Basic Linear Dynamics
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Define emittance =1, then for an upright ellipse:
\theta_o=\sqrt{1 \over \beta_o}
a_o = \sqrt{beta_o}
Drift
-----
after a drift of length z, the top of the ellipse has moved by $\zeta$:
\zeta=z \ \theta_o
The maximum width of the ellipse is the rms sum of $\zeta$ and $a_o$:
a'=\sqrt{a_o^2 + \zeta^2}
and thus
\beta'=a_o^2 + z^2 theta_o^2
\beta'=beta_o + {z^2 \over \beta_o}
Defining \alpha_o=0 at z=0, when the elipse is upright, and
\alpha'=\alpha_o +{z \over beta_o}
then
\beta'=beta_o (1 + \alpha'^2)
The angle $\eta'$ of the point of maximum size can be obtained by symmetry
considerations:
{\eta' \over \theta_o}={\zeta' \over a'}
so
\eta' = {\zeta' \theta_o \over a'}
The ratio $\kappa'$ of the angle $\eta'$ to its amplidude a' is a measure of
the divergence of the beam. It gives, for instance, the strength of a focus
system that would set the ellipse upright again:
\kappa' = {\eta' \over a'}={\theta_o^2 z \over (a')^2}={\alpha' \over \beta'}
Thin Lens
---------
At a thin lens, the size a', and thus $\beta'$, cannot change, but the angles
change in proportion to the focus strength $k$ times the amplitude a'. For a
paricle at the maximum amplitude $a'$
\eta''=\eta' + k \ a'
\kappa'' = \kappa' + k
\alpha'' = \alpha' +  {k \beta'}
\beta'' = \beta'
Another Drift
-------------
As before, we can give $\beta$ in the following drift by:
\beta'''= beta'''_o (1 + (\alpha''')^2)
where
\alpha'''=\alpha'' + {z \over beta'''_o}
and $\beta'''_o$ is the beta at the following waist, if focussed, or at an
imaginary waist before the thin lens; it can be obtained from the values of
$\alpha''$ and $\beta''$ after the thin lens:
\beta'''_o = {\beta'' \over (1 + (\alpha'')^2)}
Momentum Dependences
--------------------
At the end of a matched solenoid, i.e. one in which the beta function remains
constant,
\beta_o (solenoid) \propto p
but at the end of a matched lithium lens:
\beta_o (lens) \propto \sqrt{p}
The strength of a thin solenoid:
k_{solenoid} \propto p^2
But for a thin lithium lens or horn:
k_{lens} \propto p
Matching
--------
Using the above formulae and momentum dependences we can ask if solutions are
possible where we go from one matche slement, such as a solenoid, into
another, in such a way that matching is achieved both on momentum, and for
first order deviations from that momentum. I have not solved th equations
explicitly, but there seems no fundamental law that prohibits this, even in
the absence of chromatic correction elements (e.g. sextupoles in dispersion).
We are NOT correcting chromaticity. The phase advance through the system will
NOT be independent of momentum. But this we do not require.
EXAMPLES
========
I will consider a number of examples, the first 3 of which are motivated by
the need for a match between alternating solenoids in a cooling channel. Where
the field reverses, it must go to zero at one point, and be low over a
distance comparable with the beam pipe diameter. So in the following examples,
I approximate this situation by requiring a finite gap between a central pair
of thin solenoids.
I am given the initial $\beta$, initial ${d \beta/ dp}$, $\alpha=0$ and
$d\alpha/dp=0$; and require all four to have the same value at the entry to
the following solenoid: 4 constraints.
But ignoring the energy changes in the hydrogen and
reaccelerators, I can assume symmetry about the middle of the gap.
In this case the matching requirement reduces to just two
requirements at the symmetry plane:
\alpha= 0
{d \alpha \over dp} = 0
0) Simple solenoid-solenoid
---------------------------
In the first example place a thin lens at the exit of the solenoid,
followed by the gap.
In the solenoid, alpha=0 and the $\beta$'s are constant.
The thin lens drives $\alpha$ negative and the $\beta$ falls to a minimum and
then rises to meat the thin lens at the entrance to the opposite solenoid.
For a fixed gap I have only one variable (the thin solenoid strength). These
thin solenoid strengths can be chosen to put the minimum at the center for one
momentum, and at this momentum a good match is achieved. But {d \alpha \over
dp} can not be set to zero, and at other momenta the match is bad.
see figures Match0
1) Simple solenoid after a gap
------------------------------
Again I match from one long solenoid to another, but now with gaps between the
end of the solenoids and the thin lenses (two thin lenses and three gaps
altogether)
Now I have two variables (the first gaps and the thin solenoid strengths, and
should thus be able to satisfy both constraints.
In fact, I looked for, and found, solutions with perfect matching at two
separated momenta - probably a best solution for wide momentum acceptance.
Clearly I could have satisfied the two other constraints.
See figures Match1
  a: tracings at relative momenta of  .6 to 1.4;
  b: $\alpha$ vs. relative p
2) solenoid-solenoid with tapers
--------------------------------
The above solution is nice because it tells us that solutions are possible,
but the length of the match and is constrained and would not accept the
required acceleration.
So in the second example I keep the symmetry and again match between solenoids,
but I introduce a sequence of lenses to allow the $\beta$ to grow to a much
larger value at the symmetry point. This is close to the solution required for
alternating solenoids.
Again, I looked for, and found, solutions with perfect matching at two
separated momenta.
See Figures Match2
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The second example will be the same, but will match between lithium lenses