JUAS 2023
Course 1: The science of particle accelerators
A. Lasheen
$$ \begin{aligned} & \overrightarrow{\nabla}\cdot\overrightarrow{\mathcal{E}} = \frac{\rho_q}{\epsilon_0} && \text{Gauss' law} \\ & \overrightarrow{\nabla}\cdot\overrightarrow{\mathcal{B}} = 0 && \text{Flux/Thomson's law} \\ & \overrightarrow{\nabla}\times\overrightarrow{\mathcal{E}} = -\frac{\partial\overrightarrow{\mathcal{B}}}{\partial t} && \text{Faraday's law} \\ & \overrightarrow{\nabla}\times\overrightarrow{\mathcal{B}} = \mu_0\left(\overrightarrow{\mathcal{j}}+\epsilon_0\frac{\partial\overrightarrow{\mathcal{E}}}{\partial t}\right) && \text{Ampère's law} \end{aligned} $$
$\epsilon_0$ Vacuum permittivity , $\mu_0$ Vacuum permeability
$\rho_q$ Charge density, $\overrightarrow{j}$ Current density
$$ \begin{aligned} & \oiint_{ S}{\overrightarrow{\mathcal{E}}\cdot\overrightarrow{d S}} = \frac{1}{\epsilon_0} \iiint{\rho_q d V} && \text{Gauss' law} \\ & \oiint_{ S}{\overrightarrow{\mathcal{B}}\cdot\overrightarrow{d S}} = 0 && \text{Flux/Thomson's law}\\ & \oint_{ C}{\overrightarrow{\mathcal{E}}\cdot\overrightarrow{d z}} = -\frac{d}{dt}\iint_{ S}{\overrightarrow{\mathcal{B}}\cdot\overrightarrow{d S}} && \text{Faraday's law} \\ & \oint_{ C}{\overrightarrow{\mathcal{B}}\cdot\overrightarrow{d z}} = \mu_0\iint_{ S}{\overrightarrow{\mathcal{j}}\cdot\overrightarrow{d S}}+\mu_0\epsilon_0\iint_{ S}{\frac{\partial\overrightarrow{\mathcal{E}}}{\partial t}\cdot\overrightarrow{d S}} && \text{Ampère's law} \end{aligned} $$
$dz$ Line element, $dS$ Surface element, $dV$ Volume element
$$ \left< P_b \right> = \frac{N_{p} E_\mathrm{acc}}{T_\mathrm{acc}} $$
$$ \oint_{ C}{\overrightarrow{\mathcal{E}}\cdot\overrightarrow{d z}} = 0 $$
$$ \begin{aligned} \mathcal{B}_y\left(\rho_0\right) & = \frac{1}{2} \frac{\Phi_{S,\rho_0}}{\pi\rho_0^2} = \frac{1}{2} \left< \mathcal{B}_y\right >_{S,\rho_0} \end{aligned} $$
We will introduce the magetic flux $\Phi_{S,\rho_0}$ and an averaged magnetic field in the betatron core $\left< \mathcal{B}_y\right >_{S,\rho_0}$
$$ \begin{aligned} \Phi_{S,\rho_0} = 2\pi \int_0^{\rho_0}{\mathcal{B}_y\left(\rho\right) \rho\ d\rho} = \pi\rho_0^2 \left< \mathcal{B}_y\right >_{S,\rho_0} \end{aligned} $$
$$ \mathcal{B}_y \rho = \frac{p_{\theta}}{q}= \frac{p}{q} $$
$$ \begin{aligned} \int_{0}^{2\pi}{\mathcal{E}_{\theta}\ \rho\ d\theta} & = \frac{d}{d t} \int_0^{2\pi}\int_0^{\rho_0}{\mathcal{B}_y\left(\rho,t\right) \rho\ d\rho\ d\theta} \\ \implies 2\pi\rho_0 \mathcal{E}_{\theta} & = \frac{ d \Phi_{S,\rho_0}}{ d t} \\ \implies \mathcal{E}_{\theta} & = \frac{1}{2\pi\rho_0}\frac{ d \Phi_{S,\rho_0}}{ d t} \end{aligned} $$
