\subsection{Solid or Liquid Target?}
The muons of a muon collider will arise from the decay of pions produced in
the interaction of some $1.5 \times 10^{15}$ protons/s on a primary target.
If these protons have 16-GeV energy, the beam power is 4~MW. The target will
be about 2~interaction lengths long and will absorb about 10\% of the beam
power, \ie, about 400~kW. The beam repetition rate is 15 Hz, so each beam
pulse deposits about 30~kJ of energy in the target.
\subsubsection{Cooling the Target}
The target is in the form of a cylinder, about 1 cm in radius, 30 cm in length,
with a volume of about 100~cm$^3$. If the target material has density
10 g/cm$^3$, then the target mass is about 1~kg. Thus the beam-energy
deposition in the target averages about 30~J/g. Taking the heat capacity
of a typical metal as 0.2 J/(g-$^\circ$K), the temperature rise of a
metallic target is about 150$^\circ$K per pulse. If the target were not
cooled, it would melt (if solid) or boil (if liquid) after a dozen or so
beam pulses.
However, the goal of collecting the maximal number of muons per proton is
not consistent with typical cooling schemes in which the target material is
in immediate contact with a large thermal reservoir.
Radiative cooling of the target appears to be unfeasible. The power radiated
by a surface of area $A$ and emissivity $\epsilon$ at temperature $T$ is
$\epsilon \sigma T^4 A$, where $\sigma = 5.67 \times 10^{-12}$
J/(cm$^2(^\circ$K$)^4$s) is the Stefan-Boltzmann constant.
The emissivities of typical metals are only about
0.2 (in contrast to carbon, with $\epsilon \approx 0.8$).
The surface area of the target cylinder is about 200~cm$^2$.
At 2000$^\circ$K, the radiated power would be
only 3.6~kW, well shy of the 400-kW beam load.
Thus, we are led to consider scenarios in which the target is in motion,
with any subunit of the target exposed to only one or a few beam pulses
before being transported away from the interaction region, to be cooled by
a remote thermal reservoir. For similar reasons, targets at multimegawatt
neutron spallation sources are expected to be based on flowing liquid metal,
usually mercury
\cite{Bauer95a,Bauer95b,Schruns96,Broome96,Gabriel97,NSNStarget,Haines97b}.
\subsubsection{Thermal Shock}
The simplest option would be use of a flowing liquid metal contained within
a (static) metal pipe \cite{Hassberger,Stefanski,C2365}.
However, when a large pulse of energy is deposited in a material in a time
that is short compared to the transit time of a sound wave ($\approx
3$~$\mu$s/cm), a pressure (stress) wave results
\cite{Tang,Skala95,Schruns96,Taleya,Taleyb,Taleyc,Bauer98}. Since the
proton pulse for a muon collider is expected to be only about 2~ns long,
stress waves will be excited in the target, and consequent damage to the target
and adjacent material is a concern.
This issue is often called ``thermal shock''. The pressure wave front
propagates outwards at roughly the speed of sound in the target material.
However, thermal shock is not necessarily associated with bulk transport of
matter at the speed of sound.
\paragraph{The Initial Pressure Wave.}
If the beam-induced stress
exceeded the tensile strength of the target material, the latter would fracture.
We present a simplified model to estimate the regime in which a pressure
wave will ``tear'' the target apart, whether solid or liquid (see also
refs.~\cite{Tang,Haines97a}). When an energy density $\Delta U$ (per gram)
is deposited quickly in the target, we first calculate the temperature
change $\Delta T$ that would occur, assuming no thermal diffusion. Then,
we calculate the strain,
$\Delta l/l$ corresponding to that $\Delta T$, and evaluate the stress $P$
corresponding to that strain. Tearing is likely to occurs if the
stress exceeds the tensile strength. In this model, tearing occurs
during the phase of the pressure wave when the material is under tension,
\ie, for negative pressure. In liquids, this phenomenon is called
cavitation \cite{Trevena,Young,Leighton}.
We suppose that the effective
tensile strength of a liquid metal is similar to that of the same metal in
solid form. For example,
the tensile strength of sea ice is reported as about 8~atm
\cite{Kovacs}, while that of water can exceed 200 atm in a static
measurement \cite{Briggs50} but was measured as 8~atm by an explosive
technique \cite{Wilson} that has much in common with the present concerns;
see also \cite{Trevena,Young}.
For most metals, the tensile strength (pressure) $P$ is
about 0.002 of the modulus of elasticity $E$ (Young's modulus). Thus,
\begin{equation}
\Delta U = C\Delta T = {C \over \alpha} {\Delta l \over l} =
{C \over \alpha} {P \over E} \approx 0.002 {C \over \alpha},
\label{eq456}
\end{equation}
where $C$ is the heat capacity, $\alpha$ is the coefficient of
thermal expansion, and the approximation holds on setting the
stress equal to the tensile strength. If, for example, the heat capacity is
$C = 0.3$ J/g-$^\circ$C, and $\alpha \approx 2 \times 10^{-5}/^\circ$C, then
we expect the target material will tear when $\Delta U \approx 30$ J/g.
This is very nearly the expected energy deposition in the muon-collider
target. It is, however, somewhat smaller than a semi-empirical value of
200~J/g for the threshold for thermal-shock damage \cite{TEV1}.
If the target tears, it is possible that the fragments do damage to the
surrounding vessel. If all the deposited energy were converted to kinetic
energy and the target fragmented completely, then the velocity of the
fragments would be $\sqrt{2\Delta U} \approx 8$~m/s for $\Delta U = 30$~J/g.
A model of spray velocities in explosions of liquids \cite{Wilson} leads to
similar values. Studies of an exploding mercury jet were performed safely
inside a plastic vessel whose wall was only 1/4$''$ thick
\cite{Criss95,Ansley1,Ansley2}.
Thus, it is encouraging that if a liquid target were dispersed into
droplets by the beam, the droplets would do little damage to the surrounding
vessel. However, this should be confirmed by experiment.
\paragraph{The Reflected Pressure Wave.}
Even if target parameters are chosen such that the initial pressure wave does
not tear the material, there is another concern.
When the pressure wave reaches the surface of the target, it will, in general,
be reflected. In the case of a cylindrical target, the reflected wave
converges on the axis of the target and will typically result in higher
peak pressures than exist in the initial outgoing wave. Hence, there is
a serious prospect for localized fracture or vaporization of the target
material close to the target axis, whether the cylindrical target is solid or
liquid. In the case of a liquid, this is likely to induce localized
damage to the target pipe, particularly on the upstream wall.
\begin{figure}[htp] % h = here, t = top, b = bottom, p = new page
\begin{center}
\includegraphics [ width=4.5in, clip] {isolde5.eps}
\parbox{5.5in} % change 5.5in to \hsize for full-width caption
{\caption[ Short caption for table of contents ]
{\label{isolde1} Full-scale model of an ISOLDE target consisting of a
2-cm-diameter, 20-cm-long tantalum cylinder containing molten lead.
}}
\end{center}
\end{figure}
\begin{figure}[htp] % h = here, t = top, b = bottom, p = new page
\begin{center}
\includegraphics [ width=4.5in, clip] {isolde2.eps}
\parbox{5.5in} % change 5.5in to \hsize for full-width caption
{\caption[ Short caption for table of contents ]
{\label{isolde2} Photograph of an ISOLDE liquid-lead target that ruptured
around the upstream window after a few beam pulses \cite{Lettry}.
}}
\end{center}
\end{figure}
\begin{figure}[htp] % h = here, t = top, b = bottom, p = new page
\begin{center}
\includegraphics [ width=5.5in, clip] {silvestrov3.eps}
\parbox{5.5in} % change 5.5in to \hsize for full-width caption
{\caption[ Short caption for table of contents ]
{\label{silvestrov3} Photograph of a mercury target contained in an aluminum
cylinder that ruptured after exposure to a beam that deposited 1~kJ/g per
pulse \cite{Silvestrov}.
}}
\end{center}
\end{figure}
The destructive effects of pressure waves and
cavitation on liquid-metal targets in pipes have been demonstrated at the
CERN ISOLDE facility \cite{Lettry,Lettry98}, as shown in
Figs.~\ref{isolde1}-\ref{isolde2}, and also at the Budker Institute
\cite{Silvestrov}, as shown in Fig.~\ref{silvestrov3}.
Practical targets based on liquid metals in pipes require sufficiently long
beam pulses and/or target geometries (\eg, planar) in which imploding
pressure waves do not occur. Therefore, liquids in pipes are not considered
further for a muon-collider target. Likewise, a moving solid cylindrical
target is very problematic at a muon collider.
