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\documentclass{article}
\usepackage{graphicx}
\usepackage{color}


\begin{document}
%\pagestyle{empty}
%\setlength{\oddsidemargin}{.2in}
%\Large
%\begin{centering}
\title{\Large First order perturbative calculations for a conducting liquid jet in a solenoid\\
  DRAFT}
%\vskip.3in
\author{R. B. Palmer\\
  Brookhaven National Laboratory}

\maketitle
\begin{abstract}
  A perturbative calculation is given of the behavior of a continuous
  jet of conducting fluid as it enters and leaves a solenoidal
  magnetic field. It is assumed that the changes in direction, jet
  cross section and velocity are small.
  
  If the jet enters the field along, or close to the axis, then the
  induced forces are compressive and retarding. The jet slows, suffers
  an increase in hydrostatic pressure, and increases in diameter;
  later, the jet reaccelerates and shrinks. As the jet leaves the
  field, the hydrostatic pressure becomes negative and cavitation may
  occur.
  
  If the jet enters at an angle to the axis, there are, in addition,
  deflections and elliptical deformations of the jet.
  
  Formulae are given for these effects and numerical values given for
  the example of a solenoidal field with a Gaussian axial profile.
\end{abstract}
\section{\bb Introduction}
A mercury jet, injected at an angle respect to the axis of the solenoidal field, is the current baseline solution for the Feasibitity Study II\cite{ref1}. The interaction of the liquid-metal jet with the strong $20$~T target solenoid has as result a number of forces on the jet which potentially may affect the viability of this target.\cite{ref2},\cite{ref3},\cite{ref4} We present here perturbative calculations which confirm the findings of previous authors.  
\section{\bb Formulae}

\subsection{Introduction}

The jet is assumed to have an initial radius $r$, and be traveling at
a velocity $v$. Changes in radius, shape, direction and velocity are
all assumed to be small. The angle between the jet and solenoid axes
is also assumed to be small.

In the following formulae, the coordinate system is defined by the
jet; $z$ is along the direction of motion and $r$ is perpendicular to
$z$. If the jet is along the axis then $z$ is also the solenoid axis,
and the azimuthal field $B_\phi$ = 0.

If the jet is not directed along the solenoid axis, then we also
define $y$ in a direction perpendicular to $z$ (the jet axis) and away
from the solenoid axis; and $x$ perpendicular to $y$ and $z$; see
fig.~\ref{fig1}
%In this case, $B_x=0$.
\begin{figure}
%  \centering \includegraphics[width=3.5in,angle=-90]{jet_fig1.ps}
%\includegraphics[bb= 45 90 480 700,angle=-90]{jet_fig1.ps}
\label{fig1}
\caption{Schematic of the geometrical arrangement of solenoid and jet}
\end{figure}

\subsection{Induced azimuthal current}


The magnetic flux through a circle of radius r perpendicular to the
jet axis is
\begin{equation}
\Phi = \int_S dS\vec{n}\cdot \vec{B}\approx \pi r^2 B_z(x,y,z)
\end{equation}
As a liquid metal jet passes axially down such a field at a velocity
$v={dz / dt}$, a circumferential potential will be generated\cite{ref6}
\begin{equation}
V \equiv \oint\vec{E}\cdot \vec{dl}=-{d\Phi \over dt}= \pi r^2v {dB_z(x,y,z) \over dz} 
\end{equation}
\marginpar{this equation is not correct!!! additional term proportional to $\dot{r}$. See Ref.~\cite{ref6}}
If the metal electrical conductivity $\sigma$ is low enough so that
the resulting current has a negligible effect on the field, then the
azimuthal current density $i_{\phi}$ will be
\begin{equation}
i_{\phi}\approx {V \over 2\pi r}\sigma
\end{equation}
\begin{equation}
{\rr i_{\phi}\approx {rv\kappa\over 2} {dB_z(0,0,z) \over dz}} 
\end{equation}

