You are reading a single comment by @TheBrick(Tommy) and its replies. Click here to read the full conversation.
  • \begin{subequations}
    \label{num3}
    \beq
    U_i^\alpha-u_i^\infty(x^\alpha)=\frac{F^\alpha_i}{6\pi \mu a}+(1+\frac{1}{6}a^2\nabla^2)u'_i(x^\alpha),
    \eeq
    \beq
    \Omega_i^\alpha-\Omega_i^\infty=\frac{Li^\alpha}{8\pi \mu a^3}+\frac{1}{2}\varepsilon{ijk}\nabla_ju'k(x^\alpha),
    \eeq
    \beq
    -E
    {ij}^\infty=\frac{S{ij}^\alpha}{(\frac{20}{3})\pi \mu a^3}+\left(1+\frac{a^2}{10}\nabla^2\right)e'{ij}(x^\alpha),
    \eeq
    \label{batandg}
    \end{subequations}

About