\subsubsection{Muon-Collider Primary-Target Options}
The remaining options for the muon-collider target are a free, cylindrical
liquid-metal jet or a moving planar solid target, as have been sketched in
Figs.~\ref{capture} and \ref{bandsaw}. The baseline design to be
studied in the proposed R\&D program utilizes a free liquid-metal jet.
The moving band target presently is considered as the backup option.
\subsubsection{Additional Issues for a Solid Target}
As well as surviving the thermal-shock of each beam pulse, a solid target must
survive the long-term effects of radiation damage. These include
embrittlement, swelling and even fracture of the material due to displacement
damage and hydrogen/helium production \cite{Ullmaier95}. The immunity of a
liquid target to these effects is an argument in favor of that option.
The cross section for displacement of a target atom in a 16-30-GeV proton
beam is about $10^{-24}$~m$^2$. In an operational year of 10$^7$~s, a
beam of $1.5 \times 10^{15}$ proton/s whose rms radius is nominally 4 mm
results in about 300 displacements per atom (dpa) in a nonmoving target.
For a moving solid target, any given atom is exposed to a proton flux that is
less than the total by the factor
\begin{equation}
{L_{\rm int} N_{\rm overlap} \over L_{\rm total}},
\label{eq2.1}
\end{equation}
where $L_{\rm int}$ is the length of the interaction region
$(\approx 30$~cm), $L_{\rm total}$ is the total length of the target material
($\approx 30$~m, say), and $N_{\rm overlap}$ is the number of adjacent beam
pulses that irradiate a given region. That is,
\begin{equation}
N_{\rm overlap} = {L_{\rm int} f \over v},
\label{eq2.2}
\end{equation}
for a target moving with velocity $v$ along the beam whose pulse rate is
$f$~Hz. For example, with $v = 1$~m/s and $f = 15$~Hz, then $N_{\rm overlap}
\approx
5$, and each part of the moving target experiences only 5\% of the exposure of
a nonmoving target, namely about 15~dpa in a year of operation. This still
corresponds to significant radiation damage, suggesting that the target
band would need to be replaced several times a year.
The viability of a moving solid target needs further investigation.
There is little available evidence as to the mechanical properties of
targets after such exposures. A series of studies of
solid targets at CERN generally indicated severe mechanical damage to
intensely irradiated targets
\cite{Eaton,cern3103,cern3198,eirkt,Johnson85,OEFZS},
although the damage was due to a combination of single-pulse shock as well
as radiation. Nickel targets appear to survive well for energy depositions
up to a large fraction of the shock-damage threshold \cite{ODay}. Copper
targets appear to form excessive amounts of copper sulfate if irradiated
in contact with air \cite{Pendzickpc}.
\subsection{Beam-Induced Radioactivity}
The high flux of protons into the target will induce some radioactivity
no matter what the target consists of. A thorough study of this issue will
be needed before a final decision as to the target material is taken.
A first study will be made using a MARS simulation of the hadronic cascade
in the target.
For any target material with an atomic number greater than or equal to that
of mercury, radioactive isotopes of mercury will emerge in the vapor phase
\cite{Ravnpc}.
These are only the most prominent among other radioactive vapors that will
be produced. Hence, a high-$Z$ target at a muon-collider source must be
enclosed in a gas-tight system that
includes capture tanks for the vapors, including those exhausted from any
pumps. This proposal does not cover R\&D towards such a system, which would
be modeled after the target-containment system at the ISOLDE facility at
CERN.
Especially high levels of activity would arise for targets containing bismuth,
due to the sequence:
\begin{equation}
\mbox{Low-energy neutron} + \mbox{Bi}^{209}_{83} \to \mbox{Bi}^{210}_{83}
+ \gamma, \qquad \mbox{Bi}^{210}_{83} \to \mbox{Po}^{210}_{84} + \beta^-
\quad \mbox{(half life = 5 days)}.
\label{eq12}
\end{equation}
Po$^{210}$ has a half life of 135 days and decays primarily via a 5.3-MeV
$\alpha$, but has a 0.1\% branch to an 803-keV x-ray.
If, say, every beam proton results in one transmutted Bi atom, then the
steady-state Po population would be equal to the total flux of protons in
135 days: $135 \times 10^5 \times 10^{15} \approx 10^{22}$ atoms, assuming a
proton flux of $10^{15}$/s. The number of Po decays would be
$10^{15}$/s in the steady state, \ie, about 30,000 curies!
The $\alpha$-particles will be almost entirely absorbed in the target, but
the 800-keV x-rays present more of a problem. The steady-state strength of the
x-rays corresponds to about 30 curies (assuming each beam proton results in one
Po atom).
However, the resulting polonium would be dissolved in the liquid of the target
with very low probability of being in vapor molecules. Hence, the handling of
an activated bismuth alloy would be relatively straightforward and little more
difficult than for other target materials \cite{Bauerpc}.
%Before committing to a target containing Bi the activation of Bi by a
%high-energy proton beam should be studied further, both experimentally and
%theoretically.
%A liquid target based on an In-Pb-Sn alloy (Fig.~\ref{fig13}) might be more
%satisfactory from the point of view of induced radioactivity.
\subsection{Candidate Liquid Metals}
Liquid metals have long been used as coolants for nuclear
reactors, and a useful summary of candidate materials has been given by
Lyon \cite{Lyon54}. Table~\ref{liquids} presents a brief survey of
low-melting-temperature metals, and Table~\ref{elements} list some
relevant physical properties of the elements of candidate materials.
\begin{table}[htbp] % h = here, t = top, b = bottom, p = on a new page
\begin{center}
\parbox{5.5in} % replace 5.5in by \hsize if want full-width caption
{\caption[ Short caption for the List of Tables. ]
{\label{liquids} Survey of low-melting-temperature metals \cite{Lyon54}.
}}
\vskip6pt
\includegraphics [ width=4.5in, clip] {lyon.eps}
\end{center}
\end{table}
\begin{table}[htbp] % h = here, t = top, b = bottom, p = on a new page
\begin{center}
\parbox{5.5in} % replace 5.5in by \hsize if want full-width caption
{\caption[ Short caption for the List of Tables. ]
{\label{elements} Properties of some candidate elements for primary targets.
}}
\vskip6pt
\begin{tabular}{lccccccccc}
\hline\hline
Element & $Z$ & Density & Melting & Boiling & Heat & Heat of & Thermal &
Resist. & Thermal \\
& & & Temp. & Temp. & Cap. & Vapor. & Cond. & ($\mu\Omega$-cm) & Exp.\\
& & (g/cm$^3$) & ($^\circ$C) & ($^\circ$C) & (J/g-$^\circ$C) & (J/g) &
(W/cm-$^\circ$C) & & ($10^{-5}/^\circ$C) \\
\hline
Copper & 29 & 8.96 & 1087 & 2567 & 0.39 & 4796 & 4.01 & 1.7 & 1.7 \\
Zinc & 30 & 7.1 & 420 & 906 & 0.39 & 1733 & 1.16 & 6.0 & 3.1 \\
Gallium & 31 & 5.9 & 30 & 2204 & 0.33 & 3712 & 0.4 & 26$^\dag$ & 12 \\
%Cadmium & 48 & 8.6 & 321 & 767 & 0.23 & 886 & 0.97 & 34 & 3.0 \\
Indium & 49 & 7.3 & 156 & 2073 & 0.23 & 2016 & 0.82 & 10 & 3.2 \\
Tin & 50 & 7.3 & 232 & 2270 & 0.18 & 2487 & 0.67 & 13 & 2.2 \\
Mercury & 80 & 13.6 & $-39$ & 357 & 0.14 & 295 & 0.087 & 94$^\dag$ & 6.1 \\
Lead & 82 & 11.35 & 327 & 1750 & 0.16 & 858 & 0.35 & 80$^\dag$ & 2.9 \\
Bismuth & 83 & 9.7 & 271 & 1610 & 0.12 & 857 & 0.079 & 120 & 1.3\\
\hline\hline
$^\dag$ liquid
\end{tabular}
\end{center}
\end{table}
We are interested in high-$Z$, high-density target materials for the
most-efficient pion production. High electrical resistivity is helpful in
reducing the eddy-current problem of a jet entering a magnetic field (sec.~2.5).
In many ways mercury is an excellent
candidate, but its vapor is toxic. Also, it may be preferable to use
a material that is solid at room temperature, to simplify cleanup of spills.
Lead alloys (solder-like materials) are good candidates. Rather low
melting temperatures are obtained by alloying with indium, which, however,
wets most solid metals so much as to make it difficult to produce liquid
jets. Lead-bismuth alloys have the drawback of relatively high production of
polonium when activated by the proton beam. Gallium alloys are nontoxic and
can be liquid at room temperature, so the resulting ease of handling
indicates their use in the initial stages of the R\&D program. The relatively
low atomic number, low density and low resistivity of gallium is somewhat
undesirable for use in the final target.