\subsection{Induced axial current}

Consider a transverse field component $B_y$

The magnetic flux between transverse positions $-x$ to $x$ and $dz$ is
\begin{equation}
d \Phi_y = 2x dz B_y(z)
\end{equation}
As a liquid metal jet passes axially down such a field at a velocity
$v={dz / dt}$, axial voltage gradients will be generated
\begin{equation}
G=x {dB_y \over dz} v
\end{equation}
If the metal electrical conductivity $\kappa$ is low enough so that
the resulting current has a negligible effect on the field, then the
axial current density $i_z$ will be
\begin{equation}
i_z= G\sigma
\end{equation}
\begin{equation}
{\rr i_z~=xv\kappa{dB_y(0,0,z) \over dz}}
\end{equation}


\subsection{Axial force}

The radial field is given by
\begin{equation}
B_r(x,y,z)\approx -{r \over 2}{\partial B_z(0,0,z) \over \partial z} 
\end{equation} 
\marginpar{ CHECK THIS EQUATION: $B_z$ is mainly $\theta B_z$}
The induced axial force per unit volume $(dr r d\phi dz)$ is
\begin{equation}
f_z=-B_r i_{\phi} \approx {r^2 \over 4}v \kappa\({dB_z \over dz} \)^2 
\end{equation} 
and the average force of the disk of radius $r_o$ is
\begin{equation}
{\rr <{f_z}> \approx{r_o^2 \over 8}v \kappa\({dB_z \over dz} \)^2 }
\end{equation} 

\subsection{Radial forces and hydrostatic pressure}

The induced radial force per unit volume $(dr~r~d\phi dz)$ is
\begin{equation}
f_r=B_z i_{\phi} \approx  {r \over 2 }~v \kappa ~B_z {dB_z \over dz} 
\end{equation}
If we assume the effects of the fields are small so that the jet
radius and liquid velocities do not vary by large fractions, and if we
ignore radial inertia, then the hydrostatic pressures in a jet of
outside radius $r_o$, at radius r, will be given by
\begin{equation}
p(r,z)=\int_{r_o}^r -{f_r}dr\approx \({r_o^2-r^2 \over 4}\)~v \kappa~ B_z {dB_z \over dz}
\end{equation}
This pressure is a function of $z$, and gradients in it will exert
axial pressures $f_p$ on the liquid that must be added to the magnetic
term $f_z$.
\begin{equation}
f_p(r,z)= {dp(r,z) \over dz} \approx~-~ \({r_o^2-r^2 \over 4}\)~v \kappa~
 {d \over dz}\( B_z {dB_z \over dz}\)
\end{equation}
and the average value of this force is
\begin{equation}
<{f_p(r,z)}> \approx ~-~\({r_o^2 \over 8}\)~v \kappa~ {d \over dz}\( B_z {dB_z \over dz}\)
\end{equation}
 
\subsection{Axial acceleration}


These forces will then decelerate, or accelerate layers of the fluid,
thus inducing differences of liquid velocity as a function of radius
\begin{equation}
\rho~{dv \over dt}~=~f
\end{equation}
\begin{equation}
\Delta~ v(r,z)~=~ \int_{z_o}^z{(f_z+f_p)~{1 \over v \rho}~ dz}
\end{equation}
The average change in velocity is then
\begin{equation}
{\rr <\Delta v>(z)~=~{\kappa \over \rho}~{r_o^2 \over 8}~   \(\int_{z_o}^z{\({dB_z \over dz} \)^2 
~+~{d \over dz}\( B_z {dB_z \over dz}\)~dz} \)}
\end{equation}
and the radius as a function of z is
\begin{equation}
r(z)~=~\(1~-~{<\Delta v>(z) \over v} \)
\end{equation}
\marginpar{WRONG DIMENSIONS!!!}

