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Translation: Radiation of a charge passing through a cylindrical cavity

Radiation of a charge passing through a cylindrical cavity

O.A. Kolpakov and V.I. Kotov, Journal of Technical Physics 34, № 8 (1964)

Излучение заряда, пролетающего через цилиндрический резонатор

О. А. Колпаков и В. И. Котов, Журнал технической физики 34, № 8 (1964)

We calculate the energy emitted by a charged beam, while passing through a cavity. Is shown that the formula for the limit in the case of an infinite conducting screen with a hole, considered by Yu. N. Dnestrofsky and D.P. Kostomarov, coincide with their results obtained by other methods.

The question of the radiation passing through the various structures of the charges of conducting bodies has been discussed in a number of papers [1], [2], [3], [4], [5]. Such structures can be waveguides, fat and loaded with crystals and other conducting bodies of arbitrary shape.

It is of interest to determine the radiation produced by the passage of single charged particle beam through the cavity. This radiation is dependent on the size, speed and shape of a charged beam and in some cases may be so large that the energy lost by the passage through the cavity will be comparable to the energy acquired by the acceleration in the cavity, or even with the energy of the beam of charged particles. Therefore, the correct assessment of the radiation is the answer the question, whether it is possible in this case to produce the acceleration of the beam with the cavity.

We have a solution of problems which is more or less approaching the set. For example, Dnestrofsky and Kostomarov [1], [2], [3] determined the total radiation of a single charged beam passing through a hole through an infinite conducting screen. In the papers [4], [5] the emission of charged particles is considered which fly along the axis of the waveguide with partitions. Such a waveguides can be considered as a chain of coupled resonators. The authors limited the calculation to the radiation of the fundamental mode. If in the formulas obtained, the coupling coefficient between resonators is set equal to zero, then like in Boltovsky [4], the result will be the same as the radiation from the fundamental mode in a stand alone cavity.

A beam of charged particles passing through a cavity excites all its harmonics. In the relativistic case, the contribution of the higher harmonics can be very significant.

We have made estimates of the radiation for the individual modes and the total radiation when a charged beam through a cylindrical cavity having inlet and outlet openings. The task is to determine the energy of the electromagnetic field produced by the passage of a charged beam through a single cylindrical cavity radius \(a\) and length \(h\). Beam with respect to the geometric shape is assumed as charged “thread” of length \(h_1\) whereas the “axis” of the beam and the direction of motion coincide with the axis of the resonator.

For the solution we naturally assumed that the velocity of the beam is constant. It is also assumed that the resonator can be considered electrically isolated from the environment.

The search for the induced field is reduced to the solution of the inhomogeneous wave equation. The solution is sought in the form of a product of a vector function depending on the origin and as a function of time. This method is discussed in detail [6].

Eigenmodes of the resonator can be expressed as the sum

\begin{equation} \vec A(\vec{r}, t) = \sum_\lambda \vec A_\lambda (\vec r)q(t,\omega_\lambda), \end{equation}

where \(\omega_\lambda\) are natural frequencies of the resonator, and \(\vec A_\lambda (\vec r)\) are the eigenfunctions of the vector potential, which are the solution of equation

\begin{equation} \Delta \vec A_\lambda (\vec r) - \dfrac{\omega_\lambda^2}{c^2}\vec A_\lambda = 0. \end{equation}

\(\vec A_\lambda\) are orthogonal functions normalized to cavity volume \(V_0\) so that

\begin{equation} \dfrac{1}{V_0}\int\limits_{V_0}\vec A_\lambda^{\ 2} (\vec r)\; dV = \dfrac{1}{\varepsilon_0}, \end{equation}

where \(\varepsilon_0\) is the dielectric constant of vacuum. Normalization and the choice of the coefficients of the equations are due to the use of MKSA system. In the future, we will use only this system of units. For our task is convenient to use a cylindrical coordinate system.

