Invariance of differential operators How do we prove that del operator is invariant under any kind of change of coordinates, specifically under galilean transformations? I am getting an extra term containing the relative velocity of two frames in denominator, componentwise.
Invariance of divergence can be under stood since it is a scalar field. But how do we understand this?
 A: This is a good exercise to do, so I will only outline the steps. The problem ultimately boils down to finding out how partial derivatives transform under a coordinate transformation. For simplicity, consider two sets of independent variables $(x,t)$ and $(x', t')$ that are related to each other, and suppose you want to find how their partial derivatives are related. Once this is done, you can find out how operators like $\nabla$ and $\Box$ are related between these two "coordinate systems".
If you have an appropriately differentiable (test) function of the first pair, say $f(x,t)$, then we can write $$\text{d}f = \frac{\partial f}{\partial x}\text{d}x + \frac{\partial f}{\partial t}\text{d}t.$$
Using the multivariable chain rule, we can then calculate the derivative with respect to a new variable (say $x'$) as
\begin{aligned}
\frac{\partial f}{\partial x'} &= \frac{\partial f}{\partial x}\frac{\partial x}{\partial x'} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial x'}\\ &= \frac{\partial x}{\partial x'}\frac{\partial f}{\partial x} + \frac{\partial t}{\partial x'}\frac{\partial f}{\partial t}.
\end{aligned}
Since this relation must hold for any function $f$, we can say that
$$\boxed{\frac{\partial }{\partial x'} \equiv \frac{\partial x}{\partial x'}\frac{\partial }{\partial x} + \frac{\partial t}{\partial x'}\frac{\partial }{\partial t}.}$$
Now, if you know the explicit forward and inverse transformations, for example:
\begin{aligned} x' &= \psi(x,t)  \quad \quad \quad \quad    &x  &= \Psi(x',t')\\
                t' & =\phi(x,t)               &t  & = \Phi(x',t')
\end{aligned}
then you can calculate $$\frac{\partial x}{\partial x'} = \frac{\partial \Psi}{\partial x'} \quad \quad \quad \quad \frac{\partial t}{\partial x'} = \frac{\partial \Phi}{\partial x'}$$
and so on. Now repeat the process for the other variables. Once you have the form of the individual operators, the rest should be straightforward.

Note: Remember that $$\frac{\partial x}{\partial x'} \neq \left(\frac{\partial x'}{\partial x} \right)^{-1},$$
and so on, since the first term means you're differentiating $x$ with respect to $x'$ while keeping $t'$ constant, while the second means that you are differentiating $x'$ with respect to $x$ while keeping $t$ constant. This is the most common one made in such exercises.
A: I don't think it is supposed to be. It is supposed to return "appropriate" results in order to be consistent in equations. If the slope in front of you is 20° and you turn right 90°, the slope in front of you is no longer 20° but 0°. Geometric consistency is what is needed, not invariance.
