Spin Normal Form

Let us imagine that a spin-orbit transformation $U=(a,A)$ such that:

$$T= \left({m,S\vphantom{{e}^{\theta {L}_{y}}}}\right)= \underbrace... ...{{\left({a,A\vphantom{{e}^{\theta {L}_{y}}}}\right)}^{-1}}\limits_{{U}^{-1}}^{}$$ (11.10)

It is clear that the map $R$ can be formally computed as a Taylor map in both orbital and spin variables around the origin. It is now implemented in FPP. The type normal_spin form is given by:

  type normal_spin
     type(normalform) N   ! regular orbital normal form
     type(damapspin) a1   ! brings to fixed point
     type(damapspin) ar   ! normalises around the fixed point
     type(damapspin) as   ! pure spin map 
     type(damapspin) a_t  ! !! (a_t%m,a_t%s) = (a1%m, I ) o (I ,as%s) o (ar%m,I)
!!!  extra spin info
     integer M(NDIM,NRESO),MS(NRESO),NRES  ! orbital and spin resonances to be left in the map
     type(real_8) n0(3)     ! n0 vector
     type(real_8) theta0    !  angle for the matrix around the orbit (analogous to linear tunes)
!!!Envelope radiation stuff
     real(dp) s_ij0(6,6)  !  equilibrium beam sizes
     ! equilibrium emittances (partially well defined only for infinitesimal damping)
     real(dp) emittance(3),tune(3),damping(3)
     logical(lp) AUTO,STOCHASTIC
     real(dp)  KICK(3)   ! fake kicks for tracking stochastically
     real(dp)  STOCH(6,6)  ! Diagonalized of stochastic part of map for tracking stochastically
  end type normal_spin

If NS is of type normal_spin and DS is a damapspin, then DS can be normalized as follows:

      NS=DS

in complete analogy with the orbital normal form. In fact the analogy on the computer is complete. ^11.1

So, as I said, if you believe in the normal form of Eq. (11.10), then even a stupid person can comprehend the concept of the invariant spin axis in the normalized space.

In the normalized space, we notice that the spin vector $\vec{e}_2=(0,1,0)$ is an invariant of $R=(r,\exp(\theta L_y))$. Of course the choice of $e_2$ for the spin normal form is arbitrary. In the rarified air of mathematics, we are free to use anything for the normal form axis. We chose a ``vertical'' axis. Further more the angle of the spin rotation $\theta$ is itself invariant because it depends only on the actions and the parameters of the orbital map $r$. So the spin axis is defined.

We now intend to show the obvious: the vector $A \vec{e}_{2}$ is the spin field axis $\vec{n}$.