text:
Home News and Events Year 2010 Long-term Contraceptives Preferred in State: Experts
Long-term Contraceptives Preferred in State: Experts
The slump in the usage of condoms in Maharashtra as a preferred choice of contraception, feel experts, may be a partial depiction of the situation. While some say the simultaneous dip in the abortion figures may indicate increasing use of the morning–after pill, experts feel it may be attributed more to the growing use of sophisticated contraceptives. Also, doctors and health experts feel that there is still a huge divide in contraception preferences between urban and rural areas.
"While people are showing an increased preference for intra–uterine contraceptive devices (IUCDs) in the longterm, emergency contraceptives do have their market, but have not replaced the normal contraceptives," said an expert.
And statistics can throw results which may not be entirely representative of the exact situation and are often only indicative, say some doctors. Secretary general of the Family Planning Association of India Vishwanath Koliwad said the government should also take into consideration the sales of condoms through commercial outlets. Not everyone may buy them from a government hospital or centre, he said. But, differences between methods of contraception in rural areas and cities can make it difficult to generalize preferences.
For instance, in Mumbai, demand for IUCDs, which provide protection for 10 years, has shot up mostly among the educated. But it is not the case in rural areas, said Koliwad, who are all for vasectomy or tubectomy. Gynaecologist Dr C Purandare said the acceptance of IUCDs is little more than 5% in India as opposed to 60% in countries like China. It is however, gradually picking up, he said.
In Mumbai, the preference for oral pills has been going down. Head of gynaecology department of the JJ Hospital Dr Rekha Davar could not agree more. Oral pills have side–effects and the idea of a healthy person taking a pill every day does not go down with many, she said. For couples who have already raised a family both IUCD and sterilisation can be excellent methods, she said.
Davar added that contraceptive usage trends can vary from time to time. Condoms, however, have their dual advantage of preventing pregnancy and protecting against sexually transmitted diseases, she said. "In our state couple protection rate is way above 60% and that is a fairly good achievement," she said.
About the city, authorities from the Mumbai District AIDS Control Society (MDACS) say increased Condom usage has indeed controlled HIV/AIDS to a great extent. Condom usage figures have been about 300–400% more than desired and that way the city is doing fine, said Dr S S Kudalkar, project director, MDACS. Koliwad however, feels to ensure better fertility, oral pills or IUCDs should be promoted.
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Sutton Liberal Democrats can now dispel recent scaremongering by local Conservatives about the proposed site for a new secondary school that is needed to accommodate the borough's growing pupil numbers by 2019/20.
Following inaccurate claims about a new school on a privately owned site in Mollison Drive in Wallington, we can confirm that a detailed analysis of all possible sites across the borough has shown that the only suitable location is on the All Weather Pitch and hopefully part of the Sports Village in Rosehill Park.
The evidence has been circulated to members of the Local Plan Task & Finish Group, which is expected to recommend to the Housing, Economy & Business Committee in December that the Rosehill site is allocated for a second secondary school in the draft Local Plan.
Projections of the numbers of additional school places required identify a need for two new secondary schools in the borough.
The ex- Sutton Hospital site in Belmont has already been agreed for the first secondary school. National Planning Policy requires the use of brownfield land in preference to Metropolitan Open Land (MOL), so that had to be the first choice to achieve the required planning permission. Rosehill Park is MOL, which is why officers need robust evidence to show that it is the only remaining deliverable site with the space available for a new school.
Cllr Jayne McCoy commented on the news: " Local Conservatives seem to think you can play a game of "pin the tail on the donkey" when it comes to deciding where a new school should go - residents expect more from their Council, and also greatly value their green spaces. That is why we have taken the time to look at a range of sites across the borough to ensure that we really have no option but to locate the new school in a park."
Cllr Wendy Mathys added "The benefit of reviewing all the available sites is that we have found an option for a second secondary school that is much better than when we first looked at the site . With the change of ownership of the Sports Village we hope to be able to minimise the impact on local green space, whilst ensuring we can provide the places for Sutton children in the borough who will be needing them after 2019."


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Hockey Junior World Cup: India colts over the moon match reviews
December 19, 2016 neoprimesport hockey 0
Lucknow: Indian Junior hockey players celebrate after their victory over Belgium in the final Junior World Cup Hockey 2016 in Lucknow on Sunday.
With third entry into the Junior Hockey World Cup final, India would look to repeat the gold medal winning heroics of 2001.
A win away from rewriting history after a long gap of 15 years, Indian colts would look to showcase their best and leave nothing to chance when they take on Belgium in the summit showdown of the men's Junior Hockey World Cup on Sunday.
The wintry Sunday could turn out to be the biggest day for these Under-21 Indian hockey players as they chase a dream of winning the mega event for the second time. India had earlier won the trophy way back in 2001 in Hobart, Australia. This is India's third entry into the final of the Junior World Cup after having finished runners-up in 1997.
Gurjant Singh would have lived the scenario several times in his dreams: a World Cup final, the crowd cheering him, and he beating the goalkeeper with a fierce hit. But even in his dreams, the 21-year-old wouldn't have executed it with such precision. Varun Kumar, the scorer of India's first goal in the junior World Cup, spotted an idle Gurjant near the Belgian 'D'. You could have forgiven the European side for assuming it was harmless to leave Gurjant unmarked.
It isn't India's style, after all, to play long, aerial balls. But this Indian style has ditched several old ways. And the Belgians would realize that the hard way. That Varun dared to play the lobbed ball – that travelled half the length of the field from right to left – was a surprise in itself. Belgian defenders were caught off guard and they failed to control it. Gurjant was the first to reach. He controlled the wobbling ball with the two deft touches and took it away from the defenders.
To close Gurjant's angle, Belgium goalkeeper Loic van Doren charged towards him. Gurjant looked up, saw the tournament's best goalkeeper running towards him, then spotted faintest of gaps between him at the post, and from an acute angle, unleashed a reverse hit that flew past Van Doren.
It was one of the finest goals of the tournament, if not the finest. And it couldn't have come at a more crucial moment. That goal, in the 8th minute, changed how the final would be played. Belgium are a side who like to play patient, structured hockey. They prefer scoring first and generally rely on their technically strong defenders to see out the rest of the game. That's how they defeated India in the Olympics quarterfinals and their juniors followed the template in the World Cup.
But India coach Harendra Singh had a plan, which his team executed to perfection. On the eve of the final, Harendra said the most effective way to unlock the Belgian defence was to play diagonal passes into their 'D'. Most of India's attacks came from the flanks, and combined with the raw pace of the midfielders and forwards, Belgium looked visibly uncomfortable.
Mandeep Singh has emerged as the difference maker in India's team during the Junior World Cup.
The strategy also yielded India's both goals. While the first was scored from a diagonal lobbed pass from right to the left, the second came in the 22nd minute from India's movement from left to right. Nilakanta Sharma picked up a loose ball on the right, near the 25-yard line, and played it towards Simranjeet Singh on top of the 'D'. Simranjeet took one touch to step away from the defender, and then beat Van Doren with a powerful shot. Two reverse flicks, two goals.
If the crowd hadn't already deflated Belgium's confidence, the second goal surely did. Chaos is how Belgium coach Jeroen Baart chose described the opening half hour. "The players couldn't handle the chaos of the Indian team, and the crowd," he confessed.
Tempo-setting goals
Inside 30 minutes, India were up by two goals. Such was the confidence among the Indian players that the ball didn't touch a Belgian stick even once for almost two minutes towards the end of the first half. Roelant Oltmans, here as the manager of the junior team, had demanded a perfect performance. The players had delivered just that. But their job was only half done.
Harendra knows the consequence of being over-ambitious too well. In the semifinal of the 2005 junior World Cup, India – again coached by Harendra – were leading Australia 2-0 in the semifinals. A second consecutive final was in the sight but the coach and his players got greedy, trying to go for a higher margin. In doing so, they not just conceded their two-goal advantage but went on to lose 3-2.
A repeat of 2005 would have been disastrous, especially with the crowd already celebrating as if India had won the title. But the Indian players did not lose their focus. Dipsan Tirkey, the youngest member of the Indian team, has been rock-solid as a deep defender and he only added to his reputation by sniffing out any danger posed by the Belgian attackers.
Blunt Belgians
Chances for Belgium were few, and they weren't really threatening – largely because India dominated the midfield play. Every time India lost possession, they were quick to regain it by putting pressure on Belgians in their own half.
India lost concentration only in the final minute, when they conceded a penalty corner. By the time Belgium could take it, the hooter had sounded. India were champions but as the rules prescribe, the corner had to be taken.
When Fabrice van Bockrijck drag-flicked it, the Indian players did not even bother to rush to stop it. They ran, but towards their dugout to celebrate. The ball flew past goalkeeper Vikas Dahiya. But no one cared. India became the first host country to win a junior World Cup.
In a city that has produced a golden generation of players in 1970s and 80s, another golden generation of players has emerged.
India would look to repeat the gold medal winning heroics of 2001.
With third entry into the Junior Hockey World Cup final
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\section{Introduction}

The problem of inelastic dissipation of a freely rotating body emerges
naturally in the context of interstellar grain alignment.

It is well known that the polarization of starlight is caused by
the alignment
of interstellar dust grains. This was discovered by Hiltner and Hall
(see Hall 1949  and references therein) who showed that the polarization
was entailed by the linear dichroism,
i.e., by the differential extinction, by clouds of aligned grains, of
electromagnetic waves of different polarizations.