$$ \begin{aligned} \frac{d p_\theta}{dt} & = q \mathcal{E}_{\theta} = \frac{q}{2\pi\rho_0}\frac{ d \Phi_{S,\rho_0}}{ d t} \\ \implies p_\theta & = \frac{q}{2\pi\rho_0} \Phi_{S,\rho_0} \end{aligned} $$
$$ \begin{aligned} q \mathcal{B}_y\left(\rho_0\right) \rho_0 & = \frac{q}{2\pi\rho_0} \Phi_{S,\rho_0} \\ \implies \mathcal{B}_y\left(\rho_0\right) & = \frac{1}{2} \frac{\Phi_{S,\rho_0}}{\pi\rho_0^2} \end{aligned} $$
We introduce an averaged magnetic field in the betatron core $\left< \mathcal{B}_y\right >_{S,\rho_0}$
$$ \begin{aligned} & \overrightarrow{\nabla}\cdot\overrightarrow{\mathcal{E}} = 0 && \text{Gauss' law} \\ & \overrightarrow{\nabla}\cdot\overrightarrow{\mathcal{B}} = 0 && \text{Flux/Thomson's law} \\ & \overrightarrow{\nabla}\times\overrightarrow{\mathcal{E}} = -\frac{\partial\overrightarrow{\mathcal{B}}}{\partial t} && \text{Faraday's law} \\ & \overrightarrow{\nabla}\times\overrightarrow{\mathcal{B}} = \mu_0\epsilon_0\frac{\partial\overrightarrow{\mathcal{E}}}{\partial t} && \text{Ampère's law} \end{aligned} $$
$$ \begin{aligned} \Delta \overrightarrow{\mathcal{E}} - \frac{1}{c^2} \frac{\partial^2\overrightarrow{\mathcal{E}}}{\partial t^2} = 0 \quad , \left(c=\frac{1}{\sqrt{\mu_0\epsilon_0}}\right) \end{aligned} $$
$$ \begin{aligned} \overrightarrow{\nabla}\times\overrightarrow{\mathcal{E}} & = -\frac{\partial\overrightarrow{\mathcal{B}}}{\partial t} \\ \implies \overrightarrow{\nabla}\times\left(\overrightarrow{\nabla}\times\overrightarrow{\mathcal{E}}\right) & = -\overrightarrow{\nabla}\times\left(\frac{\partial\overrightarrow{\mathcal{B}}}{\partial t}\right) \\ \implies \overrightarrow{\nabla}\left(\overrightarrow{\nabla}\cdot\overrightarrow{\mathcal{E}}\right) - \overrightarrow{\nabla}^2 \overrightarrow{\mathcal{E}} & = -\frac{\partial}{\partial t}\left(\overrightarrow{\nabla}\times\overrightarrow{\mathcal{B}}\right) \\ \implies \Delta \overrightarrow{\mathcal{E}} -\mu_0\epsilon_0 \frac{\partial^2 \overrightarrow{\mathcal{E}}}{\partial t^2} & = 0 \quad , \left(c=\frac{1}{\sqrt{\mu_0\epsilon_0}}\right) \end{aligned} $$
$$ \overrightarrow{\mathcal{E}} = \mathcal{E}_z\left(\rho,z\right) \cos\left(\omega_r t\right)\vec{e_z} $$
where $\omega_r=2\pi f_r$ is the (angular) frequency of the field and depends on the geometry of the cavity.
$$ \begin{aligned} \delta E_\mathrm{rf} & = \int{q \mathcal{E}_z \left(\rho,z,t\right) dz} \\ & = q V_{\mathrm{rf}}\left(\rho, \tau\right) \end{aligned} $$
where $V_{\mathrm{rf}}$ is the total accelerating potential of a particle arriving at a time $\tau$ in the cavity (we will derive a relevant expression of $V_{\mathrm{rf}}$ during the next lesson!).
