If we parametrize straight lines as shown in Figure \fixme{ref}, we can
define the set of all staight lines as $\mathcal{L} = \{ (\phi, \rho) :
-\phi \in [0, 2\pi), \rho \geq 0 \}$. By defining the Lebesgue measure on
-this set $d\mathcal{L} = \dpdr$ we are ready to introduce the
+\phi \in [0, 2\pi), \rho \geq 0 \}$. We will write a line $l_{\phi,
+\rho} = l_{\nu, \rho}$ where $\nu$ is a unit vector along the line,
+i.e.\ $\nu = (\sin \phi, -\cos \phi)^T$. By defining the
+measure on this set $d\mathcal{L} = \dpdr$ we are ready to introduce the
Cauchy--Crofton formula, which gives us a way to calculate the length of
a curve by looking at the measure of the set of lines that intersect the
curve. Note that the measure $d\mathcal{L}$ is invariant under rigid
motions, meaning combinations of translations and rotations.
\begin{theorem}[The Euclidean Cauchy--Crofton formula]
- Given a curve $C$ in $\mathbb{R}^2$, the length of this curve
- $\abs{C}$ is related to the set of lines $\mathcal{L}$ as follows
+ Given a curve differentiable $C$ in $\mathbb{R}^2$, the length of
+ this curve $\abs{C}$ is related to the set of lines $\mathcal{L}$ as
+ follows
\begin{equation}
- \int_\mathcal{L} n(L \cap C) \, d\mathcal{L}(L) = 2 \abs{C},
+ \int_\mathcal{L} \#(l_{\phi, \rho} \cap C) \,
+ d\mathcal{L}(l_{\phi, \rho}) = 2 \abs{C},
\end{equation}
- where $n(L \cap C)$ is the number of times the line $L$ intersects
- with the curve $C$.
+ where $\#(l_{\phi, \rho} \cap C)$ is the number of times the line
+ $l_{\phi, \rho}$ intersects the curve $C$.
\end{theorem}
This elegant formula is very useful when we later will discretize our
energy function. The set of lines $\mathcal{L}$ is then discretized
depending on our choice of neighborhood, and the length of the curve $C$
can be approximated by a sum.
+Our space is equipped with a metric tensor $M(p)$ such that the inner
+product of two vectors in a point $p$ is calculated as $\langle a,
+b\rangle_M = \langle a, M(p) b \rangle$. The length of a curve $\gamma$
+parametrized by some parameter $t$ then becomes
+\begin{equation}
+ \abs{\gamma}_M = \int_\gamma \langle \dot{\gamma},
+ M\big(\gamma(t)\big) \, \dot{\gamma} \rangle \, dt
+\end{equation}
+We will now prove a Cauchy--Crofton formula in this case where we have a
+metric tensor in each point.
\begin{theorem}[The Riemannian Cauchy--Crofton formula]
- In the case of a Riemannian geometry, where the scalar product in
- each point depends on a metric tensor $M(p)$ varying continuously over
- our space, the Cauchy--Crofton formula becomes
+ In the case where the scalar product in each point depends on a
+ metric tensor $M(x)$ varying continuously over our space, the
+ Cauchy--Crofton formula becomes
\begin{equation}
- \abs{C}_R = \int_\mathcal{L} n(L \cap C) \, \frac{\det
- M(p)}{2 \left( u_L^T \cdot M(p) \cdot u_L \right)} \,
- d\mathcal{L}(L).
+ \abs{C}_M = \int_\mathcal{L} \sum_{p \in l_{\nu, \rho} \cap C}
+ \, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu \right)}
+ \, d\mathcal{L}(l_{\nu, \rho}).
\end{equation}
- Here, $u_L$ is a unit vector in the point $p$ along the line
- currently being integrated over.
