text:
\section{Introduction}

In Grand Unified Theories with a simple compact gauge group the series
of symmetry breakings necessary to reach the standard model inevitably
gives rise to monopoles. These monopoles interact with the other
particles of the theory and the nature of this interaction is of
interest. This interaction has been studied by a number of authors
\cite{Rubakov82}-\cite{Olsen90}. As well as scattering states
\cite{Marciano83}, bound states of bosons and fermions with monopoles
have also been found \cite{Tang82}-\cite{Ajithkumar88}. Perhaps the
most interesting finding is the baryon decay catalysis by a monopole
\cite{Rubakov82}.

Here we are most interested in the work of Sonoda \cite{Sonoda84}. In
ref.~\cite{Sonoda84} the physics of the massive gauge boson $(W^{\pm})$
and the neutral Higgs scalar $(H)$ in the presence of a 't Hooft-Polyakov
monopole \cite{tHooft74,Polyakov74} was investigated. Spherically symmetric
perturbations of the fields of the 't Hooft-Polyakov monopole were considered
and a series of bound states were found. The eigenvalue of the lowest bound
state was numerically estimated to be $\omega \simeq 0.876 m$ in the
Prasad-Sommerfield limit ($m$ is the mass of the gauge boson).
We investigate these bound states further by using a more systematic
approach to the study of general angular momentum perturbations in the fields
of the 't Hooft-Polyakov monopole. Our motivation for studying them is,
however, quite different.

In ref.~\cite{Montonen77} Montonen and Olive observed that in the Bogomol'nyi
limit \cite{Bogomolnyi76} the masses of all the particles of the theory are
given by the same universal mass formula $m^2 =\eta^2 (q^2 + g^2)$, where
$q$ is the electric and $g$ the magnetic charge of the particle. This formula
applies to both the classical non-singular monopole and dyon solutions as
well as the normal particles, and in part motivated Montonen and Olive to
conjecture that the $W$-bosons and the monopoles are dual particles.
The recent construction of supersymmetric models \cite{Seiberg94,Sen94} which
realize this electromagnetic duality has stimulated a great deal of interest.

We do not consider these supersymmetric models here, however, but consider the
non-supersymmetric model of 't Hooft and Polyakov. We note from the literature
cited above that the massive particles of this theory, whose masses are given
by the
universal mass formula, namely the monopole, dyon and $W$-boson form bound
states with one another. So here we seek to answer the question: are the bound
state energies of the various two particle systems related by electromagnetic
duality?

To answer this question we need to find the values of the bound state energies
in question. The bound state energies for the monopole-dyon system were found
in ref.~\cite{Gibbons86}. Here we try to complete the picture by using
perturbative methods to investigate the bound state energies of $W$-bosons in
the presence of a monopole, a dyon and another $W$-boson.

\section{Yang-Mills-Higgs model}

We are considering field configurations in the Yang-Mills-Higgs theory with
the lagrangian
$$
{\cal L} = -\frac{1}{4} F^a_{\mu\nu} F^{a\mu\nu} + \frac{1}{2}
(D_{\mu}{\rm \Phi}^a)(D^{\mu}\Phi^a)
-\frac{\lambda}{4}(\Phi \cdot \Phi-\eta^2)^2,
$$
where
\begin{eqnarray}
F_{\mu\nu}^a & = & \partial_{\mu}A_{\nu}^a
-\partial_{\nu}A_{\mu}^a
+q\epsilon_{abc}A_{\mu}^bA_{\nu}^c \nonumber \\
D_{\mu} \Phi^a & = & \partial_{\mu}\Phi^a
+q\epsilon_{abc}A_{\mu}^b \Phi^c, \nonumber
\end{eqnarray}
and $A_{\mu}^a$ and $\Phi^a$ are $SU(2)$ triplets. We will consider
monopole, dyon and point electric solutions to the field equations
\begin{eqnarray}
D_{\mu}D^{\mu} \Phi^a +\lambda (\Phi \cdot \Phi-\eta^2)
\Phi^a & = & 0 \nonumber \\
D_{\mu}F^{\mu\nu a} + q\epsilon_{abc}\Phi^bD^{\nu}\Phi^c
& = & 0. \nonumber
\end{eqnarray}
It will be convenient later on if we use dimensionless fields and co-ordinates,
so we make the replacements
$$
r=q\eta R, \hspace{5mm} \phi^a = \eta \Phi^a, \hspace{5mm}
W_{\mu}^a = \eta A_{\mu}^a .
$$
Note that the mass of the massive gauge bosons is $m_w=q\eta $.
With these dimensionless variables we will take the ansatz for the monopole
with unit magnetic charge to be \cite{tHooft74,Polyakov74}
$$
\phi^a = \hat{r}^a H(r), \hspace{5mm}
W_i^a = -\epsilon_{iab} \frac{\hat{r}^b}{r} W(r),\hspace{5mm}
W_0^a = 0.
$$
The profile functions are solutions to
\begin{eqnarray}
-\frac{1}{r^2}\frac{d}{dr}\left( r^2\frac{dH}{dr}\right) +2H\frac{(1-W)^2}{r^2}
+\frac{\beta}{2}(H^2 -1)H & = & 0 \nonumber \\
-\frac{d^2 W}{dr^2} + H^2(W-1)+(W-1)\frac{W(W-2)}{r^2} & = & 0, \nonumber
\end{eqnarray}
where we have defined $\beta =2\lambda /q^2$. We will only be considering the
Bogomol'nyi limit $(\beta \rightarrow 0)$ where we have the analytic
Prasad-Sommerfield solution \cite{Prasad75}
$$
H(r) = \frac{\cosh (r)}{\sinh (r)} - \frac{1}{r}, \hspace{5mm}
W(r) = 1 -\frac{r}{\sinh (r)}.
$$
For the dyon with unit magnetic charge we will take the ansatz \cite{Julia75}
$$
\phi^a = \hat{r}^a H(r), \hspace{5mm}
W_i^a = -\epsilon_{iab} \frac{\hat{r}^b}{r} W(r),\hspace{5mm}
W_0^a = \hat{r}^a J(r),
$$
where the profile functions are solutions to
\begin{eqnarray}
-\frac{1}{r^2}\frac{d}{dr}\left( r^2\frac{dH}{dr} \right) +
2H\frac{(1-W)^2}{r^2}
+\frac{\beta}{2}(H^2 -1)H & = & 0 \nonumber \\
-\frac{d^2 W}{dr^2} +(H^2-J^2)(W-1)+(W-1)\frac{W(W-2)}{r^2} & = & 0 \nonumber \\
-\frac{1}{r^2}\frac{d}{dr}\left( r^2\frac{dJ}{dr} \right)
+ 2J\frac{(1-W)^2}{r^2} & = & 0 .\nonumber
\end{eqnarray}
In the Bogomol'nyi limit we will follow Bogomol'nyi \cite{Bogomolnyi76} and
take the dyon solution to have the form
$$
\phi^a = \overline{\phi}^a (\sqrt{1-C^2}\: {\bf x} ), \hspace{5mm}
W_i^a = \sqrt{1-C^2} \: \overline{W}_i^a (\sqrt{1-C^2} \:{\bf x} ),
\hspace{5mm}
W_0^a = C \: \overline{\phi}^a (\sqrt{1-C^2} \: {\bf x} )
$$
where $(\overline{\phi}^2,\overline{W}_i^a)$ is the Prasad-Sommerfield monopole
solution given above. The parameter $C$ is given by
$$
C=\frac{e}{\sqrt{1+e^2}} \hspace{8mm} \mbox{where} \hspace{8mm}
e=\frac{qQ}{4\pi}
$$
and $Q$ is the electric charge of the dyon. The mass of the dyon in
dimensionfull units is given by
$$
E= \frac{4\pi \eta}{q} \sqrt{1+e^2},
$$
and setting $e=0$ gives the mass of the monopole. From this we can see that the
quantity $(1-C^2)^{1/2} \equiv (1-e^2)^{-1/2}$ appearing in the rescaling above
is the ratio of the monopole and dyon masses. Note that with this form of the
solution the profile function $J$ is given by $J=CH$.

To investigate the bound state energy of a $W$-boson in the presence of another
$W$-boson we will consider perturbations about the point electric solution
$$
\phi^3 = (1-\frac{Z}{r} ), \hspace{5mm} W_0^3 = -\frac{Z}{r}.
$$
Substituting this ansatz into the field equations gives
\begin{eqnarray}
D_i^2 \phi^3 & = & 4\pi Z \delta^3 (r) \nonumber \\
D_iG_{0i}^3 & = & 4\pi Z \delta^3 (r) \nonumber
\end{eqnarray}
and so we have a point electric source of strength $4\pi Z$ in rescaled units.
Note that in the rescaled units the $W$-boson has charge 1, and so $Z=1/4\pi$
corresponds to a $W$-boson source.

Now to find the equation for the perturbations
$(\delta \phi^a,\delta W_{\mu}^a)$ about a solution
$(\phi^a ,W_{\mu}^a )$ to the field equations, we take
the second functional derivative of the action. To remove the gauge degrees
of freedom we use the background gauge condition
$$
D^{\mu} \delta W^a_{\mu} + \epsilon^{abc} \phi^b \delta \phi^c = 0.
$$
Imposing this gauge condition, we find the gauge fixed perturbation equations
are
\begin{eqnarray} \label{pert}
\left( \begin{array}{c}
       (D^2)_{ab} + \delta_{ab} (\frac{\beta}{2} (\phi \cdot \phi -\eta^2 )
       +  \phi \cdot \phi ) + (\beta -1) \phi^a \phi^b \hspace{5mm}
        2\epsilon^{abc} D^{\mu} \phi^c \\ 2\epsilon^{abc} D^{\nu} \phi^c
        \hspace{5mm}
        g^{\mu\nu} ((D^2)_{ab} + \delta_{ab} \phi \cdot \phi - \phi^a \phi^b )
        + 2\epsilon^{abc} G^{\mu\nu c}
       \end{array}  \right) \left(
       \begin{array}{c}
           \delta \phi^b \\ \delta W_{\mu}^b
        \end{array} \right) & &  \\
        & = & 0,\nonumber
\end{eqnarray}
where the metric is given by $g^{\mu \nu} = diag (-1,1,1,1)$.
The gauge condition does not, however, fix the gauge completely. To remove the
remaining degrees of freedom requires the introduction of Fadeev-Popov ghosts
$\eta^a$ which satisfy
\begin{eqnarray}
\left( (D^2)_{ab} + \delta_{ab} \phi \cdot \phi - \phi^a \phi^b \right)
\eta^b  =  0 \label{FP}.
\end{eqnarray}
If we count the number of degrees of freedom we find that equation \ref{pert}
has 15 and equation \ref{FP} has 3. When calculating any quantum corrections
the Fadeev-Popov ghost contribution appears with a factor of $-2$ relative to
the contribution from the modes of equation \ref{pert}. In this way the
Fadeev-Popov ghosts cancel 6 degrees of freedom of equation \ref{pert} leaving
9 physical degrees of freedom. They correspond to 6 degrees of freedom for the
2 massive gauge bosons, 2 degrees of freedom for the massless gauge boson and
one degree of freedom for the Higgs boson.

We seek to solve equation \ref{pert} for the physical bound states for general
angular momentum in the presence of the three solutions given above. To check
that the states we find are actually physical states we shall also solve
equation \ref{FP}. If the eigenspectrums of equation \ref{FP} correspond to an
eigenspectrum of equation \ref{pert} then that eigenspectrum is not physical;
it is cancelled by the ghosts.

\section{Perturbations about the monopole}

If we substitute the monopole solution into the gauge fixed second variation
of the action, we find
\begin{eqnarray}
\delta S_{gf} & = & \frac{\eta}{q}
\int r^2 dr \int d \Omega \left( \phi^a \left[ D^2_{ab} +
 \delta^{ab} \left( \frac{\beta}{2}(H^2 -1) + H^2 \right)
 + \hat{r}^a \hat{r}^b(\beta -1)H^2 \right] \phi^b
  \right. \nonumber \\
  & - & \left. W_0^a \left[ D^2_{ab}+H^2(\delta^{ab}-\hat{r}^a \hat{r}^b)
  \right] W_0^b + W_i^a \left[ D^2_{ab}+H^2(\delta^{ab}-\hat{r}^a \hat{r}^b)
  \right] W_i^b \right. \nonumber \\
 & + & \left. 4\epsilon^{abc} \phi^a W_i^b (D_i \phi )^c
 + 2\epsilon^{abc}  W_i^a W_j^b G_{ij}^c \right),
\end{eqnarray}
and the Fadeev-Popov ghost term is
\begin{eqnarray}
\delta S_{FP} & = & \frac{\eta}{q}
\int r^2 dr \int d \Omega  \eta^a \left[ D^2_{ab}
+H^2(\delta^{ab}-\hat{r}^a \hat{r}^b) \right] \eta^b ,
\end{eqnarray}
where
\begin{eqnarray}
(D^2 )_{ab} & = & \delta_{ab} \partial_0^2 + \delta_{ab}
\left( -\nabla_r^2 + \frac{L^2}{r^2}
+ \frac{W^2}{r^2} \right)  - 2\frac{W}{r^2} \epsilon_{iab} L_i
+ \hat{r}^a \hat{r}^b \frac{W^2}{r^2}  \nonumber \\
G_{ij}^c & = & \epsilon^{kij} \left[ \hat{r}^k \hat{r}^c
\left( \frac{W^{\prime}}{r} + \frac{W}{r^2} (W-2) \right)
- \delta^{kc}\frac{W^{\prime}}{r}  \right] \nonumber \\
(D_i \phi )^c & = &  \hat{r}^i \hat{r}^c\left( H^{\prime}-\frac{H}{r}(1-W)
\right) + \delta^{ic} \frac{H}{r} (1-W), \nonumber
\end{eqnarray}
and $L$ is the angular momentum operator $L=-i\hat{r}\times \nabla$.
Note that because the monopole solution is time independent and $W_0^a =0$ we
can separate out the time derivatives and set up a standard eigenvalue problem.

