JUAS 2023
Course 1: The science of particle accelerators
A. Lasheen
$$ \begin{aligned}\delta E_{\mathrm{rf}} & = \int{q \mathcal{E}_z \left(\rho,z,t\right) dz} \\ & = q V_{\mathrm{rf}}\left(\rho, \tau\right) \end{aligned}$$
$$ \begin{aligned} \mathcal{E}_z\left(\rho, t\right) & = \mathcal{E}_0 J_0\left(\chi_0\frac{\rho}{\rho_c}\right) \cos\left(\omega_r t\right) \\ \mathcal{B}_\theta\left(\rho, t\right) & = - \frac{\mathcal{E}_0}{c} J_1\left(\chi_0\frac{\rho}{\rho_c}\right) \sin\left(\omega_r t\right) \end{aligned}$$
$J_n$ Bessel function
$\chi_0\approx 2.405$
$\omega_r = \chi_0 c/\rho_c$
$$ \begin{aligned} \Delta \overrightarrow{\mathcal{E}} - \frac{1}{c^2} \frac{\partial^2 \overrightarrow{\mathcal{E}}}{\partial t^2} & = 0 \end{aligned} $$
$$ \Delta \overrightarrow{\mathcal{E}} = \frac{\partial^2 \mathcal{E}_z}{\partial z^2} + \frac{1}{\rho}\frac{\partial \mathcal{E}_z}{\partial \rho} + \frac{\partial^2 \mathcal{E}_z}{\partial \rho^2} $$
$$ \begin{aligned} & x^2\frac{\partial^2 y}{\partial x^2} + x\frac{\partial y}{\partial x} + \left(\frac{x}{x_0}-n\right)^2 y = 0 \\ \rightarrow \quad & y = y_0 J_n\left(\frac{x}{x_0}\right) \end{aligned}$$
$$ \begin{aligned} \frac{\partial^2 \mathcal{E}_z}{\partial z^2} + \frac{1}{\rho}\frac{\partial \mathcal{E}_z}{\partial \rho} + \frac{\partial^2 \mathcal{E}_z}{\partial \rho^2} - \frac{1}{c^2} \frac{\partial^2 \mathcal{E}_z}{\partial t^2} & = 0 \end{aligned} $$
Assuming a solution of the form $\mathcal{E}_z= \mathcal{E}_{z,\rho} \left(\rho\right)\cos\left(\omega_r t\right)$ lead to
$$ \begin{aligned} & \frac{1}{\rho}\frac{\partial \mathcal{E}_{z,\rho}}{\partial \rho} \cos\left(\omega_r t\right) + \frac{\partial^2 \mathcal{E}_{z,\rho}}{\partial \rho^2} \cos\left(\omega_r t\right) - \frac{1}{c^2} \frac{\partial^2 \cos\left(\omega_r t\right)}{\partial t^2} \mathcal{E}_{z,\rho} & = 0 \\ \implies & \frac{\partial^2 \mathcal{E}_{z,\rho}}{\partial \rho^2} \cos\left(\omega_r t\right) + \frac{1}{\rho}\frac{\partial \mathcal{E}_{z,\rho}}{\partial \rho} \cos\left(\omega_r t\right) + \left(\frac{\omega_r}{c}\right)^2 \mathcal{E}_{z,\rho} \cos\left(\omega_r t\right) & = 0 \\ \implies & \rho^2\frac{\partial^2 \mathcal{E}_{z,\rho}}{\partial \rho^2} + \rho\frac{\partial \mathcal{E}_{z,\rho}}{\partial \rho} + \left(\frac{\rho\omega_r}{c}\right)^2 \mathcal{E}_{z,\rho} = 0 \end{aligned} $$
