Transformation of magnetic field under an isometry of space

Question

THE PROBLEM

Suppose we are in stationary conditions, in which case the following Maxwell Equations hold: $$ \underline{\nabla} \cdot \underline{B} \equiv 0 \\ \underline{\nabla} \times \underline{B} = \frac{\mu_0}{4\pi} \underline{J} $$ Let $\phi(\underline{x}):=R\underline{x} + \underline{a}$ be an isometry of $\Bbb R^3$ (i.e. $R$ is ortohogonal) and suppose that all the current lies in a bounded domain (so that we can have zero boundary conditions at infinity).

I would like to prove tha following claim:

The magnetic field $\underline{B}'$ generated by the current density $\underline{J}'(\underline{x}) := R\underline{J}(\phi^{-1}(\underline{x}))$ (which is the result of the so called "active transformation" of the source under the isometry) is given by: $$ \underline{B}'(\underline{x}) = \det(R)R\underline{B}(\phi^{-1}(\underline{x})) \hspace{5mm} (1) $$

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MY ATTEMPT

My idea is to use the uniqueness theorem for the magnetic field with zero boundary condition at infinity. Therefore, if I can show that (1) satisfies Maxwell's Equations I would be done. I have already verified the first equation but I have not been able to verify the second because I get stuck at:

$$[\underline{\nabla} \times \underline{B}(\underline{x})]_i = \epsilon_{ijk} \det(R) R_{km} R_{jl} \partial_l B_m(\psi^{-1}(\underline{x}))$$

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Any help would be much appreciated!

Answer 1

Perhaps you can use $$ \epsilon_{ijk}R_{ai}R_{bj}R_{ck}= \epsilon_{abc} {\rm det}(R) $$ to show that $\nabla \times {\bf B}$ transforms as a 3-(pseudo?)-vector under ${\rm O}(3)$?