Irrotational fields and change of reference frame

Question

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.

Answer 1

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.