$$ \begin{aligned} v_{\rho} & = \dot{\rho} \\ v_{\theta} & = \rho\dot{\theta} = \rho\omega \\ v_y & = \dot{y} \end{aligned}$$
$$ \begin{aligned} \dot{p_{\rho}} & = m\left(\ddot{\rho}-\rho\dot{\theta}^2\right) \\ \dot{p_{\theta}} & = m \left(\rho\ddot{\theta}+2\dot{\rho}\dot{\theta}\right) \\ \dot{p_y} & = m\ \ddot{y} \end{aligned}$$
$$ \begin{aligned} \frac{d\vec{p}}{dt}=\vec{F_{\mathcal{B}}} & = q\left(\vec{v}\times\vec{\mathcal{B}}\right) \\ \text{in}\ \vec{e_{\rho}} \implies m \left(\ddot{\rho}-\rho\dot{\theta}^2\right) & = - q v_{\theta} \mathcal{B}_y \quad ,\left(\dot{\rho}=0\right) \\ \implies m \frac{v_{\theta}^2}{\rho} & = q v_{\theta} \mathcal{B}_y \quad ,\left(v_{\theta}=\rho\dot{\theta}\right) \\ \implies p_{\theta} & = q \mathcal{B}_y \rho \\ \implies \mathcal{B}_y \rho & = \frac{p_{\theta}}{q}= \frac{p}{q} \end{aligned} $$
In the next lessons, we will focus on the synchrotron design.
The maximum energy of a circular accelerator is in principle limited by the maximum $\mathcal{B}_y$ in the bending magnets or the radial size of the accelerator (e.g. FCC 100km!). The typical range for bending magnetic field is $\sim\mathcal{O}\left(\text{1-10 T}\right)$.
In presence of synchrotron radiation (for lepton machines), the maximum energy is limited by RF power.
$$\begin{aligned} E^2 & = P^2+E_0^2 \\ \implies d\left(E^2\right) & = d\left(P^2\right)+d\left(E_0^2\right) \\ \implies 2 E dE & = 2 P dP = 2 p dp c^2 \\ \implies \frac{dE}{dp} & = \frac{p c^2}{E} \\ \implies \frac{dE}{dp} & = \beta c = v \\ \end{aligned}$$
$$\begin{aligned} E dE & = p dp c^2 \\ \implies \frac{dE}{E} & = \frac{p c^2}{E^2} dp \\ \implies \frac{dE}{E} & = \left(\frac{pc}{E}\right)^2 \frac{dp}{p} \\ \implies \frac{dE}{E} & = \beta^2 \frac{dp}{p} \\ \implies \frac{dp}{p} & = \frac{1}{\beta^2}\frac{dE}{E} = \frac{1}{\beta^2}\frac{d\gamma}{\gamma} \\ \end{aligned}$$
$$\begin{aligned} \beta^2 & = 1 - \frac{1}{\gamma^2} \\ \implies d\left(\beta^2\right) & = d\left(1 - \frac{1}{\gamma^2}\right) \\ \implies 2\beta d\beta & = 2\gamma^{-3} d\gamma \\ \implies \frac{d\beta}{\beta} & = \left(\frac{1}{\beta\gamma}\right)^2 \frac{d\gamma}{\gamma} \\ \implies \frac{d\beta}{\beta} & = \frac{1}{\gamma^2} \frac{dp}{p} \\ \end{aligned}$$
$$ \delta E = \int{q \mathcal{E}_z \left(\rho,z,t\right) dz} = q V_{\mathrm{rf}}\left(\rho, \tau\right) $$
$$ \mathcal{B}_y\rho = \frac{p}{q} \quad \rightarrow \quad p\left[\mathrm{GeV/c}\right] \approx 0.3\ Z\ \mathcal{B}_y \left[\mathrm{T}\right] \rho \left[\mathrm{m}\right] $$