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<!-- DOI: 10.1051/0004-6361/200913450 -->

<h2 class="sec">Online Material</h2>

<h2 class="sec"><a name="SECTION000100000000000000000"></a>
<A NAME="appendix_modelling"></A>
Appendix A: Modelling
</h2>

<p>Most of the calculations discussed here are limited to the 
two-dimensional (2D) plane of the binary orbit. Only in 
Sect.&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#sect_map_into_3D">A.7</a> do we rotate the results of the
2D calculations into the three-dimensional (3D) simulation volume that is used
to calculate the resulting flux.

</p><p></p><h3 class="sec2"><a name="SECTION000101000000000000000"></a>
A.1 Binary orbit</h3>

<p>Figure&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#fig_orbit">A.1</a> shows how the
two stars are positioned in the <i>XY</i>&nbsp;plane of the 3D
simulation volume.
The primary is at the centre of the volume
and the position of the secondary is
determined by the orbital phase. 
The angle of periastron (<IMG SRC="img10.png" ALT="$\omega $" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="12">)
is measured from
the intersection of the plane of the sky with the
plane of the orbit. The sight-line towards the observer is in the
<i>YZ</i>&nbsp;plane and the inclination angle <i>i</i> is the angle between the
plane of the sky and the orbital plane. The secondary rotates
clockwise in this figure (as seen from above). Periastron
corresponds to phase&nbsp;0.0.

</p><p></p><div class="inset-old">
<table>
<tr><td><!-- init Label --><A NAME="fig_orbit"></A><!-- end Label--><A NAME="1723"></A><A NAME="figure1211" HREF="img144.png"><IMG SRC="Timg144.png" ALT="\begin{figure}
\par\includegraphics[width=9cm,clip]{13450fg10.eps}
\end{figure}" HEIGHT="70" WIDTH="96"></A><!-- HTML Figure number: 10 -->
</td>
<td class="img-txt"><span class="bold">Figure A.1:</span><p>
Binary orbit in the <i>XY</i>
plane of the 3D simulation volume. The primary is positioned at the
centre of the volume. The argument of periastron is denoted <IMG SRC="img10.png" ALT="$\omega $" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="12">and the inclination angle <i>i</i>. The units on the axes are <IMG SRC="img11.png" ALT="$R_{\odot }$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="19">.
</td>
</tr><tr><td colspan="2"><a href="http://dexter.edpsciences.org/applet.php?pdf_id=10&DOI=10.1051/0004-6361/200913450" target="DEXTER">Open with DEXTER</a></td></tr>

</table></div>
<p></p><h3 class="sec2"><a name="SECTION000102000000000000000"></a>
A.2 Contact discontinuity</h3>

<p>The position of the contact discontinuity at any given orbital phase
is determined by the balance of ram pressure between the velocity components
orthogonal to the contact discontinuity (see equations in <a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2004ApJ...611..434A">Antokhin et&nbsp;al.  2004</a>).
These equations take into account that the terminal velocity of the
winds
may not have been reached yet before they collide. Specifically, in the
Cyg&nbsp;OB2 No.&nbsp;8A standard model presented in Sect.&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#section_standard_model">5.1</a>, the winds collide with 
<!-- MATH: $0.65 ~ \varv_\infty$ -->
<IMG SRC="img145.png" ALT="$0.65 ~ \varv_\infty$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="46">
(primary) and 
<!-- MATH: $0.71 ~ \varv_\infty$ -->
<IMG SRC="img146.png" ALT="$0.71 ~ \varv_\infty$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="45">
(secondary) at the apex at periastron. 
After the stellar wind material has passed through the shock, it
then moves outward parallel to the contact discontinuity (there is also a small orthogonal velocity component, see Sect.&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#sect_advection_and_cooling">A.5</a>).
We assume that the shocks and the contact discontinuity are
sufficiently close to one another that we do not need to make a
distinction in their geometric positions.
On each side of the contact discontinuity, we assume the stellar wind
to have a <IMG SRC="img147.png" ALT="$\beta=1$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="34">
velocity law and a density derived from 
the mass conservation law, using the assumed velocity law
and mass-loss rate.

</p><p></p><h3 class="sec2"><a name="SECTION000103000000000000000"></a>
A.3 Momentum distribution relativistic electrons</h3>

<p>We determine the momentum distribution of the relativistic electrons 
for a grid of points
along the two shocks. We start with a population of electrons
generated at a given point on the shock itself. We then follow the
time evolution of this population as it advects away and cools down.
The momentum distribution uses a grid that is uniform in
<IMG SRC="img148.png" ALT="$\log p$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="31">,
between <i>p</i><sub>0</sub> = 1.0&nbsp;MeV/<i>c</i> 
(for the choice of this specific value, see
     <A NAME="aaref43"></A><a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2004AetA...418..717V">Van Loo et&nbsp;al. 2004</a>)
and the upper limit <IMG SRC="img4.png" ALT="$p_{\rm c}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="16">
(defined below, Eq.&nbsp;(<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#eq_pc">A.3</a>))
These limits are sufficiently wide to
cover all of the radio emission we are interested in.
We stop following the electrons when their 
momentum falls below <i>p</i><sub>0</sub>, or when they leave the simulation
volume. (In our calculations we checked that the
emission beyond the simulation volume is negligible.)