\subsubsection{Mercury}
Studies of materials issues for use of mercury as a proton target include
refs.~\cite{Schruns96,LANLworkshop,DiStefano95,Filges95}. The tensile
strength of liquid mercury has been studied by Briggs \cite{Briggs53}.
\subsubsection{Lead Alloys}
A lead alloy of particular interest is
eutectic lead-bismuth, 45\% Pb by weight, with a melting point of 126$^\circ$C
($255^\circ$F).
Other interesting low-melting allows of lead and/or bismuth are made by adding
cadmium, indium or tin.
Some quaternary and quinternary alloys have extremely low melting temperature,
such as
alloys 117 and 136 (designated by their melting temperatures in $^\circ$F).
Table~\ref{Belmont} gives a summary of physical properties of several
commercial lead alloys.
\begin{table}[htbp] % h = here, t = top, b = bottom, p = on a new page
\begin{center}
\parbox{5.5in} % replace 5.5in by \hsize if want full-width caption
{\caption[ Short caption for the List of Tables. ]
{\label{Belmont} Lead Alloy Specifications from Belmont Metals.
See also \hfill\break
{\tt http://www.indium.com/fusiblealloys.html}
}}
\vskip6pt
\includegraphics[width=5.5in]{belmont150.eps}
\end{center}
\end{table}
We have verified that the resistivities of liquid-lead alloys are very
similar to those of the corresponding solid, as listed in Table~\ref{Belmont}
\cite{mumu97-3}.
An extensive literature exists on materials-handling issues for applications
of lead alloys at nuclear reactors \cite{Brasunas54,Cathcart56,Romano63,%
Tolsen66,Weeks71,DeStefano76}.
\subsubsection{Gallium and Gallium Alloys}
Some information on liquid gallium alloys can be found at\hfill\break
{\tt http://www.indium.com/liquidalloys.html}
The eutectic Ga-Sn alloy is particularly convenient, as it is a liquid at
room temperature.
It is easy to prepare by dissolving tin wire (but not powder)
in liquid gallium at 40-50$^\circ$C.
The binary phase diagram for Ga-Sn is shown in Fig.~\ref{gasn}.
\begin{figure}[htpb] % h = here, t = top, b = bottom, p = new page
\begin{center}
\includegraphics[width=4in]{GaSn.eps}
\parbox{5.5in} % change 5.5in to \hsize for full-width caption
{\caption[ Short caption for table of contents ]
{\label{gasn} Ga-Sn phase diagram. Minimum melting temperature = 20.5$^\circ$C
\cite{ASM}.
}}
\end{center}
\end{figure}
Liquid gallium has a viscosity of 1.6 cp/g-cm$^3$, only slightly highly than
that of water. The temperature dependence of the resistivity of gallium
has been studied in detail \cite{Yahia72}.
Liquid gallium has been used as a coolant for silicon crystals at
synchrotron light sources \cite{Smither1,Smither2}.
\subsubsection{Exotic Target Materials}
The interaction of moving metals with the 20-T capture magnet (sec.~2.5)
could be avoided by use of insulating liquids such as molten
PtO$_2$ or Re$_2$O$_3$, or by granular materials such as slurries
({\it e.g.}, Pt in water) or powders \cite{mumu98-10}.
The problem of beam-induced thermal shock (sec.~2.1.2) might be minimized by
use of materials with very low (or negative) thermal expansion
coefficients \cite{Carpenter96}.
For example, Invar
alloys have been studied for many years, and their properties are
well known. Other materials are less well understood. For example,
oxides such as ZrW$_{2}$O$_{8}$ produce a net contraction over a range of
1-1200$^\circ$K. Bismuth ice slurries have the property, like water
ice, of contracting when melting. Aqueous salt suspensions such as
iron formate and sodium chloride contract with heating due to a
molecular electrostrictive effect. Molecular liquids such as molten SiO$_{2}$
exhibit a negative thermal expansion in the region of 1700$^\circ$C due to
a molecular-bond modification.
Bubbly liquids may also have a higher
compressibility, which would make them useful \cite{Schruns96}.
%The study of these and other more exotic options can proceed in parallel
%with the rest of the program, primarily involving experienced material
%scientists who may be able to suggest other options. This effort is ongoing
%at Argonne.
%xxx
\subsection{Jet Velocity}
To be useful as a target, a pulsed liquid jet must have a minimum velocity, so
that each pulse is sufficiently distinct from the others, and so that the
jet is reasonably straight. The minimum velocity for a conductive jet to
enter a strong magnetic field is considered in sec.~2.5.
\subsubsection{Effect of 15-Hz Repetition Rate}
We consider the use of a pulsed jet, leading to a series of cylinders of
liquid, each of length $l$ and radius $r$, and moving with velocity $v$. The
frequency $f$ of the pulse is nominally 15~Hz.
If the material from one pulse is not to overlap that of the next, then
the jet velocity must obey
\begin{equation}
v > fl.
\label{eq401}
\end{equation}
For example, if the length of each pulse is be about two nuclear interaction
lengths (about 30~cm for a dense, high-$Z$ material), then with $f = 15$~Hz,
we need $v > 4.5$~m/s.
The jet velocity will have to be several times this to create gaps between
adjacent pulses, so that the proton beam interacts with only a single jet pulse.
\subsubsection{Effect of Gravity}
The trajectory of the jet will be a parabolic arc, due to the acceleration of
gravity. If the jet velocity were too low, the curvature of the jet would be
large, and the proton beam would not be able to intersect the jet pulse over
its whole length.
The ends of the jet are displaced
downward from the ideal straight trajectory by amount
\begin{equation}
\Delta y = {gt^2 \over 2} = {g l^2 \over 8 v^2},
\label{eq402}
\end{equation}
noting that the time for the center of the jet to reach its end is $t = l/2v$.
%Hence, to have offset $\Delta y$ between center and ends to the jet, we need
%\begin{equation}
%v = \sqrt{gl^2 \over 8 \Delta y}.
%\label{eq403}
%\end{equation}
For example, with $\Delta y = 1/8$~cm, as might be desired
for a jet of radius 1~cm, $l = 30$~cm, we find $v \approx 1000$~cm/s = 10~m/s.
Thus, the effects of pulse frequency and of gravitational curvature
both require the jet velocity to be at least 10~m/s.
%aaa
\subsubsection{Jet Velocity \vs\ Pressure}
Suppose the liquid metal is stored in a tank of area $A$ perpendicular to the
flow, and the pressure is $P$ above ambient.
A valve lets a jet of liquid escape through an aperture of area $a \ll A$.
Then Bernoulli's equation tells us that the flow velocity $v$ out the
aperture obeys
\begin{equation}
{1 \over 2} \rho v^2 = P + {1 \over 2} \rho V^2,
\label{eq9a}
\end{equation}
where $\rho$ is the density of the liquid and $V$ is the velocity of the
liquid surface of area $A$ in the tank.
The equation of continuity for an incompressible liquid tells us that
$av = AV$, so that
\begin{equation}
v = \sqrt{2P \over \rho (1 - (a/A)^2)} \approx \sqrt{2P \over \rho}.
\label{eq9b}
\end{equation}
For example, with a Bi-Pb alloy of density $\rho \approx 10$~g/cm$^3$,
\begin{equation}
v[\mbox{m/s}] \approx \sqrt{20P[\mbox{atm.}]}.
\label{eq9c}
\end{equation}
Thus, to reach $v = 4.5$~m/s would require 1~atm overpressure in the
storage tank; 20~atm would be required to reach a velocity of 20~m/s.
\subsection{The Interaction of a Liquid-Metal Jet with a Magnetic Field}
As a jet of liquid metal enters the magnetic field that surrounds the beam
interaction region, it will be repelled according to Lenz' law.
The effect is due to the
Lorentz force on the eddy currents induced in the moving metal.
In an extreme case the jet would not reach the center of the interaction
region.
Here we present simplified analytic estimates of the effects of eddy currents,
and reach the tentative conclusion that they would visibly alter the
trajectory of the liquid-metal jet, but would not prevent the jet from
functioning as a target. However, there appears to be little safety margin,
suggesting the need for laboratory experiments to confirm that the proposed
liquid-metal jet is viable.
Additional details of the analytic arguments can be found in
refs.~\cite{mumu97-3,Weggela,Walker79}.
It is useful to establish numerical values for some relevant parameters of
our system as a qualitative guide to its magnetohydrodynamic behavior.
