagator definition.
Covariant-gauge dependence resides in the free longitudinal sector.
Gauge-independent observables require more than examining one propagator or diagram.

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