$$ \begin{aligned} \rho^2\frac{\partial^2 \mathcal{E}_{z,\rho}}{\partial \rho^2} + \rho\frac{\partial \mathcal{E}_{z,\rho}}{\partial \rho} + \left(\frac{\rho\omega_r}{c}\right)^2 \mathcal{E}_{z,\rho} = 0 \end{aligned} $$
$$ \delta E_{\mathrm{rf}} = \int{q \mathcal{E}_z \left(\rho,z,t\right) dz} = q \int_{-g/2}^{g/2}{\mathcal{E}_0\left(\rho,z\right)\cos\left(\omega_r t\right) dz} $$
$$ \begin{aligned} & \delta E_{\mathrm{rf}}\left(\tau\right) = q V_{\mathrm{rf,0}} T_t \cos{\left(\omega_r \tau\right)} && \text{Linac convention} \\ & \delta E_{\mathrm{rf}}\left(\tau\right) = q V_{\mathrm{rf,0}} T_t \sin{\left(\omega_r \tau\right)} && \text{Synchrotron convention} \end{aligned}$$
$$ \begin{aligned} T_t = \frac{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) \cos\left(\frac{\omega_r z}{\beta c}\right) dz}{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) dz} \end{aligned} $$
$$ \delta E_{\mathrm{rf}} = q \int_{-g/2}^{g/2}{\mathcal{E}_0\left(\rho,z\right)\cos\left(\omega_r t\right) dz} $$
$$ z\left(t\right) = \int_{\tau}^{t}{\beta\left(t\right) c\ dt} \approx \beta c \left(t-\tau\right) $$
$$ \delta E_{\mathrm{rf}} = q \int_{-g/2}^{g/2}{\mathcal{E}_0\left(\rho,z\right)\cos\left[\omega_r \left(\frac{z}{\beta c}\right) - \omega_r \tau \right] dz} $$
$$ \begin{aligned} \delta E_{\mathrm{rf}} = q \int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) & \cos\left(\frac{\omega_r z}{\beta c}\right)\cos\left(\omega_r \tau\right) dz + \\ & \sin\left(\frac{\omega_r z}{\beta c}\right)\sin\left(\omega_r \tau\right) dz \end{aligned}$$
$$ \begin{aligned} \delta E_{\mathrm{rf}} = q \cos\left(\omega_r \tau\right) \int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) \cos\left(\frac{\omega_r z}{\beta c}\right) dz \end{aligned} $$
$$ V_{\mathrm{rf,0}} = \int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) dz $$
$$ \begin{aligned} T_t = \frac{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) \cos\left(\frac{\omega_r z}{\beta c}\right) dz}{V_{\mathrm{rf,0}}} \end{aligned} $$
$$ \begin{aligned} \delta E_{\mathrm{rf}}\left(\tau\right) = q V_{\mathrm{rf,0}} T_t \cos{\left(\omega_r \tau\right)} = q V_{\mathrm{rf}} \cos{\left(\omega_r \tau\right)} \end{aligned}$$
$$ T_t = \frac{\sin\left(\frac{\chi_0 g}{2\beta \rho_c}\right)}{\left(\frac{\chi_0 g}{2\beta\rho_c}\right)} $$
