TITLE: Find trajectory where work of a vector field is the lowest
QUESTION [0 upvotes]: Given a vector field $\vec{F}(P(x,y),Q(x,y))$ and both starting point $A$ and endpoint $B$ find the trajectory where the work of vector field is the lowest asuming that trajectory starts at $A$ and ends in $B$. Generally saying find such $\overparen{AB}$ that $\int_{\overparen{AB}}{Pdy+Qdx}$ is minimal.
If that is not possible then, more particularly saying, we can assume that $\vec{F}(x,y)=<x^2+y, 3y^2+x>$ and starpoint and endpoint are $(-3, 0)$ and $(3, 0)$. Would there be any solution?

REPLY [1 votes]: $$
(\text{curl } F)_3 = \partial_x F_2 - \partial_y F_1 =
1 - 1 = 0
$$
So $F$ is conservative. This has the nice consequence that all closed curves will give zero as integral value and this means that the integral value (for open curves) is curve independent.


TITLE: Eigenvalues of a bipartite graph
QUESTION [8 upvotes]: Let $X$ be a connected graph with maximum eigenvalue $k$. Assume that $-k$ is also an eigenvalue. I wish to prove that $X$ is bipartite.
Now if $\vec{x}=(x_1,\cdots ,x_n)$ is the eigenvector for $-k$ then I can show that for the vector $\vec{y}$ whose entries are $(|x_1|,\cdots ,|x_n|)$ we have $y'Ay=ky'y$. From here can I conclude that $\vec{y}$ is an eigenvector with eigenvalue $k$?
How to proceed to prove this result?
Thanks.

REPLY [5 votes]: $A$ is symmetric, nonnegative, and irreducible. By a theorem of Perron and Frobenius, $k$ is a simple eigenvalue with a positive eigenvector $u$. Now with componentwise absolute value, $k|x|=|-kx|=|Ax|\le A|x|$. Multiplication with $u^T$ shows that we must have equality. Hence $|x|$ is an eigenvector, hence a multiple of $u$. Therefore $x$ has no zero component.
Partition the nodes into $P$ (of size $p$) and $N$ (of size $n$), where $x_i>0$ if $i\in P$ and $x_i<0$ if $i\in N$. Then $v=x_P>0$ and $w=-x_N>0$. Partition $A$ conformally as $A=\pmatrix{B & C\\C^T & D}$ (of size $p+n\times p+n$, with $B$ of size $p\times p$, and note that $B,C,D$ are nonnegative. Then the equation $|Ax|= A|x|$ implies $|Bv-Cw|=Bv+Cw$ and $|C^Tv-Dw|=C^Tv+Dw$. Taking the squared norm and simplifying yields $(Bv)_i(Cw)_i=0$ for $i\in P$ and $(C^Tv)_k(Dw)_k=0$ for $k\in N$. Since $v,w>0$, $C_{ik}=1$ implies that $B_{ij}=0$ for $j\in P$ and $D_{kj}=0$ for $j\in N$.
This means that for every edge $ik$ with $i\in P$ and $k\in N$, the neighbors of $i$ must lie in $N$ and the neighbors of $k$ must lie in $P$. Growing the graph starting with some such edge implies that its connected component is bipartite. On the other hand, if there is no such edge then $P$ and $N$ are unions of connected components. Since the graph was assumed connected, it follows that it is bipartite.


TITLE: Proof of joint distribution of the order statistics for n=2
QUESTION [1 upvotes]: Theorem
Let $X_{1},X_{2}$ be i.i.d. and continuous with the common pdf $f_{X}$. Then the vector of order statistics $(X_{(1)},X_{(2)})$ has joint pdf:
\begin{equation}
f(x_{1},x_{2})=2f_{X}(x_{1})f_{X}(x_{2}),
\end{equation}
for $-\infty<x_{1}\leq x_{2}<\infty$.
Proof
Let $x\leq y$ and consider the joint cdf of $(X_{(1)},X_{(2)})$:
\begin{equation}
\begin{split}
F(x,y)&=P(X_{(1)}\leq x,X_{(2)}\leq y)\\
&=P(X_{1}\leq y, X_{2}\leq y)-P(x<X_{1}\leq y,x<X_{2}\leq y).\\
\end{split}
\end{equation}
(...)
What I can't seem to figure out, is how the last equality is derived. Thanks in advance.

REPLY [1 votes]: Note that
$$\{X_1\leqslant y, X_2 \leqslant y\} \\ =\{ X_1 \leqslant X_2,X_1 \leqslant x, X_2 \leqslant y\} \cup \{ X_1 > X_2,X_1 \leqslant y, X_2 \leqslant x \}\cup\{x < X_1 \leqslant y, x < X_2 \leqslant y\},$$
where the three sets on the RHS are disjoint, and
$$\{X_{(1)} \leqslant x, X_{(2)} \leqslant y \} = \{ X_1 \leqslant X_2,X_1 \leqslant x, X_2 \leqslant y\} \cup \{ X_1 > X_2,X_1 \leqslant y, X_2 \leqslant x \}$$
Thus,
$$P(X_1\leqslant y, X_2 \leqslant y) \\ =P( X_1 \leqslant X_2,X_1 \leqslant x, X_2 \leqslant y) + P( X_1 > X_2,X_1 \leqslant y, X_2 \leqslant x ) + P(x < X_1 \leqslant y, x < X_2 \leqslant y)\\ = P(X_{(1)} \leqslant x, X_{(2)} \leqslant y) + P(x < X_1 \leqslant y, x < X_2 \leqslant y) $$


text:
\begin{document}
\sf
\bc
\textbf{\large The gradient of potential vorticity, quaternions\\
and an orthonormal frame for fluid particles}
\par\vspace{7mm}
\textbf{\large J. D. Gibbon and D. D. Holm}
\par\vspace{7mm}
Department of Mathematics, Imperial College London SW7 2AZ, UK\\
{\small email: j.d.gibbon@ic.ac.uk and d.holm@ic.ac.uk}
\ec

