I have seen the following dispersion relation as a differential equation:
$$
\frac{ \partial \mathbf{k} }{ \partial t } + \left( \mathbf{V}_{g} \cdot \nabla \right) \mathbf{k} = 0 \tag{1}
$$
in the page number 135 of the textbook Holton, J. R., & Hakim, G. J. (2013). An introduction to dynamic meteorology (5th ed., Vol. 88). Amsterdam: Elsevier.
, and also here:
https://physics.stackexchange.com/a/381974/393802
But it does not make sense to me, so I am not sure it is a (propagated) typo or there is something I do not understand.
Let's follow the cited book and assume that we have a 2-dimensional wave and that the phase of the wave, $\phi$, is real. The wave number, $\mathbf{k}$ and the angular frequency, $\omega$, can be expressed as differential operations on the angular phase, $\phi$. The angular phase $\phi$ for a two dimensional wave can be expressed as $\phi = kx+ly- \omega t$, where $k$ and $l$ are the wave numbers in the $x$ and $y$ directions, respectively, and $\omega$ is the angular frequency. The wave vector is defined as $ \mathbf{k} = k \mathbf{i} + l \mathbf{j} $, So, we can get the wave vector from the angular phase by applying the spatial gradient:
$$
\mathbf{k} = \nabla \phi = {\partial \phi \over \partial x} \mathbf{i} + {\partial \phi \over \partial y} \mathbf{j} \tag{2}
$$
Or, in matrix notation and assuming that the gradient is a row vector:
$$
\mathbf{k} = \nabla \phi = \left(
{\partial \phi \over \partial x} ,
{\partial \phi \over \partial y}\right) \tag{3}
$$
On the other hand we can get the angular frequency by applying the time derivative on the angular phase:
$$
\omega = – {\partial \phi \over \partial t} \tag{4}
$$
Now we can take the time derivative of 3 to get:
$$
{\partial \mathbf{k} \over \partial t} =
{\partial \over \partial t} \nabla \phi =
\nabla {\partial \phi \over \partial t} =
– \nabla \omega
\tag{5}
$$
where we have used 4 in the last step.
That is to say:
$$
{\partial \mathbf{k} \over \partial t} + \nabla \omega = 0
\tag{6}
$$
We now recall that $\omega$ is a function of $\mathbf{k}$ (dispersion relation), id est, $\omega (x, y) = \omega (\mathbf{k} (x,y)) = \omega(k(x,y),l(x,y))$, so that we can rewrite the gradient of $\omega$ in "wavenmber coordinates" as:
$$
\begin{align}
\nabla \omega &= \left(
{\partial \omega \over \partial x} ,
{\partial \omega \over \partial y}\right)
\\
&= \left({\partial \omega \over \partial k} {\partial k \over \partial x} + {\partial \omega \over \partial l} {\partial l \over \partial x}
,
{\partial \omega \over \partial k} {\partial k \over \partial y} + {\partial \omega \over \partial l} {\partial l \over \partial y}
\right)
\\
&=
\left(
{\partial \omega \over \partial k} ,
{\partial \omega \over \partial l}\right)
\left(\begin{array}{cc}
\partial k \over \partial x & \partial k \over \partial y\\
\partial l \over \partial x & \partial l \over \partial y
\end{array}\right)
\\
&= \nabla _k \omega \cdot \nabla \mathbf{k}
\end{align}
\tag{7}
$$
where $\nabla \mathbf{k}$ is the jacobian matrix of $\mathbf{k}$.
Replacing this into ec 6, gives:
$$
\frac{ \partial \mathbf{k} }{ \partial t } + \nabla _k \omega \cdot \nabla \mathbf{k} = 0 \tag{8}
$$
Recalling that the group velocity is $\mathbf{V}_{g} = \nabla _k \omega$:
$$
\frac{ \partial \mathbf{k} }{ \partial t } + \mathbf{V}_{g} \cdot \nabla \mathbf{k} = 0 \tag{9}
$$
This contrasts with eq. 1, because $\mathbf{V}_{g} \cdot \nabla \mathbf{k} \ne \left( \mathbf{V}_{g} \cdot \nabla \right) \mathbf{k}$
Or, at least, this is what I think. To see this, let $\mathbf{V}_{g} = (v_x, v_y)$. Then:
$$
\begin{align}
\left( \mathbf{V}_{g} \cdot \nabla \right) \mathbf{k} &=
\left[
( v_x, v_y) \cdot
\left(
{\partial \over \partial x} ,
{\partial \over \partial y}
\right)
\right]
\left(
k, l
\right)
\\
&= \left(v_x {\partial \over \partial x} +
v_y {\partial \over \partial y}
\right) (k, l)
\\
&= \left(v_x {\partial k \over \partial x} +
v_y {\partial k \over \partial y},
v_x {\partial l \over \partial x} +
v_y {\partial l \over \partial y}
\right)
\end{align}
\tag{10}
$$
In order to compare this with $\mathbf{V}_{g} \cdot \nabla \mathbf{k}$, recall that $\mathbf{V}_{g} = \nabla _k \omega = \left({\partial \omega \over \partial k}, {\partial \omega \over \partial l} \right) = ( v_x, v_y )$, so that the second line of eq. 9 is:
$$
\begin{align}
\mathbf{V}_{g} \cdot \nabla \mathbf{k} &= \nabla \omega
\\
&= \left(
{\partial \omega \over \partial x} ,
{\partial \omega \over \partial y}\right)
\\
&= \left({\partial \omega \over \partial k} {\partial k \over \partial x} + {\partial \omega \over \partial l} {\partial l \over \partial x}
,
{\partial \omega \over \partial k} {\partial k \over \partial y} + {\partial \omega \over \partial l} {\partial l \over \partial y}
\right)
\\
&=
\left({v_x} {\partial k \over \partial x} + {v_y} {\partial l \over \partial x}
,
{v_x} {\partial k \over \partial y} + {v_y} {\partial l \over \partial y}
\right)
\end{align}
\tag{11}
$$
Which is not the same as eq. 10
What am I missing?