$$ \begin{aligned} T_t & = \frac{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) \cos\left(\frac{\omega_r z}{\beta c}\right) dz}{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) dz} \\ \implies T_t & = \frac{\mathcal{E}_0 J_0\left(\rho\right) \int_{-g/2}^{g/2}\cos\left(\frac{\omega_r z}{\beta c}\right) dz}{ \mathcal{E}_0 J_0\left(\rho\right) \int_{-g/2}^{g/2} dz} \\ \implies T_t & = \frac{1}{g} \int_{-g/2}^{g/2}\cos\left(\frac{\omega_r z}{\beta c}\right) dz \end{aligned}$$
$$ \begin{aligned} \implies T_t & = \frac{1}{g} \int_{-g/2}^{g/2}\cos\left(\frac{\omega_r z}{\beta c}\right) dz \\ \implies T_t & = \frac{\beta c}{\omega_r g} \left[\sin\left(\frac{\omega_r g}{2\beta c}\right) - \sin\left(\frac{- \omega_r g}{2\beta c}\right) \right] \\ \implies T_t & = \frac{\sin\left(\frac{\omega_r g}{2\beta c}\right)}{\left(\frac{\omega_r g}{2\beta c}\right)} = \frac{\sin\left(\frac{\chi_0 g}{2\beta \rho_c}\right)}{\left(\frac{\chi_0 g}{2\beta \rho_c}\right)} \end{aligned}$$
$$ \begin{aligned} \delta E_{\mathrm{rf}}\left(\tau\right) = q V_{\mathrm{rf}} \sin{\left(\omega_r \tau\right)} \quad \rightarrow \quad \delta E_{\mathrm{rf}}\left(\phi\right)= q V_{\mathrm{rf}} \sin{\left(\phi\right)} \end{aligned}$$
$$ \dot{p} = q\dot{\mathcal{B}}_y\rho $$
$$ \begin{aligned} \delta E_{\mathrm{b}}\left(\rho\right) = q \oint_{ C}{\overrightarrow{\mathcal{E}}\cdot\overrightarrow{d z}} = q \int_{0}^{2\pi}\int_{0}^{\rho}{\frac{\partial\mathcal{B}_y\left(\rho',\theta,t\right)}{\partial t} \rho'\ d\rho'\ d\theta} \end{aligned}$$
$$ \begin{aligned} P_{\mathrm{sr}} = \frac{q^2 c}{6\pi \epsilon_0} \frac{\left(\beta\gamma\right)^4}{\rho^2} \end{aligned} $$
$$ \begin{aligned} \delta E_{\mathrm{sr}}\left(E,\rho\right) = \frac{q^2}{3\epsilon_0} \frac{\beta^3 E^4}{ \rho\ E_0^4} \quad , \left(T_b = \frac{2\pi\rho}{\beta c}\right) \end{aligned} $$
$$ \begin{aligned} \delta E_{\mathrm{ind}}\left(\tau\right) = q V_{\mathrm{ind}}\left(\tau\right) = - q N_b \left(\lambda \ast \mathcal{W} \right) \end{aligned} $$
$$ T_0 = \frac{C}{v} = \frac{2\pi R}{\beta c} $$
$$ \omega_0 = 2\pi f_0 = \frac{2\pi}{T_0} = \frac{\beta c}{R}$$
$$ \omega_r = h\ \omega_{0,s} = h\frac{\beta_s c}{R_s} $$
where $h$ is the RF harmonic number (integer number).
$$ \begin{aligned} \delta E_{s} & = \delta E_{\mathrm{rf},s} + \delta E_{\mathrm{b,s}} \\ & + \delta E_{\mathrm{sr,s}} + \delta E_{\mathrm{ind,s}} \end{aligned} $$
$$ \begin{aligned} \delta E_{s} & = q V_{\mathrm{rf}} \sin{\left(h\omega_{0,s} \tau_s\right)} \\ & = q V_{\mathrm{rf}} \sin{\left(\phi_s\right)} \end{aligned}$$
where $\phi_s$ is the synchronous phase.