The monopole solution is invariant under rotations generated by
$K=L+T+S$, and so we seek to expand the perturbations as eigenstates of $K$.
First we consider the scalar fields $(W_{0}^a,\eta^a,\phi^a )$ which have spin
zero and so $K=L+T$. Simultaneous eigenfunctions of $K^2$, $K_3$, $L^2$, and
$T^2$ can be constructed from eigenfunctions of $L^2$, $L_3$, $T^2$ and $T_3$
by using standard Clebsch-Gordan methodology
$$
\Phi^{km}_{lt} = \sum_{m_l m_t} {\cal C}^{km}_{ltm_lm_t} Y_{lm_l} Y_{tm_t}
\hspace{5mm} \mbox{where} \hspace{5mm} m=m_l + m_t
$$
where ${\cal C}$ are Clebsch-Gordan coefficients and $Y_{lm_l}$ are the
spherical harmonics. We will, however follow ref.~\cite{Carson90} and define
vector spherical harmonics in terms of operators acting on the spherical
harmonics by
$$
Y^a_{kjm} = N_{k,j}^{-1/2} O_j^a Y_{k,m}.
$$
The vector operators $O_j$ are given by
\begin{eqnarray}
O_0 & = & iL \nonumber \\
O_1 & = & \hat{r} (\overline{L} +1) + i\hat{r}\times L \nonumber \\
O_{-1} & = & \hat{r}\overline{L} - i\hat{r}\times L \nonumber
\end{eqnarray}
where
$$
\overline{L} \equiv \left(L^2 +\frac{1}{4} \right)^{1/2} -\frac{1}{2}.
$$
The normalization constants $N_{k,j}$ are given by the orthogonality condition
$$
\int d\Omega (Y^a_{kjm})^{\dagger} Y^a_{k^{\prime}j^{\prime}m^{\prime}} =
\delta_{kk^{\prime}} \delta_{jj^{\prime}} \delta_{mm^{\prime}}.
$$
The vector spherical harmonics $Y^a_{kjm}$ defined above are eigenfunctions of
$K^2$, $K_3$, $L^2$, and $T^2$
\begin{eqnarray}
K^2 Y^{a}_{kjm} & = & k(k+1) Y^{a}_{kjm} \nonumber \\
K_3 Y^{a}_{kjm} & = & m Y^{a}_{kjm} \nonumber \\
L^2 Y^{a}_{kjm} & = & (k+j)(k+j+1) Y^{a}_{kjm} \nonumber \\
T^2 Y^{a}_{kjm} & = & 2 Y^{a}_{kjm} \nonumber
\end{eqnarray}
We now expand the scalar fields as
\begin{eqnarray}
\phi^a & = & \sum_{km} \sum_{j=-1}^{1} \phi^{km}_{j} Y^a_{kjm} \nonumber \\
\eta^a & = & \sum_{km} \sum_{j=-1}^{1} \eta^{km}_{j} Y^a_{kjm} \nonumber \\
W_0^a & = & \sum_{km} \sum_{j=-1}^{1} \tilde{W}^{km}_{j} Y^a_{kjm}. \nonumber \\
\end{eqnarray}
Note that the orbital angular momentum $l$ is given by $l=k+j$, and so for
each value of $k$ there will be three values of $l$. So the subscript $j$ on
the vector spherical harmonics denotes the three different values of $l$ that
are possible for each value of $k$.

For the spatial components of the gauge field we have non-zero spin and isospin
and so $K=L+T+S$. There are two different ways of adding three angular momenta,
differing in the order in which the addition is performed. We choose to
first construct eigenvectors of
$$
J=T+S.
$$
Note that $J$ here is not the total angular momentum operator. For spin one and
isospin one there are three possible values of $j$, $j=0,1,2$ denoting scalar,
pseudovector and tensor representations of $J$.
If we then consider eigenstates of
$$
K=J+L = T+S+L
$$
we can see that in general for each $k$; for $j=0$ there is only one value of
$l$, $l=k$; for $j=1$ there are three values of $l$, $l=|k-1|,k,k+1$;
and for $j=2$ there are five values of $l$, $l=|k-2|,|k-1|,k,k+1,k+2$.
So we introduce the variables $j_1$ and $j_2$ where $j_1$ is the
$J$ eigenvalue and $j_2$ takes the values $j_2 = -j_1, \ldots ,j_1$ so that in
general, $l$ is given by
$$
l=k+j_2.
$$
With this notation we expand $W_i^a$ in tensor spherical harmonics as
$$
W_i^a  =  \sum_{km} \sum_{j_1=0}^{2} \sum_{j_2=-j_1}^{j_1}
W^{km}_{j_1j_2} Y^{ai}_{kj_1j_2m}
$$
where the tensor spherical harmonics $Y^{ai}_{kj_1j_2m}$ are defined in terms of
operators acting on spherical harmonics by
$$
Y^{ai}_{kj_1j_2m} = N_{kj_1j_2}^{-1/2} O^{ai}_{j_1j_2} Y_{km},
$$
where the tensor operators $O_{j_1j_2}$ are constructed from the $O_j$.
The tensor spherical harmonics defined in this way are
eigenfunctions of $K^2$, $K_3$, $L^2$ and $J^2$
\begin{eqnarray}
K^2 Y^{ai}_{kj_1j_2m} & = & k(k+1) Y^{ai}_{kj_1j_2m} \nonumber \\
K_3 Y^{ai}_{kj_1j_2m} & = & m Y^{ai}_{kj_1j_2m} \nonumber \\
L^2 Y^{ai}_{kj_1j_2m} & = & (k+j_2)(k+j_2+1) Y^{ai}_{kj_1j_2m} \nonumber \\
J^2 Y^{ai}_{kj_1j_2m} & = & (T+S)^2 Y^{ai}_{kj_1j_2m} =
j_1 (j_1 +1) Y^{ai}_{kj_1j_2m}. \nonumber
\end{eqnarray}
Note that the scalar ($j_1=0$), pseudovector ($j_1=1$) and tensor ($j_1=2$)
representations of $J$ have all been incorporated into the single set of
functions $Y^{ai}_{kj_1j_2m}$.

Now the monopole is also invariant under the action of the parity operator $P$.
If we apply the parity operator to the perturbations, then from
$$
P Y^{a}_{kjm} = (-1)^{k+j} Y^{a}_{kjm} \hspace{5mm} \mbox{and} \hspace{5mm}
P Y^{ai}_{kj_1j_2m} = (-1)^{k+j_2} Y^{ai}_{kj_1j_2m}
$$
we find that the perturbations separate into two channels under parity given by
$$
\psi^+_{km} = \left( \begin{array}{c}
                W^{km}_{22} \\ W^{km}_{20} \\ W^{km}_{2-2} \\ W^{km}_{10} \\
                W^{km}_{00} \\ \phi^{km}_{0}
                \end{array} \right), \hspace{20mm}
\psi^-_{km} = \left( \begin{array}{c}
                W^{km}_{21} \\ W^{km}_{2-1} \\ W^{km}_{11} \\ W^{km}_{1-1} \\
                \phi^{km}_{1} \\ \phi^{km}_{-1}
                \end{array} \right)
$$
where the superscripts are such that
\begin{eqnarray}
+ & & \hspace{5mm} \mbox{denotes parity} \hspace{5mm} (-1)^k \nonumber \\
- & & \hspace{5mm} \mbox{denotes parity} \hspace{5mm} (-1)^{k+1}.\nonumber
\end{eqnarray}
The perturbation equations for $\psi_{km}^{\pm}$ decouple. Note that the vector
and tensor spherical harmonics are eigenstates of $K^2$, $K_3$, $L^2$ and $J^2$
and so are not eigenstates of the charge operator
$$
\hat{Q} = \hat{r}.T.
$$
Substituting these expansions into $\delta S_{gf}$ and performing the
angular integrals gives the perturbation operator
\begin{eqnarray}
M_{\alpha \beta}^{(\pm)} & = & \delta_{\alpha \beta} \left[
-\nabla_r^2 + \frac{(k+j_2 )(k+j_2 +1)}{r^2} + \frac{W^2}{r^2} + H^2
+\frac{\beta}{2} (H^2 -1) P_{\phi} \right] \nonumber \\
 & + & 2\frac{W}{r^2} C_{\alpha\beta}^{(\pm )(1)}
 + \left( \frac{W^2}{r^2} - H^2 + \beta H^2 P_{\phi} \right)
 C_{\alpha\beta}^{(\pm )(2)} \nonumber \\
 & + & 2\frac{W^{\prime}}{r} C_{\alpha\beta}^{(\pm )(3)}
  +  2 \frac{W}{r^2} (W-2) C_{\alpha\beta}^{(\pm )(4)} \nonumber \\
 & + & 4\frac{H}{r} (1-W) C_{\alpha\beta}^{(\pm)(5)}
 + 4 \left( H^{\prime} - \frac{H}{r} (1-W) \right)
 C_{\alpha\beta}^{(\pm )(6)} \nonumber
\end{eqnarray}
where $(\pm )$ denotes the two channels with different parity, the
subscripts $\alpha$ and $\beta$ denote the modes labeled by $j_1$
and $j_2$ (i.e. $\alpha = (j_1,j_2)$ and
$\beta = (j_1^{\prime},j_2^{\prime})$), and $P_{\phi}$ is a projection operator
which is 1 for a $\phi $ field and 0 for the $W$ fields. The
$C_{\alpha\beta}^{(\pm )(1-6)}$ are symmetric matrices of coefficients
resulting from the integration over angular co-ordinates. This notation is
chosen to be the same as that used in ref.~\cite{Carson90}, and the matrix
elements of $C_{\alpha\beta}^{(\pm )(1-6)}$ are taken from there.

The perturbations for the time components (and the ghosts) similarly separate
under parity
$$
\tilde{W}_{km}^+ = \left( \begin{array}{c}
                       \tilde{W}_{0}^{km}
                       \end{array} \right) \hspace{5mm}
\tilde{W}_{km}^- = \left( \begin{array}{c}
                       \tilde{W}_{1}^{km} \\ \tilde{W}_{-1}^{km}
                       \end{array} \right) \hspace{20mm}
$$
where the superscripts $\pm$ denote the parities as before.

The perturbation operator for the time components (and the ghosts) is
\begin{eqnarray}
\tilde{M}_{\alpha\beta}^{(\pm)} & = & \delta_{\alpha\beta}
\left[ -\nabla_r^2 + \frac{(k+j_2)(k+j_2+1)}{r^2} + \frac{W^2}{r^2} + H^2
\right] \nonumber \\
& +& \frac{2W}{r^2} C_{\alpha\beta}^{(\pm)(1)} + \left( \frac{W^2}{r^2} - H^2
\right) C_{\alpha\beta}^{(\pm)(2)} \nonumber
\end{eqnarray}
where the $C_{\alpha\beta}^{(\pm)(1,2)}$ elements are the same as those for
the $\phi$ terms.

