Lecture 04 - From Bishop Frame to NLSE

#math

Recall the Bishop frame:

\begin{aligned} T & = T \\ U & = c o s (θ) N - s i n (θ) B \\ V & = s i n (θ) N + c o s (θ) B \end{aligned}

With Frenet Serret Equations:

\frac{d}{d s} (\begin{matrix} T \\ U \\ V \end{matrix}) = (\begin{matrix} 0 & κ_{1} & κ_{2} \\ - κ_{1} & 0 & 0 \\ - κ_{2} & 0 & 0 \end{matrix}) (\begin{matrix} T \\ U \\ V \end{matrix})

With $κ (s) = \sqrt{κ_{1}^{2} + κ_{2}^{2}}$ , $θ (s) = a r c t a n (\frac{κ_{2}}{κ_{1}})$ .

Binormal flow in the Bishop frame:

\begin{aligned} γ_{t} & = - κ_{2} U + κ_{1} V \\ γ_{s s} & = κ_{1} U + κ_{2} V \end{aligned}

We have a compatibility condition:
$(γ_{t})_{s s} = (γ_{s s})_{t}$

\begin{matrix} (1) & \begin{aligned} (γ_{t})_{s s} & = - {κ_{2}}_{s s} U + {κ_{1}}_{s s} V + (κ_{1} κ_{2} s - κ_{2} {κ_{1}}_{s}) T \\ (γ_{s s})_{t} & = {κ_{1}}_{t} U + {κ_{2}}_{t} V + κ_{1} U_{t} + κ_{2} V_{t} \end{aligned} \end{matrix}

We need to compute $U_{t}$ and $V_{t}$ in order to contextualize $(γ_{s s})_{t}$ in our frame.

First we compute:

\begin{matrix} (2) & T_{t} = γ_{s t} = (γ_{t})_{s} = - {κ_{2}}_{s} U + {κ_{1}}_{s} V \end{matrix}

Notice that since we are working on an orthogonal frame we get the following:

\begin{matrix} (3) & \frac{d}{d t} (U \cdot T) = 0 ⟹ U_{t} \cdot T = - U \cdot T_{t}, \end{matrix}

similarly as well for $V$ .

Thus we obtain the following:

\begin{aligned} (U_{t} \cdot V)_{s} & = U_{t s} \cdot V + U_{t} \cdot V_{s} \\ V_{s} & = - κ_{2} T \\ U_{t} \cdot V_{s} & = - κ_{2} T \cdot U_{t} \\ = κ_{2} U \cdot T_{t} & (Using (3)) \\ = - κ_{2} {κ_{2}}_{s} & (Using (2)) \end{aligned}

Similarly we also compute $V \cdot U_{s t}$ :

\begin{aligned} U_{s} & = - κ_{1} T \\ ⟹ V \cdot U_{s t} & = - κ_{1} {κ_{1}}_{s} \end{aligned}

Hence:

\begin{aligned} (U_{t} \cdot V)_{s} & = - κ_{1} {κ_{1}}_{s} - κ_{2} {κ_{2}}_{s} \\ ⟹ U_{t} \cdot V & = \frac{- 1}{2} (κ_{1}^{2} + κ_{2}^{2}) + c (t) \end{aligned}

From ( $1$ ) we get that:
$- {κ_{2}}_{s s} U + {κ_{1}}_{s s} V + (κ_{1} κ_{2} s - κ_{2} {κ_{1}}_{s}) T = {κ_{1}}_{t} U + {κ_{2}}_{t} V + κ_{1} U_{t} + κ_{2} V_{t}$

Notice that $\frac{\partial}{\partial t} | U |^{2} = 2 U \cdot U_{t} = 0$ since $| U | = 1$ $\forall t$ .
We will project the equation onto $U$ first by taking the dot product with $U$ on both sides. We obtain:

\begin{aligned} - {κ_{2}}_{s s} & = {κ_{1}}_{t} + κ_{2} U \cdot V_{t} \\ - {κ_{2}}_{s s} & = {κ_{1}}_{t} + κ_{2} (\frac{1}{2} (κ_{1}^{2} + κ_{2}^{2}) - c (t)) \end{aligned}

Similarly by projecting onto $V$ we get:

{κ_{1}}_{s s} = {κ_{2}}_{t} - κ_{1} (\frac{1}{2} (κ_{1}^{2} - κ_{2}^{2}) + c (t))

Recall the definition of the complex curvature:

\begin{aligned} ψ & = κ e^{\int_{0}^{s} τ d s^{'}} \\ = κ c o s (θ) + i κ s i n (θ) \\ = κ_{1} + i κ_{2} \end{aligned}

Thus we can map $(κ_{1}, κ_{2}) \to ψ$ and from the above equations we obtain our NLSE!

- i ψ_{t} = ψ_{s s} + ψ (\frac{1}{2} | ψ |^{2} + c (t))

Geometric Interpretation of the Vortex Filament Equation AND NLSE

What have we done so far:
Let $γ$ be a curve embedded in $R^{3}$ and we have mapped $I : (R^{3}, \cdot, X) \to (SU (2), T r ., \land)$

Exposition:

SU (2) = {(\begin{matrix} α & - \bar{β} \\ β & \bar{α} \end{matrix}) : α, β \in C, | α |^{2} + | β |^{2} = 1}

with inner product defined to be $(A, B) = \frac{- 1}{2} T r (A B)$ and wedge product $A \land B = - \frac{1}{2} [A, B]$ .

This map works by:
$I : [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] \to \sum_{i = 1}^{3} x_{i} E_{i} = [\begin{matrix} i x_{1} & x_{2} + i x 3 \\ - x_{2} + i x_{3} & - i x_{1} \end{matrix}]$
Where $E_{1} = i σ_{3} [\begin{matrix} i & 0 \\ 0 & - i \end{matrix}]$ , $E_{2} = i σ_{2} [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]$ , $E_{3} = i σ_{1} [\begin{matrix} 0 & i \\ i & 0 \end{matrix}]$ , which forms a basis for $SU (2)$ .