special relativity - How did L.H. Thomas derive his 1927 expressions for an electron with an axis?

I'm looking at the 1927 paper of Thomas, The Kinematics of an Electron with an Axis, where he shows that the instantaneous co-moving frame of an accelerating electron rotates and moves with some infinitesimal velocity. He states:

At $t=t_0$ let the electron have position $\mathbf{r}_0$ and velocity $\mathbf{v}_0$, with $\beta_0=(1-{\mathbf{v}_0}^2/c^2)^{-\frac 1 2}$, in $(\mathbf{r}, t)$. Then, by (2.1), that definite system of coordinates $(\mathbf{R}_0, T_0)$ in which the electron is instantaneously at rest at the origin and which is obtained from $(\mathbf{r},t)$ by a translation and a Lorentz transformation without rotation is gven by

$$\begin{align*}\mathbf{R}_0 &= \mathbf{r} - \mathbf{r}_0 + (\beta_0 - 1)\frac{ (\mathbf{r}-\mathbf{r}_0)\cdot \mathbf{v}_0}{{\mathbf{v}_0}^2}\mathbf{v}_0-\beta_0 \mathbf{v}_0(t-t_0)\tag{3.1a}\\ T_0 &= \beta_0\left( t - t_0 - \frac {(\mathbf{r} - \mathbf{r}_0)\cdot \mathbf{v}_0}{c^2}\right)\tag{3.1b}\end{align*}$$

By eliminating $(\mathbf{r},t)$ from equations (3.1) and the similar equations for $(\mathbf{R}_1, T_1)$,

$$\begin{align*}\mathbf{R}_1 &= \mathbf{R}_0 + \frac{(\beta_0 - 1)} {{\mathbf{v}_0}^2}(\mathbf{R}_0\times (\mathbf{v}_0\times \mathbf{dv}_0)) - \beta_0 T_0(\mathbf{dv}_0 + (\beta_0 - 1)\frac{(\mathbf{v}_0\cdot \mathbf{dv}_0)}{{\mathbf{v}_0}^2}\mathbf{v}_0)\tag{a}\\ T_1 &= T_0 - \frac {\beta_0} c^2((\mathbf{R}_0\cdot(\mathbf{dv}_0 + (\beta_0 - 1)\frac{(\mathbf{v}_0\cdot \mathbf{dv}_0)}{{\mathbf{v}_0}^2}\mathbf{v}_0))) - d\tau_0\tag{3.3b} \end{align*}$$

What are the steps to get from (3.1) to (3.3)?