\subsection*{$k=0$ modes}

If we turn our attention to the $k=0$ modes we find that, because of the
constraints $l=k+j_2 \geq 0$ and $k \geq 0$, the number of channels for $k=0$
are reduced. For $k=0$ the allowed perturbations are
$$
\psi_{00}^+ = \left( \begin{array}{c}
                     W_{22}^{00} \\  W_{00}^{00}
                     \end{array} \right)  \hspace{15mm}
\psi_{00}^- = \left( \begin{array}{c}
                     W_{11}^{00} \\  \phi_{1}^{00}
                     \end{array} \right).
$$
The $\psi_{00}^+$ mode satisfies the perturbation equation
\begin{equation} \label{+zero}
\left( \begin{array}{cc}
        M_{11} & M_{12} \\ M_{21} & M_{22}
        \end{array}  \right) \left(  \begin{array}{c}
                                      W_{22}^{00} \\ W_{00}^{00}
                                     \end{array} \right) = \omega^2
        \left(  \begin{array}{c}
                W_{22}^{00} \\ W_{00}^{00}
                  \end{array} \right),
\end{equation}
where
\begin{eqnarray}
M_{11} & = & -\nabla_r^2 + \frac{6}{r^2} - 6\frac{W}{r^2} + \frac{5}{3}
\frac{W^2}{r^2} + \frac{1}{3} H^2 + \frac{8}{3} \frac{W^{\prime}}{r}
+\frac{2}{3} \frac{W(W-2)}{r^2} \nonumber \\
M_{22} & = & -\nabla_r^2 + \frac{4}{3} \frac{W^2}{r^2} + \frac{2}{3} H^2
- \frac{8}{3} \frac{W^{\prime}}{r}
+\frac{4}{3} \frac{W(W-2)}{r^2} \nonumber \\
M_{12} & = & M_{21} = \frac{\sqrt{2}}{3} \left( \frac{W^2}{r^2} -H^2 \right)
- 2 \frac{\sqrt{2}}{3} \left( \frac{W^{\prime}}{r} + \frac{W(W-2)}{r^2}
\right), \nonumber
\end{eqnarray}
and the $\psi_{00}^-$ mode satisfies the perturbation equation
$$
\left(  \begin{array}{c}
        -\nabla_r^2 + 2 \frac{(1-W)^2}{r^2} + H^2 + \frac{W(W-2)}{r^2}
        \hspace{10mm} 2\sqrt{2} \frac{H}{r} (1-W) \\
         2\sqrt{2} \frac{H}{r} (1-W)  \hspace{10mm}
         -\nabla_r^2 + 2 \frac{(1-W)^2}{r^2} + \frac{\beta}{2} (3H^2 -1)
         \end{array} \right) \left( \begin{array}{c}
                                    W_{11}^{00} \\ \phi_{1}^{00}
                                    \end{array} \right) = \omega^2
                            \left( \begin{array}{c}
                                    W_{11}^{00} \\ \phi_{1}^{00}
                                    \end{array} \right).
$$
If we substitute in a numerical solution for the profiles $H$ and $W$, these
equations can be solved numerically for general $\beta$ by standard matrix
methods. The most interesting case is the Prasad-Sommerfield limit where we
have an analytic solution. Despite the existence of the analytic solution we
have been unable to solve the perturbation equations analytically.
We can however, proceed analytically if we take the far field forms of the
profile functions
$$
H \rightarrow 1 -\frac{1}{r}, \hspace{5mm}
W \rightarrow 1 \hspace{5mm} \mbox{for} \hspace{5mm}
r \rightarrow \infty .
$$
Substituting this into  the perturbation equation for $\psi_{00}^+$ gives
for $r \rightarrow \infty $
$$
\left( \begin{array}{cc}
       -\nabla_r^2 -\frac{2}{3r} + \frac{4}{3r^2} + \frac{1}{3} &
       \frac{\sqrt{2}}{3} \left( -1 + \frac{2}{r} +\frac{2}{r^2} \right) \\
        \frac{\sqrt{2}}{3} \left( -1 + \frac{2}{r} +\frac{2}{r^2} \right)
       & -\nabla_r^2 -\frac{4}{3r} + \frac{2}{3r^2} + \frac{2}{3}
       \end{array} \right) \left( \begin{array}{c}
                                  W_{22}^{00} \\ W_{00}^{00}
                                  \end{array} \right) = \omega^2
                           \left( \begin{array}{c}
                                  W_{22}^{00} \\ W_{00}^{00}
                                  \end{array} \right),
$$
for which we can find two orthogonal eigenvectors. One of the eigenvectors is
$$
\left( \begin{array}{c}
      W_{22}^{00} \\ W_{00}^{00}
       \end{array} \right) = \left( \begin{array}{c}
                       1 \\ 1/\sqrt{2}
                    \end{array} \right) w_1
$$
where the function $w_1(r)$ satisfies
$$
\left( -\nabla_r^2 + \frac{2}{r^2} \right) w_1 = \omega^2 w_1.
$$
This has spherical Bessel function solutions for all $\omega^2 \geq 0$. The
other eigenvector is
$$
\left( \begin{array}{c}
      W_{22}^{00} \\ W_{00}^{00}
       \end{array} \right) = \left( \begin{array}{c}
                       1 \\ -\sqrt{2}
                    \end{array} \right) w_2
$$
and the function $w_2(r)$ satisfies
$$
\left( -\nabla_r^2 + 1 - \frac{2}{r} \right) w_2 = \omega^2 w_2.
$$
If we write $w_2(r) =u(r)/r$, rescale $\rho =2(1-\omega^2)^{1/2} r$ and define
$\Lambda=1/(1-\omega^2)^{1/2}$ then we obtain Whittakers form of the
hypergeometric equation
$$
\left( \frac{d^2}{d\rho^2} - \frac{1}{4} + \frac{\Lambda}{\rho} +
\frac{\frac{1}{4}-\mu^2}{\rho^2} \right) u(\rho ) =0,
$$
with $\mu^2 =1/4$. For $\omega^2 \geq 1$ this equation has Whittaker function
solutions $W_{\Lambda,\mu}(\rho)$ for all $\Lambda$ which are oscillatory at
large $\rho$. For $1 > \omega^2 \geq 0$ the Whittaker function solutions
only exist for $\Lambda = 1,2,3,\ldots$ and the solution decays exponentially
at large $\rho$. These bound states are analogous to the states of the
non-relativistic hydrogen atom, and have eigenvalues
$$
\omega_n^2 = 1 - \frac{1}{n^2} \hspace{5mm} \mbox{where} \hspace{5mm}
n=1,2,3,\ldots
$$
The Whittaker function solutions can actually be written as Laguerre
polynomials, since if we set $u=\rho {\rm e}^{-\rho/2}v$ we get Laguerre's
equation
\begin{equation}
\left( \rho \frac{d^2}{d\rho^2} + (2-\rho)\frac{d}{d\rho} + (\Lambda -1)
\right) v(\rho) =0,
\end{equation}
which for $\Lambda = n \in {\cal Z}$ has associated Laguerre polynomial
solutions $L_n^1 (\rho)$.

If we now consider the equation for $\psi_{00}^-$ in the limit
$r \rightarrow \infty$ we find
$$
\left( \begin{array}{cc}
       -\nabla_r^2 + 1 - \frac{2}{r} & 0 \\
       0 & -\nabla_r^2
       \end{array} \right) \left( \begin{array}{c}
                                  W_{11}^{00} \\ \phi_{1}^{00}
                                  \end{array} \right) = \omega^2
                           \left( \begin{array}{c}
                                  W_{11}^{00} \\ \phi_{1}^{00}
                                  \end{array} \right).
$$
The Higgs scalar field has spherical Bessel function solutions at large $r$,
while the $W_{11}^{00}$ field has the same type of bound solutions as above
with eigenvalues
$$
\omega_n^2 = 1 - \frac{1}{n^2} \hspace{5mm} \mbox{where} \hspace{5mm}
n=1,2,3,\ldots
$$

These spherical Bessel function and Whittaker function solutions are finite at
the origin and so are physically acceptable solutions for all $r$. They were
obtained using the far field limit for the profiles which ignores any effect of
the monopole core, so it is not obvious that the eigenvalues obtained will still
hold when we use the full analytic solution.
However, numerical solution of the perturbation equations with the full analytic
form of the profile functions gives numerical results which are consistent with
the analytic result obtained in the far field limit.

Table 5 shows the analytic and numerical results for the
eigenvalues of the lowest identifiable bound modes. The numerical results were
obtained by solving the $\psi_{00}^+$ equation using standard matrix methods.
The non-linear map $s=\tanh (r)$ was used to assign more lattice points to the
core of the monopole where the fields are varying most. Apart from the $n=2$
eigenvalue we can see that the agreement between the analytic and numerical
results is fairly good. In trying to argue that the numerical results actually
agree with the analytic result we have to address two principle shortcomings of
the numerical results: the difference between the $n=2$ numerical and analytic
results; and the limited number of bound modes identified.

The $n=2$ bound mode differs from the other bound modes in that the fields of
this mode vary most in exactly the same place that the monopole profiles vary
most. We would argue that this leads to the error in this mode being greater
than for the other modes. In fact we found a similar situation in
ref.~\cite{Goodband95a} where we considered perturbations about a Nielsen-Olesen
vortex \cite{Nielsen73}. There we found a similar bound mode whose fields varied
most in exactly the same place as the string profiles, and its eigenvalue was
the most sensitive of all the modes to the accuracy of the string profiles
used.

This $n=2$ mode also appears to be the one greatest affected by the omission of
the origin, which is necessary because of the $1/r^2$ centrifugal terms in the
perturbation equation. The effect that the omission of the origin can have is
best illustrated by considering the analogous case of perturbations about the
kink \cite{Rajaraman82}. In this case the perturbation equation can be solved
analytically and two discrete modes are found. One is a $\omega^2 =0$
translation mode and the other is a $\omega^2 =0.75\; m^2$ bound mode ($m$ is
the
mass of the Higgs boson in this case). If we solve this eigenproblem on a
lattice numerically we obtain agreement with the analytic values for the
discrete modes above. However, if we remove the lattice point at the origin,
there is a $0.01\; m^2$ shift in the eigenvalue of the bound mode, whereas the
$\omega^2=0$ translation mode eigenvalue is unaltered to order $0.01\; m^2$.

We would therefore argue that these affects, notably the omission of the origin,
are responsible for the disagreement between the numerical and analytic results
of the $n=2$ mode, and that its true eigenvalue is in fact given by the
analytic result.

Now the limit on the number of bound modes that could be identified in the
numerical procedure is not really surprising. In the analytic expression
the difference between the eigenvalues of adjacent modes decreases for the
higher modes, whereas in solving the eigenvalue problem numerically the
differences between eigenvalues that can be resolved is dependent on the number
of lattice points and the box size. Our use of the non-linear $\tanh$ map was
such that the number of lattice points and box size were increased
simultaneously. We can see that the eigenvalues of the $n=1,\ldots,5$ bound
modes do not change significantly as the number of lattice points is increased.
The eigenvalue of the $n=6$ bound mode, the highest bound mode identifiable with
a 100 point lattice, does change significantly as the number of lattice points
is increased; it tends towards the analytic result. The higher the eigenvalue
of the bound mode, the further from the monopole core the fields of the bound
mode extend, hence as the box size is increased (note that box size and lattice
size increase simultaneously) the number of identifiable bound modes increases.
We would therefore expect that if we continued to increase the number of lattice
points (and box size) we would continue to find more bound modes, and their
eigenvalues would agree with the analytic result. It is worth noting the $1/r$
dependence of the monopole profile $H$ for large $r$, which in a sense means
that all box sizes are too small.

As for the interpretation of the modes, for $\psi_{00}^+$ we can interpret the
two eigenspectrums as being associated with the massless gauge boson and one of
the massive gauge bosons. The massive gauge boson possesses bound states. For
$\psi_{00}^-$ we can interpret the two eigenspectrums as being associated with
the massless Higgs boson and the other massive gauge boson. It is again the
massive gauge boson which has bound states.

To check that these modes are in fact physical we solve the equation for the
ghosts. For $k=0$ the only allowed mode is
$$
\eta_{00}^- = \eta_{1}^{00}
$$
which satisfies the equation
\begin{equation} \label{Laguerre}
\left[ -\nabla_r^2 + \frac{2(1-W)^2}{r^2} \right] \eta_{1}^{00}
=\omega^2 \eta_{1}^{00}.
\end{equation}
This has no bound mode solutions and so the bound states found above are
physical. This ghost mode cancels the $\tilde{W}^{00}_{1}$ and $\phi^{00}_{1}$
massless modes.

These results agree with ref.~\cite{Sonoda84} where he only studied the
$\psi_{00}^+$ perturbations for $k=0$ and not the $\psi_{00}^-$ type. He found
an infinite number of bound states and numerically estimated the lowest to be
$\omega^2 \simeq 0.768$. This agrees well with the value we obtained numerically
for the first non-zero eigenvalue, but he did not find the bound zero mode.
This spherically symmetric zero mode is not canceled by the ghosts, and is
in fact associated with a physical degree of freedom of the monopole.
Finally no negative modes were found in agreement with the perturbative
stability analysis of ref.~\cite{Yoneya77}

\section{Perturbations about the dyon}

Substituting the dyon solution into the gauge fixed second variation
of the action gives
\begin{eqnarray}
\delta S_{gf} & = & \frac{\eta}{q}
\int r^2 dr \int d \Omega \left( \phi^a \left[ (D^2)_{ab}
+ \delta^{ab} \left(\frac{\beta}{2}(H^2 -1) + H^2 \right)
+ \hat{r}^a \hat{r}^b (\beta -1)H^2  \right] \phi^b \right. \nonumber \\
 & - & \left. W_0^a \left[ (D^2)_{ab} + H^2 (\delta^{ab} -\hat{r}^a \hat{r}^b)
 \right] W_0^b + W_i^a \left[ (D^2)_{ab} + H^2
 (\delta^{ab} -\hat{r}^a \hat{r}^b) \right] W_i^b \right. \nonumber \\
 & + & \left. 4\epsilon^{abc} \phi^a W_i^b (D_i \phi )^c
 + 4\epsilon^{abc} W_0^a W_i^b G_{0i}^c + 2\epsilon^{abc} W_i^a W_j^b G_{ij}^c
 \right),
\end{eqnarray}
and the Fadeev-Popov ghost term as
$$
\delta S_{FP}  =  \frac{\eta}{q}
\int r^2 dr \int d \Omega  \eta^a \left[ (D^2)_{ab}
+ H^2 (\delta^{ab} -\hat{r}^a \hat{r}^b) \right] \eta^b .
$$
where the key terms are given by
\begin{eqnarray}
(D^2 )_{ab} & = & \delta_{ab}\partial_0^2 +2\epsilon_{abc}\hat{r}^cJ\partial_0
-\delta_{ab}J^2+ \delta_{ab} \left( -\nabla_r^2 + \frac{L^2}{r^2}
+ \frac{W^2}{r^2} \right)  - 2\frac{W}{r^2} \epsilon_{iab} L_i
+ \hat{r}^a \hat{r}^b \frac{W^2}{r^2}  \nonumber \\
G_{ij}^c & = & \epsilon^{kij} \left[ \hat{r}^k \hat{r}^c
\left( \frac{W^{\prime}}{r} + \frac{W}{r^2} (W-2) \right)
- \delta^{kc}\frac{W^{\prime}}{r}  \right] \nonumber \\
(D_i \phi )^c & = &  \hat{r}^i \hat{r}^c\left( H^{\prime}-\frac{H}{r}(1-W)
\right) + \delta^{ic} \frac{H}{r} (1-W) \nonumber \\
G_{0i}^c & = & \hat{r}^i \hat{r}^c\left( J^{\prime}-\frac{J}{r}(1-W)
\right) + \delta^{ic} \frac{J}{r} (1-W) \nonumber \\
(D_0 \phi )^a & = & 0.  \nonumber
\end{eqnarray}
Note that in this case there is a linear
time derivative so it is not obvious that it is possible to set up a
standard eigenvalue problem.