</p><p>In the following, we refer mainly to 
<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2005PhDT.........1V">Van Loo (2005,  hereafter VL)</a>
as the source for the equations. Citations to the original papers
can be found in that reference.

</p><p></p><h3 class="sec2"><a name="SECTION000104000000000000000"></a>
A.4 New relativistic electrons</h3>

<p>At the shock, new relativistic electrons are put into the system.
Taking into account the particle acceleration mechanism
and the inverse Compton cooling, it can be shown that
their momenta follow a modified power-law distribution 
(VL, Eqs.&nbsp;(5.5), (5.6)):
<br>
</p><DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
N(p) & = & N_{\rm E} p^{-n} \left(1 - \frac{p^2}{p_{\rm c}^2} \right)^{(n-3)/2} , \\
n & = & \frac{\chi +2 }{\chi -1} \cdot
\end{eqnarray} -->
<TABLE ALIGN="CENTER" cellpadding="0" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><i>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;N</i>(<i>p</i>)
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img149.png" ALT="$\displaystyle N_{\rm E} p^{-n} \left(1 - \frac{p^2}{p_{\rm c}^2} \right)^{(n-3)/2} ,$" ALIGN="MIDDLE" BORDER="0" HEIGHT="60" WIDTH="142">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.1)
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><i>n</i>
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img150.png" ALT="$\displaystyle \frac{\chi +2 }{\chi -1} \cdot$" ALIGN="MIDDLE" BORDER="0" HEIGHT="49" WIDTH="44">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.2)
</td>
</tr>
</table></div><br clear="all"><p></p>
With (VL, Eqs.&nbsp;(5.1)-(3)):
<p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
\left( \frac{p_{\rm c}}{m_{\rm e} c} \right)^2 =
\frac{\chi \Delta u^2}{\chi^2 - 1} \frac{4\pi r^2}{\sigma_{\rm T} L_*}
\frac{eB(r)}{c} ,
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><A NAME="eq_pc"></A><IMG SRC="img151.png" ALT="\begin{displaymath}\left( \frac{p_{\rm c}}{m_{\rm e} c} \right)^2 =
\frac{\chi ...
...i^2 - 1} \frac{4\pi r^2}{\sigma_{\rm T} L_*}
\frac{eB(r)}{c} ,
\end{displaymath}" HEIGHT="75" WIDTH="184">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.3)
</td>
</tr>
</table></div><br clear="all"><p></p>
with
<IMG SRC="img110.png" ALT="$m_{\rm e}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="18">
and <i>e</i> the mass and charge of the electron,

<!-- MATH: $\sigma_{\rm T}$ -->
<IMG SRC="img152.png" ALT="$\sigma_{\rm T}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="19">
the Thomson electron scattering cross section,
<i>c</i> the speed of light,
<i>L</i><sub>*</sub> the luminosity of the star
and <i>r</i> the distance of the point on the shock to the star. 
Because there are
two stars, we should calculate the contribution of both:
<p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
\frac{L_*}{r^2} = \frac{L_{*,1}}{r_1^2} + \frac{L_{*,2}}{r_2^2} ,
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><IMG SRC="img153.png" ALT="\begin{displaymath}\frac{L_*}{r^2} = \frac{L_{*,1}}{r_1^2} + \frac{L_{*,2}}{r_2^2} ,
\end{displaymath}" HEIGHT="73" WIDTH="112">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.4)
</td>
</tr>
</table></div><br clear="all"><p></p>
with <i>r</i><sub>1</sub> the distance to the primary and <i>r</i><sub>2</sub> the distance to
the secondary.
Also (VL, Eq.&nbsp;(3.7)):
<br>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
\Delta u = u_1 - u_2 = u_1 \frac{\chi-1}{\chi} ,
\end{eqnarray} -->
<TABLE ALIGN="CENTER" cellpadding="0" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td style="text-align: center;" nowrap="nowrap"><IMG SRC="img154.png" ALT="$\displaystyle \Delta u = u_1 - u_2 = u_1 \frac{\chi-1}{\chi} ,$" ALIGN="MIDDLE" BORDER="0" HEIGHT="49" WIDTH="157">
</td>
<td>&nbsp;
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.5)
</td>
</tr>
</table></div><br clear="all"><p></p>
where <i>u</i><sub>1</sub> is the upstream velocity and <i>u</i><sub>2</sub> the downstream one
and the shock strength 
<!-- MATH: $\chi=u_1/u_2$ -->
<IMG SRC="img155.png" ALT="$\chi=u_1/u_2$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="59">.
Throughout this work we assume
strong shocks with <IMG SRC="img102.png" ALT="$\chi=4$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="34">,
but we note that higher values can be expected in radiative shocks.

