Allowance for the Self-Magnetic Field of an Equilibrium Beam in a Betatron

P. A. Cherdantsev

(Presented by the scientific seminar of the Faculty of Physics and Technology)

The magnetic field of a betatron, without taking into account the moving electron beam, is described by the equations

\[ \begin{cases} \operatorname{rot} H = 0, & \tag{1 а}\\ \operatorname{div} H = 0. & \tag{1 б} \end{cases} \]

It is assumed here that the change of the field with time is so slow that one may use the so-called adiabatic approximation and take

\[ \frac{\partial H}{\partial t}=0. \]

In the presence of an electron beam, especially for large currents and high electron velocities, it is necessary to take into account the magnetic field of this beam. The field equations in this case may be written in the form:

\[ \begin{cases} \operatorname{rot} H = \dfrac{4\pi}{c}\, j, & \tag{2 а}\\[6pt] \operatorname{div} E = 4\pi \rho, & \tag{2 б}\\ \operatorname{rot} E = 0, & \tag{2 в}\\ \operatorname{div} H = 0. & \tag{2 г} \end{cases} \]

The second pair of equations is satisfied automatically. Starting from the first pair, we obtain an equation for the potential function \(V_\rho\) [1], taking into account the self-magnetic field of the equilibrium electron beam.

For a cylindrically symmetric betatron field, the density of the equilibrium current \(\bar{j}\) has one nonzero component \(j_\varphi\), with

\[ j_\varphi = \rho v_\varphi . \]

Substituting for \(\rho\) and \(v_\varphi\) their expressions in terms of \(V_p\) [1], namely,

\[ \rho=\frac{1}{4\pi}\,\operatorname{div} E=-\frac{1}{4\pi}\nabla^2 V_p, \]

\[ \frac{v_\varphi}{c}= \frac{ \sqrt{\left(\dfrac{eV_p}{m_0c^2}\right)^2-1} }{ \dfrac{eV_p}{m_0c^2} }. \]

Then

\[ \frac{j_\varphi}{c} = -\frac{1}{4\pi}\nabla^2 V_p\, \frac{ \sqrt{\left(\dfrac{eV_p}{m_0c^2}\right)^2-1} }{ \dfrac{eV_p}{m_0c^2} }. \]

Introducing the vector potential \(A\) into equation (2a), we obtain

\[ \nabla^2 A-\frac{A}{r^2} = \nabla^2 V_p\, \frac{ \sqrt{\left(\dfrac{eV_p}{m_0c^2}\right)^2-1} }{ \dfrac{eV_p}{m_0c^2} }. \]

Passing in the left-hand side of the equation from \(A\) to \(V_p\) by the formula

\[ A=\frac{m_0c^2}{e} \sqrt{\left(\frac{eV_p}{m_0c^2}\right)^2-1}, \]

we obtain, after introducing the dimensionless quantity

\[ U=\frac{eV_p}{m_0c^2}=\frac{mc^2}{m_0c^2}=\frac{E}{E_0} \tag{3} \]

the following equation:

\[ \frac{\nabla^2 U}{U} - \frac{1}{U^2-1} \left[ \left(\frac{\partial U}{\partial r}\right)^2 + \left(\frac{\partial U}{\partial z}\right)^2 \right] - \frac{U^2-1}{r^2} =0, \tag{4} \]

where \(E=mc^2,\quad E_0=m_0c^2\).

Equation (4) makes it possible to find the potential function

\[ V_p=\frac{m_0c^2}{e}\,U \]

with allowance for the magnetic interaction of the electrons in the beam. This interaction increases the focusing forces by changing the charge density in the beam. The function \(V_p\) characterizes the distribution of electrons in space and makes it possible to find the charge density from the formula

\[ \rho=-\frac{1}{4\pi}\nabla^2 V_p = -\frac{m_0c^2}{4\pi e}\nabla^2 U. \tag{5} \]

Let us estimate the influence of the magnetic interaction on the potential function of the magnetic field. To this end we represent \(V_p\) in the form of a sum

\[ V_p=V'_p+V''_p, \]

where \(V_p'\) is the potential function of the charge-free field, and \(V_p''\) is the potential function of the magnetic field of the beam.