$$ \dot{E}_{s} \approx \frac{\delta E_{s}}{T_{0,s}} \quad \rightarrow \quad \dot{E}_{s} = \frac{q V_{\mathrm{rf}}}{T_{0,s}} \sin{\left(\phi_s\right)} $$
$$ \dot{\mathcal{B}}_y\rho_s = \frac{\dot{p}_s}{q} $$
$$ \frac{dE}{dp} = \beta c \quad \rightarrow \quad \dot{E}=\beta c \dot{p} $$
$$ \delta E_{s} = 2\pi q \rho_s R_s \dot{\mathcal{B}}_y \quad \text{and} \quad \phi_s = \arcsin\left(2\pi\rho_s R_s\frac{\dot{\mathcal{B}}_y}{V_\mathrm{rf}}\right) $$
$$ \begin{aligned} & \frac{d}{dt}\left(\mathcal{B}_y\rho_s\right)= \frac{\dot{p}_s}{q} \\ \implies & \dot{\mathcal{B}}_y\rho_s + \mathcal{B}_y \dot{\rho}_s = \frac{\dot{E}_{s} }{q\beta_s c} \end{aligned}$$
$$ \begin{aligned} \implies & \dot{\mathcal{B}}_y\rho_s = \frac{V_{\mathrm{rf}}}{\beta_s c T_{0,s}} \sin{\left(\phi_s\right)} = \frac{V_{\mathrm{rf}}}{2\pi R_s} \sin{\left(\phi_s\right)} \\ \implies & 2\pi q R_s \rho_s \dot{\mathcal{B}}_y = q V_{\mathrm{rf}} \sin{\left(\phi_s\right)} = \delta E_{s}\end{aligned}$$
$$ \omega_r\left(t\right) = \frac{h c}{R_s} \beta_s\left(t\right) $$
$$ f_r\left(t\right) = \frac{h c}{2\pi R_s} \sqrt{\frac{\mathcal{B}_y^2\left(t\right)}{\mathcal{B}_y^2\left(t\right) + \left(\frac{m_0 c}{\rho_s q}\right)^2}} $$
$$ \begin{aligned} \beta_s\left(t\right) & = \frac{p c}{E} = \frac{p c}{\sqrt{\left(pc\right)^2+E_0^2}} \\ & = \frac{\mathcal{B}_y \rho_s q c}{\sqrt{\left(\mathcal{B}_y \rho_s q c \right)^2 + \left(m_0 c^2\right)^2}} \\ & = \frac{\mathcal{B}_y}{\sqrt{\left(\mathcal{B}_y\right)^2 + \left(\frac{m_0 c^2}{\rho_s q c}\right)^2}} \end{aligned} $$
$$ \begin{aligned} \beta_s\left(t\right) & = \sqrt{\frac{\mathcal{B}_y^2}{\mathcal{B}_y^2 + \left(\frac{m_0 c}{\rho_s q}\right)^2}} \end{aligned} $$
$$ \begin{aligned} f_r\left(t\right) & = \frac{\omega_r\left(t\right)}{2\pi} = \frac{h c}{2\pi R_s} \beta_s\left(t\right) \\ & = \frac{h c}{2\pi R_s} \sqrt{\frac{\mathcal{B}_y^2\left(t\right)}{\mathcal{B}_y^2\left(t\right) + \left(\frac{m_0 c}{\rho_s q}\right)^2}} \end{aligned}$$
$$ \begin{aligned} dp_s & = q \left(d\mathcal{B}_y \rho_s+ \mathcal{B}_y d\rho_s \right) \\ & = q \mathcal{B}_y d\rho_s \end{aligned}$$
$$ \begin{aligned} x_D\left(z\right) = D_x\left(z\right) \frac{dp}{p} \end{aligned}$$
$$ \begin{aligned} \alpha_p = \frac{dR/R}{dp/p} \end{aligned}$$
$$ \begin{aligned} \eta = -\frac{d\omega_0/\omega_0}{dp/p} = \frac{dT_0 / T_0}{dp/p} \end{aligned}$$
$$ \begin{aligned} \eta = -\frac{d\omega_0/\omega_0}{dp/p} \end{aligned}$$
$$ \frac{dp}{p} = \left(\frac{\partial p}{\partial R}\right) _{\mathcal{B}_y} \frac{R}{p} \frac{dR}{R} + \left(\frac{\partial p}{\partial \mathcal{B}_y}\right) _{R} \frac{\mathcal{B}_y}{p} \frac{d\mathcal{B}_y}{\mathcal{B}_y} \\ \ \\ \implies \frac{dp}{p} = \frac{1}{\alpha_p} \frac{dR}{R} + \frac{d\mathcal{B}_y}{\mathcal{B}_y} $$