The dyon solution is also invariant under rotations generated by $K$, and so
we again expand the perturbations in eigenstates of $K$,
\begin{eqnarray}
\phi^a & = & \sum_{km} \sum_{j=-1}^{1} \phi^{km}_{j} Y^a_{kjm} \nonumber \\
\eta^a & = & \sum_{km} \sum_{j=-1}^{1} \eta^{km}_{j} Y^a_{kjm} \nonumber \\
W_0^a & = & \sum_{km} \sum_{j=-1}^{1} \tilde{W}^{km}_{j} Y^a_{kjm} \nonumber \\
W_i^a & = & \sum_{km} \sum_{j_1=0}^{2} \sum_{j_2=-j_1}^{j_1}
W^{km}_{j_1j_2} Y^{ai}_{kj_1j_2m} \nonumber
\end{eqnarray}
and again find that the perturbations separate into two channels under parity
$$
\psi^+_{km} = \left( \begin{array}{c}
                W^{km}_{22} \\ W^{km}_{20} \\ W^{km}_{2-2} \\ W^{km}_{10} \\
                W^{km}_{00} \\ \tilde{W}^{km}_{0} \\ \phi^{km}_{0}
                \end{array} \right), \hspace{20mm}
\psi^-_{km} = \left( \begin{array}{c}
                W^{km}_{21} \\ W^{km}_{2-1} \\ W^{km}_{11} \\ W^{km}_{1-1} \\
                \tilde{W}^{km}_{1} \\ \tilde{W}^{km}_{-1} \\
                \phi^{km}_{1} \\ \phi^{km}_{-1}
                \end{array} \right).
$$
Substituting these expansions into $\delta S_{gf}$ and performing the
angular integrals gives the perturbation operator
\begin{eqnarray}
M_{\alpha \beta}^{(\pm)} & = & \delta_{\alpha \beta} \left[
\partial_0^2 -\nabla_r^2 + \frac{(k+j_2 )(k+j_2 +1)}{r^2} + \frac{W^2}{r^2}
+ H^2 -J^2 +\frac{\beta}{2} (H^2 -1) P_{\phi} \right] \nonumber \\
 & + & 2\frac{W}{r^2} C_{\alpha\beta}^{(\pm )(1)}
 + 2J C_{\alpha\beta}^{(\pm )(9)} \partial_0
 + \left( \frac{W^2}{r^2} - H^2 +J^2 + \beta H^2 P_{\phi} \right)
 C_{\alpha\beta}^{(\pm )(2)} \nonumber \\
 & + & 2\frac{W^{\prime}}{r} C_{\alpha\beta}^{(\pm )(3)}
  +  2 \frac{W}{r^2} (W-2) C_{\alpha\beta}^{(\pm )(4)} \nonumber \\
 & + & 4\frac{H}{r} (1-W) C_{\alpha\beta}^{(\pm)(5)}
 + 4 \left( H^{\prime} - \frac{H}{r} (1-W) \right)
 C_{\alpha\beta}^{(\pm )(6)} \nonumber  \\
 & + & 4\frac{J}{r} (1-W) C_{\alpha\beta}^{(\pm)(7)} \epsilon_{\alpha\beta}
 + 4 \left( J^{\prime} - \frac{J}{r} (1-W) \right)
 C_{\alpha\beta}^{(\pm )(8)} \epsilon_{\alpha\beta} \nonumber
\end{eqnarray}
for the Higgs and gauge fields, and
\begin{eqnarray}
\tilde{M}_{\alpha\beta}^{(\pm)} & = & \delta_{\alpha\beta}
\left[ -\nabla_r^2 + \frac{(k+j_2)(k+j_2+1)}{r^2} + \frac{W^2}{r^2} + H^2
 -J^2 \right] \nonumber \\
& +& \frac{2W}{r^2} C_{\alpha\beta}^{(\pm)(1)}
+ 2J C_{\alpha\beta}^{(\pm )(9)} \partial_0
+ \left( \frac{W^2}{r^2} - H^2 +J^2 \right) C_{\alpha\beta}^{(\pm)(2)}
\nonumber
\end{eqnarray}
for the ghosts. We have used the same notation as for the monopole and the
non-zero elements of the matrices $C_{\alpha\beta}^{(\pm)(1-6)}$ are the same
as for the monopole. The non-zero elements of the matrices
$C_{\alpha\beta}^{(\pm)(7,8)}$ are the same as those of the matrices
$C_{\alpha\beta}^{(\pm)(5,6)}$ but with the $\alpha,\beta$ refering to the
$\tilde{W}$ fields instead of the $\phi$ fields. We have not calculated the
elements of $C_{\alpha\beta}^{(\pm)(9)}$ for general $k$, but for the case of
interest ($k=0$) the elements are all zero, and so in this case we can set up a
standard eigenvalue problem as for the monopole.

We will restrict our attention to the Bogomol'nyi limit and consider the
analytic form of the dyon solution given earlier. Substituting this solution
into the perturbation equation and rescaling
$$
x \rightarrow (1-C^2)^{-1/2} x \equiv (1+e^2)^{1/2} x, \hspace{5mm}
t \rightarrow (1-C^2)^{-1/2} t \equiv (1+e^2)^{1/2} t
$$
gives $M_{\alpha\beta}^{(\pm)} = (1+e^2)^{-1}
\overline{M}_{\alpha\beta}^{(\pm)}$
where
\begin{eqnarray}
\overline{M}_{\alpha \beta}^{(\pm)} & = & \delta_{\alpha \beta} \left[
\partial_0^2 -\nabla_r^2 + \frac{(k+j_2 )(k+j_2 +1)}{r^2} + \frac{W^2}{r^2}
+ H^2 -J^2 +\frac{\beta}{2} (H^2 -1) P_{\phi} \right] \nonumber \\
 & + & 2\frac{W}{r^2} C_{\alpha\beta}^{(\pm )(1)}
 + \left( \frac{W^2}{r^2} - H^2 +J^2 + \beta H^2 P_{\phi} \right)
 C_{\alpha\beta}^{(\pm )(2)} \nonumber \\
 & + & 2\frac{W^{\prime}}{r} C_{\alpha\beta}^{(\pm )(3)}
  +  2 \frac{W}{r^2} (W-2) C_{\alpha\beta}^{(\pm )(4)} \nonumber \\
 & + & 4\sqrt{1+e^2} \left( \frac{H}{r} (1-W)
 C_{\alpha\beta}^{(\pm)(5)}+ \left( H^{\prime} - \frac{H}{r} (1-W) \right)
 C_{\alpha\beta}^{(\pm )(6)} \right) \nonumber  \\
 & + & 4e \left( \frac{J}{r} (1-W)
 C_{\alpha\beta}^{(\pm)(7)} \epsilon_{\alpha\beta} + \left( J^{\prime}
 - \frac{J}{r} (1-W) \right)
 C_{\alpha\beta}^{(\pm )(8)} \epsilon_{\alpha\beta} \right). \nonumber
\end{eqnarray}
Note that the quantity we have used in the rescaling $(1-C^2)^{-1/2}$ is equal
to
$\sqrt{1+e^2}$, which is the ratio of dyon mass to the monopole mass.

\subsection*{$k=0$ modes}

If we now consider the $k=0$ modes we find that the allowed modes are
$$
\psi_{00}^+ = \left( \begin{array}{c}
                     W_{22}^{00} \\  W_{00}^{00}
                     \end{array} \right)  \hspace{15mm}
\psi_{00}^- = \left( \begin{array}{c}
                     W_{11}^{00} \\ \tilde{W}_{1}^{00} \\ \phi_{1}^{00}
                     \end{array} \right).
$$
$\psi_{00}^+$ satisfies the eigenvalue equation
$$
\overline{M} \psi_{00}^+ = \omega^2 \psi_{00}^+
$$
where the elements of $\overline{M}$ are the same as for the monopole, and
so this equation is the same as equation \ref{+zero}. So we have the same
bound state solutions with eigenvalues
$$
\omega_n^2 = 1 - \frac{1}{n^2} \hspace{5mm} \mbox{where} \hspace{5mm}
n=1,2,3,\ldots
$$
as for the monopole. Note that there is explicitly no $e$-dependence of the
eigenvalues.

The equation for $\psi_{00}^-$ is different for the dyon and does possess
some $e$-dependence. $\psi_{00}^-$ satisfies the eigenvalue equation
$$
\overline{M} \psi_{00}^-  = \omega^2 \psi_{00}^-
$$
where
\begin{eqnarray}
M_{11} & = &  -\nabla_r^2 + \frac{2(1-W)^2}{r^2} + H^2 + \frac{W(W-2)}{r^2}
\nonumber \\
M_{22} & = & M_{33} \hspace{3mm} = \hspace{3mm} -\nabla_r^2
+ \frac{2(1-W)^2}{r^2} \nonumber \\
M_{12} & = & -M_{21} \hspace{3mm} = \hspace{3mm}
 e \: 2\sqrt{2} \frac{H(1-W)}{r} \nonumber \\
M_{13} & = & M_{31} \hspace{3mm} = \hspace{3mm} \sqrt{1+e^2}\: 2\sqrt{2}
\frac{H(1-W)}{r} \nonumber \\
M_{23} & = & M_{32} \hspace{3mm} = \hspace{3mm} 0 \nonumber
\end{eqnarray}
However in the far field limit the perturbation equation becomes
$$
\left( \begin{array}{ccc}
   -\nabla_r^2 + 1 -\frac{2}{r} & 0 & \\
   0 & -\nabla_r^2 & \\
    0 & 0 & -\nabla_r^2
    \end{array} \right)
                    \left( \begin{array}{c}
                     W_{11}^{00} \\ \tilde{W}_{1}^{00} \\ \phi_{1}^{00}
                     \end{array} \right)
                     = \omega^2 \left( \begin{array}{c}
                     W_{11}^{00} \\ \tilde{W}_{1}^{00} \\ \phi_{1}^{00}
                     \end{array} \right)
$$
which has no $e$-dependence. So as for the monopole we find that the
$W_{11}^{00}$ field has bound solutions with eigenvalues
$$
\omega_n^2 = 1 - \frac{1}{n^2} \hspace{5mm} \mbox{where} \hspace{5mm}
n=1,2,3,\ldots
$$
The $k=0$ ghost equation for the dyon is the same as for the monopole, and
so in the far field limit the ghosts cancel the $\tilde{W}_{1}^{00}$ and
$\phi_{1}^{00}$ massless modes just as for the monopole.

If we were to solve the $\psi_{00}^-$ eigenvalue equation with the full
analytic solution, then from our results for the monopole we would expect
the eigenvalues to agree with the analytic far field result i.e. with no
$e$-dependence. On physical grounds, however, we would expect some of the
eigenvalue spectra to possess an $e$-dependence because there should be an
interaction between the electric charge of the dyon and the electric charge
of the $W$-boson. The situation is unfortunately obscured by the fact that
eigenstates of $K$ are not eigenstates of charge. The perturbation equations
for the higher $k$ modes do in general possess an $e$-dependence in the far
field limit, but we have not solved the perturbation equations in these
cases.