<p>The normalization factor <IMG SRC="img156.png" ALT="$N_{\rm E}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="19">
is related to <IMG SRC="img88.png" ALT="$\zeta$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="9">,
the fraction of
energy that gets transferred from the shock to the relativistic particles, by
(VL, Eq.&nbsp;(5.10)):
<br>
</p><DIV ALIGN="CENTER"><A NAME="eq_normalization_factor"></A>
<!-- MATH: \begin{eqnarray}
N_{\rm E} = 4 \frac{\zeta}{\displaystyle \log_{\rm e} \left(\frac{p_{\rm c}}{p_0} \right)} \rho_1 \frac{\Delta u^2}{c} \quad ({\rm for}~\chi=4),
\end{eqnarray} -->
<TABLE ALIGN="CENTER" cellpadding="0" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td style="text-align: center;" nowrap="nowrap"><IMG style="border: 0px solid ; width: 234px; height: 95px;" SRC="img157.png" ALT="$\displaystyle N_{\rm E} = 4 \frac{\zeta}{\displaystyle \log_{\rm e} \left(\frac{p_{\rm c}}{p_0} \right)} \rho_1 \frac{\Delta u^2}{c} \quad ({\rm for}~\chi=4),$" ALIGN="MIDDLE">
</td>
<td>&nbsp;
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.6)
</td>
</tr>
</table></div><br clear="all"><p></p>
where <IMG SRC="img158.png" ALT="$\rho_1$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="15">
is the upstream density and we 
assume 
<!-- MATH: $\zeta=0.05$ -->
<IMG SRC="img89.png" ALT="$\zeta=0.05$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="52">
(<a name="tex2html46" href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1987PhR...154....1B">Blandford &amp; Eichler  1987</a>; <a name="tex2html47" href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1993ApJ...402..271E">Eichler &amp; Usov  1993</a>). 

<p>The magnetic field at distance <i>r</i> to the relevant star is assumed to be given by 
(<A NAME="aaref47"></A><a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1967ApJ...148..217W">Weber &amp; Davis 1967</a>;
(VL, Eq.&nbsp;(2.48))):
<br></p><p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
B(r) = B_* \frac{\varv_{\rm rot}}{\varv_\infty} \frac{R_*}{r} ,
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><A NAME="eq_magnetic_field_simple"></A><IMG SRC="img159.png" ALT="\begin{displaymath}B(r) = B_* \frac{\varv_{\rm rot}}{\varv_\infty} \frac{R_*}{r} ,
\end{displaymath}" HEIGHT="69" WIDTH="107">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.7)
</td>
</tr>
</table></div><br clear="all"><p></p>
where <i>B</i><sub>*</sub> is the surface magnetic field (assumed to be 100 G)
and 
<!-- MATH: $\varv_{\rm rot}$ -->
<IMG SRC="img160.png" ALT="$\varv_{\rm rot}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="20">
is the equatorial rotational velocity.
This equation is only valid at larger distances from
the star, but we use it everywhere as the region close to the star will
not be visible anyway (due to free-free absorption). This equation
neglects any amplification
of the field (such as described by, e.g. <a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2004MNRAS.353..550B">Bell 2004</a>)
or any reduction by magnetic reconnection. For each of the two shocks, the corresponding
magnetic field is taken into account.

<p></p><h3 class="sec2"><a name="SECTION000105000000000000000"></a>
<A NAME="sect_advection_and_cooling"></A>
A.5 Advection and cooling
</h3>

<p>During a time step <IMG SRC="img130.png" ALT="$\Delta t$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="16">,
the relativistic electrons advect away and
cool mainly due to inverse Compton and adiabatic cooling  (see  
<A NAME="aaref9"></A><a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1992PhDT.........1C">Chen 1992</a>; and VL, p.&nbsp;96,
for a discussion on other cooling mechanisms and why they are not
important in the present situation). 

</p><p>The equation for cooling of a single relativistic electron can be written&nbsp;as:
<br>
</p><DIV ALIGN="CENTER"><A NAME="eq_cooling"></A><A NAME="eq_adiabatic_cooling"></A>
<!-- MATH: \begin{eqnarray}
\frac{{\rm d}p}{{\rm d}t} & = & -a p^2 +b p \quad\mbox{with}\\
a & = & \frac{\sigma_{\rm T}}{3 \pi m_{\rm e}^2 c^3} \frac{L_*}{r^2} , \\
b & = & \frac{1}{3 \rho} \frac{\Delta \rho}{\Delta t},
\end{eqnarray} -->
<TABLE ALIGN="CENTER" cellpadding="0" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <IMG SRC="img161.png" ALT="$\displaystyle \frac{{\rm d}p}{{\rm d}t}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="49" WIDTH="23">
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img162.png" ALT="$\displaystyle -a p^2 +b p \quad\mbox{with}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="35" WIDTH="112">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.8)
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><i>a</i>
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img163.png" ALT="$\displaystyle \frac{\sigma_{\rm T}}{3 \pi m_{\rm e}^2 c^3} \frac{L_*}{r^2} ,$" ALIGN="MIDDLE" BORDER="0" HEIGHT="49" WIDTH="75">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.9)
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><i>b</i>
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img164.png" ALT="$\displaystyle \frac{1}{3 \rho} \frac{\Delta \rho}{\Delta t},$" ALIGN="MIDDLE" BORDER="0" HEIGHT="49" WIDTH="48">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.10)
</td>
</tr>
</table></div><br clear="all"><p></p>
where <i>a p</i><sup>2</sup> gives the inverse 
Compton<A NAME="tex2html19" HREF="#foot1736"><sup><IMG ALT="[*]" SRC="/icons/foot_motif.png" align="bottom" BORDER="1"></sup></A>
cooling  (VL, Eq.&nbsp;(5.2))
and <i>b p</i> the adiabatic cooling (which can be derived from the fact that
for adiabatic cooling

