In part 1 of this discussion we had concluded that the Laplacian was a "zero-sum game", i.e. that a static modulation along one space dimension was exactly matched by a decay in the perpendicular dimension: \(q_x^2 + q_z^2 = 0\).

What happens for a time-dependent field distribution? For simplicity, let us assume a purely harmonic dependence: \(f(x,z,t) = F(x,z) \exp(i \omega t)\), with translation invariance along \(y\), where \(f\) stands for a component of the electric or magnetic field (in the scalar wave approximation).

The field now obeys:

\begin{equation}\label{eq:dalemb}

\Box \, f = \frac{1}{c^2} \partial _t^2 f(x,z,t) - \underbrace {( \partial _x^2 + \partial _z^2)}_{\Delta} f(x,z,t)= 0

\end{equation}

where \(c\) is the speed of light and \(\Delta\) is the Laplacian operator discussed in part 1. The wave operator \(\Box \) is often called d'Alembertian.

In \(z=0\), the field behaves as \(F(x,0) = \cos(q_x x) = \Re \left ( \text{e}^{i q_x x} \right ) \), but we will omit taking the real part and use the complex form in the following. The length scale of the modulation is \(\Lambda = 2 \pi / q_x\).

We will assume a similar behavior along \(z\), so that finally: \(f(x,z,t) = \exp [ i (q_x x + q_z z + \omega t) ]\). Plugging this expression into Eq. \eqref{eq:dalemb} yields:

\begin{equation}In \(z=0\), the field behaves as \(F(x,0) = \cos(q_x x) = \Re \left ( \text{e}^{i q_x x} \right ) \), but we will omit taking the real part and use the complex form in the following. The length scale of the modulation is \(\Lambda = 2 \pi / q_x\).

We will assume a similar behavior along \(z\), so that finally: \(f(x,z,t) = \exp [ i (q_x x + q_z z + \omega t) ]\). Plugging this expression into Eq. \eqref{eq:dalemb} yields:

\label{eq:balance}

-k^2 + q_x^2 + q_z^2 = 0 \Rightarrow q_z^2 = k^2 - q_x^2 = 4 \pi^2 \left ( \frac{1}{\lambda^2} - \frac{1}{\Lambda^2}\right )

\end{equation}

with \(k = \omega /c = 2 \pi / \lambda \). Since \(k\) and \(q_x\) are real, so is \(q_z\), provided \(|k| > |q_x|\). The difference with respect to the static case is the time derivative term, which ultimately contributes the \(k^2\) term in \eqref{eq:balance}. We could say that the time variation acts as a "reservoir" for space propagation:

A time-varying signal periodically modulated on a length scale \(\Lambda\) can propagate in the perpendicular direction provided its wavelength is smaller than this scale: \( \lambda < \Lambda \).

The representation below is purely schematic (on the right, only the vertically modulated components are damped) but intuitively clear (I hope).

This discussion should make it easier to understand why standard optical microscopes cannot resolve details smaller than the wavelength of light. It also explains why the metallic mesh on the door of a microwave oven stops the microwave radiation while still letting through visible light.

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