Accounting for Electron Interaction in Problems of Electron Optics

B. N. Rodimov

(Presented by the scientific seminar of the Physico-Technical Faculty)

Derivation of the density equation

We shall call a focusing system a homogeneous system with focusing axis \(z\) if the action of the magnetic or electrostatic field forming the system can be replaced by the action of the electric field of an infinite cylindrical beam of positive charges distributed with a definite density about the focusing axis. The density of the beam \(\rho_{\mathrm{eqv}}\) (the number of charges in \(1\ \mathrm{cm}^3\)) is a function of all coordinates except \(z\). The system will be inhomogeneous if \(\rho_{\mathrm{eqv}}\) is also a function of the coordinate \(z\).

In a homogeneous focusing system one may conceive of an equilibrium electron beam moving along the \(z\) axis, in which, for each electron, the repulsive forces between the electrons are balanced by the focusing forces of the system. The density of such a beam will be denoted by \(\rho_{\mathrm{eq}}\). By definition, for a homogeneous system \(\rho_{\mathrm{eqv}}=\rho_{\mathrm{eq}}\) at any point of the system.

The equations for a beam of electrons moving in a focusing system, for the stationary case, may be written in the form:

\[ \frac{d\rho}{dt} = (\vec v \nabla)\rho = -\rho(\nabla \vec v), \]

\[ \frac{d\vec v}{dt} = (\vec v \nabla)\vec v = -\frac{e}{m}\vec E, \tag{1} \]

\[ \nabla \vec E = \nabla \vec E_{\mathrm{self}}+\nabla \vec E_{\mathrm{foc}} = -4\pi e(\rho_{\mathrm{self}}-\rho_{\mathrm{eqv}}), \]

where \(\vec E_{\mathrm{self}}\) is the intensity of the electrons’ own field and \(\vec E_{\mathrm{foc}}\) is the intensity of the focusing field (the equivalent intensity if the focusing field is a magnetic field).

Introducing the notation \(L=(\vec v\nabla)\), we can obtain the expression

\[ \frac{d^2\rho}{dt^2} = -\nabla\vec v\cdot L\rho-\rho\cdot L(\nabla\vec v), \tag{2} \]

which, if the preceding relations are used, gives:

\[ \frac{d^{2}\rho}{dt^{2}}-\frac{2}{\rho}\left(\frac{d\rho}{dt}\right)^{2} +\frac{4\pi e^{2}}{m}\rho(\rho-\rho_{\mathrm{eq}})+ \]

\[ +2\rho\left[\left(\frac{\partial v_x}{\partial x}\cdot \frac{\partial v_y}{\partial y} +\frac{\partial v_x}{\partial x}\cdot \frac{\partial v_z}{\partial z} +\frac{\partial v_y}{\partial y}\cdot \frac{\partial v_z}{\partial z}\right)-\right. \]

\[ \left.-\left(\frac{\partial v_x}{\partial y}\cdot \frac{\partial v_y}{\partial x} +\frac{\partial v_x}{\partial z}\cdot \frac{\partial v_z}{\partial x} +\frac{\partial v_y}{\partial z}\cdot \frac{\partial v_z}{\partial y}\right)\right]=0. \tag{3} \]

In certain special cases the last term of equation (3) can be determined. The equation obtained in this case,

\[ \frac{d^{2}\rho}{dt^{2}}-\frac{2}{\rho}\left(\frac{d\rho}{dt}\right)^{2} +\frac{4\pi e^{2}}{m}\rho(\rho-\rho_{\mathrm{eq}})+f(t)=0 \tag{4} \]

will be called the density equation of the electron beam.

Having determined the function \(f(t)\), from the density equation we find \(\rho(t)\). Knowledge of \(\rho(t)\) makes it possible to determine the configuration of the electron beam in space. Below, the application of the method of the beam-density equation will be shown on elementary examples.