<!-- MATH: $p \propto T^{1/2} \propto \rho^{(\gamma-1)/2}$ -->
<IMG SRC="img166.png" ALT="$p \propto T^{1/2} \propto \rho^{(\gamma-1)/2}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="30" WIDTH="108">
and using 
<!-- MATH: $\gamma=5/3$ -->
<IMG SRC="img167.png" ALT="$\gamma=5/3$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="48">).

<p>Integration of Eq.&nbsp;(<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#eq_cooling">A.8</a>) gives:
<br></p><p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
p  =  \frac{p_{\rm init} \exp (b \Delta t)}{\displaystyle 1 - \frac{a}{b} p_{\rm init} \left( 1 - \exp (b \Delta t) \right)} ,
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><A NAME="eq_cooling_one_electron"></A><IMG SRC="img168.png" ALT="\begin{displaymath}p = \frac{p_{\rm init} \exp (b \Delta t)}{\displaystyle 1 - \frac{a}{b} p_{\rm init} \left( 1 - \exp (b \Delta t) \right)} ,
\end{displaymath}" HEIGHT="83" WIDTH="187">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.11)
</td>
</tr>
</table></div><br clear="all"><p></p>
where 
<!-- MATH: $p_{\rm init}$ -->
<IMG SRC="img169.png" ALT="$p_{\rm init}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="25">
is the momentum at the previous time step.
We can invert this to:
<p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
p_{\rm init} = \frac{p \exp (-b \Delta t)}{\displaystyle 1 - \frac{a}{b} p \left( 1 -\exp (-b \Delta t) \right)} \cdot
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><A NAME="eq_p-pinit_simple"></A><IMG SRC="img170.png" ALT="\begin{displaymath}p_{\rm init} = \frac{p \exp (-b \Delta t)}{\displaystyle 1 - \frac{a}{b} p \left( 1 -\exp (-b \Delta t) \right)} \cdot
\end{displaymath}" HEIGHT="83" WIDTH="197">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.12)
</td>
</tr>
</table></div><br clear="all"><p></p>
We will also need the following derivative:
<p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
\frac{{\rm d}p_{\rm init}}{{\rm d}p} = \frac{\exp (-b \Delta t)}{\displaystyle \left[1 - \frac{a}{b} p \left( 1 - \exp (-b \Delta t) \right)\right]^2} \cdot
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><IMG SRC="img171.png" ALT="\begin{displaymath}\frac{{\rm d}p_{\rm init}}{{\rm d}p} = \frac{\exp (-b \Delta ...
...{a}{b} p \left( 1 - \exp (-b \Delta t) \right)\right]^2} \cdot
\end{displaymath}" HEIGHT="88" WIDTH="225">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.13)
</td>
</tr>
</table></div><br clear="all"><p></p>
We assume that only the momentum <i>p</i> changes significantly with time; for the other
variables, we take their value mid-way between the start and the end
of the time step <IMG SRC="img130.png" ALT="$\Delta t$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="16">.
The density <IMG SRC="img116.png" ALT="$\rho$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="10">
we need in Eq.&nbsp;(<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#eq_adiabatic_cooling">A.10</a>)
is the one between the shock and the contact discontinuity. We do not have detailed hydrodynamical
models for this, so we simply approximate it by 
<!-- MATH: $\rho = \chi \rho_{\rm s}$ -->
<IMG SRC="img172.png" ALT="$\rho = \chi \rho_{\rm s}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="46">
where 
<!-- MATH: $\rho_{\rm s}$ -->
<IMG SRC="img173.png" ALT="$\rho_{\rm s}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="15">
is the density in the smooth wind at the position of the shock.
We stress once again that in these radiative shocks, the post-shock 
density increases rapidly downstream, maybe by several orders of magnitude,
rather than the <IMG SRC="img102.png" ALT="$\chi=4$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="34">
value we use. 

<p>The above equations describe what happens to a single relativistic
electron. The time evolution of the momentum distribution function over
the time step <IMG SRC="img130.png" ALT="$\Delta t$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="16">
is:
<br></p><p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
N(p) = N(p_{\rm init}) \frac{\rho (t+\Delta t)}{\rho (t)} \frac{{\rm d}p_{\rm init}}{{\rm d}p} \cdot
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><A NAME="eq_p-distribution_simple"></A><IMG SRC="img174.png" ALT="\begin{displaymath}N(p) = N(p_{\rm init}) \frac{\rho (t+\Delta t)}{\rho (t)} \frac{{\rm d}p_{\rm init}}{{\rm d}p} \cdot
\end{displaymath}" HEIGHT="70" WIDTH="192">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.14)
</td>
</tr>
</table></div><br clear="all"><p></p>
We stop following the electrons when their 
momentum falls below <i>p</i><sub>0</sub>, or when they leave the simulation
volume.