Since the geometry of the beam has an axial symmetry, the solution depends only on the coordinates r and z. Eigenfunctions of the vector potential have the form

\begin{equation} \left. \begin{array}{l} A^z_{lm}={}\dfrac{\sqrt{2}c}{\omega_\lambda a \sqrt{\pi \varepsilon_0 h}} \left(\dfrac{\nu_l}{a}\right) \dfrac{J_0\left(\nu_l\dfrac{r}{a}\right)}{J_1(\nu_l)} \cos \left ( \dfrac{m \pi}{h} z\right ),\\ \\ A^r_{lm}={}\dfrac{\sqrt{2}c}{\omega_\lambda a \sqrt{\pi \varepsilon_0 h}} \left(\dfrac{m \pi}{h}\right) \dfrac{J_1\left(\nu_l\dfrac{r}{a}\right)}{I_1(\nu_l)} \sin \left ( \dfrac{m \pi}{h} z\right ) \end{array}\right\} \begin{array}{l} m=1,2,\cdots,\\ l = 1,2,\cdots, \end{array} \label{eqn:four} \end{equation}
\begin{equation} A^z_{l0}=\dfrac{c}{\omega_\lambda d \sqrt{\pi \varepsilon_0 h}}\left( \dfrac{\nu_l}{a} \right) \dfrac{J_0\left(\nu_l\dfrac{r}{a}\right)}{J_1(\nu_l)},\ \ m=0;\ l=1,2,\cdots \tag{4-1}\label{eqn:fourone} \end{equation}

where \(\nu_l\) is the root of the equation \(J_0(\nu_l)=0\), \(\omega_\lambda = c\ \sqrt{\left( \dfrac{\nu_l}{a} \right)^2+\left(\dfrac{m \pi}{h}\right)^2}\). As a function of time \(q(t,\omega_\lambda)\) satisfies the following

\begin{equation} \left. \begin{array}{r} \ddot{q}_{lm}+\omega_\lambda^2 q_{lm}=j_{lm}(t),\\ \\ j_{lm}(t)=\int\limits_{V_0}\vec A_{lm}(\vec r)\vec j(\vec r, t)\; dV. \end{array}\right\} \label{eqn:five} \end{equation}

The current density vector \(\vec j (\vec r, t)\) is in this case

\begin{equation} \vec j(\vec r, t) = \left\{ \begin{array}{lcl} \dfrac{Qv}{2h_1\pi r}\delta(\vec r) &{\rm with}&vt-h_1 < z < vt,\\ 0 & {\rm with} &z < vt-h_1,\ z>vt. \end{array} \right. \label{eqn:six} \end{equation}

where \(v\) is velocity of the beam along the \(z\) axis and \(Q\) is its charge. \(t = 0\) corresponds to the appearance of the beam in the resonator. The solution of equation (\ref{eqn:five}) has the form \(q_{lm}\)

\begin{equation} q_{lm} = a_{lm}{\rm e}^{i\omega_\lambda t} + {\rm h.\ c.}, \label{eqn:seven} \end{equation}

\(a_{lm}\) is according to (\ref{eqn:five}) and (\ref{eqn:six}) for \(h_1\leq h\) (case \(h_1> h\) is considered similarly) and (\(t > \dfrac{h_1+h}{v}\)) is

\begin{equation} a_{lm}=\dfrac{K_{lm}Qv^2}{2i h_1 \omega_\lambda} \dfrac{1}{\omega_\lambda^2-\omega_m^2}\left(1-{\rm e}^{-i\omega_\lambda\frac{h_1}{v}}\right)\left[(-1)^m{\rm e}^{-i\omega_\lambda\frac{h}{v}}-1 \right], \label{eqn:eight} \end{equation}

where \(\omega_m= \dfrac{m\pi v}{h}\) and \(K_{lm}=\sqrt{\dfrac{2}{\pi h \varepsilon_0}}\left(\dfrac{\nu_l}{a} \right)\dfrac{c}{\omega_\lambda a J_1(\nu_l)}\).

The energy of the emitted field is given by

\begin{equation} U=\int\limits_{V_0} \sum_{l,m}\dfrac{1}{2}\left[\varepsilon_0\vec E_{lm}^2+\mu_0\vec H^{\color{red}2}_{lm}\right]\; dV, \label{eqn:nine} \end{equation}

where \(\mu_0\) is the permeability of free space.