$$ \begin{aligned} df\left(x,y\right) = \left(\frac{\partial f}{\partial x}\right) _y dx + \left(\frac{\partial f}{\partial y}\right) _x dy \end{aligned}$$
$$ \begin{aligned} \frac{\partial f}{\partial x} = \left(\frac{\partial x}{\partial f}\right)^{-1} \end{aligned}$$
$$ \left(\frac{\partial p}{\partial \mathcal{B}_y}\right) _{R}\frac{\mathcal{B}_y}{p} = \rho q \frac{\mathcal{B}_y}{p} = 1 $$
$$ \begin{aligned} \alpha_p = \frac{1}{2\pi R} \int_0^{2\pi R} \frac{D_x\left(z\right)}{\rho\left(z\right)} dz = \frac{\left< D_x \right>_\rho}{R} \end{aligned}$$
$$ \begin{aligned} dC & = 2\pi dR = \int_0^{2\pi}{x_D\left(z\right) d\theta} \end{aligned}$$
$$ \begin{aligned} dC & = 2\pi dR = \int_0^{2\pi}{x_D\left(z\right) d\theta} \\ \implies dR & = \frac{1}{2\pi} \int_0^{2\pi R}{D_x\left(z\right) \frac{dp}{p} \frac{dz}{\rho\left(z\right)}} \quad ,\left(dz=\rho\left(z\right)d\theta\right) \\ \implies \frac{dR}{R}\frac{p}{dp} & = \frac{1}{2\pi R} \int_0^{2\pi R}{\frac{D_x\left(z\right)}{\rho\left(z\right)} dz} \\ \implies \alpha_p & = \frac{1}{2\pi R} \int_0^{2\pi R}{\frac{D_x\left(z\right)}{\rho\left(z\right)} dz} \end{aligned}$$
$$ \begin{aligned} \omega_0=\frac{\beta c}{R} \quad \rightarrow \quad \frac{d\omega_0}{\omega_0} = \frac{d\beta}{\beta} - \frac{dR}{R} \left( = -\frac{dT_0}{T_0}\right) \end{aligned}$$
$$ \begin{aligned} \eta = \frac{dT_0 / T_0}{dp/p} = -\frac{d\omega_0 / \omega_0}{dp/p} = \alpha_p - \frac{1}{\gamma^2} \end{aligned}$$
$$ \eta > 0 \quad\text{if}\quad\alpha_p > \frac{1}{\gamma^2} ,\quad \gamma > \gamma_t $$
$$ \eta < 0 \quad\text{if}\quad\alpha_p < \frac{1}{\gamma^2} ,\quad \gamma < \gamma_t $$
We introduced the transition gamma $\gamma_t=1 / \sqrt{\alpha_p}$.
In the two different regimes, which particle is circulating faster/slower in the machine? What happens for very large $\gamma$?
We introduced the transition gamma $\gamma_t$.
$$ \begin{aligned} \eta = 0 \quad\text{if}\quad\alpha_p=\frac{1}{\gamma^2} ,\quad \gamma=\gamma_t \end{aligned}$$
$$ \begin{aligned} \alpha_{p_0} = \alpha_0 + \alpha_1\frac{\Delta p}{p_0} + \alpha_2\left(\frac{\Delta p}{p_0}\right)^2 + ... \end{aligned}$$
$$ \begin{aligned} p = q \mathcal{B}_y \rho = q \left< \mathcal{B}_y \right> R \end{aligned}$$
$$ \begin{aligned} \left< \mathcal{B}_y \right> = \frac{1}{2\pi R} \int_0^{2\pi R}{\mathcal{B}_y dz} \end{aligned}$$
$$ \begin{aligned} \left < n \right > = -\frac{d \left< \mathcal{B}_y \right> / \left< \mathcal{B}_y \right>}{dR / R} = 1 - \frac{1}{\alpha_p} \end{aligned}$$