Various mechanisms of grain alignment (see table~1 in Lazarian, Goodman
\& Myers 1997) depend on the internal dissipation of energy (Purcell 1979).
This dissipation aligned the axis of the maximal moment of inertia
and the angular momentum. This changes grain dynamics and should be
accounted for in the theory of grain alignment. Therefore the
grain alignment depends on the internal relaxation.
In some cases, e.g. the alignment of thermally
rotating grains, the alignment is not sensitive to the exact rate of
internal relaxation. However for grains rotating suprathermally,
more detailed knowledge of the internal relaxation is usually
required (see more
in the discussion section of this paper).

It is normally believed that the Barnett dissipation is the dominant
process of internal relaxation within interstellar grains (Purcell 1979,
Spitzer \& McGlynn 1979, Lazarian 1994, Draine 1996).
It happens due to the time-dependent (due to the
precession of angular velocity in grain body axes) remagnetisation of
the sample by means of the Barnett effect.
Indeed, the Barnett effect develops magnetization parallel to the angular
velocity of a rotating body because a share of the angular momentum gets
transferred to the spins (see Landau and Lifshitz 1984). In the cause of the
angular-velocity precession about the angular momentum (in accordance with
the Euler equations), the
magnetization direction is also precessing in the body-axes frame, and this
entails dissipation. A recent study of paramagnetic alignment of suprathermally
rotating grains (or ``Purcell alignment'') has shown that grains can
be nearly perfectly aligned by the interstellar magnetic field if the
internal dissipation is controlled by the Barnett effect (Lazarian
\& Draine 1997).

Another process responsible for dissipation, known as inelastic
relaxation happens as time-dependent stresses emerge in a body when
it rotates about an arbitrary axis. In this paper we subject this process
to scrutiny, as the existing treatment (Purcell 1979) has a number of
difficiencies that we discuss in the paper.

A rigorous treatment of internal relaxation is important: if the inelastic
relaxation is more efficient than the Barnett relaxation then it is the
former, not the latter, that determines the dynamics of crossovers
and, thereby, the degree of alignment achievable by suprathermally
rotating grains (see Lazarian \& Draine 1997).

In what follows we briefly discuss the rotation of an oblate
grain (Section~II); calculate the acceleration experienced by a point inside
the grain (Section~III). Then we compute the stresses caused by grain
precession (Section~IV), the rate of energy dissipation (Section~V) and the
rate of internal-dissipation-caused alignment of the major-inertia axis of
the grain to its angular momentum. We also discuss whether the obtained
formulae could be easily altered in the case of prolate grains (Section~VI).
Our results are discussed in Section~VII.

\section{Notations and Assumptions}

We consider a freely rotating cosmic-dust grain,
employing two Cartesian coordinate systems, each with its origin at the
center of mass of the body. The inertial frame ($X$, $Y$, $Z$), with
unit vectors $\mathbf{e}$$_{X}$, $\mathbf{e}$$_{Y}$, $\mathbf{e}$$_{Z}$,
is chosen so that its $Z$ axis is parallel to the (conserved) angular
momentum $\mathbf{J}$ (see Fig.~1). Coordinates with respect to this frame are
denoted by the same capital letters: $X$, $Y$, and $Z$.
We also use the system which we call body frame
and associate with the three principal axes of inertia: $1$, $2$, and
$3$, with coordinates $x$, $y$, and $z$ along these. The appropriate unit
vectors are $\mathbf{e}$$_{1}$, $\mathbf{e}$$_{2}$,
$\mathbf{e}$$_{3}$.

Without loss of generality, one may take $I_3 \geq  I_1$ and $I_3 \geq I_2$,
where $I_i$ are the principal moments of inertia.

Following Purcell (1979),
we denote the angular velocity by $\bf{\Omega}$ and reserve
$\bf{\omega}$ for the rate of precession.
It is a well-known fact that, as a perfectly elastic body rotates, the
extremity of the angular velocity vector $\bf{\Omega}$ describes a curve that
is the intersection of two ellipsoids. One of those comes from the
angular-momentum conservation:
\begin{equation}
I_1^2 \; \Omega_1^2 \; + \; I_2^2 \; \Omega_2^2 \; +
\; I_3^2 \; \Omega_3^2 \; = \; {\bf{J}}^2 \; = \; const~~~,
\label{2.4}
\end{equation}
where ${\Omega}_{1,2,3}$ are the body-frame-related components of
$\bf{\Omega}$). Another one, known as the Poinsot ellipsoid, is
defined by the kinetic-energy conservation:
\begin{equation}
I_1 \; \Omega_1^2 \; + \; I_2 \; \Omega_2^2 \; + \; I_3 \; \Omega_3^2 \;
= \; {\bf{J \; \Omega}} \; = 2 \: E_{rot} \; = \; const~~~.
\label{2.5}
\end{equation}
In the presence of inner dissipation, the
Poinsot ellipsoid will be distorting in the cause of time, and will eventually
degenerate into a line of length $I_3^{-1/2}$ oriented along the
axis of maximal moment of inertia. That
will happen after the body aligned its angular velocity along its
maximal-inertia axis: ${\Omega}_1 = {\Omega}_2 = 0, \; |{\Omega}_3| =
|{\bf{J}}|/I_3$. This mode minimizes the kinetic energy, with
the angular momentum $\bf{J}$ fixed.

In what follows we  model the grain by an oblate symmetric top, e.g. a disk
\begin{equation}
I_{3} \; > \; I_{1} \; = \; I_{2} \; \equiv \; I \; \; .
\label{2.6}
\end{equation}
In the grain-alignment studies such an approximation is customary (Purcell
1979, Spitzer \& McGlynn 1979). For such a body the Euler equations read:
\begin{equation}
\dot{\Omega}_1 \; + \: {\frac{I_3 - I}{I}} \; {{\Omega}_2} \;
{{\Omega}_3} \; = \; 0 ~~~,
\label{6}
\end{equation}
\begin{equation}
\dot{\Omega}_2 \; + \: {\frac{I - I_3}{I}} \; {{\Omega}_3} \;
{{\Omega}_1} \; = \; 0 ~~~,
\label{7}
\end{equation}
\begin{equation}
\dot{\Omega}_3 \; = \; 0~~~,
\label{8}
\end{equation}
where ${\Omega}_i$ is the body-frame components of the angular velocity.
As well known, the angular velocity ${\bf{\Omega}}$ nutates about
the principal axis 3 at a  constant rate
\begin{equation}
\omega \; = \; (h - 1) \Omega_3,
\; \; \; \; \; \; \; \;  h\equiv {I_3}/I~~~.
\label{9}
\end{equation}
However from the point of view of the inertial observer it is
rather the principal axis 3 that is precessing
about $\bf{J}$ (which is conserved in the inertial frame). Therefore
angle $\theta$ is constant.
Now, let $\alpha$ be the angle described by the precession of vector
${\bf \Omega}$ about axis $3$ (or of axis $3$ about ${\bf \Omega}$).
It is easy to see that
\begin{equation}
\tan {\theta} \; = \;
\; (I \; {|{\bf{\Omega}}|} \; \; \sin \; {\alpha})/(I_3 \;
{|{\bf{\Omega}}|} \; \; \cos \; {\alpha}) \; = \; h \;\tan {\alpha}~~~;
\label{fnote}
\end{equation}
hence  $\alpha$ is also constant. As $\Omega_3 = \Omega\;\cos\alpha$,
the angular velocity of precession given by (\ref{9}),
 except a special case of ${\bf{\Omega}}$ and
${\bf{J}}$  orthogonal or almost orthogonal to the maximal-inertia axis
$3$, is typically of the order of $|{\bf{\Omega}}|$.
Hence one may call the rotation and precession
``fast motions'', implying that the relaxation is  a slow
process (which means that the rate of alignment: $\; \dot{\theta}\ll\omega$).

Before pursuing with the calculational part of our article, let us once
again dwell upon the physical mechanism that changes $\Omega_3$
over the spans of time comparable with the internal relaxation time.
In terms of energy, everything is clear: the  inelastic body
aligns so as to minimize its rotational energy and still to conserve its
angular momentum. This obvious mechanism does not seem to be instilled in the
Euler equations ($\ref{6}$), ($\ref{7}$), ($\ref{8}$), for the torques remain
to be zero, and nothing seems to be able to shift the solutions
${{\Omega}_{1,2,3}}(t)$ from the form that yields $\alpha$ constant.
In reality,
however, this would-be contradiction is easily resolved if one recalls that
inelastic relaxation stems from the inner displacements of grain material.
These displacements result in changes of grain moments of inertia. Therefore
in the cause of the grain's wobbling, its principle moments of inertia
fluctuate, and terms $\Omega_i\dot{I}_i \not = 0$ should be
accounted for in Euler equations. Inelasticity results in a phase
shift between the angular velocity and $\dot{I}_i$ and this
causes internal alignment\footnote{The latter statement is equivalent
to the statement that there is a phase shift between the stresses
that act upon the grain, and the deformation of the grain.}.

Since $\dot{I}_i$ depend on the grain elasticity, while the phase
offset of $\dot{I}_i$ and $\Omega$ is determined by inelastic effects, then
the parameters that determine elasticity and  dissipation within
the material should enter our
final formulae for the inelastic relaxation.