Case of longitudinal interaction of electrons in a beam

We consider a beam of electrons moving along the \(z\)-axis. We assume that the electrons interact with one another only in the direction of motion, i.e., that there are no forces acting in the transverse direction. Then the initial equations will have the form

\[ \frac{d\rho}{dt}=v_z\frac{\partial\rho}{\partial z} =-\rho\frac{\partial v_z}{\partial z}, \]

\[ \frac{dv_z}{dt}=v_z\frac{\partial v_z}{\partial z} =-\frac{e}{m}E_z, \tag{5} \]

\[ \frac{\partial E_z}{\partial z} =-4\pi e(\rho-\rho_{\mathrm{eq}}). \]

Hence we obtain

\[ \frac{d^{2}\rho}{dt^{2}} -\frac{2}{\rho}\left(\frac{d\rho}{dt}\right)^{2} +\frac{4\pi e^{2}}{m}\rho(\rho-\rho_{\mathrm{eq}})=0. \tag{6} \]

The function \(f(t)\) for the given case is equal to zero. As an example for this case one may consider the problem of an electron beam between two plane electrodes of infinite extent. For this case

\[ E_{\mathrm{ext}}=\mathrm{const};\quad \rho_{\mathrm{eq}}=0;\quad \frac{\partial E_{\mathrm{self}}}{\partial x}=-4\pi e\rho. \tag{7} \]

The density equation will have the form:

\[ \frac{d^{2}\rho}{dt^{2}}-\frac{2}{\rho}\left(\frac{d\rho}{dt}\right)^{2} +\frac{4\pi e^{2}}{m}\rho^{2}=0. \tag{8} \]

Its solution is

\[ \rho=\frac{1}{\dfrac{2\pi e^{2}}{m}\,t^{2}+C_{1}t+C_{2}}. \tag{9} \]

Since \(i=e\rho v=\mathrm{const}\), at \(t=0\) the density \(\rho\) must be equal to infinity and \(v=0\). Thus, \(C_{2}=0\). From the condition \(\left(\dfrac{d\rho}{dt}\right)_{0}=0\) we obtain \(C_{1}=0\).

Finally,

\[ \rho=\frac{1}{\dfrac{2\pi e^{2}}{m}\,t^{2}}. \tag{10} \]

Next, from the condition of constancy of the current density we obtain

\[ v=\frac{2\pi e}{m}\,it^{2}, \]

which gives

\[ l=\frac{2}{3}\cdot\frac{\pi e}{m}\,it^{3}. \tag{11} \]

On the other hand,

\[ \frac{dE}{dt}=v\frac{\partial E}{\partial x} =-4\pi e\rho v=-4\pi i;\qquad E=-4\pi it=-\frac{\partial \varphi}{\partial x}, \]

whence

\[ \varphi=\frac{2\pi^{2}}{m}\,i^{2}\cdot e\cdot t^{4}, \tag{12} \]

where \(\varphi\) is the potential at a distance \(l\) from the cathode. Eliminating \(t\) from expressions (11) and (12), we find the known relation between the current density and the anode potential

\[ i=\sqrt{\frac{2e}{m}}\cdot\frac{\varphi_{a}^{3/2}}{9\pi l^{2}}. \tag{13} \]

The case of transverse interaction of electrons in a beam

Let us consider a beam with cylindrical symmetry and assume that the electrons interact only in the direction transverse to the motion of the beam, i.e., in the direction of the \(z\)-axis there are no accelerating or retarding forces.

