$\psi(\vec{\mathbf{r}}, t) = C_1 f(\vec{\mathbf{r}} \cdot \vec{\mathbf{k}}/k - vt) + C_2 g(\vec{\mathbf{r}} \cdot \vec{\mathbf{k}}/k + vt)$

Question

I am currently studying Optics, fifth edition, by Hecht. I was presented with the three-dimensional differential wave equation

$$\dfrac{\partial^2 \psi}{\partial x^2} + \dfrac{\partial^2 \psi}{\partial y^2} + \dfrac{\partial^2\psi}{\partial z^2} = \dfrac{1}{v^2} \dfrac{\partial^2 \psi}{\partial t^2}.$$

I am then told that it can be shown that

$$\psi(x, y, z, t) = f(\alpha x + \beta y + \gamma z - vt)$$

and

$$\psi(x, y, z, t) = g(\alpha x + \beta y + \gamma z + vt)$$

are both plane-wave solutions of the differential wave equation. Finally, I am told that a linear combination of these solutions is also a solution, and we can write this as

$$\psi(\vec{\mathbf{r}}, t) = C_1 f(\vec{\mathbf{r}} \cdot \vec{\mathbf{k}}/k - vt) + C_2 g(\vec{\mathbf{r}} \cdot \vec{\mathbf{k}}/k + vt),$$

where $C_1$ and $C_2$ are constants.

My understanding is that $\vec{\mathbf{r}} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}$ and $\vec{\mathbf{k}} = k_x \hat{\mathbf{i}} + k_y \hat{\mathbf{j}} + k_z \hat{\mathbf{k}}$, and $k$ is the magnitude of $\vec{\mathbf{k}}$. But I'm still confused about how $C_1 f(\vec{\mathbf{r}} \cdot \vec{\mathbf{k}}/k - vt)$ and $C_2 g(\vec{\mathbf{r}} \cdot \vec{\mathbf{k}}/k + vt)$ are equivalent to $\psi(x, y, z, t) = f(\alpha x + \beta y + \gamma z - vt)$ and $\psi(x, y, z, t) = g(\alpha x + \beta y + \gamma z + vt)$, respectively.

I would greatly appreciate it if someone would please take the time to clarify this.

Answer 1

If \begin{equation} \alpha = k_x/k, \beta = k_y/k, \gamma = k_z/k, \end{equation} then \begin{equation} \alpha x + \beta y + \gamma z = (k_x x + k_y y + k_z z)/k = \vec{\mathbf{k}}\cdot \vec{\mathbf{x}}/k. \end{equation}

It is worth noting that strictly speaking $f(\alpha x + \beta y + \gamma z \pm vt)$ is a solution of wave equation only if $\alpha^2 + \beta^2 + \gamma^2 = 1$, which is easy to check by substituting these solutions into the equation (left as an exercize).