是否所有波非横即纵？非纵横波的数学表示方法问询

Great question—this is a common point of confusion once you start digging beyond basic wave mechanics (like your deep dive into electromagnetic transverse wave proofs!). Let’s break this down clearly:

1. No, not all waves fit neatly into transverse or longitudinal categories #

Most introductory physics focuses on pure transverse (e.g., EM waves, string waves) or pure longitudinal (e.g., sound waves in air) examples, but there are plenty of waves that have both transverse and longitudinal components—we often call these "hybrid" or "mixed-mode" waves.

Some classic examples:

• Rayleigh waves: These are surface waves that propagate along the boundary of a solid (like seismic waves near the Earth’s surface). Particle motion is elliptical: part of the displacement is parallel to the wave’s direction of travel (longitudinal component), and part is perpendicular (transverse component).
• Waves in anisotropic elastic media: In materials where properties vary with direction (e.g., crystals), elastic waves can propagate with displacement vectors that are neither purely parallel nor perpendicular to the wave’s propagation vector.
• Certain fluid waves: For example, waves in shallow water can have both horizontal (longitudinal) and vertical (transverse) particle motion depending on depth.

2. Mathematical representation of non-pure transverse/longitudinal waves #

To formalize this, let’s start with the general description of a wave’s displacement field $\mathbf{u}(\mathbf{r}, t)$ (for mechanical waves; for EM waves, we’d use $\mathbf{E}$ and $\mathbf{B}$ fields, but the logic extends).

First, recall the definitions for pure waves:

• Pure longitudinal: Displacement is parallel to the wave vector $\mathbf{k}$ (direction of propagation). Mathematically: $\mathbf{u} \times \mathbf{k} = 0$, or $\mathbf{u} = u(\mathbf{r}, t) \hat{\mathbf{k}}$ where $\hat{\mathbf{k}}$ is the unit vector in the direction of $\mathbf{k}$.
• Pure transverse: Displacement is perpendicular to $\mathbf{k}$. Mathematically: $\mathbf{u} \cdot \mathbf{k} = 0$.

For a mixed-mode wave, we can decompose the displacement field into its longitudinal and transverse components:
$$\mathbf{u} = \mathbf{u}_l + \mathbf{u}_t$$
Where:

• $\mathbf{u}_l = (\mathbf{u} \cdot \hat{\mathbf{k}})\hat{\mathbf{k}}$: The longitudinal component (parallel to $\mathbf{k}$)
• $\mathbf{u}_t = \mathbf{u} - \mathbf{u}_l$: The transverse component (perpendicular to $\mathbf{k}$, since $\mathbf{u}_t \cdot \hat{\mathbf{k}} = 0$)

Example: Rayleigh Wave Displacement #

For a Rayleigh wave propagating along the $x$-axis (so $\mathbf{k} = k\hat{\mathbf{x}}$) in a semi-infinite solid ($z \geq 0$), the displacement field can be written as:
$$
\mathbf{u}(x, z, t) = \left( A e^{-\alpha z} + B e^{-\beta z} \right) \cos(kx - \omega t) \hat{\mathbf{x}} + \left( -A \frac{\alpha}{k} e^{-\alpha z} - B \frac{\beta}{k} e^{-\beta z} \right) \sin(kx - \omega t) \hat{\mathbf{z}}
$$
Here:

• The $\hat{\mathbf{x}}$ component is parallel to the wave propagation direction (longitudinal)
• The $\hat{\mathbf{z}}$ component is perpendicular (transverse)
• Both components are non-zero, so this wave is neither purely transverse nor purely longitudinal.

In more complex cases (like anisotropic media), you’d solve the wave equation’s eigenvalue problem to find the allowed displacement vectors $\mathbf{u}$, which will naturally have both parallel and perpendicular components relative to $\mathbf{k}$.

内容的提问来源于stack exchange，提问作者GimmeCats