A.1 Case I

We assume that $\|\boldsymbol{f}\|_{L^\infty (\mathbb{R}^3\backslash \mathcal{B}_1(0))} \lt \infty$ and

(A2) \begin{equation} \int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)}|\boldsymbol{f}(y)|\,{\textrm d}y \lt \infty . \end{equation}

We compute

(A3) \begin{eqnarray} &&\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} {\unicode{x1D642}_\gamma }(x-y)\boldsymbol{f}(y)\,{\textrm d}y\nonumber \\ &=& {\unicode{x1D642}_\gamma }(x)\left (\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} \boldsymbol{f}(y)\,{\textrm d}y\right ) +\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} ({\unicode{x1D642}_\gamma }(x-y)-{\unicode{x1D642}_\gamma }(x))\boldsymbol{f}(y)\,{\textrm d}y\nonumber \\ &=& {\unicode{x1D642}_\gamma }(x)\left (\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} \boldsymbol{f}(y)\,{\textrm d}y\right ) +\frac {1}{|x|}\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} \left ({\unicode{x1D642}_\gamma }(\hat {x}-\frac {y}{|x|})- {\unicode{x1D642}_\gamma }(\hat {x})\right )\boldsymbol{f}(y)\,{\textrm d}y\nonumber \\ &=& {\unicode{x1D642}_\gamma }(x)\left (\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} \boldsymbol{f}(y)\,{\textrm d}y\right ) + o\left (\frac {1}{|x|}\right )\!. \end{eqnarray}

In order to characterize the $o ({1}/{|x|})$ term, we assume that

(A4) \begin{equation} \boldsymbol{f}(y) \lesssim \frac {1}{|y|^4}. \end{equation}

Let $L \gg 1$ . Then we have

(A5) \begin{eqnarray} &&\left |\frac {1}{|x|}\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} \left ({\unicode{x1D642}_\gamma }(\hat {x}-\frac {y}{|x|})- {\unicode{x1D642}_\gamma }(\hat {x})\right )\boldsymbol{f}(y)\,{\textrm d}y\right |\nonumber \\ &=& \left |\frac {1}{|x|} \int _{1 \leq |y| \lt \frac {|x|}{L}} +\int _{\frac {|x|}{L} \leq |y| \lt L|x|} +\int _{L|x| \leq |y|} \left (\frac {{\unicode{x1D642}_\gamma }(\widehat {\hat {x}-\frac {y}{|x|}})} {\left |\hat {x}-\frac {y}{|x|}\right |}- {\unicode{x1D642}_\gamma }(\hat {x})\right )\boldsymbol{f}(y)\,{\textrm d}y\right | \nonumber \\ & \lesssim & \frac {1}{|x|}\int _{1 \leq |y| \lt \frac {|x|}{L}} \frac {|y|}{|x|}\frac {1}{|y|^4}\,{\textrm d}y + \frac {1}{|x|}\int _{\frac {|x|}{L} \leq |y| \lt L|x|} \frac {1}{|y|^4}\,{\textrm d}y + \frac {1}{|x|}\int _{L|x| \leq |y|} \frac {|x|}{|y|}\frac {1}{|y|^4}\,{\textrm d}y\nonumber \\ & \lesssim & \frac {1}{|x|^2}\int _{1}^{\frac {|x|}{L}} \frac {r^2}{r^3}\,{\textrm d}r +\frac {1}{|x|}\int _{\frac {|x|}{L}}^{L|x|} \frac {r^2}{r^4}\,{\textrm d}r +\int _{L|x|}^\infty \frac {r^2}{r^5}\,{\textrm d}r\nonumber \\ & \lesssim & \frac {\log |x|}{|x|^2} + \frac {1}{|x|^2}\nonumber \\ & \lesssim & \frac {\log |x|}{|x|^2}. \end{eqnarray}

Note that in the above, we have used the fact that $({1}/{|\hat {x}-({y}/{|x|})|})$ is an integrable singularity at $y\sim x$ in $\mathbb{R}^3$ .

A.2 Case II

Next, we assume that $f$ has the following large spatial asymptotic behaviour:

(A6) \begin{equation} |\boldsymbol{f}| \leq O\left (\frac {A+ B\ln r}{r^k}\right ) \,\,\,\text{for some $k\geq 3$ and all $r \gg 1$.} \end{equation}

Then we compute

(A7) \begin{align} & \quad \left |\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} {\unicode{x1D642}_\gamma }(x-y)\boldsymbol{f}(y)\,{\textrm d}y\right |\nonumber \\&\lesssim \int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)}\frac {A + B\ln |y|}{|y|^k}\frac {1}{|x-y|}\,{\textrm d}y \nonumber \\& \lesssim \frac {1}{|x|}\int _{\mathbb{R}^3\backslash \mathcal{B}_{\frac {1}{|x|}}(0)} \frac {A + B\ln |z| + B\ln |x|}{|x|^k|z|^k}\frac {1}{|\widehat {x}-z|}\,|x|^3 dz \quad \text{(where $\widehat {x} = \frac {x}{|x|}$ and $z=\frac {y}{|x|}$)} \nonumber \\& \lesssim \frac {1}{|x|^{k-2}}\left [ \int _1^\infty \frac {A + B\ln |r| + B\ln |x|}{r^{k+1}}\,r^2{\textrm d}r + \int _{\frac {1}{|x|}}^1 \frac {A + B\ln |r| + B\ln |x|}{r^{k}}\,r^2{\textrm d}r \right ] \nonumber \\&\quad \text{(note that the singularity at $\widehat {x}$ is integrable)}\nonumber \\& \lesssim \left \{ \begin{array}{ll} \dfrac {A\ln |x| + B\ln |x| + B(\ln |x|)^2}{|x|}, & \text{if $k=3$,}\\[8pt]\dfrac {A + B\ln |x|}{|x|}, & \text{if $k\gt 3$.} \end{array} \right .\\[6pt] \nonumber \end{align}

Similarly,

(A8) \begin{align} \left |\int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} - \unicode{x1D666}(x-y)\boldsymbol{\cdot }\boldsymbol{f}(y)\,{\textrm d}y\right | &\lesssim \int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)}\frac {A + B\ln |y|}{|y|^k} \frac {1}{|x-y|^2}\,{\textrm d}y \\[-12pt] \nonumber\end{align}

(A9) \begin{align} & \lesssim \left \{ \begin{array}{ll} \dfrac {A\ln |x| + B\ln |x| + B(\ln |x|)^2}{|x|^2}, & \text{if $k=3$,}\\ \dfrac {A + B\ln |x|}{|x|^2}, & \text{if $k\gt 3$.} \end{array} \right . \\[6pt] \nonumber\end{align}

Note that for $k=3$ , even if $B=0$ , a $\ln |x|$ term can appear in the bulk integrals.

A.3 Case III

In order to do a more careful analysis of the case $k=3$ of which we are concerned most, we further assume that $\boldsymbol{f}$ is homogeneous of degree $-3$ , i.e.

(A10) \begin{equation} \boldsymbol{f}(\lambda x) = \lambda ^{-3}\boldsymbol{f}(x)\,\,\,\text{so that}\,\,\, \boldsymbol{f}(x) = \frac {\boldsymbol{f}(\hat {x})}{|x|^3}. \end{equation}

Note that the integrand ${\unicode{x1D642}_\gamma }(x-\boldsymbol{\cdot })\boldsymbol{f}(\boldsymbol{\cdot }) \sim ({1}/{|y|^4})$ is integrable in $\mathbb{R}^3\backslash \boldsymbol{B}_1(0)$ . Hence, we can use Fubini’s Theorem to compute ${\textrm d}y$ iteratively as $r^2\,{\textrm d}\omega \,{\textrm d}r$ , where $r = |y|$ and $\omega =\hat {y}$ . To this end, we have

(A11) \begin{align} \int _{\mathbb{R}^3\backslash \boldsymbol{B}_1(0)} {\unicode{x1D642}_\gamma }(x-y)\boldsymbol{f}(y)\,{\textrm d}y &=\int _{\mathbb{S}^2} \left [\int _1^\infty {\unicode{x1D642}_\gamma }(x-r\omega )\frac {\boldsymbol{f}(\omega )}{r^3}r^2\,{\textrm d}r\right ] \,{\textrm d}\omega \nonumber \\[3pt]&=\frac {1}{|x|}\int _{\mathbb{S}^2} \left [\int _1^\infty {\unicode{x1D642}_\gamma }(\hat {x}-\frac {r}{|x|}\omega )\frac {1}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega \nonumber \\[3pt]&=\frac {1}{|x|}\int _{\mathbb{S}^2} \left [\int _{\frac {1}{|x|}}^\infty {\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega . \end{align}

For $|x| \gt 2$ , we write

(A12) \begin{eqnarray} \int _{\frac {1}{|x|}}^\infty {\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r &=& \int _{\frac 12}^\infty {\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r + \int _{\frac {1}{|x|}}^{\frac 12}{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r. \end{eqnarray}

Now for $\omega \neq \hat {x}$ , define

(A13) \begin{equation} \unicode{x1D643}_1(\hat {x},\omega ):=\int _{\frac 12}^\infty {\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r = O(1). \end{equation}

Then we have

(A14) \begin{equation} \int _{\mathbb{S}^2} \left [\int _{\frac 12}^\infty {\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega = \int _{\mathbb{S}^2}\unicode{x1D643}_1(\hat {x}, \omega )\boldsymbol{f}(\omega )\,{\textrm d}\omega . \end{equation}

On the other hand, we have $\displaystyle \int _{({1}/{|x|})}^{({1}/{2})}{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )({1}/{r})\,{\textrm d}r = O(\ln |x|)$ . To analyse this term, we compute

(A15) \begin{align} & \quad \int _{\mathbb{S}^2} \left [\int _{\frac {1}{|x|}}^\frac 12{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )\frac {1}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega \nonumber \\[3pt]&= \int _{\mathbb{S}^2} \left [\int _{\frac {1}{|x|}}^\frac 12\frac {{\unicode{x1D642}_\gamma }(\hat {x})}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega + \int _{\mathbb{S}^2} \left [\int _{\frac {1}{|x|}}^\frac 12\frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )-{\unicode{x1D642}_\gamma }(\hat {x})}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega \nonumber \\[3pt]&= {\unicode{x1D642}_\gamma }(\hat {x})\ln \left(\frac {|x|}{2}\right)\int _{\mathbb{S}^2}\boldsymbol{f}(\omega )\,{\textrm d}\omega + \int _{\mathbb{S}^2} \unicode{x1D643}_2(\hat {x},\omega ) \boldsymbol{f}(\omega )\,{\textrm d}\omega \nonumber \\[3pt]& \quad -\int _{\mathbb{S}^2} \left [\int _0^{\frac {1}{|x|}}\frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )-{\unicode{x1D642}_\gamma }(\hat {x})}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega \end{align}

where

(A16) \begin{equation} \unicode{x1D643}_2(\hat {x},\omega ) := \int _0^\frac 12 \frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )-{\unicode{x1D642}_\gamma }(\hat {x})}{r}\,{\textrm d}r=O(1) \end{equation}

because $({{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )-{\unicode{x1D642}_\gamma }(\hat {x})}/{r})=O(1)$ as $r\to 0^+$ . Lastly, we have

(A17) \begin{equation} \int _{\mathbb{S}^2} \left [\int _0^{\frac {1}{|x|}}\frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )-{\unicode{x1D642}_\gamma }(\hat {x})}{r}\,{\textrm d}r\right ] \boldsymbol{f}(\omega )\,{\textrm d}\omega =O\left(\frac {1}{|x|}\right). \end{equation}

