unless $E_{||} \neq 0$. Therefore, we emitted the surface charge $\sigma$ per unit area which is defined as $E_{||} = 4 \pi (\sigma - \sigma_{\mathrm {GJ}}) (\boldsymbol{n} \cdot \hat{ \boldsymbol{B}})$, where $\sigma_{\mathrm {GJ}}$ is the surface charge density when the corotational electric field exists both inside and outside of the star; $\boldsymbol{n}$ is the unit normal vector to the stellar surface; and $\hat{ \boldsymbol{B}}$ is the unit vector of the magnetic field on the stellar surface. The surface charges are emitted $\sim 230$ times in a rotation period.

We solve the equation of motion, $$\frac{{\mathrm {d}} \boldsymbol{p}_i}{{\mathrm {d}} t}
 = 
 q_i \left( \boldsymbol{E} + \boldsymbol{\beta}_i
 \times \boldsymbol{B} \right)
 + \boldsymbol{F}_{\mathrm {rad}},$$ where $\boldsymbol{\beta}_i$ is the velocity of the $i$th particle in units of the speed of light, $\boldsymbol{p}_i = \gamma_i m_i \beta_i c$, and $\gamma_i = (1 - \beta_i^2)^{-1/2}$; $\boldsymbol{F}_{\mathrm {rad}} = (2/3) (q_i^2 / R_{\mathrm c}^2) \gamma_i^4
(\boldsymbol{p}_i / |\boldsymbol{p}_i|)$ (if $\beta_i \simeq 1$) is the radiation reaction force, and $R_{\mathrm c}$ is the curvature radius. This approximate expression for the radiation drag force can be used when the force due to the external field is larger than the radiation drag force.

For pair creation, grid points $\boldsymbol{r}_{i,j} = 