The Effective Geometry Factor

The induced wakefield voltage for a particle traveling within a longitudinal bunch $\lambda(\tau)$ influenced by a longitudinal impedance $Z/n$ is given by

\[V_{W} = -\frac{1}{\omega}\frac{Z}{n}\frac{d\lambda}{d\tau},\]

where the space-charge impedance is defined by

\[\frac{Z}{n} = -j\frac{Z_0}{\beta \gamma^2}g,\]

and geometry factor $g$ for a particle within a long round bunch of radius $a$ within a conductive beam pipe of radius $b$ is given by:

\[g = \frac{1}{2} + \ln\frac{b}{a}-\frac{1}{2}\frac{r^2}{a^2}\]

Typically in longitudinal trackers, the maximum geometry factor is applied homogeneously to all particle’s tracked in a bunch. Realistically the incurred space-charge for an individual particle will vary based on it’s transverse emittance $\epsilon$ and momentum-spread $\delta$.

Analysis

Integrated around the ring, the effective geometry factor $\bar{g}$ can be computed by

\[\bar{g} = \frac{1}{2} + \ln\frac{b}{\bar{a}} -\frac{\overline{r^2}}{\bar{a}^2}\]

where:

\[\overline{r^2} = \frac{\beta}{2}(\epsilon_x+\epsilon_y)+D^2\delta^2\]
A particle's transverse position around a ring defined by optics $\beta(s) \approx \beta$ and $D(s) \approx D$ is given by $$\begin{aligned} x &= \sqrt{\beta \epsilon_x}\cos\phi_x+D\delta \cr y &= \sqrt{\beta\epsilon_y}\cos\phi_y \end{aligned}$$ where the betatron phase advance is given by: $$\phi = \frac{s}{\beta} + \phi_0.$$ A particles transverse position is given by $r^2 = x^2 = y^2$ therefore $$r^2 = \beta\epsilon_x\cos^2\phi_x + \beta\epsilon_y\cos^2\phi_y + D^2\delta^2 + 2 \sqrt{\beta\epsilon_x}\cos\phi_x D\delta.$$ Because the betatron tune $Q_x$ and $Q_y$ are large, the variation due to phase-advance averages out to zero and accordingly: $$\overline{r^2} = \frac{1}{C}\oint r^2 ds $$ since $$\overline{\cos^2x} =\lim_{x\to\infty} \frac{1}{x}\int_0^x \cos^2x'dx' = \frac{1}{2}$$ and $$\overline{\cos x} = \lim_{x\to\infty} \int_0^x \cos x' dx' = 0$$ $$\boxed{\overline{r^2} = \frac{\beta}{2}(\epsilon_x+\epsilon_y)+D^2\delta^2}$$

and

\[\bar{a}=2\sqrt[4]{(\beta\sigma_\epsilon+D^2\sigma_\delta^2)(\beta\sigma_\epsilon)}\]
For the beam width $a$, if the beta function is approximated as constant throughout the ring, the beam radius is given by: $$a = \sqrt{a_y a_y} = 2\sqrt{\sigma_x\sigma_y}$$ where: $$\begin{aligned} \sigma_x^2 &=\beta\sigma_\epsilon + D^2\sigma_\delta^2\cr \sigma_y^2 &= \beta\sigma_\epsilon \end{aligned}$$ Because the optics optics are assumed invariant along the ring, the beam width is constant nd so accordingly $$\boxed{\bar{a} = 2\sqrt[4]{(\beta\sigma_\epsilon+D^2\sigma_\delta^2)(\beta\sigma_\epsilon)}}$$