First, we note that the problem of a moving conductor in a static magnetic
field is equivalent to a moving field that encounters a conductor initially
at rest. Thus, the electric field ${\bf E}'$ in the frame of the conductor is
related to
the electric and magnetic fields {\bf E} and {\bf B} in the lab frame by
\begin{equation}
{\bf E}' = {\bf E} + {\bf v} \times {\bf B},
\label{eq201}
\end{equation}
where {\bf v} is the laboratory velocity of the conductor $(v \ll c)$,
and we use MKSA units.
Next, we recall that the penetration of a time-dependent magnetic
field into a conductor is governed by a diffusion equation. Assuming $v \ll c$
and reasonably good conductivity $\sigma$, we may
neglect the displacement current, and the basic electromagnetic
equations are
\begin{equation}
\nabla \times {\bf E} = - {\partial {\bf B} \over \partial t}, \qquad
\nabla \times {\bf B} = \mu_0 {\bf j}, \qquad \mbox{and} \qquad
{\bf j} = \sigma {\bf E}' = \sigma ({\bf E} + {\bf v} \times {\bf B}),
\label{eq202}
\end{equation}
where {\bf j} is the current density.
On eliminating {\bf j} and {\bf E} we find that
\begin{equation}
{\partial {\bf B} \over \partial t} = {\nabla^2 {\bf B} \over \mu_0 \sigma}
+ \nabla \times ({\bf v} \times {\bf B}).
\label{eq203}
\end{equation}
With the neglect of the second term (justified for low velocity), we find
the desired diffusion equation. Thus, the characteristic time for diffusion
of the magnetic field into a long conducting cylinder of radius $r$ is
\begin{equation}
\tau = \mu_0 \sigma r^2.
\label{eq204}
\end{equation}
The low-melting temperature alloys in Table~\ref{Belmont}
all have relatively low
conductivity. In particular, alloy 255 has conductivity only 2\% that of
copper (resistivity = 1.67 $\mu\Omega$-cm), \ie, about $10^6$ MKSA units.
Hence, for a cylinder of radius 1 cm the diffusion time is
\begin{equation}
\tau \approx 4 \pi \times 10^{-7} \cdot 10^6 (10^{-2})^2 \approx 10^{-4}\
\mbox{sec}.
\label{eq205}
\end{equation}
Another characteristic time in our problem is that over which the external
magnetic field varies appreciably. For a jet of velocity $v$ that enters a
solenoid of diameter $D$, this time is $D/v$. The ratio of the diffusion
time to time $D/v$ is called the magnetic Reynold's number:
\begin{equation}
{\cal R} = {\tau v \over D}.
\label{eq206}
\end{equation}
For ${\cal R} \ll 1$ the external magnetic field penetrates the conductor, but
for ${\cal R} \gg 1$ it does not.
Anticipating a jet velocity of order 10 m/s and a solenoid of diameter $D
\approx 0.3$ m, we have $D/v \approx 0.03$ s, and the magnetic Reynold's
number is ${\cal R} \approx 0.003$. We conclude that in our problem the
diffusion
is rapid enough that the external field penetrates the conductor. That is,
our candidate metals are not sufficiently ``good'' conductors to
exclude the magnetic field from their interior. This is fortunate,
as a ``good'' conductor could not enter a 20-T magnetic field unless its
initial velocity were very high.\footnote{To see this, consider a good
conductor moving along the $z$-axis of a solenoid field. Surface current $I =
B_z/\mu_0$
(per unit length) is induced so as to cancel the external solenoid field
$B_z$. This current interacts with the radial component of the external
field, $B_r \approx - (r/2)dB_z(0,z)/dz = -rB'_z/2$ to produce retarding force
$F = -2\pi r I B_r =
-2\pi r (B_z/\mu_0)(rB'_z/2) = \pi r^2 (B_z^2)'/2\mu_0$ per unit length.
But also, $F = ma = \pi r^2 \rho \dot v = \pi r^2 \rho vv' = \pi r^2 \rho
(v^2)'/2$, where $\rho \approx 10^4$ kg/m$^3$ is the mass density. This
integrates to give
$v^2(z) = v^2_{-\infty} - B_z^2/\mu_0\rho$. Thus, to enter a field of
$B_z = 20$T,
the initial velocity would need to be at least $B_z/\sqrt{\mu_0\rho} \approx
200$~m/s for our heavy metals.}
The magnetic Reynold's number can be thought of in another way. From the
point of view of the conductor, the external magnetic field is time
dependent with frequency content up to $\omega \approx v/D$. The skin
depth at this frequency is $\delta = \sqrt{2/\mu_0 \omega \sigma} =
\sqrt{2D/\mu_0 \sigma v}$. This is to be compared to the
radius $r$ of the conductor. Indeed,
\begin{equation}
{r^2 \over \delta^2} = {\mu_0 \sigma r^2 v\over 2 D} = {{\cal R} \over 2}.
\label{eq207}
\end{equation}
In our case, the low value of the magnetic Reynold's number indicates that
the conductor is much smaller than the relevant skin depth, and again we
expect the external field to penetrate the conductor.
We now give some approximate analyses of the forces on the liquid jet as it
enters a solenoid.
\subsubsection{Jet on Axis of a Solenoid}
We model the forces on a conducting jet in a magnetic field by considering only
a ring (or disc) perpendicular to the axis of the jet.
The ring has
radius $r$, radial extent $\Delta r$ and thickness $\Delta z$.
We first consider only motion along the axis of the ring, which
we call the $z$ axis, and which is also the axis of a solenoid magnet with
field {\bf B}$(r,z)$.
Then the magnetic flux through the ring at position $z$ is
\begin{equation}
\Phi \approx \pi r^2 B_z(0,z),
\label{eq1}
\end{equation}
whose time rate of change is
\begin{equation}
\dot\Phi = \pi r^2 \dot B_z = \pi r^2 B'_z v,
\label{eq2}
\end{equation}
where $\dot{\phantom{a}}$ indicates differentiation with respect to time,
$'$ is differentiation with respect to $z$, $B_z$ stands for $B_z(0,z)$, and
$v$ is the velocity of the center of mass of the ring.
%The magnetic field is evaluated at the center of the ring.
If the metal has electrical conductivity $\sigma$, then its resistance to
currents around the ring is
\begin{equation}
R = {2\pi r \over \sigma \Delta r \Delta z},
\label{eq3}
\end{equation}
so the (absolute value of the) induced current is
\begin{equation}
I = {{\cal E} \over R} = {\dot\Phi \over R} = {\sigma r B'_z v \Delta r \Delta z
\over 2}.
\label{eq4}
\end{equation}
\paragraph{Radial Pinch.}
The Lorentz force on the ring due to the interaction of this current with the
axial field pinches the jet radially, while that due to the interaction with
the radial field opposes the motion. The radial pinch can be characterized
by a radial pressure gradient,
\begin{equation}
{\Delta P_r \over \Delta r} = {\Delta F_r \over \Delta r \Delta z \Delta l} =
-{B_z I \Delta l \over \Delta r \Delta z \Delta l} = - {\sigma r B_zB'_z v \over
2}.
\label{eq4a}
\end{equation}
As the jet enters the magnet from $z = -\infty$, the axial field gradient,
$B'_z$, is initially positive, and the radial forces are inward. However,
as the jet exits the solenoid, the gradient $B'_z$ becomes negative, and the
radial force is outwards. Even if the jet has not been destabilized by the
pinch on entering the magnetic, the radially outward forces experienced on
leaving the magnet may disperse the jet.
The pinch is greatest as the ring passes the edge of the solenoid, where
$B_z \approx B_0/2$ and $B'_z \approx B_0/D$ for a solenoid of diameter $D$
and peak axial field $B_0$. That is,
\begin{equation}
{\Delta P_{r,{\rm max}} \over \Delta r} \approx
- {\sigma \over 4} {r \over D} B_0^2 v.
\label{eq4b}
\end{equation}
Integrating this over radius, the pressure gradient between the axis and
radius $r$ is
\begin{equation}
\Delta P_{r,{\rm max}} \approx - {\sigma \over 8} {r^2 \over D} B_0^2 v.
\label{eq4c}
\end{equation}
In general, the pinch will cause the jet to
shrink radially and elongate axially.
Instabilities in this process may break the jet up into droplets.
However, once the jet begins to deform, additional eddy currents are
induced that will oppose the deformation. See sec.~2.5.2.
