Spin decoherence rate in a weak focusing all-electric ring

I treat a particle of mass $m$ and charge $e$, with velocity $\bm{v}=\bm{\beta}c$ and Lorentz factor $\gamma=1/\sqrt{1-\beta^{2}}$. I shall mostly set $c=1$ below. The particle spin $s$ is treated as a unit vector and $a=\frac{1}{2}(g-2)$ denotes the magnetic moment anomaly. There is no magnetic field in the model I treat; the ring is all-electric. I shall treat a homogeneous weak focusing model below, so the reference orbit is a circle of radius $r_{0}$. The arc-length along the reference orbit is $s$ and the azimuth is $\theta=s/r_{0}$. The radial coordinate is $x$ and $r=r_{0}+x$ and the vertical coordinate is $z$. The Hamiltonian with $s$ as the independent variable is

Here $H$ is the total energy. It a dynamical invariant in this model. Note that $H=\gamma mc^{2}+\Phi$. For off-energy orbits, following ManeElectrostaticBending2012 , I shall employ the parameter $\lambda_{p}=(1/\beta_{0}^{2})(\Delta H/H_{0})$. The Thomas-BMT equation Thomas ; BMT for spin motion in prescribed external electric and magnetic fields $\bm{E}$ and $\bm{B}$, respectively, is

As stated above, I shall set $\bm{B}=0$ below and treat only motion in an electrostatic field. In the absence of a magnetic field, the equation of motion for the helicity $\bm{s}\cdot\hat{\bm{\beta}}$ is given by MSY2

Note that the right-hand side vanishes when $a=1/(\beta^{2}\gamma^{2})$; this is referred to as the ‘magic’ condition. Hence the literature employs the terms ‘magic gamma’ and ‘magic momentum’ which I shall use below. In the model I treat, the reference orbit is at the magic gamma, hence $a=1/(\beta_{0}^{2}\gamma_{0}^{2})$.

For most of this paper, I shall restrict the orbital motion to lie in the horizontal plane, so $z=p_{z}=0$. Then the electric field is radial and the problem reduces to relativistic charged particle orbital and spin motion in a central field of force. It is standard to treat central force problems using cylindrical polar coordinates $(r,\theta,z)$. I define the dimensionless variable $\xi=x/r_{0}$, so $r=r_{0}(1+\xi)$. I denote the electric field index by $n$. The field and potential in the horizontal plane are given by

The potential is normalized so that $V=0$ at $r=r_{0}$. Note that, although in principle the analytical formulas derived below are valid for all $n>0$, in practice I treated only values $0\leq n\leq 1$ in all of my numerical computations below. This is because it is well known that for a homogeneous weak-focusing all-electric ring, the horizontal betatron tune is given by ManeElectrostaticBending2012 ; Laslett_elec_ring $\nu_{x}=\sqrt{2-n-\beta_{0}^{2}}$, and this can possibly be imaginary if $n>1$, depending on the value of $\beta_{0}$. Next, the centripetal force yields

Hence $\gamma mv_{\theta}^{2}=eE_{0}r_{0}(r_{0}/r)^{n}$. On the reference orbit, $\gamma=\gamma_{0}$ and $v_{\theta}=v_{0}$, hence $eE_{0}r_{0}=p_{0}v_{0}=mc^{2}\gamma_{0}\beta_{0}^{2}$.

Note that if the orbital motion lies in the horizontal plane, then the spin precession vector is vertical: $\bm{\Omega}\parallel\bm{\beta}\times\bm{r}\parallel\hat{\bm{z}}$. Hence the spin precesses around the vertical axis, i.e. the vertical component of the spin is a constant of the motion. Hence, to analyze the spin decoherence rate, it is only the horizontal spin components of the particles which decohere. Hence I restrict the spin to also lie in the horizontal plane in the analysis below; no generality is lost by doing so. Furthermore, the initial condition in an EDM experiment is that the spins are injected along the reference axis $\bm{s}=\hat{\bm{\theta}}$, which lies in the horizontal plane, by definition. I next define the angle $\alpha$, which is the angle between the spin unit vector $\bm{s}$ and the unit vector in the direction of the velocity $\hat{\bm{\beta}}$. I follow the definition in TalmanIPAC2012 , in which we go counterclocksise from $\hat{\bm{\beta}}$ to $\bm{s}$. Then