We take \(V_p'\) accurate to terms quadratic in the deviations from the equilibrium radius \(R_0\), i.e.

\[ V_p'=V_0\left(1+\frac{m_0c^2V_{0m}}{eV_0^2}(1-n_0)\frac{x^2}{R_0^2} +\frac{m_0c^2}{eV_0^2}\,n_0V_{0m}\frac{z^2}{R_0^2}\right). \]

Here \(V_{0m}\) is the value of the nonrelativistic potential function at \(r=R_0\), \(z=0\) [2], and

\[ V_0=\frac{m_0c^2}{e}\sqrt{1+\frac{2eV_{0m}}{m_0c^2}}. \]

We write \(V_p''\) in the form of a series with undetermined coefficients, i.e.

\[ V_p''=V_0\left(k\frac{x^2}{R_0^2}+k'\frac{z^2}{R_0^2}\right). \]

In the last expression there is no constant term. This can be justified by the fact that at the center of the beam the magnetic-field strength of the electron beam is equal to zero. Thus,

\[ V_p=V_0\left[1+\left(\frac{m_0c^2V_{0m}}{eV_0^2}(1-n_0)+k\right)\frac{x^2}{R_0^2} +\left(\frac{m_0c^2V_{0m}}{eV_0^2}n_0+k'\right)\frac{z^2}{R_0^2}\right]. \tag{6} \]

Substitute \(V_p\) into equation (4). Then the relation between the coefficients \(k\) and \(k'\) is obtained in the form

\[ k+k'=\frac{1}{2}\left(\frac{eV_0}{m_0c^2}\right)^2-\frac{1}{2} -\frac{m_0c^2V_{0m}}{eV_0^2}. \]

If the beam cross section is sufficiently symmetric, which is obtained for \(n_0\simeq 0.5\) and large \(R_0\), then \(k\simeq k'\). In this case

\[ 2k=\frac{1}{2}\left(\frac{eV_0}{m_0c^2}\right)^2-\frac{1}{2} -\frac{m_0c^2V_{0m}}{eV_0^2}. \]

Now \(V_p\) takes the form

\[ \begin{aligned} V_p=V_0\Bigg\{1 &+\left[\frac{m_0c^2V_{0m}}{2eV_0^2}(1-2n_0) +\frac{1}{4}\left(\frac{eV_0}{m_0c^2}\right)^2-\frac{1}{4}\right]\frac{x^2}{R_0^2} \\ &+\left[\frac{m_0c^2V_{0m}}{eV_0^2}n_0 +\frac{1}{4}\left(\frac{eV_0}{m_0c^2}\right)^2-\frac{1}{4} -\frac{m_0c^2V_{0m}}{2eV_0^2}\right]\frac{z^2}{R_0^2}\Bigg\}. \end{aligned} \tag{7} \]

Thus, taking account of the beam’s own magnetic field, \(V_p\) gives a focusing force that is greater than the force without the self-field by factors \(f_r\) and \(f_z\) for the radial and \(z\)-directions, respectively, where

\[ f_r=1+\frac{keV_0^2}{m_0c^2V_{0m}(1-n_0)}, \qquad f_z=1+\frac{keV_0^2}{m_0c^2V_{0m}n_0}. \]

Let us estimate the charge density from equation (4) for \(U\).

Since

\[ \rho=-\frac{1}{4\pi}\Delta^2 V_p, \qquad V_p=\frac{m_0c^2}{e}U, \]

then

\[ \rho=-\frac{m_0c^2}{4\pi e}\nabla^2 U. \]

But from equation (4)

\[ \nabla^{2}U=-\frac{U}{U^{2}-1} \left[\left(\frac{\partial U}{\partial r}\right)^{2} +\left(\frac{\partial U}{\partial z}\right)^{2}\right] +\frac{U^{2}-1}{r^{2}}\,U . \]

At the center of the beam \(\dfrac{\partial U}{\partial r}=\dfrac{\partial U}{\partial z}=0\), therefore

\[ \nabla^{2}U_{0}=\frac{U_{0}^{2}-1}{R_{0}^{2}}\,U_{0}. \]

Here \(U_{0}\) is the value of \(U\) at the point \(r=R_{0},\ z=0\).