As a very crude model of the effect of the pinch on the radius of the
jet, we suppose that the jet surface accelerates inwards during the
characteristic time $D/v$ under a force approximated by the pinch pressure
$\Delta P_{r,{\rm max}}$ times the surface area. Then eq.~(\ref{eq4c})
leads to the estimate
\begin{equation}
\Delta r \approx - {\sigma \over 16} {r D \over \rho v} B_0^2
\label{eq4d}
\end{equation}
for the radial perturbation caused by the pinch.
\paragraph{Axial Retarding Force.}
The component of the Lorentz force that opposes the motion of the ring is
\begin{equation}
\Delta F_z = 2\pi r B_rI = - \pi\sigma r^2 B_rB'_z v \Delta r \Delta z
\approx - {\pi \sigma r^3 (B'_z)^2 v \Delta r \Delta z \over 2},
\label{eq5}
\end{equation}
using the approximate relation for the radial field
near the $z$-axis,
\begin{equation}
B_r(r,z) \approx -{r \over 2}{dB_z(0,z) \over dz} = -{r B'_z \over 2},
\label{eq3b}
\end{equation}
as can be deduced from the Maxwell equation $\nabla \cdot {\bf B} = 0$.
The equation of motion of a ring is then
\begin{equation}
dF_z = -{\pi\sigma r^3 (B'_z)^2 v_z \Delta r \Delta z \over 2} = m \dot v_z =
2\pi \rho r \Delta r \Delta z v'_z v_z,
\label{eq6}
\end{equation}
where $\rho$ is the mass density of the metal.
After dividing out the common factor of $\pi r \Delta r \Delta z v_z$ we find
\begin{equation}
v'_z(r) = -{\sigma r^2 (B'_z)^2 \over 4 \rho}.
\label{eq6a}
\end{equation}
Before considering a detailed model of the axial field profile, $B_z$, we note
that the peak gradient of the axial field of a solenoid of diameter $D$ is
$B_0/D$, and the gradient is significant over a region $\Delta z \approx D$.
Hence, we estimate that on entering a solenoid the jet velocity is reduced by
an increment
\begin{equation}
\Delta v_z(r) \approx {\sigma r^2 B_0^2 \over 4 \rho D}.
\label{eq6b}
\end{equation}
On leaving the solenoid, the jet velocity is reduced by a second increment
$\Delta v_z$. (Since the effect depends on $(B'_z)^2$, the force is
retarding on both entering and exiting, as predicted by Lenz' law.)
The jet velocity cannot actually go negative whatever the
magnetic field . If the velocity reaches zero, the jet stops (falls).
Note that we divided eq.~(\ref{eq6}) by $v_z$ before integrating;
once $v_z$ becomes zero, $F$ goes to zero and stays there.
The reduction of velocity (\ref{eq6b}) is zero on the axis of the jet, and grows
quadratically with radius. If the jet were a rigid body, $\Delta v_z$
would be one half the value given by eq.~(\ref{eq6b}) at the outer radius.
If the change in velocity is small compared to the initial velocity,
$v_{-\infty}$, we estimate the distance $\Delta z(r)$ by which the
material in the jet at radius $r$ is retarded compared to the material on
axis as
\begin{equation}
\Delta z(r) \approx \Delta v_z(r) \Delta t
\approx \Delta v_z(r) {D \over v_{-\infty}}
\approx {\sigma r^2 B_0^2 \over 4 \rho v_{-\infty}}.
\label{eq6c}
\end{equation}
We desire this to be small compared to the length of the jet. Indeed,
it will be awkward if $\Delta_z$ exceeds the radius of the jet.
We now consider a more specific model. See \cite{mumu97-3} for
discussion of a finite solenoid.
\paragraph{Semi-Infinite Solenoid.}
The field on the axis of a semi-infinite solenoid
is amenable to analytic calculation. Indeed, for a solenoid of radius
$D/2$ with windings from $z = 0$ to $+\infty$, the axial field is
\begin{equation}
B_z(0,z) = {B_0 \over 2} \left( 1 + {z \over \sqrt{(D/2)^2 + z^2}} \right),
\label{eq101}
\end{equation}
whose derivative is
\begin{equation}
B'_z = {dB_z(0,z) \over dz} = {B_0 \over 2} {(D/2)^2 \over
[(D/2)^2 + z^2]^{3/2} }.
\label{eq102}
\end{equation}
Using eq.~(\ref{eq102}) in eq.~(\ref{eq6a}) and integrating the equation of
motion from $-\infty$ to $z$, we find
\begin{equation}
v_z(r,z) = v_{-\infty} - {3 \sigma r^2 B_0^2 \over 64 \rho D}
\left( {\pi \over 2 } + \tan^{-1} w + {w \over 1 + w^2} + {2w \over 3
(1 + w^2)^2} \right),
\label{eq103}
\end{equation}
where $D$ is the diameter of the solenoid and $w = 2z/D$.
The semi-infinite solenoid is meant to approximate a finite solenoid of length
$L = \alpha D$. Since the semi-infinite coil begins at $z = 0$, the
center of the finite solenoid it approximates is at $z = \alpha D/2$, \ie,
at $w = \alpha$. For $\alpha \gsim 1$, as is reasonable for an actual
magnet, there is little difference between the result of eq.~(\ref{eq103})
at $w = \alpha$ and at $+\infty$, so we estimate the change in velocity as
\begin{equation}
\Delta v_z(r) \approx - {3\pi \sigma r^2 B_0^2 \over 64 \rho D}.
\label{eq103a}
\end{equation}
The retardation relative to the center of the jet is related by
\begin{equation}
\Delta\dot z(r) = \Delta v_z(r) = \Delta z' v_z \approx \Delta z' v_{-\infty},
\label{eq103b}
\end{equation}
where the approximation holds if $\Delta v_z \ll v_{-\infty}$. In this
approximation, we integrate eq.~(\ref{eq103}) to find
\begin{equation}
\Delta z(r) \approx - {3 \sigma r^2 B_0^2 w \over 128 \rho v_{-\infty}}
\left( {\pi \over 2 } + \tan^{-1} w - {1 \over 3w (1 + w^2)} \right).
\label{eq103c}
\end{equation}
This diverges for large $w$, but at $w = \alpha \approx 1$, corresponding
to the center of a real magnet, we have
\begin{equation}
\Delta z(r)
\approx - {3 \pi \sigma r^2 B_0^2 \alpha \over 128 \rho v_{-\infty}}.
\label{eq103d}
\end{equation}
\paragraph{Numerical Examples.}
We consider the lead-bismuth alloy 255, whose
conductivity is about $10^6$ MKSA units and whose density is about
10~gm/cm$^3$, \ie, $10^4$~kg/m$^3$.
Then, eq.~(\ref{eq103a}) leads to the requirement
\begin{equation}
v_{-\infty} > 60\ \mbox{m/s} \left[ {r \over \mbox{1 cm}} \right]
\left[ {r \over D} \right] \left[ {B_0 \over \mbox{20 T}} \right]^2.
\label{eq9}
\end{equation}
It is thought that the jet radius must be 0.5-1 cm to match the proton beam,
and that the inside diameter of the solenoid will be about 20 cm. In this
case we need $v_{-\infty} > 0.75$-3 m/s for $B_0 = 20$ T.
Again, if the jet is to exit the magnet, $v_{-\infty}$ must be twice the
minimum given in (\ref{eq9}).
In the approximation of eq.~(\ref{eq103d}), the shear in the jet profile between
its axis and radius $r$ is
\begin{equation}
{\Delta z(r) \over r} \approx
- 3 \alpha \left[ {r \over \mbox{1 cm}} \right]
\left[ {B_0 \over \mbox{20 T}} \right]^2
\left[ {10\ \mbox{m/s} \over v_{-\infty}} \right] .
\label{eq7a}
\end{equation}
For, say, $r$ = 1 cm, $v_{-\infty}$ = 10 m/s, $\alpha = 2$ and $B_0 = 20$ T,
we would have $\Delta z(r) \approx 6r$, which is a fairly severe distortion
of the jet.
Returning to the issue of the radial pinch, we can now cast eq.~(\ref{eq4c})
in the form
\begin{equation}
\Delta P_{r,{\rm max}} \approx 50\ \mbox{atm}
\left[ {r \over \mbox{1 cm}} \right]
\left[ {r \over D} \right] \left[ {B_0 \over \mbox{20 T}} \right]^2
\left[ {v_{-\infty} \over \mbox{10 m/s}} \right].
\label{eq8a}
\end{equation}
For, say, $r$ = 1 cm, $v_{-\infty}$ = 10 m/s, $D = 20$ cm and $B_0 = 20$ T,
the maximum radial pressure
is 2.5 atmospheres. This may be enough to perturb the
shape of the jet as it enters the magnet.