\vspace{5mm}

\bc
\textit{Dedicated to Raymond Hide on the occasion of his 80th birthday.}
\ec

\vspace{5mm}

\begin{abstract}
The gradient of potential vorticity (PV) is an important quantity because
of the way PV (denoted as $q$) tends to accumulate locally in the oceans
and atmospheres. Recent analysis by the authors has shown that the vector
quantity $\bdB = \bnabla q\times \bnabla\theta$ for the three-dimensional
incompressible rotating Euler equations evolves according to the same
stretching equation as for $\bom$ the vorticity and $\bB$, the magnetic
field in magnetohydrodynamics (MHD). The $\bdB$-vector therefore acts
like the vorticity $\bom$ in Euler's equations and the $\bB$-field in MHD.
For example, it allows various analogies, such as stretching dynamics,
helicity, superhelicity and cross helicity. In addition, using quaternionic
analysis, the dynamics of the $\bdB$-vector naturally allow the construction
of an orthonormal frame attached to fluid particles\,; this is  designated
as a quaternion frame. The alignment dynamics of this frame are particularly
relevant to the three-axis rotations that particles undergo as they traverse
regions of a flow when the PV gradient $\bnabla q$ is large.
\end{abstract}

\vspace{1cm}

\newpage

\section{\large\textsf{Introduction}}\label{intro}

The ideas in this paper weave together two strands of research on the Euler fluid
equations recently pursued by the authors. Both of these required the use of
Ertel's Theorem \cite{Er42} -- always a favourite topic with Raymond -- and
were discussed at length with him during their development.

The first and latest strand of research concerns the evolution of the gradient
of potential vorticity ($\bnabla q$) and the gradient of potential temperature
($\bnabla\theta$) \cite{GH10}. In the case of the Euler equations, while both
$q$ and $\theta$ are material constants, the evolution of their gradients
involves the strain and rotation rates of the flow. Physically, understanding
the behaviour of $\bnabla q$ in the atmosphere and the oceans is of paramount
importance because potential vorticity tends to accumulate into localised spatial
regions (patches) with sharp edges, where the magnitude $|\bnabla q|$ is much
larger than its average value \cite{HMR85}. In this regard, the divergenceless
vector combination $\bdB = \bnabla q \times\bnabla \theta$ is a natural choice,
because it leads to an evolution equation identical to that for either the
vorticity in the incompressible three-dimensional Euler equations, or the
magnetic field in an electrically conducting fluid. As a consequence, $\bdB$
may undergo the same violent stretching and twisting associated with the
vorticity field in three-dimensional turbulence, or with magnetic field
lines in magnetohydrodynamics (MHD), particularly if $\bdB$ were to align with
an eigenvector of the $3\times3$ strain-rate matrix $S$ associated with the
fluid motion.

The second strand of research involves the use of quaternions in identifying an
ortho-normal frame attached to fluid particles in an Euler flow and whose dynamics
represent the tumbling of the particle as it undergoes three-dimensional rotations
during its flight  \cite{GH06,GHKR}. The vector $\bdB$ turns out to be an ideal
candidate for the construction of this ortho-normal frame. This would be applicable
to the dynamics of particles travelling through regions of the oceans or atmospheres
which have a high values of $|\bnabla q|$.

\subsection{\textsf{\large Potential vorticity gradient for the incompressible Euler}}\label{qintro}

Consider the dimensionless form of the Euler equations for incompressible, stratified
and rotating flow
\bel{compeul1}
\frac{D\bu}{Dt} + 2\bcapom\times\bu + a_{0}\bk\theta = -\bnabla p\,,\qquad\qquad
\frac{D~}{Dt} = \frac{\partial~}{\partial t} + \bu\cdot\bnabla\,,
\ee
in which the potential temperature $\theta(\bx,\,t)$ evolves passively according to
\bel{compeul2}
\frac{D\theta}{Dt} = 0\,.
\ee
In (\ref{compeul1}), the vectors $\bcapom$ and $\bk$ are the rotation rate and vertical
direction, respectively, and the scalar $a_{0}$ is a dimensionless constant. Information
about $\bnabla\theta$ would be needed to discuss how $\theta(\bx,\,t)$ might accumulate
into large local concentrations. This is best studied in the context of potential vorticity
defined by
\bel{qdef}
q := \bom_{rot}\cdot\bnabla\theta
\ee
where $\bom_{rot}$ is defined as $\bom_{rot} := \bom + 2\bcapom$ and $\bom :={\rm curl}\,\bu$
denotes the fluid vorticity. Ertel's theorem says that the material time derivative $D/Dt$
and the vector field $\bom_{rot}\cdot\bnabla$ operating on a scalar function commute with
each other under an Euler flow \cite{Er42}
\bel{q1A}
\frac{Dq}{Dt} =
\left(\frac{D\,\bom_{rot}}{Dt}
- \bom_{rot}\cdot\bnabla\bu \right)\cdot\bnabla \theta
+ \bom_{rot}\cdot\bnabla\left(\frac{D\theta}{Dt}\right)\,.

\ee
When $\bom_{rot}$ obeys the incompressible Euler equations $(\bnabla^{\perp} =
\left(\partial_{y},\,-\partial_{x},\,0\right))$
\bel{q1B}
\frac{D\,\bom_{rot}}{Dt} = \bom_{rot}\cdot\bnabla\bu - a_{0}\bnabla^{\perp}\theta
\ee
then $q$ is a also material constant because
\bel{q1C}
\frac{Dq}{Dt} = 0\,.
\ee

\par\vspace{-30mm}
\bc
\begin{minipage}[htb]{11cm}
\setlength{\unitlength}{.7cm}
\begin{picture}(11,11)(0,0)

\qbezier[400](.5,5)(5,3)(8,3)
\qbezier[400](.5,5)(.75,5.5)(.5,6)
\qbezier[400](.5,6)(2,5.5)(3,5)
\qbezier[400](3,5)(6,5.8)(9,6.5)
\qbezier[400](8,3)(7,3.5)(7,4)
\qbezier[400](7,4)(6.5,4.1)(5.3,4.4)

