Irrotational fields and change of reference frame Given a reference frame $(x_1,x_2,x_3)$ and a vector field $\overrightarrow{V}(x_1,x_2,x_3)$, in this frame, if $\overrightarrow{\nabla}\times\overrightarrow{ V}(x_1,x_2,x_3)=0$ the field is conservative and irrotational. Changing the reference frame $(x_1,x_2.x_3)\to (x_1^{'},x_2{'},x_3{'})$ is the field still irrotational and so, conservative in the new reference frame? Thanks.
 A: Suppose the frame of reference transformation can be represented by a change of basis in $\mathbb{R}^3$,  and plus a translation:
$$\newcommand{\b}{\mathbf} 
T: \b{x} = (\b{e}_1,\b{e}_2,\b{e}_3)\begin{pmatrix}x_1\\x_2\\x_3 \end{pmatrix}\mapsto \mathbf{x}' = (\b{e}_1',\b{e}_2',\b{e}_3')\begin{pmatrix}x_1'\\x_2'\\x_3' \end{pmatrix} + \b{x}_0,
$$
where basis got transformed by $ \b{e}_i' = S\b{e}_i$, then the coordinates is $(x_i') = S^{-1} ( x_i - x_{0,i})$.
$$
\b{V} = \sum_{i=1}^3 v_i \b{e}_i = \sum_{i=1}^3 v_i' \b{e}_i',
$$
using transformation map:
$$
\sum_{i=1}^3 v_i' \b{e}_i' = \sum_{i=1}^3 v_i'S \b{e}_i.
$$
The curl in the new reference frame: 
$$
\nabla' \times \b{V} :=\begin{vmatrix} \b{e}_1' & \b{e}_2' & \b{e}_2' \\ 
\partial_{x_1'} & \partial_{x_2'} & \partial_{x_3'}\\  v_1' & v_2' & v_3' \end{vmatrix}
= \sum_{i=1}^3 \nabla' \times (v_i'S \b{e}_i)
\\
= \sum_{i=1}^3 \nabla' v_i' \times S \b{e}_i = \sum_{i=1}^3 \left(\nabla  v_i\Big|_{\b{x} = T^{-1}(\b{x}')} S^t\right ) \times S  \b{e}_i.
$$
Now by an idenity:
$$
(M\b{a}) \times (M\b{b}) = (\det M) M^{-t}(\b{a} \times \b{b})
$$
Above is:
$$
\nabla' \times \b{V} = \det(S)S^{-t} \nabla\times \b{V}.
$$
Hence if the new reference frame is an affine linear transformation of the old one, zero curl will be preserved.