\section{Acceleration of a Point inside an Oblate Grain
$ \; \; ( \; \; \bf{I_3 \; \geq \; I_1 \; = \; I_2 \; \; })$}

Our goal is to calculate $\dot{\theta}$, the rate of the
maximal-inertia axis' approaching the direction of angular momentum $\bf{J}$.
 To achieve this goal, one has to know the rate of energy losses entailed
by the inelastic deformation. To calculate the deformation,
we shall have to know
the acceleration experienced by a particle located inside the grain at a
point ($x$, $y$, $z$). Note that we address the proper acceleration, i.e.
that with respect to the inertial frame $(X,Y,Z)$, but we shall express it in
terms of coordinates $x$, $y$ and $z$ of the body frame $(1,2,3)$.
The fast processes (revolution and precession of a symmetric
oblate grain) are described by the Euler equations ($\ref{6}$ -
$\ref{8}$)  whose solution, in neglect of the slow relaxation, will read
\begin{equation}
{\Omega}_1 \; \; = \; \; {\Omega}_{\perp} \cos {\omega}t~~,~~~
{\Omega}_2 \; \; = \; \; {\Omega}_{\perp} \sin {\omega}t~~,~~~
{\Omega}_3 \; \; = \; \; const
\label{rotation}
\end{equation}
where
\begin{equation}
{{\Omega}_{\perp}} \; \; = \; \; {\Omega} \; \; \sin \; {\alpha}~~, ~~~~~
{{\Omega}_{3}} \; \; = \; \; {\Omega} \; \; \cos \; {\alpha}
\label{24}
\end{equation}
and
\begin{equation}
{{\Omega}_{\perp}}/{{\Omega}_{3}} \; \; = \; \; \tan \; \alpha \; \; = \; \;
h \; \; \tan \; {\theta}~~~.
\label{25}
\end{equation}

Besides, we shall need formulae connecting the components of $\bf{\Omega}$
with the absolute values of the angular momentum:
\begin{equation}
{\Omega}_3 \; \; = \; \; \frac{J_3}{I_3} \; \; = \; \; \frac{J}{I_3} \; \;
\cos \; \theta~~, ~~~~
{\Omega}_{\perp} \; \; = \; \; \frac{J}{I_3} \; h \; \; \sin \; \theta
\label{om4}
\end{equation}
We denote the position, velocity and acceleration relative to the
body frame as: ${\bf{r}}$, ${\bf{v}}$, ${\bf{a} \;}$, while those
related to the body frame $ \; (1,2,3) \; $ will be called ${\bf{r''}}$,
${\bf{v''}}$, ${\bf{a''} \;}$, where we shall keep in mind that $\;{\bf{r}} =
{\bf{r''}}$.
The acceleration ${\bf{a} \;}$ in the inertial frame looks:
\begin{equation}
{\bf{a}} \; \; = \; \; {\bf{a''}} \; \; + \; \;
{\bf{\dot{\Omega}}} \; \times \; {\bf{r''}} \; \; + \; \;
2 \; \; {\bf{\Omega}} \; \times \; {\bf{v''}} \; \; + \; \;
{\bf{\Omega}} \; \times \; ({\bf{\Omega}} \; \times \; {\bf{r''}})~~~,
\label{anonimous}
\end{equation}
where the first and the third terms vanish, as the values of relative
deformations $\delta l/l$ are minute in a solid body. The values
of ${\bf v}''\approx \delta l/\tau$ and ${\bf a}'' \approx \delta l/\tau^2$,
where $\tau$ is the period of rotation,
are negligible compared to velocities and accelerations in the inertial
system of reference that by the order of magnitude are, respectively
$l/\tau$ and $l/\tau^2$. The second and the fourth term in
Eq.~(\ref{anonimous}), referred to the grain frame with unit vectors
(${\bf e}_1$,${\bf e}_2$, ${\bf e}_3$), will read:
\begin{eqnarray}
{\bf{\dot{\Omega}}} \; \times \; {\bf{r''}}
& = &
{{\bf{e}}_1} \; ({\dot{\Omega}}_2 \; z \; \; -
\; \;  {\dot{\Omega}}_3 \; y) \; \; + \; \;
{{\bf{e}}_2} \; ({\dot{\Omega}}_3 \; x \; \; -
\; \;  {\dot{\Omega}}_1 \; z) \; \; + \; \;
{{\bf{e}}_3} \; ({\dot{\Omega}}_1 \; y \; \; -
\; \;  {\dot{\Omega}}_2 \; x)  \nonumber  \\
& = &
{{\bf{e}}_1} \; {\omega} \; z \; {\Omega_1} \; \; + \; \;
{{\bf{e}}_2} \; {\omega} \; z \; {\Omega_2} \; \; + \; \;
{{\bf{e}}_3} \; (\; - \; {\omega} \; y \; {\Omega_2} \; - \;
{\omega} \; x \; {\Omega_1}) \;\;\; ,
\label{26}
\end{eqnarray}
and
\begin{eqnarray}
{\bf{\Omega}} \; \times \; ({\bf{\Omega}} \; \times \; {\bf{r''}}) \; \; =\; \;
{\bf{\Omega}} \; ({\bf{\Omega}} \; \cdot \; {\bf{r''}}) \; \; - \; \;
{\bf{r''}} \; {\bf{\Omega}}^2 \; \; = \nonumber \\
{{\bf{e}}_1} \; \{
{\Omega_1} \; ( \; {\Omega_1} \; x \; + \; {\Omega_2} \; y + \; {\Omega_3} \; z
\; ) \; \; - \; \; x \; {\Omega}^2
\} \; \; + \nonumber  \\
{{\bf{e}}_2} \; \{
{\Omega_2} \; ( \; {\Omega_1} \; x \; + \; {\Omega_2} \; y + \; {\Omega_3} \; z
\; ) \; \; - \; \; y \; {\Omega}^2
\} \; \; + \nonumber  \\
{{\bf{e}}_3} \; \{
{\Omega_3} \; ( \; {\Omega_1} \; x \; + \; {\Omega_2} \; y + \; {\Omega_3} \; z
\; ) \; \; - \; \; z \; {\Omega}^2
\}~~~.
\label{27}
\end{eqnarray}
All in all,
\begin{eqnarray}
{\bf{a}} \; \; = \; \;
{{\bf{e}}_1} \; \{
\; - \; x \; {\Omega_2}^2 \; \; - \; \; x \; {\Omega_3}^2 \; \; + \; \;
y \; {\Omega_2} \; {\Omega_1} \; \; + \; \; z \; {\Omega_3} \; {\Omega_1}
\; \; + \; \; z \; \omega \; {{\Omega}_{\perp}} \; \cos \; {\omega}t \;
\}  \nonumber  \\
+ \; \;
{{\bf{e}}_2} \; \{
\; x \; {\Omega_2} \; {\Omega_1} \; \; - \; \; y \; {\Omega_1}^2 \; \; - \; \;
y \; {\Omega_3}^2 \; \; + \; \; z \; {\Omega_3} \; {\Omega_2}\; \;\
 + \; \; z \; \omega \; {{\Omega}_{\perp}} \; \sin \; {\omega}t \;
\}  \nonumber  \\
+ \; \;
{{\bf{e}}_3} \; \{
\; x \; {\Omega_1} \; {\Omega_3} \; \; + \; \; y \; {\Omega_3} \; {\Omega_2}
\; \; - \; \; z \; {\Omega_{\perp}}^2 \; \; -
\; \; \omega \; {\Omega}_{\perp}
\; (x \; \cos \; {\omega}t \; + \; y \; \sin \; {\omega}t) \;
\}~~~.
\label{28}
\end{eqnarray}
We had to present our calculation in great detail because the latter formula
considerably differs from the one in (Purcell 1979), eq.~28.
To be doubly sure of our formula being correct, we derived
it also in the intermediate, ``Eulerian'', system of reference
(see Appendix~\ref{A}).

\section{Stresses and Strains Caused by the Precession}

In this section we derive the stresses and strains produced
in the rotating body by the time-dependent terms in ($\ref{28}$), as only
these terms influence the rate of energy dissipation in the grain.

With aid of (\ref{rotation}) one can easily split ($\ref{28}$)
into a time-independent and time-dependent parts:
\begin{equation}
{\bf{a}} \; \; = \; \; {\bf{a}}_{0} \; \; + \; \; {\bf{a}}_{\mathsf{t}}~~~,
\label{4.1}
\end{equation}
where
\begin{equation}
{\bf{a}}_{0} \; \; = \; \; - \; ( \; {{\bf{e}}_1} \; x \; \; + \; \;
                                     {{\bf{e}}_2} \; y \; ) \;
( \; {\Omega}_3^2 \; \; + \; \; {\frac{1}{2}} \; {\Omega}_{\perp}^2 \; ) \; \;
- \; \; {{\bf{e}}_3} \; z \; {\Omega}_{\perp}^2~~~,
\label{4.2}
\end{equation}
and
\begin{eqnarray}
{\bf{a}}_{\mathsf{t}} \; \; = \; \; {{\bf{e}}_1} \; \{ \;
{\frac{1}{2}} \; {\Omega}_{\perp}^2 \; x \; \; \cos \; 2{\omega}t \; \; + \; \;
{\frac{1}{2}} \; {\Omega}_{\perp}^2 \; y \; \; \sin \; 2{\omega}t \; \; + \; \;
z \; {\Omega}_{\perp} \; {\Omega}_3 \; h \; \; \cos \; {\omega}t
\; \}  \nonumber  \\
+ \; \;
{{\bf{e}}_2} \; \{
{\frac{1}{2}} \; {\Omega}_{\perp}^2 \; x \; \; \sin \; 2{\omega}t \; \; - \; \;
{\frac{1}{2}} \; {\Omega}_{\perp}^2 \; y \; \; \cos \; 2{\omega}t \; \; + \; \;
z \; {\Omega}_{\perp} \; {\Omega}_3 \; h \; \; \sin \; {\omega}t \;
\; \}  \nonumber  \\
+ \; \;
{{\bf{e}}_3} \; \{
{\Omega}_{\perp} \; {\Omega}_3 \; (2 \; - \; h) \; (x \; \; \cos \; {\omega}t
\; \; + \; \; y \; \; \sin \; {\omega}t \; )
\; \}~~~.
\label{4.3}
\end{eqnarray}
Now we see that dissipation of energy will be taking place at two modes one
of which will be of double frequency. Hereafter we shall be taking into
consideration only time-dependent, i.e. produced by
${\mathbf{a}}_{\mathsf{t}}$, inputs in the stresses and strains.

