### explain how Q-Tensors solve more problems than just head-tail-symmetry

parent 2947ceb5
 ... ... @@ -131,8 +131,7 @@ Let $g$ be the metric on $M$ and $m = \dim M$.% Since the director field is modelling a direction the obvious choice is a vector and therefore the obvious choice for second moment tensor is contravariant as well. } \subsection{Constrained Landau-de Gennes model} \label{sec:fs-vs-ldg} { As the introduction mentions this thesis and \textcite{q4-Ball2011} are only concerned with the special case of uniaxial liquid crystals witch constant order parameter. % concerned with the special case of uniaxial liquid crystals witch constant order parameter. In order to formulate this in the $Q$-tensor theory consider the spectral representation of a symmetric trace-free $Q ∈ T^{(2,0)}T_pM$ ($p ∈ M$) \begin{align*} Q_p = λ^i E_i ⊗ E_i \text{ with } λ_m = - \sum_{i = 1}^{m - 1} λ^i \text{ and } E_i \text{ an orthonormal basis.} ... ... @@ -167,9 +166,15 @@ Let $g$ be the metric on $M$ and $m = \dim M$.% $s = 0$ means that the liquid is isotropic, and $s = 1$ means that all molecules are perfectly aligned with $n$. According to \textcite[A Measures of orientational order]{q50-Q-tensor-introduction-mottram} a typical liquid crystal could have $s = 0.6$ (here $m = 3$.) Throughout the thesis we assume $s ∈ \left[-\frac1{m-1}, 1 \right] \without \{0\}$ to be a constant. While we only consider the case of uniaxial nematic crystals with constant order parameter $s$ we see that the $Q$-tensor model in the general form fixes two other shortcoming of the Frank-Oseen model. For once the director field does not take into account how close the molecules are to the average direction. This is also related to the fact that liquid crystals are in some cases forced to have \enquote{defects} where no director can be properly defined. That would be modeled with $s = 0$. Secondly the Frank-Oseen model assumes uniaxiality but biaxial nematic liquid crystals were found. Note that in the generality of this section $m$ is any number of dimensions but since we live in three dimensions, the maximum of independent eigenvalues of $Q$ can be $2$, so there cannot be triaxial nematic liquid crystals. \begin{definition}[Constrained $Q$-tensors] \label{def:constrained_q_tensors} We introduce the following fiber bundles over $M$. Let $p ∈ M$. ... ... @@ -199,14 +204,14 @@ Let $g$ be the metric on $M$ and $m = \dim M$.% \begin{note}[Line fields] \label{note:line_fields} We see that $P(-n) = P(n)$ since $(-n) ⊗ (-n) = (-1)^2 n ⊗ n$. By identifying two opposite unit vectors we define a line. As the introduction explains, the identification of two opposite unit vectors defines a line. Therefore we will call the constrained $Q$-tensor fields $Γ_R(𝒬^{𝕊}M)$ \newTerm{line fields}. Here $R$ is any regularity class like $C$ (continuous), $W^{1,q}$ (weakly differentiable), $C^∞$ (smooth) or $ℒ$ (Lebesgue-measurable). \end{note} % end note Line fields \begin{definition}[Orientable line field] \label{def:orientable_line_field} A line field $Q ∈ Γ_R(𝒬^{𝕊}(M))$ is called \newTerm{orientable} if there exists $n ∈ Γ_R(𝕊^*M)$ such that $P(n) = Q$. A line field $Q ∈ Γ_R(𝒬^{𝕊}(M))$ is called \newTerm{orientable} if there exists $n ∈ Γ_R(𝕊M)$ such that $P(n) = Q$. \end{definition} % end definition of Orientable line field % When we want to consider $Q$-tensors an directions fields on the boundary, ... ...
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