5.4. Ambiguities in the Hanle effect in the saturation regime

In the saturation regime of the Hanle effect, Stokes \(Q\) and \(U\) are insensitive to the field strength, but are sensitive to the geometry of the field. For a \(J=0 \to J=1\) transition, the linear polarization can be written as:

\[\begin{split}\begin{aligned} Q &=& \frac{q}{2} \left( 3 \cos^2 \theta_B-1 \right) \sin^2\Theta_B \cos 2\Phi_B \nonumber \\ U &=& \frac{q}{2} \left( 3 \cos^2 \theta_B-1 \right) \sin^2\Theta_B \sin 2\Phi_B.\end{aligned}\end{split}\]

These expressions contain a mixture of angles to make it clear that the polarization amplitude depends on both the angle between the vertical and the magnetic field and between the magnetic field and the line-of-sight (LOS).

The coordinates of the magnetic field vector \(\mathbf{B}\) in the reference system of the vertical and the reference system of the LOS are:

\[\begin{split}\begin{aligned} \mathbf{B} &=& B \left(\sin \theta_B \cos \phi_B \mathbf{i}+\sin \theta_B \sin \phi_B \mathbf{j}+\cos \theta_B \mathbf{k} \right) \nonumber \\ \mathbf{B} &=& B \left(\sin \Theta_B \cos \Phi_B \mathbf{i}'+\sin \Theta_B \sin \Phi_B \mathbf{j}'+\cos \Theta_B \mathbf{k}' \right),\end{aligned}\end{split}\]

where the unit vectors are related by a simple rotation:

\[\begin{split}\begin{aligned} \mathbf{i}' &=& \cos \theta \mathbf{i} - \sin \theta \mathbf{k} \nonumber \\ \mathbf{k}' &=& \sin \theta \mathbf{i} + \cos \theta \mathbf{k}.\end{aligned}\end{split}\]

Introducing these relations on the expression for the magnetic field, we find that the following has to be fulfilled, given that the magnetic field vector is the same in both reference systems:

\[\begin{split}\begin{aligned} \sin \theta_B \cos \phi_B &=& \sin \Theta_B \cos \Phi_B \cos \theta + \cos \Theta_B + \sin \theta \nonumber \\ \sin \theta_B \sin \phi_B &=& \sin \Theta_B \sin \Phi_B \nonumber \\ \cos \theta_B &=& \cos \Theta_B \cos \theta - \sin \Theta_B \cos \Phi_B \sin \theta.\end{aligned}\end{split}\]

Solving the previous three equations in the two directions, we find the following transformations between the angles in the vertical reference system and the LOS reference system:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \cos \Theta_B &=& \cos\theta \cos\theta_B + \sin\theta \sin\theta_B \cos\phi_B \nonumber \\ \sin \Theta_B &=& +\sqrt{1-\cos^2\Theta_B} \nonumber \\ \cos \Phi_B &=& \frac{\cos\theta \sin\theta_B \cos\phi_B - \cos\theta_B \sin\theta}{\sin \Theta_B} \nonumber \\ \sin \Phi_B &=& \frac{\sin\theta_B \sin\phi_B}{\sin\Theta_B}\end{aligned}\end{split}\\and\end{aligned}\end{align} \]

\[\begin{split}\begin{aligned} \cos \theta_B &=& \cos\theta \cos\Theta_B - \sin\theta \sin\Theta_B \cos\Phi_B \nonumber \\ \sin \theta_B &=& +\sqrt{1-\cos^2\theta_B} \nonumber \\ \cos \phi_B &=& \frac{\cos\theta \sin\Theta_B \cos\Phi_B + \cos\Theta_B \sin\theta}{\sin \theta_B} \nonumber \\ \sin \phi_B &=& \frac{\sin\Theta_B \sin\Phi_B}{\sin\theta_B}.\end{aligned}\end{split}\]

Note that, since \(\Theta_B \in [0,\pi]\), we can safely use the square root and take the positive value. In order to transform from one reference system to the other, we can compute the inclination easily by inverting the sinus or the cosinus. However, the situation is different for the azimuth, because the range of variation is \([-\pi,\pi]\). Therefore, one has to compute the cosinus and the sinus separately and the decide which is the correct quadrant fo the angle in terms of the signs of both quantities.