\end{theorem}
\begin{proof}
Assume first that our space is equipped with at constant metric
- tensor $M$. The length of our curve using this Riemannian metric can
- then be calculated by transforming the curve back into euclidean
- space and applying the Euclidean Cauchy--Crofton formula
+ tensor $M$. The length of our curve using this tensor can be
+ calculated by transforming the curve and applying the Euclidean
+ Cauchy--Crofton formula
\begin{align}
- \abs{C}_M &= \abs{M^{\sfrac{1}{2}}C} \\
- &= \int n(L \cap M^{\sfrac{1}{2}}C) \,
- d\mathcal{L}(L) \\
- &= \int n(M^{-\sfrac{1}{2}}L \cap C) \,
- d\mathcal{L}(L) \\
- &= \int n(P \cap C) \abs{J(\Mhalf)}\,
- d\mathcal{L}(P)
+ \abs{C}_M &= \int_C \sqrt{\langle \dot{C}, M \, \dot{C} \rangle} \, dt
+ = \int_C \sqrt{\langle \Mhalf \dot{C}, \Mhalf
+ \dot{C} \rangle} = \abs{M^{\sfrac{1}{2}}C} \\
+ &= \int_\mathcal{L} \#(l_{\phi, \rho} \cap M^{\sfrac{1}{2}}C) \,
+ d\mathcal{L}(l_{\phi, \rho}) \\
+ &= \int_\mathcal{L} \#(M^{-\sfrac{1}{2}}l_{\phi, \rho} \cap C) \,
+ d\mathcal{L}(l_{\phi, \rho}) \\
+ &= \int_\mathcal{L} \#(m_{\phi, \rho} \cap C) \abs{J_M(l_{\phi, \rho})}\,
+ d\mathcal{L}(m_{\phi, \rho})
\end{align}
- where $J(M^{\sfrac{1}{2}})$ is the Jacobian of our coordinate
- transformation $P = M^{-\sfrac{1}{2}} L$, which is what we need to
- find next.
+ where $J_M(l_{\phi, \rho})$ is the Jacobian of our coordinate
+ transformation $F : \mathcal{L} \to \mathcal{L}$, which maps
+ $l_{\phi, \rho} \mapsto \Mhalf l_{\phi, \rho}$.
As $M\in \mathbb{R}^{2\times2}$ is symmetric, so is
- $M^{\sfrac{1}{2}}$, and it emits an eigendecomposition $\Mhalf
- = U\Lambda U^T$ where the components correspond to the following
+ $M^{\sfrac{1}{2}}$, and it admits an eigendecomposition $\Mhalf
+ = U\Sigma U^T$ where the components correspond to the following
coordinate transformations
\begin{align}
- U(\alpha,\rho) &= (\alpha + \xi, \rho), \\
- U^T(\alpha,\rho) &= (\alpha - \xi, \rho), \\
- \Lambda &= \begin{pmatrix}
+ %U(\phi,\rho) &= (\phi + \xi, \rho), \\
+ %U^T(\phi,\rho) &= (\phi - \xi, \rho), \\
+ U(l_{\nu,\rho}) &= l_{\phi + \xi, \rho} = l_{U\nu, \rho} \\
+ U^T(l_{\nu,\rho}) &= l_{\phi - \xi, \rho} = l_{U^T\nu, \rho} \\
+ \Sigma &= \begin{pmatrix}
\sigma_1 & 0 \\
0 & \sigma_2
\end{pmatrix}.
