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the rotational component of the vector field. The symmetric part can in turn be decomposed in a multiple of the identity matrix, the divergence part that describe the dilation component of the vector field, and a symmetric matrix with zero trace. The remaining symmetric component of the matrix can be described in terms of the deformation, i.e. an area preserving stretching in one direction combined with shrinking in the orthogonal direction, and the direction $\phi$ of the stretching relative to the direction of $\partial_u$.

6.2 Choice of Gauge

For Galilean 2 + 1 dimensional geometry isophotes are typically 2 dimensional surfaces. There are two generic cases: points where the isophote surface cuts the spatial surface through the point, and points where the isophote surface is tangent to the spatial surface through the point. The first case can be interpreted as motion of isophote curves in the image, and the second case as creation, annihilation or saddle points.

6.3 Tangent Gauge

Our next task is to define an adapted frame for points where the isophote surface cuts the spatial surface. For the spatial plane we can reuse the tangent gauge for $E_2$ in Section 4.1. Starting from an arbitrary frame $i$, we first adapt the spatial sub frame ${\partial_x, \partial_y}$, to the gradient and tangent direction in the spatial plane:

(tuv)=1fx2+fy2(1000fyfx0fxfy)(txy)=Ai.(24) \begin{pmatrix} \partial_t \\ \partial_u \\ \partial_v \end{pmatrix} = \frac{1}{\sqrt{f_x^2 + f_y^2}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & f_y & -f_x \\ 0 & f_x & f_y \end{pmatrix} \begin{pmatrix} \partial_t \\ \partial_x \\ \partial_y \end{pmatrix} = Ai. \quad (24)

The spatio-temporal vector $\partial_s$ must have unit length in time to be part of a Galilean frame. By requiring $\partial_s$ to lie in the spatio-temporal tangent plane, i.e. $f_s = 0$, it is constrained in one direction. The adapted spatio-temporal direction must have the form:

s=t+βu+γv, \partial_s = \partial_t + \beta\partial_u + \gamma\partial_v,

in terms of the new frame. Using $0 = f_s = f_t + \gamma f_v$ and solving for $\gamma$ we get that $\gamma = -f_t/f_v$. Still we have one undetermined degree of freedom $\beta \in \mathbb{R}$. For each choice of $\beta$ we have a plane spanned by ${\partial_s, \partial_v}$. The image restricted to such a plane is a function on $\Gamma_2$ and can be studied by the methods from Section 5.1. From Theorem 2 there are two scalar invariants: acceleration $a = -f_{ss}/f_v$ and divergence $\delta = -f_{sv}/f_v$. We can see that acceleration becomes a quadratic function of $\beta$ and thus the gauge can be fixed by finding a $\beta$ s.t. $a(\beta)$ is an extremum, i.e. by solving $\partial_\beta a(\beta) = 0$ for $\beta$, which gives:

βa=ftfuvfvfuuftufuu.(25) \beta_a = \frac{f_t f_{uv}}{f_v f_{uu}} - \frac{f_{tu}}{f_{uu}}. \quad (25)

Which is defined as long as $f_{uu} \neq 0$, i.e. as long as the isophote curvature in the spatial plane is non-vanishing. It can be shown that requirement of an