Appendix 1: Retarded and Advanced Radiation Field
The retarded Liénard–Wiechert potential corresponding to a charged particle of position of \({\mathbf{r}}_{s}(t)=z_0 e^{-i\omega t}{\hat{z}}\) is the following:
$$\begin{aligned} \varPhi ({\mathbf{r}},t)_{ret}= & {} \frac{1}{4\pi \varepsilon _0} \Big \{ \frac{q}{(1-\hat{{n}}\cdot \varvec{\beta }_s)|{\mathbf{r}}-{\mathbf{r}}_s|}\Big \}_{ret}, \end{aligned}$$
(25)
$$\begin{aligned} {\mathbf{A}}({\mathbf{r}},t)_{ret}= & {} \frac{\mu _0 c}{4\pi }\Big \{ \frac{q\varvec{\beta }_s}{(1-\hat{n}\cdot \varvec{\beta }_s)|{\mathbf{r}}-{\mathbf{r}}_s|} \Big \}_{ret}, \end{aligned}$$
(26)
where \({\hat{n}}=\frac{{\mathbf{r}}-{\mathbf{r}}_s}{|{\mathbf{r}}-{\mathbf{r}}_s|}\). And \(\frac{1}{(1-\hat{n}\cdot \varvec{\beta }_s)|{\mathbf{r}}-{\mathbf{r}}_s|}\biggr |_{\begin{array}{c} ret \end{array}}\) becomes the following (\(|{\mathbf{r}}-{\mathbf{r}}_s|= r+\delta r\)):
$$\begin{aligned} \begin{aligned} \frac{1}{(1-\hat{n}\cdot \varvec{\beta }_s)|{\mathbf{r}}-{\mathbf{r}}_s|}\Biggr |_{\begin{array}{c} ret \end{array}}=\frac{1}{r}\frac{1}{1+\frac{\delta r}{r}-\cos \theta \beta _s+\frac{r_s}{r}\beta _s\cos \theta }\biggr |_{\begin{array}{c} ret \end{array}}, \end{aligned} \end{aligned}$$
(27)
where \(\beta _s=-\frac{i\omega z_0}{c}e^{-i\omega t}\), \(\theta\) is the polar angle of \({\mathbf{r}}\), and \(\delta r=-z_0 \cos (\omega t)\cos \theta\). Then, using \(t_{ret}=t-\frac{r}{c}-\frac{\delta r_{ret}}{c}\), \(\delta r_{ret}\) becomes the following:
$$\begin{aligned} \delta r_{ret}= -z_0 \cos \theta \cos \big \{\omega \left( t-\frac{r}{c}\right) \big \}. \end{aligned}$$
(28)
And \({\beta _s}_{ret}\) becomes the following (\(\varepsilon =\frac{U_0}{\sqrt{2}c\omega }\)):
$$\begin{aligned} \begin{aligned} {\beta _s}_{ret}=\sqrt{2}\varepsilon \sin \left\{ \omega \left( t-\frac{r}{c}\right) \right\} -2\varepsilon ^2\cos \theta \cos ^2\{\omega (t-\frac{r}{c})\}. \end{aligned} \end{aligned}$$
(29)
Then from (25), \(\varPhi ({\mathbf{r}},t)_{ret}\) becomes the following:
$$\begin{aligned} \varPhi ({\mathbf{r}},t)_{ret}= \frac{q}{4\pi \varepsilon _0 r} \Big \{1+\frac{iU_0}{c\omega }\big (1+\frac{i}{u}\big )\cos \theta e^{-i\omega \left( t-\frac{r}{c}\right) } \Big \}, \end{aligned}$$
(30)
where \(u\equiv kr\), and \(U_0\equiv -\omega ^2 z_0\). And from (26), \({\mathbf{A}}({\mathbf{r}},t)_{ret}\) becomes the following:
$$\begin{aligned} {\mathbf{A}}({\mathbf{r}},t)_{ret}\simeq \frac{q\mu _0 U_0 i}{4\pi c u}e^{-i\omega \left( t-\frac{r}{c}\right) }{\hat{z}}. \end{aligned}$$
(31)
Then from \({\mathbf{E}}=\nabla \varPhi -\frac{\partial {\mathbf{A}}}{\partial t}\) and \({\mathbf{B}}=\nabla \times {\mathbf{A}}\), corresponding electromagnetic field is the following (\(\varphi\) is the azimuthal angle of \({\mathbf{r}}\)):
$$\begin{aligned} {\mathbf{E}}({\mathbf{r}},t)_{ret}= & {} \frac{q}{4\pi \varepsilon _0 r^2}{\hat{r}} +\frac{q U_0}{4\pi \varepsilon _0 c^2 r}e^{-i(\omega t-u)}\Big \{2\left( \frac{i}{u}-\frac{1}{u^2}\right) \cos \theta {\hat{r}}\nonumber \\&\quad +\,\big (1+\frac{i}{u}-\frac{1}{u^2}\big )\sin \theta {\hat{\theta }} \Big \},\nonumber \\ {\mathbf{B}}({\mathbf{r}},t)_{ret}= & {} \frac{q U_0}{4\pi \varepsilon _0 c^3r} e^{-i(\omega t-u)}\left( 1+\frac{i}{u}\right) \sin \theta {\hat{\varphi }}. \end{aligned}$$
(32)
Therefore, using (32), the radiated energy flux over the boundary of a sphere of radius \(R\rightarrow \infty\) centered at the origin becomes the following:
$$\begin{aligned} \begin{aligned} \lim \limits _{R\rightarrow \infty }P(t,R)&\equiv \lim \limits _{R\rightarrow \infty }\frac{1}{\mu _0}\oiint _{\partial V} (Re[{\mathbf{E}}_{ret}]\times Re[{\mathbf{B}}_{ret}])\cdot d{\mathbf{A}}\\ {}&=\lim \limits _{R\rightarrow \infty }\frac{q^2U_0^2}{6\pi \varepsilon _0 c^3}\Big [\cos ^2\{\omega \left( t-\frac{R}{c}\right) \}+\frac{\sin \{2\omega \left( t-\frac{R}{c}\right) \}}{kR}\\&\quad -\frac{\cos \{2\omega \left( t-\frac{R}{c}\right) \}}{(kR)^2}-\frac{\sin \{2\omega \left( t-\frac{R}{c}\right) \}}{2(kR)^3}\Big ]\\&=\lim \limits _{R\rightarrow \infty }\frac{q^2U_0^2}{6\pi \varepsilon _0 c^3}\cos ^2\{\omega (t-\frac{R}{c})\}. \end{aligned} \end{aligned}$$
(33)
The advanced Liénard–Wiechert potential corresponding to the periodically oscillating charged particle is the following:
$$\begin{aligned} \varPhi ({\mathbf{r}},t)_{adv}= & {} \frac{1}{4\pi \varepsilon _0} \Big \{ \frac{q}{(1+{{\hat{n}}}\cdot {\varvec{\beta} }_s)|{\mathbf{r}}-{\mathbf{r}}_s|}\Big \}_{adv}, \end{aligned}$$
(34)
$$\begin{aligned} {\mathbf{A}}({\mathbf{r}},t)_{adv}= & {} \frac{\mu _0 c}{4\pi }\Big \{ \frac{q\varvec{\beta }_s}{(1+{{\hat{n}}}\cdot {\varvec{\beta }}_s)|{\mathbf{r}}-{\mathbf{r}}_s|} \Big \}_{adv}. \end{aligned}$$
(35)
Then, using
$$\begin{aligned} \delta r_{t_{adv}}= -z_0 \cos \theta \cos \{\omega \left( t+\frac{r}{c}\right) \}, \end{aligned}$$
(36)
\(\varPhi ({\mathbf{r}},t)_{adv}\) becomes the following:
$$\begin{aligned} \varPhi ({\mathbf{r}},t)_{adv}= \frac{q}{4\pi \varepsilon _0 r} \Big \{1+\frac{iU_0}{c\omega }\big (-1+\frac{i}{u}\big )\cos \theta e^{-i\omega (t+\frac{r}{c})} \Big \}, \end{aligned}$$
(37)
and \({\mathbf{A}}({\mathbf{r}},t)_{adv}\) becomes the following:
$$\begin{aligned} {\mathbf{A}}({\mathbf{r}},t)_{adv}=\frac{q\mu _0 U_0 i}{4\pi r \omega }e^{-i\omega \left( t+\frac{r}{c}\right) }{\hat{z}}. \end{aligned}$$
(38)
Then, the advanced electromagnetic field is the following:
$$\begin{aligned} \begin{aligned} {\mathbf{E}}({\mathbf{r}},t)_{adv}&=\frac{q}{4\pi \varepsilon _0 r^2}{\hat{r}}\\&\quad +\frac{q \omega U_0}{4\pi \varepsilon _0 c^3}e^{-i(\omega t+u)}\frac{1}{u}\Big \{2 \big (-\frac{i}{u}-\frac{1}{u^2}\big )\cos \theta {\hat{r}}\\&\quad +\,\big (1-\frac{i}{u}-\frac{1}{u^2}\big )\sin \theta {\hat{\theta }}\Big \}, \\ {\mathbf{B}}({\mathbf{r}},t)_{adv}&=\frac{q\omega U_0}{4\pi \varepsilon _0 c^4} e^{-i(\omega t+u)}\frac{1}{u}\left( -1+\frac{i}{u}\right) \sin \theta {\hat{\varphi }}. \end{aligned} \end{aligned}$$
(39)
Appendix 2: Advanced Field from the Absorber of Wheeler–Feynman Absorber Theory
\({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) [portion of \({\mathbf{E}}_{abs}({\mathbf{r}},t)\) perpendicular to \({\mathbf{U}}(t)\)] is the following in the coordinate system of which \({\hat{z}}\) is along \({\mathbf{r}}\), according to WFAT (\(u\equiv kr\)):
$$\begin{aligned} \begin{aligned} {\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)&=-\frac{i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}\int _{-1}^{1} \Bigg [ \frac{1}{2}d(\cos ({\mathbf{r}},{\mathbf{r}}_k))\\&\quad \times \,\int _{0}^{2\pi } \frac{d\phi _d}{2\pi }\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))e^{iu\cos ({\mathbf{r}},{\mathbf{r}}_k)} \\ {}&\quad \times \Big [\{-\sin ({\mathbf{r}},{\mathbf{r}}_k)\cos \phi '+\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))\sin ({\mathbf{r}},{\mathbf{U}}(t))\cos \phi \}{\hat{x}} \\ {}&\quad +\{-\sin ({\mathbf{r}},{\mathbf{r}}_k)\sin \phi ' +\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))\sin ({\mathbf{r}},{\mathbf{U}}(t))\sin \phi \}{\hat{y}}\\&\quad +\{-\cos ({\mathbf{r}},{\mathbf{r}}_k)+\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))\cos ({\mathbf{r}},{\mathbf{U}}(t))\}{\hat{z}}\Big ]\Bigg ]; \end{aligned} \end{aligned}$$