Combining (A13), (A15), (A16), we define

(A18) \begin{align} \unicode{x1D643}(\hat {x},\omega ) &:=\unicode{x1D643}_1(\hat {x},\omega )+\unicode{x1D643}_2(\hat {x},\omega ) - {\unicode{x1D642}_\gamma }(\hat {x})\ln (2)\nonumber \\[3pt]&= \int _0^\frac 12 \frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )-{\unicode{x1D642}_\gamma }(\hat {x})}{r}\,{\textrm d}r +\int _{\frac 12}^\infty \frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )}{r}\,{\textrm d}r - {\unicode{x1D642}_\gamma }(\hat {x})\ln (2). \end{align}

Note that the above does not depend on the choice of $(1/2)$ . Hence, it can also be equivalently written as

(A19) \begin{equation} \unicode{x1D643}(\hat {x},\omega ) = \lim _{\epsilon \to 0}\left [ \int _{\epsilon }^\infty \frac {{\unicode{x1D642}_\gamma }(\hat {x}-r\omega )}{r}\,{\textrm d}r +{\unicode{x1D642}_\gamma }(\hat {x})\ln (\epsilon ) \right ]. \end{equation}

Then we have for $|x|\gg 1$ that

(A20) \begin{align} \int _{\varOmega } {\unicode{x1D642}_\gamma }(x-y)\boldsymbol{f}(y)\,{\textrm d}y & = {\unicode{x1D642}_\gamma }(x)\ln |x|\int _{\mathbb{S}^2}\boldsymbol{f}(\omega )\,{\textrm d}\omega \nonumber \\& \quad +\frac {1}{|x|}\int _{\mathbb{S}^2} \unicode{x1D643}(\hat {x},\omega ) \boldsymbol{f}(\omega )\,{\textrm d}\omega + O\left (\frac {1}{|x|^2}\right )\!. \end{align}

Hence, if $\displaystyle \int _{\mathbb{S}^2}\boldsymbol{f}(\omega )\,{\textrm d}\omega =0$ , then

(A21) \begin{equation} \int _{\varOmega } {\unicode{x1D642}_\gamma }(x-y)\boldsymbol{f}(y)\,{\textrm d}y =\frac {1}{|x|}\int _{\mathbb{S}^2} \unicode{x1D643}(\hat {x},\omega ) \boldsymbol{f}(\omega )\,{\textrm d}\omega + O\left (\frac {1}{|x|^2}\right )\!. \end{equation}

A.4 Mean zero condition of the inhomogeneous term

The conclusion of the previous section, in particular (A21), will be used to handle two terms. One is the dominant term $\textrm {div}\,\mathcal{A}_\gamma (x)$ in $f_\gamma (\boldsymbol{v})$ (see (5.11)). Another is $D^2\unicode{x1D640}$ in the analysis of $I_1$ in (5.34).

For $\textrm {div}\,\mathcal{A}_\gamma (x)$ , recall its form (4.28), $\displaystyle \textrm {div}\,\mathcal{A}_\gamma (x) = \gamma ({\boldsymbol{F}(\hat {x})}/{r^3}) + O ({1}/{r^4} ).$ We will show that

(A22) \begin{equation} \int _{{\mathbb S}^2}\boldsymbol{F}(\hat {x})\,{\textrm d}\sigma = 0. \end{equation}

To this end, we write $\mathcal{A}_\gamma$ explicitly as

(A23) \begin{eqnarray} \mathcal{A}_\gamma &=& \gamma _1\mathcal{A}_1 + \gamma _2\mathcal{A}_2 + \gamma _3\mathcal{A}_3 + \gamma _9\mathcal{A}_9 \nonumber \\&=& \gamma _1\big (\unicode{x1D64C}_*(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}) - (\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\unicode{x1D64C}_* \big ) +\gamma _2\big (\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}\big )\nonumber \\&&+ \frac {\gamma _3}{2}\left (\unicode{x1D64C}_*(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}) + (\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\unicode{x1D64C}_*\right ) +\gamma _9\big ((\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\boldsymbol{\cdot }\unicode{x1D64C}_*\big )\unicode{x1D64C}_*. \end{eqnarray}

By considering only the dominating term $({1}/{r})$ in the expression (3.4) of $\unicode{x1D64C}$ , we have

(A24) \begin{eqnarray} \unicode{x1D64C} - \unicode{x1D64C}_*& = & \left ({-}\frac {w\unicode{x1D64C}_*}{1+w}\right )\frac {1}{r} + O\left (\frac {1}{r^2}\right )\!,\nonumber \\\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C} & = & \left ({-}\frac {w\unicode{x1D64C}_*}{1+w}\right ) \left \langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right \rangle + O\left (\frac {1}{r^3}\right )\!,\nonumber \\\unicode{x1D64C}_*(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}),\,\,\, (\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\unicode{x1D64C}_* & = & \left ({-}\frac {w\unicode{x1D64C}_*^2}{1+w}\right ) \left \langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right \rangle + O\left (\frac {1}{r^3}\right )\!, \nonumber \\\big ((\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\boldsymbol{\cdot }\unicode{x1D64C}_*\big )\unicode{x1D64C}_* & = & \left ({-}\frac {w|\unicode{x1D64C}_*|^2\unicode{x1D64C}_*}{1+w}\right ) \left \langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right \rangle + O\left (\frac {1}{r^3}\right )\!. \end{eqnarray}

Hence, we can write for some constant $3\times 3$ matrix $\boldsymbol{M}_* = (m_{\textit{ij}})$ that

(A25) \begin{equation} \textrm {div}\,\mathcal{A}_\gamma = \text{div}\left ( \left \langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right \rangle \boldsymbol{M}_* \right ) + O\left (\frac {1}{r^4}\right )\!. \end{equation}

Note that the term in $\mathcal{A}_\gamma$ multiplied by $\gamma _1$ completely vanishes.

We next claim that

(A26) \begin{equation} \int _{\mathbb S^2} \left .\text{div}\left ( \left \langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right \rangle \boldsymbol{M}_* \right )\right |_{r=1}\,{\textrm d}\sigma = 0. \end{equation}

To see this, let $\displaystyle g(r) = ({1}/{r})$ , and we compute for $i=1,2,3$ ,

(A27) \begin{equation} \text{div}\left ( \left \langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right \rangle \boldsymbol{M}_* \right )_i = \partial _{j}\big ({\boldsymbol{v}_*}_k \partial _{k}g(r)m_{\textit{ij}}\big ) ={\boldsymbol{v}_*}_km_{\textit{ij}}\partial ^2_{kj}g(r), \end{equation}

and

(A28) \begin{eqnarray} \partial ^2_{kj}g(r) &=& g''(r)\partial _{k}r\partial _{j}r + g'(r)\partial ^2_{kj}r \nonumber \\ &=& g''(r)\frac {x_kx_j}{r^2} + \frac {g'(r)}{r}\left (\delta _{kj} - \frac {x_kx_j}{r^2}\right ) \nonumber \\ &=& \frac {g'(r)}{r}\delta _{kj} + \left (g''(r)-\frac {g'(r)}{r}\right )\frac {x_kx_j}{r^2} \nonumber \\ & = & -\frac {1}{r^3}\left ( \delta _{kj} - 3\frac {x_kx_j}{r^2} \right ) \end{eqnarray}

where we have used the facts that $\displaystyle g'(r) = -({1}/{r^2})$ and $\displaystyle g''(r) = ({2}/{r^3})$ . As

(A29) \begin{align} \int _{\mathbb{S}^2}3\hat {x}\otimes \hat {x}\,{\textrm d}\sigma = 4\pi \unicode{x1D644}, \end{align}

we have for all $k,j$ that

(A30) \begin{align} \int _{\mathbb{S}^2}\partial ^2_{kj}g(r)\,{\textrm d}\sigma = 0. \end{align}

Thus, (A26) holds and so does (A22).

For $D^2\unicode{x1D640}$ , to show $\displaystyle \int _{{\mathbb S}^2}D^2\unicode{x1D640}(\hat {x})\,{\textrm d}\sigma =0$ , by (B39), we have

(A31) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma & = & 4\pi \left (-\delta _{\textit{ij}}\delta _{kl} + \delta _{ik}\delta _{jl} + \delta _{jk}\delta _{il}\right ) -4\pi \left ( -\delta _{\textit{ij}}\delta _{kl} +\delta _{ik}\delta _{jl} +\delta _{jk}\delta _{il} \right )\nonumber \\ & & -4\pi \left ( \delta _{il}\delta _{jk} +\delta _{jl}\delta _{ik} +\delta _{kl}\delta _{\textit{ij}} \right ) + 15\int _{\mathbb{S}^2}\hat {x}_i\hat {x}_j\hat {x}_k\hat {x}_l\,{\textrm d}\sigma \nonumber \\ & = & -4\pi \left ( \delta _{il}\delta _{jk} +\delta _{jl}\delta _{ik} +\delta _{kl}\delta _{\textit{ij}} \right ) + 15\int _{\mathbb{S}^2}\hat {x}_i\hat {x}_j\hat {x}_k\hat {x}_l\,{\textrm d}\sigma . \end{eqnarray}

Using spherical coordinates and considering symmetry, it suffices to check the following cases:

1. (a) $i\neq j, k\neq l$ , $(k,l) = (i,j)$ or $(j,i)$

   (A32) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma & =& -4\pi + 15\int _{\mathbb{S}^2}\hat {x}_1^2\hat {x}_2^2\,{\textrm d}\sigma \nonumber \\ &=& -4\pi + 15\int _0^\pi \int _0^{2\pi } (\sin \phi \cos \theta )^2(\sin \phi \sin \theta )^2 \sin \phi \,{\textrm d}\phi \,{\textrm d}\theta \nonumber \\ & = & -4\pi + 15\int _0^\pi \sin ^5\phi \,{\textrm d}\phi \int _0^{2\pi } \cos ^2\theta \sin ^2\theta \,{\textrm d}\theta = 0; \end{eqnarray}

2. (b) $i\neq j, k\neq l$ , $(k,l) \neq (i,j), (j,i)$

   (A33) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma = 0; \end{eqnarray}

3. (c) $i\neq j, k=l=i$ or $k=l=j$

   (A34) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma = + 15\int _{\mathbb{S}^2}\hat {x}_1^3\hat {x}_2\,{\textrm d}\sigma = 0; \end{eqnarray}

4. (d) $i\neq j, k=l,\neq i, j$

   (A35) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma = 0; \end{eqnarray}

5. (e) $i=j, k\neq l$

   (A36) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma = 0; \end{eqnarray}

6. (f) $i=j, k=l,\neq i,j$

   (A37) \begin{eqnarray} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma = -4\pi + 15\int _{\mathbb{S}^2}\hat {x}_1^2\hat {x}_2^2\,{\textrm d}\sigma = 0, \quad \text{as in}\ (a); \end{eqnarray}

7. (g) $i=j=k=l$

   (A38) \begin{align} \int _{\mathbb{S}^2} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x)\,{\textrm d}\sigma = -12\pi + 15\int _{\mathbb{S}^2}\hat {x}_3^4\,{\textrm d}\sigma = -12\pi + 15\int _0^\pi \int _0^{2\pi }\cos ^4\phi \sin \phi \,{\textrm d}\theta \,{\textrm d}\phi = 0. \end{align}

Thus, $D^2\unicode{x1D640}$ does also satisfy the mean zero condition.