\end{align}
As $U$ and $U^T$ correspond to rotations they do not contribute to
- the Jacobian, except for changing the input angle of the $\Lambda$
- operator. Given a line
+ the Jacobian, except for changing the input angle of the operator
+ $\Lambda$ such that $J_M(l_{\phi, \rho}) = J_{\Sigma^2}(U^T l_{\phi,
+ \rho})$. Given a line
\begin{equation}
- L_{\alpha, \rho} = \begin{pmatrix}
- \rho\cdot \cos \alpha \\
- \rho\cdot \sin \alpha
+ l_{\phi, \rho} = \begin{pmatrix}
+ \rho\cdot \cos \phi \\
+ \rho\cdot \sin \phi
\end{pmatrix}
+ \mathbb{R} \begin{pmatrix}
- -\sin \alpha \\
- \cos \alpha
+ -\sin \phi \\
+ \cos \phi
\end{pmatrix},
\end{equation}
- the $\Lambda$ operator transforms it into
+ the operator $\Sigma$ transforms it into
\begin{equation}
- \Lambda L_{\alpha, \rho} = \begin{pmatrix}
- \sigma_1 \rho\cdot \cos \alpha \\
- \sigma_2 \rho\cdot \sin \alpha
+ \Sigma l_{\phi, \rho} = \begin{pmatrix}
+ \sigma_1 \rho\cdot \cos \phi \\
+ \sigma_2 \rho\cdot \sin \phi
\end{pmatrix}
+ \mathbb{R} \begin{pmatrix}
- -\sigma_1 \sin \alpha \\
- \sigma_2 \cos \alpha
+ -\sigma_1 \sin \phi \\
+ \sigma_2 \cos \phi
\end{pmatrix},
\end{equation}
- which is has parameters
+ which is the line $l_{\theta, \eta}$ with
\begin{align}
- \beta &= \arctan \left( \frac{\sigma_1}{\sigma_2} \tan \alpha
+ \theta &= \arctan \left( \frac{\sigma_1}{\sigma_2} \tan \phi
\right) \\
- \nu &= \left\langle \begin{pmatrix}
- \sigma_1 \rho\cdot \cos \alpha \\
- \sigma_2 \rho\cdot \sin \alpha
+ \eta &= \left\langle \begin{pmatrix}
+ \sigma_1 \rho\cdot \cos \phi \\
+ \sigma_2 \rho\cdot \sin \phi
\end{pmatrix}, \begin{pmatrix}
- \cos \beta \\
- \sin \beta
+ \cos \theta \\
+ \sin \theta
\end{pmatrix}
- \right\rangle.
+ \right\rangle = \sigma_1 \rho \cdot \cos \phi \cdot \cos \theta
+ + \sigma_2 \rho \cdot \sin \phi \cdot \sin \theta.
\end{align}
- As $\partial_\rho \beta = 0$, the Jacobian becomes
- $\abs{J} = \partial_\alpha\beta \cdot \partial_\rho \nu$. Differentiation
- yields
+ As $\partial_\rho \theta = 0$, the Jacobian becomes
+ $\abs{J_{\Sigma^2}(l_{\phi, \rho})} = \partial_\phi\theta \cdot
+ \partial_\rho \eta$. Differentiation yields
\begin{align}
- \partial_\alpha \beta &= \frac{ \frac{\sigma_1}{\sigma_2}
- \sec^2 \alpha}{1 + \frac{\sigma_1^2}{\sigma_2^2}
- \tan^2 \alpha}, \\
- \partial_\rho \nu &= \sigma_1 \cos \alpha \cdot \cos \beta + \sigma_2
- \sin \alpha \cdot \sin \beta
- = \frac{\sigma_1 \cos \alpha \left( 1 + \tan^2 \alpha \right)
- }{\sqrt{1 + \frac{\sigma_1^2}{\sigma_2^2} \tan^2 \alpha}}
+ \partial_\phi \theta &= \frac{ \frac{\sigma_1}{\sigma_2}
+ \sec^2 \phi}{1 + \frac{\sigma_1^2}{\sigma_2^2}
+ \tan^2 \phi}
+ = \frac{\sigma_1 \sigma_2}{\sigma_1^2 \sin^2 \phi + \sigma_2^2
+ \cos^2 \phi}, \\
+ \partial_\rho \eta &= \sigma_1 \cos \phi \cdot \cos \theta +
+ \sigma_2 \sin \phi \cdot \sin \theta.