\section{Perturbations about the point electric solution}

Substituting the point electric solution into the perturbation equations
\ref{pert} gives, in the Bogomol'nyi limit
\begin{equation} \label{massless}
\left( \begin{array}{ccc}
 \overline{D}^2 & 0 & 0 \\ 0 & \overline{D}^2 & 0 \\
 0 & 0 & \overline{D}^2
 \end{array} \right) \left( \begin{array}{c}
                \phi^3 \\ W_0^3 \\ W_i^3
                \end{array} \right) = 0,
\end{equation}
and
\begin{equation} \label{massive}
\left( \begin{array}{ccc}
 D^2 +(\phi^3)^2 & 0 & \pm 2iD_i\phi^3 \\
 0 & D^2 +(\phi^3)^2 & \mp 2iG_{0i}^3 \\
 \mp 2iD_i\phi^3 & \mp 2iG_{0i}^3 & D^2 +(\phi^3)^2
 \end{array} \right) \left( \begin{array}{c}
                \phi^{\pm} \\ W_0^{\pm} \\ W_i^{\pm}
                \end{array} \right) = 0
\end{equation}
where
\begin{eqnarray}
\overline{D}^2 & = & -\omega^2 - \nabla_r^2 + \frac{L^2}{r^2} \nonumber \\
D^2 & = & -\left( \omega \mp \frac{Z}{r} \right)^2 - \nabla_r^2
+ \frac{L^2}{r^2} \nonumber \\
D_i\phi^3 & = & Z\frac{\hat{r}_i}{r^2} \nonumber \\
G_{0i}^3 & = & Z\frac{\hat{r}_i}{r^2} .\nonumber
\end{eqnarray}
The equations for the gosts $(\eta^3,\eta^{\pm})$ are
\begin{eqnarray}
\overline{D}^2 \eta^3 & = & 0 \nonumber \\
\left( D^2 +(\phi^3)^2 \right) \eta^{\pm} & = & 0. \nonumber
\end{eqnarray}
The point electric solution is invariant under the total angular momentum
operator $J=L+S$, and so we expand the fields in terms of spherical harmonics
$Y_{jm}$ and vector spherical harmonics $Y^a_{jsm}$ as
\begin{eqnarray}
\phi & = & \sum_{jm} \phi_{jm} Y_{jm} \nonumber \\
W_0 & = & \sum_{jm} \tilde{W}_{jm} Y_{jm} \nonumber \\
\eta & = & \sum_{jm} \eta_{jm} Y_{jm} \nonumber \\
W_i & = & \sum_{jm} \sum_{s=-1}^1 W_s^{jm} Y^i_{jsm} \nonumber
\end{eqnarray}
where we have suppressed the $\pm,3$ isospin labels. The angular momentum $l$
is given by $l=j+s$.

Unlike the cases of the monopole and dyon, in this case because the total
angular momentum operator is independent of the isospin generator $T$ we can
construct  simultaneous eigenstates of $J$ and the charge operator $Q=T_3$.
The $(\phi^3,W^3_{\mu})$ perturbations therefore have zero charge and so the
physical modes of equation \ref{massless} involve the neutral massless gauge
boson and the neutral Higgs scalar. The $(\phi^{\pm},W^{\pm}_{\mu})$
perturbations have charges $\pm 1$ and so the physical modes of equation
\ref{massive} involve the massive charged gauge bosons.

\subsection*{$j=0$ modes}

If we consider the $j=0$ modes then we find that the only allowed mode is
$$
\left( \begin{array}{c}
                 \phi^{00} \\ \tilde{W}^{00} \\ W^{00}_{1}
                 \end{array} \right),
$$
which satisfies the perturbation equation
\begin{equation} \label{epert}
\left( \begin{array}{c}
-\left(\omega -\frac{Z}{r}\right)^2 -\nabla_r^2 +\left(1-\frac{Z}{r} \right)^2
\hspace{13mm} 0 \hspace{13mm} 2iZ/r^2 \\
0 \hspace{11mm}
-\left(\omega -\frac{Z}{r}\right)^2 -\nabla_r^2 +\left(1-\frac{Z}{r} \right)^2
\hspace{11mm} -2iZ/r^2 \\
-2iZ/r^2 \hspace{3mm} -2iZ/r^2 \hspace{3mm}
-\left(\omega -\frac{Z}{r}\right)^2 -\nabla_r^2 + \frac{2}{r^2}
+\left(1-\frac{Z}{r} \right)^2
  \end{array} \right) \left( \begin{array}{c}
                       \phi^{00} \\ \tilde{W}^{00} \\ W^{00}_{1}
                 \end{array} \right)   = 0,
\end{equation}
where we have taken the matrix elements for $(\phi^+,W_{\mu}^+)$ in equation
\ref{massive}. This is not of the form of a standard eigenvalue problem as it
stands, but if we rescale $\rho =2(1-\omega^2)^{1/2} r$ and define
$$
\Lambda = Z \frac{(1-\omega)}{\sqrt{1-\omega^2}}
$$
then we obtain
\begin{equation}
\left( \begin{array}{c}
-\nabla_{\rho}^2 +\frac{1}{4}-\frac{\Lambda}{\rho}
\hspace{14mm} 0 \hspace{14mm} 2iZ/\rho^2 \\
0 \hspace{11mm} -\nabla_{\rho}^2 +\frac{1}{4}-\frac{\Lambda}{\rho}
\hspace{11mm} -2iZ/\rho^2 \\
-2iZ/\rho^2 \hspace{3mm} -2iZ/\rho^2 \hspace{3mm}
-\nabla_{\rho}^2 +\frac{1}{4}-\frac{\Lambda}{\rho} +\frac{2}{\rho^2}
  \end{array} \right) \left( \begin{array}{c}
                       \phi^{00} \\ \tilde{W}^{00} \\ W^{00}_{1}
                 \end{array} \right)   = 0,
\end{equation}
which is now an eigenvalue problem with eigenvalues $\Lambda$. If we rewrite
this as
$$
\left[ \left( \nabla_{\rho}^2 -\frac{1}{4}+\frac{\Lambda}{\rho} \right)
\underline{I}_3 +\frac{2}{\rho^2} \underline{A} \right] \left(
\begin{array}{c}
  \phi^{00} \\ \tilde{W}^{00} \\ iW^{00}_{1}
    \end{array} \right)   = 0,
$$
where $\underline{I}_3$ is the $3\times 3$ identity and the matrix
$\underline{A}$ is given by
$$
\underline{A} = \left( \begin{array}{ccc}
  0 & 0 & -Z \\ 0 & 0 & Z \\ -Z & -Z & -1
  \end{array} \right),
$$
then the problem of diagonalizing the perturbation matrix has been reduced to
diagonalizing the matrix $\underline{A}$, which is easy. The eigensolutions
of
$$
\underline{A}\: e = \Omega \: e
$$
are
$$
\Omega_1=0 \hspace{5mm} e_1=\left(
\begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right), \hspace{5mm}
\Omega_2=0 \hspace{5mm} e_2=\left(
\begin{array}{c} 1 \\ -1 \\ 0 \end{array} \right), \hspace{5mm}
\Omega_3=-1 \hspace{5mm} e_3=\left(
\begin{array}{c} Z \\ -Z \\ 1 \end{array} \right) .
$$
The corresponding eigenvectors of the perturbation matrix are
$$
\left( \begin{array}{c}
\phi^{00} \\ \tilde{W}^{00} \\ iW^{00}_{1} \end{array} \right)
\hspace{3mm} =  \hspace{3mm}
\left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) w_1(\rho),\hspace{5mm}
\left( \begin{array}{c} 1 \\ -1 \\ 0 \end{array} \right) w_2(\rho),\hspace{5mm}
\left( \begin{array}{c} Z \\ -Z \\ 1 \end{array} \right) w_3(\rho),
$$
where the functions $w_i(\rho)$ satisfy
\begin{eqnarray}
\left( \nabla_{\rho}^2 -\frac{1}{4}+\frac{\Lambda}{\rho} \right) w_i(\rho)
& = & 0 \hspace{5mm} \mbox{for} \hspace{3mm} i=1,2 \nonumber \\
\left( \nabla_{\rho}^2 -\frac{1}{4}+\frac{\Lambda}{\rho} -\frac{2}{\rho^2}
\right) w_3(\rho) & = & 0. \nonumber
\end{eqnarray}
If we write $w_i(\rho) =u_i(\rho)/\rho$, then the equations for the $u_i$ are of
the form of Whittakers equation
$$
\left( \frac{d^2}{d\rho^2} -\frac{1}{4} + \frac{\Lambda}{\rho}
+\frac{\frac{1}{4} -\mu^2}{\rho^2} \right) u_i(\rho )=0.
$$
where $\mu =1/2$ for $u_1$ and $u_2$, and $\mu =3/2$ for $u_3$. As we stated
earlier, for $\omega^2 \geq 1$ (imaginary $\rho$) these equations have Whittaker
function solutions $W_{\Lambda,\mu}(\rho)$ for all $\Lambda$ which are
oscillatory
at large $\rho$. For $1 > \omega^2 \geq 0 $ (real $\rho$) the Whittaker fuction
solutions only exist for
$$
\Lambda = n_r + \mu + \frac{1}{2} \hspace{10mm} n_r = 0,1,2,3,\ldots
$$
and the solutions decay exponentially at large $\rho$.

At this point it is worth considering the ghosts.
The corresponding equation for the $j=0$ mode of the charged ghosts is
\begin{equation} \label{eghosts}
\left[ -\left(\omega -\frac{Z}{r}\right)^2 -\nabla_r^2
+\left(1-\frac{Z}{r} \right)^2 \right] \eta_{00} =0.
\end{equation}
If we similarly rescale $\rho = 2(1-\omega^2)^{1/2} r$ and define
$$
\Lambda = Z \frac{(1-\omega )}{\sqrt{1-\omega^2}}
$$
then we can rewrite the ghost equation as
$$
\left( \nabla_{\rho}^2 - \frac{1}{4} + \frac{\Lambda}{\rho} \right) \eta_{00}
=0,
$$
which is the same as the equation for the $w_1$ and $w_2$ eigenmodes. The ghosts
therefore cancel the $w_1$ and $w_2$ eigenmodes leaving the $w_3$ eigenmode as
the physical mode.

For the $w_3$ mode we have $\mu=3/2$ and so there are physical bound modes for
$$
\Lambda = n = 2,3,4,\ldots
$$
Substituting this into the expression
$$
\Lambda = Z \frac{(1-\omega )}{\sqrt{1-\omega^2}}
$$
and squaring it gives a quadratic equation in $\omega$ with the two solutions
$$
\omega_+=1, \hspace{5mm} \mbox{and} \hspace{5mm}
\omega_- = -\frac{(1 - Z^2/n^2)}{(1 + Z^2/n^2)} \hspace{10mm} n=2,3,4,\ldots
$$
For our point electric solution to correspond to a $W^+$ boson we take
$Z=1/4\pi $ in our rescaled units. Since then $Z^2 =1/16\pi^2 < 1$ we can
approximate $\omega_-$ as
$$
\omega_- \simeq -\left( 1-\frac{2Z^2}{n^2} \right) \hspace{5mm} n=2,3,4,\ldots
$$
Now the perturbations are eigenstates of charge and so we have
$$
W_{\mu}^{\pm} (\underline{x},t)=W_{\mu}^{\pm} (\underline{x})
{\rm e}^{\pm i\omega t}.
$$
So the $\omega_+$ solution therefore corresponds to the $W^+$ boson
and the $\omega_-$ solution corresponds to the $W^-$ boson.
Our result therefore agrees with the result of ref.~\cite{Manton77} where
it is shown that the force between two gauge bosons of the same charge is
zero, and hence no bound states, whilst the force between gauge bosons
of opposite charges is twice the electrostatic coulomb force.

We note that the point electric solution used here has no perturbative
instabilities ($\omega$ imaginary), no matter how large the charge $4\pi Z$.
This differs from the point electric solution
$$
W_0^3 = - \frac{Z}{r}, \hspace{5mm} \phi^3 = \eta
$$
which possesses perturbative instabilities above a critical charge
\cite{Mandula77}.

\section{Discussion}

We have found the eigenvalue spectra for spherically symmetric perturbations
about the monopole, dyon and point electric solutions. For the perturbative
modes of positive parity about both the monopole and dyon we argue that the
eigenvalue spectra of bound modes are the same using both the far field form
and the full form of the Prasad-Sommerfield solution. These bound states are
interpreted as bound states of the massive $W$-bosons in the presence of the
monopole and the dyon. The force between the monopole and the massive gauge
bosons consists of two components. One part is the electromagnetic interaction
between the monopole magnetic charge and the magnetic dipole of the gauge
boson. However it was shown in ref.~\cite{Olsen90} that this interaction by
itself is not strong enough to give bound states. The other part of the
monopole-gauge boson force arises from the Higgs field of the monopole.
Because the gauge boson acquires its mass from the Higgs field expectation
value, the gauge boson has a lower energy when it is in the presence of the
monopole core and this gives rise to a force which is responsible for confining
the gauge boson to the core of the monopole.
The force between the dyon and the massive gauge bosons has an electrostatic
force in addition to the monopole-gauge boson force. The effect of this
electrostatic force is unfortunately obscured by the fact that
the perturbative modes about the dyon (and monopole) are not in eigenstates of
charge. This is not due to any failings of our perturbative method but is an
intrinsic property of both the dyon and monopole solutions.

The perturbations about the point electric solution are in eigenstates of
charge. The force between two charged gauge bosons consists of an attractive
Higgs force and an electrostatic force of equal magnitude. For gauge bosons of
equal charge the two forces cancel and in this case we find no bound states,
whereas for oppositely charged bosons the two forces add and our results for
the bound states in this case are consistent with this.

We set out to answer the question: are the bound state energies of the various
two particle systems of the Yang-Mills-Higgs model related by electromagnetic
duality? We are specifically considering the $SL(2,{\cal Z})$ symmetry of Sen
\cite{Sen94}. The original conjecture of Montonen and Olive \cite{Montonen77}
was that there are two equivalent choices of action, either with the
$W$-bosons as the gauge particles and the monopoles as the solitons, or with
the monopoles as the gauge particles and the $W$-bosons as the solitons. The
spectrum of dyons does not, however, fit nicely into this picture. In the
electromagnetic duality of Sen \cite{Sen94} the electric $(q)$ and magnetic
$(g)$ charges of a particle state are given by
$$
q+ig = q_0 (m\tau + n) \hspace{10mm} m,n \in {\cal Z}
$$
where $q_0$ is the minimum electric charge and
$$
\tau = \frac{\theta}{2\pi} + \frac{i}{q_0^2}.
$$
$\theta$ is a new parameter of the theory. Plotting $q+ig$ in the complex
plane gives the charge lattice $(q,g)$, and the single particle states
correspond to the primitive vectors of this charge lattice. A basis for the
charge lattice is given by any two non-collinear primitive vectors, and choices
of basis correspond to choices of equivalent actions. Invariance of the theory
under electromagnetic duality is now realised as invariance of the theory
to changes of basis of the charge lattice.

