Phase velocity definition and direction in Thorne/Blandford Optics

Question

Following the book's very introduction to optics:

Suppose we have a monochromatic wave described by $$\psi = A e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} = A e^{i \varphi}.$$

The apparent intuition given to the definition of the wave velocity is this: given a fixed value for $\varphi$, say, $\varphi_0$, the geometrical (3D) locus of the "phase of the wavefront" associated with $\varphi_0$ is propagating on space (as time passes), this displacement described by: $$ \varphi_0 = \mathbf{k} \cdot \mathbf{x} - \omega t. $$

Now it makes sense to wonder what is the velocity of said propagation of geometrical locus as it evolves with time. Therefore, we might implicitly differentiate the above expression w.r.t. time and get: $$ 0 = \mathbf{k} \cdot \left(\frac{\partial{\mathbf{x}}}{\partial t}\right)_{\varphi = \varphi_0} - \omega \;\; \Longrightarrow \;\; \frac{\omega}{k} = \left(\frac{\partial{\mathbf{x}}}{\partial t}\right)_{\varphi = \varphi_0} \cdot\mathbf{\hat{k}} $$

If we define $\mathbf{v}_\varphi = \left(\frac{\partial{\mathbf{x}}}{\partial t}\right)_{\varphi = \text{constant}}$ we get from above: $$ \frac{\omega}{k} = \mathbf{v_\varphi} \cdot \mathbf{\hat{k}} $$

we get the usual textbook definition of phase velocity $v_\varphi = \frac{\omega}{k}$ only if $\mathbf{v_\varphi}$ is parallel to $\mathbf{\hat{k}}$. My question is:

Where exactly does that supposition ($\mathbf{v_\varphi}$ is parallel to $\mathbf{\hat{k}}$) come from? Are there cases where it fails and, if it does fail, does the above dot product relation ($\frac{\omega}{k} = \mathbf{v_\varphi} \cdot \mathbf{\hat{k}}$) and definition of phase velocity still hold?

Answer 1

If your phase is given by $$ \phi(\vec{x},t)=\vec{k}\vec{x}-\omega t $$ the phase fronts are by definition propagating along the direction of $\vec{k}$.

This is apparent as $\phi$ only changes in direction $\vec{k}$. Independent of time the phase is invariant under translations orthogonal to $\vec{k}$ and therefore definitely not propagating in these directions. To easily see this choose your coordinate system such that $\vec{k}\parallel\hat{z}$, that is $$ \phi=kz-\omega t \tag{1} $$

and observe how $\partial_x \phi=\partial_y \phi=0$.

So when we talk about propagation velocity of the phase it only makes sense to speak about phase propagation (and therefore a phase propagation velocity vector) in direction of $\vec{k}$

$$\vec{v}_\phi\parallel\vec{k}$$

---

To me it also does not make much sense just to apply the time derivative on the space coordinate and define this to be phase velocity. This is not inutitve to me. I think better is to ask the question by how much $\delta z$ does a phase front $\phi(z,t)=\phi_0$ travel during a time $\delta t$.

$$ \phi_0=\phi(z+\delta_z,t+\delta_t)=k(z+\delta z)-\omega(t+\delta t)=\phi(z,t)+k \delta z-\omega \delta t $$

$$ \rightarrow \frac{\delta z}{\delta t}=\frac{\omega}{k} $$ Interpreting this ratio as phase velocity is now obvious, as the phase travels the distance $\delta z$ in ime $\delta t$.