When the jet leaves the magnet, the radial pressure goes negative. This
pressure is small compared to the tensile strength of the jet material,
so the jet will not necessarily tear apart. However, the rapid change of
pressure from positive to negative may excite oscillations of the jet which
lead to breakup into macroscopic droplets. This would occur after the
proton beam interacted with the jet, so is more of a nuisance for the
liquid-collection system than a fundamental flaw.
The longitudinal effects, (\ref{eq9}) and (\ref{eq7a}), are suppressed
at higher jet velocities, which, however, enhance the radial pinch
(\ref{eq8a}).
\subsubsection{Magnetic Damping of Radial Perturbations}
If the liquid jet deforms, either due to the Lorentz forces on the eddy
currents or due to the beam-induced pressure wave that was discussed in
sec.~2.1.2, further eddy currents will arise, leading to further Lorentz
forces that damp the deformation.
To estimate this, we follow the argument of sec.~2.5.1 for a conducting ring
of radius $r$ perpendicular to a magnetic field $B_0$. Suppose the ring
is being deformed with a radial velocity $v_r$, either inwards or outwards.
Then the rate of change
of magnetic flux through the ring is $\dot\Phi = 2 \pi r v_r B_0$, and the
eddy current induced in a ring of cross section $\Delta r \Delta z$ is
$I = \sigma \Delta r \Delta z v_r B_0$, where $\sigma$ is the conductivity.
The Lorentz force on this current leads to a radial pressure gradient
of magnitude
\begin{equation}
{\Delta P_r \over \Delta r} = {\Delta F_r \over \Delta r \Delta z \Delta l} =
{B_0 I \Delta l \over \Delta r \Delta z \Delta l} = {\sigma v_r B_0^2}.
\label{eq4z}
\end{equation}
The total damping pressure at radius $r$ is therefore
\begin{equation}
\Delta P_{r,\rm damp} \approx \sigma r v_r B_0^2.
\label{eq4y}
\end{equation}
As a first example, consider inward radial motion caused by the pinch pressure
(\ref{eq4c}).
In the first approximation, the jet surface accelerates inwards during the
characteristic time $D/v_z$ under a force approximated by the pinch pressure
$\Delta P_{r,\rm max}$ times the surface area. Then eq.~(\ref{eq4c})
leads to the estimate for the radial velocity:
\begin{equation}
v_r \approx - {\sigma \over 4} {r \over \rho } B_0^2.
\label{eq4w}
\end{equation}
Combining this with eq.~(\ref{eq4y}), the damping pressure is
\begin{equation}
\Delta P_{r,\rm damp} \approx {\sigma^2 r^2 B_0^4 \over 4 \rho}.
\label{eq4x}
\end{equation}
For a liquid-metal jet of conductivity $10^6$ MKSA, radius 1~cm, and
inside a 20-T field, the damping pressure is about 4,000~atm. This greatly
exceeds the pinch pressure of 2.5~atm estimated above, and suggests that the
radial motion due to the pinch is highly damped.
As a second example, consider the same jet that is now being blown apart
with a radial velocity $v_r = 1,000$~m/s by the beam-induced shock.
Then the damping pressure (\ref{eq4y}) is 4~GPa, which equals the tensile
strength of steel, and so might hold the jet together.
The damping pressure in the second example is about 20 times the magnetic
pressure, $B_0^2/2\mu_0$.
Thus, it is encouraging that significant damping of all radial perturbations
will occur in a strong magnetic field.
\subsubsection{Jet at an Angle to the Axis of a Solenoid}
To improve the yield of pions in the interaction of the proton beam with the
liquid jet, it is desirable that the jet axis make a small angle $\theta \approx
0.1$ to the axis of the solenoid. In this case the motion of the jet includes
a component perpendicular to the magnetic field.
The induced eddy currents will flow in loops that are roughly perpendicular
to the magnetic field lines. As a simplification, we suppose that the current
loops are circles perpendicular to the axis of the solenoid. This approximation
should be reasonable for small $\theta$.
The (unperturbed) trajectory of the conductor of radius $a$
is taken to be along the line
\begin{equation}
x = z\theta,
\label{eq240}
\end{equation}
where $z$ increases with increasing time. To simplify the calculations, we
suppose the trajectory follows eq.~(\ref{eq240}), even though the velocity of
the jet is perturbed by the magnetic field (impulse approximation).
Following the same line of argument as in sec.~2.5.1 we again calculate
the drag force induced on the conducting disc.
The force element on a piece of the disk at radius $r = \epsilon a$ and
azimuth $\phi$ is found to be \cite{mumu97-3}
\begin{eqnarray}
& & d {\bf F} = I d{\bf l} \times {\bf B}
\label{eq251} \\
%= I\epsilon a d\phi \left[ \hat\rho B_z(r,z) + \hat {\bf z} {B'_z \over 2}
%( \epsilon a + z \theta \cos\phi) \right]. \\
& = & I\epsilon a d\phi \left[
( \hat {\bf x} \cos\phi + \hat {\bf y} \sin\phi )
\left( B_z - {(z\theta)^2 + (\epsilon a)^2 + 2\epsilon a z \theta \cos\phi
\over 4} B''_z \right) + \hat {\bf z} {B'_z \over 2}
(\epsilon a + z \theta \cos\phi) \right], \nonumber
\end{eqnarray}
where $\hat\rho$ is the unit vector pointing radially outwards from the
axis of the jet ($\rho = \epsilon a$).
The transverse force can be decomposed into a radial pinch (or expansion)
as discussed in sec.~2.5.1 plus
a drag in the $x$ direction. The longitudinal ($z$) force vanishes on the
axis of the jet, has a drag that is independent of azimuth, and
another component that varies with azimuth, causing a torque (or shear).
\paragraph{Drag Forces.}
We first ignore the radial pinch and the shear by
integrating eq.~(\ref{eq251}) over $\phi$ and using eq.~(\ref{eq4}) for $I$
to obtain
\begin{equation}
d{\bf F} = {\pi \sigma a^4 v_z B'_z \epsilon^3 \Delta\epsilon
\Delta z \over 2} \left(
\hat {\bf x} {z \theta B''_z \over 2} - \hat {\bf z} B'_z \right).
\label{eq252}
\end{equation}
The $z$ component of the drag force is the same as found previously in
eq.~(\ref{eq5}).
The retarding force vanishes on the jet axis, and increases as the cube of
the radius within the jet. As a result, the core of the jet will move
ahead of the outer regions.
%ppp
In turn, we integrate (\ref{eq252}) over $\epsilon$ to obtain the total force
on a disc of thickness $\Delta z$:
\begin{equation}
{\bf F} = {\pi \sigma a^4 v_z B'_z \Delta z \over 8} \left(
\hat {\bf x} {z \theta B''_z \over 2} - \hat {\bf z} B'_z \right).
\label{eq253}
\end{equation}
%The $z$ component of eq.~(\ref{eq253}) was previously deduced as
%eq.~(\ref{eq6}).
In the equation of motion, we again replace differentiation by time with
that by z:
\begin{equation}
{\bf F} = m \dot{\bf v} = \pi a^2 \Delta z \rho v_z {\bf v}'.
\label{eq254}
\end{equation}
The components of the equation of motion of the conducting jet are thus,
\begin{equation}
v'_x = { \sigma a^2 z \theta B'_z B''_z \over 16 \rho},
\qquad \mbox{and} \qquad
v'_z = - { \sigma a^2 (B'_z)^2 \over 8 \rho }.
\label{eq255}
\end{equation}
We use the example of a semi-infinite solenoid to illustrate the
effect of the eddy currents on the jet velocity, because the needed field
derivatives have simple analytic forms. The form of the trajectory,
(\ref{eq240}), assumes that the center of the magnet is at the origin.
Suppose the length of the physical magnet is $\alpha$ times its diameter $D$,
so that the coil extends over $-\alpha D/2 \leq z \leq \alpha D/2$.
Then the field of the physical magnet can be represented by the field of
a semi-infinite solenoid beginning at $z = -\alpha D/2$.
From the derivatives of eq.~(\ref{eq101}) we see that $v'_z$
is always negative, but that $v'_x$ is negative only until the jet enters the
magnet $(z = -\alpha D/2)$. Integrating (\ref{eq255})
from $-\infty$ to $z$, we find that
the velocity components of the jet are
\begin{equation}
v_x = v_{x,-\infty} - {3\sigma a^2 B_0^2 \theta \over 1024 \rho D}
\left[ {\pi \over 2} + \tan^{-1} w + {w \over 1 + w^2}
+ {2 w \over 3 (1 + w^2)^2}
- {16z \over 3 D (1 + w^2)^3} \right],
\label{eq259}
\end{equation}
and
\begin{equation}
v_z = v_{z,-\infty} - {3\sigma a^2 B_0^2 \over 128 \rho D}
\left[ {\pi \over 2} + \tan^{-1} w + {w \over 1 + w^2}
+ {2 w \over 3 (1 + w^2)^2}
\right],
\label{eq260}
\end{equation}
where $w = 2z/D +\alpha$.