Substitution of \(\nabla^{2}U_{0}\) into the equation for \(\rho\) gives the charge density at the center of the equilibrium beam. We have

\[ \rho_{0}=-\frac{m_{0}c^{2}}{4\pi e R_{0}^{2}}\,(U_{0}^{2}-1)U_{0}. \tag{8} \]

Or, taking (3) into account, we obtain

\[ \rho_{0}= -\frac{m_{0}c^{2}}{4\pi e R_{0}^{2}} \left[\left(\frac{E}{E_{0}}\right)^{2}-1\right]\cdot \frac{E}{E_{0}} . \tag{9} \]

For ultrarelativistic velocities \(E\gg E_{0}\), and from (9) we obtain

\[ \rho_{\mathrm{rel}}= -\frac{m_{0}c^{2}}{4\pi e R_{0}^{2}}\cdot \left(\frac{E}{E_{0}}\right)^{3}. \tag{10} \]

For nonrelativistic velocities \(E=E_{0}+E_{\mathrm{kin}}=E_{0}+\dfrac{m_{0}v^{2}}{2}\), and

\[ \rho_{\mathrm{n.r.}}= -\frac{V_{\mathrm{o.m.}}}{2\pi e R_{0}^{2}} \left(1+\frac{1}{2}\frac{v^{2}}{c^{2}}\right). \tag{11} \]

Let us compare the charge densities computed above with allowance for the self-magnetic field with the corresponding densities found without taking it into account. Equation (1a) for a charge-free field makes it possible to obtain an equation analogous to equation (4), of the form:

\[ U\nabla^{2}U-\frac{1}{U^{2}-1} \left[\left(\frac{\partial U}{\partial r}\right)^{2} +\left(\frac{\partial U}{\partial z}\right)^{2}\right] -\frac{U^{2}-1}{r^{2}}=0. \tag{12} \]

From this equation we obtain

\[ \rho_{0}^{(0)}= -\frac{m_{0}c^{2}}{4\pi e R_{0}^{2}}\cdot \frac{U_{0}^{2}-1}{U_{0}} = -\frac{m_{0}c^{2}}{4\pi e R_{0}^{2}}\, \frac{\left(\dfrac{E}{E_{0}}\right)^{2}-1}{\dfrac{E}{E_{0}}}. \tag{13} \]

At high velocities

\[ \rho_{\mathrm{rel}}^{(0)}= -\frac{m_{0}c^{2}}{4\pi e R_{0}^{2}}\cdot \frac{E}{E_{0}} . \tag{14} \]

At low velocities

\[ \rho_{\mathrm{n.r.}}^{(0)}= -\frac{V_{\mathrm{o.m.}}}{2\pi e R_{0}^{2}} \left(1-\frac{1}{2}\frac{v^{2}}{c^{2}}\right). \tag{15} \]

Let us find the differences \(\rho_0\) and \(\rho_0^{(1)}\) for both cases. We obtain

\[ \Delta \rho_{\mathrm{orb.l}} = \rho_{\mathrm{orb.l}}-\rho_{\mathrm{orb.l}}^{(1)} = -\frac{m_0 c^2}{4\pi e R_0^2}\cdot \frac{E}{E_0} \left[\left(\frac{E}{E_0}\right)^2-1\right], \]

\[ \Delta \rho_{\mathrm{ok.l}} = \rho_{\mathrm{ok.l}}-\rho_{\mathrm{ok.l}}^{(1)} = -\frac{V_{\mathrm{om}}}{4\pi e R_0^2}\cdot \frac{v^2}{c^2}. \]

The last differences of densities show that at relativistic velocities one cannot compute the density of the equilibrium charge by a formula that does not take into account the magnetic field of the beam. In the classical case, however, for

\[ \frac{v}{c}\ll 1 \]

the difference between \(\rho_{\mathrm{ok.l}}\) and \(\rho_{\mathrm{ok.l}}^{(1)}\) is very small, and the use of formulas that do not take the charge into account is permissible.

Thus, the intrinsic magnetic field of the beam plays an essential role at high electron velocities. The charge density depends strongly on the electron energy (as \(E^3\)). This dependence may explain the behavior of the maximum radiation intensity of a betatron for different values of the injection energy \(E_i\). Indeed, for a given betatron, the volume occupied by the beam at different \(E_i\) may be regarded as constant. Then the radiation intensity is proportional to \(\rho_0\), if \(E\) is understood as the electron energy at injection. At small injection energies the beam intensity should be proportional to \(E_i\), while at large energies it should be proportional to \(E_i^2\).

For a more exact study of the density and total charge in the magnetic field of a betatron it is necessary to find a more accurate solution of equation (4), which is not a simple problem and is the subject of the author’s further investigation.

LITERATURE

1. Cherdantsev P. A. Relativistic potential function. Izvestiya TPI, vol. 87, 1957.

2. Cherdantsev P. A. On the stability of an equilibrium electron beam in a betatron. Izvestiya TPI, vol. 87, 1957.