Symplectic Integrator

If we use position along a curve as our our coordinate, the Hamiltonian is given by p_z.

(1)
\begin{align} H=1+\delta -(1+hx)\frac{A_s}{B\rho}-(1+hx)\sqrt{(1+\delta)^2-p_x^2-p_y^2} \end{align}

Expanding and splitting, we get $H=H_1+H_2$

(2)
\begin{align} H_1 = (1+hx)\frac{p_x^2+p_y^2}{2(1+\delta)} \end{align}
(3)
\begin{align} H_2 = -(1+hx)\frac{A_s}{B\rho}-(1+\delta)hx \end{align}

4th order Symplectic integrator

Now, suppose we can solve H_1 and H_2 independently.

Define

(4)
\begin{align} d_1 = d_4 = \frac{1}{2-2^{1/3}} L \end{align}
(5)
\begin{align} d_2 = d_3 = \frac{1-2^{1/3}}{2(2-2^{1/3})} L \end{align}
(6)
\begin{align} k_1 =k_3 = \frac{1}{2-2^{1/3}} L \end{align}
(7)
\begin{align} k_2 = -\frac{2^{1/3}}{2-2^{1/3}} L \end{align}

Now, suppose we can solve $H_1$ (drift) and $H_2$ (kick) independently, and we notate

(8)
\begin{equation} e^{:H_1 d:} = D(d) \end{equation}
(9)
\begin{equation} e^{:H_2:k} = K(k) \end{equation}

Then the 4th order integrator is

(10)
\begin{equation} D(d_1) K(k_1) D(d_2) K(k_2) D(d_2) K(k_1) D(d_1) \end{equation}

Idea for a fast symplectic integrator
we want to define a grid of initial conditions, z1 … zn. We then define the one turn map via the mapped valus of these initial conditions: ringpass(ring,[z1 z2 … zn])

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