Amplitude and Polarization Transport

Once the eikonal equation has selected a null ray, the next Maxwell order determines what the wave carries along it. Expansion dilutes the amplitude so that leading energy flux is conserved, while the physical polarization is parallel transported on the two-dimensional screen transverse to the ray. Both statements hold away from sources and caustics, at leading order in the wavelength-to-background-scale ratio $\varepsilon$.

Required background. Classical Fields and Their Actions supplies Maxwell gauge freedom; The Wave Equation and Characteristics supplies transport equations in a WKB expansion; Affine Connections and Covariant Derivatives supplies parallel transport.

Helpful background. Events, Intervals, and Minkowski Geometry reviews null frames; Tensor Fields and Natural Operations reviews complexified tensor amplitudes; Timelike Congruences and Spatial Projection prepares expansion and transverse projection.

Transport of the complex amplitude

Write the leading potential coefficient as a complex one-form $a_\mu$. In Lorenz gauge the eikonal hierarchy gives

$$k^2=0, \qquad k^\mu a_\mu=0, \qquad 2k^\nu\nabla_\nu a_\mu+\Theta a_\mu=0,$$

where $\Theta\equiv\nabla_\mu k^\mu$ and $k^\nu\nabla_\nu k^\mu=0$. Define a nonnegative magnitude by $\mathcal A^2=a_\mu^*a^\mu$ after choosing a transverse screen representative. Contracting the transport equation with $a^{*\mu}$ and adding its complex conjugate yields

$$k^\mu\nabla_\mu\mathcal A =-\frac12\Theta\mathcal A,$$

or equivalently

$$\nabla_\mu\left(\mathcal A^2k^\mu\right)=0.$$

This conserved current is the leading wave-action or photon-number flux, up to normalization. The cycle-averaged stress-energy has the null-fluid form

$$\langle T_{\mu\nu}\rangle \propto \frac{\mathcal A^2}{\varepsilon^2}k_\mu k_\nu,$$

so its conservation follows from the flux law and the affine geodesic equation. The normalization depends on whether $\mathcal A$ was assigned to the potential or directly to $F_{\mu\nu}$; the transport law does not.

Beam area and brightness dilution

Consider a narrow, twist-free bundle of neighboring rays and let $\mathscr S(\lambda)$ be its infinitesimal screen area. Away from a caustic,

$$\frac{d}{d\lambda}\ln\mathscr S=\Theta.$$

Combining this with the amplitude equation gives

$$\frac{d}{d\lambda} \left(\mathcal A^2\mathscr S\right)=0.$$

Thus $\mathcal A\propto\mathscr S^{-1/2}$ and $\mathcal A^2\propto\mathscr S^{-1}$. The latter is wave-action dilution, not by itself the energy flux measured by an arbitrary observer. If $u^\mu$ is an observer four-velocity and $\omega=-u\cdot k/\varepsilon$ is the measured leading frequency, then

$$\langle T_{\mu\nu}\rangle u^\mu u^\nu \propto \mathcal A^2\omega^2,$$

with an additional incidence-angle projection for flux through a specified detector surface. Measured intensity therefore shares the $\mathscr S^{-1}$ factor only when the endpoint frequencies and detector geometry are held fixed; gravitational or kinematic redshift contributes its own factor. This is a local bundle statement, not a claim that brightness always follows an inverse coordinate-radius law. Curvature and shear determine $\mathscr S$ through the Jacobi equation. The area relation follows independently from a pair of deviation fields spanning the beam in Dolan 2018, §4.1.3, p. 10, PDF.

At a conjugate point $\mathscr S\to0$, the formula predicts $\mathcal A\to\infty$. The divergence diagnoses failure of the single-phase approximation, not infinite electromagnetic energy in an exact solution. A uniform wave treatment resolves the caustic by diffraction and interference over a wavelength-dependent neighborhood.

Polarization as a screen equivalence class

Choose an auxiliary null vector $n^\mu$ with $k\cdot n=-1$. The screen metric is

$$q_{\mu\nu}=g_{\mu\nu}+k_\mu n_\nu+n_\mu k_\nu.$$

It projects onto the two physical transverse directions. A leading gauge transformation can shift $a_\mu$ by $\alpha k_\mu$, so the physical polarization is the equivalence class

$$[a_\mu]=\{a_\mu+\alpha k_\mu\}.$$

Select a representative $a_\mu=\mathcal A f_\mu$ satisfying

$$k\cdot f=0, \qquad n\cdot f=0, \qquad f^*\cdot f=1.$$

The amplitude transport equation then leaves

$$k^\nu\nabla_\nu f_\mu=0$$

in a parallel-transported screen gauge. More invariantly, the screen class $[f_\mu]$ is parallel transported even if another gauge adds a longitudinal multiple of $k_\mu$. A complex circular basis transports the two helicities in the same way in vacuum Maxwell theory, so leading geometric optics has no frequency-dependent birefringence. Dolan 2018, §3.2, pp. 7–8, PDF derives transversality, flux conservation, and polarization transport directly from the Faraday tensor.

Rotation, gauges, and finite-wavelength limits

Parallel transport does not mean that every observer assigns a constant polarization angle. An observer projects $f_\mu$ into a chosen screen basis; that basis may rotate relative to parallel transport along a curved path. The resulting gravitational rotation is a comparison between transported frames and must state the observer congruence and screen convention. It is not a local rotation caused by a refractive medium.

The equations neglect relative $O(\varepsilon)$ corrections to the field. Those corrections can shift the energy centroid in a helicity-dependent way and can mix phase and polarization near sharp gradients, but they are not obtained by reinterpreting the leading parallel-transport law. Material dispersion, plasma Faraday rotation, absorption, and nonlinear optical response require constitutive data beyond vacuum Maxwell theory. The leading area and polarization equations also cease to be uniform at caustics, even when the wavelength remains short compared with the curvature radius.

Exercise

In Minkowski spacetime, consider an outgoing narrow bundle subtending a fixed solid angle $\Delta\Omega$. Inertial observers at rest with the source use detector surfaces orthogonal to the rays, so the measured frequency and incidence factor are the same at every radius. Determine the radial scaling of $\mathcal A$ and of the leading measured intensity.

Solution

The beam area is

$$\mathscr S(r)=r^2\Delta\Omega.$$

Since $\mathcal A^2\mathscr S$ is constant along the bundle,

$$\mathcal A(r)\propto\frac1r, \qquad I(r)\propto\mathcal A(r)^2\propto\frac1{r^2}.$$

The intensity conclusion uses the stated observer and detector choices: the factors $\omega^2$ and the incidence projection do not change, so only $\mathcal A^2$ supplies radial dependence. This is the inverse-square law in the special case of a flat, radially expanding bundle. A collimated plane-wave bundle has constant $\mathscr S$ and therefore constant leading amplitude; changing observer frequency or detector orientation changes the measured flux even at fixed $\mathscr S$. Coordinate distance alone is therefore not the governing quantity.

References

• Dolan, Sam R. “Geometrical Optics for Scalar, Electromagnetic and Gravitational Waves on Curved Spacetime.” International Journal of Modern Physics D 27 (2018): 1843010. DOI. Open PDF.