The eikonal equation and the ray limit
Derive phase geometry and amplitude transport from a high-frequency wave ansatz, then identify where the ray approximation fails.
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The question for today
What you’ll understand
- Obtain the eikonal equation from a scalar Helmholtz ansatz.
- Recover ray direction and optical path from a phase gradient.
- Use transport conservation while recognizing caustics and wavelength-scale failures.
Ray optics is a short-wavelength approximation to a wave problem, not a competing fundamental theory. A rapidly oscillating phase can carry the local propagation direction while a slowly varying amplitude describes how the beam spreads. Separating these roles leads to the eikonal equation and explains why optical path appears in both variational ray tracing and interference phase. The same derivation identifies terms that were discarded, giving us a principled way to recognize boundaries, caustics, and structures where a ray calculation needs a wave treatment.
Separate rapid phase from slow amplitude
Begin with the scalar Helmholtz model in a transparent isotropic medium of real index n at fixed angular frequency. Write the complex field as , with real leading amplitude A, vacuum wavenumber , and eikonal S measured in metres. The dimensionless phase is . This scalar starting point already omits vector polarization effects and requires care near material boundaries; it is not a complete Maxwell derivation for every spatially varying medium.
Substitution into the Helmholtz equation produces terms proportional to the square, first power, and zeroth power of . The leading square term vanishes when the gradient magnitude of S equals n. Slow amplitude variation means that the neglected amplitude-curvature term is small compared with the leading oscillatory terms, typically when the wavelength in the medium is much shorter than the relevant variation length. A short vacuum wavelength alone is insufficient if an aperture edge or a sharp feature introduces a competing small scale.
Phase normals become rays
Surfaces of constant S are wavefronts. Their normal is its gradient, so the unit ray tangent in an isotropic medium is . Along a ray, the rate of change of S with arclength is n. Integrating shows that the difference in eikonal values equals optical path along that ray, provided a consistent local phase branch is chosen. The local wavevector is , with magnitude and wavelength .
Taking a derivative of the eikonal relation along its gradient recovers , joining phase geometry to the GRIN ray equation. Initial or boundary wavefront data are still required; specifying only n does not identify which family of rays a source launches. The eikonal may need several branches where distinct ray families reach the same point. A single-valued smooth phase field cannot describe all possible interference from those branches without additional construction.
Transport and its failure have physical meaning
The next-order imaginary terms give , equivalent to the displayed divergence law. It conserves scalar flux through a ray tube. In a uniform medium with cross sections perpendicular to the rays, amplitude therefore scales as the inverse square root of tube area. This is why a spreading beam grows dimmer without absorption. In a varying index, the factor n enters that flux, so the same area-only rule cannot be transferred unchanged.
At a caustic the geometric tube cross section can vanish, making this leading amplitude diverge. The actual wave field remains finite after an appropriate diffraction or uniform asymptotic treatment. Turning points, interference fringes, narrow apertures, wavelength-scale guiding, and abrupt interfaces also demand supplementary wave analysis or boundary matching. These failures do not mean every ray direction is useless; they mean the approximation is nonuniform in a particular region. Compare scales and observables before deciding whether a ray, scalar wave, or vector electromagnetic model is adequate.
Work it through
Read a plane wave from its eikonal
In the x-z plane let , with coordinates in metres and a constant real amplitude. Determine the compatible uniform index, ray direction, local wavelength for , and phase cycles accumulated over 3 mm along a ray.
The gradient is , whose magnitude is . The eikonal equation therefore requires uniform index n = 2.
Divide the gradient by n to obtain the unit tangent . The local wavelength is . Constant amplitude satisfies transport because this gradient is constant and the ray tube is parallel.
Over arclength , the eikonal increase is . Dividing by vacuum wavelength gives phase cycles, equivalently radians.
The field describes a plane wave in index 2 traveling along (0.6, 0.8). Its 3 mm geometric journey accumulates 6 mm of optical path and 12000 cycles.
The living notebook
Explore the model
Interactive lab · vectors
Gradient magnitude and direction
Interpret the displayed vector as a possible local eikonal gradient. Reset, set Angle = 0° and Magnitude = 1.5, and predict projections (1.5, 0). For a fresh run Reset, set Angle = 90° and Magnitude = 1.5, and predict projections (0, 1.5). The same gradient magnitude can represent different directions in one uniform index.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
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 forThe lab displays one vector and its projections. It does not solve a spatial eikonal equation, draw wavefronts, propagate a ray tube, or regularize caustics.
Pause & reason
Check your understanding
Your turn
Try it yourself
A smooth ray tube in a uniform lossless medium expands from perpendicular area to . Its initial leading scalar amplitude is 6 arbitrary units. Find the final amplitude and intensity ratio, assuming the tube remains away from a caustic and the scalar high-frequency model applies.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Integrate the transport flux over a tube cross section; the uniform index cancels in the ratio.
Compare with the solution
Flux conservation gives , where the calligraphic symbol denotes tube area. Thus arbitrary units, and intensity ratio is . The total transported power is unchanged because area increased by four. The result would need the index factor retained in a nonuniform medium and would fail as a leading approximation at a caustic.
Keep the idea
Take it with you
- The eikonal is a length-valued phase function whose gradient magnitude is the refractive index.
- Rays are wavefront normals in isotropic media, and optical path is the eikonal change along a ray.
- Amplitude transport is useful away from caustics and other regions where short-wavelength approximations fail.
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