If we introduce the notation \(r v_r = X\) and \(r E_r = Y\) and the new independent variable \(x = r^2/2\), then the original equations will have the form

\[ \frac{d \rho}{dt} = v_z \frac{\partial \rho}{\partial z} = \rho \cdot \frac{1}{r} \cdot \frac{\partial}{\partial r}(r v_r) = -\rho \frac{\partial X}{\partial x}, \tag{14} \]

\[ \frac{d v_r}{dt} = v_r \frac{\partial v_r}{\partial r} = X \frac{\partial v_r}{\partial x}, \tag{15} \]

\[ -\frac{1}{r}\cdot \frac{\partial}{\partial r}(r E_r) = -4\pi e(\rho-\rho_{\mathrm{ion}}) = -\frac{\partial Y}{\partial x}. \tag{16} \]

Differentiating the first equation and using the relation

\[ \frac{dX}{dt} - v_r^2 = X \frac{\partial X}{\partial x} - v_r^2 = -\frac{e}{m}Y, \tag{17} \]

we obtain the density equation for a cylindrical beam:

\[ \frac{d^2 \rho}{dt^2} - \frac{2}{\rho}\left(\frac{d\rho}{dt}\right)^2 + \frac{4\pi e^2}{m}\rho(\rho-\rho_{\mathrm{ion}}) + \frac{2\rho}{r}\cdot v_r \cdot \frac{\partial v_r}{\partial r} = 0. \tag{18} \]

Since the last term contains the quantity \(v_r\), which can be found once \(\rho(t)\) is already known, for a first approximation we take the value of this term for the case of noninteracting electrons. In this case it is equal to zero. Having found \(\rho(t)\) from the resulting equation and using the relation

\[ \rho r^2 = \rho_0 r_0^2, \tag{19} \]

we can find the beam radius at any instant of time and the corresponding values of \(v_r\) and \(\partial v_r/\partial r\). Substituting, as the last term in equation (18), the expression obtained from the first approximation, we can solve equation (18) in the second approximation. In the same way the third approximation is obtained, and so on.

As an example, let us consider the divergence of a straight free electron beam. For the moment we shall restrict ourselves to the first approximation. The density equation reduces to the equation

\[ \frac{d^2 \rho}{dt^2} - \frac{2}{\rho}\left(\frac{d\rho}{dt}\right)^2 + \frac{4\pi e^2}{m}\rho^2 = 0. \tag{21} \]

Replacing \(\rho\) in this equation by \(1/u\), we obtain for \(u\) the equation

\[ \frac{d^2 u}{dt^2} = \frac{4\pi e^2}{m}. \tag{22} \]

Its solution is

\[ u = \frac{2\pi e^2}{m}\, t^2 + C_1 t + C_2. \]

Or, returning to \(\rho\),

\[ \rho = \frac{1}{ \frac{2\pi e^2}{m}t^2 + C_1 t + C_2 }. \tag{23} \]

The constants \(C_1\) and \(C_2\) are determined from the initial conditions at \(t=0\), when, for example,

\[ \rho(0)=\rho_0,\qquad \left(\frac{d\rho}{dt}\right)_0=0. \tag{24} \]

In this case

\[ \rho=\frac{\rho_0}{\dfrac{2\pi e^2}{m}\rho_0 t^2+1}. \tag{25} \]

Knowing \(\rho\), one can find the trajectories of the outer electrons of the beam. If the beam density in the cross section is assumed constant, we obtain for the beam radius in any cross section

\[ \rho r^2=\rho_0 r_0^2, \]

whence

\[ \frac{r^2}{r_0^2}=1+\frac{2\pi e^2}{m}\rho_0 t^2. \tag{26} \]

For small values of the term in comparison with unity, we obtain

\[ \frac{r}{r_0}=1+\frac{\pi e^2}{m}\rho_0 t^2. \tag{27} \]

Replacing the beam density by the current density according to the formula \(\rho_0=\dfrac{i_0}{e v_0}\) and the time by \(t=\dfrac{z}{v_0}\), we shall have

\[ \frac{r}{r_0}=1+2.41\cdot 10^4\,\frac{i_0 z^2}{V_0^{3/2}}, \tag{28} \]

where \(z\) is expressed in cm, \(i_0\)—in \(\dfrac{a}{cm^2}\), and \(V_0\)—in volts. The same formula is also obtained by the usual method under the condition that the beam density over the cross section is constant and the velocity in the \(z\) direction does not change \([1, 3]\).

