text:
Cancer-fighting viruses win approval
Heidi Ledford
1704 Altmetric
US regulators clear a viral melanoma therapy, paving the way for a promising field with a chequered past.
Killer T cells (orange) are recruited to attack malignant cells (mauve) in the viral-based cancer therapy T-VEC. Credit: Dr. Andrejs Liepins/SPL
An engineered herpesvirus that provokes an immune response against cancer has become the first treatment of its kind to be approved for use in the United States, paving the way for a long-awaited class of therapies. On 27 October, the US Food and Drug Administration (FDA) approved a genetically engineered virus called talimogene laherparepvec (T-VEC) to treat advanced melanoma. Four days earlier, advisers to the European Medicines Agency had endorsed the drug.
With dozens of ongoing clinical trials of similar 'oncolytic' viruses, researchers hope that the approval will generate the enthusiasm and cash needed to spur further development of the approach. "The era of the oncolytic virus is probably here," says Stephen Russell, a cancer researcher and haematologist at the Mayo Clinic in Rochester, Minnesota. "I expect to see a great deal happening over the next few years."
Cancer treatment: The killer within
Many viruses preferentially infect cancer cells. Malignancy can suppress normal antiviral responses, and sometimes the mutations that drive tumour growth also make cells more susceptible to infection. Viral infection can thus ravage a tumour while leaving abutting healthy cells untouched, says Brad Thompson, president of the pharmaceutical-development firm Oncolytics Biotech in Calgary, Canada.
Early attempts
The strategy builds on a phenomenon that has been appreciated for more than a century. Physicians in the 1800s noted that their cancer patients sometimes unexpectedly went into remission after experiencing a viral infection. These case reports later inspired doctors, particularly in the 1950s and 1960s, to raid nature's viral cupboard. Clinicians injected cancer patients with a menagerie of viruses. Sometimes the therapy destroyed the tumour, and sometimes it killed the person instead.
Unlike the wild viruses used in those mid-twentieth-century experiments, some of today's anti-cancer viruses are painstakingly engineered. T-VEC, for example, has been altered to drastically reduce its ability to cause herpes. Researchers also inserted a gene encoding a protein that stimulates the immune system, which makes the virus even more potent against cancer (see 'Going viral against cancer').
As more researchers entered the field and initiated small clinical tests, they began to produce enticing anecdotes. Russell recalls the case of an individual with myeloma who remained sick after undergoing two stem-cell transplants. A tumour on the left side of her forehead had degraded the bone underneath and was putting pressure on her brain. Yet treatment with an experimental virus sent her into complete remission (S. Russell et al. Mayo Clin. Proc. 89, 926–933; 2014). "She's a star patient who convinced us that this oncolytic paradigm can really work," he says.
HPV: Sex, cancer and a virus
But statistics — not anecdotes — rule over drug approvals. In 2005, regulators in China approved an oncolytic adenovirus called H101 to treat head-and-neck cancer, after evidence showed that the treatment could shrink tumours. Those trials stopped short of assessing improvements in patient survival — a measure often required for FDA approval. Since then, a medical-tourism industry has built up in China for people who cannot get the therapy in their home countries.
Then, in May this year, a team supported by biotechnology giant Amgen of Thousand Oaks, California, published promising results from a large clinical trial of T-VEC (R. H. Andtbacka et al. J. Clin. Oncol. 33, 2780–2788; 2015). The virus both shrank tumours in people with advanced melanoma and extended patient survival by a median of 4.4 months. Yet statistically, survival benefits fell just a hair's breadth of significance. "That raised the question, 'Well, what is statistical significance? Is this an active agent or not?'" Russell says.
He and others note that the therapy — which must be injected directly into tumours — seemed to rein in cancer elsewhere in the body as well. This is a sign that results are real and that the virus sparked an immune response as intended, Thompson says.
Administering T-VEC in combination with cancer immunotherapy could prove particularly effective, notes Stephen Hodi, an oncologist at the Dana-Farber Cancer Institute in Boston, Massachusetts. In June 2014, a small clinical trial by Amgen suggested that this combination may boost effectiveness over that of the immunotherapies alone.
And researchers continue to look for ways to improve T-VEC. In particular, they would like to be able to deliver the therapy systemically, so that the virus could target tumours in organs that are difficult to reach with an injection. This would require a technique to prevent the body from mounting an immune response to the virus prematurely, which would disable it before it could reach and kill tumour cells, says Howard Kaufman, a cancer researcher at Rutgers Cancer Institute of New Jersey.
Cancer: The Ras renaissance
To that end, those in the field are experimenting with a smorgasbord of viruses — from poxviruses to vesicular stomatitis virus, which does not normally infect humans but causes a blistering disease in cattle. Oncolytics Biotech is studying a virus that hitch-hikes through the body on certain blood cells, camouflaged from the immune system.
If cancer-killing viruses could be delivered to their targets through the bloodstream, rather than via injection directly into the tumour, they could be used to treat a greater range of cancers. Thompson envisions a day when physicians will be able to peruse a menu of oncolytic viruses and select the best fit. "Each virus interacts with the immune system differently," he says. "They could have a role in pretty much all cancer therapy."
Tweet Follow @NatureNews
Related links in Nature Research
World's largest cancer charity lays out field's grand challenges 2015-Oct-12
Cancer treatment: The killer within 2014-Apr-02
HPV: Sex, cancer and a virus 2013-Nov-20
Immunotherapy's cancer remit widens 2013-May-28
Nature Outlook: Cancer immunotherapy
Related external links
US National Cancer Institute: Biotherapies for Cancer
Ledford, H. Cancer-fighting viruses win approval. Nature 526, 622–623 (2015). https://doi.org/10.1038/526622a
Issue Date: 29 October 2015
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\section{Introduction}

At very low temperatures, glasses and other amorphous systems show
similar thermal, acoustic and dielectric properties \cite{Esq98},
which are in turn very different from those of crystalline solids.
Below $1 \rm{K}$, the specific heat $C_v$ of dielectric glasses is
much larger and the thermal conductivity $\kappa$ orders of
magnitude lower than the corresponding values found in their
crystalline counterparts. $C_v$ depends approximately linearly and
$\kappa$ almost quadratically on temperature\cite{Zel71}, respectively. This is
in clear contrast to the cubic dependence observed in crystals for
both properties, well understood in terms of Debye's theory of
lattice vibrations. Above $1 \rm{K}$, $C_v$ still deviates
strongly from the expected cubic dependence, exhibiting a hump in
$C_v/T^3$ which is directly related to the so-called boson peak
observed by neutron or Raman vibrational
spectroscopies\cite{Ahm86}. To explain these, it was considered
that atoms, or groups of atoms, are tunnelling between two
equilibrium positions, the two minima of a double well potential.
The model is known as the two level system (TLS) \cite{Phi72,
And72}. In the standard TLS model, these tunnelling excitations
are considered as independent, and some specific assumptions are
made regarding the parameters that characterize them (for a review
see for example Ref.~\onlinecite{Esq98}).

In recent years an intriguing magnetic-field dependence of the
dielectric and coherent properties of some insulating glasses was
reported. In 1998 Strehlow et al. observed a sharp kink at $T=5.84
\; mK$ in the dielectric constant of the multi-component glass
$BaO-Al_2O_3-SiO_2$ \cite{Str98} measured in very weak magnetic
field, of the order of $10 \; \mu} \def\n{\nu} \def\x{\xi} \def\p{\pi T$. The effect was several orders
of magnitudes larger than what is expected, considering the absence of
magnetic impurities in this insulator. Later studies carried on in
magnetic field ranging up to $25 \; T$ and temperatures well below
$100 \; mK$ revealed for several materials a complex and strong
dependence of the dielectric response on the external magnetic
field, on the applied voltage and on temperature \cite{Str98,
Str00, Woh01, Lec02, Hau04}. Even more surprising phenomena were
observed in the spontaneous polarization echoes experiments on
$BaO-Al_2O_3-SiO_2$ \cite{Lud02}.
The increase of the echo amplitude by a factor of $4$
was reported when varying the magnetic field
in the range of $0$ to $200 \; mT$. Similar results were later
reported in the case of amorphous mixed crystal $KBr_{1-x}CN_x$
\cite{Ens02}. Until now a unified theory does not exist while some
contradictions are present in the recent works.
We have  clear evidence that such
intriguing magnetic properties are in connection with the low-energy
tunnelling excitations present in almost all amorphous solids.
These tunnelling states are known to be proved very well by
the echo experiments \cite{Wue97}.
On the other hand these
excitations have been reported as showing up at very low temperatures
(usually $T<100 \; mk$), where the TLS are responsible for the
thermal and dynamical properties of glasses \cite{Ens02, Hun00}.

Several generalizations of the standard TLS model have been reported after
the anomalous behavior of glasses in a magnetic field.
The main question for such models is how a TLS should
interact with the magnetic field, being not clear how the
tunnelling entity would acquire a finite magnetic moment.
According to the proposed solutions, the models can be divided into
"orbital" and "spin" models.

In the orbital models, the tunnelling entities are not simple
two-state systems, but performing some kind of circular motions.
Due to the presence of the magnetic field, a charged particle
moves on a loop encloses a magnetic flux and thus can acquire an
Aharonov-Bohm phase. In order to obtain such a closed trajectory,
a "Mexican-hat potential" was proposed by Kettemann, instead of
the usual double-well potential \cite {Ket99}. The resulting flux
$\phi =\pi r^{2}B$, with $r$ being the hat radius, proves to be by
several orders of magnitudes smaller than the flux $\phi
_{0}=h/e$. Even though the existence of cluster configurations of
atoms or molecules containing up to $N=200$ units which contribute
in a collective tunnelling were reported, (see for example
\cite{Lub01} and references therein) it is improbable that
such an effect can be extended to an amount of the order of $N
\sim 10^5$. A detailed discussion of this aspect has been
presented in Ref.\onlinecite{Lan02}. The combination of
the Aharonov-Bohm effect with a flip-flop configuration of the two
interacting TLS has been proposed
to explain the dependence of the echo amplitudes on the applied magnetic field
\cite{Akb05}. A different modality to consider the occurrence of
an apparent flux phase having the correct order of magnitude was
proposed by W\"{u}rger \cite{Wue02}. The mechanism proposed
consist of pair of coupled TLS and a non-linear coupling to the
external voltage. Still, the required degeneracy of the nearby TLS
is not in accordance with the known distribution for the
tunnel splitting. Another possibility for the formation of
closed loops was proposed by Le Cochec \cite{Lec02}, namely a
jagged and uneven potential landscape between the two wells.

To conclude, the "orbital" models can provide an explanation for
some of the magnetic field effects by considering the flux
dependance of the tunnelling splitting. Unfortunately, some
assumptions have been made which cannot be reconciled with the standard
features of the tunnelling model.

The spin models\cite{Wue02a} provide an alternative mechanism for
the observed magnetic properties, especially for the polarization
echoes\cite{Wue04}. A direct coupling between the nuclear spin of
the tunnelling entities and the applied magnetic field is
considered. We can easily notice that the multi-component glasses
used in the echo experiments contain one or several kind of atoms
that carry a nuclear quadrupole. For example in the case of
$BaO-Al_2O_3-SiO_2$ we find the abundant isotopes $^{27}Al(I=5/2)$
and less frequent $^{137}Ba(I=3/2)$; for the boro-silicate we find
the abundant isotope $^{11}B(I=3/2)$. However, the echo
experiments convincingly put into evidence the role of the nuclear
quadrupole moments in glasses. According to this model the
magnetic properties should not be measured in materials whose
nuclei contains no finite quadrupole moment. Until now no
counter-example has been reported. Moreover, recently the role of the
quadrupole moments have been systematically studied and confirmed
by the echo experiments in ordinary glycerol $C_3H_8O_3$ and
deuterated one $C_3D_8O_3$ \cite{Nag04}.