\qbezier[400](1,3)(3,4)(8,5)
\qbezier[400](1,3)(1,3.5)(.5,4)
\qbezier[400](.5,4)(1,4.2)(1.6,4.5)
\qbezier[200](9,6.5)(8,5.8)(8,5)
\thicklines
\qbezier[400](3.45,3.9)(3.2,4.5)(3,5)
\put(3.75,4.2){\vector(-1,2){.4}}
\put(4.1,4.4){\makebox(0,0)[b]{\scriptsize$\bdB$}}
\put(8.4,3.5){\makebox(0,0)[b]{\scriptsize$\theta = \mbox{const}$}}
\put(-0.2,3.2){\makebox(0,0)[b]{\scriptsize$q = \mbox{const}$}}
\put(5.7,3.4){\makebox(0,0)[b]{\scriptsize$\bnabla\theta\!\nearrow$}}
\put(7,5.1){\makebox(0,0)[b]{\scriptsize$\nwarrow\!\!\bnabla q$}}
\end{picture}
\end{minipage}
\ec
\par\vspace{-15mm}\noindent
{\textbf{Figure 1\,:} \textit{\small The vector $\bdB = \bnabla q\times\bnabla\theta$
is tangent to the curve defined by the intersection of the two surfaces $q = \mbox{const}$
and $\theta = \mbox{const}$ in three dimensions.}\label{fig1}}
\par\vspace{5mm}\noindent
To achieve this, the following divergence-free flux vector is constructed \cite{GH10}
\bel{Bdef}
\bdB = \bnabla Q(q)\times\bnabla \theta \,,
\ee
where $Q(q)$ is any smooth function of $q$. The vector $\bdB$, as in Figure 1, lies along
the intersection of iso-surfaces of $q$ and $\theta$ and could be thought of as pointing
along the tangent to an iso-PV curve on a level set of temperature, or vice versa. Thus,
the PV function $Q(q)$ is the stream function for the flux vector $(\bdB)$ on a level set
of potential temperature $(\theta)$. This observation results in the remarkably simple
evolution equation
\beq{stretch1}
\frac{\partial\bdB}{\partial t} - \mbox{curl}\,(\bu\times\bdB) =
-\bnabla(qQ'\mbox{div}\,\bu)\times\bnabla\theta\,.
\eeq
The cross-product combination is special\,: an appendix in \cite{GH10} contains a proof
of (\ref{stretch1}) performed using differential geometry and also by conventional vector
analysis. It follows from (\ref{stretch1}) that $\bdB$ satisfies the stretching relation
\bel{sreuler}
\frac{D\bdB}{Dt} = \bdB\cdot\bnabla\bu  - \bdB\,\mbox{div}\,\bu - \bnabla(qQ'\mbox{div}\,\bu)\times\bnabla\theta\,,
\ee
whose properties will be discussed in the next section\footnote{\sf In \cite{GH10} the original references
have been discussed (\cite{KurgTat87,KurgPis00,Kurg02}) in which (\ref{stretch1}) had been derived for the
incompressible Euler equations where the right hand side turned out to be zero because $\mbox{div}\,\bu = 0$.
In \cite{GH10} the argument is extended to both the Navier-Stokes and hydrostatic viscous primitive equations.}.

\subsection{\textsf{Stretching, helicity, superhelicity and cross helicity}}\label{super}

In the incompressible case where $\mbox{div}\,\bu = 0$, equation (\ref{sreuler})
simplifies so that the divergenceless vector $\bdB$ satisfies the same stretching
equation as that for the vorticity $\bom$\,; namely
\bel{Bstretch}
\frac{D\bdB}{Dt} = \bdB\cdot\bnabla\bu
\qquad\hbox{with}\qquad
\mbox{div}\, \bdB = 0\,.
\ee
This leads immediately to
\bel{ev1}
\frac{D~}{Dt}|\bdB|^{2} = \bdB\cdot S\bdB = \lambda^{\tiny (S)}|\bdB|^{2}\,,
\ee
where $\lambda^{\tiny(S)}(\bx,\,t)$ is an estimate for an eigenvalue of the rate of strain
matrix $S$ and lies within its spectrum.  Alignment of $\bdB$ with a positive (negative)
eigenvector of $S$ may produce violent growth (decay) thus reproducing the stretching
mechanism that produce the very large vorticity intensities that can develop locally in
the early and intermediate stages of turbulence.

Moffatt's analogy between vorticity and magnetic field \cite{HKM1}, and his detailed
discussion of the topology of magnetic field lines, is based on the concept of helicity
which for us requires the existence of a vector potential $\bA$ defined by
\bel{Adef}
\bA = \shalf\big(Q\bnabla\theta - \theta\bnabla Q\big) + \bnabla\psi\,,
\ee
in terms of the dynamical quantities $Q$, $\theta$ and a potential $\psi$. Then the
helicity $H$, defined by
\bel{heldef}
H = \int_{V}\bA\cdot\bdB\,dV
= \oint_{\partial V} \psi \bdB\cdot\bn\,dS\,,
\ee
measures the number of linkages of the field lines of $\bdB$ with themselves.
The time derivative of helicity under the flow of the Euler equations is given by
\bel{hel-dot}
\frac{dH}{dt}
= \oint_{\partial V} \bigg[-(\bA\cdot\bdB)\, \bu\cdot\bn
+ \Big(\frac{D \psi}{Dt}\Big)\,  \bdB\cdot\bn\bigg]\,dS\,,
\ee

which would vanish for either homogeneous or periodic boundary conditions.
For the Euler equations, $\bu\cdot\bn = 0$ is imposed on a fixed boundary.
However, $H\neq 0$ and $dH/dt \ne 0$ would be possible for topography in
which $\bdB\cdot\bn \ne 0$, so the boundaries play the only role in
allowing linkages in the $\bdB$-field as there is no other source of
helicity.

Hide's intriguing concept of a \textit{super-helicity}, which measures the
linkages of the field lines of $\bdJ:=\mbox{curl}\,\bdB$ with itself,
may be introduced for the $\bdJ$-vector as in for MHD \cite{Hide1989}.
Super-helicity is defined as
\bel{superhel}
\mathcal{S}= \int_{V}\bdB\cdot\bdJ\,dV\,.