If we choose the primitive vector $q_0$ to correspond to the electric gauge
boson and the primitive vector $q_0\tau $ to correspond to the magnetic
monopole, then a change of basis to $(q_0^{\prime},q_0^{\prime}\tau^{\prime})$
is given by \cite{Olive95}
$$
\left( \begin{array}{c}
 q_0^{\prime}\tau^{\prime} \\ q_0^{\prime}
 \end{array} \right) = A \left( \begin{array}{c}
 q_0\tau \\ q_0
 \end{array} \right) = \left( \begin{array}{cc}
              a & b \\ c & d
              \end{array} \right)  \left( \begin{array}{c}
                                 q_0\tau \\ q_0
                                  \end{array} \right),
$$
where $a,b,c,d \in {\cal Z}$ and $ad - bc =1$, so the matrices $A$ form the
group $SL(2,{\cal Z})$. Under this transformation $\tau$ becomes
$$
\tau^{\prime} = \frac{a\tau + b}{c\tau +d}.
$$
We now consider whether our results are consistent with this picture. Our
$W$-boson-monopole system corresponds to choosing the primitive vectors
$q_0$ as the $W$-boson and $q_0\tau=i/q_0$ as the monopole. In this case we
find that the spherically symmetric bound perturbative modes about the monopole
have energies
$$
\omega^2_n = \left(1-\frac{1}{n^2}\right)
\equiv \left(1-\frac{qg}{n^2}\right) \hspace{5mm}
n \in {\cal Z}
$$
where $q$ is the charge of the $W$-boson and $g$ is the charge of the monopole.
If we now apply the $SL(2,{\cal Z})$ transformation
\begin{eqnarray}
\left( \begin{array}{c}
 q_0^{\prime}\tau^{\prime} \\ q_0^{\prime}
 \end{array} \right) & = & \left( \begin{array}{cc}
              1 & 1 \\ 0 & 1
              \end{array} \right)  \left( \begin{array}{c}
                                 q_0\tau \\ q_0
                                  \end{array} \right) =
                                 \left( \begin{array}{c}
                                 q_0\tau + q_0 \\ q_0
                                  \end{array} \right) \nonumber \\
\tau^{\prime} & = & \tau +1 \nonumber
\end{eqnarray}
then the $W$-boson still corresponds to
the point $q_0$, but the monopole in the new basis corresponds to the point
$q_0\tau + q_0$, which is the dyon of the old basis. The product of the
$W$-boson charge $q$ and the monopole charge $g$ is invariant under this
transformation, so the bound state energies of the two particle system should
be invariant under this transformation. We did indeed find that the
$W$-boson-dyon bound state energies are given by the same expression, despite
our reservations about the $W$-field perturbations not being in eigenstates of
charge.

The $SL(2,{\cal Z})$ transformation
\begin{eqnarray}
\left( \begin{array}{c}
 q_0^{\prime}\tau^{\prime} \\ q_0^{\prime}
 \end{array} \right) & = & \left( \begin{array}{cc}
              0 & -1 \\ 1 & 0
              \end{array} \right)  \left( \begin{array}{c}
                                 q_0\tau \\ q_0
                                  \end{array} \right) =
                                 \left( \begin{array}{c}
                                  - q_0 \\ q_0\tau
                                  \end{array} \right)  \nonumber \\
\tau^{\prime} & = & -\frac{1}{\tau} \nonumber
\end{eqnarray}
applied to the original $W$-boson-monopole basis $(q_0,i/q_0)$ corresponds
to the duality transformation of Montonen and Olive \cite{Montonen77}. The
transformation displays the interchange of weak and strong coupling
$\tau \rightarrow -1/\tau$.
This weak-strong coupling interchange also occurs for the transformation
\begin{eqnarray}
\left( \begin{array}{c}
 q_0^{\prime}\tau^{\prime} \\ q_0^{\prime}
 \end{array} \right) & = & \left( \begin{array}{cc}
              1 & -1 \\ 1 & 0
              \end{array} \right)  \left( \begin{array}{c}
                                 q_0\tau \\ q_0
                                  \end{array} \right) =
                                 \left( \begin{array}{c}
                                  q_0\tau- q_0 \\ q_0\tau
                                  \end{array} \right)  \nonumber \\
\tau^{\prime} & = & 1 -\frac{1}{\tau}, \nonumber
\end{eqnarray}
where the new gauge boson again corresponds to the old monopole but the new
monopole corresponds to the old dyon with charges $(-q_0,i/q_0)$ in the
original basis. The product $qg$ is again invariant under this transformation
and so we would expect the monopole-dyon bound state energies to again be
given by
$$
\omega^2_n = \left(1-\frac{1}{n^2}\right)
\equiv \left(1-\frac{qg}{n^2}\right).
$$
The monopole-dyon two particle system was studied in ref.~\cite{Gibbons86}
where they found physical bound states for
\begin{eqnarray}
\epsilon & = & \frac{1}{2} (n^2-s^2)^{1/2} (n-(n^2-s^2)^{1/2}) \nonumber \\
    & \simeq & \frac{s^2}{4} \left( 1-\frac{s^2}{4n^2} \right) \hspace{5mm}
    \mbox{for} \hspace{5mm} n^2>s^2 ,\nonumber
\end{eqnarray}
where $s,n$ are integers and $s$ is the relative electric charge between the
monopole and dyon. This is obviously different from what we have above.

Both our results and those of ref.~\cite{Gibbons86} are only approximations,
and under the $SL(2,{\cal Z})$ duality transformations they correspond to
different limits. Our approach was perturbative, and we found the harmonic
oscillator states about the monopole and dyon. Although in the quantum theory
these quantized harmonic oscillator states are interpreted as the gauge bosons
of the theory, this approach only accurately describes the $W$-boson-soliton
two particle system in the limit $q^2 \ll 1$.

In ref.~\cite{Gibbons86} they considered the dynamics of two non-relativistic
BPS monopoles using the Atiyah-Hitchin metric on the space of collective
co-ordinates of the monopoles. These collective co-ordinates consist of the
three spatial positions and a phase. The spherically symmetric zero mode
we obtained for both the monopole and dyon corresponds to changes in this
phase. In the limit of large monopole separation they used the Taub-NUT
metric as an approximation to the Atiyah-Hitchin metric, and solved the
non-relativistic Schr\"odinger equation in Taub-NUT space for the relative
motion of the two monopoles.

The result of ref.~\cite{Gibbons86} is non-relativistic and was obtained in
the limit of weak electric coupling. If we consider the energy of the dyon
given earlier in the limit $q^2 \ll 1$ we find
$$
E_d =  \frac{4\pi\eta}{q} \sqrt{1+\left(\frac{Qq}{4\pi}\right)^2}
\simeq  \frac{4\pi\eta}{q} \left( 1+\frac{1}{2}
\left(\frac{Qq}{4\pi}\right)^2 \right).
$$
So the total energy of the static monopole-dyon system in the limit of
infinite separation for $q^2 \ll 1$ is
$$
E=E_d+E_m \simeq 4\pi \frac{\eta}{q}  \left( 2+\frac{1}{2}
\left(\frac{Qq}{4\pi}\right)^2 \right).
$$
If we explicitly include factors of $\hbar$ so that the gauge coupling is
$\hbar q$, then the weak coupling condition is $\hbar q \ll 1$. Now we have
taken $\hbar =1$, and so weak coupling condition is $q \ll 1$, whereas in
ref.~\cite{Gibbons86} they use units $q=1$ and so the weak coupling condition
is given by $\hbar \ll 1$. With the units $q=1$, $\eta=1$ of
ref.~\cite{Gibbons86} the energy given above becomes
$$
E \simeq 2 \left( 4\pi + \frac{Q^2}{16\pi} \right),
$$
and in the centre of motion frame with the mass and length in units of $4\pi$,
bound states occur for
$$
0 < \epsilon < \frac{Q^2}{4}.
$$
Now the electric charge of a dyon in the presence of a monopole is not
a good quantum number, but the relative electric charge $(s)$ is. If, in the
centre of motion frame, we assign electric charges $(Q_1=-s/2)$ to the
monopole and $(Q_2=s/2)$ to the dyon, then the result
$$
\epsilon \simeq \frac{s^2}{4} \left( 1-\frac{s^2}{4n^2} \right)
$$
is consistent with an electrostatic interaction of the form $Q_1Q_2/r$
between the two particles \cite{Manton85}.

So both here and in ref.~\cite{Gibbons86} bound states due to a Coulomb
interaction are found in the limit $q^2 \ll 1$. However, the
$SL(2,{\cal Z})$ transformation between the $W$-boson-monopole and
monopole-dyon systems involves a weak-strong coupling interchange. So our
result for the $W$-boson-monopole system, which holds for $q^2 \ll 1$,
would correspond to the limit $q^2 \gg 1$ for the monopole-dyon system,
and the result of ref.~\cite{Gibbons86}, which holds for $q^2 \ll 1$ for the
non-relativistic monopole-dyon system, would correspond to the limit
$q^2 \gg 1$ for the non-relativistic $W$-boson-monopole system.

In ref.~\cite{Manton85} Manton showed that the monopole-dyon interaction
could be understood in terms of the motion of classical point particles with
electric and magnetic charges. If we consider a dyon with electric charge $q$
and magnetic charge $g$, then the scalar charge must be $(g^2+q^2)^{1/2}$
\cite{Manton85}. So the Coulomb interaction between two static particles
with charges $(q_1,g_1)$ and $(q_2,g_2)$ is \cite{Manton85}
$$
V = \left[ g_1g_2 + q_1q_2 - (g_1^2+q_1^2)^{1/2}(g_2^2+q_2^2)^{1/2} \right]
\frac{1}{r}.
$$
Now if we consider a monopole $(0,g)$ and a $W$-boson $(q,0)$ we find
$$
V = -\frac{qg}{r},
$$
and for a dyon $(q,g)$ and a $W$-boson $(q,0)$ we find
\begin{eqnarray}
V & = & \left[ q^2 - q(g^2+q^2)^{1/2} \right] \frac{1}{r} \nonumber \\
  & \simeq & -\frac{qg}{r} \hspace{5mm} \mbox{for} \hspace{5mm}
  g \gg q ,\nonumber
\end{eqnarray}
which is the same as the $W$-boson-monopole interaction for $q^2 \ll 1$
(note $g=1/q$). The monopole-dyon interaction is given by
\begin{eqnarray}
V & = & \left[ g^2 - g(g^2+q^2)^{1/2} \right] \frac{1}{r} \nonumber \\
  & \simeq & -\frac{q^2}{2r} \hspace{5mm} \mbox{for} \hspace{5mm}
  g \gg q ,\nonumber
\end{eqnarray}
which gives the result for $n^2 \gg s^2$ of ref.~\cite{Gibbons86}.
For $q \gg g$ we find
$$
V \simeq -\frac{qg}{r},
$$
which is the same as for the $W$-boson-monopole system.
This simple argument supports the view that our result for the bound state
energy
$$
\omega_n^2 = \left( 1-\frac{qg}{n^2} \right),
$$
applies for $q \ll 1$ for the relativistic $W$-boson-monopole and $W$-boson-dyon
systems, and also for the relativistic monopole-dyon system for $q^2 \gg 1$.

If the bound state energies of these three two-particle systems are indeed
related
by $SL(2,{\cal Z})$ transformations on the charge lattice, then the result is in
fact far more general. This is because any two non-collinear primitive vectors
of
the charge lattice can be transformed to the basis $(q_0,i/q_0)$. Our result
then
gives the relativistic bound state energies in the weak coupling limit
$(q^2 \ll 1)$, and the result of ref.~\cite{Gibbons86} gives the
non-relativistic
bound state energies in the strong coupling limit ($q^2 \gg 1$).

If we consider two general single particle states with the same charges $(g,q)$,
they exert no static Coulomb forces on each other since
$$
V = \left[ g^2 + q^2 - (g^2 +q^2)^{1/2}(g^2 +q^2)^{1/2} \right] \frac{1}{r} =0,
$$
and so they form no bound states with one another. Two general single particle
states with opposite charges, however, have a static Coulomb force
$$
V = \left[ -g^2 - q^2 - (g^2 +q^2)^{1/2}(g^2 +q^2)^{1/2} \right] \frac{1}{r}
= -2 \frac{g^2 +q^2}{r},
$$
and can form bound states. By an $SL(2,{\cal Z})$ transformation we can
transform this two particle system to the $W^+$--$W^-$ boson system, which we
found to have bound state energies
$$
\omega = -\frac{(1-q^2/n^2)}{(1+q^2/n^2)} \hspace{5mm} \mbox{for} \hspace{5mm}
q \ll 1.
$$

As to our original question of whether the bound states of the various two
particle systems are related by electromagnetic duality, we have no definitive
answer but our results would appear to be consistent with it.