The question, wether or not the nuclear quadrupole should be
responsible for the magnetic field dependence of the dielectric
constant seems to be natural. Relevant experiments involve
temperature where the thermal energy is significantly larger than
the quadrupole splitting. For thermal tunnelling systems, with TLS
splitting ($E$) much larger than the quadrupole splitting, it was
proven already that the resonant, or van Vleck susceptibility is
rather insensitive to the nuclear quadrupole motion \cite{Dor04}.
For such systems  the relaxation or Debye part of the
susceptibility shows a large magnetic field dependence
\cite{Dor06}. By considering TLS with small splitting comparable
to the nuclear quadrupole energies and using a numerical
estimation, a pronounced dependence of the electric permittivity
of the applied magnetic field was obtained \cite{Pol05}.

In this article we have studied the magnetic field dependence of
the dielectric susceptibility and the heat capacity in cold
glasses taking into account the quadrupole effects. The general expressions
in the mentioned cases are complex, however, we have obtained these
thermodynamic behaviour in the small and large magnetic fields limit.
 In Section II we introduce the nuclear
spins in the frame of the TLS. The resulting magnetic field
dependent part of the susceptibility is given in the Section III,
using the perturbation expansion  for two different regimes,
namely small and large magnetic fields. We end this section with a
discussion of our results and a comparison with available
experimental data and previous calculations.
In section IV, the specific heat has been
calculated in  three separate terms, namely TLS part; angular
momentum and nuclear spin parts. Finally the conclusion is
presented with a discussion of our results and a comparison with
available experimental data\cite{Str04}.

\section{TLS with a nuclear spin}

The standard TLS can be described as a particle or a small group
of particles moving in an effective double-well potential (DWP).
At very low temperatures, only the ground states corresponding to
the two wells are relevant. Using a pseudo-spin representation the
Hamiltonian of such a TLS is written
\begin{equation}
H_{TLS} \; = \; \frac{1}{2}\Delta _{0}\sigma _{x} +
\frac{1}{2}\Delta \sigma _{z},
\label{eq1}
\end{equation}
where $\sigma _{z}$ is the reduced two-state coordinate,
\begin{displaymath} \sigma _{z} \; = \; \mid L \rangle \langle L \mid - \mid R \rangle \langle
R \mid.
\end{displaymath}
The eigenvalues of $\sigma _{z}$ are $\pm 1$, labelling the localized states in the
two wells (left well $L$, right well $R$), while the tunnelling matrix is
taken into account by
\begin{displaymath} \sigma _{x} \; = \; \mid L \rangle \langle R \mid + \mid R \rangle \langle
L \mid. \end{displaymath}
We denoted by $\Delta $ the energy off-set at the bottom of the
wells, and by $\Delta _{0}$\ the tunnel matrix element. According
to the randomness of the glassy structure, the energy difference
between the two wells
\begin{equation}
E \; = \; \sqrt{\Delta _{0}^{2}+ \Delta ^{2}},
\label{eq2}
\end{equation}
have a broad distribution. The energy off-set and the tunnelling
matrix element obey a distribution law
\begin{equation}
{\cal P}\left( \Delta ,\Delta _{0}\right) \; = \;
\frac{P_{0}}{\Delta _{0}},
\label{eq3}
\end{equation}
where $P_0$ is a constant. It is useful to define new spin
operators using the relations
\begin{eqnarray} \sigma_x \; = \; u \sigma_E - w \sigma_A, \nonumber\\ \sigma_z \; = \;
w \sigma_E + u \sigma_A,
\label{eq4} \end{eqnarray}
such that the TLS Hamiltonian becomes diagonal
\begin{equation} H^D_{TLS} \; = \; \frac{E}{2} \; \sigma_E.
\label{eq5} \end{equation}
We used the notations
\begin{equation} u \; = \; \frac{\Delta_0}{E} \; \; , \; \; w \; = \;
\frac{\Delta}{E}
\label{eq6}
\end{equation}
which satisfy $u^2 + w^2 =1$.

For the moment there is no rigorous theory for tunnelling in
glasses. It is assumed that atoms or groups of atoms participate
in one TLS. As we mentioned before, in the case of the
multi-component glasses, one or several of the tunnelling atoms
carry a nuclear magnetic dipole and an electric quadrupole. When
the system moves from one well to another, the atoms change their
positions by a fraction of an \AA ngstr\"{o}m.

We can describe the internal motion of the nuclei by a nuclear
spin ${\bf I}$ of absolute value ${\bf I}^2=\hbar^2I(I+1)$. This
is related to a magnetic dipole moment $g \mu_N {\bf I}/\hbar$,
where $g$ is the Land\'{e} factor and $\mu _ N \approx 5\times
10^{-27}J/T$ is the nuclear magneton. The magnetic dipole couples
to  an external magnetic field ${\bf B}=B e_B$ and give rise to
a Zeeman term
\begin{equation} H_Z^{(1)} \; = \; - \epsilon_z e_B\cdot I
\label{eq7}
\end{equation}
where the frequency $\epsilon_z = g \mu_N B /\hbar$ is directly
proportional to $B$.

For a nucleus with spin quantum number $I \geq 1$ the charge
distribution $\rho({\bf r})$ is not isotropic. Besides the charge
monopole, an electric quadrupole moment can be defined with
respect to an axis ${\bf e}$
\begin{equation}
Q=\int d^{3}r\left[ 3({\bf r\cdot e})^{2}-{\bf r}^{2}\right] \rho
({\bf r}).
\label{eq8}
\end{equation}
This can couple to an electric field gradient (EFG) at the nuclear
position, expressed by the curvature of the crystal field
potential. The potential describing this coupling reads as
\cite{Abr89}
\begin{equation}
V_Q \; = \;
\frac{-eQ}{I(2I-1)}[V_{11}I_{1}^{2}+V_{22}I_{2}^{2}+V_{33}I_{3}^{2}].
\label{eq9}
\end{equation}
The bases used here $(e_1,e_2,e_3)$ are the principal axes of the
tensor $V_{ij}$ which describes the electric field gradient, and
$e$ is the electron charge. According to the Laplace equation the
potential obey $V_{11}+V_{22}+V_{33}=0$. If we define the
asymmetry parameter $\eta = \frac{V_{22}-V_{33}}{V_{11}}$, the
quadrupole potential can be expressed as:
\begin{equation}
V_Q \; = \;\epsilon_q[3I_{1}^{2}+\eta (I^{2}_{2}-I_{3}^{2})-I^{2}]
\label{eq10}
\end{equation}
where we denote by $\epsilon_q=\frac{-eQ V_{11}}{4I(2I-1)}$ the
quadrupole coupling constant.

\section{Static dielectric susceptibility}

Our purpose is to determine the magnetic field dependent part of
the static dielectric susceptibility for a TLS coupled with a
nuclear quadrupole.
Therefore we have to consider the interaction of the dipole
operator $(1/2)P \sigma_z$ with an external electric field $F$.
The dipole moment arises from the relative motion of partial
charges related to the atoms forming the tunnel system. This
interaction will be described by
\begin{equation}  V_F=(P \cdot F)\sigma_z.
\label{eq11}
\end{equation}
 If we denote
\begin{equation}
Z \; = \; {\rm Tr}\left\{ e^{-\beta H}\right\},
\label{eq12}
\end{equation}
the partition function of the system, where $H$ is the total
Hamiltonian; the trace, denoted later by $<...>$ involves
two-state and spin variables. Here
 $\beta =
\frac{1}{ K_BT}$, $K_B$ is the Boltzmann constant and $T$ is
temperature. We can express the susceptibility as
\begin{equation}
\widehat{\chi}\; = \; \overline{-\left\langle \frac{\partial
^{2}f}{\partial F^{2}}\right\rangle},
\label{eq13}
\end{equation}
where $f=-\frac{1}{\beta }\ln Z$ represents the free energy. The
statistical average implies also an integration over the two
parameters of the TLS, the energy off-set and the tunnelling
matrix element, according to the distribution law given by Eq.
(\ref{eq3}), as well as an integration over the nuclear quadrupole
parameters; this step, denoted by $\overline{(...)}$, will be
discussed at the end of the calculation. We can write the partition
function in terms of the energy levels $E_{i,m}$ (eigenvalues of
the total Hamiltonian)
\begin{equation} Z=\sum_{i=L/R} \sum_{m=-I}^Ie^{-\beta E_{i,m}}.
\label{eq14}
 \end{equation}
Therefore, the susceptibility contains three terms:
\begin{eqnarray} \chi^{(1)} &=& -\frac{1}{Z}\sum_{i,m}e^{-\beta
E_{i,m}}\frac{\partial ^{2}}{\partial F^{2}} E_{i,m}, \nonumber\\
\chi^{(2)} &=& \frac{\beta}{Z}\sum_{i,m}e^{-\beta E_{i,m}}
\left[\frac{\partial }{\partial F} E_{i,m}\right]^{2}, \nonumber\\
\chi^{(3)} &=&-\frac{\beta}{Z^2}\left[\sum_{i,m}e^{-\beta E_{i,m}}
\frac{\partial }{\partial F} E_{i,m}\right]^{2}.
\label{eq15} \end{eqnarray}
The eigen energies ($E_{i,m}$) of the full Hamiltonian can not be obtained analytically
in the general case because the Zeeman and nuclear quadrupole terms do not commute.
In this respect we will consider two different regimes where we can obtain the
analytic results in a perturbation frame work.
In order to develop a perturbation expansion we will consider
two separate cases, a small external magnetic field
and a large one, respectively.

\subsection{Small magnetic field}
We begin with the case in which the Zeeman term is smaller than
the nuclear quadrupole potential, $\varepsilon_z\ll
\varepsilon_q$, thus the Zeeman term is  treated as a weak
perturbation. If we consider the simplest symmetric case,
$\eta=0$, then the nuclear quadrupole potential in the $I_1$ basis
has the following representation
\begin{eqnarray} H_q=\left[ \varepsilon_q^L{\cal
V}_{I,m}(\frac{1+\sigma_z}{2})+ \varepsilon_q^R{\cal
V}_{I,m}(\frac{1-\sigma_z}{2})\right]|I,m\rangle \langle I,m|, \nonumber\\
\label{eq16}
\end{eqnarray}
where we have defined  ${\cal V}_{I,m}=3m^2-I(I+1)$, and
$\varepsilon_q^{L/R}$ as the quadrupole coupling constant in the
left and the right well.
Therefore, in the $\sigma_E$ basis we can write
\begin{eqnarray} H_q=\left[ \varepsilon_q{\cal V}_{I,m} + \varepsilon_q'{\cal
V}_{I,m}(w \sigma_E+u\sigma_A)\right]|I,m\rangle \langle I,m|.
\label{eq17}
\end{eqnarray}
Here we have defined $\varepsilon_q=\frac{1}{2}({ \varepsilon_q^L+
\varepsilon_q^R})$ and $\varepsilon_q'=\frac{1}{2}({
\varepsilon_q^L- \varepsilon_q^R})$.
We can split the total Hamiltonian in two parts,
\begin{eqnarray} H=H^{(0)}+\delta H,
\label{eq18}
\end{eqnarray}
where the unperturbed part is
\begin{eqnarray} H^{(0)}=\left[\frac{E}{2}\sigma_E+\varepsilon_q{\cal
V}_{I,m}-m\varepsilon_z\cos\theta'\right]|I,m\rangle \langle I,m|,
\label{eq19}
\end{eqnarray}
and the perturbation term is
\begin{eqnarray} \delta H&=&\left[( \varepsilon_q'{\cal V}_{I,m}+PF)(w
\sigma_E+u\sigma_A)\right]|I,m\rangle \langle I,m|
\nonumber\\ &-& \alpha_{-}^{m'}\varepsilon_z |I,m\rangle \langle I,m+1|
\nonumber\\&-&\alpha_{+}^{m'}\varepsilon_z |I,m\rangle \langle I,m-1|.
\label{eq20}
\end{eqnarray}
We have defined
\begin{eqnarray} \alpha_{\pm}^{m'}=\sqrt{I(I+1)-m(m\pm 1)}\sin\theta'e^{-[\pm
i \phi' ]},
\label{eq21}
 \end{eqnarray}
where $\theta'$ and $\phi'$ are the direction of magnetic field
in the basis  $(e_1, e_2, e_3)$; here we have assumed that $e_1=e_z$.