Four possible kinds of ambiguities can exist for the Stokes \(Q\) and \(U\) parameters. The idea is that \(\Phi_B\) can be modified and still obtain the same \(Q\) and \(U\) by properly adjusting the value of \(\Theta_B\). It is clear that, given that the term that can be used to compensate for the change in the azimuth on the LOS reference system is the same for Stokes \(Q\) and \(U\), we can only compensate for changes in the sign. Therefore, we have the following potential ambiguities:

\[\begin{split}\begin{aligned} \Phi_B' &=& \Phi_B \nonumber \\ \Phi_B' &=& \Phi_B -\pi/2 \nonumber \\ \Phi_B' &=& \Phi_B + \pi/2 \nonumber \\ \Phi_B' &=& \Phi_B + \pi.\end{aligned}\end{split}\]

For each case, we have to compute the value of \(\Theta_B'\) that keeps the value of \(Q\) and \(U\) unchanged. Therefore, once we find a solution to the inversion problem in the form of the pair \((\theta_B,\phi_B)\), we can find the remaining solutions in the saturation regime following the recipes that we present now. Remember that, unless one knows the polarity of the field, or in other words, the sign \(\cos\Theta_B\), the number of potential ambiguous solutions is 8. If the polarity of the field is known, the number is typically reduced to 4 (or 2 if no 90\(^\circ\) ambiguity is present).

5.4.1. \(\Phi_B’=\Phi_B\)

Under this change, we have that

\[\cos 2\Phi_B' = \cos 2\Phi_B, \quad \sin 2\Phi_B' = \sin 2\Phi_B, \quad \cos \Phi_B' = \cos \Phi_B, \quad \sin \Phi_B' = \sin \Phi_B.\]

Making use of the previous relations between the angles wrt to the vertical and the LOS, we have to solve the following equation:

\[\left( 3 \cos^2\theta_B'-1 \right) \sin^2 \Theta_B' = \left( 3 \cos^2\theta_B-1 \right) \sin^2 \Theta_B,\]

which can be written as:

\[\left[ 3 \left( \cos \Theta_B' \cos \theta - \sin\theta \sin\Theta_B' \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B' = \left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\]

After some algebra and doing the substitution \(t=\sin\Theta_B'\), we end up with the following equation to be solved:

\[A t^4 + Bt^2 + C t^3 \sqrt{1-t^2} = K,\]

where

\[\begin{split}\begin{aligned} A &=& -3\cos^2 \theta + 3\sin^2 \theta \cos^2 \Phi_B \nonumber \\ B &=& 3\cos^2 \theta - 1 \nonumber \\ C &=& -6 \cos\theta \sin\theta \cos \Phi_B \nonumber \\ K &=& \left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\end{aligned}\end{split}\]

The previous equation can be solved if we make the change of variables \(t=\pm \sqrt{Z}\), resulting in:

\[(C^2+A^2) Z^4 + (-C^2+2AB) Z^3 + (-2AK+B^2) Z^2 - 2BKZ + K^2 = 0.\]

This polynomial of 4-th order can have four different solutions. From these solutions, we have to take only the real solutions which are larger than 0, given the range of variation of \(\Theta_B\):

\[t \in \mathbb{R}, \qquad 0 \leq t \leq 1.\]

Once the solutions for \(t\) are found, we make \(\Theta_B' = \arcsin t\). Note that, for a fixed value of \(t\), two values of \(\Theta_B'\) are possible. We choose the correct one by evaluating the expressions for \(Q\) and \(U\) and testing which of the two possible choices give the values equal (or very similar) to the original ones.