Then, using eq. (3) and $v_{\theta}=c\beta_{\theta}=r\dot{\theta}$,

Hence, using $eE_{0}r_{0}=mc^{2}\gamma_{0}\beta_{0}^{2}$,

This is applicable for any value of the reference momentum but the case of interest below is when the reference is the magic gamma

Let us make some general observations using this expression. On the reference orbit, the term in parentheses vanishes, by definition of the magic gamma. If we define ‘$\epsilon$’ as a generic ‘small parameter’ for an off-axis orbit (for example the amplitude of a betatron oscillation or an energy offset) then we can write

Here $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are quantities which do not depend on $\epsilon$, and their internal details do not matter. The other terms outside the parentheses are of $O(1)$ on the reference orbit, say $\mathcal{D}_{0}(1+\mathcal{D}_{1}\,\epsilon+\cdots)$ (here $\mathcal{D}_{0}$ and $\mathcal{D}_{1}$ are also quantities which do not depend on $\epsilon$). Then we may write, overall

Hence we require the term in parentheses to $O(\epsilon^{2})$, as expected, but we also require the other terms to $O(\epsilon)$. We cannot approximate the outside terms as $O(1)$ constants; to do so loses the contribution of the term in $\mathcal{C}_{1}\mathcal{D}_{1}$.

Returning to the main thread, let us write $\gamma=\gamma_{0}+\Delta\gamma$ to obtain

Next we expand $r=r_{0}(1+\xi)$ to obtain

Then, noting that both $\Delta\gamma/\gamma_{0}$ and $\xi$ are of the first order in small quantities

Next, we decompose the motion into betatron oscillations and energy offset terms. I do not say ‘synchrotron oscillations’ because $\Delta H/H_{0}$ is a constant of the motion in this model. Using the subscripts ‘$\beta$’ to denote ‘betatron oscillation’ and ‘$d$’ for ‘dispersion,’ we write

(Note that $\Delta\gamma_{\beta}=0$ in an all-magnetic ring.) Because the terms in the betatron motion and the energy offsets are statistically independent variables, all averages over cross terms with ‘$\beta$’ and ‘$d$’ subscripts will vanish. Hence we can group the terms into separate contributions from the betatron motion and the energy offset

I therefore determine $\langle d\alpha/d\theta\rangle_{\beta}$ and $\langle d\alpha/d\theta\rangle_{d}$ in separate calculations below (in different sections).

This problem can be solved exactly. In this model, the off-energy (dispersion) orbits are circles. Following ManeElectrostaticBending2012 , I define the parameter $\lambda_{p}=(1/\beta_{0}^{2})(\Delta H/H_{0})$, hence the energy is $H=H_{0}(1+\Delta H/H_{0})=H_{0}(1+\beta_{0}^{2}\lambda_{p})$. For simplicity of the notation, $\xi$ and $\gamma$ will denote $\xi_{d}$ and $\gamma_{d}$, respectively, in the calculations below. The centripetal force yields (see eq. (5))

We know that for $n=0$ (logarithmic potential) $\gamma\beta^{2}=\gamma_{0}\beta_{0}^{2}$, and the solution is $\gamma=\gamma_{0}$ independent of the radius. The kinetic energy is the same for all the off-energy dispersion orbits. Then from eq. (9) we have $d\alpha_{d}/d\theta=0$. This is the exact solution for off-energy motion in this model for $n=0$. The orbit radius $r_{d}$ is fixed by the value of the potential energy as follows

For $n>0$ we proceed as follows (where we use eq. (19) to replace $r_{0}^{n}/r^{n}$ in terms of $\gamma$)

We want the solution of the quadratic equation for which $\gamma_{d}>1$. Hence

However, the above yields a $0/0$ expression for $n=1$. For $n=1$, note from eq. (24) that

Then for all $n>0$, we have using eq. (19)

This expression was used as an initial condition in tracking studies to be reported below. For the spin precession, from eq. (9) and using eq. (30) (here $\beta_{d}=\sqrt{1-1/\gamma_{d}^{2}}$ is defined in the obvious way)

This works also for $n=0$ because $\gamma_{d}=\gamma_{0}$ so the right-hand side vanishes. Hence this is the exact solution for all $n\geq 0$.

