Please refer to R. K. Luneburg, Mathematical Theory of Optics (U. California Press, Berkeley, Calif., 1966), Chap. 14, p. 64 for this question.

Consider a set of wavefronts described by a characteristic hypersurface $\phi(t,\vec{x})=0$ in a spacetime $\mathfrak{M}_4 = \mathbb{R} \otimes \mathbb{E}_3$. The medium $\mathbb{E}_3$ hosts a variable refractive index $n(\vec{x})$.

If we write $\phi(t,\vec{x}) =: \psi(\vec{x}) - ct$, then the normal to the spatial wavefronts $\psi= constant$ defines the so-called quasimomentum $\vec{P} := \vec{\nabla} \psi$ of the propagating wave.

It is claimed that because $\phi = \psi -ct$, therefore $\psi$ must be a continuous solution of the equation

$$ |P|^2=n^2 \,. $$

I don't understand how.

Why is the gradient of the spatial wavefront equal to the refractive index (times a unit vector)?


You are asking about the derivation of the Eikonal Equation.

We have a phase field $\psi(\vec{x})$ describing the phase lag in a steady state light field relative to some phase reference.

Let's remove some scaling constants to make it clearer and, for a start, define $\psi$ to be the phase in radians; then the surface $\psi = 2\,\pi\,\omega\,t$ is the surface comprising all points lagging by time $t$ relative to the reference.

The maximum rate of increase of $\psi$ with respect to distance is $|\nabla\psi|$ and it is in the direction of $\nabla\psi$. The neighboring surface $\psi = 2\,\pi\,\omega\,(t+\mathrm{d} t)$ lies a constant, normal distance $\mathrm{d}r$ from the surface $\psi = 2\,\pi\,\omega\,t$; the phase difference between them is $2\,\pi\,\mathrm{d}r/\tilde{\lambda}$, where $\tilde{\lambda}(\vec{x})$ is the wavelength in the medium at the point in question. So the maximum rate $|\nabla\psi|$ is $2\,\pi/\tilde{\lambda}$ radians per meter, whence:

$$|\nabla \psi|^2 = k^2 = n(\vec{x})^2\,k_0^2$$

where $k_0=2\,\pi/\lambda_0$ is the freespace wavenumber and $\lambda_0$ the freespace wavelength.

Now if we divide $\psi,\,\phi$ by the constant $k_0$ and use these as the phase fields instead of measuring the phase in radians, we get the result claimed.

So we're actually measuring the phase in cycles and distance in wavelengths.

I don't have Luneburg before me so I am not fully aware of your question's background, but, to answer your title question, the pseudomomentum is a momentum in the sense of Hamiltonian mechanics and the Hamiltonian treatment of optics. I discuss this in my answer here as well as in the answers linked therein.

Further background that may be relevant here is the fact that the Eikonal equation, which is what is being derived here, is equivalent to Fermat's least time principle. I sketch how to prove the equivelence of all these notions in my answer here.

  • 1
    $\begingroup$ Thanks. Luneburg uses this fact to derive Snell's law by imposing boundary conditions. $\endgroup$ Jan 28 '18 at 11:40

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