Some personal notes PT decomposition

Here are notes on PT decomposition.

The PT decomposition is defined as follows, \(\mathbf{B}=\nabla\times(\nabla\eta\times\nabla\Phi)+\nabla\eta\times\nabla\Psi,\)

\(\eta\) is a coordinate parameter in Cartesian or Spherical coordinates, it can be chosen as \(z\) and \(r\), respectively. The electric current is defined as follows, here we ignore the vacuum permeability.

\[\begin{aligned} \mathbf{J}=\nabla\times\mathbf{B} \end{aligned}\]

Thus we will have the following equations

\[\begin{aligned} \nabla^2_{||}\Phi=B_n,\\ \nabla^2_{||}\Psi=J_n,\\ \end{aligned}\]

where the subscript \(||\) indicates the operator only works in the directions parallel to the plane, which is perpendicular to \(\nabla\eta\). As the gradient operator along \(\eta\) commutates with the parallel Laplace operator, \([ \partial_{\eta}, \nabla^2_{||}]=0\). We will have,

\[\begin{align} \nabla^2_{||}\partial_{\eta}\Phi=\partial_{\eta}B_n,\\ \nabla^2_{||}\partial_{\eta}\Psi=\partial_{\eta}J_n.\\ \end{align}\]

Adopting the divergence-free and charge-free conditions, we can obtain,

\[\begin{align} \nabla^2_{||}\partial_{\eta}\Phi=-\nabla_{||}\cdot\mathbf{B}_{||},\\ \nabla^2_{||}\partial_{\eta}\Psi=-\nabla_{||}\cdot\mathbf{J}_{||}.\\ \end{align}\]

The parallel current components can be obtained using the force-free condition \(\mathbf{J}=\alpha\mathbf{B}\).

Thus in each layer, we can obtain the variable \((\Phi,\Psi)^\top\) and their \(\eta\) derivative \((\partial_{\eta}\Phi,\partial_{\eta}\Psi)^\top\). Therefore, we can make an integral upward along \(\eta\), just like the integration of an ordinary equation.

We can obtain the physical variable on a staggered grid, described in the following section.

Cartesian Case

We choose \(\eta=z\), and the above equations become,

\[\begin{align} \nabla^2_{x,y}\Phi&=B_z,\\ \nabla^2_{x,y}\Psi&=J_z,\\ \nabla^2_{x,y}\partial_z\Phi&=-\partial_x B_x -\partial_y B_y,\\ \nabla^2_{x,y}\partial_z\Psi&=-\partial_x J_x -\partial_y J_y. \end{align}\]

Therefore, the \(\mathbf{B}\) can be obtained directly from observations, and \(\mathbf{J}\) can be obtained by combining the force-free condition \(\mathbf{J}=\alpha\mathbf{B}\), where \(\alpha=J_z/B_z\). In the following discussion, we assume that \(\mathbf{B}\), \(J_z\), and \(\alpha\) have already been provided on the bottom boundary.

Combining the Eq. 1, we can express the transverse components as,

\[\begin{align} \mathbf{B}_{x,y}&=\hat{e}_x(\partial_x\partial_z\Phi)+\hat{e}_y(\partial_y\partial_z\Phi)\\ &+\hat{e}_x(\partial_y\Psi)+\hat{e}_y(-\partial_x\Psi). \end{align}\]

In the \(\hat{z}\) direction, the transverse components of \(\mathbf{B}\) only contain the first-order derivative.

Here, we design a numerical scheme for the upward integral along the \(\hat{z}\) direction. Suppose that we already have \((\Phi,\Psi)^\top\) and \((\dot{\Phi},\dot{\Psi})^\top\) on the location of \(z_i\), where \((\dot{})\) over the variable indicates the derivative along \(\hat{z}\), hereafter we denote them as \((\Phi_i,\Psi_i)^\top\) and \((\dot{\Phi}_i,\dot{\Psi}_i)^\top\). With \((\dot{\Phi}_i,\dot{\Psi}_i)^\top\), we can deal it like the integral of an ODE.

Lame parameters