Following Purcell, we shall model the oblate grain by a prism of sizes
$2a \times 2a \times 2c$, ($c < a$).
The stresses vanishing on the boundaries of our rectangular prism
read\footnote{Our formulae
for the stresses differ from those in Purcell (1979).
Purcell's expressions do not include the double-frequency
terms, and they do not vanish on the boundaries of the
grain, which they should do.}:
\begin{equation}
\sigma_{xx} = \frac{\rho \Omega_\perp^2}{4}
(x^2 - a^2) \; \cos 2 {\omega} t  \; \; , \; \; \;
\sigma_{yy} = - \frac{\rho \Omega_\perp^2}{4} (y^2 - a^2) \;
\cos 2 {\omega} t \; \; , \; \; \;
\sigma_{zz} = 0~,
\label{4.5}
\end{equation}
\begin{equation}
\sigma_{xy} \; \; = \; \; \frac{\rho}{4} \; \Omega_\perp^2 \;
(x^2 \; \; + \; \; y^2 \; \; - \; \; 2a^2)
\; \; \sin \; 2 {\omega} t \; \; \; ,
\label{4.6}
\end{equation}
\begin{equation}
\sigma_{xz} \; \; = \; \; \frac{\rho}{2} \; {\Omega_\perp} \; {\Omega_3} \;
\left[ \; h \; (z^2 \; - \; c^2)
\; \; + \; \; (2 \; - \; h) \; (x^2 \; - \; a^2) \; \right] \; \;
\cos \; {\omega} t \; \; \; ,
\label{4.7}
\end{equation}
\begin{equation}
\sigma_{yz} \; \; = \; \; \frac{\rho}{2} \; {\Omega_\perp} \; {\Omega_3} \;
\left[ \; h \; (z^2 \; - \; c^2)
 \; \; + \; \; (2 \; - \; h) \; (y^2 \; - \; a^2) \; \right] \; \;
\sin \; {\omega} t \; \; \; ,
\label{4.8}
\end{equation}
which leads to forces per unit volume:
\begin{equation}
F_x  = {{\partial}_{i}} {{\sigma}_{xi}}  \; \; =
\frac{\rho}{2} {\Omega_\perp}^2 x \; \cos  2 {\omega} t \; \; + \; \;
\frac{\rho}{2} {\Omega_\perp}^2 y \; \sin  2 {\omega} t \; \; +  \; \;
\rho {\Omega_\perp} {\Omega_3} h  z \; \cos  {\omega} t \; \; \; ,
\label{4.9}
\end{equation}
\begin{equation}
F_y  = {{\partial}_{i}} {{\sigma}_{yi}} \; \; =
\frac{\rho}{2} {\Omega_\perp}^2 x  \; \sin 2 {\omega} t \; \; - \; \;
\frac{\rho}{2} {\Omega_\perp}^2 y \; \cos 2 {\omega} t  \; \; + \; \;
\rho {\Omega_\perp}  {\Omega_3} h z \; \sin  {\omega} t \; \; \; ,
\label{4.10}
\end{equation}
\begin{equation}
F_z  = {{\partial}_{i}} {{\sigma}_{zi}} =
\rho {\Omega_\perp} {\Omega_3}  (2  -  h) x  \; \cos \omega t \; \; + \; \;
\rho {\Omega_\perp} {\Omega_3}  (2  -  h) y  \; \sin \omega t \; \; \; ,
\label{4.11}
\end{equation}
in full accordance with ($\ref{4.3}$). Here $\rho$ is the density of the
grain material. For a $\; 2a \times 2a \times 2c \;$ prism, the moment of
inertia $I_3$ and the parameter $h$ are function of the half-sizes $a$ and $c$:
\begin{equation}
I_3 \;  = \; \frac{16}{3} \; \rho \; a^4 \; c
\label{4.4}
\end{equation}
and
\begin{equation}
h \; \; \equiv \; \; \frac{I_3}{I} \; \; = \; \; \frac{2}{1 \; + \; (c/a)^2}
\; \; \; \; \; , \; \; \; \; \; \; \; \; \; \; \; \; 2 \; - \; h \; \; =
\; \; \left( \frac{c}{a} \right)^2 \; \frac{2}{1 \; + \; (c/a)^2} \;
\; \; .
\label{4.12}
\end{equation}
According to (3.12), (3.13),
\begin{equation}
\Omega_3 \; \; = \; \; \Omega_0 \; \cos \theta
\; \; \; \; \; , \; \; \; \; \; \; \; \; \; \; \; \;
\Omega_{\perp} \; \; = \; \; \Omega_0 \; h \; \sin \theta \; \; \; ,
\label{4.13}
\end{equation}
$\Omega_0 \; \equiv \; J/I_3$ being the typical angular velocity
of a grain. Our knowledge of the stresses should be sufficient for computing
the rate of energy losses in the body.

Our expressions for the stress-tensor components differ very significantly
from the appropriate expressions derived in (Purcell 1979). There are two
reasons for it. The first reason is the afore-mentioned Purcell's
miscalculation of the acceleration experienced by a point inside the wobbling
body. (Because of that Purcell completely missed the double-frequency
contribution to the stress tensor. Later we shall show that this input is of
the leading order.) The second reason is that Purcell forgot
to impose proper boundary conditions upon the stress-tensor components: his
stresses fail to vanish on the boundaries of the body.

Our goal now is to calculate the average deformation-caused energy, stored in
the tumbling body, and to estimate the energy dissipation, using an
appropriate quality factor. We shall take into account that the deformation of
the grain is neither purely elastic nor purely plastic, but is a superposition
of the former and the latter. It is then to be described by the tensor
$\epsilon_{\it{ij}}$ of {\textit {viscoelastic}} strains and by the velocity
tensor consisting of the time-derivatives $\dot{\epsilon}_{\it{ij}}$. The
stress tensor will now be separated into two components: the elastic stress
and the plastic (viscous) stress:
\begin{equation}
\sigma_{\it{ij}} \; \; = \; \; \sigma_{\it{ij}}^{(e)} \; \; +
\sigma_{\it{ij}}^{(p)}~~~,
\label{4.18}
\end{equation}
where the components of the elastic stress tensor are interconnected with
those of the strain tensor (Landau and Lifshitz 1976):
\begin{equation}
\epsilon_{ij} \; \; = \; \; \delta_{ij} \; \; \frac{Tr \;
\sigma^{(e)}}{9 \; K} \; \; + \; \;
\left( \; \sigma_{\it{ij}}^{(e)} \; \; - \; \; \frac{1}{3} \; \; \delta_{ij}
\; \; Tr \; \sigma^{(e)} \right) \; \frac{1}{2 \; \mu} \; \; \; ~~~,
\label{4.19}
\end{equation}
\begin{equation}
\sigma_{ij}^{(e)} \; \; = \; \; K \; \delta_{ij} \; \; Tr \; \epsilon \; \; +
\; \; 2 \; \mu \; \left(\epsilon_{ij} \; - \; \frac{1}{3} \; \delta_{ij} \;
\; Tr \; \epsilon \right)~~~,
\label{4.20}
\end{equation}
$\mu$ and $K$ being the {\textit{isothermal}} shear and bulk moduli, and $Tr$
standing for the trace of a tensor. Components of the plastic stress are
connected with the strain derivatives as
\begin{equation}
\dot{\epsilon}_{ij} \; \; = \; \; \delta_{ij} \;
\frac{Tr \; \sigma^{(p)}}{9 \zeta} \; \; + \; \; \left(\sigma_{ij}^{(p)} \; -
\; \frac{1}{3} \; \delta_{ij} \;\; Tr \; \sigma^{(p)} \right) \;
\frac{1}{2 \eta}~~~,
\label{4.21}
\end{equation}
\begin{equation}
\sigma_{\it{ij}}^{(p)} \; \; = \; \; \zeta \; \delta_{ij} \; \;
Tr \; {\dot{\epsilon}} \; \; + \; \; 2 \; \eta \; \left(\dot{\epsilon}_{ij}
\; - \; \frac{1}{3} \; \delta_{ij} \; \; Tr \; \dot{\epsilon} \right)~~~,
\label{4.22}
\end{equation}
where  $\; \eta \;$ and $\; \zeta \;$ are the shear and stretch viscosities.

