Coupling From One Turn Map Matrix

(1)
\begin{align} \vec Z_2 = M \vec Z_1 \end{align}

Now, find the eigenvectors of $M$:

(2)
\begin{align} M \vec v_j = \lambda_j \vec v_j \end{align}

Next, we normalize the $\vec v_j$:

(3)
\begin{align} n_j = \frac{\sqrt{-i \vec v_j^T J \vec v_j}}{\sqrt{2}} \end{align}

with

(4)
\begin{align} J = \pmatrix{ 0 & 1 & 0 & 0 & 0 & 0\cr -1 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1 & 0 & 0\cr 0 & 0 & -1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0 & -1 & 0} \end{align}

and define

(5)
\begin{align} \vec V_j = \frac{\vec v_j}{n_j} \end{align}

The matrix $A$ is now given by taking columns as the real and imaginary parts of the
$V_j$:

(6)
\begin{align} A = \pmatrix{Re(V_1) & Im(V_1) & Re(V_3) & Im(V_3) & Re(V_5) & Im(V_5)} \end{align}

One can check that $A$ is symplectic:

(7)
$$A^T J A = J$$

and that $A$ converts $M$ into three uncoupled rotations:

(8)
\begin{align} A^{-1} M A = \pmatrix{ R_1 & 0 & 0\cr 0 & R_2 & 0\cr 0 & 0 & R_3} \end{align}

The $A$ matrix acts in the normalized space $\vec U = (u_{1...6})$to transform back to $\vec Z = (x,px,y,py,z,pz)$ coordinates:

(9)
\begin{align} \vec Z = A \vec U \end{align}

From the $A$ matrix, one can construct the so-called betaoids and etaoids:

(10)
\begin{align} \beta^i_{jk} = A^{-1}_{(2i-1)j} A_{k(2i)} - A^{-1}_{(2i)j}A_{k(2i-1)} \end{align}

and

(11)
\begin{align} \eta^i_{jk} = A^{-1}_{(2i-1)j} A_{k(2i-1)} + A^{-1}_{2i j} A_{k(2i)} \end{align}

The etaoids can be related to the betaoids via

(12)
\begin{align} \eta^j_{ba} = \sum_{c,n} \beta^j_{bc} \beta^n_{ca} \end{align}

A simpler way to write down these etaoids and betaoids is

(13)
\begin{align} \eta^a = A I_a A^{-1} \end{align}

and

(14)
\begin{align} \beta^a = A J I_a A^{-1} \end{align}

where $I_a$ is the ath identity submatrix, e.g.:

(15)
\begin{align} I_1 = \pmatrix{I_2 &0&0\cr 0 & 0 & 0\cr 0 & 0 & 0} \end{align}

# Dispersion from etaoids

We compute the $H_{1,2,3}$ from the one turn map.
The dispersion is given by
$\vec\eta = H_3 \cdot \pmatrix{{0}\cr {0}\cr{0}\cr{0}\cr{1}\cr{0}}$
We compute it in Matlab with the function
disp = get_dispersion_from_etaoids(esrf(:),1:length(esrf(:))
and we plot with
svals = findspos(esrf(:),1:length(esrf(:))
plot(svals,disp(1,1:n),'-r')

# Transforming to a new position

Let $\vec Z^n_{s}$ be the phase space vector on the nth turn at position s.
Suppose we know the one turn map at position s1, and the transfer matrix from s1 to s2.

(16)
\begin{align} \vec z^n_{s2} = T_{s1,s2} \vec z^n_{s1} \end{align}

and

(17)
\begin{align} \vec z^{n+1}_{s} = M_s \vec z^n_s \end{align}

From these, we can derive the transformation of the one turn map matrix to a different position

(18)
$$M_{s_i} = T^{-1}_{s_i,s_j} M_{s_j} T_{s_i,s_j}$$

# Transformation properties of etaoids and betaoids

The betaoids transform as quadratic forms, whereas the etaoids are basically transformations of phase space. This implies the following transformation properties under a transformation $M$:

(19)
\begin{align} \eta^a\rightarrow M \eta_a M^{-1} \end{align}

whereas

(20)
\begin{align} \beta_a \rightarrow M \beta_a M^T \end{align}

If A is given from the one turn map M as described above, then one can show that the etaoids and betaoids are invariant under these transformations.

