Solution

In August 1931 C. J. Davisson and C. J. Calbick of Bell Labs reported that a hole in a charged plate between two uniform fields gathers electrons like a lens [1]. A year later they corrected the factor, and for a round hole $f = 4V/(E_2 - E_1)$ [2]. This is formula (12) below, while their earlier $2V$ holds for a slit (14). Davisson had by then discovered electron diffraction, for which he shared the 1937 Nobel Prize.

In the same months M. Knoll and E. Ruska at the high-voltage laboratory of the Technische Hochschule Berlin were building geometrical electron optics [3, 4]. In a note added in proof they cite the Davisson and Calbick abstract [3, p. 625], and in the second part Ruska describes exactly the effect of this problem, writing about mesh electrodes [4, p. 654]

Es werden daher Strahlen, die in großer Nähe der Drähte durch die Netzelektroden hindurchgehen, stärker abgelenkt, als Strahlen, die das Netz mehr in der Mitte einer Masche passieren.

that is, rays near the wires are deflected more than rays through the middle of a cell. This is (11). Their first microscope focused the beam with magnetic coils, after an idea of H. Busch, not with the electrostatic field of charged wires as here. For electron optics and the electron microscope Ruska received the 1986 Nobel Prize in Physics, 55 years after this work [5].

A real mesh is more complicated than charged wires. Read and co-workers computed meshes of round wires and of strips and found that the effective potential of the mesh plane differs from the potential of the wires themselves, which changes the field far away on both sides [6, 7]. The potential of a grating of parallel line charges, one of the two families of wires in (5), (6), is written out in their appendix [6], and Williams, Read and Bowring treat the defocusing of a beam by the cells [8]. Today the lens (12) sits in every cell of the mesh in the wide-angle electrostatic lenses of photoelectron spectrometers [9]. A mesh in the beam corrects spherical aberration, as O. Scherzer proposed [10], and an ellipsoidal mesh accepts electrons up to $\pm 60^\circ$. The price is the blur that every cell adds, with different focal lengths in the tangential and sagittal planes at oblique incidence [9].

Here the electrons fly through vacuum independently, and the mesh imposes the period $d$ on them. In clean graphene the electrons collide with one another and flow like a viscous liquid, and the current can choose a period for itself. Above a critical drift velocity the uniform flow becomes unstable and breaks into running waves, an electronic analogue of Kapitsa's waves on a falling film, which radiate at the drift velocity divided by the wave period [11].

SI units. The $z$ axis is along the electrons' velocity (they fly from the field $E_1$ into the field $E_2$), the mesh lies in the plane $z=0$, the wires along $y$ are at $x = jd$ and the wires along $x$ at $y = id$. The cell is $0 \le x < d$, $0 \le y < d$.

$E_1$ and $E_2$ are positive when the field accelerates the electron (points against $\vec v$, as in the figure)

$$E_z(\mp\infty) = -E_{1,2} \tag{1}$$

The field of one wire

$$E(r) = \frac{\lambda}{2\pi\varepsilon_0 r} \tag{2}$$

Wires along $x$ give only $E_y$ and $E_z$, so $E_x$ comes only from the wires along $y$. Summing over all wires

$$E_x = \frac{\lambda}{2\pi\varepsilon_0}\sum_{j=-\infty}^{\infty}\frac{x - jd}{(x - jd)^2 + z^2} \tag{3}$$

$$E_z = \frac{\lambda}{2\pi\varepsilon_0}\sum_{j=-\infty}^{\infty}\left[\frac{z}{(x - jd)^2 + z^2} + \frac{z}{(y - jd)^2 + z^2}\right] - \frac{E_1 + E_2}{2} \tag{4}$$

Let $w = (x + iz)/d$

$$\frac{x - jd}{(x - jd)^2 + z^2} = \frac{1}{d}\mathrm{Re}\frac{1}{w - j}$$

$$\frac{z}{(x - jd)^2 + z^2} = -\frac{1}{d}\mathrm{Im}\frac{1}{w - j}$$

$$\sum_{j=-\infty}^{\infty}\frac{1}{w - j} = \pi \cot(\pi w)$$

$$\cot(a + ib) = \frac{\sin 2a - i \sinh 2b}{\cosh 2b - \cos 2a}$$

$$E_x = \frac{\lambda}{2\varepsilon_0 d} \cdot \frac{\sin\frac{2\pi x}{d}}{\cosh\frac{2\pi z}{d} - \cos\frac{2\pi x}{d}} \tag{5}$$

$$E_z = \frac{\lambda}{2\varepsilon_0 d}\left[\frac{\sinh\frac{2\pi z}{d}}{\cosh\frac{2\pi z}{d} - \cos\frac{2\pi x}{d}} + \frac{\sinh\frac{2\pi z}{d}}{\cosh\frac{2\pi z}{d} - \cos\frac{2\pi y}{d}}\right] - \frac{E_1 + E_2}{2} \tag{6}$$