<p>In our simplifying assumptions the contact discontinuity
and the shock are at the same geometrical position. Taken literally, this
would result in a synchrotron emitting region that has zero thickness and
zero synchrotron emission. In the true hydrodynamical situation, particles
move through the shock (which is at some distance from the contact
discontinuity) and then move slowly towards the contact discontinuity.
At the same time, they are being advected outward with a velocity
parallel to the contact discontinuity that is much larger than
the corresponding orthogonal velocity. The thickness of the
synchrotron emitting region is set by the combination of these
two velocities.
In the absence of hydrodynamical information, we use the following
procedure to determine these velocities.
The parallel component (
<!-- MATH: $\varv_{\|}$ -->
<IMG SRC="img175.png" ALT="$\varv_{\Vert}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="14">)
at the contact discontinuity
is determined by 
projecting the smooth wind velocity vector on to the contact discontinuity. 
At other positions in the wind, we interpolate in the grid of

<!-- MATH: $\varv_{\|}$ -->
<IMG SRC="img175.png" ALT="$\varv_{\Vert}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="14">
as a function of the <i>y</i>-coordinate.
The orthogonal component (
<!-- MATH: $\varv_{\bot}$ -->
<IMG SRC="img133.png" ALT="$\varv_{\bot}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="17">)
at the contact discontinuity is expected to be of the order of 
the orthogonal component of the material coming into the shock, with a minimum
value corresponding to the thermal expansion of the gas. As a 
simplification, we set it exactly equal to the incoming orthogonal component,
and we use the same type of interpolation
as for the parallel velocity to determine it at other positions.
We reverse the orthogonal velocity direction
to have the gas moving away from the contact discontinuity.
Note that this situation is ``inside out'' compared to the
true hydrodynamical situation where the particles move towards
the contact discontinuity. The thickness of the synchrotron emitting
region is so small however that this inversion has negligible influence on
our results.

</p><p></p><h3 class="sec2"><a name="SECTION000106000000000000000"></a>
<A NAME="section_emissivity"></A>
A.6 Emissivity
</h3>

<p>The equation for the emissivity is given by (VL, Eq.&nbsp;(B.1)):
<br>
</p><DIV ALIGN="CENTER"><A NAME="equation_emissivity"></A>
<!-- MATH: \begin{eqnarray}
j_\nu(r) & = & \frac{1}{4 \pi} \int_{p_0}^{p_{\rm c}} {\rm d}p N(p,r)
\int_0^\pi \frac{{\rm d}\theta}{4\pi} 2\pi \sin \theta
\frac{\sqrt{3} e^3}{m_{\rm e} c^2} \nonumber \\
& & \times  B f(\nu,p) \sin \theta
F \left(\frac{\nu}{f^3(\nu,p)\nu_{\rm s}(p,r) \sin\theta} \right) ,
\end{eqnarray} -->
<TABLE ALIGN="CENTER" cellpadding="0" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<IMG SRC="img176.png" ALT="$\displaystyle j_\nu(r)$" ALIGN="MIDDLE" BORDER="0" HEIGHT="28" WIDTH="33">
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img177.png" ALT="$\displaystyle \frac{1}{4 \pi} \int_{p_0}^{p_{\rm c}} {\rm d}p N(p,r)
\int_0^\pi \frac{{\rm d}\theta}{4\pi} 2\pi \sin \theta
\frac{\sqrt{3} e^3}{m_{\rm e} c^2}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="58" WIDTH="246">
</td>
<td ALIGN="RIGHT" WIDTH="10">
&nbsp;
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap">&nbsp;
</td>
<td>&nbsp;
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img178.png" ALT="$\displaystyle \times B f(\nu,p) \sin \theta
F \left(\frac{\nu}{f^3(\nu,p)\nu_{\rm s}(p,r) \sin\theta} \right) ,$" ALIGN="MIDDLE" BORDER="0" HEIGHT="53" WIDTH="250">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.15)
</td>
</tr>
</table></div><br clear="all"><p></p>
with (VL, Eq.&nbsp;(2.51), (2.50)):
<br>
<DIV ALIGN="CENTER"><A NAME="eq_Razin"></A><A NAME="eq_nu_s"></A>
<!-- MATH: \begin{eqnarray}
f (\nu,p) & = & \left[ 1+ \frac{\nu_0^2}{\nu^2} \left( \frac{p}{m_{\rm e} c} \right)^2 \right]^{-1/2}\\
\nu_0 & = & \sqrt{\frac{n_{\rm e} e^2}{\pi m_{\rm e}}} \\
\nu_{\rm s} (p,r) & = & \frac{3}{4 \pi} \frac{eB}{m_{\rm e} c}
\left( \frac{p}{m_{\rm e} c} \right)^2\\
F(x) & = & \int_x^\infty {\rm d}\eta K_{5/3} (\eta) .