We apply the perturbation theory to get the second order correction for the
energy levels of the system, $E_{i,m}$. Therefore, the first
order correction is
\begin{eqnarray} \delta E^{(1)}_{i,m}\; = \; w(P\cdot F+\varepsilon_q'{\cal
V}_{I,m})\langle i |\sigma_E|i\rangle.
\label{eq22}
\end{eqnarray}
For the sake of simplicity we use $\langle i |\sigma_E|i\rangle$
instead of $\langle i,m |\sigma_E|i,m\rangle$, which is independent of the
magnitude of $m$. The second order correction of the energy reads
\begin{eqnarray} &&\delta E_{i,m}^{(2)}= \sum_{j,n|(j,n)\neq(i,m)} \frac{|\langle
i,m |\delta H|j,n\rangle|^2}{E^{(0)}_{i,m}-E^{(0)}_{j,n}}=
 \nonumber\\&&\epsilon_z^2
\left[
 \frac{|\alpha^{'m}_{+}|^2}{(2m-1)\varepsilon_q-\varepsilon_z\cos\theta'}\right.-
\left.\frac{|\alpha^{'m}_{-}|^2}{(2m+1)\varepsilon_q-\varepsilon_z\cos\theta'}\right]
\nonumber\\&&+ \frac{u^2}{E}(P\cdot F+\varepsilon_q'{\cal V}_{I,m})^2\langle
i |\sigma_E|i\rangle,
\label{eq23}
\end{eqnarray}
where $E^{(0)}_{i,m}$ is the eigenvalue of the unperturbed part of
the Hamiltonian $H^{(0)}$. Taking into account the second order
correction of energy, the susceptibility is calculated by
Eq.(\ref{eq15}). By using the following relations
\begin{displaymath} \frac{\partial }{\partial F} E_{i,m}\vert_{F=0} \; = \; wP\langle
i |\sigma_E|i\rangle,
\label{eq24}
\end{displaymath}
and
\begin{eqnarray} \frac{\partial ^2}{\partial F^2} E_{i,m}\vert_{F=0}=
\frac{2P^2u^2}{E}\langle i |\sigma_E|i\rangle,
\label{eq24b}
\end{eqnarray}
we can immediately express the second term of the susceptibility
\begin{equation} \chi_{2}\;=\;\beta P^2w^2.
\label{eq25}
\end{equation}
The third term is
\begin{equation} \chi_{3}\;=\; -P^2w^2\beta t^2,
\label{eq26}
\end{equation}
where we defined  $t=\tanh(\beta E/2)$. The first
term of the susceptibility contain the second derivative of the
energy levels with respect to the electric field. After some simple
algebra $\chi_1$, can be written as
\begin{eqnarray} \chi_{1}&=& \frac{2P^2u^2}{ZE} \sum_{m} \left[e^{-\beta
E_{d,m}}-e^{-\beta E_{u,m}}\right],
\label{eq27}
\end{eqnarray}
where $Z=\sum_{m} \left[e^{-\beta E_{d,m}}+e^{-\beta
E_{u,m}}\right]$. Defining $\Lambda_m$ as
\begin{eqnarray} &&\Lambda_m=\varepsilon_q{\cal
V}_{I,m}-\varepsilon_zm\cos\theta' + \nonumber\\
&&
 \epsilon_z^2
\left(
 \frac{|\alpha^{'m}_{+}|^2}{(2m-1)\varepsilon_q-\varepsilon_z\cos\theta'}
 -\frac{|\alpha^{'m}_{-}|^2}{(2m+1)\varepsilon_q-\varepsilon_z\cos\theta'}\right),
\nonumber
\label{eq27b}
\end{eqnarray}
one can write that
\begin{eqnarray} E_{i,m}\vert_{F=0}=\left[\frac{E}{2}+
 \frac{u^2}{E}(\varepsilon_q'{\cal
V}_{I,m})^2\right]\langle i |\sigma_E|i\rangle+\Lambda_m. \;\;\;\;\;\;%
\label{eq28}
\end{eqnarray}
The partition function can be written as $ Z=\sum_{m}
 \cosh
\left[\frac{\beta E}{2}+
 \frac{u^2\beta}{E}(\varepsilon_q'{\cal
V}_{I,m})^2\right]
 e^{-\beta \Lambda_m}$.
Then the first term of the susceptibility becomes
\begin{eqnarray} \chi_{1}&=& \frac{2P^2u^2}{ZE} \sum_{m}
 \sinh
\left[\frac{\beta E}{2}+
 \frac{u^2\beta}{E}(\varepsilon_q'{\cal
V}_{I,m})^2\right]
 e^{-\beta \Lambda_m}.
 \nonumber\\
 %
\label{eq29}
 \end{eqnarray}
If at this point we assume $\varepsilon'_q=0$ and $B=0$ we will simply
recover the result of standard TLS model, $\chi_{1}=
\frac{2P^2u^2t}{E}$.

\subsubsection{Thermal expansion}

The experiments we are addressing are performed at temperatures
between few tens and few hundreds of milliKelvin, where the
thermal energy $k_BT \sim 10^{-24} J$ is few orders of magnitude
larger than the Zeeman term (at $B=1T$ for example $\mu_N B/k_B
\sim 0.4 mK$) or the quadrupole coupling (echo experiments
suggested that $\epsilon_q \sim \epsilon_z$ at $200mT$, so smaller
than one $mK$). We can conclude that the parameters $\beta
\epsilon_z$, $\beta \epsilon_q$  are small, which justify an
expansion of $\chi_{1}$ and the exponential factors like $e^{\pm
\beta \epsilon_{z,q}}$. Carrying out such an expansion in the
thermal energy,
 we can
easily show that
\begin{eqnarray}
&&\Delta\chi_{1}(B)\simeq\frac{2P^2u^4}{(2I+1)E^2}
(1-t^2)\varepsilon_q'^2
 \beta^2 \epsilon_z^2\times
 %
 \nonumber\\
 &&\sum_{m}
\left[
 \frac{|\alpha^{'m}_{+}|^2}{(2m-1)\varepsilon_q-\varepsilon_z\cos\theta'}
 -\frac{|\alpha^{'m}_{-}|^2}{(2m+1)\varepsilon_q-\varepsilon_z\cos\theta'}\right].
 \nonumber\\
 %
\label{eq30}
 \end{eqnarray}

In order to complete the calculation of the magnetic field
dependent part of the static susceptibility we need to perform the
integral over the TLS parameters, considering the distribution
function defined in Eq. (\ref{eq3}), for this regime
($\varepsilon_z\ll \varepsilon_q$), which can be rewritten as
\begin{eqnarray} \Delta\chi_{1}(B)\simeq \frac{2P^2\kappa\xi}{(2I+1)} \beta^3
\varepsilon_q'^2 \frac{\epsilon_z^2}{\varepsilon_q}
\label{eq31}
\end{eqnarray}
 where $\kappa=\left[\frac{1}{2I+1}
(\sum_{m}{\cal V}_{I,m}^2)(\sum_{m}\alpha_m )- \sum_{m}{\cal
V}_{I,m}^2\alpha_m\right]$, $\alpha_m=\sum_{m}[
\frac{|\alpha^{'m}_{+}|^2}{(2m-1)}
 -\frac{|\alpha^{'m}_{-}|^2}{(2m+1)}]$, and $\xi$
 is a constant (nonmagnetic-dependent) which
  comes from averaging over the TLS parameters which depends on
 the magnitude of  the lower
and upper bound of the two-level splitting, $\Delta_{0min}$ and
$E_{max}$. It can be easily shown that in the case of same EFG in
the two wells ($\e_q^L = \e_q^R$), the dependence of the static
dielectric susceptibility on the magnetic field will vanish.

\subsection{Large magnetic field}

 Let us denote by $(e_1, e_2, e_3)$ the basis
in the left well and by $(e_1^{\prime}, e_2^{\prime},e_3^{\prime})$ the
corresponding one in the right well. If we suppose that
$e_3^{\prime}\parallel e_3\parallel \hat{z}$ and $e_1^{\prime},e_1$
make an angle $\theta$ with each other \cite{Pol05}, we can
write the total Hamiltonian in the basis of the operator $I_3=I_z$
in the following form
\begin{eqnarray}
 \left(\begin{array}{cc} H^L+ \left( \frac{\Delta}{2}+PF
\right)\cdot\textbf{1
}&\frac{\Delta_0}{2}\cdot\textbf{1 }\\
\frac{\Delta_0}{2}\cdot\textbf{1 }&H^R-\left( \frac{\Delta}{2}+PF
\right)\cdot\textbf{1 }\end{array}\right) \vspace{0.5cm}
\label{eq32}
\end{eqnarray}
where $\textbf{1}$ is the unit matrix of rank $(2I+1)$.
$H^L=V_Q^L+H_Z^{(1)L}$ and $H^R=V_Q^R+H_Z^{(1)R}$ are defined for
particles in the left and right wells, respectively. The state in
each well (L or R) can be characterized by $\mid
\psi^{L(R)}_{I,m}\rangle=\mid L(R)\rangle\bigotimes \mid I,m\rangle$ and
consequently the matrix element of the Hamiltonian is defined by
the following equation,
\begin{equation} \label{aaaa} H^{L(R)}_{m,n}=\langle
\psi^{L(R)}_{I,m}\mid H^{L(R)}\mid \psi^{L(R)}_{I,n}\rangle.
\label{eq33}
\end{equation}
After some simple algebra (see for example
Ref.~\onlinecite{Akb06}) we can easily show that for $H^{R}_{m,n}$
we obtain
\begin{eqnarray} H^{R}_{m,n}&=&
-\epsilon_z\cos\theta^{''}
m\delta_{m,n}+\epsilon_q^R\delta_{m,n}\Upsilon_{m,m} \nonumber\\ &
-&\epsilon_z\sin\theta^{''}
\left[e^{i\phi^{''}}\alpha_{-}^m\delta_{m+1,n}+e^{-i\phi^{''}}\alpha_{+}^m\delta_{m-1,n}\right]
  \nonumber\\
&+&\frac{\epsilon_q^R}{4}\delta_{m-2,n}
\Upsilon_{m,m-2}e^{2i\theta}
+\frac{\epsilon_q^R}{4}\delta_{m+2,n}\Upsilon_{m,m+2}e^{-2i\theta},
\nonumber\\
\label{eq34}
\end{eqnarray}
where $\theta^{''}$ and $\phi^{''}$ define the direction of
magnetic field in the basis  $(e_1, e_2, e_3)$; $\alpha_{-}^m$ and
$\alpha_{+}^m$ are denoting
\begin{eqnarray} \alpha_{\pm}^m=\sqrt{I(I+1)-m(m\pm 1)},
\label{eq35}
 \end{eqnarray}
and
\begin{eqnarray} \Upsilon_{m,m}&=&\left[ \frac{1}{2} I(I+1)-\frac{3}{2}m^2
+\frac{\eta }{2} [I(I+1) -3m^2]\right],
\nonumber\\ \Upsilon_{m,m-2}&=&\frac{1}{4}\left[(3-\eta)
[I(I+1)-(m-1)(m-2)]^{\frac{1}{2}}\right.
\nonumber\\&\times&\left.[I(I+1)-m(m-1)]^{\frac{1}{2}}\right], \nonumber\\
\Upsilon_{m,m+2}&=& \frac{1}{4}\left[(3-\eta)
[I(I+1)-(m+1)(m+2)]^{\frac{1}{2}}\right. \nonumber\\
&\times&\left.[I(I+1)-m(m+1)]^{\frac{1}{2}}\right].
\label{eq36}
\end{eqnarray}

In a similar manner we can obtain the matrix element of the
Hamiltonian for the particles in the left well $H^{L}_{m,n}$ by
taking $\theta=0$ and changing
$\epsilon_q^R\longmapsto\epsilon_q^L$ in Eq.~(\ref{eq34}).