\subsection{Transverse forces and deflections}

From above, the radial force per unit volume $(dr~r~d\phi dz)$ is
\begin{equation}
f_r=B_z i_{\phi} \approx  {r \over 2 }~v \kappa ~B_z {dB_z \over dz} 
\end{equation}
If $B_z$ varies with a transverse distance $y$, then the component of
this radial force in the $y$ direction is
\begin{equation}
f_y = f_r~\sin{\phi}
\end{equation}
and the net deflective force per unit length $dF_y$ is
\begin{equation}
{dF_y \over dz}~ =~ \int_o^r ~\int_o^{2\pi} 
   {r \over 2 }~v \kappa ~{dB_z \over dy}~r \sin{\phi}^2 {dB_z \over dz} ~r~ dr d\phi 
\end{equation}
\begin{equation}
= ~{v \kappa \over 2} ~{dB_z \over dy}~ {dB_z \over dz} ~\int_o^{2\pi}{ \sin{\phi}^2 d \phi}
\int_o^r {r^3~dr }
 \end{equation}
\begin{equation}
= ~{\pi \over 8}v \kappa ~r^4 ~{dB_z \over dy}~ {dB_z \over dz} 
\end{equation}
and the change in transverse velocity
\begin{eqnarray}
{d v_y \over dz} & = & {1 \over v}{d v_y \over dt}\nonumber \cr
&=& {{dF_y \over dz}dz\over v \rho \pi  r^2  dz}\nonumber \cr
&=&{\kappa r^2 \over 8 \rho}  {dB_z \over dy} {dB_z \over dz}\cr 
\end{eqnarray}
and the inverse radius of bend is {\rr \begin{equation} {d \theta
      \over dz}~=~{\kappa~r^2 \over 8~v~\rho} ~{dB_z \over dy}~ {dB_z
      \over dz}
\end{equation}}


\subsection{Transverse elliptical distortion}

If the jet is not on the solenoid axis, the axial induced currents
interacting with the transverse fields will generate distorting forces
on the jet. These transverse forces per unit volume $dx~ dy~ dz$ are
\begin{equation}
f_x~=~ i_z~B_y~
\end{equation}
\begin{equation}
=~x~v~\kappa~B_y~ {dB_y \over dz}
\end{equation}
Ignoring surface tension, there is no restoring force, so the shape
will change. The sides will go in and the height increase. A
conservative estimate of the rate of width reduction can be obtained
assuming that the only inertial resistance comes from the elements
themselves. Then

\begin{equation}
 {d^2x \over dz^2}~<~{ f_x \over v~\rho } =~~{x~\kappa \over \rho}~B_y~ {dB_y \over dz}\end{equation}

In reality the inertial constraint will be greater, because the inward
motion of elements on the sides must generate equal outward motions at
the top and bottom, and azymuthal motions between the two. Including
these effects we can guess {\rr
\begin{equation} {d^2y \over dz^2}~\approx~{1\over3}~{x~\kappa \over \rho}~B_y~ {dB_y \over dz}\end{equation}
} If this velocity is sufficiently small, then surface tension will
restrain it. The total inward force on each side is
\begin{eqnarray}
{dF_B \over dz}&\approx &{1\over3}v\kappa B_y{dB_y \over dz} \int_{-\pi}^\pi \int_o^r  r \cos{\phi} dr r d\phi \nonumber \cr 
%\end{equation}
%\begin{equation}
&\approx &{1\over3}{2\over3}r^3\kappa B_y {dB_y \over dz}
\end{eqnarray}
and this is opposed by the surface tension difference for a finite
ellipticity $\epsilon=y/x$
\begin{equation}
{dF_T \over dz}~\approx~2~T~\epsilon
\end{equation}
where $T$ is the surface tension.

