Least squares method to determine the position of the eccentric dipole

In 1930's Adolf Schmidt derived formulas of the position of the eccentric dipole by minimizing the sum of squares of \(n\)=2 terms. In the following explanation, the radius of the earth \(a\) is set to \(a=1\) for brevity.

The position \((x_c,y_c,z_c)\) of the eccentric dipole is determined by minimizing the next \(S\), \begin{eqnarray*} S & = & \left(g_2^0-2g_1^0z_c+g_1^1x_c+h_1^1y_c\right)^2 \\ & & + \left(g_2^1-\sqrt{3}(g_1^0x_c+g_1^1z_c)\right)^2 + \left(h_2^1-\sqrt{3}(g_1^0y_c+h_1^1z_c)\right)^2 \\ & & + \left(g_2^2-\sqrt{3}(g_1^1x_c-h_1^1y_c)\right)^2 + \left(h_2^2-\sqrt{3}(h_1^1x_c+g_1^1y_c)\right)^2. \end{eqnarray*} By setting \(\partial S/\partial z_c = 0\), \begin{eqnarray*} \partial S/\partial z_c & = & -4g_1^0\left(g_2^0-2g_1^0z_c+g_1^1x_c+h_1^1y_c\right) \\ & & - 2\sqrt{3}g_1^1\left(g_2^1-\sqrt{3}(g_1^0x_c+g_1^1z_c)\right) - 2\sqrt{3}h_1^1\left(h_2^1-\sqrt{3}(g_1^0y_c+h_1^1z_c)\right) = 0. \end{eqnarray*} This is simplified to, \begin{equation} g_1^0g_1^1x_c + g_1^0h_1^1y_c + \left(4(g_1^0)^2+3(g_1^1)^2+3(h_1^1)^2\right)z_c \\ = 2g_1^0g_2^0 + \sqrt{3}g_1^1g_2^1 + \sqrt{3}h_1^1h_2^1. \label{eq01} \end{equation} Similarly, \begin{eqnarray*} \partial S/\partial x_c & = & 2g_1^1\left(g_2^0-2g_1^0z_c+g_1^1x_c+h_1^1y_c\right) - 2\sqrt{3}g_1^0\left(g_2^1-\sqrt{3}(g_1^0x_c+g_1^1z_c)\right) \\ & & - 2\sqrt{3}g_1^1\left(g_2^2-\sqrt{3}(g_1^1x_c-h_1^1y_c)\right) - 2\sqrt{3}h_1^1\left(h_2^2-\sqrt{3}(h_1^1x_c+g_1^1y_c)\right) = 0, \end{eqnarray*} which becomes, \begin{equation} \left(3(g_1^0)^2+4(g_1^1)^2+3(h_1^1)^2\right)x_c + g_1^1h_1^1y_c + g_1^0g_1^1z_c \\ = -g_1^1g_2^0 + \sqrt{3}g_1^0g_2^1 + \sqrt{3}g_1^1g_2^2 + \sqrt{3}h_1^1h_2^2. \label{eq02} \end{equation} Setting \(\partial S/\partial y_c = 0\) gives, \begin{eqnarray*} \partial S/\partial y_c & = & 2h_1^1\left(g_2^0-2g_1^0z_c+g_1^1x_c+h_1^1y_c\right) - 2\sqrt{3}g_1^0\left(h_2^1-\sqrt{3}(g_1^0y_c+h_1^1z_c)\right) \\ & & + 2\sqrt{3}h_1^1\left(g_2^2-\sqrt{3}(g_1^1x_c-h_1^1y_c)\right) - 2\sqrt{3}g_1^1\left(h_2^2-\sqrt{3}(h_1^1x_c+g_1^1y_c)\right) = 0, \end{eqnarray*} which becomes, \begin{equation} g_1^1h_1^1x_c + \left(3(g_1^0)^2+3(g_1^1)^2+4(h_1^1)^2\right)y_c + g_1^0h_1^1z_c \\ = -h_1^1g_2^0 + \sqrt{3}g_1^0h_2^1 - \sqrt{3}h_1^1g_2^2 + \sqrt{3}g_1^1h_2^2. \label{eq03} \end{equation}

