3. Twiss Parametrization

3.1. Hill’s Equation

We start with the Hill’s equation of both transverse plane.

(3.1)\[\begin{align} &x''\left(s\right) +\left(\frac{1}{\rho^2} + k_x\left(s\right) \right) x\left(s\right)=0\\ &y''\left(s\right) + k_y\left(s\right)y\left(s\right)=0 \end{align}\]

In general, we need solve:

(3.2)\[\begin{align} &x''\left(s\right) + K\left(s\right)x\left(s\right)=0 \end{align}\]

For a ring accelerator or a linac with periodical structure, we have \(K(s)=K(s+C)\). To have a stable motion, we try to express the solution of the Hill’s equation as:

(3.3)\[\begin{align} x\left(s\right)=aw\left(s\right)e^{i\psi\left(s\right)} \end{align}\]

where \(a\) is a constant, \(w(s)\) is the envelope function and \(\psi(s)\) is the phase function. Then the derivatives are:

(3.4)\[\begin{split}\begin{align} x'\left(s\right)&=aw'e^{i\psi}+iawe^{i\psi}\psi'\\ x''\left(s\right)&=aw''e^{i\psi}+2iaw'e^{i\psi}\psi'-awe^{i\psi}\psi'^2+iawe^{i\psi}\psi'' \end{align}\end{split}\]

Combine Equation (3.4) and (3.2), then the real and imaginary part gives:

(3.5)\[\begin{split}\begin{align} &2aw'\psi'+aw\psi''=a\frac{\left(w^2\psi'\right)'}{w}=0\\ &aw''+Kw-w\psi'^2=0 \end{align}\end{split}\]

Obviously, we are seeking for non-trivial solution (\(w\ne 0\)). The equation (3.5) gives:

(3.6)\[\begin{align} w^2\psi'=1 \end{align}\]

Then by choose the constant a as 1, the equations for \(w\) and \(\psi\) become

(3.7)\[\begin{align} &w''+Kw-1/w^3=0\\ &\psi'-1/w^2=0 \end{align}\]

This shows that the phase function \(\psi(s)\) is linked with the envelope function \(w(s)\).

3.2. Twiss Parameter

We define the beta function (represent the envelope of the oscillation)

\[\beta(s)=w^2(s)\]

and alpha function (the slope of the envelope)

\[\alpha(s)=-\beta'(s)/2=-w(s)w'(s)\]

The phase advance from location \(s_0\) to \(s\) becomes:

(3.8)\[\begin{align} \psi(s,s_0)=\int_{s_0}^s\frac{1}{\beta(s')}ds' \end{align}\]

The derivatives of beta function are:

(3.9)\[\begin{align} \beta'&=2ww'\\ \beta''&=2w'^2+2ww'' \end{align}\]

which satisfy:

(3.10)\[\begin{align} \frac{1}{2}\beta''+K\beta-\frac{1}{\beta}\left[1+\frac{\beta'^2}{4}\right]=0 \end{align}\]

Using the beta function, we can write the transverse motion as:

(3.11)\[\begin{align} x(s)=a\sqrt{\beta(s)}\cos\left(\psi\left(s\right)\right) \end{align}\]

Its derivative becomes

(3.12)\[\begin{align} x'(s)&=-a\sqrt{\beta}\psi'\sin\psi+a\frac{\beta'}{2\sqrt{\beta}}\cos\psi\\ &=-\frac{x}{\beta}\left(\tan\psi-\frac{\beta'}{2}\right)\\ &=-\frac{x}{\beta}\left(\tan\psi+\alpha\right) \end{align}\]

An interesting relation can be revealed as:

(3.13)\[\begin{align} \tan\psi(s)=-\frac{\beta(s)x'(s)+\alpha(s)x(s)}{x(s)} \end{align}\]

Using the triganular relations \(\cos^{-2}\alpha=1+\tan^2\alpha\) for any angle \(\alpha\), we can combine equation (3.11) and (3.13) and get:

(3.14)\[\begin{align} \left(\frac{\beta(s)x'(s)+\alpha(s)x(s)}{x(s)}\right)^2+1&=\left(\frac{a\sqrt{\beta(s)}}{x(s)}\right)^2\\ \frac{\left(\beta(s)x'(s)+\alpha(s)x(s)\right)^2+x^2(s)}{\beta (s)}&=a^2 \end{align}\]

3.3. Action-Angle Variable

Since the transverse Hamiltonian

(3.15)\[\begin{align} H(x,p_x,y,p_y,s)=\frac{p_x^2}{2}+\frac{p_y^2}{2}+\frac{x^2}{2\rho(s)^2}+k(s)\left(x^2-y^2\right)/2 \end{align}\]

is not a constant of motion,

(3.16)\[\begin{align} \frac{dH}{ds}&=\frac{\partial H}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial H}{\partial p_x}\frac{\partial p_x}{\partial s} + \frac{\partial H}{\partial s}\\ &=\frac{\partial H}{\partial s}\ne 0 \end{align}\]