B.1 Formula for $\textrm {div}\,\mathcal{A}_\gamma$

By (A25), (A27) and (A28), we have for some matrix $\boldsymbol{M}_* = (m_{\textit{ij}})$ that

(B1) \begin{align} \big (\textrm {div}\,(\mathcal{A}_\gamma )\big )_i = \textrm {div} \left (\left\langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right\rangle \boldsymbol{M}_*\right )_i = -m_{\textit{ij}}\big(\delta _{kj} - 3\hat {x}_k\hat {x}_j \big){\boldsymbol{v}_*}_k = -m_{\textit{ij}}{\boldsymbol{v}_*}_j + 3m_{\textit{ij}}\hat {x}_j\langle \boldsymbol{v}_*, \hat {x}\rangle \end{align}

so that

(B2) \begin{align} \textrm {div}\,(\mathcal{A}_\gamma ) = -\boldsymbol{M}_*(\unicode{x1D644} - 3\hat {x}\otimes \hat {x})\boldsymbol{v}_*. \end{align}

Hence

(B3) \begin{align} \textrm {div}\,\left (\left\langle \boldsymbol{v}_*, \boldsymbol{\nabla }\frac {1}{r} \right\rangle \boldsymbol{M}_*\right ) = \left \{ \begin{array}{ll} \big ({-}\langle \boldsymbol{v}_*, {n_*}\rangle + 3\langle {n_*}, \hat {x}\rangle \langle \boldsymbol{v}_*, \hat {x}\rangle \big ) {n_*}, & \text{if}\,\,\,\boldsymbol{M}_* = {n_*}\otimes {n_*},\\[5pt]-\boldsymbol{v}_* + 3\langle \boldsymbol{v}_*, \hat {x}\rangle \hat {x}, & \text{if}\,\,\,\boldsymbol{M}_* = \unicode{x1D644}. \end{array} \right . \end{align}

Using the form of $\unicode{x1D64C}_*$ from (3.5), we have

(B4) \begin{align} \boldsymbol{M}_* = -\frac {w}{1+w}\left [ \left (\gamma _2s_* + \frac {\gamma _3}{3}s_*^2 + \frac {2\gamma _9}{3} s_*^3 \right ){n_*}\otimes {n_*} - \frac 13\left ( \gamma _2s_* -\frac {\gamma _3s_*^2}{3} + \frac {2\gamma _9s_*^3}{3} \right )\unicode{x1D644} \right ]\!. \end{align}

Hence

(B5) \begin{eqnarray} \textrm {div}\,(\mathcal{A}_\gamma ) &=& -\frac {w}{1+w}\left [ \left (\gamma _2s_* + \frac {\gamma _3}{3}s_*^2 + \frac {2\gamma _9}{3}s_*^3 \right )\big ( -\langle \boldsymbol{v}_*, {n_*}\rangle + 3\langle {n_*},\hat {x}\rangle \langle \boldsymbol{v}_*, \hat {x}\rangle \big ){n_*} \right .\nonumber \\ &&\hspace {50pt}\left . - \frac 13\left ( \gamma _2s_* -\frac {\gamma _3s_*^2}{3} + \frac {2\gamma _9s_*^3}{3} \right ) \big ( -\boldsymbol{v}_* + 3\langle \boldsymbol{v}_*, \hat {x}\rangle \hat {x} \big ) \right ]\nonumber \\ &=& -\frac {w}{1+w}\left [ \left (\gamma _2s_* + \frac {\gamma _3}{3}s_*^2 + \frac {2\gamma _9}{3}s_*^3 \right ) \big (-{n_*}\otimes {n_*} + 3\langle {n_*},\hat {x}\rangle {n_*}\otimes \hat {x} \big ) \right .\nonumber \\ &&\hspace {50pt}\left . - \left ( \gamma _2s_* -\frac {\gamma _3s_*^2}{3} + \frac {2\gamma _9s_*^3}{3} \right ) \big ( \hat {x}\otimes \hat {x} - \frac 13\unicode{x1D644} \big ) \right ]\boldsymbol{v}_*. \end{eqnarray}

B.2 Formula for $\textrm {div}\,[\mathcal{B}_\gamma[\boldsymbol{\nabla}\textbf{v}]]$

In this section, we will identify $\mathcal{M}_{i,j;k,l}$ which appear in (4.24) and (4.25). We note the following symmetry of $\mathcal{M}_{i,j;k,l}$ with respect to $k$ and $l$ ,

(B6) \begin{equation} \mathcal{M}_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j = \mathcal{M}_{i,j;l,k}\partial _{kl}\boldsymbol{v}_j. \end{equation}

Furthermore, we emphasize that $\mathcal{M}_\gamma$ will only act on incompressible vector fields $\boldsymbol{v}$ : $\partial _j\boldsymbol{v}_j = 0$ .

Before proceeding, using (3.5), we record that

(B7) \begin{equation} {\unicode{x1D64C}_*}_{\textit{ij}} = s_*\left ({n_*}_i{n_*}_j - \frac 13\delta _{\textit{ij}}\right ),\,\,\, {\unicode{x1D64C}_*^2}_{\textit{ij}} = \frac {s_*^2}{3}\left ({n_*}_i{n_*}_j + \frac 13\delta _{\textit{ij}}\right ),\,\,\, |\unicode{x1D64C}_*|^2 = \frac {2s_*^2}{3}. \end{equation}

With that, we compute

1. (a) $\textrm {div}(\mathcal{B}_1[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B8) \begin{align} & \quad \textrm {div}\,\big (\big [\unicode{x1D64C}_*^2\unicode{x1D652}+\unicode{x1D652}\unicode{x1D64C}_*^2-2\unicode{x1D64C}_*\unicode{x1D652}\unicode{x1D64C}_*\big ]_i\big ) =\partial _j\big [\unicode{x1D64C}_*^2\unicode{x1D652}+\unicode{x1D652}\unicode{x1D64C}_*^2-2\unicode{x1D64C}_*\unicode{x1D652}\unicode{x1D64C}_*\big ]_{\textit{ij}}\nonumber \\&=\partial _j\Big [ \frac {1}{2}\big (\unicode{x1D64C}_*^2\big )_{ik}(\partial _j\boldsymbol{v}_k - \partial _k\boldsymbol{v}_j) +\frac {1}{2}\big (\unicode{x1D64C}_*^2\big )_{kj}\big (\partial _k\boldsymbol{v}_i - \partial _i\boldsymbol{v}_k\big ) - {\big (\unicode{x1D64C}_*\big )}_{ik} (\partial _l\boldsymbol{v}_k - \partial _k\boldsymbol{v}_l) {\big (\unicode{x1D64C}_*\big )}_{lj} \Big ] \nonumber \\&=\frac {1}{2}\big (\unicode{x1D64C}_*^2\big )_{ik}\partial _{jj}\boldsymbol{v}_k+\frac {1}{2}\big (\unicode{x1D64C}_*^2\big )_{kj}\big (\partial _{kj}\boldsymbol{v}_i - \partial _{\textit{ij}}\boldsymbol{v}_k\big )-{\big (\unicode{x1D64C}_*\big )}_{ik}{\big (\unicode{x1D64C}_*\big )}_{lj} \big (\partial _{jl}\boldsymbol{v}_k - \partial _{jk}\boldsymbol{v}_l\big )\nonumber \\&=\left (\frac {1}{2}\delta _{k,l}\left (\unicode{x1D64C}_*^2\right )_{\textit{ij}}+\frac {1}{2}\delta _{i,j}\big (\unicode{x1D64C}_*^2\big )_{kl} - \frac {1}{2}\delta _{i,l}\big (\unicode{x1D64C}_*^2\big )_{jk}-{\big (\unicode{x1D64C}_*\big )}_{\textit{ij}}{\big (\unicode{x1D64C}_*\big )}_{lk} + {\big (\unicode{x1D64C}_*\big )}_{ik}{\big (\unicode{x1D64C}_*\big )}_{lj}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\ &=:\mathcal{M}^1_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j, \end{align}
   where
   (B9) \begin{align} \mathcal{M}^1_{i,j;k,l}&=s_*^2\left [\frac {1}{6}\delta _{k,l}\left ({n_*}_i{n_*}_j+\frac 13\delta _{i,j}\right )+\frac {1}{6}\delta _{i,j}\left ({n_*}_k{n_*}_l+\frac 13\delta _{k,l}\right )\right .\nonumber \\ & \quad \left . - \frac {1}{6}\delta _{i,l}\left ({n_*}_k{n_*}_j+\frac 13\delta _{k,j}\right )-{\left ({n_*}_i{n_*}_j-\frac 13\delta _{i,j}\right )}{\left ({n_*}_k{n_*}_l-\frac 13\delta _{k,l}\right )} \right .\nonumber \\ & \quad \left . + {\left ({n_*}_k{n_*}_i -\frac 13\delta _{k,i}\right )}{\left ({n_*}_j{n_*}_l-\frac 13\delta _{j,l}\right )}\right ]\nonumber \\&=\frac {s_*^2}{2}\left ( \delta _{kl}{n_*}_i{n_*}_j + \delta _{\textit{ij}}{n_*}_k{n_*}_l -\delta _{ik}{n_*}_l{n_*}_j \right ). \end{align}
   In the above, we have used the symmetry property (B6) of $\mathcal{M}$ and the incompressibility of $\boldsymbol{v}$ :
   (B10) \begin{equation} \delta _{il}{n_*}_k{n_*}_j\partial _{kl}\boldsymbol{v}_j = \delta _{ik}{n_*}_l{n_*}_j\partial _{kl}\boldsymbol{v}_j \,\,\,\text{and} \,\,\, \delta _{il}\delta _{kj}\partial _{kl}\boldsymbol{v}_j = \delta _{ik}\delta _{lj}\partial _{kl}\boldsymbol{v}_j = \partial _{\textit{ij}}\boldsymbol{v}_j = 0. \end{equation}
   The above property will be used in several places in what follows.