We will use the perturbation approach assuming that $\varepsilon_z
\gg \varepsilon_q$ (here for satisfying the perturbation condition
we consider the special magnetic direction in a way that
$\theta^{''}\ll \frac{\pi}{2}$). It proves to be useful to change
the representation of the Hamiltonian into the basis of the spin
operators $\sigma_E$ and $\sigma_A$. In this basis we split the
total Hamiltonian of system as $H_0^D+\delta V$, where $H_0^D$,
the unperturbed part, will now contain the diagonal matrix
elements of the total Hamiltonian except those coming from $V_F$;
therefore the perturbation, $\delta V $, will contain the
non-diagonal terms of the Hamiltonian together with $V_F$. We find
that
\begin{displaymath} H_0^D \; = \; \sum_{m=-I}^I \left[\frac{E}{2}\; \sigma_E
 -(m\epsilon_z\cos\theta{''}) \textbf{1} \right] |I,m\rangle \langle I,m|,
\label{eq36b}
\end{displaymath}
and
\begin{eqnarray} \langle m |\delta V|n\rangle \; = \; \varepsilon _{mn}+\eta _{mn}w
\sigma _{E}+\eta _{mn}u\sigma _{A}.
\label{eq37}
\end{eqnarray}
We denote $\varepsilon _{mn}^{L/R}=\langle m|\delta
V^{L/R}|n\rangle $ and
\begin{eqnarray} \varepsilon _{mn} = \frac{1}{2}\left( \varepsilon
_{mn}^{L}+\varepsilon _{mn}^{R}\right) ,\;\;\; \eta _{mn}=
\frac{1}{2}\left( \varepsilon _{mn}^{L}-\varepsilon
_{mn}^{R}\right).
\label{eq38}
 \end{eqnarray}
The diagonal part of the perturbation term is
\begin{equation} V^D_F \; = \; (w\sigma_E+u\sigma_A) P \cdot F.
\label{eq39}
 \end{equation}
The first order correction of the energy level will be obtained as
\begin{equation}  \delta E_{i,m}^{(1)}\;=\; w  P \cdot F\langle i |\sigma_E|i\rangle+
\varepsilon_{mm}+\eta_{mm} w \langle i |\sigma_E|i\rangle,
\label{eq40} \end{equation}
and the second order one will be
\begin{eqnarray}
 &&\delta E_{i,m}^{(2)}=
 P^2 F^2 u^2\sum_{j\neq i}\frac{1}{E^{(0)}_{i,m}-E^{(0)}_{j,m}} \nonumber\\ &+&
PFu^2\sum_{j\neq i}\left[\frac{|\eta_{mm}\langle j |
\sigma_A|i\rangle|^2}{E^{(0)}_{i,m}-E^{(0)}_{j,m}} +C.C. \right]
 \nonumber\\ &+&
 \sum_{j\neq i}
 \frac{|\eta_{mm}|^2u^2}{E^{(0)}_{i,m}-E^{(0)}_{j,m}}
 +\sum_{n\neq m}
 \frac{|\langle i |
\sigma_E|i\rangle\eta_{mn}w+\varepsilon_{mn}\textbf{1}|^2}{E^{(0)}_{i,m}-E^{(0)}_{i,n}}.
\nonumber\\
\label{eq41}
\end{eqnarray}
Since we have assumed that $\delta V$ do not contain any diagonal
term, $\varepsilon_{mm}=\eta_{mm}=0$, the corrections to the
energy levels simplify to
\begin{equation}  \delta E_{i,m}^{(1)}\;=\; w  P \cdot F\langle i |\sigma_E|i\rangle,
\label{eq42} \end{equation}
and
\begin{eqnarray}
 \delta E_{i,m}^{(2)}&=& \frac{ P^2 F^2 u^2}{E}\langle i |\sigma_E|i\rangle
 \nonumber\\ &+  & \sum_{n\neq m}
 \frac{|\langle i |
\sigma_E|i\rangle\eta_{mn}w+\varepsilon_{mn}\textbf{1}|^2}{E^{(0)}_{i,m}-E^{(0)}_{i,n}}
.
\label{eq43}
\end{eqnarray}

For the large  magnetic field regime we calculate the first and
the second derivatives of $E_{i,m}$ with respect to the electric
field
\begin{eqnarray} \frac{\partial }{\partial F} E_{i,m}\vert_{F=0}= w  P \langle i
|\sigma_E|i\rangle,\nonumber
\label{eq44a}
\end{eqnarray}
and
\begin{eqnarray} \frac{\partial ^2}{\partial F^2} E_{i,m}\vert_{F=0}=
\frac{2P^2 u^2}{E}\langle i |\sigma_E|i\rangle.
\label{eq44b}
\end{eqnarray}
We can easily express the susceptibility
\begin{eqnarray} \chi^{(1)} &=&\frac{2P^2 u^2 }{ZE}\sum_{m}[e^{-\beta
E_{d,m}}-e^{-\beta E_{u,m}}].
\label{eq45}
 \end{eqnarray}
Here  $Z = \sum_{m}[e^{-\beta E_{d,m}}+e^{-\beta E_{u,m}}]$ is the
partition function, and $E_{d,m}$ and $ E_{u,m}$ are defined by
\begin{eqnarray} E_{i,m}|_{F=0}& =& \frac{E}{2}\langle i|\sigma_E|i\rangle
-m\cos\theta^{''} \varepsilon_{z} \nonumber\\&+& \sum_{n\neq m}
 \frac{|\langle i |
\sigma_E|i\rangle\eta_{mn}w+\varepsilon_{mn}\textbf{1}|^2}{E^{(0)}_{i,m}-E^{(0)}_{i,n}}.
\label{eq46}
 \end{eqnarray}

\subsubsection{Thermal expansion}
Let us remember here that the thermal energy exceeds  few
orders of magnitude the Zeeman energy and the nuclear quadrupole
potential. For this reason we can carry out a thermal expansion in
terms of $\beta \varepsilon_q$, $\beta \varepsilon_z$ and find
\begin{eqnarray} &&e^{-\beta E_{i,m}}=e^{-\beta ( \frac{E}{2}\langle
i|\sigma_E|i\rangle)}\times \nonumber\\&&[1+m\cos\theta^{''}
\beta\varepsilon_{z} - \beta\sum_{n\neq m}
 \frac{|\langle i |
\sigma_E|i\rangle\eta_{mn}w+\varepsilon_{mn}\textbf{1}|^2}{-\cos\theta^{''}
\varepsilon_{z}(m-n)}]\nonumber\\&&+O(\beta^2\varepsilon^2),
\label{eq47}
 \end{eqnarray}
After some algebra, we can easily write
\begin{eqnarray}
 \chi^{(1)}&=&\frac{2P^2 u^2}{E}t +
 \frac{2\beta P^2
u^2}{E(2I+1)}(1-t^2)\times
 \nonumber\\&&
\sum_{m,n\neq m}
 w\frac{\eta_{mn}^{*}\varepsilon_{mn}+\eta_{mn}\varepsilon_{mn}^{*}}{\cos\theta^{''}
\varepsilon_{z}(m-n)}+ O(\beta^2\varepsilon^2).
\label{eq48}
\end{eqnarray}

Similar to the small magnetic field regime, from Eq.~(\ref{eq15})
and Eq.~(\ref{eq44b}) we can straightforwardly show that
\begin{eqnarray}
 \chi^{(2)}+ \chi^{(3)}&=&P^2w^2\beta(1-t^2),
\label{eq49}
\end{eqnarray}
The magnetic field dependent correction of the susceptibility will
be
\begin{eqnarray}
 \delta\chi(B)&=&
 \frac{4\beta P^2
}{E(2I+1)}(1-t^2)wu^2 \sum_{m,n\neq m}
 \frac{\rm{Re}[\eta_{mn}^{*}\varepsilon_{mn}]}{\cos\theta^{''}
\varepsilon_{z}(m-n)}
\nonumber\\&+& O(\beta^2\varepsilon^2).
\label{eq50}
 \end{eqnarray}
If we set $ \varepsilon_q=0$, we can assume that $I=0$, we
recover once more the result of standard TLS model,
\begin{eqnarray}
 \chi_{TLS}&=&\frac{2P^2 u^2}{E}t+P^2w^2\beta(1-t^2).
\label{eq51}
\end{eqnarray}

By averaging over the TLS parameters according to the distribution
function (\ref{eq3}), we obtain:
\begin{eqnarray}
 \delta\chi(B)
&=&
 \frac{4\beta P^2\xi'
}{(2I+1)} \sum_{m,n\neq m}
 \frac{\rm{Re}[\eta_{mn}^{*}\varepsilon_{mn}]}{\cos\theta^{''}
\varepsilon_{z}(m-n)}
\nonumber\\&+& O(\beta^2\varepsilon^2).
\label{eq52}
 \end{eqnarray}
where $\xi'$ is the numeric constant  (nonmagnetic-dependent)
which depends on the magnitude of $E_{max}$ and $\Delta_{0min}$.
We can easily see that $\delta\chi(B) \sim \frac{\beta
\bar{\varepsilon_q^2}}{\varepsilon_z}$, where
$\bar{\varepsilon_q^2}\propto
[(\varepsilon^L_q)^2-(\varepsilon^R_q)^2]$.

As we expected   from the beginning  if we neglect the phase
difference between the nuclear moment in the wells and if we
assume that the EFG in the both wells are the same
$(\varepsilon_q^L=\varepsilon_q^R)$, then we do not find any
magnetic dependence in the second order perturbation
(Eq.~\ref{eq31} and Eq.~\ref{eq52}).

\subsection{Discussion on the field dependence of dielectric susceptibility }

In this section we have addressed the question whether or not the
coupling of the two-state coordinate and the nuclear spin
variables through the nuclear quadrupole potential in an
inhomogeneous crystal field can been taken into account as the
source of the magnetic field dependence of the dielectric
susceptibility.

Analyzing the existent data regarding the dielectric
susceptibility as a function of the magnetic field of different
oxide glasses and mixed crystals
\cite{Str98,Str00,Woh01,Lec02,Hau04}, we can notice along with a
pronounced bump around $200 mT$, some irregular oscillations. The
curvature of the susceptibility changes its sign few times up to
$5 T$ of the magnetic field. The amplitude of the real part of the
susceptibility seems to be about 10 percent of the TLS
susceptibility, corresponding to approximately $10^{-4}$ of the
dielectric constant. Relevant experiments involve temperatures
from tens of milliKelvin to a few hundreds. We can observe that in
this range, the dielectric constant vary with the inverse
temperature, $1/T$.