Best Answer
Here is my way of proceeding. Most of it agrees with you, but I can't quite pin down where we differ. Perhaps the mixed-partial trick of interchanging the $i$ and $j$?
The essential feature of a wave train is that at any particular point of space and time, ${\bf x}$ and $t$, it has a definite phase $\Theta({\bf x},t)$. Once we know this phase, we can define the local frequency $\omega$ and wave-vector ${\bf k}$ by $$ \omega= -\left(\frac{\partial\Theta}{\partial t}\right)_{\bf x},\qquad k_i = \left(\frac{\partial\Theta}{\partial x_i }\right)_t. $$ These definitions are motivated by the idea that $$ \Theta({\bf x},t)\sim {\bf k}\cdot {\bf x} -\omega t, $$ at least locally.
The subscripts on the partial derivatives indicate what is being left fixed when we differentiate. We must be careful about this, because we want to use the dispersion equation to express $\omega$ as a function of ${\bf k}$ and ${\bf x}$, and the wave-vector ${\bf k}$ will itself be a function of ${\bf x}$ and $t$.
We wish to understand how ${\bf k}$ changes as the wave propagates through a slowly varying medium. We introduce the inhomogeneity by assuming that the dispersion equation $\omega=\omega ({\bf k})$, which is initially derived for a uniform medium, can be extended to $\omega =\omega({\bf k}, {\bf x})$, where the $\bf x$ dependence arises, for example, as a result of a position-dependent refractive index. This assumption is only an approximation, but it is a good approximation when the distance over which the medium changes is much larger than the distance between wavecrests.
Applying the equality of mixed partials to the definitions of ${\bf k}$ and $\omega$ gives us $$ \left(\frac{\partial \omega}{\partial x_i }\right)_t= -\left(\frac{\partial k_i}{\partial t }\right)_{\bf x},\quad \left(\frac{\partial k_i}{\partial x_j }\right)_{x_i}= \left(\frac{\partial k_j}{\partial x_i }\right)_{x_j}. $$ Again the subscripts indicate what is being left fixed when we differentiate.
Now use the chain rule $$ \left(\frac{\partial \omega}{\partial x_i }\right)_t= \left(\frac{\partial \omega}{\partial x_i }\right)_{\bf k} + \left(\frac{\partial \omega}{\partial k_j }\right)_{\bf x}\left(\frac{\partial k_j}{\partial x_i }\right)_t. $$ We now use one of the previous mixed-partial identities to interchange the $i$, $j$ labels in the fixed-$t$ partial derivative, and so get $$ \left(\frac{\partial k_i}{\partial t }\right)_{\bf x}+ \left(\frac{\partial \omega}{\partial k_j }\right)_{\bf x}\left(\frac{\partial k_i}{\partial x_j }\right)_t = -\left(\frac{\partial \omega}{\partial x_i }\right)_{\bf k}. $$ Interpreting the left hand side as a convective derivative $$ \frac{dk_i}{dt}\equiv \left(\frac{\partial k_i}{\partial t }\right)_{\bf x}+({\bf V}_g\cdot\nabla) k_i, $$ we see that we percive a time rate of change of the wavevector as
$$ \frac {dk_i}{dt}= -\left(\frac{\partial \omega}{\partial x_i }\right)_{\bf k} $$ provided we are moving at velocity $$ \frac {dx_i}{dt}=({\bf V}_g)_i= \left(\frac{\partial \omega}{\partial k_i }\right)_{\bf x}. $$ Since this is the group velocity, the packet of waves is actually travelling at this speed. The last two equations therefore tell us how the orientation and wavelength of the wave train evolve if we ride along with the packet as it is refracted by the inhomogeneity.
The formulae $$ \dot {\bf k} = -\frac{\partial \omega}{\partial {\bf x} }, \\ \dot {\bf x}= \phantom -\frac{\partial \omega}{\partial {\bf k} }, $$ are Hamilton's ray equations. Their similarity to the Hamilton equations of mechanics should be manifest.