\section{Dynamics of a tumbling body}

The kinetic energy of a rotating body reads, according to (\ref{2.5}),
(\ref{2.6}), (\ref{rotation}), and (\ref{om4}):
\begin{eqnarray}
E_{rot} \; \; = \; \; \frac{1}{2} \; (I_1 \; \Omega_1^2 \; + \; I_2 \; \Omega_2^2 \; + \; I_3 \; \Omega_3^2) \; \;
= \; \; \frac{1}{2} \;
[I \; \Omega_{\perp}^2 \; + \; I_3 \; {\Omega_3}^2] \; \; =
\nonumber\\
= \; \; \frac{1}{2} \; \; \left[ \; \frac{1}{I} \; \; \sin^2 \; \theta \; \; + \; \;
\frac{1}{I_3} \; \; \cos^2 \; \theta \; \right] \; J^2
\label{5.1}
\end{eqnarray}
wherefrom
\begin{equation}
\frac{dE_{rot}}{d\theta} \; \; = \; \;
\frac{J^2}{I_3} \; (h \; - \; 1) \; \; \sin \; \theta \; \; \cos \; \theta
\; \; =  \; \; \omega \; J \; \; \sin \; \theta \; \; \; .
\label{5.2}
\end{equation}
Formula (\ref{5.2}) provides an insight in how the change of rotational energy
yields a change of $ \; \theta$. If we calculate the rate of
energy losses, $\dot{E}_{rot}$, it will be easy to find the
rate of alignment
$\; \dot{\theta} \;$ as $\; ({dE_{rot}}/{d\theta})^{-1}\dot{E}_{rot} \;$.
How to compute $\dot{E}_{rot}$?
The rotational energy changes via the inelastic dissipation, so that
\begin{equation}
\dot{E}_{rot} \; = \; \dot{W}
\label{5.3}
\end{equation}
Hence what we have to find is the rate of the elastic-energy losses $\;
\dot{W}$. Then, with aid of (\ref{5.2}) we shall calculate the rate of
alignment:
\begin{equation}
\frac{d\theta}{dt} \; = \; \left(\frac{dE_{rot}}{d\theta}\right)^{-1}
\frac{dE_{rot}}{dt} \; = \; \left( \omega \; J \; \; \sin \; \theta
\right)^{-1} \; \dot{W}
\label{5.4}
\end{equation}
In our case, dissipation is taking place at two modes:
\begin{equation}
\dot{W} \; = \; \dot{W^{({\omega})}} \; + \; \dot{W^{(2{\omega})}} \; = \;
\omega \; \frac{W^{({\omega})}}{Q^{({\omega})}} \; + \; 2 \; \omega \;
\frac{W^{({2\omega})}}{Q^{({2\omega})}} \; \approx \; \frac{\omega}{Q} \;
\left\{ W^{({\omega})} \; + \; 2 W^{({2\omega})} \protect\right\}
\label{5.5}
\end{equation}
where we used the fact that the
quality factor is almost frequency-independent: $\; Q^{({2\omega})} \,
\approx \, Q^{({\omega})} \, \approx \, Q \;$.

\section{Rate of Energy Dissipation at low temperatures}

Before pursuing to our calculations of the energy-dissipation rate $\;
\dot{W} \;$, several prefatory notes will be in order. As well known, at low
temperatures materials are fragile: when the deformations exceed some critical
threshold, the body will rather break than flow. At the same time, at these
temperatures the materials are elastic, provided the deformations are
{\textit{beneath}} the said threshold: the sound absorption, for example, is
almost exclusively due to the thermal conductivity rather than to the
viscosity. These facts may be summarized like this: at low temperatures, the
viscosity coefficient $\; \eta \;$ has, effectively, two values: one value -
for small deformations (and this value is almost exactly zero); another value
- for larger-than-threshold deformations (and that value is
high\footnote{Effectively it may be put infinity because, as
explained above, the body will rather crack than demonstrate fluidity.}).

At high temperatures materials become plastic, which means that the shear
viscosity $\; \eta \;$ gets its single value, deformation-independent in the
first approximation. On the one hand, this value will be far from zero (so
that the scattering of vibrations will now be predominantly due to the
viscosity terms, not due to the thermal conductivity). On the other hand, this
value will not be that high: a plastic body will rather yield than break.
All this is certainly valid for the stretch viscosity $\; \zeta \;$ as well.