We will follow the action-angle variable treatment in classical mechanics to simplify the transverse motion formula of one transverse direction using the Hamiltonian:

(3.17)\[\begin{align} H(x,x',s)=\frac{x'^2}{2}+\frac{K(s)}{2}{x^2} \end{align}\]

the transverse motion formula, by changing the coordinate from \((x,p_x)\) to \((J,\psi)\). We will use the generating function of the first kind

(3.18)\[\begin{align} F_1(x,\psi)&=\int_0^x x'(\tilde x,\psi)d\tilde x \\ &= -\frac{x^2}{2\beta}\left(\tan\psi-\frac{\beta'}{2}\right) \end{align}\]

The action is calculated by:

(3.19)\[\begin{align} J&=-\frac{\partial F_1}{\partial\psi}=\frac{x^2}{2\beta}\sec^2\psi\\ &=\frac{x^2}{2\beta}\left(1+\left(\frac{\beta x'}{x}+\alpha\right)^2\right)\\ &=\frac{1}{2\beta}\left(x^2+\left(\beta x'+\alpha x\right)^2\right) \end{align}\]

The new Hamiltonian is:

(3.20)\[\begin{align} \tilde{H}&=H+\frac{\partial F_1}{\partial s} \\ &=\frac{1}{2}\frac{x^2}{\beta^2}\left(\tan\psi+\alpha\right)^2+\frac{1}{2}Kx^2 +\\ &\quad\frac{x^2}{2\beta^2}\beta'\left(\tan\psi-\frac{\beta'}{2}\right)-\frac{x^2}{2\beta}\left(\psi'\sec^2\psi-\frac{\beta''}{2}\right) \nonumber\\ &=\frac{J}{\beta} \end{align}\]

Obviously, \(J\) is constant since:

(3.21)\[\begin{align} \frac{dJ}{ds}=-\frac{\partial \tilde{H}}{\partial \psi}=0 \end{align}\]

Now, we can revisit the coordinate transformation \((x,x')\) to \((\psi, J)\)

(3.22)\[\begin{split}\begin{align} J&=\frac{1}{2}\left(\left(\frac{x}{\sqrt{\beta}}\right)^2+\left(\sqrt\beta x'+\frac{\alpha}{\sqrt{\beta}} x\right)^2\right) = \text{constant} \\ \psi&=-\arctan\frac{\sqrt\beta x'+\alpha x/\sqrt{\beta}}{x/\sqrt{\beta}} \end{align}\end{split}\]

However, the Hamiltonian is still not a constant since the beta function is not a constant in the lattice. In a ring, We can ‘force’ to eliminate the variation by defining the canonical transformation

(3.23)\[\begin{align} \bar{\psi}&=\psi - \int_0^s\frac{1}{\beta(s')}ds' + 2\pi\nu\frac{s}{C} \\ \bar{J}&=J \end{align}\]

derived from the generation of second type:

(3.24)\[\begin{equation} F_2(\psi, \bar{J}, s) = \left(\psi - \int_0^s\frac{1}{\beta(s')}ds' + 2\pi\nu\frac{s}{C}\right) \bar{J} \end{equation}\]

The factor \(2\pi/C\) is introduced to make the last term be unitless and approximately an angle (\(\theta = 2\pi s/C\)) that advances \(2\pi\) per revolution. Then the new Hamiltonian \(\bar{H}\) is simply:

(3.25)\[\begin{equation} \bar{H} = \tilde{H} + \frac{\partial F_2(\psi, \bar{J})}{\partial s} = \frac{2\pi}{C} \nu \bar{J} \end{equation}\]

The factor can be removed by scale the new Hamiltonian by the same factor. Therefore the original Hamiltonian (3.17) will be simplified to \(H(\bar\psi, \bar J) = \nu \bar{J}\), by

(3.26)\[\begin{split}\begin{align} x&=\sqrt{2\beta \bar{J}} \cos(\bar\psi+ \int_0^s\frac{1}{\beta(s')}ds'-\nu \theta)\\ \alpha x+\beta x'&=-\sqrt{2\beta \bar{J}} \sin(\bar\psi+ \int_0^s\frac{1}{\beta(s')}ds'-\nu \theta) \end{align}\end{split}\]

3.4. Normalized Coordinates

Therefore we can define the normalized coordinate:

(3.27)\[\begin{align} \mathcal{X}&=\frac{x}{\sqrt{\beta}} \\ \mathcal{P_x}&=\sqrt\beta x'+\frac{\alpha}{\sqrt{\beta}}x \end{align}\]

In matrix form, we have

(3.28)\[\begin{align} \left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)=\left(\begin{array}{cc} 1/\sqrt{\beta} & 0\\ \alpha/\sqrt{\beta} & \sqrt{\beta} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right) \end{align}\]

or reversely:

(3.29)\[\begin{align} \left(\begin{array}{c} x\\ x' \end{array}\right) =\left(\begin{array}{cc} \sqrt{\beta} & 0\\ -\alpha/\sqrt{\beta} & 1/\sqrt{\beta} \end{array}\right)\left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right) \end{align}\]

For location 1 to location 2, the normalized coordinate follows:

(3.30)\[\begin{align} \left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)_2=\left(\begin{array}{cc} \cos\psi_{12} & \sin\psi_{12}\\ -\sin\psi_{12} & \cos\psi_{12} \end{array}\right)\left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)_1 \end{align}\]

The original coordintate has the map:

(3.31)\[\begin{align} \left(\begin{array}{c} \mathcal{x}\\ x' \end{array}\right)_2= \left(\begin{array}{cc} \sqrt{\beta_2} & 0\\ -\alpha_2/\sqrt{\beta_2} & 1/\sqrt{\beta_2} \end{array}\right) \left(\begin{array}{cc} \cos\psi_{12} & \sin\psi_{12}\\ -\sin\psi_{12} & \cos\psi_{12} \end{array}\right)\left(\begin{array}{cc} 1/\sqrt{\beta_1} & 0\\ \alpha_1/\sqrt{\beta_1} & \sqrt{\beta_1} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right)_1 \end{align}\]

Therefore we rederived the transfer map of two arbitrary location in usual form:

(3.32)\[\begin{split}\begin{equation} M\left(s_{2}\mid s_{1}\right) =\left(\begin{array}{cc} \sqrt{{\frac{\beta_{2}}{\beta_{1}}}}\left(\cos\psi+\alpha_{1}\sin\psi\right) & \sqrt{\beta_{1}\beta_{2}}\sin\psi\\ -\frac{1+\alpha_{1}\alpha_{2}}{\sqrt{\beta_{1}\beta_{2}}}\sin\psi+\frac{\alpha_{1}-\alpha_{2}}{\sqrt{\beta_{1}\beta_{2}}}\cos\psi & \sqrt{{\frac{\beta_{1}}{\beta_{2}}}}\left(\cos\psi-\alpha_{2}\sin\psi\right) \end{array}\right) \end{equation}\end{split}\]

When the two locations have same optical functions, the one turn map is rederived:

(3.33)\[\begin{align} \left(\begin{array}{c} \mathcal{x}\\ x' \end{array}\right)_2&= \left(\begin{array}{cc} \sqrt{\beta} & 0\\ -\alpha/\sqrt{\beta} & 1/\sqrt{\beta} \end{array}\right) \left(\begin{array}{cc} \cos\psi & \sin\psi\\ -\sin\psi & \cos\psi \end{array}\right)\left(\begin{array}{cc} 1/\sqrt{\beta} & 0\\ \alpha/\sqrt{\beta} & \sqrt{\beta} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right)_1\\ &= \left(\begin{array}{cc} \cos\psi+\alpha\sin\psi & \beta \sin\psi\\ -\frac{1+\alpha^2}{\beta}\sin\psi & \cos\psi-\alpha\sin\psi \end{array}\right) \left(\begin{array}{c} x\\ x' \end{array}\right)_1 \end{align}\]

3.5. Uniqueness of the Parametrization

Apparently, the solution of the Hills equation is unique except two initial conditions, one scales the envelope function \(w(s)\) and the other one defines the initial phase of \(\psi(s)\) when \(s=0\).

However, is the parametrization a unique answer?

From the action’s formula (3.22), it is easy to see that the definition of the Parametrization can have infinite sets. In general, we can define the the normalized coordinate as:

(3.34)\[\begin{align} \left(\begin{array}{c} \mathcal{X}\\ \mathcal{P}_{x} \end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right) \end{align}\]

The four independent parameter only need to satisfy three conditions:

(3.35)\[\begin{align} a^2+c^2&=\gamma=\frac{1+\alpha^2}{\beta}\\ b^2+d^2&=\beta\\ ab+cd&=\alpha \end{align}\]

The widely used parametrization version, takes the special case \(b=0\). An ‘equally good’ choice can be found by forcing \(c=0\). Then the normalized coordinate becomes:

(3.36)\[\begin{align} \left(\begin{array}{c} \tilde{\mathcal{X}}\\ \tilde{\mathcal{P}_{x}} \end{array}\right)=\left(\begin{array}{cc} \sqrt{\gamma} & \alpha/\sqrt{\gamma}\\ 0 & 1/\sqrt{\gamma} \end{array}\right)\left(\begin{array}{c} x\\ x' \end{array}\right) \end{align}\]

The corresponding phase advance becomes:

(3.37)\[\begin{align} \psi&=-\arctan\frac{x'/\sqrt\gamma }{\sqrt{\gamma}x+\alpha x'/\sqrt{\gamma}} \end{align}\]