Given that \(E_{lm}=-\dot q_{lm}(t)\vec A_{lm}(\vec r), \vec H_{lm}=\dfrac{1}{\mu_0}q_{lm}(t)\ \vec\nabla\times\vec A_{lm}(\vec r)\) and using (\ref{eqn:seven}) and (\ref{eqn:eight}) we finally obtain U

\begin{equation} U=\sum_{l,m}\dfrac{4Q^2v^4c^2}{\pi\varepsilon_0hh_1^2a^2J_1^2 \left (\nu_l \right )} \dfrac{1}{\omega_\lambda^2(\omega_\lambda^2-\omega_m^2)^2}\left (\dfrac{\nu_l}{a}\right )^24\sin^2 \left( \omega_\lambda\dfrac{h_1}{2v}\right ) \left\{ \begin{array}{lcl} \sin^2\left ( \omega_\lambda\dfrac{h}{2v}\right ),& m {\rm -even,}\\\\ \cos^2\left ( \omega_\lambda\dfrac{h}{2v}\right ) ,& m {\rm -odd.} \end{array} \right\} \label{eqn:ten} \end{equation}

In the limit of \(h_1\rightarrow0\) we have the formula for a point charge

\begin{equation} U_{h_1=0}=\sum_{l,m}\dfrac{4Q^2v^4c^2}{\pi\varepsilon_0ha^2J_1^2(\nu_l)} \dfrac{1}{(\omega_\lambda^2-\omega_m^2)^2} \left\{ \begin{array}{lcl} \sin^2\left(\omega_\lambda\dfrac{h}{2v}\right ),& m {\rm -even,}\\\\ \cos^2\left(\omega_\lambda\dfrac{h}{2v}\right ),& m {\rm -odd.} \end{array} \right\} \label{eqn:eleven} \end{equation}

Hence, for the radiated energy of the fundamental mode (m = 0, l = 1) we obtain

\begin{equation} U_{10}=\dfrac{8Q^2v^2}{4\pi\varepsilon_0\omega_\lambda^2ha^2J_1^2(\nu_1)}\sin^2 \left ( \dfrac{h\omega_\lambda}{2v} \right ). \label{eqn:twelve} \end{equation}

If (\ref{eqn:twelve}) is multiplied by a factor of \(v / h\) (the number of cavities, passing charge per unit of time), then the result will be exactly the same as the expression found for a chain of coupled resonators whenever the coupling coefficient of the latter is set equal to zero [4], [5].

From (\ref{eqn:ten}) we see that the contribution to the intensity of the individual harmonics is inversely proportional to \(m\)

\begin{equation} U_{lm}\sim\dfrac{1}{m^6} \label{eqn:thirteen} \end{equation}

while keeping other factors constant.

Therefore by determining the total radiation of only dominant modes (i.e. \(m = 0\)), we can estimate the total radiation in the resonator.

In a practical cavity resonator with an inlet and outlet opening of radius \(R_0\), those waves are emitted that are radiated by the passage of charge through the hole. It follows that in the sum over \(l\), we must consider the terms for such modes with wavelengths greater than the radius of the opening. Assume that the shortest wavelength is \textcrlambda\(_{min}\). Keeping in mind that \textcrlambda\(_{min}=2\pi R_0\) and if \(m = 0\), we have \(\omega_\lambda=c \dfrac{\nu_l}{a}\) and \textcrlambda\(=\dfrac{2\pi c}{\omega_\lambda}\) then

\begin{equation} \dfrac{\nu_{max}}{a}R_0 = 1. \label{eqn:fourteen} \end{equation}

Furthermore we assume that in the cavity we have \(a \gg h\) and \(a \gg h_1\).