We introduce \(m^2\) as below and put the right-hand side of the equations \eqref{eq01}, \eqref{eq02}, and \eqref{eq03} as \(L_0\), \(L_1\), and \(L_2\), respectively. \begin{eqnarray*} m^2 & = & (g_1^0)^2 + (g_1^1)^2 + (h_1^1)^2, \\ L_0 & = & 2g_1^0g_2^0 + \sqrt{3}\left(g_1^1g_2^1 + h_1^1h_2^1\right), \\ L_1 & = & -g_1^1g_2^0 + \sqrt{3}\left(g_1^0g_2^1 + g_1^1g_2^2 + h_1^1h_2^2\right), \\ L_2 & = & -h_1^1g_2^0 + \sqrt{3}\left(g_1^0h_2^1 - h_1^1g_2^2 + g_1^1h_2^2\right). \end{eqnarray*} Using these notations, equations \eqref{eq01}, \eqref{eq02}, and \eqref{eq03} are written as, \begin{eqnarray} g_1^0g_1^1x_c + g_1^0h_1^1y_c + \left((g_1^0)^2+3m^2\right)z_c = L_0, \label{eq04} \\ \left((g_1^1)^2+3m^2\right)x_c + g_1^1h_1^1y_c + g_1^0g_1^1z_c = L_1, \label{eq05} \\ g_1^1h_1^1x_c + \left((h_1^1)^2+3m^2\right)y_c + g_1^0h_1^1z_c = L_2. \label{eq06} \end{eqnarray} Subtraction, \(\eqref{eq05}\times\left((g_1^0)^2+3m^2\right)-\eqref{eq04}\times g_1^0g_1^1\), gives, \begin{equation} 3m^2\left((g_1^0)^2+(g_1^1)^2+3m^2\right)x_c + 3m^2g_1^1h_1^1y_c = \left((g_1^0)^2+3m^2\right)L_1 - g_1^0g_1^1L_0. \label{eq07} \end{equation} Subtraction, \(\eqref{eq05}\times h_1^1-\eqref{eq06}\times g_1^1\), gives, \begin{equation} 3m^2h_1^1x_c - 3m^2g_1^1y_c = h_1^1L_1 - g_1^1L_2 \label{eq08} \end{equation} Solving equations \eqref{eq07} and \eqref{eq08} for \(x_c\) and noting \((g_1^0)^2+(g_1^1)^2+(h_1^1)^2=m^2\), \begin{equation} 3m^2x_c = L_1 - g_1^1\left.\left(L_0g_1^0+L_1g_1^1+L_2h_1^1\right)\right/(4m^2). \label{eq09} \end{equation} Similarly, \(y_c\) is obtained from \eqref{eq07} and \eqref{eq08} as, \begin{equation} 3m^2y_c = L_2 - h_1^1\left.\left(L_0g_1^0+L_1g_1^1+L_2h_1^1\right)\right/(4m^2). \label{eq10} \end{equation} For \(z_c\), substituting \eqref{eq09} and \eqref{eq10} to \eqref{eq05} and after rather tedious modification, we obtain, \begin{equation} 3m^2z_c = L_0 - g_1^0\left.\left(L_0g_1^0+L_1g_1^1+L_2h_1^1\right)\right/(4m^2). \label{eq11} \end{equation}

Restoring the earth's radius \(a\) and introducing another notation \(E\), Schmidt's formulas of the position of the eccentric dipole \((x_c,y_c,z_c)\) are summarized as the followings. \begin{eqnarray*} x_c & = & a\left.\left(L_1 - g_1^1E\right)\right/(3m^2), \\ y_c & = & a\left.\left(L_2 - h_1^1E\right)\right/(3m^2), \\ z_c & = & a\left.\left(L_0 - g_1^0E\right)\right/(3m^2), \end{eqnarray*} where, \begin{eqnarray*} m^2 & = & (g_1^0)^2 + (g_1^1)^2 + (h_1^1)^2, \\ E & = & \left.\left(L_0g_1^0+L_1g_1^1+L_2h_1^1\right)\right/(4m^2), \\ L_0 & = & 2g_1^0g_2^0 + \sqrt{3}\left(g_1^1g_2^1 + h_1^1h_2^1\right), \\ L_1 & = & -g_1^1g_2^0 + \sqrt{3}\left(g_1^0g_2^1 + g_1^1g_2^2 + h_1^1h_2^2\right), \\ L_2 & = & -h_1^1g_2^0 + \sqrt{3}\left(g_1^0h_2^1 - h_1^1g_2^2 + g_1^1h_2^2\right). \end{eqnarray*}

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