So the equilibrium ellipticity is {\rr \begin{equation}
    \epsilon_{equilib}~\approx~{1\over3}~{r^3\over 3~T}~\kappa~B_y~
    {dB_y \over dz}
\end{equation}}

Again, this is only an order of magnitude calculation since the factor
1/3 is only an estimate for the inertial effects in distorting the
ellipse. With the factor removed, however, the equation gives a harder
maximum value.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bb Gaussian axial example}

As an example we can consider a field that varies as a Gaussian in $z$
The fields of the solenoid and coordinates are denoted with primes $({}^{\prime})$, ie.
\begin{eqnarray}
B_z'(r',z') &\approx &B_o e^{-{z'^2\over 2\sigma_z^2}}-{1\over 4}r'^2{\partial^2B_z'(0,z')\over \partial z'^2}\\
B'_r(r',z') &\approx &-{1\over 2}r' {\partial B_z'(0,z')\over \partial z'} 
\end{eqnarray}
In the coordinate system of the jet, see Fig.~\ref{fig1}, assuming a very small angle $\theta$ then $x'=x$, $y'=y+z\theta$, $r'^2=r^2+2zy\theta$ and $z'=z-y\theta$ and the fields are
\begin{eqnarray}
B_z(x,y,z) &\approx &\left[B_z'(r',z')-{1\over 2}\theta y{\partial B_z'(0,z')\over \partial z'}\right]\Bigg|_{r'=r+{y\theta z\over r}, s'=s-y\theta} \nonumber \\
B_x(x,y,z) &\approx &-{1\over 2}x{\partial B_z'(0,z')\over \partial z'}\Bigg|_{s'=s-y\theta} \nonumber \\
B_y(x,y,z) &\approx & -\left[{1\over 2} (y+z\theta){\partial B_z'(0,z')\over \partial z'}-\theta B_z'(r',z')\right]\Bigg|_{r'=r+{y\theta z\over r}, s'=s-y\theta}\nonumber \\
{\partial B_z(x,y,z)\over \partial z}&\approx &\left[{\partial B_z'(0,z')\over \partial z'} -{1\over 2}y\theta {\partial^2 B_z'(0,z')\over \partial z'^2} -{1\over 4} (r^2+2zy\theta){\partial^3B_z'(0,z') \over \partial z'^3}\right]\Bigg|_{s'=s-y\theta} \nonumber \\
{dB_z(x,y,z)\over dy}&\approx &-\left[\theta {\partial B_z'(0,z')\over \partial z'}+{1\over 2}(z\theta+y){\partial^2 B_z'(0,z')\over \partial z'^2}-{1\over 4}r^2\theta {\partial^3B_z'(0,z') \over \partial z'^3}   \right]\Bigg|_{s'=s-y\theta} \nonumber  \\
{dB_y(x,y,z)\over dz}&\approx & -\left[{3\over 2}\theta{\partial B_z'(0,z')\over \partial z'}+{1\over 2}(y+z\theta){\partial^2 B_z'(0,z')\over \partial z'^2}\right]\Bigg|_{s'=s-y\theta} 
\end{eqnarray}
Therefore we obtain,
\begin{eqnarray}
f_r&\approx &-{\kappa v \over 2 \sigma_z^2}rzB_o^2 e^{-{z^2\over \sigma_z^2}}\nonumber \cr
f_z&\approx &{\kappa v \over 4 \sigma_z^4}r^2z^2B_o^2  e^{-z^2/\sigma_z^2}\nonumber \cr
p_r&\approx & { \kappa v \over 4 \sigma_z^2}( r_o^2-r^2 )B_o^2z e^{-z^2/\sigma_z^2}\nonumber \cr
f_p &\approx &{\kappa v \over 4 \sigma_z^2}\( r_o^2-r^2 \)\(1-{2z^2 \over \sigma_z^2}\)B_o^2e^{-z^2/\sigma_z^2}
\end{eqnarray}

Defining $R=r/r_o$ and $Z=z/\sigma_z$, the change of velocities
starting from points outside the field are
\begin{equation}
\Delta v(R,Z) \approx {\kappa r_o^2 \over 4  \rho \sigma_z}B_o^2
\left[  
~-~R^2\int_{-\infty}^Z {Z^2e^{-Z^2}dZ} + (1-R^2)\(\int_{-\infty}^Z{e^{-Z^2}dZ} 
- ~2~\int_{-\infty}^Z {Z^2e^{-Z^2}dZ}     \)  
\right]      
\end{equation}
where $\rho $ is the density and the average velocity change is
\begin{equation}
<\Delta v>(Z)~ \approx~ {\kappa~ r_o^2 \over 8~ \rho~\sigma_z}~B_o^2~\(
 \int_{-\infty}^Z{e^{-Z^2}dZ} 
- \int_{-\infty}^Z {Z^2e^{-Z^2}dZ}      \)
\end{equation}

%\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Plots}

The following plots show calculated values of some of the functions.