Allowance for simultaneous longitudinal and transverse electron interactions

To allow for longitudinal and transverse electron interactions in a beam, one may represent the beam density as the geometric mean of the density taking into account only the transverse interaction and the density taking into account only the longitudinal interaction, i.e.

\[ \rho=\sqrt{\rho_+\cdot\rho_-}. \tag{29} \]

In the approximation in which the “transverse” density satisfies the same equation as the “longitudinal” one, allowance for the existence of transverse and longitudinal interactions reduces only to estimating the particle velocity from the direction of the \(z\) axis. When only the transverse interaction is taken into account, these

the velocity is constant and equal to \(v_{z0}=\dfrac{z}{t}\). If only the longitudinal interaction is taken into account, this velocity is found from the equation

\[ \frac{d^{2}v_z}{dt^{2}}=\frac{4\pi e^{2}}{m}\,(\rho-\rho_{\text{eqv}})\,v_z , \tag{30} \]

which is solved by expansion in a series, if the time dependence of \((\rho-\rho_{\text{eqv}})\) is known. The simultaneous existence of longitudinal and transverse interactions can be taken into account by calculating the distance at which the beam density has the given value. When only the transverse interaction is taken into account, this distance is \(z_{+}=v_{z0}\,t\). When only the longitudinal interaction is taken into account, this distance is \(z_{-}=\int_{0}^{t} v_z dt\). When both interactions are taken into account, one may take

\[ z=\sqrt{z_{+}z_{-}}. \tag{31} \]

For the longitudinal interaction of a free electron beam without an external field, from equation (30) we obtain, in the first approximation,

\[ v_z=v_{z0}\left(1+\frac{2\pi e^{2}}{m}\rho_0 t^{2}\right). \tag{32} \]

The density for a straight beam, when the longitudinal interaction is taken into account, will be expressed by the previous formula (20). Only the time \(t\) through the distance \(z\) must be replaced from the formula

\[ z \simeq v_{z0}\,t\left(1+\frac{\pi e^{2}\rho_0 t^{2}}{3m v_{z0}}\right). \tag{33} \]

It is clear that for small \(\rho_0\) or \(t\), or for large velocities \(v_{z0}\), there is no point in taking the longitudinal interaction into account.

Electron Beam in a Focusing Field

As an example, let us take an electrostatic lens described by a potential function of the form [2]

\[ \varphi(z,r)=p\left(z^{2}-\frac{r^{2}}{2}\right). \tag{34} \]

To begin with, we neglect the dependence of \(\varphi\) on \(z\), i.e., the forces in the direction of the \(z\)-axis. The equivalent density is calculated from the expression

\[ \frac{\partial^{2}\varphi}{\partial r^{2}}+\frac{1}{r}\,\frac{\partial\varphi}{\partial r}=-4\pi e\rho_{\text{eqv}}. \tag{35} \]

We obtain

\[ \rho_{\text{eqv}}=-\frac{p}{2\pi e}. \tag{36} \]

The solution of the density equation gives:

\[ \rho_{+}= \frac{\rho_{\mathrm{eqv}}^{+}} {1-\left(1-\dfrac{\rho_{\mathrm{eqv}}^{+}}{\rho_{0}}\right) \cos \sqrt{\dfrac{4\pi e^{2}}{m}\rho_{\mathrm{eqv}}^{+}\,t}} \tag{37} \]

under the initial conditions \(\rho(0)=\rho_{0}\) and \(\left(\dfrac{d\rho}{dt}\right)_{0}=0\).

