Vector Wavenumber

Mathematically, a sinusoidal plane wave, as in Fig.E.3 or Fig.E.4, can be written as

$$p(t,\underline{x}) = p_0 + A\cos\left(\omega t - \underline{k}^T\underline{x}+ \phi\right), \quad \underline{x}\in{\bf R}^3 \protect$$ (E.9)

where the amplitude $A$, phase $\phi$, and radian frequency $\omega$ are ordinary sinusoid parameters [456], $\underline{x}$ is position in 3D space, and $\underline{k}$ is the vector wavenumber:

$$\underline{k}= \left[\begin{array}{c} k_x \\ [2pt] k_y \\ [2pt] k... ...] \cos{\beta} \\ [2pt] \cos{\gamma}\end{array}\right] \isdef k\,\underline{u},$$

where

• $\underline{u}=$ (unit) vector of direction cosines
• $k = \frac{2\pi}{\lambda} =$ (scalar) wavenumber along travel direction

Thus, the vector wavenumber $\underline{k}= k\,\underline{u}$ contains

• wavenumber along the travel direction in its magnitude $k=\left\Vert\,\underline{k}\,\right\Vert$
• travel direction in its orientation $\underline{u}= \underline{k}/k$

Note that wavenumber units are radians per meter (spatial radian frequency).

To see that the vector wavenumber $\underline{k}= k\,\underline{u}$ has the claimed properties, consider that the orthogonal projection of any vector $\underline{x}$ onto a vector collinear with $\underline{u}$ is given by $(\underline{u}^T\underline{x})\underline{u}$ [456].^E.11Thus, $(\underline{u}^T\underline{x})\underline{u}$ is the component of $\underline{x}$ lying along the direction of wave propagation indicated by $\underline{u}$. The norm of this component is $\vert\vert\,(\underline{u}^T\underline{x})\underline{u}\,\vert\vert =\vert\underline{u}^T\underline{x}\vert$, since $\underline{u}$ is unit-norm by construction. More generally, $\underline{u}^T\underline{x}$ is the signed length (in meters) of the component of $\underline{x}$ along $\underline{u}$. This length times wavenumber $k$ gives the spatial phase advance along the wave, or, $\theta(\underline{x})=k(\underline{u}^T\underline{x}) \isdeftext \underline{k}^T\underline{x}$.

For another point of view, consider the plane wave $\cos(\underline{k}^T\underline{x})$, which is the varying portion of the general plane-wave of Eq.$\,$(E.9) at time $t=0$, with unit amplitude and zero phase $\phi$. The spatial phase of this plane wave is given by

$$\theta(\underline{x}) \;\isdef \; \underline{k}^T\underline{x}\;=\; k_x x + k_y y + k_z z.$$

Recall that the general equation for a plane in 3D space is

$$\alpha x + \beta y + \gamma z =$$   constant

where $\alpha$, $\beta$, and $\gamma$ are real constants, and $x$, $y$, and $z$ are 3D spatial coordinates. Thus, the set of all points $\underline{x}^T=(x,y,z)$ yielding the same value $\theta(\underline{x})=\theta_0$ define a plane of constant phase $\theta_0$ in ${\bf R}^3$.

As we know from elementary vector calculus, the direction of maximum phase advance is given by the gradient of the phase $\theta(\underline{x})$:

$$\underline{\nabla }\theta(\underline{x}) \;\isdef \; \left[\begi... ...}{c} k_x \\ [2pt] k_y \\ [2pt] k_z\end{array}\right] \;\isdef \; \underline{k}$$

This shows that the vector wavenumber $\underline{k}$ is equal to the gradient of the phase $\theta(\underline{x})$, so that $\underline{k}$ points in the direction of maximum spatial-phase advance.

Since the wavenumber $k$ is the spatial frequency (in radians per meter) along the direction of travel, we should be able to compute it as the directional derivative of $\theta(\underline{x})$ along $\underline{k}$, i.e.,

$$k \;\isdef \; d_{\underline{\nabla \theta}}\theta(\underline{x}) ... ...ta(\underline{x})}{\delta \left\Vert\,\underline{\nabla \theta}\,\right\Vert}.$$

An explicit calculation yields

$$\left\Vert\,\underline{\nabla \theta}\,\right\Vert = \sqrt{k_x^2+k_y^2+k_z^2} \;\isdef \; \left\Vert\,\underline{k}\,\right\Vert$$

as needed.

Finally, scattering of plane waves is discussed in §G.8.1.