$z \to \pm\infty$ in (6) and (1)

$$\pm\frac{\lambda}{\varepsilon_0 d} - \frac{E_1 + E_2}{2} = -E_{2,1} \rightarrow \frac{2\lambda}{\varepsilon_0 d} = E_1 - E_2 \tag{7}$$

Inside the cell $v \approx \mathrm{const}$, $dt = dz/v$

$$\Delta v_x = -\frac{e}{m}\int E_x dt = -\frac{e}{mv}\int_{-\infty}^{\infty} E_x dz \tag{8}$$

$\alpha = 2\pi x/d$, $t = e^{2\pi z/d}$ $\rightarrow$ $dz = \frac{d}{2\pi}\frac{dt}{t}$, $\cosh\frac{2\pi z}{d} = \frac{1}{2}\left(t + \frac{1}{t}\right)$

$$\int_{-\infty}^{\infty}\frac{dz}{\cosh\frac{2\pi z}{d} - \cos\alpha} = \frac{d}{\pi}\int_0^{\infty}\frac{dt}{t^2 - 2t\cos\alpha + 1} = \frac{d}{\pi}\int_0^{\infty}\frac{dt}{(t - \cos\alpha)^2 + \sin^2\alpha}$$

$$= \frac{d}{\pi\sin\alpha}\left[\frac{\pi}{2} + \arctan(\cot\alpha)\right] = \frac{d}{\pi\sin\alpha}\left[\frac{\pi}{2} + \frac{\pi}{2} - \alpha\right] = \frac{d(\pi - \alpha)}{\pi\sin\alpha} \tag{9}$$

for $0 < \alpha < \pi$, and for $\pi < \alpha < 2\pi$ the same after $\alpha \to 2\pi - \alpha$. Substituting (5) and (9) into (8), $\xi = x - d/2$

$$\int_{-\infty}^{\infty} E_x dz = \frac{\lambda}{2\varepsilon_0 d} \sin\alpha \cdot \frac{d(\pi - \alpha)}{\pi\sin\alpha} = \frac{\lambda}{\varepsilon_0}\left(\frac{1}{2} - \frac{x}{d}\right) = -\frac{\lambda}{\varepsilon_0 d}\xi \tag{10}$$

$\sin\alpha$ cancels and $y$ does not enter $\rightarrow$ the kick is linear in $\xi$ over the whole cell, not only near the axis. With (7)

$$\Delta v_x = -\frac{e(E_2 - E_1)}{2mv}\xi \tag{11}$$

$\Delta v_y$ is the same with $\eta = y - d/2$ in place of $\xi$.

For $E_2 > E_1$ every electron turns towards the cell axis with a speed $\propto$ its distance from it $\rightarrow$ all of them meet at one point.

With no field behind the mesh, $mv^2 = 2eV$

$$f_0 = \frac{v\xi}{|\Delta v_x|} = \frac{2mv^2}{e(E_2 - E_1)} = \frac{4V}{E_2 - E_1} \tag{12}$$

Behind the mesh $E_2$ accelerates along $z$, $a = eE_2/m$, $v_x = \mathrm{const}$. The time to the axis is $\tau = \xi/|\Delta v_x| = f_0/v$

$$f = v\tau + \frac{eE_2\tau^2}{2m} = f_0 + \frac{eE_2}{2mv^2}f_0^2 = f_0\left(1 + \frac{E_2 f_0}{4V}\right) \tag{13}$$

$$f = \frac{4V}{E_2 - E_1}\cdot\frac{2E_2 - E_1}{E_2 - E_1} = \frac{4V(2E_2 - E_1)}{(E_2 - E_1)^2}$$

Keeping only the wires along $y$ (a slit), the whole jump (7) falls on them, $\lambda/(\varepsilon_0 d) = E_1 - E_2$, and the kick (11) doubles

$$f_0 = \frac{2V}{E_2 - E_1} \tag{14}$$

This is the first Davisson and Calbick formula [1]. In a square cell the jump is shared equally between the two families of wires, hence $4V$ in (12).

For the especially attentive, $v$ in (8) is constant while $eE_{1,2}d \ll eV$, i.e. $f \gg d$. Fig. 1 below shows the field (5), (6), drawn the same way as in [12], and Fig. 2 computes the trajectories in it without this approximation.