The angles \((\theta_B,\phi_B)\) are obtained by doing the transformation from \((\Theta_B',\Phi_B)\) to the vertical reference system.

5.4.2. \(\Phi_B’=\Phi_B+\pi\)

Under this change, we have:

\[\cos 2\Phi_B' = \cos 2\Phi_B, \quad \sin 2\Phi_B' = \sin 2\Phi_B, \quad \cos \Phi_B' = -\cos \Phi_B, \quad \sin \Phi_B' = -\sin \Phi_B.\]

Following the same approach, we have to solve for \(\Theta_B'\) in

\[\left[ 3 \left( \cos \Theta_B' \cos \theta + \sin\theta \sin\Theta_B' \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B' = \left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\]

The solution are obtained as the roots of the same equations as before but now

\[\begin{split}\begin{aligned} A &=& -3\cos^2 \theta + 3\sin^2 \theta \cos^2 \Phi_B \nonumber \\ B &=& 3\cos^2 \theta - 1 \nonumber \\ C &=& 6 \cos\theta \sin\theta \cos \Phi_B \nonumber \\ K &=& \left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\end{aligned}\end{split}\]

The angles \((\theta_B,\phi_B)\) are obtained by doing the transformation from \((\Theta_B',\Phi_B+\pi)\) to the vertical reference system.

5.4.3. \(\Phi_B’=\Phi_B+\pi/2\)

Under this change, we have:

\[\cos 2\Phi_B' = -\cos 2\Phi_B, \quad \sin 2\Phi_B' = -\sin 2\Phi_B, \quad \cos \Phi_B' = -\sin \Phi_B, \quad \sin \Phi_B' = \cos \Phi_B.\]

Following the same approach, we have to solve for \(\Theta_B'\) in

\[\left[ 3 \left( \cos \Theta_B' \cos \theta + \sin\theta \sin\Theta_B' \sin\Phi_B\right)^2-1 \right] \sin^2 \Theta_B' = \left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\]

The solution are obtained as the roots of the same equations as before but now

\[\begin{split}\begin{aligned} A &=& -3\cos^2 \theta + 3\sin^2 \theta \sin^2 \Phi_B \nonumber \\ B &=& 3\cos^2 \theta - 1 \nonumber \\ C &=& 6 \cos\theta \sin\theta \sin \Phi_B \nonumber \\ K &=& -\left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\end{aligned}\end{split}\]

The angles \((\theta_B,\phi_B)\) are obtained by doing the transformation from \((\Theta_B',\Phi_B+\pi/2)\) to the vertical reference system.

5.4.4. \(\Phi_B’=\Phi_B-\pi/2\)

Under this change, we have:

\[\cos 2\Phi_B' = -\cos 2\Phi_B, \quad \sin 2\Phi_B' = -\sin 2\Phi_B, \quad \cos \Phi_B' = \sin \Phi_B, \quad \sin \Phi_B' = -\cos \Phi_B.\]

Following the same approach, we have to solve for \(\Theta_B'\) in

\[\left[ 3 \left( \cos \Theta_B' \cos \theta + \sin\theta \sin\Theta_B' \sin\Phi_B\right)^2-1 \right] \sin^2 \Theta_B' = \left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\]

The solution are obtained as the roots of the same equations as before but now

\[\begin{split}\begin{aligned} A &=& -3\cos^2 \theta + 3\sin^2 \theta \sin^2 \Phi_B \nonumber \\ B &=& 3\cos^2 \theta - 1 \nonumber \\ C &=& -6 \cos\theta \sin\theta \sin \Phi_B \nonumber \\ K &=& -\left[ 3 \left( \cos \Theta_B \cos \theta - \sin\theta \sin\Theta_B \cos\Phi_B\right)^2-1 \right] \sin^2 \Theta_B.\end{aligned}\end{split}\]

The angles \((\theta_B,\phi_B)\) are obtained by doing the transformation from \((\Theta_B',\Phi_B-\pi/2)\) to the vertical reference system.