\end{eqnarray} -->
<TABLE ALIGN="CENTER" cellpadding="0" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <IMG SRC="img179.png" ALT="$\displaystyle f (\nu,p)$" ALIGN="MIDDLE" BORDER="0" HEIGHT="28" WIDTH="43">
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img180.png" ALT="$\displaystyle \left[ 1+ \frac{\nu_0^2}{\nu^2} \left( \frac{p}{m_{\rm e} c} \right)^2 \right]^{-1/2}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="65" WIDTH="127">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.16)
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><IMG SRC="img181.png" ALT="$\displaystyle \nu_0$" ALIGN="MIDDLE" BORDER="0" HEIGHT="28" WIDTH="16">
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img182.png" ALT="$\displaystyle \sqrt{\frac{n_{\rm e} e^2}{\pi m_{\rm e}}}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="68" WIDTH="50">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.17)
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><IMG SRC="img183.png" ALT="$\displaystyle \nu_{\rm s} (p,r)$" ALIGN="MIDDLE" BORDER="0" HEIGHT="28" WIDTH="47">
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img184.png" ALT="$\displaystyle \frac{3}{4 \pi} \frac{eB}{m_{\rm e} c}
\left( \frac{p}{m_{\rm e} c} \right)^2$" ALIGN="MIDDLE" BORDER="0" HEIGHT="59" WIDTH="98">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.18)
</td>
</tr>
<tr VALIGN="MIDDLE"><td ALIGN="RIGHT" nowrap="nowrap"><i>F</i>(<i>x</i>)
</td>
<td ALIGN="CENTER" nowrap="nowrap">=
</td>
<td ALIGN="LEFT" nowrap="nowrap"><IMG SRC="img185.png" ALT="$\displaystyle \int_x^\infty {\rm d}\eta K_{5/3} (\eta) .$" ALIGN="MIDDLE" BORDER="0" HEIGHT="51" WIDTH="96">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.19)
</td>
</tr>
</table></div><br clear="all"><p></p>
We use Eq.&nbsp;(<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#eq_magnetic_field_simple">A.7</a>) for the magnetic field.
Note that this is an approximation, as the shock may compress the
magnetic field as well, leading to higher values than given by
Eq.&nbsp;(<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#eq_magnetic_field_simple">A.7</a>). The expression for <IMG SRC="img90.png" ALT="$j_\nu (r)$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="29">
is evaluated at position
<i>r</i> and frequency <IMG SRC="img109.png" ALT="$\nu$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="10">.
The <i>f</i> function corrects for the Razin effect and
<i>K</i><sub>5/3</sub> is the modified Bessel function. To calculate <IMG SRC="img85.png" ALT="$n_{\rm e}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="16">,
the number density of the <EM>thermal</EM> electrons (i.e. those in the non-relativistic plasma),
we use:
<p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
n_{\rm e}=\frac{\chi \rho_{\rm s}}{m_{\rm H}} \frac{1.2}{1.4} ,
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><A NAME="eq_ne"></A><IMG SRC="img186.png" ALT="\begin{displaymath}n_{\rm e}=\frac{\chi \rho_{\rm s}}{m_{\rm H}} \frac{1.2}{1.4} ,
\end{displaymath}" HEIGHT="69" WIDTH="85">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.20)
</td>
</tr>
</table></div><br clear="all"><p></p>
where 
<!-- MATH: $\rho_{\rm s}$ -->
<IMG SRC="img173.png" ALT="$\rho_{\rm s}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="15">
is the smooth wind density and <IMG SRC="img187.png" ALT="$m_{\rm H}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="21">
the mass
of a hydrogen atom. The factor 1.2/1.4 takes into account a 
fully ionized hydrogen+helium solar composition.
In these radiative shocks the post-shock density may increase
by several orders of magnitude rather than by <IMG SRC="img102.png" ALT="$\chi=4$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="34">,
which will influence the value of&nbsp;<IMG SRC="img85.png" ALT="$n_{\rm e}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="16">.

<p>To reduce
the computing time we set 
<!-- MATH: $\sin \theta =1$ -->
<IMG SRC="img188.png" ALT="$\sin \theta =1$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="51">
in the 
<!-- MATH: $f \sin \theta$ -->
<IMG SRC="img189.png" ALT="$f \sin \theta$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="39">
factors of Eq.&nbsp;(<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#equation_emissivity">A.15</a>) and we also use:
<br></p><p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
\int_0^\pi \frac{{\rm d}\theta}{4\pi} {2\pi \sin \theta} = 1 .
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><IMG SRC="img190.png" ALT="\begin{displaymath}\int_0^\pi \frac{{\rm d}\theta}{4\pi} {2\pi \sin \theta} = 1 .
\end{displaymath}" HEIGHT="70" WIDTH="122">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.21)
</td>
</tr>
</table></div><br clear="all"><p></p>
The <i>F</i> function is precalculated on a grid of <i>x</i>-values; the evaluation
of the function then reduces to an interpolation.