The present work reconsider the static dielectric susceptibility of a glassy
systems in an external magnetic field implementing a perturbation approach for the
energy levels . Starting from the coupling of
the nuclear quadrupole with the tunnelling system, calculations
have been performed for two different regimes, i.e. small and
large external magnetic fields with respect to the nuclear
quadrupole potential. We have found that the magnetic field dependent part
of the susceptibility in both regimes is as the following
\begin{eqnarray} \Delta \chi_{small}(B) &\sim& \beta^2 \epsilon_z^2
\frac{\epsilon_q'^2}{\epsilon_q}, \nonumber\\
\Delta \chi_{large}(B)
&\sim& \beta \frac{\bar{\epsilon_q^2}}{\epsilon_z}. \nonumber
\label{eq53a}
 \end{eqnarray}
As it is obvious from the above results the dielectric
susceptibility depends on the magnetic field in the second order
correction of the perturbation scheme via the Zeeman and the
nuclear quadrupole terms. This correction is directly the result
of different electric field gradients (EFG) in the two wells. In
the small magnetic field regime the correction increases with
$B^2$ while it decreases inversely with field ($1/B$) in the large
magnetic field. The dependence on the temperature was put into
evidence in agreement with the existent experimental data. The
magnetic field dependence will disappear in the second order
corrections if we take the same EFG in both wells. We would like
to remind that the previous existing perturbation calculations
\cite{Dor04} which has been considered the same EFG provides a
magnetic field dependence of the dielectric susceptibility only on
the fourth order of expansion.

The theoretical approach we have used do not allow us to calculate
the static dielectric susceptibility for the case where the
applied magnetic field has such values so that the Zeeman energy
and the nuclear quadrupole have similar magnitudes, $\epsilon_z
\simeq \epsilon_q$. But for this regime we know that \cite{Dor04}
the magnetic field dependent part of the static susceptibility
$\chi$ is proportional to the small ratio $[(\varepsilon_z
\varepsilon_q)/T^3]^2$,
 this ratio being of the order of
$10^{-6}$ (one million part) for $T \sim 1 K$.

We have to mention here that our calculation provides only the
static limit ($\omega = 0$) of this contribution.

The dynamical susceptibility of a simple TLS was expressed by
J\"{a}ckle \cite{Jac72}
\begin{eqnarray} \chi_{TLS}^{rel}(\omega) \; = \; \left< P^2 w^2
\frac{\beta}{\cosh^2(\beta E/2)} \frac{i \gamma}{\omega + i
\gamma} \right>,
\label{eq53}
\end{eqnarray}
where $\gamma$ is the relaxation rate. Now, by considering the
coupling of the nuclear quadrupole to the tunnelling motion, the
dynamic susceptibility proves to be more complicated, but it can be
generally expressed as \cite{Dor06}
\begin{eqnarray} \chi^{rel}(\omega) \; = \; \left< P^2 w^2
\frac{\beta}{\cosh^2(\beta E/2)} \sum_m A_m \frac{i r_m}{\omega +
i r_m} \right>
\label{eq54} \end{eqnarray}
where $r_m$ represents  the eigenvalues of the relaxation rate
matrix, and $A_m$ the corresponding amplitudes ($m=0, \ldots ,2I+1
$). In the limit of $\omega \rightarrow 0$, an approximation of
the susceptibility is provided by Eqs. (\ref{eq31}) and
(\ref{eq52}) for the two considered regimes. Reminding that the
relevant experiments are performed on frequencies in the range of
kilohertz, and that the relaxation rates should be extremely small
quantities (details will be give elsewhere \cite{Dor06}). We can
conclude that such a limiting case cannot provide a proper
explanation for the existent experimental data.

The method we developed cannot provide a plausible amount for the
relative magnitude of the magnetic field dependence part with
respect to the simple TLS susceptibility. We do not find
oscillations of the dielectric susceptibility at magnetic fields
of the order of $1T$.

\section{The heat capacity}
Our purpose is to find the magnetic field dependence of the
heat capacity in multicomponent glasses. To the best of our knowledge there
are not many experimental data for heat capacity of (nonmagnetic)
glasses in the presence of magnetic field. Therefore, we
assume that the tunneling particles have an angular momentum, $J$,
(intrinsic angular momentum such as electron spin). Then, we
must add an additional term to the Hamiltonian which in the presence
of magnetic field leads to
\begin{equation} H_Z^{(2)} \; = \; - \epsilon_z' J_z.
\label{eq55}
 \end{equation}
\noindent The energy scale is  $\epsilon_z' = g_E \mu_B B /\hbar$,
$\mu_B$ is the Bohr magneton and $g_E$ is the electronic
Land\'{e} factor. Here we have neglected the spin-spin interaction
in the Hamiltonian. We have also assumed that the quantization
axes of angular momentum, $J_z$,  is in the direction of $e_B$
(for more detail, see the Appendix).
The total Hamiltonian of the system is then
\begin{displaymath} H \; = \; H_{TLS}\; + \; H_{Z} + \;H_Q . \end{displaymath}
Here $H_Z= H_Z^{(1)}+H_Z^{(2)}$ (defined in Eqs.~(\ref{eq7},
\ref{eq55})) and \begin{equation}
H_Q=(\frac{1+\sigma_z}{2})V_Q^L+(\frac{1-\sigma_z}{2})V_Q^R,
\label{eq56}
 \end{equation}
where $V_Q^{R(L)}$ is defined in Eq.~(\ref{eq10}) for the
particles in right (left) well \cite{Dor04}.
The heat capacity is expressed by the following
relation
\begin{equation}
C_v =\frac{1}{K_BT^2}\left\langle \frac{\partial ^{2}\ln
Z}{\partial \beta^{2}}\right\rangle,
\label{eq58}
\end{equation}
where $Z$ is the partition function which is a sum over the TLS, spin and
angular momentum degrees of freedom.
If we  neglect the phase difference between the nuclear moments in
the two wells (taking $\theta=0$) and  assume that the EFG in
both wells are the same ($\epsilon_q^R=\epsilon_q^L=\epsilon_q$),
then the partition function can be written in the following form
\begin{equation}
Z \; = \; Z_{TLS}SS',
\label{eq59}
\end{equation}
where
\begin{eqnarray} Z_{TLS}&=& \; {\rm Tr}\left\{ e^{-\beta
H_{TLS}}\right\}=2\cosh(\beta E/2),
 \nonumber\\ S&=& \; {\rm Tr}\left\{
e^{-\beta (H_z^{(1)}+H_Q)}\right\},
 \nonumber\\  S'&=& \; {\rm
Tr}\left\{ e^{-\beta H_z^{(2)}}\right\}.
\label{eq60}
 \end{eqnarray}

Therefore, according to Eq.~(\ref{eq58}) and Eq.~(\ref{eq59}),  we
 find that
 %
 \begin{eqnarray} C_v& =&\frac{1}{K_BT^2}\left\langle
\frac{\partial ^{2}\ln Z_{TLS}}{\partial \beta^{2}}+\frac{\partial
^{2}\ln S}{\partial \beta^{2}}+\frac{\partial ^{2}\ln S'}{\partial
\beta^{2}}\right\rangle \nonumber\\ &=&C_{TLS}+C^{(I)}+C^{(II)}.
\label{eq61}
\end{eqnarray} By some calculations and after averaging over TLS parameters,
we can easily show that [see for example
Ref.~(\onlinecite{Esq98})]
\begin{eqnarray} C_{TLS}&=&\frac{\pi^{2}}{6}P_0K_{B}^{2}T
 \propto T,
 %
\label{eq62}
 %
 \end{eqnarray}
which is independent of the magnetic field. The Zeeman
contribution of the tunneling particle's angular momentum ($J$) in
the partition function is simplified to the following form,
\begin{eqnarray} S'=\sum_{m'=-J}^Je^{\beta m'\epsilon'_z}= \frac{\sinh[\beta
\epsilon'_z (J+\frac{1}{2})] }{\sinh[ \frac{\beta
\epsilon'_z}{2}]}.
\label{eq63}
\end{eqnarray}
Thus the Zeeman contribution of the angular momentum in the
specific heat is given by
\begin{eqnarray} &&C^{(II)}=\frac{P'}{K_BT^2}\left\langle \frac{\partial
^{2}\ln S'}{\partial \beta^{2}}\right\rangle= \frac{K_B \beta^2
\varepsilon_z'^2}{4}\times P'
 \nonumber\\&&\left[ csch^2\left(\frac{\beta
\varepsilon_z'}{2}\right)-(2J+1)^2csch^2\left(\frac{\beta
\varepsilon_z'(2J+1)}{2}\right)\right]. \nonumber\\
\label{eq64}
\end{eqnarray}
where $P'$ is the concentration of intrinsic angular momentum times a
constant which comes from the averaging over the TLS parameters.

Finally, the contribution of the remaining term of the Heat Capacity
to the specific heat, $C^{(I)}$,  is given by

\begin{eqnarray} C^{(I)}=K_B\beta^2\left\langle\frac{\partial ^{2}\ln
S}{\partial \beta^{2}}\right\rangle,
\label{eq65}
\end{eqnarray}
where $S=Tr\{e^{-\beta H'}\}$ and $H'=H_z^{(1)}+H_Q$.

As we have mentioned in previous section the thermal energy exceeds
few orders of magnitude the Zeeman energy and the nuclear
quadrupole potential. Therefore, $\beta \epsilon_z$ and $\beta
\epsilon_q$ are very small quantities, consequently $\beta  H'
\ll 1$. Therefore, Eq. (\ref{eq65}) is approximated by
\begin{eqnarray} C^{(I)}&=&K_B\beta^2\frac{\partial ^{2} }{\partial
\beta^{2}}\ln\left[ Tr( 1-\beta H'+\frac{\beta^2}{2}H'^2+\ldots
)\right]
\nonumber\\
&=&K_B\frac{\beta^2}{2I+1}\left[Tr(H'^2)-\frac{Tr(H')^2}{2I+1})
\right]+O(\beta^3\varepsilon^3),\nonumber\\
\label{eq66}
\end{eqnarray}
and finally we arrive at the following expression
\begin{eqnarray}
C^{(I)}&=&P''K_B\beta^2\left[\gamma_1\varepsilon_z^2+\gamma_2\varepsilon_q^2\right]+O(\beta^3\varepsilon^3).
\label{eq67}
 \end{eqnarray}
where $P''$ is the concentration of the nuclear spin times a constant
which comes from the averaging over the TLS parameters. Here
$\gamma_1$ and $\gamma_2$ are numerical constant which have been
defined by~(see appendix A),
\begin{eqnarray} \gamma_1&=&\frac{1}{2I+1}\sum_m\left[m^2\cos^2\theta^{'}
+\sin^2\theta^{'}[I(I+1)-m^2] \right],
\nonumber\\
\gamma_2&=&\frac{1}{2I+1}\sum_m[\Upsilon_{m,m+2}^2+\Upsilon_{m,m-2}^2-\frac{\Upsilon_{m,m}^2}{2I+1}].
\label{eq68}
 \end{eqnarray}

Eq.~(\ref{eq67})  shows a quadratic dependence on the magnetic
field. In the large field regime $C^{(II)}$ contributes as a
constant value to the whole specific heat. Thus, the field
dependence of the specific heat  in the high magnetic fields comes
from $C^{(I)}$ which shows a quadratic behaviour versus the
magnetic field. However, for small field regime we have
$\varepsilon_z\ll\varepsilon_z'$ which makes $C^{II}$ to have the
dominant effect ($C^{(I)}\ll C^{(II)}$). It can be shown that when
the magnetic field is small the leading term of $C^{II}$ is given
by the following expression
\begin{equation}  C^{(II)}\simeq P'K_B \beta^2
\varepsilon_z'^2\frac{J(J+1)}{3}, \end{equation}
which shows  $B^2$ behaviour for low magnetic fields.

\begin{figure}[t]
\vspace{0.5cm} \centering \scalebox{0.35}[0.35]{
\includegraphics*{fig1.eps}
}
\caption{ Magnetic field (thermal) variation of the $C^{(II)}$.}
\label{fig1}
\end{figure}

\subsection{Discussion on the field dependence of the Heat Capacity}

In this section, we have studied the magnetic field dependence of
the Heat capacity of glasses at low temperature. We have assumed
that the tunneling particles carry both an angular momentum and a
nuclear spin. So, the magnetic field couple to both terms.