2. (b) $\textrm {div}(\mathcal{B}_2[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B11) \begin{eqnarray} \textrm {div}\,\left [\unicode{x1D64C}_*\boldsymbol{\nabla }\boldsymbol{v}-(\boldsymbol{\nabla }\boldsymbol{v})\unicode{x1D64C}_*\right ]_i & = & \partial _j\left [{\left (\unicode{x1D64C}_*\right )}_{ik}\partial _j\boldsymbol{v}_k-\partial _k\boldsymbol{v}_i{\left (\unicode{x1D64C}_*\right )}_{kj}\right ]\nonumber \\ & = & {\left (\unicode{x1D64C}_*\right )}_{ik}\partial _{jj}\boldsymbol{v}_k- {\left (\unicode{x1D64C}_*\right )}_{kj}\partial _{kj}\boldsymbol{v}_i\nonumber \\ & = & \left (\delta _{k,l}{\left (\unicode{x1D64C}_*\right )}_{\textit{ij}}- \delta _{i,j}{\left (\unicode{x1D64C}_*\right )}_{kl}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\ &=:&\mathcal{M}^2_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B12) \begin{equation} \mathcal{M}^2_{i,j;k,l}=s_*\left (\delta _{k,l}{n_*}_i{n_*}_j- \delta _{i,j}{n_*}_k{n_*}_l\right ). \end{equation}

3. (c) $\textrm {div}(\mathcal{B}_3[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B13) \begin{eqnarray} \frac {1}{2}\textrm {div}\,\left [\unicode{x1D64C}_*^2\boldsymbol{\nabla }\boldsymbol{v}-(\boldsymbol{\nabla }\boldsymbol{v})\unicode{x1D64C}_*^2\right ]_i & = & \frac {1}{2}\left ({\big (\unicode{x1D64C}_*^2\big )}_{ik}\partial _{jj}\boldsymbol{v}_k- \big ({\unicode{x1D64C}_*^2}\big )_{kj}\partial _{kj}\boldsymbol{v}_i\right )\nonumber \\& = & \frac {1}{2}\left (\delta _{k,l}{\big (\unicode{x1D64C}_*^2\big )}_{\textit{ij}}- \delta _{i,j}\big ({\unicode{x1D64C}_*^2}\big )_{kl}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\&=:&\mathcal{M}^3_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B14) \begin{equation} \mathcal{M}^3_{i,j;k,l}=\frac {s_*^2}{6}\left (\delta _{k,l}{n_*}_i{n_*}_j- \delta _{i,j}{n_*}_k{n_*}_l\right ). \end{equation}

4. (d) $\textrm {div}(\mathcal{B}_4[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B15) \begin{eqnarray} \textrm {div}\,\left [\unicode{x1D64C}_*\unicode{x1D63C}+\unicode{x1D63C}\unicode{x1D64C}_*\right ]_i & = & \frac {1}{2}\left (\unicode{x1D64C}_*\right )_{ik}\partial _{jj}\boldsymbol{v}_{k}+\frac {1}{2}\left (\partial _{\textit{ij}}\boldsymbol{v}_k+\partial _{kj}\boldsymbol{v}_i\right )\left (\unicode{x1D64C}_*\right )_{kj}\nonumber \\ & = & \frac {1}{2}\left (\delta _{k,l}\left (\unicode{x1D64C}_*\right )_{\textit{ij}}+\delta _{i,l}\left (\unicode{x1D64C}_*\right )_{jk}+\delta _{i,j}\left (\unicode{x1D64C}_*\right )_{kl}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\ &=:&\mathcal{M}^4_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B16) \begin{align} \mathcal{M}^4_{i,j;k,l}& =\frac {s_*}{2}\left (\delta _{k,l}\left ({n_*}_i{n_*}_j-\frac 13\delta _{i,j}\right )+\delta _{i,l}\left ({n_*}_j{n_*}_k-\frac 13\delta _{j,k}\right ) \right. \nonumber \\& \left. \quad +\delta _{i,j}\left ({n_*}_k{n_*}_l-\frac 13\delta _{k,l}\right )\right )\nonumber \\& = \frac {s_*}{2}\left ( \delta _{kl}{n_*}_i{n_*}_j + \delta _{\textit{ij}}{n_*}_k{n_*}_l + \delta _{ik}{n_*}_l{n_*}_j -\frac 23\delta _{kl}\delta _{\textit{ij}}\right ). \end{align}

5. (e) $\textrm {div}(\mathcal{B}_5[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B17) \begin{align} \textrm {div}\,\left [\unicode{x1D64C}_*^2\unicode{x1D63C}+\unicode{x1D63C}\unicode{x1D64C}_*^2\right ]_i & = \frac {1}{2}\big (\unicode{x1D64C}_*^2\big )_{ik}\partial _{jj}\boldsymbol{v}_{k}+\frac {1}{2}\left (\partial _{\textit{ij}}\boldsymbol{v}_k+\partial _{kj}\boldsymbol{v}_i\right )\big (\unicode{x1D64C}_*^2\big )_{kj}\nonumber \\& = \frac {1}{2}\left (\delta _{k,l}\big (\unicode{x1D64C}_*^2\big )_{\textit{ij}}+\delta _{i,l}\big (\unicode{x1D64C}_*^2\big )_{jk}+\delta _{i,j}\big (\unicode{x1D64C}_*^2\big )_{kl}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\& =: \mathcal{M}^5_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{align}
   Here
   (B18) \begin{align} \mathcal{M}^5_{i,j;k,l} &=\frac {s_*^2}{6}\left (\delta _{k,l}\left ({n_*}_i{n_*}_j+\frac 13\delta _{i,j}\right )+\delta _{i,l}\left ({n_*}_j{n_*}_k+\frac 13\delta _{j,k}\right )\right.\nonumber \\& \left. \quad +\delta _{i,j}\left ({n_*}_k{n_*}_l+\frac 13\delta _{k,l}\right )\right )\nonumber \\&=\frac {s_*^2}{6}\left ( \delta _{kl}{n_*}_i{n_*}_j + \delta _{\textit{ij}}{n_*}_k{n_*}_l + \delta _{ik}{n_*}_l{n_*}_j +\frac 23\delta _{\textit{ij}}\delta _{kl} \right )\!. \end{align}

6. (f) $\textrm {div}(\mathcal{B}_6[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B19) \begin{eqnarray} \textrm {div}\,\left [(\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_*)\unicode{x1D64C}_*\right ]_i & = & \frac {1}{2}\left (\unicode{x1D64C}_*\right )_{\textit{ij}}\left (\unicode{x1D64C}_*\right )_{kl}\left (\partial _{lj}\boldsymbol{v}_{k}+\partial _{kj}\boldsymbol{v}_l\right )\nonumber \\ & = & \frac {1}{2}\left (\left (\unicode{x1D64C}_*\right )_{ik}\left (\unicode{x1D64C}_*\right )_{jl}+\left (\unicode{x1D64C}_*\right )_{il}\left (\unicode{x1D64C}_*\right )_{kj}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\ &=:&\mathcal{M}^6_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B20) \begin{eqnarray} \mathcal{M}^6_{i,j;k,l}&=&\frac {s_*^2}{2}\left (\left ({n_*}_i{n_*}_k-\frac 13\delta _{i,k}\right )\left ({n_*}_j{n_*}_l-\frac 13\delta _{j,l}\right )\right .\nonumber \\&&\left .+\left ({n_*}_i{n_*}_l-\frac 13\delta _{i,l}\right )\left ({n_*}_j{n_*}_k-\frac 13\delta _{j,k}\right )\right )\nonumber \\ &=& s_*^2\left ( {n_*}_i{n_*}_j{n_*}_k{n_*}_l - \frac 13\delta _{ik}{n_*}_l{n_*}_j \right )\!. \end{eqnarray}

7. (g) $\textrm {div}(\mathcal{B}_7[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B21) \begin{eqnarray} \textrm {div}\,\left [|\unicode{x1D64C}_*|^2\unicode{x1D63C}\right ]_i =\textrm {div}\,\left [\frac {2s_*^2}{3}\unicode{x1D63C}\right ]_i =:\mathcal{M}^7_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B22) \begin{equation} \mathcal{M}^7_{i,j;k,l}=\frac {s_*^2}{3}\delta _{k,l}\delta _{i,j}. \end{equation}

8. (h) $\textrm {div}(\mathcal{B}_{10}[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B23) \begin{eqnarray} \textrm {div}\,\left [(\unicode{x1D64C}_*^2\boldsymbol{\cdot }\unicode{x1D63C})\unicode{x1D64C}_* + (\unicode{x1D64C}_*\boldsymbol{\cdot }\unicode{x1D63C})\unicode{x1D64C}^2_*\right ]_i & = & \frac {1}{2}\left (\big (\unicode{x1D64C}_*^2\big )_{kl}\left (\unicode{x1D64C}_*\right )_{\textit{ij}}+\left (\unicode{x1D64C}_*\right )_{kl}\big (\unicode{x1D64C}_*^2\big )_{\textit{ij}}\right )\left (\partial _{lj}\boldsymbol{v}_{k}+\partial _{kj}\boldsymbol{v}_{l}\right )\nonumber \\ & = & \left (\big (\unicode{x1D64C}_*^2\big )_{jk}\left (\unicode{x1D64C}_*\right )_{il}+\left (\unicode{x1D64C}_*\right )_{jk}\big (\unicode{x1D64C}_*^2\big )_{il}\right )\partial _{kl}\boldsymbol{v}_j\nonumber \\ &=:&\mathcal{M}^{10}_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B24) \begin{eqnarray} \mathcal{M}^{10}_{i,j;k,l} &=&\frac {s_*^3}{3}\left (\left ({n_*}_j{n_*}_k+\frac 13\delta _{j,k}\right )\left ({n_*}_i{n_*}_l-\frac 13\delta _{i,l}\right )\right .\nonumber \\&&\left .+\left ({n_*}_j{n_*}_k-\frac 13\delta _{j,k}\right )\left ({n_*}_i{n_*}_l+\frac 13\delta _{i,l}\right )\right )\nonumber \\ &=&\frac {2s_*^3}{3} {n_*}_i{n_*}_j{n_*}_k{n_*}_l. \end{eqnarray}

9. (i) $\textrm {div}(\mathcal{B}_{11}[\boldsymbol{\nabla }\boldsymbol{v}])$ :

   (B25) \begin{eqnarray} \textrm {div}\,\big [|\unicode{x1D64C}_*|^2(\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_*)\unicode{x1D64C}_*\big ]_i = \frac {2s_*^2}{3}\left (\unicode{x1D64C}_*\right )_{il}\left (\unicode{x1D64C}_*\right )_{jk}\partial _{kl}\boldsymbol{v}_j=:\mathcal{M}^{11}_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}
   Here
   (B26) \begin{eqnarray} \mathcal{M}^{11}_{i,j;k,l}&=&\frac {2s_*^4}{3}\left ({n_*}_i{n_*}_l-\frac 13\delta _{i,l}\right )\left ({n_*}_j{n_*}_k-\frac 13\delta _{j,k}\right )\nonumber \\ & = & \frac {2s_*^4}{3}\left ( {n_*}_i{n_*}_j{n_*}_k{n_*}_l - \frac 13\delta _{ik}{n_*}_l{n_*}_j \right ) . \end{eqnarray}

Hence, the fourth-order tensor $\mathcal{M}_\gamma$ can be decomposed as

(B27) \begin{equation} \mathcal{M}_{i,j;k,l} = \sum _{p=1,\ldots , 7, 10,11}\gamma _p\mathcal{M}^p_{i,j;k,l}. \end{equation}

From (B9) to (B26), the $\mathcal{M}^p_{i,j;k,l}$ are effectively given by a linear combination of the following tensors:

(B28) \begin{equation} \delta _{kl}\delta _{\textit{ij}},\,\,\, \delta _{kl}{n_*}_i{n_*}_j,\,\,\, \delta _{\textit{ij}}{n_*}_k{n_*}_l,\,\,\, \delta _{ik}{n_*}_l{n_*}_j\,\,\,(\text{or equivalently, $\delta _{il}{n_*}_k{n_*}_j$}),\,\,\, {n_*}_i{n_*}_j{n_*}_k{n_*}_l \end{equation}

where $\delta _{pq}$ is the delta-function $\delta _{pq} =1$ for $p=q$ and $\delta _{pq} =0$ for $p\neq q$ .

B.3 Formula for $\mathcal{M}_\gamma :D^2\boldsymbol{F}$

Here we list the formulae for $\mathcal{M}_\gamma :D^2\boldsymbol{F}$ where $\boldsymbol{F}$ is some divergence free vector field with homogeneous degree $-k$ , for $k=1,3,5$ .