$$ \begin{aligned} \left< \mathcal{B}_y \right> & = \frac{1}{2\pi R} \int_0^{2\pi R}{\mathcal{B}_y dz} \\ & = \frac{1}{2\pi R} \int_0^{2\pi R}{\frac{p}{q\rho} dz} \\ & = \frac{1}{2\pi R} \frac{p}{q} \int_0^{2\pi R}{\frac{dz}{\rho\left(z\right)}} \\ & = \frac{1}{2\pi R} \frac{p}{q} \frac{2\pi \rho}{\rho} && \text{$\rho\to\infty$ in straigt sections}\\ \left< \mathcal{B}_y \right> R & = \frac{p}{q} && \text{otherwise constant} \end{aligned}$$
$$ \begin{aligned} & p = q \left< \mathcal{B}_y \right> R \\ \implies & \frac{dp}{p} = \frac{d \left< \mathcal{B}_y \right>}{\left< \mathcal{B}_y \right>} + \frac{dR}{R} \\ \implies & \frac{dp/p}{dR/R} = \frac{d \left< \mathcal{B}_y \right>/ \left< \mathcal{B}_y \right>}{dR/R} + 1 \\ \implies & \frac{1}{\alpha_p} = - \left < n \right > + 1 \\ \implies & \left < n \right > = 1 - \frac{1}{\alpha_p} \\ \end{aligned}$$
$$ \begin{aligned} \frac{dp}{p} = \gamma^2 \frac{d\beta}{\beta} \quad \text{and} \quad \frac{d\beta}{\beta} = \frac{df_0}{f_0} + \frac{dR}{R} \end{aligned}$$
$$ \begin{aligned} \delta E_{\mathrm{rf}}\left(\tau\right) = q V_{\mathrm{rf},0} T_t \sin{\left(\omega_r \tau\right)} \quad\rightarrow\delta E_{\mathrm{rf}}\left(\phi\right) = q V_{\mathrm{rf}} \sin{\left(\phi\right)} \end{aligned}$$
$$ \begin{aligned} T_t\left(\rho,\beta\right) = \frac{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) \cos\left(\frac{\omega_r z}{\beta c}\right) dz}{\int_{-g/2}^{g/2}\mathcal{E}_0\left(\rho,z\right) dz} \end{aligned} $$
$$ \begin{aligned} \mathcal{E}_z\left(\rho, t\right) = \mathcal{E}_0 J_0\left(\chi_0\frac{\rho}{\rho_c}\right) \cos\left(\omega_r t\right) \end{aligned}$$
$J_n$ Bessel function, $\chi_0\approx 2.405$, $\omega_r = \chi_0 c/\rho_c$
$$ \begin{aligned} T_t = \frac{\sin\left(\frac{\chi_0 g}{2\beta \rho_c}\right)}{\left(\frac{\chi_0 g}{2\beta\rho_c}\right)} \end{aligned}$$
$$ \begin{aligned} \delta E_{\mathrm{b}}\left(\rho\right) = q \int_{0}^{2\pi}\int_{0}^{\rho}{\frac{\partial\mathcal{B}_y\left(\rho',\theta,t\right)}{\partial t} \rho'\ d\rho'\ d\theta} \end{aligned}$$
$$ \begin{aligned} \delta E_{\mathrm{sr}}\left(E,\rho\right) = \frac{q^2}{3\epsilon_0} \frac{\beta^3 E^4}{ \rho\ E_0^4} \end{aligned} $$
$$ \delta E_{s} = 2\pi q \rho_s R_s \dot{\mathcal{B}}_y \quad \rightarrow \quad \phi_s = \arcsin\left(2\pi\rho_s R_s\frac{\dot{\mathcal{B}}_y}{V_\mathrm{rf}}\right) $$
$$ \begin{aligned} \alpha_p = \frac{dR/R}{dp/p} = \frac{\left< D_x \right>_\rho}{R} = \frac{1}{\gamma_t^2} \approx \frac{\Delta R/R_0}{\Delta p/p_0} \approx \frac{\Delta R/R_s}{\Delta p/p_s} \end{aligned}$$
$$ \begin{aligned} \eta = -\frac{d\omega_0/\omega_0}{dp/p} = \frac{dT_0/T_0}{dp/p} = \alpha_p - \frac{1}{\gamma^2} \approx - \frac{\Delta \omega_{0,0}/\omega_{0,0}}{\Delta p/p_0} \approx - \frac{\Delta \omega_{0,s}/\omega_{0,s}}{\Delta p/p_s} \end{aligned}$$