Since \(\dfrac{\omega_\lambda h}{2v} = \dfrac{\nu_l}{a}\dfrac{h}{2 \beta}\), \(\dfrac{\omega_\lambda h_1}{2v}=\dfrac{\nu_l}{a}\dfrac{h_1}{2\beta}\) and the relativistic case where \(\sin \left (\dfrac{\omega_\lambda h}{2 v}\right ) \simeq \dfrac{\omega_\lambda h}{2v}\), \(\sin \left ( \dfrac{\omega_\lambda h_1}{2 v}\right ) \simeq \dfrac{\omega_\lambda h_1}{2v}\). Taking into account these considerations and using (10), we can separate the fundamental mode

\begin{equation} U_{l0}=\dfrac{h}{2\pi\varepsilon_0}\dfrac{Q^2}{a^2J_1^2(\nu_l)} \label{eqn:fifteen} \end{equation}

and to all trapped modes

\begin{equation} U_{0}=\sum_{l=1}^{l_{max}}\dfrac{h}{2\pi\varepsilon_0}\dfrac{Q^2}{a^2J_1^2(\nu_l)}. \label{eqn:sixteen} \end{equation}

Since \(a \gg h\) we can assume that it is large enough and use the asymptotic representation

$$ \dfrac{1}{J_1^2(\nu_l)}=\dfrac{\pi\nu_l}{2},\ \nu_{l+1}-\nu_l=\pi .$$

Introducing the notation \(\dfrac{\pi}{a}R_0=d\xi\), \(\dfrac{\nu_l}{a}R_0=\xi\) we change the sum to an integral (by using (14) changing the limits of \(\xi\) from 0 to 1)

\begin{equation} U_0=\int_0^1\dfrac{hQ^2}{4\pi\varepsilon_0R_0^2}\xi d\xi = \dfrac{h}{4\pi\varepsilon_0}\dfrac{Q^2}{2R_0^2}. \label{eqn:seventeen} \end{equation}

Since \(R_0 \ll a\), then by comparing (\ref{eqn:seventeen}) and (\ref{eqn:fifteen}) we can easily find that the integrated radiation into the trapped modes is higher than into the fundamental mode alone.

In [2] Dnestrofsky and Kostomarov show a formula that determines the radiation of a charge while passing through a hole in an infinite screen by solving the integral equation. We can show this result in a very simple way. Indeed, if in (\ref{eqn:eleven}) \(h\rightarrow\infty\), \(a\rightarrow\infty\), we get the formula for radiation of a passing point charge through the hole in the screen. Introducing the notation \(\dfrac{\nu_l}{a}R_0=\xi\), \(\dfrac{\pi}{a}R_0=d\xi\), \(\dfrac{m\pi R_0}{h}=y\),\(\dfrac{2\pi R_0}{h}=dy\), letting \(a\rightarrow\infty\), \(h\rightarrow\infty\) and substituting in (\ref{eqn:eleven}) the amount to the integral, we find

\begin{equation} U=\int_0^1\left[\int_0^\infty \dfrac{4Q^2\beta^2}{4\pi^2\varepsilon_0R_0}\dfrac{\xi^3\gamma^4}{[(\gamma\xi)^2+y^2]^2}dy\right]d\xi = \dfrac{\beta^2Q^2\gamma}{4\pi\varepsilon_0R_0}, \label{eqn:eighteen} \end{equation}

where \(\gamma = \dfrac{1}{\sqrt{1-\beta^2}}\). It is easy to see that (\ref{eqn:eighteen}) coincides with the formula obtained in [2].

In conclusion, the authors thank B. M. Bolotovsky for useful hints when discussing the issues raised in the paper.

References

  • [1] Yu. N. Dnestrofsky und D. P. Kostomarov. DAN USSR, 116, 377 (1957).
  • [2] Yu. N. Dnestrofsky und D. P. Kostomarov. DAN USSR, 124, 1026 (1959).
  • [3] Yu. N. Dnestrofsky und D. P. Kostomarov. Radiotechnika i Elektronika, 4, 303 (1959).
  • [4] B. M. Bolotofsky, UFN, 57, 259 (1961).
  • [5] A. I. Ahiezer und G. Y. Lubarski und Y. B. Feinberg. JTF, XXV, 2526 (1955).
  • [6] W. M. Lopuxin, Excitation of electromagnetic fluctuations and waves via electronic current. Gostehizdat, M., (1953).