If only the longitudinal interaction is considered, we obtain:

\[ \frac{\partial^{2}\varphi}{\partial z^{2}}=-4\pi e\,\rho_{\mathrm{eqv}} \tag{38} \]

and

\[ \rho_{\mathrm{eqv}}=-\frac{p}{2\pi e}=-\left|\rho_{\mathrm{eqv}}^{+}\right|. \tag{39} \]

The solution of the density equation will have the form:

\[ \rho_{-}= \frac{2\left|\rho_{\mathrm{eqv}}^{+}\right|} {e^{\sqrt{b_{1}t+b_{2}}}+e^{-\sqrt{b_{1}t}}-2}, \tag{40} \]

where

\[ b_{1}=\frac{4\pi e^{2}}{m}\left|\rho_{\mathrm{eqv}}^{+}\right|. \]

The constant \(b_{2}\) has the value \(\ln\left(1+\dfrac{\left|\rho_{\mathrm{eqv}}^{+}\right|}{\rho_{0}}\right)\). The initial conditions are the same, i.e. \(\rho(0)=\rho_{0}\) and \(\left(\dfrac{d\rho}{dt}\right)_{0}=0\).

The “total” density will be expressed rather complicatedly:

\[ \rho= \frac{\sqrt{2}\left|\rho_{\mathrm{eqv}}^{+}\right|} {\left[e^{\sqrt{b_{1}t+b_{2}}}+e^{-\sqrt{b_{1}t}}-2\right]^{1/2} \left[1-\left(1-\dfrac{\left|\rho_{\mathrm{eqv}}^{+}\right|}{\rho_{0}}\right) \cos \sqrt{b_{1}t}\right]^{1/2}} \tag{41} \]

For \(\dfrac{r}{r_{0}}\) one obtains the equation

\[ \frac{r^{4}}{r_{0}^{4}}= \frac{\rho_{0}^{2}}{2\left|\rho_{\mathrm{eqv}}^{+}\right|^{2}} \left[ e^{\sqrt{b_{1}\dfrac{z}{v_{0}}+b_{2}}} + e^{-\sqrt{b_{1}\dfrac{z}{v_{0}}}} -2 \right] \times \]

\[ \times \left[ 1-\left(1-\frac{\left|\rho_{\mathrm{eqv}}^{+}\right|}{\rho_{0}}\right) \cos \sqrt{b_{1}\frac{z}{v_{0}}} \right]. \tag{42} \]

Here it has been set that

\[ t=\frac{z}{v_{0}}. \]

Passage of an Electron Beam through a Diverging Electron Lens

Let us consider an electron lens formed by a cloud of electrons with a density along the \(z\)-axis expressed by the formula

\[ \rho_{\text{lens}} = n \cdot \rho_0 e^{-\frac{(z-z_l)^2}{\alpha^2}}, \tag{43} \]

where \(\rho_0\) is the initial density of the beam, and \(n\) is a number indicating how many times the greatest density of the electron cloud forming the lens exceeds the initial density of the electron beam. The density equation for this case reduces to

\[ \frac{d^2 u}{dt^2} - \frac{4\pi e^2}{m}\rho_{\text{lens}}\cdot u = \frac{4\pi e^2}{m}, \tag{44} \]

where \(\rho = \dfrac{1}{u}\). The equation is solved by the method of successive approximations. In the second approximation we obtain for \(u\) (under the condition that \(\left(\dfrac{du}{dt}\right)_0 = 0\)) the expression

\[ u = u_0 + \frac{a t^2}{2} + a u_0 \int_0^t \varphi\,dt. \tag{45} \]

For the density itself we shall have

\[ \rho = \frac{\rho_0}{ 1 + \dfrac{a\rho_0 t^2}{2} + a\displaystyle\int_0^t \varphi\,dt }. \tag{46} \]

Here

\[ a = \frac{4\pi e^2}{m}, \qquad \varphi = n\rho_0 \int_0^t e^{-\frac{v_0^2}{\alpha^2}(t-t_0)^2}\,dt. \]

Further, in the usual way, we find the expression for the beam radii. The method described may prove useful in calculating electron guns for large currents.

Literature

1. E. Brüche and Recknagel, Electron Devices, Gosenergoizdat, Moscow, 1949.
2. B. A. Vainrib, V. I. Milgotin, Electron Optics, Gosenergoizdat, Moscow, 1951.
3. M. D. Gabovich, UFN, vol. LVI, no. 2, 1955.