\ee
After a short computation, the super-helicity dynamics for the $\bdJ$-vector
comes out to be
\bel{superheldyn}
\frac{d\mathcal{S}}{dt}
= \int_{V} 2\bdB\cdot \mbox{curl}^2(\bu\times\bdB)\,dV
+ \oint_{\partial V} \Big[(\bu\cdot\bdJ)\,  \bdB\cdot\bn
- (\bdJ\cdot\bdB)\, \bu\cdot\bn\Big]\,dS\,,
\ee
which, unlike the helicity $H$, has both volume and surface sources. Likewise the
\textit{cross helicity} for the $\bdB$-vector can be introduced
\bel{crosshel-def}
\mathcal{C} = \int_{V}\bu\cdot\bdB\,dV
\ee
in analogy with the corresponding quantity in MHD.  Another short computation produces
the dynamics of the cross helicity ($\mathbf{R}$ is the vector potential such that
$\mbox{curl}\,\mathbf{R} = 2\bcapom$)
\bel{cross-heldyn}
\frac{d\,\mathcal{C}}{dt}
= -a_{0}\int_{V} ( \theta\,\bdB\cdot\bk) \,dV + \oint_{\partial V}
\Big(-p + \bu\cdot\mathbf{R} + \shalf u^{2}\Big)\,\bdB\cdot\bn \,dS\,,
\ee
which again has both volume and surface sources.

\rem{Finally it may be tempting to extend the parallel between $\bom$ and $\bdB$ further by
defining the sequence of scalar functions $q_{n} = \bdB_{n-1}\cdot\bnabla\theta$ with
$\bdB_{0} = \bom$, and then constructing an infinite set $\bdB_{n} = \bnabla Q_{n}(q_{n})
\times\bnabla\theta$ which all obey the same stretching relation (\ref{sreuler}). However,
this construction, while elegant, contains no new information and is equivalent to computing
the higher terms in a Taylor series expansion of an arbitrary function of $q$ in the initial
conditions.}

\subsection{\textsf{Quaternions and an attached orthonormal frame}}\label{quatintro}

\par\vspace{-6mm}\noindent
\bc
\begin{minipage}[c]{.75\textwidth}
\begin{pspicture}
\psframe(0,0)(5,5)
\thicklines
\qbezier(0,1)(4,2.5)(0,4)
\thinlines
\put(2.01,2.43){\makebox(0,0)[b]{$\bullet$}}
\put(1.4,2.5){\makebox(0,0)[b]{\scriptsize$(\bx_{1},t_{1})$}}
\thinlines
\put(2,2.5){\vector(0,1){1}}
\put(2,3.7){\makebox(0,0)[b]{$\bhB$}}
\put(2,2.5){\vector(-2,-1){1}}
\put(.7,1.8){\makebox(0,0)[b]{$\bhchi_{a}$}}
\put(2,2.5){\vector(1,0){1}}
\put(3.8,2.4){\makebox(0,0)[b]{$\bhB\times\bhchi_{a}$}}

\thicklines
\qbezier(7,1)(6,2.5)(8,4)
\thinlines

\put(6.77,2.45){\makebox(0,0)[b]{$\bullet$}}
\put(6.2,2.5){\makebox(0,0)[b]{\scriptsize$(\bx_{2},t_{2})$}}
\thicklines
\qbezier[50](2,2.5)(3,.5)(6.7,2.5)
\thinlines
\put(6.73,2.5){\vector(1,4){.3}}
\put(7,3.8){\makebox(0,0)[b]{$\bhB$}}
\put(6.7,2.5){\vector(4,-1){1}}
\put(8.4,2.8){\makebox(0,0)[b]{$\bhB\times\bhchi_{a}$}}
\put(6.7,2.5){\vector(4,1){1}}
\put(8.1,2.1){\makebox(0,0)[b]{$\bhchi_{a}$}}
\put(3,1.2){\vector(1,0){.6}}
\put(4,.7){\makebox(0,0)[b]{\scriptsize trajectory of tracer particle}}
\put(4.5,1.4){\vector(4,1){.6}}
\thinlines
\end{pspicture}
\end{minipage}
\ec
\bc
\vspace{-3mm}
\begin{minipage}[r]{\textwidth}
\textbf{Figure 2\,:} \textit{\small  The solid curves represent characteristic curves
$\bhB = d\bx/ds$ ($s$ is arc length) to which $\bhB$ is a unit tangent vector.
The quaternion-frame orientation $(\bhB,\,\bhchi_{a},~\bhB\times\bhchi_{a})$
is shown at the two space-time points $(\bx_{1},t_{1})$ to $(\bx_{2},t_{2})$;
note that this is not the Frenet-frame corresponding to the particle path but
to curves $\bhB = d\bx/ds$. The dotted line represents the tracer particle
$(\bullet)$ path.}
\end{minipage}
\ec
More than one hundred and fifty years ago William Rowan Hamilton invented quaternions
as a means of representing a composition of rotations. For most of this period they
have been under-appreciated, yet they have recently undergone a spectacular renaissance
due to their efficacy in certain applications in avoiding the difficulties incurred
at the north and south poles\footnote{\sf Computations with Euler angles often suffer
from ``gyro-lock'' because of singularities at the poles of the spherical angular
coordinate system where the azimuthal angle is undefined.}  when Euler angles are used
in computing the dynamics of objects undergoing three-axis rotations \cite{Ha06}. In
particular, quaternions now lie at the heart of many modern inertial guidance systems
where tracking the paths and the orientation of aircraft and satellites is of importance
\cite{Ku99}. They are also used in the graphics community to control the orientation of
tumbling objects in computer animations \cite{Ha06}.

A natural question is whether quaternions are useful in tracking the angular velocity and
orientation of Lagrangian particles in fluid dynamics. Experiments in turbulent flows
have now reached the stage where the trajectories of tracer particles can be detected
at high Reynolds numbers \cite{La01,Vo02,RMCB05,Eck06}. Numerical differentiation of
these trajectories gives information about the Lagrangian velocity and acceleration
of the particles and also the curvature of the particle paths \cite{Eck06}.