Since at low temperatures the bodies manifest, for small displacements, no
viscosity ($\; \omega \eta \, \sim \, \omega \zeta \; \ll \;  \mu \, \sim \, K
\; $), the stress tensor will be approximated, to a high accuracy, by its
elastic part: instead of the system (\ref{4.18}) - (\ref{4.22}) we shall
simply write:
\begin{equation}
\epsilon_{ij} \; \; = \; \; \delta_{ij} \; \; \frac{Tr \;
\sigma}{9 \; K} \; \; + \; \;
\left( \; \sigma_{\it{ij}} \; \; - \; \; \frac{1}{3} \; \; \delta_{ij}
\; \; Tr \; \sigma \right) \; \frac{1}{2 \; \mu} \; \; \; ~~~,
\label{6.1}
\end{equation}
This will enable us to derive an expression for the elastic energy stored in
a unit volume of the precessing body:
\begin{eqnarray}
dW/dV \; = \; \frac{1}{2} \; \epsilon_{ij} \; \sigma_{ij} \; =\;\frac{1}{4} \;
\left\{ \left(\frac{2 \; \mu}{9\; K} \; - \; \frac{1}{3} \protect\right) \,
\left(Tr \; \sigma \protect\right)^2 \; + \; \sigma_{ij} \,  \sigma_{ij}
\protect\right\} \; \approx
\nonumber \\
\nonumber \\
\approx \; \; \frac{1}{4\mu}\;\left\{\; - \; \frac{1}{5} \, \left(Tr \; \sigma
\protect\right)^2 \; + \; \sigma_{xx}^2\; + \; \sigma_{yy}^2 \; + \;
\sigma_{zz}^2 \; + \; 2 \, \left( \sigma_{xy}^2 \; + \; \sigma_{yz}^2 \; + \;
\sigma_{zx}^2 \protect\right) \protect\right\}
\label{6.2}
\end{eqnarray}
where we made use of the expressions connecting the shear and bulk moduli with
the Young modulus $E$ and Poisson's ratio $\sigma$: since $\; K \, = \, E/[3(1
-2\sigma] \;$ and $\; \mu \, = \, E/[2(1+\sigma] \;$ then $\; 2 \mu/(9K) \, -
1/3 \, = \, - \, \sigma /(1+\sigma ) \;$. As for frozen solids Poisson's
ratio $\sigma$ is typically about $0.25$, we put  $\; 2 \mu/(9K) \, - 1/3 \;
\approx \; - \, 1/5 \;$. Mind that the body is trembling at two frequencies:
$\; \omega \;$ and $\; 2 \, \omega \;$. Anticipating the different rates of
dissipation at these two modes, we shall split the total elastic energy into
two parts:
\begin{equation}
dW/dV \; = \; dW^{(\omega)}/dV \; + \; dW^{(2\omega)}/dV
\label{6.3}
\end{equation}
where, according to (\ref{4.5}) - (\ref{4.8}),
\begin{equation}
dW^{(\omega)}/dV \; = \; \frac{1}{2\mu}\;\left\{ \sigma_{yz}^2 \; + \;
\sigma_{zx}^2 \protect\right\}
\label{6.4}
\end{equation}
and
\begin{equation}
dW^{(2\omega)}/dV \; = \; \frac{1}{4\mu} \; \left\{\; - \; \frac{1}{5} \,
\left(Tr \; \sigma
\protect\right)^2 \; + \; \sigma_{xx}^2\; + \; \sigma_{yy}^2 \; + \;
2 \sigma_{xy}^2  \protect\right\} \; \; \; .
\label{6.5}
\end{equation}
>>From now on we shall be interested in the energies averaged over several
periods of the precession. Therefore we shall substitute $\; \sin^2... \;$ and
$\; \cos^2... \;$ by $1/2$, and shall omit expressions $\; \sin... \, \cos...\;
$, $\; \sin \omega t \, \sin 2 \omega t \;$ and $\; \cos \omega t \, \cos 2
\omega t \;$. With the above reservation being beared in mind,
expressions (\ref{4.5}) - (\ref{4.8}) for the stresses will yield:
\begin{equation}
\left(Tr \; \sigma \protect\right)^2 \; = \; \left(\frac{\rho \;
\Omega^2_{\perp}}{4}\protect\right)^2 \; a^4 \; \left(x_1^2 \; - \;
y_1^2\protect\right)^2 \; \frac{1}{2} \; \; \; ,
\label{6.6}
\end{equation}
\begin{equation}
\sigma^2_{xx} \; = \;
\left(\frac{\rho \; \Omega^2_{\perp}}{4}\protect\right)^2 \; a^4 \;
\left(x_1^2 \; - \; 1\protect\right)^2 \; \frac{1}{2} \; \; \; ,
\label{6.7}
\end{equation}
\begin{equation}
\sigma^2_{yy} \; = \;
\left(\frac{\rho \; \Omega^2_{\perp}}{4}\protect\right)^2 \; a^4 \;
\left(y_1^2 \; - \; 1\protect\right)^2 \; \frac{1}{2} \; \; \; ,
\label{6.8}
\end{equation}
\begin{equation}
\sigma^2_{xy} \; = \;
\left(\frac{\rho \; \Omega^2_{\perp}}{4}\protect\right)^2 \; a^4 \;
\left(x_1^2 \; + \; y^2_1 \; - \; 2\protect\right)^2 \; \frac{1}{2} \; \; \; ,
\label{6.9}
\end{equation}
\begin{equation}
\sigma^2_{xz} \; = \; 2 \; \gamma \;
\left(\frac{\rho \; \Omega^2_{\perp}}{4}\protect\right)^2 \; a^4 \;
\left(x_1^2 \; + \; z^2_1 \; - \; 2\protect\right)^2 \; \; \; ,
\label{6.10}
\end{equation}
\begin{equation}
\sigma^2_{yz} \; = \;  2 \; \gamma \;
\left(\frac{\rho \; \Omega^2_{\perp}}{4}\protect\right)^2 \; a^4 \;
\left(y_1^2 \; + \; z^2_1 \; - \; 2\protect\right)^2 \; \; \; ,
\label{6.11}
\end{equation}
where
\begin{equation}
x_1 \; \; \equiv \; \; x/a \; \; \; , \; \; \; \; \; \; y_1 \; \; \equiv \;
y/a \; \; \; \; \; , \; \; \; \; z_1 \; \; \equiv \; \; z/a \; \; \; ,
\label{6.12}
\end{equation}
and
\begin{equation}
\gamma \; \; \equiv \; \; \left( \frac{\Omega_{3}}{\Omega_{\perp}}
\protect\right)^2 \; (2 \; - \; h)^2 \; \; = \; \;
\left( \frac{\Omega_{3}}{\Omega_{\perp}}
\protect\right)^2 \; h^2 \; \left(\frac{c}{a} \protect\right)^4 \; \; =
\; \; \frac{(c/a)^4}{\tan^2 \theta} \; \; \; .
\label{6.13}
\end{equation}
According to (\ref{5.5}), what we shall need is the sum $\; W^{({\omega})} \; +
\; 2 \, W^{({2\omega})}\;$. To get it, we shall integrate over the volume:
\begin{eqnarray}
W^{({\omega})} \; + \; 2 W^{({2\omega})} \; = \; \int^{a}_{-a} \, dx \;
\int^{a}_{-a} \, dy \; \int^{c}_{-c} \, dz \; \left\{ dW^{({\omega})}/dV \; +
\; 2 \; dW^{({2\omega})}/dV  \protect\right\} \; =
\nonumber \\
= \; a^2 \; c \; \int^{1}_{-1} \, dx_1 \; \int^{1}_{-1} \, dy_1 \;
\int^{1}_{-1} \, dz_1 \; \frac{1}{2\mu}\left\{
\; - \; \frac{1}{5} \,
\left(Tr \; \sigma \protect\right)^2 \; + \; \sigma_{xx}^2\; + \;
\sigma_{yy}^2 \; + \; 2 \sigma_{xy}^2 \; + \; \sigma_{yz}^2 \; + \;
\sigma_{zx}^2 \protect\right\} \; \; \; .
\label{6.14}
\end{eqnarray}
Plugging of (\ref{6.6}) - (\ref{6.11}) in the above expression entails:
\begin{eqnarray}
W^{({\omega})}\;+\;2 W^{({2\omega})} \; =\;2^{-5}\;\; \frac{a^6 \; c}{\mu} \;
\left(\rho \; \Omega^2_{\perp}\protect\right)^2 \;
\int^{1}_{-1} \, dx_1 \; \int^{1}_{-1} \, dy_1 \;
\int^{1}_{-1} \, dz_1 \; \left\{ \right.
\; - \; \frac{1}{10} \; \left(x_1^2 \; - \; y_1^2 \protect\right)^2 \; +
\nonumber \\
+ \; \frac{1}{2} \; \left(x_1^2 \; - \; 1 \protect\right)^2 \; +
\; \frac{1}{2} \; \left(y_1^2 \; - \; 1 \protect\right)^2 \; +
\; \left(x_1^2 \; + \; y_1^2 \; - \; 2 \protect\right)^2 \; +
\nonumber \\
+ \; 2 \; \gamma \; \left(x_1^2 \; + \; z_1^2 \; - \; 2 \protect\right)^2 \; +
\; 2 \; \gamma \; \left(y_1^2 \; + \; z_1^2 \; - \; 2 \protect\right)^2
\left. \protect\right\} \; \approx
\nonumber \\
\approx \; \frac{a^6 \; c}{\mu} \; \left(\rho \;
\Omega^2_{\perp}\protect\right)^2 \; 2^{-5} \; (63 \; \gamma \; + \; 20)
\label{6.15}
\end{eqnarray}
where the numerics was performed by means of the Maple software.
The product $\;\rho^2\Omega_{\perp}^4\;$ emerging in the above expression may
be cast in a form wherein its temperature-dependence
becomes manifest.
According to  ($\ref{4.4}$), ($\ref{4.12}$), and ($\ref{4.13}$),
\begin{eqnarray}
\rho^2 \; \Omega_{\perp}^4 \; \; =
\frac{a^{-8} \; c^{-2}}{ \left[1+(c/a)^2  \protect\right]^4} \;
\frac{9}{4} \; \left( \beta \; kT_{\rm gas} \protect\right)^2 \; \sin^4
\theta \; \; \; \; \; \; \;
\label{6.16}
\end{eqnarray}
where $k$ signifies the Boltzmann constant and $\beta$ stands for the
parameter of suprathermality:
\begin{equation}
\frac{I_3 \; \Omega_0^2}{2} \; \; = \; \; \frac{J^2}{2I_3} \; \; = \; \;
\beta \; kT_{\rm gas}
\label{6.17}
\end{equation}
The cosmic dust is thermal when $\beta = 1$. This way of writing the
energies enables one to express these in terms of the temperature of
the surrounding gas, $T_{\rm gas}$:
\begin{equation}
W^{({\omega})} \; + \; 2 \; W^{({2\omega})} \; = \; \frac{3^2 \; 2^{-7} \;
(63 \; \gamma \; + \; 20)}{\mu} \;
\frac{a^{-2} \; c^{-1}}{ \left[1+(c/a)^2  \protect\right]^4} \;
\left( \beta \; kT_{\rm gas} \protect\right)^2 \; \sin^4 \theta
\label{6.18}
\end{equation}
Together, (\ref{5.4}), (\ref{5.5}) and (\ref{6.18}) give:
\begin{equation}
d \theta/dt \; = \; - \; \frac{3^2 \; 2^{-7} \;
(63 \; \gamma \; + \; 20)}{\mu \; Q \; J} \;
\frac{a^{-2} \; c^{-1}}{ \left[1+(c/a)^2  \protect\right]^4} \;
\left( \beta \; kT_{\rm gas} \protect\right)^2 \; \sin^3 \theta
\label{6.19}
\end{equation}
According to (\ref{6.17}) and (\ref{4.4}),
\begin{equation}
J \; = \; \left(2 \; \beta \; k \; T_{gas} \; I_3 \protect\right)^{1/2} \;
= \; \left( \frac{32}{3} \; \beta \; k \; T_{gas} \; \rho \; a^4 \; c
\protect\right)^{1/2} \; = \; 2^{5/2} \; 3^{-1/2} \; a^{2} \; c^{1/2} \;
\rho^{1/2} \; \left( \beta \; k \; T_{gas} \protect\right)^{1/2}
\label{6.20}
\end{equation}
Substitution of the latter in the former will give us the final expression for
the alignment rate:
\begin{equation}
d \theta/dt \; = \; - \; \frac{a^{-5.5}}{[1+(c/a)^2]^4} \;
\left(\frac{a}{c}\protect\right)^{3/2} \;
\left( \beta \; kT_{\rm gas} \protect\right)^{3/2} \;
\frac{\sin^3 \theta}{\mu \; Q \; \rho^{1/2}} \;
 3^{2.5} \; 2^{-9.5} \; \left(63 \; (c/a)^4 \; \cot^2 \theta \; + \; 20
\protect\right)
\label{6.21}
\end{equation}
Simply from looking at this formula one can conclude that the
major-inertia axis slows down its alignment for $\; \theta \; $ approaching
zero. This feature looks physically reasonable.

On physical grounds, one may also expect that the alignment rate vanishes
for $\; h \;$ approaching unity (i.e., when the body lacks an axis
of maximal inertia). Besides, one may expect the major axis of the grain
to linger in the position $\; \theta \; \approx \; \pi /2 \;$,
as if ``hesitating'' to whether to start aligning along or opposite its
angular momentum. However, none of the latter two features seems to be
instilled
into (\ref{6.21}): it may seem from this formula that $\; d \theta /dt \;$ is
$\; h-$independent, and that the major axis leaves the initial position $\;
\theta =
\pi/2 \;$ with a finite angular velocity. To dispel these illusions, simply
recall that our treatment is valid only for as long as the rotation and
precession are fast motions compared to the alignment: $\; {\dot{\theta}} \ll
\omega \;$.

All these subtleties are anyway irrelevant when one merely wants to
estimate the relaxation time, i.e., the time required for the maximal-inertia
axis of an oblate cosmic-dust grain to be considerably shifted toward
alignment with the angular momentum:
\begin{equation}
t_{i} \; \; \simeq \; \;
\left( \; \left\langle \frac{d\theta}{dt} \right\rangle \; \right)^{-1}
\; \; \simeq \; \; - \int_{\pi/2}^{\delta} \;
\frac{d \theta}{d \theta/dt} \; \; \; .
\label{6.22}
\end{equation}
where $\delta$ is introduced to avoid the divergence associated with the
``slow finish''. (One can take, for example, $\; \delta \; = \; \pi/8 \;$.) A
particular choice of $\delta$ will bring into the expression for
$t_{\mathsf{i}}$ some numerical factor of order unity. Since we want nothing
more but a rough estimate for $t_{\mathsf{i}}$, we shall approximate
$t_{\mathsf{i}}$ simply by the inverse $\; d \theta /dt \;$ evaluated in the
middle of the interval, at $\theta \; = \; \pi/4$. Then
\begin{eqnarray}
t_{i} \;
&\approx & a^{5.5} \; [1+(c/a)^2]^4 \;
\left(\frac{c}{a}\protect\right)^{3/2} \;
\left( \beta \; kT_{\rm gas} \protect\right)^{-3/2} \;
\mu \; Q \; \rho^{1/2} \; \frac{2^{11} \; 3^{-2.5}}{63(c/a)^4+20}
\label{6.23}
\end{eqnarray}

The quantity $\; (c/a)^{3/2} \; [1+(c/a)^2]^4 \; [63 \, (c/a)^4 \, + \,
20]^{-1} \;$ is a steep function of parameter $\; (c/a) \;$. For $\; c/a = 1/2
\;$ it equals to $\; 4.3 \, \times \, 10^{-2} \;$, while for $\; c/a = 1/10 \;$ it will be  $\; 1.6 \, \times \, 10^{-3} \;$.