We have examined our results numerically for arbitrary angle of
EFG and different phases between the quadrupole moments on each
well. The final results are not influenced if we only consider the
simple case of $\theta=0$ and  considering that
$\epsilon_q^R=\epsilon_q^L=\epsilon_q$. Thus,  we have reported
our results in the mentioned special case
 where we can find an analytical expression for the specific heat behaviour.
In this respect,   the total partition function of system can be
splitted  into the
 multiplication of  three separate terms.
 Therefore, the Heat capacity is
composed of three terms, the TLS contribution ($C_{TLS}$), the
nuclear ($C^{(I)}$) and the angular momentum ($C^{(II)}$) parts.
The magnetic field dependence will arise from $C^{(I)}$ and $C^{(II)}$.

We have introduced the generalized Hamiltonian which takes into
account the mentioned degrees of freedom. The direction of
external magnetic ($\theta', \phi'$) field has been considered
arbitrary with respect to the EFG direction of each TLS. However,
the outcome will be averaged over this spherical angle which can
be contracted as a constant to the final results. We have
calculated the contribution of $C_{TLS}$ and $C^{(II)}$ exactly
while the effect of nuclear spin ($C^{(I)}$) has been treated in a
thermal expansion up to the second order corrections.

The different scale of Zeeman energy for the nuclear moment and
the angular momentum of the tunneling entity  is roughly
$\epsilon_z/\epsilon_z'\approx10^{-3}$. Thus, for low magnetic
fields $\beta \epsilon_z'<10$ the dominant effect comes from the
angular momentum part ($C^{(II)}$). Our calculation shows that at
low magnetic field this contribution is proportional to the
square of magnetic field, $C^{(II)}\simeq B^2$ (Eq.(\ref{eq68}).
However, $C^{(II)}$ reaches a maximum at $\beta \epsilon_z'\simeq
3$ and then decreases to zero for $\beta \epsilon_z' >20$ (see
Fig.(\ref{fig1})).

For low magnetic field ($B<10 T$) our result predict a Schottky
like peak in the specific heat of glasses which have nonzero
angular momentum for the tunneling entity. Or at least the sample
has some impurity with nonzero $J$ (angular momentum). This result
is in agreement with the recent experimental observations for
Duran, AlBaSi \cite{strehlow-web} and Suprasil \cite{Str04}. The
data show explicitly an upward increase of the specific heat
versus the magnetic filed for
 $B<0.5 \mbox{T}$ and monotonic decrease
for $B>1 \mbox{T}$ at $T<1 \mbox{K}$.

We have also predicted that for high magnetic fields ($\beta
\epsilon_z' >20$) the dominant contribution to the specific heat
comes from the nuclear moments which shows a quadratic dependence
on the field (Eq.~\ref{eq67}).

In this paper we have studied the special case that the impurities
have only the intrinsic angular momentum part. For the general
case of impurities (for example electrons) with angular momentum,
$J>\frac{1}{2}$ (spin + orbit), we should take into account the
effect of spin - orbit interaction. Which means that at zero
magnetic field the levels $m' = -J,...,J-1,J$  may have different
energies due to the crystal field splitting.

However, if we consider  $J=0$  which means to have a pure glass
without magnetic impurities like BK7 the nuclear quadrupole term
is the only dominant term in the specific heat. So, the  present
calculations predict a quadratic dependence on the magnetic field,
and $1/T^2$ dependence on temperature. As far as we know there is
no experimental result for this case.  Therefore it might be a
good suggestion for future experiments.

\section{Summary}

Based on  the polarization echoes experiments \cite{Nag04} we
believe that the nuclear quadrupole model should
provide the rightful explanation for the intriguing behavior of
the dielectric properties of glasses. A relaxation spectrum rather
sensitive to the orientation of the nuclear quadrupoles is
providing a relaxation susceptibility strongly dependent on the
applied magnetic field \cite{Dor06}, driving us to the conclusion
that the observed magnetic properties of glasses might have a
relaxation origin. Other effects like non-linearities with the
voltage \cite{Bur05}, or cooperative TLS behavior might as well be
found contributing.

\begin{acknowledgments}
A. A. and D. B. would like to acknowledge first of all the
hospitality of the Max Plank Institute for the Physics of Complex
Systems (Dresden, Germany) where this work was carried out, and to
thank Prof. Fulde and Prof W\"{u}rger for stimulating discussions
and valuable comments. A. A. also would like to thank Prof.
Strehlow  for fruitful discussions and useful comments.
\end{acknowledgments}

\begin{appendix}
\section{Generalized TLS Hamiltonian with both nuclear spin and
angular momentum}
The aim of this appendix is to find the generalized TLS
Hamiltonian with both nuclear spin and angular momentum. In the
similar way leading to the Eq.~\ref{eq32}, we can write the total
Hamiltonian as:
\begin{eqnarray}
 \left(\begin{array}{cc} H^L+ \frac{\Delta}{2}
\cdot\textbf{1
}&\frac{\Delta_0}{2}\cdot\textbf{1 }\\
\frac{\Delta_0}{2}\cdot\textbf{1 }&H^R- \frac{\Delta}{2}
\cdot\textbf{1 }\end{array}\right) \vspace{0.5cm}
\label{eq69}
\end{eqnarray}
where $\textbf{1}$ is the unit matrix of rank
$(2I+1)\times(2J+1)$. $H^L=V_Q^L+H_Z^L$ and $H^R=V_Q^R+H_Z^R$ are
defined for particles in the left and right wells, respectively.
The state in each well (L or R) can be characterized by $\mid
\psi^{L(R)}_{[I,m],[J,m']}\rangle=\mid L(R)\rangle\bigotimes \mid
I,m\rangle\bigotimes \mid J,m'\rangle $ and consequently the matrix
element of the Hamiltonian is defined by the following equation,
\begin{equation}
\label{eq70}
H^{L(R)}_{mm^{\prime},nn^{\prime}}=\langle
\psi^{L(R)}_{[I,m],[J,m']}\mid H^{L(R)}\mid
\psi^{L(R)}_{[I,n],[J,n']}\rangle.
\end{equation}
After some simple algebra
we can easily show that for
$H^{R}_{mm',nn'}$ we obtain
\begin{widetext}
\begin{eqnarray} H^{R}_{mm',nn'}&=&\left\{
\epsilon_q^R\left[\Upsilon_{m,m-2}e^{2i\theta}\delta_{m-2,n} +
\Upsilon_{m,m+2}e^{-2i\theta}\delta_{m+2,n}\right]
-\epsilon_z\sin\theta^{'}
\left[e^{i\phi^{'}}\alpha_{-}^m\delta_{m+1,n}+e^{-i\phi^{'}}\alpha_{+}^m\delta_{m-1,n}\right]\right.
\nonumber\\ &+&\left.\left[-m'\epsilon_z'  -\epsilon_z\cos\theta^{'} m +
\epsilon_q^R\Upsilon_{m,m}\right]\delta_{m,n} \right\}
\delta_{m',n'} .
\label{eq71}
\end{eqnarray}
\end{widetext}
where $\theta^{'}$ and $\phi^{'}$ define the direction of magnetic
field in the basis  $(e_1, e_2, e_3)$; $\alpha_{\pm}^m$ and
$\Upsilon_{m,m'}$  are given by Eqs.~(\ref{eq35}~\&~\ref{eq36})  .
 In a similar manner we can obtain the matrix
element of the Hamiltonian for the particles in the left well
$H^{L}_{m,n}$ by taking $\theta=0$ and changing
$\epsilon_q^R\longmapsto\epsilon_q^L$ in the above equation.

\end{appendix}

redpajama_set_name: RedPajamaArXiv


Whitetail deer and racoons! What side of the Atlantic are you on anyway?
I sometimes like to think about how long it might take for an invasive species, isolated from it's mother population, to evolve into a unique 'endemic' subspecies. Odocoileus virginianus var. nordica? Its only inevitable.
I faithfully read your blog updates. I admire your efforts in sharing as much as your efforts in the field. I think what you're doing is tremendously important, because modern people need as many examples of being bold and different that they can get. Ecspecialy young people. The writings of primitive practioners gave me enormous comfort as a youth, letting me know I was not alone, and that I could cary my passions further than I ever dreamed. I would never have dared come as far as I have without the writings of people like Tom Elpel, Jim Riggs, and Brent Ladd to encourage me. You never know who your work is influencing. The extra inspiration might be all somebody needs to tip the scales between a life of mediocrity and a lifelong adventure.
(of over two hundred) I've been able to find just by keeping my eyes open.
I appreciate you keeping us posted. I'd realy like to see how that coppiced hazel looks in the seasons after you burn it. I may do the same to my hazels.
When the alarm goes off in my earth lodge, and I blow out the fat lamp, and I hike through the snow, to my car, to work at the plywood plant- it's nice to know of a certain SOB in faraway Scandanavia whose at least as crazy me.
Waint a minute… I've confused Thomas with Tojus. Oops. Thomas, you are also an inspiration. I will repost my comments for Torjus under one of his articles.
And Thomas, could you elaborate on how pitch is blended with fat? I'm assuming they are just mixed together over heat?..


text:
Meet Our Members: Richard Hamilton
• November 15, 2021
Lubbock Monterey Chapter of AMBUCS is fortunate to have Richard Hamilton as a member since April, 1983. Richard was sponsored by Fred Bryant. Richard is a go-getter and has a big heart for this chapter of AMBUCS, and the Big Hat Club, which is a club within this chapter. Richard has been a "Big Hatter" member since November, 1984.
His dedication to the Lubbock Monterey Chapter of AMBUCS was a spin-off from his days when he attended Texas Tech and was a member of Alpha Phi Omega, a service fraternity focusing on community projects for the University and surrounding community. To continue his desire for community service, Richard found after the AMBUCS mission was explained to him, AMBUCS "was the right fit for him to continue serving the community."
As a member of the Lubbock Monterey Chapter of AMBUCS, Richard's accomplishments has been many and include the following:
Chapter President on two different occasions
District Governor two times
Regional Director (2 year term)
Regional Big Hat Director
Served on the National Level AmTryke Board
Served as the National Membership Chairman
Served on a special AMBUC Committee, with the responsibility raising monies for the HUB City Playground and the Challenger Little League Field
Richard believes his greatest AMBUC accomplishment was being the "National AMBUC of the Year" on two different occasions--1999 and 2000. He is also a William White member, a Life member, and has sponsored 65 people who chose to become an AMBUC member. This speaks volumes about Richard's dedication as a member of Lubbock Monterey Chapter of AMBUCS. In addition, he is involved in other organizations: Commander of the Lubbock County Reserve Deputy's (part of the Sheriff's Office), Lubbock Challenger Little League, and Texas South Plains Honor Flight (Finance Chairman).
He carried his dedication over into his life-long career as Chief of Police for the Texas Tech University System--Texas Tech University Health Sciences Center, which covered Lubbock, Amarillo, El Paso and Odessa, and oversaw 100+ employees. He retired after 34 years of service. Currently, Richard is the Operations Manager for South Plains Mall and oversees construction and remodeling projects, working with various in-house contractors in making sure the building and grounds meet the mall's standards.
When it comes to his family, he is married to his wonderful friend, companion, and bride, Patty Hamilton. Patty is also a member of Lubbock Monterey AMBUCS. He is proud to say he has been married for 53 years. Richard and Patty have two children (Robert and Kristi), 4 grand kids (Aaron, Jordan, Caleb, and Eamon), and 5 great grand kids (Lucas, Ella, Maddi, Hannah and Sam).
So what does Richard do when not working? He enjoys fishing, hunting, watching and attending Texas Tech sporting events, golfing, and most of all entertaining his grand- and great-grand kids.
We are so grateful to have Richard Hamilton as a member of the Lubbock Monterey Chapter of AMBUCS' Team, and whether you're a long-time Lubbock resident or a newcomer, if you too, would like to learn more about our mission and see about joining our Team, just JOIN US at one of our regular Monday Lunches at the McKenzie/Merket Alumni Center on the Tech Campus. You'll be glad you did!
← Newsletter 11-15-2021
AmTryke Recipient: Jacqueline Lamug →