Conventional practice has been to consider the Frenet-frame of a trajectory. This consists
of a unit tangent vector, a normal and a bi-normal, which are used to represent the pitch,
yaw and roll of the motion. While the Frenet-frame describes the path, it ignores the
rotational dynamics of the particle. To account for this, another ortho-normal
frame associated with the motion of a Lagrangian fluid particle -- designated the
\textit{quaternion-frame} -- has been introduced by the authors \cite{GH06}. This frame
moves with the particles, but its evolution derives from the fluid equations of motion.
In the context of the incompressible Euler equations, the idea of the quaternion frame
depends on the existence of vectors for which there exists a Lagrangian equation of
motion\footnote{\sf The eigenvectors of the rate of strain matrix $S$ are excluded
for this reason.}. The natural triplet is $\{\bom,\,\bhchi,\,\bom\times\bhchi\}$
where $\bchi = \bom\times S\bom$ -- see \cite{GH06,GHKR}. For the incompressible Euler
equations the natural candidate for the $\bdB$-stretching equation (\ref{Bstretch}) is
the triplet
\bel{triplet1}
\{\bhB,\,\bhchi_{a},\,\bhB\times\bhchi_{a}\}\qquad\mbox{where}\qquad
\bchi_{a} = \bhB\times \ba
\ee
where the $\ba$-label has its origins in the definition $\ba := \bdB\cdot\bnabla\bu$.
The Lagrangian
equations of motion for this triplet is derivable through a quaternionic formulation of
the Lagrangian equation of motion for $\bdB$ given in (\ref{Bstretch}).

\section{\large\textsf{Quaternions, rigid body rotations and their properties}}\label{quatprop}

The material in this section provides the reader with a definition of quaternions,
together with a pr\`ecis of their multiplication rules and properties.

The literature on rotations in rigid body mechanics is replete with explicit formulae
relating the Euler angles and what are called the Cayley-Klein parameters of a
rotation \cite{Whitt44}. The complicated inter-relations that are unavoidable when Euler
angle formulae are used can be avoided when quaternions are used \cite{Klein04,Ta1890}\,:
for a more modern context see Holm \cite{Holmgm1,Holmgm2} and Marsden and Ratiu \cite{MaRa1994}.
\par\smallskip
In terms of any scalar $p$ and any 3-vector $\bq$, the 4-vector quaternion $\bfq = [p,\,\bq]$
is defined as (Gothic fonts denote quaternions)
\bel{quatdef1}
\bfq = [p,\,\bq] = p{\sf I} - \sum _{i=1}^{3}q_{i}\sigma_{i}\,,
\ee
where $\{\sigma_{1},\,\sigma_{2},\,\sigma_{3}\}$ are the three Pauli spin-matrices defined by
\bel{psm1}
\sigma_{1} = \left(\begin{array}{rr}
0 & i\\
i & 0
\end{array}\right)\,,
\hspace{1cm}
\sigma_{2} = \left(\begin{array}{rr}
0 & 1\\
-1 & 0
\end{array}\right)\,,
\hspace{1cm}
\sigma_{3} = \left(\begin{array}{rr}
i & 0\\
0 & -i
\end{array}\right)\,,
\ee
and ${\sf I}$ is the $2\times 2$ unit matrix. A rule (denoted as $\cast$)
for multiplication of quaternions can be derived from the multiplication
rule for the Pauli matrices $\sigma_{i}\sigma_{j} =  -
\delta_{ij}{\sf I}-\epsilon_{ijk}\sigma_{k}$, as
\bel{quatdef2}
\bfq_{1}\cast\bfq_{2} = [p_{1}p_{2} - \bq_{1}\cdot\bq_{2},\,p_{1}\bq_{2} + p_{2}\bq_{1} +
\bq_{1}\times\bq_{2}]\,.
\ee
Thus, the multiplication of quaternions is associative, but it is not commutative.

Let $\bhfp = [p,\,\bq]$ be a unit quaternion, satisfying $p^2 + q^2 = 1$. Its
inverse $\bhfp^{*} = [p,\,-\bq]$ satisfies $\bhfp \cast\bhfp^{*} = [p^2 + q^2,\,0]
= [1,0]$. A pure quaternion has a zero scalar entry, such as $\bfr = [0,\,\br]$.
Hamilton called his pure quaternions \emph{vectors}. Pure quaternions transform
among themselves as $\bfr = [0,\,\br] \to \bfR = [0,\,\bR]$ under
\bel{r1}
\bfR = \bhfp\cast\bfr\cast\bhfp^{*}\,.
\ee
This associative product can be written as
\bel{r2}
\bfR = \bhfp\cast\bfr\cast\bhfp^{*}
= [0,\, (p^{2}-q^{2})\br + 2p(\bq\times\br)+ 2\bq(\br\cdot\bq)]\,.
\ee
Choosing $p=\pm\cos\shalf\theta$ and $\bq = \pm\,\bhn\sin\shalf\theta$, where $\bhn$
is the unit normal to $\br$, we find that
\bel{r3}
\bfR = \bhfp\cast\bfr\cast\bhfp^{*} = [0,\, \br\cos\theta + (\bhn\times\br) \sin\theta]
\equiv O(\theta\,,\bhn)\br\,,
\ee
where
\bel{r4}
\bhfp = \pm [\cos\shalf\theta,\,\bhn\sin\shalf\theta]\,.
\ee
Equation (\ref{r3}) is the Euler-Rodrigues formula for the rotation $O(\theta\,,\bhn)$
by an angle $\theta$ of the 3-vector $\br$ about its normal $\bhn$ and the quantities
$\theta\,,\bhn$ are the Euler parameters. The elements of the unit quaternion $\bhfp$
are the Cayley-Klein parameters which are related to the Euler angles \cite{Whitt44}.
When $\bhfp$ is time-dependent, the Euler-Rodrigues formula in (\ref{r3}) is
\bel{rot2}
\bfR(t) = \bhfp\cast\bfr\cast\bhfp^{*}
\hspace{1cm}\Rightarrow\hspace{1cm}
\bfr = \bhfp^{*}\cast\bfR(t)\cast\bhfp\,.
\ee
It is necessary to use the property of the pure quaternion $\bfR^{*} = -\bfR$
to obtain the time derivative of $\dot{\bfR}$
\beq{rot3}
\dot{\bfR}(t)

&=& (\dot{\bhfp}\cast\bhfp^{*})\cast\bfR - ((\dot{\bhfp}\cast\bhfp^{*})\cast\bfR)^{*}\,.
\eeq
The quaternion $\bhfp = [p,\,\bq]$ is of unit length and so $p\dot{p} + q\dot{q} = 0$, which
means that $\dot{\bhfp}\cast\bhfp^{*}$ is also a pure quaternion
\bel{rot4}
\dot{\bhfp}\cast\bhfp^{*} = [0,\,\shalf\bcapom_{0}(t)]\,.
\ee
The 3-vector entry in (\ref{rot4}) defines the angular frequency $\bcapom_{0}(t)$ as
$\bcapom_{0} = 2(-\dot{p}\bq +\dot{\bq}p - \dot{\bq}\times\bq)$ thereby giving the
well-known formula for the rotation of a rigid body
\bel{rot5}
\dot{\bR} = \bcapom_{0}\times\bR\,.
\ee
For a Lagrangian particle, the equivalent of $\bcapom_{0}$ is the Darboux vector $\bD_{a}$
in Theorem \ref{abthm} in the next section.