\vskip.5in

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%\newpage
Plots of the magnetic fields and derivatives needed in this calculations including the relative tilded of $\theta=0.1$ between  jet and solenoid axis are shown in Figs.\ref{fig2} and \ref{fig3}
\begin{figure}
%  \centering \includegraphics[width=4in,angle=90]{fields_vs_s.ps}
\label{fig2}
\caption{$B_z$ and $B_y$ vs z for a gaussian target magnet at $y=-4,-2,0,2,4$ cm }
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4in,angle=90]{deriv_vs_s.ps}
\label{fig3}
\caption{${dB_z\over dz}~,{dB_y\over dz}\textrm{and} {dB_z\over dy}$ vs z for a gaussian target magnet at $y=-4,-2,0,2,4$ cm}
\end{figure} 
Using a more realistic hibrid target magnet as suggested by 
B.~Weggel\cite{ref5} yields fields and relevant derivatives shown in Figs.\ref{fig4},\ref{fig7}
\begin{figure}
% \centering \includegraphics[width=4.5in,angle=0]{steve_fig1.eps}
%  \centering \includegraphics{steve_fig1.eps}
\label{fig4}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig2.eps}
\label{fig5}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig3.eps}
\label{fig6}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig4.eps}
\label{fig7}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig5.eps}
\label{fig8}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig6.eps}
\label{fig9}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig7.eps}
\label{fig10}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
\begin{figure}
%  \centering \includegraphics[width=4.5in,angle=0]{steve_fig9.eps}
\label{fig11}
\caption{${dB_y\over dz}$ vs z for the hybrid 20~T target magnet}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Numerics}

Beyond the magnet center ($z=0$), the radial force is outward. The
pressure in the liquid is now negative and cavitation can occur,
causing the jet to break-up. If it does, however, the circulating
currents stop, reducing further dispersal.

The maximum reduction of surface velocity prior to reaching the center
$\Delta v (out)$ is seen to be given by
\begin{equation}
\Delta v~\approx~ 0.8~{\kappa~ r_o^2 \over 8~ \rho~\sigma_z}~B_o^2
\end{equation}
and at the end it is approximately twice this value.

For three possible liquids, we obtain

\begin{center}
\begin{tabular}{|l|cccc|}
\hline
material & conductivity & density & $\Delta v$ (o)& $\Delta v$ (tot)\\
      & ohm m & kg/m$^3$ & m/s & m/s \\
\hline
gallium & 3.8 $10^6$ & 6~$10^3$ & 2.15 & 4.3 \\ 
mercury & 4~$10^6$ & 13.5~$10^3$& .95 & 1.9\\
Pb(81) In(19) & 2.6~$10^6$ & 10.3~$10^3$ & .8 & 1.6 \\
\hline
\end{tabular}
\end{center}