text:
Doing Well By Doing Good
by Alan | Mar 24, 2010 | Big Ideas, Compelling Value, Engaging Employees, Unlocking Genius | 1 comment
Greetings. Most companies talk about the importance of being good corporate citizens. And many are regularly involved in efforts to improve the communities they operate in. But I've been particularly excited over the last two years to watch one of our customers make social responsibility an innovative and "strategic" part of its business. The company is CGI Federal, a leading government contractor based in Fairfax, Virginia. It's core business is providing enterprise solutions and a wide range of essential managed services to Federal agencies, and it consistently earns high marks from customers for the quality of its work and the commitment of its people. But it's also earning high marks for its growing commitment to the community, based on a clear and engaging strategy for volunteering and giving that is driven by the interests of its employees–or "members" as it calls them.
At the heart of this effort is a belief that people want to make a difference and that a company can and should be a catalyst for creating meaningful opportunities to get involved and "give where they live." It's not only the right thing to do, but sound business practice. To do this, CGI Federal forged a unique partnership with the Catalogue for Philanthropy, a very special organization operating in the Washington, D.C. area, that identifies and promotes remarkable smaller nonprofits that are making a compelling difference in their communities. These organizations often operate under the radar of big corporations, individual donors, and potential volunteers. And, with limited resources, often have trouble getting the word out about their valuable work. So the Catalogue acts as their "seal of approval" and provides tools to help them raise funds and build their marketing and organizational capacity. In partnering with the Catalogue, CGI Federal made a financial contribution, but also organizes volunteer events for the Catalogue's "grantees," and has provided some of its top talent to redesign the Catalogue's website and many of the tools that are vital to these nonprofits. In addition, the company provides support to members involved in community organizations.
It's worth noting that CGI Federal sees community involvement as "a way of life" that is an important requirement for a new generation of workers who want to be part of improving their communities. The type of thoughtful members that smart companies would love to hire. And, the bottom-line benefits of corporate social responsibility are equally significant. They include:
Enhancing Recruitment Efforts–As word of CGI Federal's program spreads, it is becoming a real differentiator in attracting highly-motivated and concerned employees. And current members are even more eager to recruit their friends.
Aiding Retention–Current employees are more likely to stay in the company longer when they see that it cares about its clients and its community, and encourages them to be involved in local initiatives that matter.
Greater Employee Engagement–Increased job and personal satisfaction leads to greater engagement and an even greater commitment to customer success.
We win in business and in life by caring about those around us. And giving our members a chance to make a real difference in the lives of those in need. It's really no surprise that we do well by doing good.
Kathy Jankowski, Catalogue for Philanthropy: Greater Washinton	on March 25, 2010 at 2:39 pm
Alan-Thank you for highlighting CGI Federal. It's been a privilege to partner with George Schindler and his thoughtful, passionate, engaged members. CGI's commitment to giving back is benefitting the entire Washington region. Our innovative website made possible by CGI Federal just won "Gold" from the Council on Foundation's Wilmer Shields Rich Awards program. The judges remarked, "An incredibly brilliant site. . . . It overlays the ease of online browsing and shopping onto a true philanthropic mission." It's got great tools for donors that want to make a difference, like a Gift Registry and Gift Cards. And here's an idea that I think you will love: The building services manager worked with Art Enables to bring artwork into the building. This afternoon, Exec. Dir. Joyce Muis will install an exhibition featuring the artwork of emerging artists with developmental and/or mental disabilities. Colorful "outsider art" will be displayed in the entrances and on five floors at CGI Federal. There are so many possibilities when the environment one works in fosters creative thinking.


text:
TsAGI Outlines Challenges of RPAS Certification at International Conference
What will be the future of urban air mobility? How will RPAS (remotely piloted air systems) help a country solve its problems? When could RPAS be integrated intro international air space? Such questions were discussed at 'Unmanned Aviation — 2021', the 8th International Conference which was held late April in Moscow.
TsAGI (a part of NRC "Zhukovsky Institute") delivered two reports.
Mark Ovsyannikov, Junior Researcher for Remotely Piloted Air Systems (TsAGI) delivered a report, 'Challenges of Light RPAS Certification'. He spoke about TsAGI scientists who elaborate national airworthiness requirements for UAVs within civil RPAS; and work on the issues related to interaction between UAV ground control station and airborne equipment. A key safety factor for RPAS is absence of people on board. Thus, RPAS airworthiness requirements could be eased, as compared to manned aircraft; so UAV developers would have a wider choice for providing safety, considering the specified operational tasks.
In their work, TsAGI specialists rely on international working-groups' experience of JARUS (Joint Authorities for Rulemaking on Unmanned Systems) and ICAO (International Civil Aviation Organization) related to RPAS.
'In Russia, the prospects of unmanned civil aviation sector are closely connected to UAV integration into single air space,' says Mark Ovsyannikov, Junior Researcher for Remotely Piloted Air Systems (TsAGI). 'A key solution to the problem is to gradually implement new technologies which improve UAV control automation and provide information exchange with other air traffic participants; basing on years-long experience of providing safety for structures and civil aircraft systems'.
At that conference, Vladimir Yershov, Engineer of Center for Remotely Piloted Air Systems (TsAGI), reported on research into computer-vision systems enabling safe automatic landing for RPAS. The equipment should have such capabilities as recognizing landing-strip, forming recommendations for off-board crew, and deciding automatically to land and to choose landing method. Yershov also presented modeling results for the device's operating principle and its suggested characteristics.
'Unmanned Aviation — 2021' is an international conference and exhibition; it is the largest annual discussion platform for all regulatory aspects related to UAV production and application; for presenting trends and innovative solutions.
In 2021, the Conference was attended by Russian federal and regional authorities, by UAV developers and producers; by forestry, hunting and agricultural sectors; by environmental organizations, by search-and-rescue organizations; by energy, freight-carrier, geodesic, leasing and insurance companies; by meteorological services; by investment groups and banks; as well as by British and Chinese scientists.


text:
THE MAKING OF MAGNOTTA
September 22, 2022 September 22, 2022 Myles ShaneToday's Top StoriesCNN, Cover Guy, Luka Magnotta, Sex Fame and Murder: The Luka Magnotta Story
At the start of my filmmaking career my wife, Naomi and my brother-in-law Jonathan were presented the opportunity to produce several shows for OUT TV, Canada's official LGBTQ channel. Early on we shot a dating series entitled "Fairy Tale", a same sex wedding series and a male modelling competition show called, "Cover Guy". Cover Guy saw male models from all over Canada compete to become Canada's top model.
One of the contestants who appeared on the program was a young man by the name of Luka Magnotta. Luka was scrawny, had a deep voice and argued with judges he could easily gain weight if needed. One of the judge's Luka had actually dated- Nina Arsenault (a transexual artist who appeared on CNN discussing Magnotta.) Luka answered questions, posed in front of several judges and explained why he should become Canada's top male model. However he ultimately didn't make the cut.
Almost a decade later Naomi and I had two toddlers. I remember we were changing their diapers and watching CNN, specifically Anderson Cooper 360. For some reason there was a clip of Cover Guy with Luka presenting himself to the judge's. Over top of the picture you could hear Anderson announcing Luka Magnotta was the most wanted man in the world and Interpol was essentially trying to capture him. His last known whereabouts was Paris, France.
After watching the footage I was shocked. A contestant on our little modeling show was the most wanted man in the world! How was that even possible? Our male model who didn't make the cut was wanted by the international police. According to Anderson, Luka had taken a male Chinese university student to his apartment in Montreal where he drugged him, killed him with an ice pick, then preceded to eat parts of his body. Luka taped the entire ordeal, then downloaded the atrocity to the internet for anyone to watch with the click of a mouse. He called the video, "1 lunatic, 1 icepick."
I stared at my wife and she glanced back at me. My first phone call was to Jonathan, who immediately called the players at A&E and told them the most wanted man in the world had been on our television show. Thirty minutes later we had a deal to produce, direct and write: SEX, FAME & MURDER: THE LUKA MAGNOTTA STORY.
For the next year and a half, my main priority was to track down the various people who knew Luka and ultimately arrange interviews with them. I found Luka's ex transexual girlfriend Barbie Swallows, who wanted to tell everyone and anyone she'd dated the most wanted man in the world. She'd already done interviews with the CBC, The Canadian Press and a number of other media outlets. When she wasn't charging people to see her naked online, she had jumped on the Magnotta bandwagon. She had no problem sharing with the world Luka almost killed her. She wanted every bit of media attention she could attract.
Later, I located Luka's pimp who had sent him all over the world to satisfy other men and women. The pimp revealed Luka was taking a bunch of prescription medications but sometimes he forgot to or didn't feel he needed the drugs. Eventually we'd determine Luka had been prescribed a number of medications for schizophrenia which he wasn't always taking. Since his teenage years he'd been hospitalized a number of times for having both auditory and visual hallucinations. As we dug deeper, we discovered Luka's father had also been diagnosed with schizophrenia and his mother was a battered woman and was dealing with a range of her own psychiatric issues.
At some point I was able to discover the company which produced many of the pornographic films he'd starred in. During my eighteen months of digging I located Alex West, a journalist who worked for the Sun in London, England who had seen several clips online of Luka abusing kittens. He wrote about the disturbing videos saying that the first sign of a killer was often torturing and killing animals.
Once the kitten killings went viral, former Playboy Playmates the Barbie twins became involved. I got to know them very well as they tried to track Luka before he killed the Chinese student but couldn't locate him in time. They also believed killing animals was a precursor to killing people. The sisters led me to friends of theirs who lived on Long Island that had a television show called Rescue Inc, where they saved animals that were in trouble.
The head of their group was a guy named Joe from New Jersey. Joe and his friends were former cops and criminals who became involved in protecting animals.
After all of our investigating, interviews, shooting and editing the film was submitted the film to The Montreal World Film Festival where it was accepted.
A day before the film was supposed to premiere we received a call from the festival director saying it was canceled. The film was on trial. It had become essentially a witness in the soon to be tried Magnotta case. The Crown Prosecutor in Quebec had a copy of the print which was taken from the festival and sent to his office. All the posters and publicity surrounding the film had been torn down and the caveat to the craziness —— our film had been canceled, not just for the festival but as well for television.
At this point the film's Executive Producer, Garry Blye, said we needed to take action. He suggested we hold a press conference in Montreal at the same time our film was supposed to premiere. A day later we stood in front of over a thousand journalists with cameras, microphones and iphone's ready to hear our side of the story.
I wanted to tell them how unfair canceling the film was. We talked about the privilege it was working with Bell Canada, A&E and the rest of our broadcasters. We assured the press we were in full agreement that holding the film back until after the trial was a good idea. It was never our intent to sway a jury. Naomi told the media our experts were all amazing and the people who knew Luka were incredibly co-operative. The next day there were over a 1000 stories about the film off and online from publications around the planet including – Turkey, China, Russia, etc…
These stories were seen by film acquisition managers all over the world resulting in hundreds of calls from interested channels, networks, streamers and distributors. In the end, the film was sold to over 100 territories around the globe.
As for Luka, he was found guilty of first degree murder and will most likely spend the rest of his life in jail. Recently he actually found love in prison and got married.
Who knew a show we produced during the early days of OUT TV, looking for Canada's top model, could change our lives so dramatically.
Steven Spielberg's The Fablemans Wins the Coveted People's Choice Award at TIFF
Ezra Comes out of the Cave and into the Inferno of Passion – Interview with Luke Hutchie


I haven't updated you on our poopy situation because I was so fed up that I didn't want to talk about it. The lab results came eventually and revealed that they had oocysts in their poop. As far as I understood, this means that they have/had Apicomplexa, a parasite. Like Giardia this parasite doesn't always cause symptoms, so they've most likely had it for a longer time. The vet assumes that the stress of moving to my place has triggered the outburst. This diagnosis also explains why the diarrhea actually stopped by itself, because cats can sometimes handle this parasite on their own.