\section{\large\textsf{An ortho-normal frame and particle trajectories}}\label{ortho-normal}

Having set the scene in \S\ref{quatprop} by describing some of the essential properties
of quaternions, it is now time to apply them to the Lagrangian relation (\ref{Bstretch})
between the two vectors $\bdB$ and $\ba$
\bel{Bequn}
\frac{D\bdB}{Dt} = \ba : = \bdB\cdot\bnabla\bu\,.
\ee
It will turn out below that a knowledge of $D\ba/Dt$ is needed. Ertel's Theorem is applicable
and the result becomes a version of Ohkitani's relation \cite{Ohk93}
\beq{ertelex1}
\frac{D~}{Dt}\big(\bdB\cdot\bnabla\bu\big) &=& \bdB\cdot\bnabla
\left(\frac{D\bu}{Dt}\right)\\
&=& - P\bdB - \bdB\cdot\bnabla\left(2\bcapom\times\bu + a_{0}\bk\theta\right)
\eeq
where the Hessian matrix of the pressure $p$ is defined as
\bel{Hessdef}
P = \frac{\partial^{2}p}{\partial x_{i}\partial x_{j}}\,.
\ee
Thus we can define a vector $\bdb$ such that
\bel{ertelex2}
\frac{D\ba}{Dt} = \bdb := - P\bdB -
\bdB\cdot\bnabla\left(2\bcapom\times\bu + a_{0}\bk\theta\right)\,.
\ee
Through the  multiplication rule in (\ref{quatdef2}) quaternions appear in the
decomposition of the 3-vector $\ba$ into parts parallel and perpendicular to another
vector, which we choose to be $\bdB$. This decomposition is expressed as
\bel{decom1}
\ba = \alpha_{a}\bdB + \bchi_{a}\times\bdB = [\alpha_{a},\,\bchi_{a}]\cast[0,\,\bdB]\,,
\ee
where the scalar $\alpha_{a}$ and 3-vector $\bchi_{a}$ are defined as
\bel{la2a}
\alpha_{a} = \mathcal{B}^{-1}(\bhB\cdot\ba)\,,\hspace{2cm}
\bchi_{a} = \mathcal{B}^{-1}(\bhB\times\ba)\,.
\ee
Equation (\ref{decom1}) thus shows that the quaternionic product is summoned in
naturally.\footnote{\sf With reference to \S\ref{quatprop}, the Cayley-Klein parameters
of the quaternion $\bfq = [\alpha,\,\bchi]$ are
$$
\bhfq = \left[\frac{\alpha}{\alpha^2 + \chi^2},\, \frac{\bchi}{\alpha^2 + \chi^2}\right]\,.
$$}
It is now easily seen that $\alpha_{a}$ is the \emph{growth rate} of the scalar magnitude
($\mathcal{B} =|\bdB|$) which obeys
\bel{la3}
\frac{D\mathcal{B}}{Dt} = \alpha_{a}\mathcal{B}\,,
\ee
while $\bchi_{a}$, the \emph{swing rate} of the unit tangent vector $\bhB = \bdB \mathcal{B}^{-1}$,
satisfies

\bel{la5}
\frac{D\bhB}{Dt} = \bchi_{a}\times \bhB\,.
\ee
Now define the two quaternions
\bel{ls6}
\bfq_{a}
= [\alpha_{a},\,\bchi_{a}]\,,
\hspace{2cm}
\bfB = [0,\,\bdB]\,,

\ee
so (\ref{Bequn}) can automatically be re-written in the quaternion form
\bel{lem1}
\frac{D\bfB}{Dt}

= \bfq_{a}\cast\bfB\,.
\ee
Moreover, because of (\ref{ertelex2}), exactly as for $\bfq_{a}$, a quaternion $\bfq_{b}$
can be defined which is based on the variables
\bel{alphachibdef}
\alpha_{b} = \mathcal{B}^{-1}(\bhB\cdot\bdb)\,,
\hspace{3cm}
\bchi_{b} = \mathcal{B}^{-1}(\bhB\times\bdb)\,,
\ee
where
\bel{qbdef}
\bfq_{b} = [\alpha_{b},\,\bchi_{b}]\,.
\ee
The 3-vector $\bdb = D\ba/Dt$ admits a decomposition similar to that for $\ba$ as in (\ref{decom1})
\bel{la9}
\frac{D^{2}\bfB}{Dt^{2}}
= [0,\,\bdb]
= [0,\, \alpha_{b}\bdB + \bchi_{b}\times\bdB ]
= \bfq_{b}\cast\bfB \,.
\ee
Using the associativity property, compatibility of (\ref{la9})
and (\ref{lem1}) implies that ($\mathcal{B} = |\bdB| \neq 0$)
\bel{la10}
\left(\frac{D\bfq_{a}}{Dt} + \bfq_{a}\cast\bfq_{a}
-\bfq_{b}\right)\cast\bfB = 0\,.
\ee
This establishes a Riccati relation between $\bfq_{a}$ and $\bfq_{b}$
\bel{Ric1}
\frac{D\bfq_{a}}{Dt} + \bfq_{a}\cast\bfq_{a} = \bfq_{b}
\,,
\ee
with components
\bel{Ric1-comp}
\frac{D}{Dt} [\alpha_{a},\,\bchi_{a}]
+ [\alpha_{a}^2-\chi_{a}^2,\,2\alpha_{a}\bchi_{a}]
= [\alpha_{b}\,,\chi_{b}]
\,,
\ee
where $\chi_{a} = |\bchi_{a}|$. There follows a Theorem on a Lagrangian particle
undergoing fluid motion that is equivalent to the well-known formula (\ref{rot5})
for a rigid body undergoing rotation about its center of mass\,:

\begin{theorem}[Alignment dynamics]
\label{abthm}
The ortho-normal quaternion-frame $\big(\bhB,\,\bhchi_{a},\,\bhB\times\bhchi_{a}\big)$
has Lagrangian time derivatives
\beq{abframe3}
\frac{D\bhB}{Dt}&=& \bD_{a}\times\bhB\,,\\
\frac{D(\bhB\times\bhchi_{a})}{Dt} &=& \bD_{a}\times(\bhB\times\bhchi_{a})\,,\label{abframe4}
\\
\frac{D\bhchi_{a}}{Dt} &=& \bD_{a}\times\bhchi_{a}\,,\label{abframe5}
\eeq
where the Darboux angular velocity vector $\bD_{a}$ is defined as
\bel{abframe6}
\bD_{a} = \bchi_{a} + \frac{c_{b}}{\chi_{a}}\bhB\,,\hspace{2cm}
c_{b} = \bhB\cdot(\bhchi_{a}\times\bchi_{b})\,,
\ee
and the quantities $[\alpha_{a},\bchi_{a}]$ and $[\alpha_{b},\bchi_{b}]$ are defined in
(\ref{la2a}) and (\ref{alphachibdef}).
\end{theorem}

\par\smallskip\noindent
\textbf{Remark\,:} The Darboux vector $\bD_{a}$ is driven by the 3-vector $\bdb=D\ba/Dt$ which sits
in $c_{b}$ in (\ref{abframe6}). The analogy with rigid body rotation expressed in (\ref{rot5})
is clear.
\par\smallskip\noindent
\textbf{Proof\,:} To find an expression for the Lagrangian time derivatives of the components of
the frame  $(\bhB,\,\bhchi_{a},\,\bhB\times\bhchi_{a})$ requires the derivative of $\bhchi_{a}$.
To find this, it is necessary to use the fact that the 3-vector $\bdb$ can be expressed in this
ortho-normal frame as the linear combination
\bel{b1}
\mathcal{B}^{-1}\bdb = \alpha_{b}\,\bhB + c_{\,b}\bhchi_{a} + d_{\,b}(\bhB\times\bhchi_{a})\,.
\ee
where $c_{\,b}$ is defined in (\ref{abframe6}) and $d_{\,b} = -\,(\bhchi_{a}\cdot\bchi_{b})$.
The 3-vector product $\bchi_{b} = \mathcal{B}^{-1}(\bhB\times\bdb)$ yields
\bel{bchibdef}
\bchi_{b} = c_{\,b}\,(\bhB\times\bhchi_{a}) - d_{\,b}\bhchi_{a}\,.
\ee
When split into components, equation (\ref{Ric1-comp}) becomes
\bel{abframe0}
\frac{D\alpha_{a}}{Dt} = \chi_{a}^{2} - \alpha_{a}^{2} + \alpha_{b}\,
\ee
and
\bel{abframe1}
\frac{D\bchi_{a}}{Dt} = - 2\alpha_{a}\bchi_{a} + \bchi_{b}\,.
\ee
A little more working gives the \emph{alignment dynamics} in
equations (\ref{abframe3})-(\ref{abframe6}).
\hspace{10mm}$\blacksquare$

\section{\large\textsf{Conclusions}}\label{concl}

When the quaternion approach to rotations outlined in \S\ref{quatprop} is
applied to the Euler equations it demonstrates that quaternions are a
natural way of calculating the orientation of Lagrangian particles in
motion through the concept of ortho-normal quaternion-frames attached
to each particle. In this particular context, where the $\bdB$-field
is the vector that helps us understand the evolution of $\bnabla q$ and
$\bnabla\theta$, knowledge of the quartet of 3-vectors $(\bu,\,\bdB,\,\ba,\,\bdb)$
is sufficient for the application of Theorem \ref{abthm}. The complexity
of the $3D$ Euler equations comes through the ortho-normal dynamics via
the pressure field. In the present state of knowledge, the projection
$P\bdB$ that is part of $\bdb$ would have to be found by computational means.

A natural question is whether these ideas can be applied to the Navier-Stokes
equations? It turns out that the evolution equation for $\bdB$ in the
incompressible case is the same as that in (\ref{sreuler}) with $\bu$ replaced
by $\bU$ \cite{GH10}.  $\bU$ is a new transport velocity calculated using the
method of Haynes and McIntyre \cite{HMc90}
\bel{HM1}
q(\bU - \bu) = - \left\{\big[ Re^{-1}\Delta\bu + 2\bcapom\times\bu +
a_{0}\bk\theta\big]\times\bnabla\theta +
(\sigma Re)^{-1}\bom\Delta\theta\right\}
\ee
The quaternion procedure can only be pursued to a certain point for the Navier-Stokes
equations, after which  a difficulty  appears when $D\bU/Dt$ is needed, and we have
no proper knowledge of this. This is consistent with the objections that $\bU$ is not
a genuine \textit{physical} velocity but merely a mathematical construction
\cite{Dan90,Viudez99}. For the Euler equations we know that this is the point
where the Hessian matrix of the pressure is introduced in equation (\ref{ertelex1}).
Moreover, the stretching relation (\ref{sreuler}) in this case becomes
\bel{sreuler-Bee}
\frac{D\bdB}{Dt} = \bdB\cdot\bnabla\bU  - \bdB\,\mbox{div}\,\bU
- \bnabla(qQ'\mbox{div}\,\bU)\times\bnabla\theta\,,
\ee
and $\mbox{div}\,\bU\ne0$, which allows richer and potentially more singular alignment
dynamics than those for the incompressible case discussed here in equations
(\ref{abframe3})-(\ref{abframe6}). As well as the Navier-Stokes alluded to above, the
case of the hydrostatic primitive equations have been discussed in this context in \cite{GH10}.