Of course, $v_{x,-\infty} = \theta v_{z,-\infty}$ by assumption. The
velocity components of the jet when it reaches the center of the magnet
($z = 0,\ w = \alpha)$ are
\begin{equation}
v_{x,0} = \theta v_{z,-\infty} - {3\sigma a^2 B_0^2 \theta \over 1024 \rho D}
\left[ {\pi \over 2} + \tan^{-1}\alpha + {\alpha \over 1 + \alpha^2} +
{2 \alpha \over 3 (1 + \alpha^2)^2} \right],
\label{eq261}
\end{equation}
and
\begin{equation}
v_{z,0} = v_{z,-\infty} - {3\sigma a^2 B_0^2 \over 128 \rho D}
\left[ {\pi \over 2} + \tan^{-1}\alpha + {\alpha \over 1 + \alpha^2} +
{2 \alpha \over 3 (1 + \alpha^2)^2} \right],
\label{eq262}
\end{equation}
Thus, while
both $v_x$ and $v_z$ are reduced on entering the solenoid, the relative
reduction in the $x$ velocity is only 1/8 that of the $z$ velocity.
As a consequence, the angle $\theta$ of the trajectory to the axis of the solenoid
actually increases as the jet enters the magnet.
For example, suppose that $v_{z,-\infty}$ is 3 times the loss of velocity
on entering the magnet. Then
\begin{equation}
v_{z,0} = {2 \over 3} v_{z,-\infty}, \qquad
v_{x,0} = {23 \over 24} \theta v_{z,-\infty},
\label{eq263}
\end{equation}
and the angle of the trajectory at the center of the magnet is
\begin{equation}
\theta_0 = {v_{x,0} \over v_{z,0}} = {69 \over 48} \theta = 1.44 \theta.
\label{eq264}
\end{equation}
Figure \ref{jet} illustrates the variation of $v_x$, $v_z$ and $\theta$ of
the jet as a function of $z$.
\begin{figure}[htp] % h = here, t = top, b = bottom, p = new page
\begin{center}
\includegraphics[width=4in]{jet.eps}
\parbox{5.5in} % change 5.5in to \hsize for full-width caption
{\caption[ Short caption for table of contents ]
{\label{jet} $v_x$, $v_z$ and $\theta$ of the jet as a function of $z$ as
it enters a solenoid of aspect ratio $\alpha = 2$, according to
eqs.~(\ref{eq259}-\ref{eq260}). The initial $z$ velocity
of the jet is taken to be 3 times the loss of velocity on entering the solenoid.
}}
\end{center}
\end{figure}
We are greatly encouraged by these idealized calculations that the effect of
eddy currents on the transverse velocity of the jet will not be too severe.
The above analysis is for the drag force on the jet as a whole.
Recall that the force varies with radius within the jet, and so leads
to longitudinal distortions as discussed in sec.~2.5.1. The variation
of the drag force in $x$ leads to additional torques and shears, which we
now discuss.
\paragraph{Torque and Shear.}
The magnetic forces on the eddy currents also produce a torque
that will twist the jet about the axis perpendicular to the plane of the
jet motion, the $y$ axis in our example.
The torque $d{\bf N}$ on a small element of a current ring about its center
can be calculated from eq.~(\ref{eq251}).
% as
%\begin{eqnarray}
%d{\bf N} & = & a \epsilon (\hat{\bf x} \cos\phi + \hat{\bf y} \sin\phi) \times
%(Id{\bf l} \times {\bf B})
%\nonumber \\
%& = & {a^2 \epsilon^2 I B'_z \over 2} d\phi
%(\hat{\bf x} \sin\phi - \hat{\bf y} \cos\phi)
%( \epsilon a + z\theta \cos\phi).
%\label{eq271}
%\end{eqnarray}
On integrating over $\phi$ and $\epsilon$
we find that $N_x = 0$, and
the total torque on a disc of radius $a$ and thickness $\Delta z$ is
\begin{equation}
N_y = {\pi \over 16} z \theta \sigma a^4 v_z B^{'2}_z \Delta z
\approx - {x F_z \over 2}.
\label{eq273}
\end{equation}
The sense of rotation is opposite to that of the deflection of the jet
trajectory as it enters the magnetic field.
The moment of inertia of the disc about a diameter is $ma^2/4 = \pi \rho a^4
\Delta z/4$. Hence the angular acceleration of the azimuthal angle $\varphi$
of the disc about the $y$ axis is
\begin{equation}
\ddot \varphi = v_z {d \dot\varphi \over dz} = {\theta \sigma z B^{'2}_z v_z
\over 4 \rho}.
\label{eq274}
\end{equation}
Note that this is independent of the radius of the jet.
Using $B'_z$ from eq.~(\ref{eq101}), we have
\begin{equation}
{d \dot\varphi \over dz} = {\theta \sigma B_0^2 \over 4 \rho D^2}
{z \over (1 + w^2)^3},
\label{eq275}
\end{equation}
where, as before, $w = 2z/D + \alpha$, and $\alpha = L/D$. This can be
integrated once to give
\begin{equation}
\dot\varphi = v_z {d\varphi \over dz} =- {3 \theta \sigma B_0^2 \over
128 \rho} \left[ {\alpha \pi \over 2} + \alpha \tan^{-1} w +
{\alpha w \over 1 + w^2} + {2 (1 + \alpha w) \over 3 (1 + w^2)^2} \right].
\label{eq276}
\end{equation}
If we ignore the variation in $v_z$ with position, this can be
integrated once more to yield
\begin{equation}
\varphi(z) = - {3 \theta \sigma B_0^2 D \over
256 \rho v_z} \left[ \left( \alpha w + {1 \over 3} \right)
\left( {\pi \over 2} + \tan^{-1} w \right) + {\alpha - w \over 3( 1 + w^2)}
\right].
\label{eq277}
\end{equation}
At the center of the magnet, $w = \alpha$, the total angle of rotation of
the disc is
\begin{equation}
\varphi_{\rm center} = - {3 \theta \sigma B_0^2 D \over
256 \rho v_z} \left[ \left( \alpha^2 + {1 \over 3} \right)
\left( {\pi \over 2} + \tan^{-1} \alpha \right) \right]
\approx - {3 \pi \alpha^2 \theta \sigma B_0^2 D \over
256 \rho v_z},
\label{eq278}
\end{equation}
where the approximation holds for $\alpha$ somewhat larger than 1.
For example, if $\alpha = L/D = 2$,
$D = 0.2$ m, $B_0 = 20$ T and $v_z = 10$ m/s, we find $\varphi_{\rm center}
\approx - 4 \pi \theta$. With $\theta = 0.1$ rad, then $\varphi_{\rm center}
\approx - 0.4
\pi$.
Equation (\ref{eq276}) indicates the interesting result that the rate of
change of rotation is independent of the velocity, so the total rotation
can be suppressed by increasing the jet velocity, thereby lowering the transit
time.
A liquid jet would presumably not rotate as a rigid body.
Rather, there would be a shear in
which $v_z$ of that portion of the jet closer to the magnet axis actually
increases, while $v_z$ decreases for material farther away (and by a larger
absolute amount).
Our estimate that $\varphi_{\rm center} \approx 90^\circ$ can
perhaps be reinterpreted as indicating that the shear distance along the jet
axis will amount to roughly the jet radius when the jet reaches the center
of the magnet. This would not be troublesome.
\subsubsection{The Rayleigh Instability of the Jet}
\paragraph{Zero Magnetic Field.}
Following earlier work by Plateau, Rayleigh deduced that a cylindrical
jet is unstable against perturbations of wavelength (along the jet axis)
greater than the circumference of the jet \cite{Rayleigh}. The
result of the instability is the breakup of the jet into droplets,
commonly seen as water exits a nozzle. The characteristic time for onset of
the instability is
\begin{equation}
\tau = 3 \sqrt{r^3 \rho \over T},
\label{eq9d}
\end{equation}
where the jet has radius $r$, mass density $\rho$ and surface tension $T$.
The distance travelled by a jet before breakup is then $v\tau$,
where $v$ is the jet velocity.
An example of breakup of a 0.5-mm-diameter mercury jet is shown in
Fig.~\ref{Rosejet} \cite{Criss95,Ansley1,Ansley2}.