Compared with a jet velocity of 20 m/sec, these are all relatively
low. Earlier optimizations had called for a jet radius of 1 cm, which
gave values of $\Delta v$ 4 times higher, and thus more significant.
%\section{References}
\begin{thebibliography}{}
\bibitem{ref1}\begin{verbatim}http://www.cap.bnl.gov/mumu/studyii/\end{verbatim}
\bibitem{ref2}K.~McDonald (Ed), \textsl{An R\&D Program for Targetry and Capture at a Muon-Collider Source}, Proposal to the BNL AGS Division, p.23
\bibitem{ref3}R.~Weggel, \textsl{Behavior of conducting solid or liquid jet moving in magnetic field: 1) Paraxial; 2) Transverse; 3) Oblique}, BNL report BNL-65611/CAP-220-Muon-98R
\bibitem{ref4}\textbf{Phys. Today}, Feb 2000, p.29
\bibitem{ref5}R.~Weggel, \textsl{ 20~T Hybrid magnet for Study II}
\bibitem{ref6} K.~McDonald, \textsl{Damping and Radial Pinching Effects},
\begin{verbatim}
http://www.hep.princeton.edu/~mcdonald/mumu/target/radialpinch.ps\end{verbatim}  
\end{thebibliography}
\end{document}



%%%%%%%%%%%%%%%%%%%%%%
Elliptical distortion can result from a second order variation of the $B_z$ field.
\vskip.2in

The hydrostatic pressure will restrain the average inward radial forces, and the net transverse force results in a deflection. Any remaining difference in the  inward forces at $\phi=0$ and $\phi=\pi$ will result in inward motion of one and  equal outward motion of the other. The driving force for these asymmetric motions, per unit of volume, is 

$$\Delta f_r~= ~ {f_r(\phi=0)~-~f_r(\phi=\pi) \over 2}    $$

$$\Delta f_r~=~  {r^3 \over 4 }~v \kappa~{d_2 B_z \over dy^2} {dB_z \over dz} $$

Ignoring surface tension, there is no restoring force, so the shape will change at a rate

$$ {dy \over dz}~=~-{dx \over dz}~=~{\Delta f_r \over v~\rho } $$

$$ =~{r^3 \over 4~\rho }~ \kappa~{d_2 B_z \over dy^2} {dB_z \over dz} $$



\begin{eqnarray}
B_z'(r',z') &\approx &B_o e^{-{z'^2\over 2\sigma_z^2}}-{1\over 4}r'^2{\partial^2B_z'(0,z')\over \partial z'^2}\\
B'_r(r',z') &\approx &-{1\over 2}r' {\partial B_z'(0,z')\over \partial z'} 
\end{eqnarray}
In the coordinate system of the jet, see Fig.~\ref{fig1}, assuming a very small angle $\theta$ then $x'=x$, $y'=y+z\theta$, $r'^2=r^2+2zy\theta$ and $s'=s-y\theta$ and the fields are
\begin{eqnarray}
B_z(x,y,z) &\approx &B_o e^{-{z^2\over 2\sigma_z^2} (1-{2y\theta \over z})}\left[1-{r^2\over 4\sigma_z^2}(1+{z^2\over \sigma_z^2}) +{\theta y z\over 2\sigma_z^2} \{1+{1\over \sigma_z^2}(r^2-z^2)\}\right] \nonumber \\
B_x(x,y,z) &\approx &B_o e^{-{z^2\over 2\sigma_z^2}(1-{2y\theta \over z})} {x(z-y\theta)\over 2 \sigma_z^2}\nonumber \\
B_y(x,y,z) &\approx &B_o e^{-{z^2\over 2\sigma_z^2}}\left[ {yz\over 2 \sigma_z^2} -\theta \{1 -{r^2\over 4\sigma_z^2}(1-{z^2\over \sigma_z^2}) +{(y^2-z^2)\over 2 \sigma_z^2}\}\right]\nonumber \\
{dB_z\over dz}(0,0,z)&\approx &-B_o {z\over \sigma_z^2} e^{-{z^2\over 2\sigma_z^2}}\nonumber \\
{dB_z\over dy}(0,0,z)&\approx & B_o {\theta z \over \sigma_z^2}({7\over 2}-{2z^2\over \sigma_z^2})e^{-{z^2\over 2\sigma_z^2}}\nonumber \cr
{dB_y\over dz}(0,0,z)&\approx & B_o {\theta z \over \sigma_z^2}\left[ 2-{z^2\over 2 \sigma_z^2}\right]e^{-{z^2\over 2\sigma_z^2}}
\end{eqnarray}





