Let us rearrange the image formation model (
3) for diffuse polarisation (
i.e \(\alpha _{spec}=0\)) using the following trigonometric formula
$$\begin{aligned} \cos (2\theta _{pol}-2\overline{\theta }) = \cos (2\theta _{pol})\cos (2\overline{\theta })+\sin (2\theta _{pol})\sin (2\overline{\theta }) \end{aligned}$$
(7)
and the duplication formulas which lead to the following equality
$$\begin{aligned} \begin{aligned} \cos (2\theta _{pol}-2\overline{\theta })&= \cos (2 \theta _{pol})\big (2\cos ^2(\overline{\theta })-1\big )\\&\quad + 2\sin (2\theta _{pol})\sin (\overline{\theta })\cos (\overline{\theta }). \end{aligned} \end{aligned}$$
(8)
By substituting (
8) into the image formation (
3) we get
$$\begin{aligned} \begin{aligned} I(\theta _{pol})&= I_++I_- \Big (\cos (2 \theta _{pol})\big (2\cos ^2(\overline{\theta })-1\big )\\&\quad + 2\sin (2\theta _{pol})\sin (\overline{\theta })\cos (\overline{\theta })\Big ). \end{aligned} \end{aligned}$$
(9)
The key step to describe the previous polarisation image formation as a partial differential equation is to consider the consistency of shape information from the angle
\(\overline{\theta }\) with respect to the surface normal. In other terms, after substituting (
5) and (
6) in (
9), we obtain the following partial differential equation
$$\begin{aligned} \begin{aligned} I(\theta _{pol})&= I_++I_- \bigg (\cos (2\theta _{pol})\bigg (2g^2(f)\frac{z^2_x}{\Vert {\overline{\mathbf{n}}}(\mathbf {x})\Vert ^2\sin ^2(\phi )} -1\bigg ) \\&\quad +2\sin (2\theta _{pol})g^2(f)\frac{z_xz_y}{\Vert {\overline{\mathbf{n}}}(\mathbf {x})\Vert ^2\sin ^2(\phi )} \bigg ) \end{aligned} \end{aligned}$$
(10)
that after some algebra becomes
$$\begin{aligned} \begin{aligned}&I(\theta _{pol}) - I_++I_-\cos (2\theta _{pol}) \\&\quad =I_-\Big (\cos (2\theta _{pol})z_x+\sin (2\theta _{pol})z_y\Big )\frac{2g^2(f)z_x}{\Vert {\overline{\mathbf{n}}}(\mathbf {x})\Vert ^2\sin ^2(\phi )}. \end{aligned} \end{aligned}$$
(11)
With the aim of simplifying the non-linear part depending on the focal length
f, the zenith angle
\(\phi \) and the normalized normal vector
\(\Vert {\overline{\mathbf{n}}}(\mathbf {x})\Vert \), we consider the ratio of the previous equations obtained with two polariser angles
\(\theta _{pol_1}\) and
\(\theta _{pol_2}\), we get
$$\begin{aligned}&\frac{I(\theta _{pol_1}) - I_++I_-\cos (2\theta _{pol_1})}{I(\theta _{pol_2}) - I_++I_-\cos (2\theta _{pol_2})} \nonumber \\&\quad =\frac{\cos (2\theta _{pol_1})z_x+\sin (2\theta _{pol_1})z_y}{\cos (2\theta _{pol_2})z_x+\sin (2\theta _{pol_2})z_y} \end{aligned}$$
(12)
that leads to the following homogeneous linear PDE
$$\begin{aligned}&\Big (\big (I(\theta _{pol_1}) - I_++I_-\cos (2\theta _{pol_1})\big )\cos (2\theta _{pol_2})\nonumber \\&\qquad -\big (I(\theta _{pol_2})- I_++I_-\cos (2\theta _{pol_2})\big )\cos (2\theta _{pol_1})\Big ) z_x \nonumber \\&\qquad +\Big (\big (I(\theta _{pol_1}) - I_++I_-\cos (2\theta _{pol_1})\big )\sin (2\theta _{pol_2}) \nonumber \\&\qquad -\big (I(\theta _{pol_2})- I_++I_-\cos (2\theta _{pol_2})\big )\sin (2\theta _{pol_1})\Big ) z_y=0 \nonumber \\ \end{aligned}$$
(13)
that we refer in the following as
$$\begin{aligned} \mathbf {b}_{pol}(\mathbf {x})\cdot \nabla z(\mathbf {x}) = 0, \end{aligned}$$
(14)
where the components of the bi-dimensional vector field
\(\mathbf {b}_{pol}(\mathbf {x})=(b_{pol}^1(\mathbf {x}),b_{pol}^2(\mathbf {x}))\) are
$$\begin{aligned} b_{pol}^1= & {} \big (I(\theta _{pol_1}) - I_++I_-\cos (2\theta _{pol_1})\big )\cos (2\theta _{pol_2}) \nonumber \\&-\big (I(\theta _{pol_2}) - I_++I_-\cos (2\theta _{pol_2})\big )\cos (2\theta _{pol_1}) \end{aligned}$$
(15)
and
$$\begin{aligned} b_{pol}^2= & {} \big (I(\theta _{pol_1}) - I_++I_-\cos (2\theta _{pol_1})\big )\sin (2\theta _{pol_2}) \nonumber \\&-\big (I(\theta _{pol_2}) - I_++I_-\cos (2\theta _{pol_2})\big )\sin (2\theta _{pol_1}). \end{aligned}$$
(16)
Let us point out that the previous equation can be substantially simplified by taking
\(\theta _{pol_1}=0\) and
\(\theta _{pol_2}=\frac{\pi }{4}\) yielding to the following equation
$$\begin{aligned} \big (-I_{\frac{\pi }{4}}+I_+\big )z_x+\big (I_0-I_++I_-\big )z_y=0. \end{aligned}$$
(17)
As a first remark, we notice that (
14) is invariant with respect to lighting and albedo. Most importantly, it describes the geometry of the surface through its isocontours circumventing the ambiguity of the SfP problem.