\end{align}
- and assuming that $v = (v_x, v_y)^T$ is a unit vector along the line
- parametrized by the angle $\alpha$ we obtain
+ In the expression for $\partial_\rho \eta$ we use that $\sin
+ (\arctan( x )) = x / \sqrt{1 + x^2}$ and that $\cos ( \arctan ( x ))
+ = 1 / \sqrt{1 + x^2}$ to obtain
+ \begin{equation}
+ \partial_\rho \eta = \frac{\sigma_1 \cos \phi + \sigma_2 \sin
+ \phi \, \frac{\sigma_1}{\sigma_2} \tan \phi}
+ {\sqrt{1 + \frac{\sigma_1^2}{\sigma_2^2}
+ \tan^2 \phi}}
+ = \frac{\sigma_1 \sigma_2 \cos^2 \phi + \sigma_1 \sigma_2
+ \sin^2 \phi}{\sqrt{\sigma_1^2 \sin^2 \phi + \sigma_2^2
+ \cos^2 \phi}}.
+ \end{equation}
+ If $\nu = (\nu_x, \nu_y)^T$ is a unit vector along the line
+ $l_{\phi, \rho} = l_{\nu, \rho}$ then
+ \begin{equation}
+ \abs{J_{\Sigma^2}(l_{\nu, \rho})}
+ = \frac{\sigma_1^2 \sigma_2^2}
+ {\left(\sigma_1^2 \sin^2 \phi + \sigma_2^2 \cos^2 \phi
+ \right)^{\sfrac{3}{2}}}
+ = \frac{\sigma_1^2 \sigma_2^2
+ }{\left(\sigma_1^2 \nu_x^2 + \sigma_2^2
+ \nu_y^2\right)^{\sfrac{3}{2}}}
+ = \frac{\det \Sigma^2}{(\nu^T \, \Sigma^2 \,
+ \nu)^{\sfrac{3}{2}}}.
+ \end{equation}
+ We are interested in the Jacobian of the whole transformation, so
+ all that is left to do is insert $U^T l_{\nu, \rho}$ to obtain
+ \begin{equation}
+ \abs{J_M(l_{\nu, \rho})} = \abs{J_{\Sigma^2}(U^T l_{\nu, \rho})}
+ = \frac{\det M}{\left(\nu^T U \Sigma^2 U^T \nu\right)^{\sfrac{3}{2}}}
+ = \frac{\det M}{\left(\nu^T M \nu\right)^{\sfrac{3}{2}}}
+ \end{equation}
+ We have now proven that for a constant metric tensor $M$, the length
+ of the differentiable curve $C$ with regards to this tensor can be
+ calculated as
\begin{equation}
- \abs{J} = \frac{\frac{\sigma_1}{\sigma_2} \cdot \sigma_1 \left(
- 1 + \tan^2 \alpha \right)^{\sfrac{3}{2}}}{\left( 1 +
- \frac{\sigma_1^2}{\sigma_2^2} \tan^2
- \alpha \right)^{\sfrac{3}{2}}}
- = \frac{\sigma_1^2 \sigma_2^2 \left( v_x^2 +
- v_y^2\right)^{\sfrac{3}{2}}}{\left(\sigma_1^2 v_x^2 + \sigma_2^2
- v_y^2\right)^{\sfrac{3}{2}}}.