\rem{
\appendix
\section{\large\textsf{Proof of (\ref{stretch1})}}

\par\noindent
In ??? a proof based on differential forms is based on the advective
transport equations for $\theta$
\bel{a1a}
\frac{D\theta}{Dt} = 0\,,
\ee
and for potential vorticity,
\bel{a1b}
\frac{D}{Dt}\big(q\,d^{3}x\big) = \big(\partial q + \bU\cdot\bnabla q +
q\,\mbox{div}\,\bU\big)\,d^{3}x = 0\,.
\ee
Then the evolution equation (\ref{stretch1}) for $\bdB = \bnabla Q(q)
\times\bnabla\theta$ may be derived by using the notation of the exterior
derivative $(\,d\,)$ and the wedge product $(\,\wedge\,)$
\bel{a4}
\bdB\cdot d\mathbf{S} = \big(\bnabla Q(q) \times\bnabla\theta\big)\cdot d\mathbf{S}
 = dQ(q)\wedge d\theta\,.
\ee
The advective time derivative of the leftmost term in this relation yields
\bel{a5}
\frac{D~}{Dt}\big(\bdB\cdot d\mathbf{S}\big)
= \left[\partial\bdB - \mbox{curl}\,(\bU\times\bdB) \right]\cdot d\mathbf{S}
\qquad\hbox{along}\qquad \frac{D\bx}{Dt} = \bu
\ee

Using equations (\ref{a1a}) and (\ref{a1b}), the advective time derivative of the
rightmost term in (\ref{a4}) yields
\begin{eqnarray}\label{comp-a}
\frac{D~}{Dt}\Big(dQ(q)\wedge d\theta \Big) &=& d\left(\frac{DQ(q)}{Dt}\wedge d\theta\right)
+ dQ(q)\wedge d\left(\frac{D \theta}{Dt} \right)\non\\
&=& -\,d\left(qQ'\,\mbox{div}\,\bu\right)\wedge d\theta

= \bdD\cdot d\mathbf{S}\,,
\end{eqnarray}
also along ${D\bx}/{Dt} = \bu$. Equation (\ref{stretch1}) arises by equating the rightmost
terms in (\ref{a5}) and (\ref{comp-a}). When $\bu$ is replaced by $\bU$ for the Navier-Stokes
equations, this yields a non-zero right hand side because in this case $\mbox{div}\,\bU \neq 0$.
\par\smallskip\noindent
An alternative version of the proof, using the notation $\bom_{U}= \mbox{curl}\,\bU$, is performed
by conventional vector identities\,:
\beq{stretch3}
\bdB_{t} &=& (\nabla Q)_{t}\times(\nabla\theta) + (\nabla Q)\times(\nabla\theta)_{t}\non\\
&=& -\nabla\big[(qQ'\,\mbox{div}\,\bU) + \bU\cdot\nabla Q)\big]\times(\nabla\theta)
- (\nabla Q)\times\left[\nabla(\bU\cdot\nabla\theta)\right]\non\\
&=& -\left\{\nabla(qQ'\,\mbox{div}\,\bU) + \bU\cdot\nabla(\nabla Q) + (\nabla Q)\cdot\nabla\bU
+ (\nabla Q)\times\bom_{U}\right\}\times(\nabla\theta)\non\\
&\quad& -\,(\nabla Q)\times\left\{\bU\cdot\nabla(\nabla\theta) +
(\nabla\theta)\cdot\nabla\bU + (\nabla\theta)\times\bom_{U}\right\}\non\\
&=& -\nabla(qQ'\,\mbox{div}\,\bU)\times\nabla\theta - \bU\cdot\nabla\bdB
+ (\nabla Q)(\bom_{U}\cdot\nabla\theta) - (\nabla\theta)(\bom_{U}\cdot\nabla Q)\non\\
&\qquad&  +\,(\nabla\theta)\times(\nabla Q\cdot\nabla\bU) - (\nabla Q)\times(\nabla\theta\cdot\nabla\bU)\non\\
&=& \mbox{curl}\,(\bU\times\bdB) -\nabla(qQ'\,\mbox{div}\,\bU)\times\nabla\theta\,.
\eeq

}

\par\vspace{-2mm}

\bibliographystyle{unsrt}

config: arxiv


TITLE: How does the moon reflect light?
QUESTION [1 upvotes]: We can see the moon in the night because it reflects sunlight. But the light is incident on the opposite side of moon with respect to the observer in the night.
In this case, how does the moon reflect light?
I am not sure if I have articulated my question correctly.

REPLY [1 votes]: Although this question is answered, I will say this: the orbital plane of moon is not in the same orbital plane of earth, there is about 5 degrees of inclination.
Example, if earth and sun was in the floor of a room, then the moon is on the opposite of sun, but raised from the floor . Now since sun is larger than the earth, the umbra of the shadow cone is diminishing behing the earth, see here: https://celestrak.org/columns/v03n01/fig-1.gif
Usually the moon is also outside the Pen-Umbra zone. That is why the mon gets illuminated, even being at the opposite side of sun (looking from top). Sometimes however, the moon, while in the opposite side of sun wrt earth, stays in the ascending or descending node, that is called a lunar eclipse.


TITLE: Weighted Averages from the world of GMAT..
QUESTION [1 upvotes]: I am really confused about how this question has been solved. I've given the question, author's approach and then mine. Can you help me figure out why/ why not I have gone wrong?
Thanks in advance,
Suraj
Question
A milkman has 20 L of milk. He mixes 5 L of water in 20 L of pure milk. If the cost of pure milk is $18/L, then the profit of the milkman, when he sells the diluted milk at cost price is:
A) 20%   B) 25%   C) 33.33%   D) 18%
The Author's approach
% Profit= 5/20=0.25=25%
My approach
The weighted percentage of milk in the new mixture=
{(1*20)+ (0*5)}/25
=0.8
=80% milk
% profit=
{(25)- (25*0.80)}/(25*0.80) where (25*0.80)= Actual price of diluted milk
=25%

REPLY [1 votes]: You made an arithmetic error when you calculated $\frac{1\cdot 20+0\cdot 5}{25}=\frac{20+0}{25}=\frac{4}{5}=0.8\neq 0.83$
Indeed, making the correction, you should arrive at:
$\frac{25-25\cdot 0.8}{25\cdot 0.8} = \frac{25-20}{20}=\frac{5}{20}=0.25$ as expected.