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You are here: Home / Arts, Culture, Entertainment, Meetings, Events / Live Oak Theatre adds two additional performances of The Best Christmas Pageant Ever
Live Oak Theatre adds two additional performances of The Best Christmas Pageant Ever
November 27, 2019 by Press Release
Brooksville, Florida (November 26, 2019) – Due to the high demand for tickets, Live Oak Theatre has added two additional performances to its original run. The Best Christmas Pageant Ever will now be presented on December 12, 13,14, and 15.
Showtimes are at 7:30 pm on Thursday & Friday; and 2:30 pm and 7:30 pm on Saturday & Sunday.
Adult tickets are only $20; children ages 13 and younger are $5 when purchased with an adult ticket. All tickets include coffee, tea and dessert.
For more information and to purchase tickets call 352-593-0027 or email LiveOakBoxOffice@GMail.com or visit www.LiveOakTheatre.org.
Follow us on Facebook at www.Facebook.com/LiveOakTheatre.
About The Best Christmas Pageant Ever
The Best Christmas Pageant Ever is a classic Christmas story that's sure to give audiences the warm and fuzzies. It certainly did for directors Lindsay McTague and Katie Moore. (A tear or two may have been shed during the read through of the script.) This contemporary story begins with an ordinary family- The Bradleys. They go to an ordinary church, which puts on the same, ordinary Christmas pageant every year. That is, until, Mrs. Bradley is suddenly put in charge of the pageant. But who should come in and literally steal the main roles? The Herdmans- "the WORST kids in the whole history of the world", as young Beth Bradley puts it. The whole town is sure the pageant will be a disaster. But Mrs. Bradley is determined to put on the BEST Christmas Pageant Ever. The Herdmans, who've never even cracked the spine of Bible, hear the Christmas Story for the first time, and are tasked with bringing it to life, "ordinary" isn't the word ANYONE would use. But will this be the worst or the best Christmas pageant ever? The stage play was adapted from the children's book, and later adapted into a movie. No matter how you first hear the story, your sure to laugh at all the wild antics of the Herdmans, and smile as you ponder the Christmas story through new eyes.
About Live Oak Theatre Company (LOT),
Live Oak Theatre Company (LOT) is a not-for-profit 501 (C) (3) company, located at 21030 Cortez Boulevard, Brooksville, FL 34601. The Live Oak Theatre Company exists to enrich families, individuals and the community by providing positive artistic experiences in the Performing Arts – including excellent, affordable, and edifying family friendlyentertainment, performance, and educational opportunities for Theatre patrons and participants of all ages.
For more information about the Live Oak Theatre Company, including sponsorship and audition opportunities, call 352-593-0027, email liveoakboxoffice@gmail.com , or go to www.LiveOakTheatre.org. Visit us on Facebook at https://www.facebook.com/LiveOakTheatre.
Filed Under: Arts, Culture, Entertainment, Meetings, Events


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2019 Women's World Cup Group A
The 2019 Women's World Cup kicks off on the 7th June, with host nation having been draw in Group A alongside South Korea, Norway and Nigeria. In what promises to be one of the most competitive groups in the entire tournament, many are already predicting that it will be a shootout between South Korea and Norway as to who joins France in the knockout phase. With that being said, let's take a closer look at the teams involved in Group A, including the full schedule and player news.
2019 Women's World Cup: Group A Table & Schedule
2019 Women's World Cup: Group A Schedule
7 June 2019 20:00 France vs South Korea Parc des Princes, Paris
8 June 2019 20:00 Norway vs Nigeria Stade Auguste-Delaune, Reims
12 June 2019 14:00 Nigeria vs South Korea Stade des Alpes, Grenoble
12 June 2019 20:00 France vs Norway Allianz Riviera, Nice
17 June 2019 20:00 Nigeria vs France Roazhon Park, Rennes
17 June 2019 20:00 South Korea vs Norway Stade Auguste-Delaune, Reims
France have moved up to No.4 in the world rankings over recent times, with the host nation being among the favourites for this summer's World Cup as a result. Head Coach Corinne Diacre took over back in 2017, leading Les Bleus to three straight friendly victories ahead of the tournament, including a 4-0 win over Denmark in April. Having qualified for the event directly due to their status as the host nation, a lack of competitive action may hinder their progress, however they are likely to have the quality to make it through to the Round of 16 with relative ease.
Vice-Captain Eugénie Le Sommer will lead the line for the French over the coming weeks, with the Lyon striker having scored an impressive 74 international goals since her debut back in 2009. The hosts have a hugely experienced playing squad to select from, with Gaëtane Thiney, Élise Bussaglia and Wendie Renard all having over 100 caps their name. Despite the fact that France's best finish at a World Cup was when they came in fourth position in 2011, whilst they have never progressed past the quarter-finals of a European Championship, France are certain to be a threat this summer, with their opening match against South Korea potentially deciding who will finish on top spot in Group A.
Having made it through to the Round of 16 four years ago in Canada, South Korea will be hoping for an even better showing in France, with Yoon Deok-yeo's team having only qualified for the Finals back in April following their 5th place finish in the 2018 AFC Women's Asian Cup. Coming into the tournament 14th in the world rankings, South Korea are not in the best of form of late, having registered one draw and one defeat in two recent friendly clashes against Iceland.
Chelsea forward Ji So-yun is undoubtedly the star of the South Korean team, with the 28-year-old having scored 49 goals at the international level to date. Elsewhere, midfielder Cho So-hyun will look to provide the creativity to unlock the defences of their Group A opponents, with the recent West Ham United signing having 120 caps to her name. Despite facing a tough start against host nation France, many are backing South Korea to progress to the Round of 16 once again.
Former World Cup winners Norway may not currently be the threat in which they once were, however the Scandinavian side still has the potential to be a threat in Group A. Head Coach Martin Sjögren has been in charge since 2016, leading Norway to an excellent qualifying campaign, in which they won seven of their eight matches, in a group that included the Netherlands, Republic of Ireland and Slovakia. The Grasshopppers conceded just four goals during qualifying, scoring 22 in the process.
Norway will be without 2019 Ballon d'Or winner Ada Hegerberg, who decided to make herself unavailable for the national team back in 2017 due to a dispute over how the NFF treat the women's game. As a result, the pressure will be on striker Isabell Herlovsen to fill the void left by Hegerberg, with the 30-year-old having scored 55 goals for Norway to date. An opening round win over Nigeria is a must for Norway if they are to have any hope of making it through to the knockout phase in France.
Nigeria may be found down in 38th place in the world rankings, the African Cup of Nations champions may well surprise supporters in France. Making their eight appearance in the Finals, the Super Falcons best finish to date came back in 1999, where they made it through to the quarter finals. Head Coach Thomas Dennerby took over in 2018 following spells with a range of teams in his native Sweden, with the boss having many talented players to select from. Barcelona forward Asisat Oshoala and Brighton & Hove Albion's Ini-Abasi Umotong could well form a formidable strike partnership in France. Nigeria come into the tournament off the back of a 2-1 friendly defeat at the hands of Canada.
2019 Women's World Cup: Group A Odds
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To Win Group A:
France @1/4
Norway @5/1
South Korea @7/1
Nigeria @33/1
World Cup Winner:
France @10/3
Norway @33/1
South Korea @50/1
Nigeria @500/1
Player of the Tournament:
Amandine Henry (France) @4/1
Eugénie Le Sommer (France) @6/1
Ji So-yun (South Korea) @33/1
Caroline Graham Hansen (Norway) @66/1


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Firms which have been "crippled" by Tier four are demanding extra authorities assist
By Elton Drewry On Dec 21, 2020
Woman wearing protective UK flag anti coronavirus face mask in front of Big Ben, London, England
Companies are calling for further assistance to prevent the collapse following the government shock from Tier 4 on Saturday.
In particular, corporate lobby groups are calling for business tariff vacation to be extended by an additional 12 months from January, sales tax breaks and additional direct support for businesses forced to close.
According to real estate consultant Altus Group, business rates are set to resume from April, which will cost powerful companies £ 12.8 billion in 2021.
> See also: The bounce-back loan program was extended to the end of March
Non-essential stores, hairdressers, and leisure and entertainment venues were forced to close on Saturday afternoon yesterday after Boris Johnson's abrupt announcement to enforce new Tier 4 restrictions amid concerns about a more virulent strain of coronavirus spreading rapidly in London and the southeast.
Health Secretary Matt Hancock yesterday suggested keeping Tier 4 until the vaccination program has adequately immunized the population – which could take months. The talk now is that Tier 4 will be imposed by Easter.
In the meantime, all non-essential businesses in Wales will be closed from Boxing Day onwards, subject to England's Tier 4 restrictions, while Scotland and Northern Ireland will have their own three-week and six-week lockdowns from December 26th, respectively.
> See also: Europe's small businesses are calling for a transition period of three months after Brexit
The harsh new rules have dashed hopes for independent business recovery over the Christmas period. Economic forecaster CEBR estimates that retailers alone will lose £ 2 billion in sales for every week of Tier 4 lockdown.
Adam Marshall of the UK Chambers of Commerce wondered in the Financial Times if there would be any help for companies that have "paid for stocks they can't sell now."
"What support will there be for companies whose cash flow forecasts are again in chaos?" he asked.
Helen Dickinson, executive director of the British Retail Consortium, said the consequences of Tier 4 would be "severe" for a sector expected to lose around £ 2 billion in sales per week. "For businesses, the government's stop-start approach is deeply of little help. That decision is made just two weeks after the last national lockdown ended and in the midst of peak trading that so many depend on to fuel their recovery," she said .
Dickinson said the government "needs to provide additional financial assistance to help these companies get back on track" and that extending interest rate relief in 2021 "is the best place to start".
Richard Lim, head of retail economics consultancy, told The Times that closing non-essential retail stores before Christmas would be a "crippling blow" to businesses in London and the Southeast.
Last week, the government extended the vacation program to the end of April for assistance available. The vacation covers up to 80 percent of the staff wages for hours not worked.
Companies that are forced to shout can also apply for grants of up to £ 3,000 per month. However, local authorities are still working to hand out grants from the November national lockdown, let alone prepare for another round of handout distribution.
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