+ \abs{C}_M = \int_C \langle \dot{C}, M \dot{C} \rangle \, dt
+ = \int_\mathcal{L} \# (l_{\nu, \rho} \cap C)
+ \frac{\det M}{\left(\nu^T M \nu \right)^{\sfrac{3}{2}}} \,
+ d\mathcal{L} (l_{\nu, \rho})
\end{equation}
- The angle of the line that $\Lambda$ takes as input has already been
- rotated by the $U^T$ operator, such that if $u$ is a unit vector
- along the line being integrated over, then $v = U^Tu$ and
+ We now argue that this also holds for a non-constant but continuous
+ metric tensor $M(x)$. By partitioning the domain into disjoint sets
+ $U_i$ such that $\Omega = \cup U_i$, we make a piecewise constant
+ approximation $M_\pi(x)$ such that if $x \in U_i$ then $M_\pi(x) =
+ M(x_i)$ for some fixed $x_i \in U_i$. We then approximate the right
+ side of \fixme{ref} by
\begin{equation}
- \abs{J} = \frac{\det M}{\left(u^T M u\right)^{\sfrac{3}{2}}}
+ \abs{C}_{M_\pi} = \sum_i \int_\mathcal{L} \#(l_{\nu, \rho} \cap
+ C \cap U_i) \, w_i(\nu) \, d\mathcal{L}(l_{\nu, \rho})
+ \label{eq:mpi_approx}
\end{equation}
- And now we argue that this also holds for a non-constant but
- continuous metric tensor $M(x)$. By partitioning the domain into
- disjoint sets $U_i$ such that $\Omega = \cup U_i$, we make a
- piecewise constant approximation such that $M_\pi(x) = M(x_i)$ for
- some fixed $x_i \in U_i$. We then approximate the right side of
- \fixme{ref} by
+ where $w_i$ is the weight-function used in the set $U_i$ using the
+ constant tensor $M_\pi(x_i)$ and can be written
+ \begin{equation}
+ w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T M(x_i) \nu
+ \right)^{\sfrac{3}{2}}}.
+ \end{equation}
+ We further simplify the approximation by introducing the global
+ weight-function $w_\pi(\nu, x)$ which is equal to
+ $w_i(\nu)$ when $x \in U_i$. It can be written
+ \begin{equation}
+ w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T M_\pi(x) \nu
+ \right)^{\sfrac{3}{2}}}.
+ \end{equation}
+ Using this weight in \eqref{eq:mpi_approx} we can get rid of the sum
+ over the partition $i$ and form a sum of all intersection point of
+ $C$ and the line $l_{\nu, \rho}$ currently being integrated over.
+ The approximation becomes
\begin{align}
- \abs{C}_{M_\pi} &= \sum_i \int_\mathcal{L} n_{C\cap U_i} w_i \,
- d\mathcal{L} \\
- &= \sum_i \int_\mathcal{L} \sum_{x\in C\cap U_i \cap
- \mathcal{L}} w_\pi(\phi, p) \, d\mathcal{L} \\
- &= \int_\mathcal{L} \sum_{x\in C\cap \mathcal{L}} w_\pi(\phi, p)
- \, d\mathcal{L}.
+ \abs{C}_{M_\pi} &=
+ \sum_i \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C \cap
+ U_i} w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) \\
+ &= \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C}
+ w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}).
\end{align}
As our partition $\pi$ is refined the weight $w_\pi(x)$ converges
- pointwise to the continously varying weight $w(x)$. We now want to
- bound the expression for $w_\pi$.
+ pointwise to the continously varying weight $w(x)$. Now it only
+ remains to show that the integral converges, and that the left side
+ converges to the left side of \fixme{ref}.
+
+ Recall from \fixme{ref} that the left hand side is calculated as
\begin{equation}
- w_\pi = \frac{\det M(p)}{2 \left(u_L^T \cdot M(p) \cdot
- u_L\right)^{\sfrac{3}{2}}}
+ \abs{C}_{M_\pi} = \int_C \sqrt{ \dot{C}^T M_\pi \dot{C} } \, dt.
\end{equation}
+ We know that $M_\pi(x)$ converges pointwise to $M(x)$ and using the
+ fact that the largest eigenvalue of $M$ is equal to 1, we know that
+ the integrand is bounded and we can apply Lebesgue's dominated
+ convergence theorem to see that $\abs{C}_{M_\pi} \to
+ \abs{C}_{M(x)}$.
+
+ We apply the same theorem to show that the right hand side of
+ \fixme{ref} converges, but showing that $w_\pi(\nu, x)$ is bounded
+ is a bit more involved.
+
If we manage to bound the singular values and keep them away from
$0$, that would be great. The singular values \fixme{or eigenvalues}
of our tensor is
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