My task is to show that for infinitesimal rotation $R$ around an axis, denoted together by d$\boldsymbol{\phi}$, the vector field $\boldsymbol{A}(\boldsymbol{r})$ transforms as

$$ \boldsymbol{A}'(\boldsymbol{r}) = R\boldsymbol{A}(R^{-1}\boldsymbol{r}) = (1 - i(\boldsymbol{l} + \boldsymbol{s})\cdot \mathrm{d}\boldsymbol{\phi})\boldsymbol{A}(\boldsymbol{r}). $$

So I first shown that an arbitrary vector transforms under $R$ as $v_i\mapsto v'_i = v_i + \varepsilon_{ijk}\mathrm{d}\phi_jr_k$ and I define $M_{ik} := \varepsilon_{ijk}\mathrm{d}\phi_j$, which coincides with the known $\boldsymbol{v} \mapsto \boldsymbol{v} + \mathrm{d}\boldsymbol{\phi} \times \boldsymbol{v}$. Then $\boldsymbol{A}$ transforms as

$$ \boldsymbol{A}'(\boldsymbol{r}) = \boldsymbol{A}(\boldsymbol{r} - d\boldsymbol{\phi}\times\boldsymbol{r}) + d\boldsymbol{\phi}\times\boldsymbol{A}(\boldsymbol{r} - d\boldsymbol{\phi}\times\boldsymbol{r}), $$

which is in first order in Taylor expansion approximately

$$ \boldsymbol{A}'(\boldsymbol{r}) = \boldsymbol{A}(\boldsymbol{r}) - D_{\boldsymbol{r}}\boldsymbol{A} \cdot (d\boldsymbol{\phi}\times \boldsymbol{r}) + d\boldsymbol{\phi}\times\boldsymbol{A}(\boldsymbol{r}), $$

where $D_{\boldsymbol{r}}\boldsymbol{A}$ is the Jacobian at $\boldsymbol{r}$. And in terms of indices:

$$ A'_i = A_i - \partial_j A_i (d\boldsymbol{\phi}\times \boldsymbol{r})_j + d\phi_j\varepsilon_{ijk}A_k = A_i - \partial_j A_i \varepsilon_{jlk} d\phi_l r_k + d\phi_j\varepsilon_{ijk}A_k. $$

Now I have trouble putting this result in the form $A'_i = (\dots)_{ij}A_j$


1 Answer 1


Assuming you're right, a little index relabelling helps:$$\begin{align}A_i^\prime&=A_i-\epsilon_{mlk}d\phi_l r_k\partial_mA_i+d\phi_k\epsilon_{ikj}A_j\\&=(\delta_{ij}(1-\epsilon_{mlk}d\phi_lr_k\partial_m)+d\phi_k\epsilon_{ikj})A_j.\end{align}$$In these three terms, the second (third) has been relabelled with $j\leftrightarrow m$ ($j\leftrightarrow k$).

  • $\begingroup$ Yes I tried adding a delta to change the indices, but it intuitively seems wrong, as now I'd have two terms which are proportional to identity (or at least diagonal). It's actually quite surprising that this result isn't somewhat more widely known, I can't find any literature to check my result. $\endgroup$
    – mb28025
    Oct 18, 2021 at 19:16

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