Finally we note the possible existence of an unexpected coincidence. Our
numerical solution of the perturbations about the monopole gave the lowest
massive bound mode to have an eigenvalue of $\omega^2 \simeq 0.768\; m^2$ in the
Bogomol'nyi limit. We, however, argued that the omission of the origin in our
numerical method is responsible for it being different from the analytical
far field result of $\omega^2 = 0.75\; m^2$. In ref.~\cite{Goodband95a} we
numerically solved for the perturbations about the local string, and in the
Bogomol'nyi limit ($m_H = m_W$) we found a bound boson with eigenvalue
$\omega^2 \simeq 0.775\;m^2$. Our numerical method in this case also
necessitates
the omission of th origin due to a $1/r^2$ centrifugal term. In addition the
string profiles had to be solved for numerically because there is no analytic
solution. The fields of this bound mode vary in exactly the same place that the
string profiles vary most, and so the value of this eigenvalue is exceptionally
sensitive to the accuracy of the string profiles. So we argue that our numerical
result of $\omega^2 \simeq 0.775\; m^2$ would be consistent with the true
eigenvalue being $\omega^2 = 0.75\; m^2$. If this is the case then the kink,
string and monopole all possess bound bosons with an eigenvalue of
$\omega^2 = 0.75\; m^2$ when they are in their respective Bogomol'nyi limits
\cite{Bogomolnyi76} ($\beta = m_H^2/m_W^2 \rightarrow 0$ for the monopole;
$\beta = m_H^2/m_W^2 \rightarrow 1$ for the local string;
and all values of the Higgs self coupling for the kink). It is precisely when
these theories are in their respective Bogomol'nyi limits that supersymmetric
extensions can be constructed where the Bogomol'nyi bounds \cite{Bogomolnyi76}
can be derived algebraically from the supersymmetry algebra. So it may be that
this apparently common bound of $\omega^2 \geq 0.75\;m^2$ for the bosons of the
theory in the presence of a defect with topological charge one, could be
similarly derived in the supersymmetric extensions of the respective theories.

\subsection*{Acknowlodgements}
We are grateful to Mark Hindmarsh for useful discussions.
This work was supported by PPARC: studentship number 93300941.

redpajama_set_name: RedPajamaArXiv


Brook I. Acute and chronic bacterial sinusitis. Infect Dis Clin North Am. 2007 Jun. 21(2):427-48, vii. [Medline].
Brook I. Bacteriology of chronic sinusitis and acute exacerbation of chronic sinusitis. Arch Otolaryngol Head Neck Surg. 2006 Oct. 132(10):1099-101. [Medline].
Finegold SM, Flynn MJ, Rose FV, et al. Bacteriologic findings associated with chronic bacterial maxillary sinusitis in adults. Clin Infect Dis. 2002 Aug 15. 35(4):428-33. [Medline].
Lusk R. Pediatric chronic rhinosinusitis. Curr Opin Otolaryngol Head Neck Surg. 2006 Dec. 14(6):393-6. [Medline].
Brook I. Microbiology and antimicrobial treatment of orbital and intracranial complications of sinusitis in children and their management. Int J Pediatr Otorhinolaryngol. 2009 Sep. 73(9):1183-6. [Medline].
Brook I. Treatment modalities for bacterial rhinosinusitis. Expert Opin Pharmacother. 2010 Apr. 11(5):755-69. [Medline].


Home / / Poppy – aged 7!
Heaters heaters they are warm and cosy, you can put them in your house and it will make the house nice and snug. There are heaters all different shapes and sizes, circular, square, triangular and cylinder. They can be big, small, medium sized and titchy.
You can paint them or put stickers on them. You can get all different colours, pink, blue, black, gold, silver, green, orange, rosey red and lilac. You can have them horizontal and vertical! You can have old ones and modern ones. You can have ones that you can hang your clothes on. You can have ones that can put things in and you can even sit on some ones. You can get ones that are mirrors and you can get ones with lights in or on. Heaters are great!


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HC at Hofstra
Camille Brodard ~ Kmile Feminine Creative Designer/Unsplash
Meet the Associate Editor of RealBeauty.com, Ariana Finlayson; She's a Hofstra Alum!
Kaitlin Cubria
Despite the increase in unemployment, Ariana Finlayson, Hofstra'08, shows that a Hofstra grad can make it in this economy. Finlayson, a former Print Journalism major, became the Associate Editor of RealBeauty.com for Hearst Digital Media (the same company that brings you Seventeen.com, CosmoGIRL.com, TeenMag.com, etc.) in August 2009 – just a year and a half after graduating from Hofstra! Her Campus Hofstra sat down with the successful alum and asked her about her career, her college experience, and even about her boyfriend who she met at Hofstra!
What were the main reasons why you decided to come to Hofstra?
In 11th grade, I had an English teacher, who became my mentor, and they went to Hofstra for undergrad, so that was an influence. I was also looking for a college with print journalism at the time, though I thought about majoring in graphic design, pr, and even thought about being a nutritionist. I wanted a place that was far enough away from home, but was still able to go home. I also wanted to go to a college where nobody from my high school was going. When I visited the Hofstra campus, I instantly loved it! I loved the campus, loved campus living…
I read that you used to be an RA at Hofstra, too. What is one of the most important things that you learned from your RA experience that applies to your life now?
You have to be flexible and go with the flow. A lot of things were thrown at me that was unexpected, which helps me in the real world. I was taught that there are a lot of different people out there. I had to get an understanding of certain people, which helps in the workplace because I learned that you have to explain things differently for different people.
I also saw that you met your boyfriend as an RA…
Yes! I love telling this story. [Shortened version] I was an RA on the 7th floor and he was on the 6th floor. One night, I was the RA on Duty, and [one of the boys] that lived next door to him called me and told me they were being loud next door. So I went to his room, knocked on the door and told them to "Please, keep it down." He said, "Okay," then slammed the door in my face. I was like, "I hate that kid." Then a few hours later, the same guy called and said they were being loud next door still. I went downstairs again, told them that I wasn't looking to write paperwork on them, I just needed them to keep it down. Again, said the same thing and slammed the door in my face. Once again, I said, "I hate that kid."Then, one of my friends told me that a person on their floor had a crush on me. I said, "Please tell me it's not that Steven kid". It was. I said, "I hate that kid." Plus, I wasn't ready to date at that time. A few months passed and I would see him on campus, see him playing basketball while I ran around the track. Then one day, I felt I was ready to date again. I asked a friend of mine about Steven, and she knew I didn't like him before, but since she was a mutual friend, she had us really meet, and yea. They've been together ever since!
What is the most valuable piece of advice or information you learned at Hofstra? Are there any professors you can look back on and say "thanks"?
[Professor Mo] Krochmal knows his stuff. When I was in his class, and this was before a big deal, he kept talking about this thing called "Twitter" and these other things called "BlogSpot" and "Tumbler." He would say, "Ari, you gotta learn this stuff." I was like "Pshh, Twitter" and "Pshh, what is this BlogSpot?" And, as it turns out, his class was really useful for my career. Another professor, he left the school, Steven Knowlton, I had heard horror stories about. I had him for Journalism 001 (Journalism Ethics), and he was vicious. I got a 'D' on my midterm, went to talk to him about it, and as it turns out, I deserved it. But I ended up doing great on my final, and he gave a really good learning experience. Then there was also [Carol] Fletcher, and also [Victoria] Geyer for PR, she was great. Even though I wasn't a Public Relations major, I thought about doing PR at one point, so I took PR100 [Fundamentals of Public Relations] with her to see the other side of the industry. Working at Admissions also helped me hone my working skills.
And now you're the Associate Editor of RealBeauty.com…How?!
Well, after I graduated in May '08, I applied for EA (Editorial Assistant) positions at Allure, Glamour, Cosmo, Seventeen – but I had no job for 3 months, so I worked for a stylist. Then, I went on MediaBistro.com and saw a job for Hearst Digital Media, and worked as a Gift Guide Coordinator. It was a full-time freelance gig, with a 40 hour work week and no benefits. This was in November/December '08, and then they kept me on for sweepstakes until February. Then in March, around Hearst they kept mentioning the launch of a beauty site. So from March '09 I was working full-time on this "beauty site" that had no name until, finally in August 2009, I became the editor of RealBeauty.com. I didn't think after a year of college I would be assigning, managing, hiring, etc. So after I didn't get all of those EA positions, I ended up in a better place than I thought I wanted.
Can you provide any of your own personal advice to achieve success?
Be persistent – "Find Your Edge" (Hofstra slogan, FYI). You have to find what drives you. Hey, no one says you have to find out what you want to be for the rest of your life, now! You'll find that some experiences won't be for you or what you expected, but you'll be surprised that you'll find that you learn something from each experience. Also, don't put so much pressure on yourself. Be optimistic; not naïve.
Kaitlin Cubria is a senior Public Relations major at Hofstra University, double minoring in Journalism, and Speech Communications & Rhetorical Studies. Born and raised in Staten Island, NY, Kaitlin always loved going into Manhattan on the weekends. She is a member of the Hofstra chapters of both Ed2010 and Public Relations Student Society of America (PRSSA), the vice president of She's the First Hofstra (non-profit) and acts as a Resident Assistant for Hofstra's Residential Programs. Outside of her academics, Kaitlin recently finished internships at Hearst Magazines Public Relations, The CW Television Network Affiliate WPIX11 in the Creative Services division AND at New York Magazine. This semester, she has her dream internship at Seventeen magazine in the Web/Features department, where she is the Blog Manager for Seventeen.com's Freshman 15 blog! She is obsessed with pop culture; if you don't see her reading the newest issue of Seventeen, you'll probably catch her browsing PerezHilton.com for celebrity gossip. After having a love for both creativity and writing for a majority of her life, Kaitlin is excited to continue her communications/journalism experience at Her Campus! After college, she plans on pursuing a career in either entertainment public relations or magazine journalism.
Her Campus Hofstra's Weekly Pick: The Sims 4
Her Campus Hofstra's Weekly Pick: Red (Taylor's Version)
Fun Facts About 1920s Shanghai That You Didn't Know Before from New York Times Bestselling Author Chloe Gong
Self Care Sunday's: Tis the Season To Decorate Your Space!
Hofstra's Spotlight on Female Artists Reflects Gender Disparity in the Art World
Author Spotlight with Amanda Foody and Christine Lynn Herman

redpajama_set_name: RedPajamaCommonCrawl


Almost all copiers made since 2002 have a hard drive buried inside, which stores images of every document that is scanned or copied.
THOUSANDS of documents ARE and can be stored and later, easily retrieved. PLEASE inform your local hospital, bank, Post Office and any other place where you've made copies - so they'll know to have the information removed/deleted before returning the machine!