The above formula for the typical time of alignment is the main
result of our article. Now we must think of the possible values
for the material parameters involved. The values may depend both on the
temperature $\; T_{\rm grain} \;$ of the cosmic-dust grains, and on the frequency
of the precession. To start with, temperature- and frequency-caused variations
of the density $\; \rho \;$ may be neglected, as they are to be small anyway.
This way, we can use the (static) densities appropriate to the room
temperature and pressure.

Now, consider the isothermal shear modulus $\; \mu \;$. The tables of
physical quantities would provide its values for the room temperature and
atmospheric pressure, and for quasistatic regimes solely.
As for the possible frequency-related effects in materials (the so-called
ultrasonic attenuation), these become noticeable only at
frequencies higher than $\; 10^8 \; Hz \;$ (see
section 17.7 in Nowick and Berry 1972). Another
fortunate circumstance is that the pressure-dependence of
the elastic moduli is known to be weak (Ahrens 1995). Besides, the elastic
moduli of solids are known to be insensitive to temperature variations, as
long as these variations are far enough from the melting point. The value of
$\; \mu \;$ may increase by several percent when the temperature is
drops from the room temperature to 10 K.  Dislocations don't affect the
elastic moduli either. Solute elements have very little effect on
moduli in quantities up to a few percent\footnote{Beyond that, one might
assume that the moduli vary linearly with substitutional impurities (in
which the atoms of the impurity replace those of the hosts). However hydrogen
is not like that: it enters the interstices between the atoms of the host,
and has marginal effect on modulus.}. As for the role of the possible
porosity, the elastic moduli scale  as the square of the relative density.
For porosities up to about $20 \; \% ~\,$, this is not of much relevance
for our estimates\footnote{We are deeply thankful to Michael Ashby and Michael
Aziz, who consulted us on all these subtle topics.}.

According to (Ryan and Blevins 1987),
at $\; T \; \simeq \; 20 \, K \;$, the share modulus value for silicate
$\mu \; \approx \; 10^{10} \; Pa \; \; $
and density is $\rho^{\rm silicate} \approx \; 2500 \; kg \;
m^{-3} $.
The temperature of subthermal rotating that according
to Lazarian and Draine (1997) is important during crossovers
is $T_{\rm rot} \; \simeq \; 20 \, K \,$.

Now, several words on the choice of values of the $\; Q-$factor will be in
order. Purcell refers to (Krause 1973) wherein a review of acoustic
dissipation in silicates is presented. According to (Krause 1973), for
frequencies varying from $\; 50 \; kHz \;$ to $\; 27 \; GHz \;$, and
temperatures varying from $\; 10 \; K \;$ to $\; 50 \; K \;$, the measured
values of $\; Q \;$ range between $\; 400 \;$ and $\; 2000 \;$.
 We would tend to believe that interstellar grains
are almost certainly have plenty of cracks and
defects. As known from seismology, for real silicate rocks the $\;
Q-$factor is typically between $\; 150 \;$ and $\; 300 \;$.
Therefore we believe  that
$Q^{({\rm sil})} \; < \; 400$.
We are not familiar with the measurements of $Q$ for carbonatious materials,
but it is likely that $Q$ for them will be lower than for silicates.

Therefore, for the silicate grains with axis ratio 1:2 we get:
\begin{equation}
t_{i} \; \approx \;  8 \; \times \; 10^9 \; \frac{1}{\beta^{3/2}} \; \left( \frac{a}{10^{-7} \; m}\protect\right)^{5.5} \; \;,
\label{6.26}
\end{equation}
while for graphite grain we assume that the axis ratio is 1:10 $Q=100$
and obtain
\begin{equation}
t_{i} \; \approx \; 7 \; \times \; 10^6 \; \frac{1}{\beta^{3/2}} \;
\left( \frac{a}{10^{-7} \; m}\protect\right)^{5.5} \; \;,
\label{6.27}
\end{equation}
where the values of $\; a \;$ are supposed to be in $metres$ while $t$ is in
$seconds\;$ (so that for example for $\; a \, = \, 10^{-7} \, m \;$, $\, c/a
\, = \, 1/2 \,$ and $\, \beta \, = \, 1 \,$ the typical time will be  $\,
8 \, \times \, 10^{9} \, s\,$ ) .

We have not attempted to calculate the dissipation rates for grains formed by
loose aggregates of smaller particles.
The inelastic relaxation within such grains may be orders of magnitude
more efficient due to friction between parts of it (``effective viscosity'').
It can well dominate the Barnett relaxation even for much larger grains.

\protect\section{Dynamics of Prolate Cosmic-Dust Grains: Libration.}

At firts glance, the dynamics of a freely-spinning prolate body obeys the same
principles as the dynamics of an oblate one: the axis of maximal inertia will
tend to align itself parallel to the angular momentum. If we model a prolate
body with a symmetric top, it will be once again convenient to choose it be a
prism of dimensions $2a \times 2a \times 2c$, though this time half-size $\, c
 \,$ are larger than $\, a \,$, and therefore $\; I_3 \; = \; I_2 \; > \; I_1
\;$. Then all our calculations {\textit{formally}} remain in force,
up to formula (\ref{5.2}): since now the factor $h-1=[1-(c/a)^2]/[1+(c/a)^2]$
becomes negative, the right-hand side in (\ref{5.2}) will change its sign:
\begin{equation}
\frac{dE_{rot}}{d\theta} \; \; = \; \;
\frac{J^2}{I_3} \; (h \; - \; 1) \; \; \sin \; \theta \; \; \cos \; \theta
\; \; =  \; - \; \omega \; J \; \; \sin \; \theta \; \; \; .
\label{7.1}
\end{equation}
Thereby formula (\ref{5.4}) will also acquire a ``minus'' sign in its
right-hand side:
\begin{equation}
\frac{d\theta}{dt} \; = \; \left(\frac{dE_{rot}}{d\theta}\right)^{-1}
\frac{dE_{rot}}{dt} \; = \; - \; \left( \omega \; J \; \; \sin \; \theta
\right)^{-1} \; \dot{W}
\label{7.2}
\end{equation}
Formulae (\ref{5.3}) and (\ref{5.5}) will remain unaltered.
Eventually, by using (\ref{7.2}) and (\ref{5.5}), we shall arrive to a formula
that differs from (\ref{6.21}) only by a sign, provided we keep the notation
$\theta$ for the angle between $\mathbf{J}$ and the body-frame axis 3
(parallel to dimension $2c$):
\begin{equation}
d \theta/dt \; = \; - \; \frac{a^{-5.5}}{[1+(c/a)^2]^4} \;
\left(\frac{a}{c}\protect\right)^{3/2} \;
\left( \beta \; kT_{\rm gas} \protect\right)^{3/2} \;
\frac{\sin^3 \theta}{\mu \; Q \; \rho^{1/2}} \;
 3^{2.5} \; 2^{-9.5} \; \left(63 \; (c/a)^4 \; \cot^2 \theta \; + \; 20
\protect\right)
\label{7.3}
\end{equation}
This looks like axis 3 tends to stand orthogonal to
$\mathbf{J}$, which seems to be so natural since axis 3 is now not the
maximal-inertia but the minimal-inertia axis.

Alas, all this nice extrapolation of the oblate-body-applicable
approach to a prolate-body case is of no practical interest, bacause
in reality an infinitesimally small deviation between the values of $\; I_2
\;$ and $\; I_3 \;$ leads to a considerably different type of wobble:
the so-called libration (Synge \& Griffith 1959). This phenomenon will
be comprehensively discussed in our next article.

\protect\section{Discussion}

\subsection{Comparison with the Barnett relaxation}

Another important mechanism of internal relaxation, i.e. the Barnett
relaxation, dissipates the energy via oscillating magnetization that
arising from angular velocity precession in grain body coordinates.
Lazarian \& Draine (1997) provide an estimate
$t_B\approx A/\omega^2$, where
$A\approx 7.1 \times 10^{17}(a/10^{-5}~{\rm cm})^2
$~s$^{-1}$ for an oblate grain with 2:1 axis ratio.
For such grains the Barnett dissipation dominates
for grains with $a<2\times 10^{-6}$~m. However, this is not true
for a larger axis ratio. Indeed, according to Lazarian \& Draine (1997)
for $a/c\gg 1$ $t_{\rm B}$ scales as $(a/c)^6$, while we found
above that $t_i$ scales as $(c/a)^{3/2}$. Therefore fore grains
with 4:1 axis ratio the inelastic relaxation dominates if $a>10^{-7}$~m.
 For grains less than $3\times
10^{-8}$~m the
inelastic relaxation becomes more important.
It also dominates for grains constituted by loose aggregates
of smaller particles. The exact value of the $Q$-factor and therefore of
relaxation rates would depend
on the structure and properties of the aggregate.

