Longitudinal space charge forces in long bunches

These notes calculate the longitudinal electric field of a long bunch within a perfectly conducting cylindrical boundary. Following Ferrario, Migliorati, and Palumbo.

General expression of the field

Start by applying Faraday’s Law, which relates the curl of the electric field to the time derivative of the magnetic field.

\[ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}. \tag{1}\]

In integral form, Faraday’s law relates the “electromotive force” around a closed loop to the change in magnetic flux through the loop:

\[ \oint \mathbf{E} \cdot d\mathbf{l} = - \frac{\partial}{\partial t} \oint \mathbf{B} \cdot d\mathbf{A}. \tag{2}\]

Choose a rectangular loop with one edge running parallel to the beam for a distance \(\Delta z\) at radius \(r\), and the opposite edge at radius \(b\) (at the pipe boundary). For a long, axially symmetric charge distribution, Faraday’s law gives

\[ \left( E_z(r, z) - E_z(b, z) \right) \Delta z + \int_{r}^{b} \left( E_r(r, z + \Delta z) - E_r(r, z) \right) dr = -\Delta z \frac{\partial}{\partial t} \int_{r}^{b} B_\phi(r) dr. \tag{3}\]

Taking the limit \(\Delta z \rightarrow 0\), we can define

\[ \frac{\partial E_r(r, z)}{\partial z} = \frac{E_r(r, z + \Delta z) - E_r(r, z)}{\Delta z}. \tag{4}\]

Substituting Equation 4 into Equation 3 gives

\[ E_z(r, z) = E_z(b, z) - \int_{r}^{b} \left( \frac{\partial E_r(r, z)}{\partial z} + \frac{\partial B_\phi(r, z)}{\partial t} \right) dr . \tag{5}\]

Equation 5 relates the longitudinal electric field at radius \(r\) to the longitudinal field at radius \(b\) (the pipe boundary), as well as the averaged transverse electric and magnetic fields between \(r\) and \(b\).

Let’s do a bit more work to simplify this expression. First define \(z = -\beta c t\), where \(c\) is the speed of light, so that \(\partial / \partial t = -\beta c \partial / \partial z\) and

\[ E_z(r, z) = E_z(b, z) - \frac{\partial}{\partial z} \int_{r}^{b} \left( E_r(r, z) - \beta c B_\phi(r, z) \right) dr . \tag{6}\]

Next, relate \(B_\phi\) to \(E_r\):

\[ B_\phi(r, z) = \frac{\beta}{c} E_r(r, z). \]

Equation 6 simplifies to

\[ E_z(r, z) = E_z(b, z) - \frac{1}{\gamma^2} \frac{\partial}{\partial z} \int_{r}^{b} E_r(r, z) dr , \tag{7}\]

where \(\gamma = (1 - \beta^2)^{-1/2}\) is the Lorentz factor. Finally, since the boundary at \(r = b\) is perfectly conducting, \(E_z(b, z) = 0\) and

\[ E_z(r, z) = - \frac{1}{\gamma^2} \frac{\partial}{\partial z} \int_{r}^{b} E_r(r, z) dr , \tag{8}\]

Exact formula for uniform transverse density

To calculate the longitudinal field \(E_z(r, z)\) in Equation 8, we first need to calculate the transverse field \(E_r(r, z)\) and then average the field over the radial coordinate. We could do this numerically by solving the two-dimensional Poisson equation within each \(z\) slice. But there’s a special case that admits an exact solution.

Assume the bunch has line-charge density \(\lambda(r, z)\) [C / m]. As long as the density varies slowly along \(z\), we can approximate the electric field at fixed \(z\) as purely transverse, i.e., generated by a set of infinite line charges (threads). We’ll apply Gauss’s Law to relate the charge density to the field. Gauss’s Law says that the electric flux through a closed surface is proportional to the charge enclosed by the surface:

\[ \oint \mathbf{E} \cdot d\mathbf{A} = Q_{enc} / \epsilon_0 . \tag{9}\]

Choose the surface to be a cylinder of radius \(r\) and length \(\ell\).

\[ \begin{align} E_r(r, z) \cdot 2 \pi r l &= \frac{\lambda(r, z) l}{\epsilon_0} , \\ E_r(r, z) &= \frac{\lambda(r, z)}{2 \pi \epsilon_0 r} . \\ \end{align} \tag{10}\]

Substituting into Equation 8 gives \[ E_z(r, z) = - \frac{1}{\gamma^2} \frac{\partial}{\partial z} \int_{r}^{b} \frac{\lambda(r, z)}{2 \pi \epsilon_0 r} dr , \tag{11}\]

Note that the field scales inversely with \(\gamma^2\), so it vanishes at high energies. (This is maybe less intuitive than the transverse case.)

The integral in Equation 11 can be evaluated exactly if the transverse density is uniform within radius \(r\) and zero outside:

\[ \lambda(r, z) = \lambda_0(z) (r / a)^2. \tag{12}\]

Note that \(\lambda\) is the line density, which is nonzero at \(r > a\). The transverse field becomes

\[ \begin{numcases} {E_r(r, z)=} \dfrac{\lambda_0(z)}{2 \pi \epsilon_0 a^2} r, & $r \le a$ \\\\ \dfrac{\lambda_0(z)}{2 \pi \epsilon_0 r } , & $r \ge a$. \end{numcases} \tag{13}\]

The radial averaging gives a field proportaionl to the density gradient:

\[ \begin{align} E_z(r, z) &= -\frac{1}{4 \pi \epsilon_0 \gamma^2} \left[ 1 - \frac{r^2}{a^2} + 2 \ln{(b / a)} \right] \frac{\partial \lambda_0(z)}{\partial z} \\ &= -\frac{g(r)}{4 \pi \epsilon_0 \gamma^2} \frac{\partial \lambda_0(z)}{\partial z}, \end{align} \tag{14}\]

where \(g(r; a, b)\) captures the dependence on the transverse distribution:

\[ g(r; a, b) = 1 - \frac{r^2}{a^2} + 2 \ln{(b / a)} \]

This model of longitudinal space charge is often called a g-factor model. It can be a good model in accelerator rings, where bunches are typically very long compared to the transverse dimensions.