Nevertheless, the above solution is not easily visualizable as a function of $\lambda_{d}$ for small $|\lambda_{d}|$. For $n=1$, we can derive the exact solution using eqs. (28) and (29)

For $n\neq 0$ and $n\neq 1$ we expand in powers of $\lambda_{p}$. We set $\gamma_{d}=\gamma_{0}(1+\Delta\gamma_{d}/\gamma_{0})$. We employ eq. (24) which we write in the form

We write this in the form

Then approximately, to the first order,

Next, to the second order,

Then for the spin precession we obtain, using eqs. (31) and (36),

Hence $d\alpha_{d}/d\theta=0$ for $n=0$, as we know from the exact solution. Since the energy $H$ is a dynamical invariant for this model (hence also $\Delta H/H_{0}$ and $\lambda_{p}$) we can leave the results in the above form. If we average over a statistical distribution of energies, we obtain

For the record, let us calculate the orbit radius $r_{d}$ perturbatively. To save tedious algebra, I calculate to $O(\lambda_{p})$ only. Using eq. (35) in eq. (30),

I conclude this section with a brief commentary on synchrotron oscillations in this model. Note very carefully that synchrotron oscillations are variations of the total energy, i.e. $\Delta H/H_{0}$ and not $\Delta\gamma/\gamma_{0}$. An rf cavity changes the value of $H$, which may or may not change the value of $\gamma$. When electrostatic guiding and focusing fields are present, the value of $\gamma$ may in fact not vary at all in a synchrotron oscillation. In particular, it is well known (and we have seen above) that for $n=0$ (logarithmic potential), $\gamma_{d}=\gamma_{0}$ on all the dispersion orbits, and does not depend on the total energy $H$. Hence we see that $\Delta\gamma_{d}/\gamma_{0}=0$ for synchrotron oscillations in a logarithmic potential.

N.B.: In this model, the dispersion is uniform around the circumference, hence there is synchrobetatron coupling. Hence to justify the above statements rigorously, we require the synchrotron tune to be very small, i.e. $\nu_{s}\ll 1$, to minimize the synchrobetatron coupling. The quasistatic approximation is also required to justify that we can continue to express the motion in $x$ and $p_{x}$ as a sum of betatron oscillations and dispersion orbits, instead of a general fully coupled symplectic formalism.

Next I treat the betatron oscillations. I fix $H=H_{0}=\gamma_{0}mc^{2}$ in the calculations below. For simplicity of the notation, $\xi$ and $\gamma$ will denote $\xi_{\beta}$ and $\gamma_{\beta}$, respectively, in the calculations below. We must expand $r$ and $\gamma$ for the off-axis betatron motion. Note that

The final expression is applicable also for $n=0$ (logarithmic potential). Note also that $\gamma=H/(mc^{2})-\Phi/(mc^{2})$ and as stated above, I fix $H=H_{0}$. Then, using eq. (40),

Then

Substituting in eq. (17) yields

To average over orbits, we need to derive expressions for $\langle\xi\rangle$ and $\langle\xi^{2}\rangle$. This will be done below using canonical transformations to diagonalize the Hamiltonian.

Henceforth I set $c=1$. I shall work with $\xi=x/r_{0}$ below. To preserve the Hamiltonian structure of the equations, we scale the independent variable to $\theta=s/r_{0}$:

Next, we scale the momentum $p_{x}=p_{0}\,p_{\xi}$. To preserve the Hamiltonian structure of the equations, we divide $K$ by $p_{0}$:

Hence we define $\bar{K}=K/p_{0}$

Define $\kappa=p_{s}/p_{0}$. Note that $K$ is invariant along an orbit, hence $p_{s}$ and hence $\kappa$ are also invariant. Hence $p_{s}=p_{s0}$ and $\kappa=\kappa_{0}$. Note that the value of $\kappa$ must be precomputed using the initial data. We can employ the equivalent Hamiltonian

The partial derivatives of all dynamical variables are the same using $K_{1}$ and $\bar{K}$. We work with $K_{1}$ below. I shall treat only the case $H=\gamma_{0}m$ below. Then there is no off-energy dispersion orbit: the orbital motion in $(x,p_{x})$, or $(\xi,p_{\xi})$, consists purely of betatron oscillations. I expand $K_{1}$ in a Taylor series in powers of $\xi$, up to the fourth power. The first step is to expand the potential in a Taylor series to the fourth power in $\xi$ (note that the Taylor series works also for a logarithmic potential, i.e. $n=0$)

Define

Then