For grains with axis ratio 2:1 and radii $10^{-7}$~m $< a< 2\times 10^{-6}$~m
the Barnett relaxation
is dominant when grains rotate thermally. Suprathermally rotating grains
stochastically undergo spin-ups and spin-downs and during crossovers
rotate with subthermal velocities.
Although short, in terms of grain alignment,
crossovers, are the most important moments of grain dynamics. Grains
are marginally susceptible to the randomization via gaseous bombardment
when they rotate suprathermally (Purcell 1979). Our calculations of
the inelastic relaxation efficiency show that the Barnett relaxation
is the dominant process that determines internal dissipation during
crossovers within grains with $a>10^{-7}$~m. Therefore results
on paramagnetic and mechanical alignment obtained for such grains
in Lazarian \& Draine (1997) stay unaltered provided that $Q$ factors
of interstellar grains are as high as they were chosen in this paper.

In this paper we assumed that grains are axially symmetric (and
oblate). For grains
of arbitrary shape accelerations will be higher as it tumbles. Therefore
the inelastic relaxation is bound to increase and a relaxation time to
decrease by a numerical factor that will depend on the grain shape.
This subject is beyond the scope of the present paper.

Our treatment of the inelastic relaxation ignored thermal fluctuations in
grain material. In reality, for finite grain temperatures, $\theta$
fluctuates and the thermal distribution proportional to the
Boltzmann factor $\exp(-E_{kin}(\theta)/kT)$ is established as
$t\rightarrow \infty$. To describe the transient processes of alignment
one can solve Fokker-Planck equation as it is done in Lazarian
\& Roberge (1997) in the case of the Barnett relaxation, but to use
coefficients derived in Appendix $E$.

\subsection{Alignment and Internal dissipation}

Several mechanisms of cosmic-dust alignment are known. They constitute
three major types: mechanical, paramagnetic and via radiative torques.
All of them appeal to internal relaxation that enables the alignment
of the angular momentum with the axis of the maximal moment of inertia,
henceforth the axis of major inertia. The degree of achievable alignment
depends on whether grains rotate thermally
\footnote{In Landau and Lifshitz (1969, section 26) it was shown that the
equilibrium position of an inelastic body corresponds to its
maximal-inertia axis being parallel to the angular momentum. However in
reality this statement remains true only up to thermal fluctuations: the
$\it rms$ of the angular deviation is $\Delta \theta =
\sqrt{\it{I}kT/\bf{J}^{2}}$, $\bf J$ and $\it {I}$ being the angular
momentum and the moment of inertia (Lazarian 1994, Lazarian and Roberge
1997).}  (the average rotational energy of a grain is of the order of
the kinetic energy of the surrounding gas)  or
suprathermally (spinning with energies much exceeding $kT$).

We note that all the types of mechanisms below provide the alignment of
the grain angular momentum in respect to the magnetic field.
This happens because grains swiftly precess about magnetic field lines.
This precession is
called into being by the interaction of the grain's magnetic moment with the
field. The said magnetic moment is generated by the Barnett effect
and is thereby parallel to the angular velocity. For a
typical cosmic-dust grain ($\sim~10^{-5}$~cm) in a typical interstellar
magnetic field (5~$\mu$G) the period of Larmor precession is less
than a week (Purcell 1979),
which is much less than the typical time of alignment.

To relate the alignment
of angular momentum to the polarization produced by grains, one has
to know the alignment of grain axis, which is determined
by the internal dissipation.
 However, the alignment of angular momentum, in its turn also
depends on the alignment of grain axis. In the cases of
thermally rotating grains, e.g. Davis \& Greenstein (1951) and Gold (1952)
alignment mechanisms  it is sufficient to know that
the rate of internal dissipation exceeds
the rate of alignment. In cases of suprathermally rotating grains
more precise estimates
of the internal relaxation time are necessary. The latter include
paramagnetic alignment of suprathermally rotating grains (Purcell 1979,
Spitzer \& McGlynn 1979, Lazarian \& Draine 1979),
crossover and cross sectional mechanical alignment (Lazarian 1995,
Lazarian \& Efroimsky 1996, Lazarian, Efroimsky \& Ozik 1996)
and the alignment via radiative torques (Draine \& Weingartner 1996, 1997).
Our finding that for some grains inelastic relaxation dominates is important
and will be accounted in the quantitative studies of alignment elsewhere.

\subsection{Comparison with the earlier work}

Our treatment differs from that in (Purcell 1979) in a number of points.

I. We have obtained different expressions for stresses. In particular, our
stresses do obey the boundary conditions: they vanish on the grain boundary
(while the stresses in (Purcell 1979) fail to do so).

II. We have found energy dissipation at the double frequency. It is possible
to show that the double-frequency dissipation provides a leading contribution
\footnote{For example, in formula (\ref{7.3}) the term $\; 63(c/a)^4 \,
\cot^2 \theta \;$ in the sum in brackets is due to damping at the precession
frequency $\; \omega \;$, while the term $\; 20 \;$ is due to damping at the
double frequency. Evidently, even for $\; c/a \approx 1/2 \;$ the double
frequency provides the leading input, while for $\; c/a \approx 1/10 \;$
its input becomes absolutely overwhelming}. The dissipation at the double
frequency was not considered in (Purcell 1979).

As a result, our final expression for the same $\; Q$-factor predicts higher
inelastic relaxation efficiency than the corresponding one in (Purcell 1979).
To make a comparison
we rewrite our formulae in a form similar to that in Purcell (1979).
Then our expression (\ref{6.21}) would read:
\begin{equation}
\left( \frac{d \theta}{dt} \protect\right)_{_{\mathit{our \; \; result}}} \;
\approx \; - \; \frac{8}{5} \; \frac{\Omega^3_0 \; \rho \;
a^2}{ \mu \; Q \; \left[1 \; + \; (c/a)^2 \protect\right]^4} \;
\left(0.4 \; (c/a)^4 \; \cos^2 \theta \; \sin \theta \; +\;0.1 \protect\right)
\label{8.1}
\end{equation}
In the appropriate expression presented in Purcell's article, the second term
in brackets was missing (because Purcell missed the contribution from the
dissipation at the double-frequency mode), while the first term is about 2.5
times larger (because Purcell's calculation of the stress tensor ignored the
boundary conditions upon stresses).
One may see that the our calculations reveal that
the inelastic relaxation being much more efficient than predicted in
Purcell (1979).
For example, if we define a typical alignment time $\; t_i \;$ as the value of
$\; (d \theta/ dt)^{-1} \;$ at $\; \theta \, = \, \pi/4 \;$ then, for silicate
grains with $\; c/a \; \approx \; 1/2 \;$, ``our'' typical time derived from
(\ref{8.1}) will be 23 times less than the appropriate estimate\footnote{
Comparing our results with that of Purcell, mind that the value of $\; t_i \;$
obtained in (Purcell 1979) was for the degree of suprathermality $\; \beta \,
= \, 100 \;$.} derived from Purcell's formula. For graphite
grains with $\; c/a \; \approx \; 1/10 \;$, ``our'' typical time will be
almost 1200 times shorter than the estimate in Purcell (1979).

III. In reality, the difference between our calculation and the one by Purcell
will be even bigger: due to imperfectness of the grain material, we would
choose for the $\; Q-$factor lower values than the one suggested by Purcell.
(See the end of Section VI for discussion.)

\section{Conclusions}

The principal results of this paper are as follows:\\

I. The stresses arising from tumbling of a grain deform it and
 change its moment of inertia, but this change lags behind grain
angular velocity due to non-elastic effects. This process causes
alignment of grain angular momentum with the grain axis of
maximal moment of inertia.

II. Deformations in tumbling grain happen both at the frequency of
precession and at a double frequency.

III. Inelastic relaxation dominates the Barnett relaxation
for (1) grains with large axis ratio, (2) grains with low $Q$ factor,
i.e. fractal grains produced by coagulation, (3) grains with
$a>2$~$\mu$m, (4) suprathermally rotating grains.

{\bf Acknowledgments}

The authors are deeply thankful to Eric Heller for his encouragement
and attention to the work, and to Patrick Thaddeus for moving
discussions. A.L. would like to acknowledge numerous stimulating
conversations with Bruce Draine. M.E. wishes to express his thanks to Frans
Spaepen, Michael Aziz, Martin Bazant and other participants of the Harvard
Condensed Matter Theory weekly seminar, whose advises on the properties of
silicates were of an obvious importance for the afore presented study. We also
want to acknowledge a very useful month-long e-mail conversation on the
properties of materials, that Michael Ashby kindly and so patiently had with
us.

\pagebreak

redpajama_set_name: RedPajamaArXiv


A.J. Finberg, A Complete Inventory of the Drawings of the Turner Bequest, London 1909, vol.I, p.171 LXIX 14 (as 'Sheep grazing under walls of a ruined castle' c.1800–2).
Drawn with the page turned horizontally, this is apparently one of the series of ink studies adumbrating a view of Flint Castle; see folios 9 recto, 15 recto, 16 recto (D04003, D04009, D04010), and 114 recto (D04132; Turner Bequest LXIX 115).
Andrew Wilton, 'A Shepherd with Sheep and Cows near the Walls of a Ruined Castle c.1799–1802 by Joseph Mallord William Turner', catalogue entry, May 2013, in David Blayney Brown (ed.), J.M.W. Turner: Sketchbooks, Drawings and Watercolours, Tate Research Publication, April 2016, https://www.tate.org.uk/art/research-publications/jmw-turner/joseph-mallord-william-turner-a-shepherd-with-sheep-and-cows-near-the-walls-of-a-ruined-r1177886, accessed 24 April 2019.