# Interpretation and connection to what I did before

There are three invariant matrices $G_{1,2,3}$ which give three invariant quadratic forms:

(21)
\begin{align} g_a = \vec z^T G_a \vec z \end{align}

That this is invariant means that if one applies the one turn map (or partial map for that matter), it should remain unchanged. This leads to the condition:

(22)
$$G_a = M^TG_a M$$

The invariants can be written in terms of the $A$ matrix by

(23)
$$G_a = A^{-1T}I_a A^{-1}$$

There is the paper I wrote. In this paper, its shown that the beam distribution may be written in the form

(24)
\begin{align} f(\vec z)=\frac{1}{\epsilon_1\epsilon_2\epsilon_3}e^{-\frac{g_1}{\epsilon_1}-\frac{g_2}{\epsilon_2}-\frac{g_3}{\epsilon_3}} \end{align}

The moments are given by

(25)
\begin{align} \Sigma = \epsilon_1 JG_1J + \epsilon_2 JG_2J + \epsilon_3 JG_3J \end{align}

Now, given the second moment matrix $\Sigma$, the emittances can be computed via
the eigenvalues $\lambda_a$ of $\Sigma J$:

(26)
\begin{align} \epsilon_j = -i \lambda_a \end{align}

Now, suppose that we have the second moments matrix and we want to get the emittances. Now if the beam is indeed a perfect a function of the invariants, then we can construct

(27)
\begin{align} A^T J \Sigma J A = \epsilon_1 I_1 + \epsilon_2 I_2 + \epsilon_3 I_3 \end{align}

To compare the emittance computed by this method to that in the modemit variables from atx, use
lindat=atx(esrftilt)
epsmodemit=cat(1,lindat.modemit);
plot(spos,eps1,'.r',spos,eps2,'.b',spos,epsmodemit(:,1),'-r',spos,epsmodemit(:,2),'-b')

And to compare the dispersion computed from the etaoids to that computed via lindat, we use
disp=disp_from_etaoids(esrftilt);
and
dispLindat=cat(2,lindat.Dispersion);
and
plot(spos,disp(1,:),'-r',spos,dispLindat(1,:),'-b')
—-
Compare_eigenemittances_dispersion script

esrf1c=build_simple_onecell('s13s20thick.str');
tilt(j)=rand*.002;
end
[eps1,eps2,eps3] = getemittances(esrftilt);
lindat=atx(esrftilt);
epsmodemit=cat(1,lindat.modemit);
disp=disp_from_etaoids(esrftilt);
dispLindat=cat(2,lindat.Dispersion);

plot(spos,eps1,'-r',spos,epsmodemit(:,1),'r')
plot(spos,eps2,'-b',spos,epsmodemit(:,2),'
b')
plot(spos,disp(1,:),'-r',spos,dispLindat(1,:),'-b')
plot(spos,disp(3,:),'-r',spos,dispLindat(3,:),'-b')
—-
Try to understand dispersion, and how to transform the one turn map matrix with the dispersion matrix:
b*m66*inv(b)
Is this uncoupled?
Did I define b wrong?

# References

E. Forest, Unified View of Lattice Functions as Representations of Complex Numbers
ccdb5fs.kek.jp/tiff/2010/1027/1027038.pdf

"Dispersive Lattice Functions in a 6-D Pseudo Harmonic Oscillator"
http://pre.aps.org/abstract/PRE/v58/i2/p2481_1

De Moivre's formula: are Sands' H-functions the same as Chao's?
http://iopscience.iop.org/1748-0221/7/07/P07012