<p></p><h3 class="sec2"><a name="SECTION000107000000000000000"></a>
<A NAME="sect_map_into_3D"></A>
A.7 Map into 3D grid
</h3>

<p>Using the equations from the previous sections, we can make a 
2D map of the synchrotron emission
(part of which is shown in Fig.&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#fig_synchrotron_emission">6</a>).
Note that the synchrotron emission extends on both sides of the
contact discontinuity, because there is a shock on either side.
Based on the cylindrical symmetry around the line
connecting the two stars (at the given orbital phase),
we then
``rotate'' this 2D map into the 3D
simulation volume.

</p><p>It is important that during this procedure we do not lose part
of the emission due to interpolation or reduced resolution.
To achieve this, we use volume-integrated emissivities.
While doing the 2D calculation we store the volume integrated
emissivity 
<!-- MATH: $\overline{j_\nu(r)}$ -->
<IMG SRC="img191.png" ALT="$\overline{j_\nu(r)}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="33" WIDTH="30">
defined by:
<br></p><p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
\overline{j_\nu(r)} \Delta V = \int_{\Delta V} j_\nu(r) {\rm d} V .
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><IMG SRC="img192.png" ALT="\begin{displaymath}\overline{j_\nu(r)} \Delta V = \int_{\Delta V} j_\nu(r) {\rm d} V .
\end{displaymath}" HEIGHT="70" WIDTH="145">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.22)
</td>
</tr>
</table></div><br clear="all"><p></p>
The volume <IMG SRC="img193.png" ALT="$\Delta V$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="22">
used in this definition is determined as
follows. We consider the 2D surface covered by (1) the step in the
coordinate along the contact discontinuity
and (2) 
the distance travelled by the relativistic electrons 
between two consecutive time steps.
We then rotate this surface around the
line connecting the two stars
over a small angle <IMG SRC="img194.png" ALT="$\alpha$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="11">,
thereby defining the emissivity
volume <IMG SRC="img193.png" ALT="$\Delta V$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="22">.

<p>When we have finished the full 2D calculation,
we rotate the 2D plane into the 3D simulation volume,
using steps of angle <IMG SRC="img194.png" ALT="$\alpha$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="11">(as defined above). 
     For each of the emissivity volumes <IMG SRC="img193.png" ALT="$\Delta V$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="22">
defined above, and for
a given rotation angle, we determine
what fraction of the rotated volume overlaps with the 
cells in the 3D simulation volume. This is done by 
subdividing each dimension of the <IMG SRC="img193.png" ALT="$\Delta V$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="22">
volume into three.
For each of the resulting 3<sup>3</sup> points 
we determine in which 
3D simulation cell they end up after rotation.
A running sum for that cell is then updated with
the fraction 3<sup>-3</sup> 
of the volume integrated emissivity 
<!-- MATH: $\overline{j_\nu}$ -->
<IMG SRC="img195.png" ALT="$\overline{j_\nu}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="33" WIDTH="15">of the rotated volume.
At the end of this procedure, the accumulated emission in each
3D simulation cell is divided by its volume, thus obtaining
<IMG SRC="img6.png" ALT="$j_\nu $" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="14">
for that cell.

</p><p>In this procedure we lose some resolution, specifically
around the apex, where the 2D resolution is quite high but where
the 3D resolution is much lower. However, because of the use of 
volume-integrated quantities, we do not lose any of the synchrotron
emission. Furthermore, the apex of the wind is well 
hidden by the free-free absorption
and will therefore not contribute to the resulting radio flux at
the wavelengths we consider. If our model were to be applied to X-rays, 
optical emission or radio
emission at much shorter wavelengths, a higher-resolution 3D grid would 
be required. 

</p><p>In the 3D model. we also include the thermal free-free opacity and 
emissivity, which is due to the ionized wind material. 
To calculate
these we use the <a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1975MNRAS.170...41W">Wright &amp; Barlow (1975)</a> equations.
The winds of both stars are assumed to be smooth.
For the Gaunt factor at frequency <IMG SRC="img109.png" ALT="$\nu$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="10">
we use the equation from
<A NAME="aaref23"></A><a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1991ApJ...377..629L">Leitherer &amp; Robert (1991)</a>:
<br></p><p></p>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{equation}
g_\nu = 9.77 \left( 1 + 0.13 \log_{10} \frac{T_{\rm e}^{3/2}}{Z \nu} \right),
\end{equation} -->
<TABLE ALIGN="CENTER" WIDTH="100%">
<tbody><tr VALIGN="MIDDLE"><td ALIGN="CENTER" nowrap="nowrap"><IMG SRC="img196.png" ALT="\begin{displaymath}g_\nu = 9.77 \left( 1 + 0.13 \log_{10} \frac{T_{\rm e}^{3/2}}{Z \nu} \right),
\end{displaymath}" HEIGHT="76" WIDTH="199">
</td>
<td ALIGN="RIGHT" WIDTH="10">
(A.23)
</td>
</tr>
</table></div><br clear="all"><p></p>
where <IMG SRC="img197.png" ALT="$T_{\rm e}$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="17">
is the thermal electron temperature in the wind
and <i>Z</i> the charge of the ions.
The stars themselves are introduced into the simulation
volume as high opacity spheres.
We then solve the radiative transfer equation in Cartesian coordinates
along the line of sight. We use
Adam's (<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1990AetA...240..541A">Adam  1990</a>) finite volume method because of its
simplicity. By repeating the whole procedure for a number of orbital
phases and wavelengths, we calculate the radio light curve at these
wavelengths and determine how the fluxes and spectral index change
with phase.

