Transverse Beam Dynamics
The sollutions of the Hill’s equations are generalized as follows:
\[\begin{aligned} u &= \sqrt{\epsilon\beta}\cos\phi\\ u' &=-\sqrt{\frac{\epsilon}{\beta}}(\alpha\cos\phi+\sin\phi) \end{aligned}\]Where:
\[\phi = \mu + \mu_0\] \[\alpha = -\frac{1}{2}\beta'\] \[\gamma = \frac{1+\alpha^2}{\beta}\] \[\mu = \int_{0}^s \frac{ds}{\beta} \to \phi' = \frac{1}{\beta}\]If we include dispersion we have:
\[\begin{aligned} u &= \sqrt{\epsilon \beta} \cos \phi + D\delta\\ u' &= \sqrt{\frac{\epsilon}{\beta}}(\alpha\cos\phi+\sin\phi) + D'\delta + D\dot{\delta} \end{aligned}\]Bunch Distributions
Elliptical Distributions
An ellipse with semi-axes a, b and area $A=\pi a b$ is defined by:
\[\begin{aligned} x(\varphi) &= a\cos\varphi\\ y(\varphi) &= b \sin\varphi \end{aligned}\]Upon rotation by angle $\phi$ our equations yield:
\[\begin{aligned} x(\varphi) &= a\cos\varphi\cos\phi-b\sin\varphi\sin\phi\\ y(\varphi) &= a\cos\varphi\sin\phi+b\sin\varphi\cos\phi \end{aligned}\]Transforming Offset Elliptical Phase-Space Distributions
We can generate our elliptical distributions by defining the following covariance matrix:
\[\Sigma=\epsilon\Omega = \epsilon\begin{bmatrix}\beta & -\alpha \\ -\alpha & \gamma\end{bmatrix}\]This distribution follows such that:
\[\epsilon = \sqrt{\det{\Sigma}}\]Said particle will normally follow an elliptical trajectory in $(u, u’)$ phase-space given by:
\[\boxed{\epsilon = \gamma u^2 + 2\alpha u u' + \beta u'^2}\]We can further deduce from $\beta(s)$, given by our optics, the following:
\[\alpha(s) = -\frac{1}{2}\beta'(s)\\ \gamma(s) = \frac{1+\alpha(s)^2}{\beta(s)}\]This elliptical distribution appears rotated by $\phi$ as defined by:
\[\tan(2\phi) = \frac{-2\alpha(s)}{\beta(s) -\gamma(s)}\]Particles $(\epsilon, \varphi)$ injected with an offset $\Delta u$ will follow enlarged elliptical trajectories given by a transformation $\varphi\to\tilde{\varphi}$ and emittance $\epsilon\to\tilde{\epsilon}$ transform.
From:
\[\begin{aligned} u = r\cos\varphi\\ u' = r\sin\varphi \end{aligned}\]Substituing this into our ellipse equation gives:
\[\epsilon = r^2\left(\gamma\cos^2\varphi+2\alpha\cos\varphi\sin\varphi+\beta\sin^2\varphi\right)\]Our particle’s radial position is therefore defined via $\epsilon$, $\varphi$, and it’s optics $\alpha, \beta, \gamma$:
\[\boxed{r^2 = \frac{\epsilon}{\gamma\cos^2\varphi+2\alpha\cos\varphi\sin\varphi+\beta\sin^2\varphi}}\]To transform from $u\to\tilde{u}$ we solve the SAS triangle first by law of cosines:
\[\boxed{\tilde{r}^2=\Delta u^2 + r^2 - 2\Delta u r \cos(\pi-\varphi)}\]We solve for $\tilde{\varphi}$ using law of sines:
\[\frac{\sin\tilde{\varphi}}{r} = \frac{\sin(\pi-\varphi)}{\tilde{r}}\]Alternatively:
\[\boxed{\tilde{\varphi} = \arcsin\left(\frac{r\sin(\pi-\varphi)}{\tilde{r}}\right)}\]Considering that:
\[\begin{aligned} \tilde{u} &= \tilde{r}\cos\tilde{\varphi}\\ \tilde{u}' &= \tilde{r}\sin\tilde{\varphi} \end{aligned}\]Then from the new phase-space trajectory defined by:
\[\begin{aligned} \tilde{\epsilon} &= \gamma \tilde{u}^2+2\alpha\tilde{u}\tilde{u}'+\beta\tilde{u^2}\\ &=\tilde{r}^2\left(\gamma\cos^2\tilde{\varphi}+2\alpha\cos\tilde{\varphi}\sin\tilde{\varphi}+\beta\sin^2\tilde{\varphi}\right) \end{aligned}\]Particle emittance $\tilde{\epsilon}$ will increase and phase $\tilde{\varphi}$ shift when offset in phase-space by $\Delta u$. These new coordinates will follow standard betatron motion as defined by:
\[\boxed{\tilde{u}(s) = \sqrt{\tilde{\epsilon}\beta(s)}\cos\tilde{\varphi}(s)}\]Compute semi-axes
For a rotated ellipse by $\phi$ we have:
\[\begin{aligned} 1 &= \frac{(u\cos\phi-u'\sin\phi)^2}{a^2}+\frac{(u\sin\phi+u'\cos\phi)^2}{b^2}\\ &=\frac{c^2u^2-2uu'cs+s^2u'^2}{a^2}+\frac{s^2u^2+2uu'cs+c^2u'^2}{b^2}\\ &= \left(\frac{c^2}{a^2}+\frac{s^2}{b^2}\right)u^2+2cs\left(\frac{1}{b^2}-\frac{1}{a^2}\right)uu'+\left(\frac{s^2}{a^2}+\frac{c^2}{b^2}\right)u'^2 \end{aligned}\]Where:
\[\begin{aligned} s = \sin\phi\\ c = \cos\phi\\ t = \tan\phi \end{aligned}\]Considering the area of an ellipse is defined by:
\[A = \pi a b = \pi \epsilon \to \epsilon = ab\]We can derive our twiss parameters as:
\[\begin{aligned} \gamma &= \epsilon \left(\frac{c^2}{a^2}+\frac{s^2}{b^2}\right) &=&\frac{1}{\epsilon} (s^2a^2+c^2b^2)\\ \alpha &= \epsilon \left(\frac{1}{b^2}-\frac{1}{a^2}\right)cs &=&\frac{cs}{\epsilon}(a^2-b^2) \\ \beta &= \epsilon \left(\frac{s^2}{a^2}+\frac{c^2}{b^2}\right) &=&\frac{1}{\epsilon}(c^2a^2+s^2b^2) \end{aligned}\]Yielding the familiar:
\[\epsilon = \gamma u^2 + 2\alpha u u' + \beta u'^2\]Therefore:
\[\begin{bmatrix}s^2& c^2 \\ cs &-cs \\ c^2& s^2 \end{bmatrix} \begin{bmatrix}a^2\\b^2\end{bmatrix}=\epsilon\begin{bmatrix} \gamma \\ \alpha \\ \beta \end{bmatrix}\]Which simplifies yielding:
\[\boxed{\begin{aligned} a^2 &= \epsilon(\alpha t+\beta)\\ b^2 &= \epsilon(\alpha t+\gamma) \end{aligned}}\]Polar to Phase
Our polar angle $\theta$ in phase-space given by $\tan\theta=\frac{x’}{x}$ requires transformation to yield the phase advance $\varphi$
We have:
\[\begin{aligned} u &= \sqrt{\beta\epsilon}\cos\phi\\ u'&=-\sqrt{\frac{\epsilon}{\beta}}(\alpha\cos\phi+\sin\phi) \end{aligned}\]Accordingly we can generate:
\[\frac{u'}{u} = -\frac{1}{\beta}(\alpha+\tan\varphi)\]Therefore:
\[\boxed{\varphi = \arctan\left(-\beta\frac{u'}{u}-\alpha\right)}\]Our polar angle $\theta$ in u-u’ phase-space is therefore is transformed to phase advance $\varphi$ via:
\[\boxed{\tan\varphi + \alpha + \beta\tan\theta = 0}\]