As a preparation, following (A28) for $g(r) = \frac {1}{r},\,\frac {1}{r^3},\,\frac {1}{r^5}$ , we compute

(B29) \begin{align} \partial _{k}g(r) & = g'(r)\frac {x_k}{r}, \\[-12pt] \nonumber\end{align}

(B30) \begin{align} \partial _{kl}g(r) &= \frac {g'(r)}{r}\delta _{kl} + \left (g''(r)-\frac {g'(r)}{r}\right )\frac {x_kx_l}{r^2}, \\[-12pt] \nonumber\end{align}

(B31) \begin{align} \partial _{kl}\big (g(r)\delta _{\textit{ij}}\big ) &= \left (\frac {g'(r)}{r}\delta _{kl} + \left (g''(r)-\frac {g'(r)}{r}\right )\frac {x_kx_l}{r^2}\right )\delta _{\textit{ij}}, \\[-12pt] \nonumber\end{align}

(B32) \begin{align} \partial _{k}\big (g(r)x_ix_j\big ) & = g(r)\big (\delta _{ik}x_j + \delta _{jk}x_i\big ) +\frac {g'(r)}{r}x_ix_jx_k, \\[-12pt] \nonumber\end{align}

(B33) \begin{align} \partial _{kl}\big (g(r)x_ix_j\big ) &= g(r)\big (\delta _{ik}\delta _{jl} + \delta _{jk}\delta _{il}\big )\nonumber \\& \quad +rg'(r)\Big (\frac { \delta _{il}x_jx_k + \delta _{jl}x_ix_k + \delta _{kl}x_ix_j + \delta _{ik}x_jx_l + \delta _{jk}x_ix_l }{r^2} \Big )\nonumber \\& \quad + (r^2g''(r) - rg'(r))\frac {x_ix_jx_kx_l}{r^4}. \\[6pt] \nonumber\end{align}

Hence,

(B34) \begin{align} \partial _{kl}\left (\frac {\delta _{\textit{ij}}}{r}\right ) & = \frac {1}{r^3}\big ({-}\delta _{kl} + 3\hat {x}_k\hat {x}_l\big )\delta _{\textit{ij}}, \\[-12pt] \nonumber\end{align}

(B35) \begin{align} \partial _{kl}\left (\frac {\delta _{\textit{ij}}}{r^3}\right ) & = \frac {1}{r^5}\big ({-}3\delta _{kl} + 15\hat {x}_k\hat {x}_l\big )\delta _{\textit{ij}}, \\[-12pt] \nonumber\end{align}

(B36) \begin{align} \partial _{kl}\left (\frac {x_ix_j}{r^3}\right ) &= \frac {1}{r^3}\big [\big (\delta _{ik}\delta _{jl} + \delta _{jk}\delta _{il}\big )\nonumber \\& \quad-3\big ( \delta _{il}\hat {x}_j\hat {x}_k + \delta _{jl}\hat {x}_i\hat {x}_k + \delta _{kl}\hat {x}_i\hat {x}_j + \delta _{ik}\hat {x}_j\hat {x}_l + \delta _{jk}\hat {x}_i\hat {x}_l \big )\nonumber \\& \quad+ 15\hat {x}_i\hat {x}_j\hat {x}_k\hat {x}_l\big ], \\[-12pt] \nonumber\end{align}

(B37) \begin{align} \partial _{kl}\left (\frac {x_ix_j}{r^5}\right ) &= \frac {1}{r^5}\big [\big (\delta _{ik}\delta _{jl} + \delta _{jk}\delta _{il}\big )\nonumber \\& \quad-5\big ( \delta _{il}\hat {x}_j\hat {x}_k + \delta _{jl}\hat {x}_i\hat {x}_k + \delta _{kl}\hat {x}_i\hat {x}_j + \delta _{ik}\hat {x}_j\hat {x}_l + \delta _{jk}\hat {x}_i\hat {x}_l \big )\nonumber \\& \quad+ 35\hat {x}_i\hat {x}_j\hat {x}_k\hat {x}_l\big ]. \\[6pt] \nonumber\end{align}

The above are applied to $\unicode{x1D640}, \unicode{x1D641}$ and $\unicode{x1D64C}$ :

1. (i) $\displaystyle \unicode{x1D640}(x) = \frac{1}{8\pi} [ \frac{\unicode{x1D644}}{r} + \frac{x\otimes x}{r^3}]$ ,

   (B38) \begin{align} \partial _{k}\unicode{x1D640}_{\textit{ij}}(x) &= \frac {1}{8\pi r^2}\big [{-}\delta _{\textit{ij}}\hat {x}_k + \delta _{ik}\hat {x}_j + \delta _{jk}\hat {x}_i - 3\hat {x}_i\hat {x}_j\hat {x}_k\big ], \\[-12pt] \nonumber\end{align}
   (B39) \begin{align} \partial _{kl}\unicode{x1D640}_{\textit{ij}}(x) & = \frac {1}{8\pi r^3}\big [ \left (-\delta _{\textit{ij}}\delta _{kl} + \delta _{ik}\delta _{jl} + \delta _{jk}\delta _{il}\right ) \big.\nonumber \\& \big. \quad -3\left ( -\delta _{\textit{ij}}\hat {x}_k\hat {x}_l +\delta _{ik}\hat {x}_j\hat {x}_l +\delta _{jk}\hat {x}_i\hat {x}_l +\delta _{il}\hat {x}_j\hat {x}_k +\delta _{jl}\hat {x}_i\hat {x}_k +\delta _{kl}\hat {x}_i\hat {x}_j \right ) \big.\nonumber \\& \big. \quad + 15\hat {x}_i\hat {x}_j\hat {x}_k\hat {x}_l\big ];\\[6pt] \nonumber\end{align}

2. (ii) $\displaystyle \unicode{x1D641}(x) = \frac{3x\otimes x - r^2\unicode{x1D644}}{r^5}$ ,

   (B40) \begin{align} \partial _{k}\unicode{x1D641}_{\textit{ij}}(x) & = \frac {1}{r^4}\big [ 3\big (\delta _{\textit{ij}}\hat {x}_k + \delta _{ik}\hat {x}_j + \delta _{jk}\hat {x}_i\big ) - 15\hat {x}_i\hat {x}_j\hat {x}_k \big ], \\[-12pt] \nonumber \end{align}
   (B41) \begin{align} \partial_{kl}\unicode{x1D641}_{ij}(x)&=\frac{1}{r^5}\Big[3\big(\delta_{ik}\delta_{jl} + \delta_{jk}\delta_{il}+ \delta_{kl}\delta_{ij}\big)\nonumber\\&\quad-15\big(\delta_{ij}\hat{x}_k\hat{x}_l+\delta_{il}\hat{x}_j\hat{x}_k+\delta_{jl}\hat{x}_i\hat{x}_k+\delta_{kl}\hat{x}_i\hat{x}_j+\delta_{ik}\hat{x}_j\hat{x}_l+\delta_{jk}\hat{x}_i\hat{x}_l\big)\nonumber\\&\quad+ 105 \hat{x}_i\hat{x}_j\hat{x}_k\hat{x}_l\Big] \\[6pt] \nonumber\end{align}

3. (iii) $\displaystyle \unicode{x1D64C}(x) = \left(1 - \frac{w}{1+w} \frac{1}{r} \right)\unicode{x1D64C}_* + \frac{w}{3+w} \frac{1}{r^3}\unicode{x1D64C}_b = \left(1 - \frac{w}{1+w}\frac{1}{r} \right)\unicode{x1D64C}_* + \frac{ws_*}{3(3+w)} \unicode{x1D641}(x)$ , where $\displaystyle {\unicode{x1D64C}_*}_{\textit{ij}} = {n_*}_i{n_*}_j - \frac{1}{3}\delta _{\textit{ij}}$ ,

   (B42) \begin{align}\partial _k\unicode{x1D64C}_{\textit{ij}}(x) = \frac {w}{(1+w)r^2}\hat {x}_k{\unicode{x1D64C}_*}_{\textit{ij}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\nonumber \\+ \frac {ws_*}{(3+w)r^4} \big [ \delta _{\textit{ij}}\hat {x}_k + \delta _{ik}\hat {x}_j + \delta _{jk}\hat {x}_i - 5\hat {x}_i\hat {x}_j\hat {x}_k \big ],\quad\quad\quad\quad\quad \\[-12pt] \nonumber\end{align}
   (B43) \begin{align} \partial _{kl}\unicode{x1D64C}_{\textit{ij}}(x) & = - \frac {w}{(1+w)r^3}\big [-\delta _{kl} + 3\hat {x}_k\hat {x}_l\big ]{\unicode{x1D64C}_*}_{\textit{ij}}\nonumber \\& \quad+ \frac {ws_*}{(3+w)r^5}\big [ \big (\delta _{ik}\delta _{jl} + \delta _{jk}\delta _{il} + \delta _{kl}\delta _{\textit{ij}}\big ) \big.\nonumber \\& \big. \quad -5\big ( \delta _{\textit{ij}}\hat {x}_k\hat {x}_l +\delta _{il}\hat {x}_j\hat {x}_k +\delta _{jl}\hat {x}_i\hat {x}_k +\delta _{kl}\hat {x}_i\hat {x}_j +\delta _{ik}\hat {x}_j\hat {x}_l +\delta _{jk}\hat {x}_i\hat {x}_l \big ) \big.\nonumber \\& \big. \quad + 35 \hat {x}_i\hat {x}_j\hat {x}_k\hat {x}_l \big ]. \\[6pt] \nonumber\end{align}

We now recall the operation (4.24), (4.25) and (B27) for $\mathcal{B}_\gamma$ :

(B44) \begin{eqnarray} \big (\textrm {div}\,\big (\mathcal{B}_\gamma [\boldsymbol{\nabla }\boldsymbol{v}]\big )\big )_i = \big (\mathcal{M}_\gamma :D^2\boldsymbol{v}\big )_i = \sum _{p = 1,\ldots 7, 10, 11}\gamma _p\mathcal{M}^p_{i,j;k,l}\partial _{kl}\boldsymbol{v}_j. \end{eqnarray}

Given any matrix valued function $\boldsymbol{F}(x) = \{\boldsymbol{F}_{\textit{ij}}(x)\}_{1\leq i,j\leq 3}$ , upon introducing the contraction

(B45) \begin{equation} \big (\mathcal{M}^p:D^2{\textbf{F}}big)_{\textit{ij}} = \mathcal{M}^p_{i,m;,k,l}\partial _{kl}{\textbf{F}}_{mj}, \end{equation}

we have

(B46) \begin{eqnarray} \big (\mathcal{M}^p:D^2(\boldsymbol{F}\boldsymbol{v}_*)\big )_i = \big (\mathcal{M}^p_{i,m;k,l}\partial _{kl}\boldsymbol{F}_{mj}\big ){\boldsymbol{v}_*}_j. \end{eqnarray}

Note that we will only consider those $\textbf{F}$ such that ${\textbf{F}}\boldsymbol{v}_*$ is divergence free for all $\boldsymbol{v}_*$ , i.e.

(B47) \begin{equation} \partial _i{\textbf{F}}_{ik} = 0\,\,\,\text{for all $k$.} \end{equation}

This condition is indeed satisfied by $\boldsymbol{F} = \unicode{x1D640}$ (3.9), $\unicode{x1D640}_S$ (3.38), and hence $\unicode{x1D641}$ (3.36).