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Irish study confirms babies born alive after botched abortions left to die
Posted by admin | Nov 19, 2020 | Irish News, Latest | 0 |
ProLife Campaign
New study goes way beyond confirming worst fears of pro-life campaigners, says Mulroy.
A new study published in the British Journal of Obstetrics and Gynaecology, based on a series of interviews with doctors carrying out abortions in Ireland makes for truly devastating reading, according to the Pro Life Campaign.
The study confirms that under Ireland's new abortion law unborn babies are sometimes born alive after botched abortions and left to die. Some of the doctors performing these abortions talk about the "internal conflict" they experience and how ending the lives of unborn babies can be "brutal", "awful" and "emotionally difficult". The study quotes one doctor referring to what they do as "stabbing the baby in the heart." Another doctor interviewed for the study said: "I remember getting sick out in the corridors afterwards because I thought it (feticide) was such an awful procedure and so dreadful."
Yet, despite all this, some of the doctors interviewed believe the new abortion law is too restrictive and want all remaining restrictions removed.
Responding to the study, conducted by researchers in UCC, Eilís Mulroy of the Pro Life Campaign said:
"The findings in this study are truly heart-breaking. It goes way beyond confirming the worst fears that pro-life campaigners expressed before the 2018 referendum about what would happen in the event of legalised abortion. Doctors in this study are openly talking about the grotesque life-ending procedures they engage in, yet in the same breath they want to see the legal grounds for abortion expanded even more. It's clear from reading the study that many of these same doctors are more preoccupied with concerns about the threat of litigation than the horror of what they are doing in ending lives. The study highlights something else very clearly – the abortions being carried out have nothing whatsoever to do with genuine healthcare."
Ms Mulroy continued:
"Reading the study also brings home the way in which politicians denied, deflected and muddied the waters when the reality was pointed out to them of what would happen in the event of an abortion regime being introduced in this country. TDs also refused to accept an amendment to the new law in 2018 that would have ensured babies born alive after botched abortions were given medical care. If there is any consolation to take from the study, it is the revelation from several of the abortion performing doctors that they are encountering significant opposition from colleagues who don't regard what they are doing as healthcare and are conscientiously objecting to assisting with the abortions. Hopefully some good will also come from the study in awakening people to the horrific reality of what is now happening in our country on a daily basis in the name of 'choice' and 'freedom'. The revelations in this study coupled with the news earlier this year of a 60% increase in the number of abortions since repeal, reminds us all that we must work even harder to inform and educate the public about the reality of the new law."
The study* interviewed 10 doctors involved in performing abortions. The researchers confined the interviews to doctors carrying out abortions on babies with a life-limiting condition or as the study refers to the babies having a 'fatal foetal anomaly'.
*Fetal medicine specialist experiences of providing a new service of termination of pregnancy for fatal fetal anomaly: a qualitative study -S Power S Meaney, K O'Donoghue, The Irish Centre for Maternal and Child Health Research (INFANT), University College Cork, Cork, Ireland b Pregnancy Loss Research Group, Department of Obstetrics and Gynaecology, University College Cork, Cork, Ireland c National Perinatal Epidemiology Centre, University College Cork, Cork, Ireland Correspondence: K O'Donoghue, Department of Obstetrics and Gynaecology, Cork University Maternity Hospital, Wilton, Cork, Ireland. Accepted 3 September 2020
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Playing with the Devil Eau de Parfum
50ml - €199.00 EUR
Fiery, flirtatiously fierce, Playing with the Devil is unforgettable. Tart blood orange essence and rose-tinged lychee fruit is set ablaze with peppery pimento berries essence. Creamy sandalwood on the drydown gives a smoldering quality, for a playful twist on a scandalizing encounter.
Key notes: Blood Orange, Piment Berries, Sandalwood.
Eau de Parfum: Women.
50ml, Refillable.
Each KILIAN bottle is transformed into a precious object. The white-lacquered flacon is engraved on its various sides with a ceramic fresco representing the original sin and decorated with a golden plate, where the name of the fragrance has been carved and manually filled with white enamel.
The bottle comes in a white clutch inspired by a slim cigarette case deeply rooted in Art Deco, ornamented with a gold-toned snake—the ultimate symbol of temptation— waiting with flickering crystal eyes. Carry the clutch as an evening bag, it stores essential items safely, with a satisfying click-to-close clasp.
Alcohol Denat., Fragrance (Parfum), Water/Aqua/Eau, Ethylhexyl Methoxycinnamate, Butyl Methoxydibenzoylmethane, Ethylhexyl Salicylate, Coumarin, Hydroxycitronellal, Limonene, Linalool, Geraniol, Benzyl Salicylate, Citronellol, Eugenol, Benzyl Benzoate, Isoeugenol, Cinnamyl Alcohol, Citral, Farnesol, Benzyl Alcohol, Bht, Tocopherol.
PERFUMER: CALICE BECKER
Calice Becker was born in France to a pre-revolution Russian immigrant family, from whom she received a love of and reverence for fragrance, art, and traditional craftsmanship. Blessed from an early age with a keen sense of smell and the good taste to accompany it, she entered the largest perfumery school in Grasse, the Roure-Bertrand-Dupont (now the Givaudin Perfumery School). Entering Quest International upon graduation, Becker found herself under the tutelage of the famous Francoise Caron, who encouraged Becker to be a pioneer of a new generation of female noses and creators. Becker responded by creating J'Adore Christian Dior, one of the world's most beloved and bestselling perfumes.
Never one to limit herself to one world or another, Becker comes onto our radar via her work as the primary nose on most of Kilian Hennessy's By Kilian line. Though By Kilian features many stunning natural ingredients and rare, expensive distillations, Becker is equally proficient with modern synthetics, as she still creates scents for niche and mass market alike. We would say she doesn't fit into any mold at all, but we can think of one: master perfumer.
ABOUT BY KILIAN
Heir to a long line of cognac makers who were pioneers in luxury, Kilian decided to take up the torch of family tradition. Creating a new luxury brand was definitely a challenge worthy of his predecessors. Many of Kilian's fragrances today carry this childhood memory as they are reminiscent of the sugar in the alcohol and the wood of the cognac barrels.


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Dutch chef Richard Ekkebus will be returning to cook in the Netherlands at Rijks Restaurant on 26 and 27 January with Joris Bijdendijk. The restaurant is housed in the Rijks Museum.
Ekkebus is chef of the world acclaimed Hong Kong restaurant Amber which is housed in the Landmark Mandarin Oriental Hotel.
The Dutch chef from Vlissingen has trained with some of the greatest three- star chefs in France including Pierre Gagnaire, Alain Passard and Guy Savoy.
After an impressive career he became executive chef of Amber. The restaurant, which ranked 56th in the World's 50 Best restaurants list 2018 and 7th in Asia's 50 Best Restaurants 2019 list is currently closed for refurbishment.
At Amber, Ekkebus strives to use the best and freshest ingredients that mainly come from Japan and Europe.
For the last weekend of January, chef Ekkebus and Bijdendijk will create a special four-hands menu for lunch and dinner. Richard said that it was inspiring for him and his team to cook in the restaurant of one of the most iconic museums in the world. "It is a location with so much history of my homeland. It will enable us to come into contact with the rich Dutch history and in particular the Golden Age where Dutch roamed the work in search of exotic spices. I am a true culinary gypsy and this tour through Europe, Africa and Asia is a true journey in the search for inspiration for Amber 2.0.
Amber is currently closed for refurbishment. In an interview with Food and Wine Gazette Ekkebus had said that for him and his team it is going to be like reopening.
The collaboration takes place on Saturday and Sunday 26 and 27 January. Bijdendijk and Ekkebus will prepare a special menu during the weekend. The lunch and dinner menu consists of 8 courses. The price is EUR 180 per person. Reservations can be made on the following link.


We are not ashamed to say that our festival would not be what it is today without the help of all the volunteers and sponsors who make the smooth running of the day possible. If you would like to work for us, please get in touch. All steward positions are unpaid, however you will receive free passes to the event itself and/or the benefit of our sponsorship package as appropriate. Shift times will be agreed prior to the event.
STALL HOLDERS: We have now added a small market to the event and would like to hear from local traders or small businesses who may want to hold a stall. See below for further information.
Showcase your wares to our music loving crowd!


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You are here: Home » Ministry of health recruits ghost workers
Ministry of health recruits ghost workers
Reliable sources at the Ministry of Health have disclosed to us how Dr Chitalu Chilufya is lying to the nation that the government has sent 1500 letters to newly recruited health care providers to help boost man power during the COVID-19 pandemic.
Following the outbreak of the Corona virus, the PF government announced that it will be recruiting about 1500 health workers to meet the demand in fighting the deadly pandemic. But this whole thing has been mared with corruption just like most of the the things that are organised by the PF indeed donations, we are told, have already been stolen.
According to our sources at the Ministry, they informed us that money was allocated for the recruitment exercise but no one has been employed yet. This is despite Dr Chitalu Chilufya boasting during COVID-19 press briefings when asked by journalists how far the recruitment process had gone.
In one of the message sent to us by health sources which reads "Lusaka Provincial Health Office has received your application for employment and therefore updating the database kindly send your scanned copy of your practicing licence to LPHO@moh.gov.zm."
This email was sent to all successful applicants early this month, and they were told they will be given offer letters and up to date there is nothing, we are at the end of the month and the pandemic is getting worse.
Meanwhile, senior health officers in the Ministry have expressed their concerns that the government is not fully providing them with protective gears a situation which has seen a good number of our health care providers contracting the virus and they were wondering why the Health Minister keeps on misleading the nation that these are available at all health centers. they have complained that man power is not enough owing to the government's ineptitude and they have seen no new employees and no one has been deployed yet.
They have further disclosed that there could be high abuse of donated moneys going on because ghost workers have been positioned to be collecting the said money.
We are very saddened with this news and we wonder why Dr Chitalu Chilufya and this government are showing this attitude towards citizens health.
It is very unfortunate that this pandemic in Zambia has been turned into a serious fundraising venture for the PF government.
We are also concerned about the donation that came from the government of China where Dr Chitalu Chilufya was asked to sign a Memorandum of Understanding. We are surprised to learn that a donation now requires a memorandum of understanding.
If Dr Chitalu Chilufya is refuting this information we challenge him to publish all the names of the people recruited within 48 hrs.
We doubt if at all there will be transparency in this so called recruitment process. The same thing that has happened with the donations meant to fight COVID-19 is the same thing is happening here. Theft on a grand scale.
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Zambian Watchdog Blog at WordPress.com.


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What's on TV in 2005?
With many old favourites coming to an end, TV stations are under pressure to hold their audiences in 2005.
Sex and the City, gone, Friends, gone, The Sopranos, question mark, ditto The West Wing. Six Feet Under? Well it's going as well after five seasons of the hit show (see more on Six Feet Underhere).
Suddenly the Nine Network has something of a problem on Monday nights
that it needs to plug pretty quickly in 2005. And it has problems
elsewhere in it 2005 schedule.
Friday night, Don Burke has gone (or goes after this Friday night). The
pressure is on for Rugby League and AFL to fill the spot from Sydney
and Melbourne respectively.
There are more question marks than ever for Nine this year and more
important decisions than for the competition. Why, because these
are decisions Nine must get substantially right to stay on top.
Seven's decisions just have to be right, otherwise it's the plank for
many senior executives. There are no more excuses allowed at Kerry
Stokes's network.
And for Ten, the decisions are easy. More of 2004 please, but there has
to be a sneaking worry about there being too many Big Events next year
to continue to make them special.
But Ten looks like having another strong year . It will be "Bigness" and Big Events – a lot of relying on the big winners like Idol, Big Brother, Law and Order, with the X Factor thrown in.
And Seven will do better than they have this year. More like the
network has performed in the past five weeks. But Nine is the
market leader and has the share that others, such as Seven, want.
After promising to lift that share this year, Seven knows 2004's
failure will count heavily if there are signs of a similar trend next
year. Instant oblivion for those responsible.
That's why Nine spinners and programmers are constantly comparing and
bagging Seven. Seven doesn't fixate as much on Nine or Ten. Ten just
runs its own race, ploughing ahead, confident of doing better each year.
Besides Burke's Backyard, This is Your Life will be lucky to be renewed after a short, but pretty boring existence these past few weeks. RPA is looking tired, but might still have to go round again. Getaway is being revamped to try and give it renewed strength, McLeod's is moseying a long with no growth in its audience.
CSI, CSI Miami and Without a Trace are still doing well. Cold Case looks good. The big imponderable is CSI New York, doing well in the US, but an expensive buy after a bidding way with Seven earlier this year.
ER is down more than 12% in US viewers this season. Meanwhile, Joey is no match for Friends and will struggle to fully fill the shoes of the fab six next year.
The Apprentice's third series will be at Nine by the end of
January, early February, an under performer in its second series, it
will not set the world on fire next year.
Nine News and A Current Affair are struggling in Sydney,
as Nine and Seven take 40 wacks a night out of each other. They do well
in Melbourne and Brisbane, but Seven's more than occasional wins
nationally in the 6.30pm slot are having a toll.
60 Minutes has been Idolised, and next year will be up against the X Factor, Idol, American Idol and Big Brother.
That's a tough year for the mob from tick tick tick, but what is the
overseer, John Westacott to do. He has to keep his eye on the main game
which is the battle between ACA and TT. To the finish? Viewers are underwhelmed by both programs.
Backyard Blitz has looked a bit repetitive and tired in format (how many more gardens can it do for disabled children?)
And if The World Around Us can push it, what will happen next year when the show comes back for more back yard renovations. Will it be "Burked"?
Frasier will be seen in its final series next year, so what then? More repeats? (It reminds me of the way that Ten used to repeat MASH, before it was snapped up by Seven).
Ten's reality type programs will probably pull a few more people next year, but the big interest is X Factor and whether Nine runs Starstruck up against it.
Nine might have an afternoon talk program, but that won't solve the holes in the evening time slots.
Sport will probably intrude earlier into Friday evenings in winter with NRL and AFL going live or near live.
Scotty's Place will probably replace Burke's Backyard in some way. Renovation Rescue will be there and the Today Show, breakfast news and ACA will be in a new studio.
The tiredness seen in the original Law and Order has spread to Criminal Intent. SVU looks good in new episodes, but both will have to be run heavily repeated next year. The new series, Trial By Jury won't be seen until after mid year.
Ten's lack of success in dramas will be another test in 2005. How many more Cooks can be made to spoil that broth.
Mary Bryant is supposed to be a local production, but it has lots of UK money in it, so its really and 'international' co-production.
For Seven the questions are many. Will Desperate Housewives and Lost from the US translate well into the Australian domestic viewing audience? Will Dancing with the Stars maintain its Tuesday night success?
Seven has Campus and Last Man Standing next year. Higher hopes for them than from local drama from Ten or Nine, after the successful revamping of All Saints and Blue Heelers.
Boston Legal will be another US program Seven has hopes for.
But unfortunately the Seven Network has a history of badly positioning successful programs sometimes.
But the network and the others know next year there will be no more chances if there's a repeat of this year's underperformance.
That's why you find there will be few promises from Seven, except to
say they know 2005 is time to perform and to execute the strategy
better than others.
Otherwise the execution will be their own. And Kerry Stokes won't miss.


Ms. McLaughlin (or Ms. M) has been working in education since 1997 when she started as a tutor at an alternative school. She taught in Salem, Massillon, and Youngstown before starting at TRECA as an English Teacher. Ms. M. loves teaching reading and writing to students of all ages. When she's not working online, she enjoys painting, photography, and traveling. She especially enjoys flying to California to spend time with her daughter and son-in-law!
First job: At an auto parts store, behind the counter. I loved working in the garage on cars with my dad growing up, so it was the perfect job for me.
Now playing: Heavy metal and classical - an awesome combination!
Favorite activity: I like taking cooking classes so I can prepare better meals!