<p></p><h2 class="sec"><a name="SECTION000110000000000000000"></a>
<A NAME="appendix_data_reduction"></A>Appendix B: Data reduction
</h2>

<p>The data reduction is accomplished using the Astronomical Image
Processing System (AIPS), developed by the NRAO. We follow
the standard procedures of antenna gain calibration, absolute flux
calibration, imaging and deconvolution. Where possible, we apply
self-calibration to the observations (see notes to 
Table&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#table_radio_data">B.1</a>). We measure the fluxes and error bars by fitting
elliptical Gaussians to the sources on the images. The error bars listed
in Table&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#table_radio_data">B.1</a> include not only the rms noise
in the map, but also an estimate of the systematic errors that 
were evaluated using a jackknife technique. This technique drops
part of the observed data and redetermines the fluxes, giving some
indication of systematic errors that can be present.
Upper limits are three times the uncertainty as derived above. 
For details of the reduction, we refer
to the previous papers in this series 
(<A NAME="aaref8"></A><a name="tex2html48" href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2007AetA...464..701B">Blomme et&nbsp;al.  2007</a>; <a name="tex2html49" href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2008AetA...483..585V">Van Loo et&nbsp;al.  2008</a>; <a name="tex2html50" href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2005AetA...436.1033B">Blomme et&nbsp;al.  2005</a>).
We exclude from Table&nbsp;<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#table_radio_data">B.1</a>
those observations with upper limits higher than 6&nbsp;mJy (which
corresponds to about 3&nbsp;times the highest detection at all wavelengths).

</p><p>A comparison with values previously published in the literature
shows that in most cases the error bars overlap with ours.
There are just a few exceptions, which we discuss further. 
<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1989ApJ...340..518B">Bieging et&nbsp;al. (1989)</a> list only an upper limit for the AC116 2&nbsp;cm observation, while we find
a detection (though marginal at just above 3 sigma). Our detection
is quite consistent in all variants we consider in the jackknife test
and is at the exact position of Cyg&nbsp;OB2 No.&nbsp;8A, so we consider it a detection.
For the AS483 6&nbsp;cm, <A NAME="aaref46"></A><a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1998ApJS..118..217W">Waldron et&nbsp;al. (1998)</a>
find a value of 
<!-- MATH: $1.04 \pm 0.08$ -->
<IMG SRC="img198.png" ALT="$1.04 \pm 0.08$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="67">&nbsp;mJy, which is <IMG SRC="img30.png" ALT="$\sim$" align="bottom" BORDER="0" HEIGHT="14" WIDTH="12">20% lower than our value.
A possible explanation is that they did not correct for the
decreasing sensitivity of the primary beam. This effect is 
important when the target is not in the centre of the image, and
is approximately 20% for the position of No.&nbsp;8A on the image.
<A NAME="aaref35"></A><a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#2006AetA...454..625P">Puls et&nbsp;al. (2006)</a> list an upper limit of 0.54 mJy for the AS786 6&nbsp;cm observation,
but we clearly detect Cyg&nbsp;OB2 No.&nbsp;8A on the image we made.
<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1989ApJ...340..518B">Bieging et&nbsp;al. (1989)</a> find a slightly lower value of 
<!-- MATH: $1.0 \pm 0.1$ -->
<IMG SRC="img199.png" ALT="$1.0 \pm 0.1$" ALIGN="MIDDLE" BORDER="0" HEIGHT="26" WIDTH="54">&nbsp;mJy for the
AC116 1984-12-21, 20&nbsp;cm observation. We have no explanation for these last two discrepancies.

</p><p>We also take the opportunity to correct some clerical errors
in the literature:
<a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1989ApJ...340..518B">Bieging et&nbsp;al. (1989)</a> date the CHUR observations 
as 1980-Mar-22, but it should be 1980-May-23.
The <a href="/articles/aa/full_html/2010/11/aa13450-09/aa13450-09.html#1998ApJS..118..217W">Waldron et&nbsp;al. (1998)</a>
1991 observation was made at 3.6&nbsp;cm, not at 6&nbsp;cm
and the 1992 observation was made in January, not June.

</p><p><A NAME="table_radio_data"></A></p><p class="inset-old"><a href="/articles/aa/full_html/2010/11/aa13450-09/tableB.1.html"><span class="bold">Table B.1:</span></a>&#160;&#160;
Reduction of the Cyg&nbsp;OB2 No.&nbsp;8A VLA data. </p><br>

</div></body></html>