With the above, we proceed to find the following expressions,

1. (i) Formula for $(\mathcal{M}^p:D^2\unicode{x1D640})_{\textit{ij}}(x) = \mathcal{M}^p_{i,m;k,l}\partial _{kl}\unicode{x1D640}_{mj}(x)$ . From (B28), we need to compute the following:

   (B48) \begin{eqnarray} &&\big \{ \delta _{kl}\delta _{im},\,\,\, \delta _{kl}{n_*}_i{n_*}_m,\,\,\, \delta _{im}{n_*}_k{n_*}_l,\,\,\, \delta _{ik}{n_*}_l{n_*}_m,\,\,\, {n_*}_i{n_*}_m{n_*}_k{n_*}_l\big \}\times \partial _{kl}\unicode{x1D640}_{mj}(x)\nonumber \\& = & \big \{ \delta _{kl}\delta _{im},\,\,\, \delta _{kl}{n_*}_i{n_*}_m,\,\,\, \delta _{im}{n_*}_k{n_*}_l,\,\,\, \delta _{ik}{n_*}_l{n_*}_m,\,\,\, {n_*}_i{n_*}_m{n_*}_k{n_*}_l\big \}\times \nonumber \\&& \frac {1}{8\pi r^3}\big [\left (-\delta _{mj}\delta _{kl} + \delta _{mk}\delta _{jl} + \delta _{jk}\delta _{ml}\right )\nonumber \\&& \hspace {35pt}-3 \big ({-}\delta _{mj}\hat {x}_k\hat {x}_l +\delta _{mk}\hat {x}_j\hat {x}_l +\delta _{jk}\hat {x}_m\hat {x}_l +\delta _{ml}\hat {x}_j\hat {x}_k +\delta _{jl}\hat {x}_m\hat {x}_k +\delta _{kl}\hat {x}_m\hat {x}_j \big )\nonumber \\&&\hspace {35pt} + 15\hat {x}_m\hat {x}_j\hat {x}_k\hat {x}_l\big ]. \end{eqnarray}
   We tabulate the result in the following:
   (B49) \begin{align} \delta _{kl}\delta _{im}\partial _{kl}\unicode{x1D640}_{mj}(x) & = \frac {1}{8\pi r^3}\big [2\delta _{\textit{ij}} - 6\hat {x}_i\hat {x}_j\big ]\nonumber \\\delta _{kl}{n_*}_i{n_*}_m\partial _{kl}\unicode{x1D640}_{mj}(x) & = \frac {1}{8\pi r^3}\big [ 2{n_*}_i{n_*}_j - 6{n_*}_i\hat {x}_j\langle n, \hat {x}\rangle \big ]\nonumber \\\delta _{im}{n_*}_k{n_*}_l\partial _{kl}\unicode{x1D640}_{mj}(x) & = \frac {1}{8\pi r^3}\big [{-}\delta _{\textit{ij}} + 2{n_*}_i{n_*}_j +3\delta _{\textit{ij}}\langle {n_*},\hat {x}\rangle ^2 \nonumber \\& \quad -6({n_*}_i\hat {x}_j + {n_*}_j\hat {x}_i)\langle {n_*},\hat {x}\rangle -3\hat {x}_i\hat {x}_j + 15\hat {x}_i\hat {x}_j\langle {n_*},\hat {x}\rangle ^2\big ]\nonumber \\\delta _{ik}{n_*}_l{n_*}_m\partial _{kl}\unicode{x1D640}_{mj}(x) & = \frac {1}{8\pi r^3}\big [ \delta _{\textit{ij}} - 3\delta _{\textit{ij}}\langle {n_*}, \hat {x}\rangle ^2 -6{n_*}_ix_j\langle {n_*},\hat {x}\rangle -3\hat {x}_i\hat {x}_j \nonumber \\ & \quad + 15\hat {x}_i\hat {x}_j\langle a,\hat {x}\rangle ^2\big ]\nonumber \\{n_*}_i{n_*}_m{n_*}_k{n_*}_l\partial _{kl}\unicode{x1D640}_{mj}(x) & = \frac {1}{8\pi r^3}\big [{n_*}_i{n_*}_j - 3{n_*}_i{n_*}_j\langle {n_*}, \hat {x}\rangle ^2 -9 {n_*}_i\hat {x}_j\langle {n_*},\hat {x}\rangle \nonumber \\ & \quad + 15{n_*}_i\hat {x}_j\langle {n_*},\hat {x}\rangle ^3\big ]. \end{align}
   With the above, we have explicitly,
   (B50) \begin{align} \mathcal{M}^1:D^2\unicode{x1D640}(x)&= \frac {s_*^2}{8\pi r^3} \big [-\big (1 - 3{\langle {n_*},\hat {x}\rangle ^2}\big )\unicode{x1D644} + 2{n_*}\otimes {n_*} -3\langle {n_*},\hat {x}\rangle \big ({n_*}\otimes \hat {x} + \hat {x}\otimes {n_*}\big ) \big ]\nonumber \\\mathcal{M}^2:D^2\unicode{x1D640}(x)&= \frac {s_*}{8\pi r^3} \big [ \big (1 - 3\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} +3\big (1-5\langle {n_*},\hat {x}\rangle ^2\big )\hat {x}\otimes \hat {x} + 6\langle {n_*},\hat {x}\rangle \hat {x}\otimes {n_*} \big ]\nonumber \\\mathcal{M}^3:D^2\unicode{x1D640}(x)&= \frac {1}{8\pi r^3}\frac {s_*^2}{6} \big [ \big (1 - 3\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} +3\big (1-5\langle {n_*},\hat {x}\rangle ^2\big )\hat {x}\otimes \hat {x} + 6\langle {n_*},\hat {x}\rangle \hat {x}\otimes {n_*} \big ]\nonumber \\\mathcal{M}^4:D^2\unicode{x1D640}(x)&= \frac {s_*}{8\pi r^3}\big [-\frac 23\unicode{x1D644} + 2{n_*}\otimes {n_*} -\big (1 - 15\langle {n_*},\hat {x}\rangle ^2\big )\hat {x}\otimes \hat {x} \nonumber \\&\hspace {35pt} -9 \langle {n_*},\hat {x}\rangle {n_*}\otimes \hat {x} -3 \langle {n_*},\hat {x}\rangle \hat {x}\otimes {n_*} \big ]\nonumber \\\mathcal{M}^5:D^2\unicode{x1D640}(x)&= \frac {1}{8\pi r^3}\frac {s_*^2}{6}\Big [ \frac 43\unicode{x1D644} + 4{n_*}\otimes {n_*} -10\big (1- 3\langle {n_*},\hat {x}\rangle ^2\big )\hat {x}\otimes \hat {x} \nonumber \\&\hspace {45pt} -18 \langle {n_*},\hat {x}\rangle {n_*}\otimes \hat {x} -6 \langle {n_*},\hat {x}\rangle \hat {x}\otimes {n_*} \Big ]\nonumber \\\mathcal{M}^6:D^2\unicode{x1D640}(x)&= \frac {s_*^2}{8\pi r^3}\Big [-\left (\frac 13 - \langle {n_*},\hat {x}\rangle ^2\right ) \unicode{x1D644} + \big (1 - 3 \langle {n_*},\hat {x}\rangle ^2\big ) {n_*}\otimes {n_*} \nonumber \\&\hspace {35pt} + \big (1 -10 \langle {n_*},\hat {x}\rangle ^2\big ) \hat {x}\otimes \hat {x} -7 \langle {n_*},\hat {x}\rangle {n_*}\otimes \hat {x} + 15 \langle {n_*},\hat {x}\rangle ^3 n\otimes \hat {x} \Big ]\nonumber \\\mathcal{M}^7:D^2\unicode{x1D640}(x)&= \frac {s_*^2}{8\pi r^3}\Big [\frac 23\unicode{x1D644} -2\hat {x}\otimes \hat {x} \Big ]\nonumber \\\mathcal{M}^{10}:D^2\unicode{x1D640}(x)&= \frac {1}{8\pi r^3}\frac {2s_*^3}{3}\big [ \big (1- 3 \langle {n_*},\hat {x}\rangle ^2\big ) {n_*}\otimes {n_*} -9 \langle {n_*},\hat {x}\rangle {n_*}\otimes \hat {x} \big. \nonumber \\ & \big. \quad + 15 \langle {n_*},\hat {x}\rangle ^3 {n_*}\otimes \hat {x} \big ]\nonumber \\\mathcal{M}^{11}:D^2\unicode{x1D640}(x)&= \frac {1}{8\pi r^3}\frac {2s_*^4}{3} \big [-\left (\frac 13 - \langle {n_*},\hat {x}\rangle ^2\right) \unicode{x1D644} + \big (1 - 3 \langle {n_*},\hat {x}\rangle ^2\big ) {n_*}\otimes {n_*} \nonumber \\&\quad + \big (1 -10 \langle {n_*},\hat {x}\rangle ^2\big ) \hat {x}\otimes \hat {x} -7 \langle {n_*},\hat {x}\rangle {n_*}\otimes \hat {x} + 15 \langle {n_*},\hat {x}\rangle ^3 {n_*}\otimes \hat {x} \big ] . \end{align}
   We note that $\mathcal{M}^3 = ({s_*}/{6})\mathcal{M}^2$ , $\mathcal{M}^{11} = ({2s_*^2}/{3})\mathcal{M}^6$ .

2. (ii) Formula for $\mathcal{M}_\gamma :D^2\unicode{x1D641}(x)$ . Similarly, we need to compute the following:

   (B51) \begin{eqnarray} && \big \{ \delta _{kl}\delta _{im},\,\,\, \delta _{kl}{n_*}_i{n_*}_m,\,\,\, \delta _{im}{n_*}_k{n_*}_l,\,\,\, \delta _{ik}{n_*}_l{n_*}_m,\,\,\, {n_*}_i{n_*}_m{n_*}_k{n_*}_l\big \} \times \partial _{kl}\unicode{x1D641}_{mj}(x)\nonumber \\& = & \big \{ \delta _{kl}\delta _{im},\,\,\, \delta _{kl}{n_*}_i{n_*}_m,\,\,\, \delta _{im}{n_*}_k{n_*}_l,\,\,\, \delta _{ik}{n_*}_l{n_*}_m,\,\,\, {n_*}_i{n_*}_m{n_*}_k{n_*}_l\big \}\times \nonumber \\&&\frac {1}{r^5}\big [ 3\big (\delta _{mk}\delta _{jl} + \delta _{jk}\delta _{ml} + \delta _{kl}\delta _{mj}\big )\nonumber \\&&\hspace {20pt} -15\big ( \delta _{mj}\hat {x}_k\hat {x}_l +\delta _{ml}\hat {x}_j\hat {x}_k +\delta _{jl}\hat {x}_m\hat {x}_k +\delta _{kl}\hat {x}_m\hat {x}_j +\delta _{mk}\hat {x}_j\hat {x}_l +\delta _{jk}\hat {x}_m\hat {x}_l \big )\nonumber \\ &&\hspace {20pt} + 105 \hat {x}_m\hat {x}_j\hat {x}_k\hat {x}_l \big ]. \end{eqnarray}
   We again tabulate the result in the following:
   (B52) \begin{eqnarray} \delta _{kl}\delta _{im}\partial _{kl}\unicode{x1D641}_{mj}(x) & = & 0\nonumber \\\delta _{kl}{n_*}_i{n_*}_m\partial _{kl}\unicode{x1D641}_{mj}(x) & = & 0\nonumber \\\delta _{im}{n_*}_k{n_*}_l\partial _{kl}\unicode{x1D641}_{mj}(x) & = & \frac {1}{r^5}\big [3\delta _{\textit{ij}} -15\delta _{\textit{ij}}\langle {n_*},\hat {x}\rangle ^2 + 6{n_*}_i{n_*}_j -15\hat {x}_i\hat {x}_j \nonumber \\&&\hspace {20pt} -30({n_*}_i\hat {x}_j + {n_*}_j\hat {x}_i)\langle a,\hat {x}\rangle + 105\hat {x}_i\hat {x}_j\langle {n_*},\hat {x}\rangle ^2\big ]\nonumber \\\delta _{ik}{n_*}_l{n_*}_m\partial _{kl}\unicode{x1D641}_{mj}(x) & = &\frac {1}{r^5}\big [3\delta _{\textit{ij}} - 15\delta _{\textit{ij}}\langle {n_*}, \hat {x}\rangle ^2 +6{n_*}_i{n_*}_j -15\hat {x}_i\hat {x}_j \nonumber \\&&\hspace {20pt} -30({n_*}_ix_j+ {n_*}_j\hat {x}_i)\langle a,\hat {x}\rangle + 105\hat {x}_i\hat {x}_j\langle {n_*},\hat {x}\rangle ^2\big ]\nonumber \\{n_*}_i{n_*}_m{n_*}_k{n_*}_l\partial _{kl}\unicode{x1D641}_{mj}(x) & = & \frac {1}{r^5}\big [9{n_*}_i{n_*}_j - 45{n_*}_i{n_*}_j\langle a, \hat {x}\rangle ^2 -45 {n_*}_i\hat {x}_j\langle {n_*},\hat {x}\rangle \nonumber \\ && \quad + 105{n_*}_i\hat {x}_j\langle {n_*},\hat {x}\rangle ^3\big ]. \end{eqnarray}
   Similar to $\mathcal{M}_\gamma :D^2\unicode{x1D640}$ , we have the following:
   (B53) \begin{eqnarray} \mathcal{M}^1:D^2\unicode{x1D641}(x)&=& 0\nonumber \\\mathcal{M}^2:D^2\unicode{x1D641}(x)&=& -\frac {s_*}{r^5}\big [ 3\big (1 - 5\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} + 6{n_*}\otimes {n_*} - 15(1-7\langle {n_*},\hat {x}\rangle ^2) \hat {x}\otimes \hat {x}\nonumber \\&&\hspace {30pt} -30\langle {n_*},\hat {x}\rangle \big ({n_*}\otimes \hat {x} + \hat {x}\otimes {n_*}\big ) \big ]\nonumber \\\mathcal{M}^3:D^2\unicode{x1D641}(x)&=& -\frac {s_*^2}{6r^5}\big [ 3\big (1 - 5\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} + 6{n_*}\otimes {n_*} - 15(1-7\langle {n_*},\hat {x}\rangle ^2) \hat {x}\otimes \hat {x}\nonumber \\&&\hspace {35pt} -30\langle {n_*},\hat {x}\rangle \big ({n_*}\otimes \hat {x} + \hat {x}\otimes {n_*}\big ) \big ]\nonumber \\\mathcal{M}^4:D^2\unicode{x1D641}(x)&=& \frac {s_*}{r^5} \big [ 3\big (1 - 5\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} + 6{n_*}\otimes {n_*} - 15(1-7\langle n,\hat {x}\rangle ^2) \hat {x}\otimes \hat {x}\nonumber \\&&\hspace {35pt} -30\langle {n_*},\hat {x}\rangle \big ({n_*}\otimes \hat {x} + \hat {x}\otimes {n_*}\big ) \big ]\nonumber \\\mathcal{M}^5:D^2\unicode{x1D641}(x)&=& \frac {s_*^2}{3r^5} \big [ 3\big (1 - 5\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} + 6{n_*}\otimes {n_*} - 15(1-7\langle {n_*},\hat {x}\rangle ^2) \hat {x}\otimes \hat {x}\nonumber \\&&\hspace {35pt} -30\langle {n_*},\hat {x}\rangle \big ({n_*}\otimes \hat {x} + \hat {x}\otimes {n_*}\big ) \big ]\nonumber \\\mathcal{M}^6:D^2\unicode{x1D641}(x)&=& \frac {s_*^2}{r^5}\big [{-}\big (1-5\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} + \big (7 - 45\langle {n_*},\hat {x}\rangle ^2\big ){n_*}\otimes {n_*} \nonumber \\&& \qquad + 5\big (1-7\langle {n_*},\hat {x}\rangle ^2\big )\hat {x}\otimes \hat {x} +35\big (-\langle {n_*},\hat {x}\rangle +3\langle {n_*},\hat {x}\rangle ^3\big ){n_*}\otimes \hat {x} \nonumber \\&& \qquad +10\langle {n_*},\hat {x}\rangle \hat {x}\otimes {n_*} \big ]\nonumber \\\mathcal{M}^7:D^2\unicode{x1D641}(x)&=&0\nonumber \\\mathcal{M}^{10}:D^2\unicode{x1D641}(x)&=& \frac {2s_*^3}{3r^5}\big [ 9\big (1 - 5\langle {n_*},\hat {x}\rangle ^2\big ){n_*}\otimes {n_*} +\big (-45\langle {n_*},\hat {x}\rangle +105\langle {n_*},\hat {x}\rangle ^3\big ){n_*}\otimes \hat {x} \big ]\nonumber \\\mathcal{M}^{11}:D^2\unicode{x1D641}&=& \frac {2s_*^4}{3r^5} \big [{-}\big (1-5\langle {n_*},\hat {x}\rangle ^2\big )\unicode{x1D644} + \big (7 - 45\langle {n_*},\hat {x}\rangle ^2\big ){n_*}\otimes {n_*} \nonumber \\&& \qquad + 5\big (1-7\langle {n_*},\hat {x}\rangle ^2\big )\hat {x}\otimes \hat {x} +35\big (-\langle {n_*},\hat {x}\rangle +3\langle {n_*},\hat {x}\rangle ^3\big ){n_*}\otimes \hat {x} \nonumber \\ && \qquad +10\langle {n_*},\hat {x}\rangle \hat {x}\otimes {n_*} \big ]. \end{eqnarray}

From the above, we can conclude that $\mathcal{M}:D^2\unicode{x1D640}$ and $\mathcal{M}:D^2\unicode{x1D641}$ are linear combinations of the following matrices:

(B54) \begin{align} \unicode{x1D644},\,\,\, {n_*}\otimes {n_*},\,\,\, \hat {x}\otimes \hat {x},\,\,\, {n_*}\otimes \hat {x}, \,\,\, \hat {x}\otimes {n_*}, \end{align}

with coefficients given by $\langle n, \hat {x}\rangle ^k$ for $k=0,1,2,3$ . Concisely, we can write

(B55) \begin{eqnarray} &&\mathcal{M}_\gamma :D^2\unicode{x1D640}(x)\nonumber \\&=& \frac {1}{r^3}\big [ f_1(\langle {n_*}, \hat {x}\rangle )\unicode{x1D644} +f_2(\langle {n_*}, \hat {x}\rangle ) {n_*}\otimes {n_*} +f_3(\langle {n_*}, \hat {x}\rangle ) \hat {x}\otimes \hat {x} +f_4(\langle {n_*}, \hat {x}\rangle ) {n_*}\otimes \hat {x}, \nonumber \\ && \quad +f_5(\langle {n_*}, \hat {x}\rangle )\hat {x}\otimes {n_*} \big ] \end{eqnarray}

and

(B56) \begin{eqnarray} &&\mathcal{M}_\gamma :D^2\unicode{x1D641}(x)\nonumber \\&=& \frac {1}{r^5}\big [ g_1(\langle {n_*}, \hat {x}\rangle )\unicode{x1D644} +g_2(\langle {n_*}, \hat {x}\rangle ) {n_*}\otimes {n_*} +g_3(\langle {n_*}, \hat {x}\rangle ) \hat {x}\otimes \hat {x} +g_4(\langle {n_*}, \hat {x}\rangle ) {n_*}\otimes \hat {x}, \nonumber \\ && +g_5(\langle {n_*}, \hat {x}\rangle )\hat {x}\otimes {n_*} \big ] \end{eqnarray}

where the $f_i$ and $g_i$ ’s are polynomials of degree at most three.

B.4 Formula for $\mathcal{C}_\gamma$ and $\mathcal{D}_\gamma$

Here we express $\mathcal{C}_\gamma$ and $\mathcal{D}_\gamma$ in more transparent forms.

1. (a) For $\mathcal{C}_\gamma$ , from (4.21), we compute

   (B57) \begin{align} \mathcal{C}_\gamma \,\,= \,\,& \unicode{x1D64F}_{{SV}}^{\boldsymbol{v}} - \unicode{x1D63C} - \mathcal{B}_\gamma [\boldsymbol{\nabla }\boldsymbol{v}]] - \mathcal{A}_\gamma (x)\nonumber \\\,\,= \,\,& \gamma _1 \Big [ \big (\unicode{x1D64C}\left (\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}\right ) -\left (\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}\right )\unicode{x1D64C}\big ) +\big (\unicode{x1D64C}^2\unicode{x1D652} - 2\unicode{x1D64C}\unicode{x1D652}\unicode{x1D64C} + \unicode{x1D652}\unicode{x1D64C}^2\big )\nonumber \\& \hspace {15pt} - \big (\unicode{x1D64C}_*(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}) - (\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\unicode{x1D64C}_*\big ) - \big (\unicode{x1D64C}_*^2\unicode{x1D652} - 2\unicode{x1D64C}_*\unicode{x1D652}\unicode{x1D64C}_* + \unicode{x1D652}\unicode{x1D64C}^2_*\big ) \Big ]\nonumber \\&+\gamma _2\Big [ \big (\boldsymbol{v}-{\boldsymbol{v}_*}\big )\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C} + \big (\unicode{x1D64C}-\unicode{x1D64C}_*\big )\boldsymbol{\nabla }\boldsymbol{v} -(\boldsymbol{\nabla }\boldsymbol{v})\big (\unicode{x1D64C}-\unicode{x1D64C}_*\big ) \Big ]\nonumber \\&+\frac {\gamma _3}{2} \Big [ \left (\unicode{x1D64C}(\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})+ (\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\unicode{x1D64C} +\unicode{x1D64C}^2\boldsymbol{\nabla }\boldsymbol{v}-(\boldsymbol{\nabla }\boldsymbol{v})\unicode{x1D64C}^2\right )\nonumber \\&\hspace {25pt} - \left (\unicode{x1D64C}_*(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}) + (\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\unicode{x1D64C}_*\right ) - \left (\unicode{x1D64C}_*^2(\boldsymbol{\nabla }\boldsymbol{v}) - (\boldsymbol{\nabla }\boldsymbol{v})\unicode{x1D64C}_*^2\right ) \Big ]\nonumber \\& +\gamma _4\Big [(\unicode{x1D64C}-\unicode{x1D64C}_*)\unicode{x1D63C}+\unicode{x1D63C}(\unicode{x1D64C}-\unicode{x1D64C}_*)\Big ]\nonumber \\& +\gamma _5\Big [(\unicode{x1D64C}^2-\unicode{x1D64C}_*^2)\unicode{x1D63C}+\unicode{x1D63C}(\unicode{x1D64C}^2-\unicode{x1D64C}_*^2)\Big ]\nonumber \\& +\gamma _6\Big [ (\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C})\unicode{x1D64C} - (\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_*)\unicode{x1D64C}_* \Big ]\nonumber \\& +\gamma _7\Big [\left (|\unicode{x1D64C}^2| - |\unicode{x1D64C}^2_*|\right )\unicode{x1D63C}\Big ]\nonumber \\& + \gamma _9\Big [ \big ((\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\boldsymbol{\cdot }\unicode{x1D64C}\big )\unicode{x1D64C} -\Big ((\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})\boldsymbol{\cdot }\unicode{x1D64C}_*\Big )\unicode{x1D64C}_* \Big ]\nonumber \\& + \gamma _{10}\Big [ \big(\unicode{x1D64C}^2\boldsymbol{\cdot }\unicode{x1D63C}\big)\unicode{x1D64C} + (\unicode{x1D64C}\boldsymbol{\cdot }\unicode{x1D63C})\unicode{x1D64C}^2 -(\unicode{x1D64C}_*^2\boldsymbol{\cdot }\unicode{x1D63C})\unicode{x1D64C}_* - (\unicode{x1D64C}_*\boldsymbol{\cdot }\unicode{x1D63C})\unicode{x1D64C}^2_* \Big ]\nonumber \\& + \gamma _{11}\big [ |\unicode{x1D64C}|^2(\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C})\unicode{x1D64C} - |\unicode{x1D64C}_*|^2(\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_*)\unicode{x1D64C}_* \big ]\nonumber \\\,\,\sim \,\,& O\left (\frac {1}{r^3}\right ). \end{align}