The density of mercury is $\rho = 13.5$ g/cm$^3$, and the surface tension is
$T = 470$ dyne/cm. Then eq.~(\ref{eq9d}) gives $\tau = 0.002$ s for $r =
0.025$ cm. At 40 psi, the jet velocity was $v = 5$ m/s, so the characteristic
length before breakup is predicted to be 1 cm, in good agreement with the
reported value of 1.4 cm.
It thus appears that Rayleigh's formula is valid for liquid-metal jets.
\begin{figure}[htp] % h = here, t = top, b = bottom, p = new page
\begin{center}
\includegraphics[width=6in]{jetplots.eps}
\parbox{5.5in} % change 5.5in to \hsize for full-width caption
{\caption[ Short caption for table of contents ]
{\label{Rosejet} Length before breakup, and velocity of a mercury jet of
radius 0.025 cm.
}}
\end{center}
\end{figure}
Turning to parameters relevant to the muon-collider target, consider a
gallium jet of radius 1 cm. The density is $\rho = 6$ g/cm$^2$, and the
surface tension is $T = 360$ dyne/cm. Then the instability time is
$\tau = 0.4$ s. For a jet velocity of 10 m/s, the breakup length would be
about 4 m, which is satisfactory. If the jet radius is reduced to 0.5 cm,
the breakup length drops to 1.4 m.
\paragraph{Nonzero Axial Magnetic Field.}
The effect of a uniform axial magnetic field on the Rayleigh instability
has been considered by Chandrasekhar \cite{Chandra}.
There is no change in the instability time for a nonconducting
liquid, unless its permeability is significantly greater than one. For a
conducting liquid, Chandrasekhar introduces a quality factor,
\begin{equation}
Q = {\mu B^2 \over 4 \pi \eta} \sqrt{r^3 \over \rho T},
\label{eq9e}
\end{equation}
where $\mu$ is the permeability, $B$ is the axial magnetic field strength,
and $\eta$ is called the ``resistivity'' ($\eta = c^2/4\pi\sigma$,
where $\sigma$ is the electrical conductivity). For $Q > 20$, the
Rayleigh instability is suppressed in the first approximation.
For mercury, $\eta = 7.5 \times 10^3$ cm$^2$/s, and $Q = 1.33 \times 10^{-7}
B^2 r^{3/2}$ in Gaussian units. Thus for a 20-T axial magnetic field and
$r = 1$ cm, $Q \approx 5000$, and the Rayleigh instability should be
almost completely damped. This conclusion is little changed by variations in
the radius or resistivity by factors of 2, and should be valid for all
liquid-metal jets under consideration here.
The quality factor does not drop to 20 until the field has fallen to slightly
over 1~T (for $r = 1$ cm).
Hence, unless the jet travels more than a meter in a region
where the magnetic field is less than 1~T, the Rayleigh instability will be of
little concern for us.
%It remains that the jet may be significantly perturbed by the
%nonuniform magnetic field as discussed in sec.~2.5, and/or by the thermal shock
%mentioned in sec.~2.1.2.
\subsection{The Operation of an RF Cavity near a Target}
Once the pions have been collected at the target, it is necessary to
shape the collected pion/muon beam into a longitudinal phase space suitable
for eventual acceleration in the following cooling section. The goal
is to capture and maintain the longitudinal phase space such that the
energy spread $dE/E$ is reduced to near 10\%, and the pulse is contained
within a pulse length of $c\tau <$ 6 m. Since we are interested in
capturing low-energy pions (kinetic energy 50-400 MeV), these pions will have a
large initial velocity spread ($\beta$ between 0.7 and 0.98). It is therefore
important to begin compressing the energy spread in the phase-rotation section
as soon a practical, with an accelerating gradient as high as possible.
The wavelengths
of the rf cavities are constrained by the requirement that
$\lambda_{\rm rf}/2$ be greater
than the bunch length of the collected pions/muons. This bunch length in turn
is determined by the initial bunch length of the proton beam impinging on the
target and the drift distance to the initial rf cavity. For an initial
proton bunch length $\tau_{rms}$ of 1 ns and a drift distance of 3~m this
total bunch length will be 4~ns (1.2~m). This constrains the
wavelength of the initial rf cavity to be greater than $\approx 2.4$~m
(frequency $< 125$~MHz. As the bunch moves down the phase-rotation channel,
it will continue
to elongate, and therefore the rf wavelength of subsequent rf cells will
need to increase, and a complete phase-rotation scenario will require many
rf cells of different frequencies.
A further constraint is placed on the frequencies of
the rf cells if we add the requirement that the same phase-rotation channel must
be capable of handling both positive and negative bunches in separate beam
spills. This constraint can be satisfied simply by requiring the set of
cavities to be odd multiples of a fundamental harmonic,
%\eg, 20, 60, 100, 140~MHz, \etc, or
10, 30, 50, 70, 90, 110~MHz, \etc\ \
%We chose for convenience the series based on the fundamental harmonic of
%10~MHz.
In such a scenario, the beginning rf frequency
could be 110~MHz, descending to a final rf frequency of 10~MHz. Parameters for
a possible complete scenario are given in the Table~\ref{linacparms}.
\begin{table}
\begin{center}
\parbox{5.5in} % replace 5.5in by \hsize if want full-width caption
{\caption[ dummy ]
{\label{linacparms} Low-Energy Collection-Linac Parameters.
}}
\vskip6pt
\begin{tabular}{lccc}
\hline\hline
rf Frequency (MHz) & 90 & 50 & 30 \\
Cavity Length (cm) & 120 & 120 & 120 \\
Full Gap Length (cm) & 36 & 36 & 36 \\
Cavity Radius (cm) & 90 & 106 & 126 \\
Beam Pipe Aperture (cm) & 30 & 30 & 30 \\
$Q/1000$ (from SFISH) & 53.4 & 71.1 & 16.8 \\
Av.\ Gradient (MV/m) & 4.2 & 4.0 & 2.1 \\
RF Peak Power (MW) & 1.8 & 1.2 & 4.8 \\
Av.\ Power @ 15 Hz (kW) & 2.4 & 7.8 & 12.6 \\
Stored Energy (J) & 166 & 260 & 418 \\
Linac Segment (m) & 6 & 18 & 18 \\
Total Power @ 15Hz (kW) & 12 & 118 & 190 \\
\hline\hline
\end{tabular}
\end{center}
\end{table}
For the \mumu collection system to be viable, the rf cells must perform
satisfactorily at high gradients in a magnetic field, and in the high-radiation
environment immediately following the target. Since experience operating an
rf cell in a high-radiation environment generated by a beam of 10$^{14}$
protons/pulse is limited
\cite{Kustom68,Shalnov95,Kishek95,Brown97,%
Grudiev97,Padamsee97,Xu97,Kishek97,Grobner97},
we propose
to establish a proof-of-principle demonstration of this issue by constructing
and operating at high gradients an rf cavity with a frequency suitable for
a muon-collider collection system. Preparation for the study will include
a MARS simulation of the flux of charged particles onto the rf cavity that
emerges from the target and surrounding material.
We sketch our limited understanding of the problem. Because the first rf
cavity is located near the target, of order one high-energy particle passes
through the cavity wall per beam proton. Secondary emission of electrons
by high-energy particles occurs with probability of order 1\% \cite{Brown97}.
These secondary electrons gain various amounts of energy before hitting the
cavity wall. This could produce undesirable heating of the cavity, and
it could also initiate resonant secondary emission (multipactoring) that
results in cavity breakdown \cite{Grobner97}. It is believed that
secondary emission of ions is of less significance \cite{Kustom68}.
For a pulse of $10^{14}$ protons on target, there will be of order $10^{12}$
secondary-emission electrons liberated in the first rf cavity. If these
gained of order 1~MeV before hitting the cavity wall, the resulting energy
deposition is about 0.1~J. In general, this energy deposition will be
spread over a large surface area, and therefore appears to be of minor
concern.
When the $10^{12}$ secondary electrons hit the cavity wall, they lead to
tertiary electrons, \etc\ If these electrons have energy such that the
secondary-emission coefficient exceeds unity, an instability occurs.
For metals typically used in cavity walls, the secondary-emission coefficient
exceeds unity only for electron energies of order 1 keV
\cite{Kishek95,Kishek97}. Hence, the troublesome electrons are those that
acquire only of order $10^{-3}$ of the nominal energy gain of the cavity.
Multipactoring discharges are more often observed at low-power operation of
cavities, and are suppressed at higher power where secondary electrons
gain higher energy.
However, since we begin with $10^{12}$ secondary electrons, it is hard to
predict whether a tiny fraction of them might lead to cavity breakdown.
Experiments must be performed in realistic configurations to study this
critical issue.