2. (b) For $\mathcal{D}_\gamma$ , from (4.2), we have

   (B58) \begin{eqnarray} \mathcal{D}_\gamma &=& -\Big [\gamma _1\big ( \boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C} +\unicode{x1D64C}_*\unicode{x1D652} - \unicode{x1D652}\unicode{x1D64C}_* +(\boldsymbol{v}-\boldsymbol{v}_*)\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C} +(\unicode{x1D64C}-\unicode{x1D64C}_*)\unicode{x1D652} - \unicode{x1D652}(\unicode{x1D64C}-\unicode{x1D64C}_*) \big )\nonumber \\&&\hspace {10pt}+\gamma _2\unicode{x1D63C} + \frac {\gamma _3}{2}\big (\unicode{x1D63C}\unicode{x1D64C}_* + \unicode{x1D64C}_*\unicode{x1D63C} + \unicode{x1D63C}(\unicode{x1D64C}-\unicode{x1D64C}_*) + (\unicode{x1D64C}-\unicode{x1D64C}_*)\unicode{x1D63C}\big ) \nonumber \\&&\hspace {10pt}+ \gamma _9\big ( (\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_* + \unicode{x1D63C}\boldsymbol{\cdot }(\unicode{x1D64C}-\unicode{x1D64C}_*))\unicode{x1D64C}_* +(\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_* + \unicode{x1D63C}\boldsymbol{\cdot }(\unicode{x1D64C}-\unicode{x1D64C}_*))(\unicode{x1D64C}-\unicode{x1D64C}_*) \big )\! \Big ] \boldsymbol{\cdot }\partial _i\unicode{x1D64C}\textbf{e}_i \nonumber \\&\sim &O\left (\frac {1}{r^4}\right ). \end{eqnarray}

Note that given $\unicode{x1D64C}$ , both $\mathcal{C}_\gamma$ and $\mathcal{D}_\gamma$ are linear in $\boldsymbol{v}$ .

From the form of $\mathcal{C}_\gamma$ and $\mathcal{D}_\gamma$ , it can be seen that they involve multiplications between the following matrices:

(B59) \begin{align} \unicode{x1D63C},\,\,\unicode{x1D652},\,\,\unicode{x1D64C}_*,\,\,\unicode{x1D64C},\,\,\,\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C},\,\, \boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}. \end{align}

1. (i) For $\unicode{x1D63C}, \unicode{x1D652}$ , note that $\boldsymbol{v} = \unicode{x1D640}_S\boldsymbol{v}_*$ so that $\partial _j\boldsymbol{v}_i = \partial _j ({\unicode{x1D640}_S}_{il}{\boldsymbol{v}_*}_l )$ and $\partial _i\boldsymbol{v}_j = \partial _i ({\unicode{x1D640}_S}_{jl}{\boldsymbol{v}_*}_l )$ . From (3.38), we have

   (B60) \begin{align} {\unicode{x1D640}_S}_{il} &\in \left [1, \frac {1}{r}, \frac {1}{r^3}\right ] \big \{\delta _{il}, \hat {x}_i\hat {x}_l\big \} \nonumber \\[-12pt]\end{align}
   (B61) \begin{align} \partial _j{\unicode{x1D640}_S}_{il},\,\, \partial _i{\unicode{x1D640}_S}_{jl} &\in \left [\frac {1}{r^2}, \frac {1}{r^4}\right ] \big \{ \delta _{il}\hat {x}_j,\,\, \delta _{\textit{ij}}\hat {x}_l,\,\, \delta _{jl}\hat {x}_i,\,\, \hat {x}_i\hat {x}_j\hat {x}_l \big \} . \\[6pt] \nonumber\end{align}

   Thus,

   (B62) \begin{eqnarray} \unicode{x1D63C},\,\,\unicode{x1D652} &\in & \big \{ \partial _j{\unicode{x1D640}_S}_{il}{\boldsymbol{v}_*}_l,\,\, \partial _i{\unicode{x1D640}_S}_{jl}{\boldsymbol{v}_*}_l \big \}\nonumber \\ & = & \left [\frac {1}{r^2}, \frac {1}{r^4}\right ] \big \{ \delta _{il}\hat {x}_j{\boldsymbol{v}_*}_l,\,\, \delta _{\textit{ij}}\hat {x}_l{\boldsymbol{v}_*}_l,\,\, \delta _{jl}\hat {x}_i{\boldsymbol{v}_*}_l,\,\, \hat {x}_i\hat {x}_j\hat {x}_l{\boldsymbol{v}_*}_l \big \}\nonumber \\ & = & \left [\frac {1}{r^2}, \frac {1}{r^4}\right ] \big [1,\,\,\langle \hat {x},\boldsymbol{v}_*\rangle \big ] \big \{ \unicode{x1D644},\,\,\boldsymbol{v}_*\otimes \hat {x},\,\,\hat {x}\otimes \boldsymbol{v}_*,\,\,\hat {x}\otimes \hat {x} \big \} . \end{eqnarray}

2. (ii) For $\unicode{x1D64C}$ , from (3.4)–(3.6), we have $\unicode{x1D64C}_*,\unicode{x1D64C}\in [1, \frac {1}{r}, \frac {1}{r^3} ] \big \{\unicode{x1D644},\,\,\,{n_*}\otimes {n_*},\,\,\,\hat {x}\otimes \hat {x}\big \}$ . Hence,

   (B63) \begin{equation} \unicode{x1D64C}_*, \unicode{x1D64C}_*^2, \unicode{x1D64C}, \unicode{x1D64C}^2 \in \left [1, \frac {1}{r}, \ldots , \frac {1}{r^6}\right ] \big [1, \langle \hat {x},{n_*}\rangle \big ] \big \{\unicode{x1D644},\,\,\,{n_*}\otimes {n_*},\,\,\,{n_*}\otimes \hat {x},\,\,\,\hat {x}\otimes {n_*},\,\,\, \hat {x}\otimes \hat {x} \big \}. \end{equation}
   Furthermore, as $\partial _l\unicode{x1D64C}_{\textit{ij}} = [({1}/{r^2}), ({1}/{r^4}) ] \{ {n_*}_i{n_*}_jx_l,\,\,\,\delta _{\textit{ij}}x_l,\,\,\,x_ix_jx_l,\,\,\,\delta _{il}x_j,\,\,\, \delta _{jl}x_i \}$ , $(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})_{\textit{ij}} = \partial _l\unicode{x1D64C}_{\textit{ij}}{\boldsymbol{v}_*}_l$ , and $(\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C})_{\textit{ij}} = \partial _l\unicode{x1D64C}_{\textit{ij}}\boldsymbol{v}_l$ , we have
   (B64) \begin{eqnarray} &&\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C} ,\,\,\,\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}\nonumber \\ &\in & \left [\frac {1}{r^2}, \frac {1}{r^4}\right ] \big \{ {n_*}_i{n_*}_jx_l,\,\,\,\delta _{\textit{ij}}x_l,\,\,\,x_ix_jx_l,\,\,\,\delta _{il}x_j,\,\,\, \delta _{jl}x_i \big \}\big \{ {\boldsymbol{v}_*}_l, \boldsymbol{v}_l \big \} \nonumber \\ &\in & \left [\frac {1}{r^2}, \frac {1}{r^4}\right ] \big [1, \langle \hat {x},\boldsymbol{v}_*\rangle \big ] \big \{\unicode{x1D644},\,\,\,{n_*}\otimes {n_*},\,\,\, \boldsymbol{v}_*\otimes \hat {x},\,\,\,\hat {x}\otimes \boldsymbol{v}_*,\,\,\, \hat {x}\otimes \hat {x} \big \} .\end{eqnarray}

3. (iii) From the above, we have

   (B65) \begin{eqnarray} && \unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_*,\,\,\,\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}_*^2,\,\,\, \unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C},\,\,\,\unicode{x1D63C}\boldsymbol{\cdot }\unicode{x1D64C}^2,\,\,\, |\unicode{x1D64C}_*|^2,\,\,\,\,|\unicode{x1D64C}|^2,\,\,\, \unicode{x1D64C}_*\boldsymbol{\cdot }(\boldsymbol{v}_*\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}),\,\,\, \unicode{x1D64C}\boldsymbol{\cdot }(\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{\nabla }\unicode{x1D64C}) \nonumber \\ &\in & \left [ 1,\frac {1}{r^2},\ldots \frac {1}{r^6} \right ] \big [ 1,\,\,\, \langle x,\boldsymbol{v}_*\rangle ,\,\,\, \langle x,{n_*}\rangle ,\,\,\, \langle \boldsymbol{v}_*, {n_*}\rangle \big ]. \end{eqnarray}

Hence, taking appropriate products of all the above, we have

(B66) \begin{align} \mathcal{C}_\gamma &\in \left [\frac {1}{r^3},\ldots ,\frac {1}{r^6}\right ] \big [1,\,\, \langle \hat {x},{n_*}\rangle ,\,\, \langle \hat {x},\boldsymbol{v}_*\rangle ,\,\, \langle {n_*},\boldsymbol{v}_*\rangle \big ]\times \nonumber \\& \big \{\unicode{x1D644},\,\,{n_*}\otimes {n_*},\,\, {n_*}\otimes \boldsymbol{v}_* ,\,\,\boldsymbol{v}_*\otimes {n_*},\,\, {n_*}\otimes \hat {x},\,\,\hat {x}\otimes {n_*},\,\,\, \boldsymbol{v}_*\otimes \hat {x},\,\,\hat {x}\otimes \boldsymbol{v}_*,\,\, \hat {x}\otimes \hat {x} \big \}. \end{align}

For $\mathcal{D}_\gamma$ , note that $\unicode{x1D64C}_{\textit{ij}} \in [1,({1}/{r}),({1}/{r^3}) ] \{\delta _{\textit{ij}}, {n_*}_i{n_*}_j, \hat {x}_i\hat {x}_j\}$ , we have

(B67) \begin{align} \partial _k\unicode{x1D64C}_{\textit{ij}} \in \left [\frac {1}{r^2},\frac {1}{r^4}\right ] \big \{\delta _{\textit{ij}}x_k,\,\, {n_*}_i{n_*}_jx_k,\,\, \delta _{ik}\hat {x}_j,\,\, \delta _{jk}\hat {x}_i,\,\,\hat {x}_i\hat {x}_j\hat {x}_k\big \}. \end{align}

Hence

(B68) \begin{equation} \mathcal{D}_\gamma \in \left [\frac {1}{r^4},\ldots ,\frac {1}{r^9}\right ] \big [1,\,\,\, \langle \hat {x},{n_*}\rangle ,\,\,\, \langle \hat {x},\boldsymbol{v}_*\rangle ,\,\,\, \langle {n_*},\boldsymbol{v}_*\rangle \big ]\big \{ \hat {x},\,\,{n_*},\,\,\boldsymbol{v}_* \big \}. \end{equation}