(40)
\({\mathbf{r}}_k\) is location of the particle in the absorber, \(\phi\) and \(\phi '\) are the azimuthal angles of \({\mathbf{U}}(t)\) and \({\mathbf{r}}_k\), respectively, and \(\phi _d\) is the dihedral angle between \(({\mathbf{r}},{\mathbf{U}})\) plane and \(({\mathbf{r}},{\mathbf{r}}_k)\) plane. The \(x\)-component of \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) of (40) becomes the following [\(\theta\) and \(\theta '\) are polar coordinates of \({\mathbf{U}}(t)\) and \({\mathbf{r}}_k\), respectively]:
$$\begin{aligned} \begin{aligned} E_{{abs,\perp }_x}({\mathbf{r}},t)&=\frac{-i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}\int _{-1}^{1}\Big [ \frac{1}{2}d(\cos \theta ')\int _{0}^{2\pi } \frac{d\phi '}{2\pi }\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))e^{iu\cos \theta '}\\&\quad \times \{-\sin \theta ' \cos \phi '+\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))\sin \theta \cos \phi \}\Big ]. \end{aligned} \end{aligned}$$
(41)
The first term of \(E_{{abs,\perp }_x}({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \begin{aligned}&\frac{i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t} \int _{-1}^{1}\Bigg [ \frac{1}{2}d(\cos \theta ')\\&\qquad \times\int _{0}^{2\pi } \frac{d\phi '}{2\pi }\Big [\{\sin \theta \sin \theta ' \cos (\phi -\phi ')+\cos \theta \cos \theta '\}e^{iu\cos \theta '}\sin \theta ' \cos \phi ' \Big ]\Bigg ] \\ {}&\quad =\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}C_1\sin \theta \cos \phi , \end{aligned} \end{aligned}$$
(42)
where \(C_1\equiv \int _{-1}^{1}(1-x^2)e^{iux}dx=4\frac{\sin u-u\cos u}{u^3}\), and the second term of \(E_{{abs,\perp }_x}\) becomes the following:
$$\begin{aligned} -\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}C_2 \sin \theta \cos \phi , \end{aligned}$$
(43)
where \(C_2\equiv \frac{1}{\pi }\int _{-1}^{1}d(\cos \theta ')\int _{0}^{2\pi }d\phi ' \cos ^2 (\mathbf{r_k},{\mathbf{U}}(t))e^{iu\cos \theta '}\). Hence, \(E_{{abs,\perp }_x}({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} E_{{abs,\perp }_x}({\mathbf{r}},t)=\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}(C_1-C_2)\sin \theta \cos \phi . \end{aligned}$$
(44)
The \(y\)-component of \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \begin{aligned} E_{{abs,\perp }_y}({\mathbf{r}},t)&=-\frac{i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t} \int _{-1}^{1} \Bigg [\frac{1}{2}d(\cos \theta ') \int _{0}^{2\pi } \Big [\frac{d\phi '}{2\pi }\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))e^{iu\cos \theta '}\\&\quad \times \{-\sin \theta ' \sin \phi '+\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))\sin \theta \sin \phi \}\Big ]\Bigg ]. \end{aligned} \end{aligned}$$
(45)
The first term of \(E_{{abs,\perp }_y}({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \begin{aligned}&\frac{i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}\int _{-1}^{1}\Big [ \frac{1}{2}d(\cos \theta ')\\&\qquad \times\int _{0}^{2\pi } \frac{d\phi '}{2\pi }\{\sin \theta \sin \theta ' \cos (\phi -\phi ')+\cos \theta \cos \theta '\}e^{iu\cos \theta '}\sin \theta ' \sin \phi '\Big ] \\ {}&\quad =\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}C_1\sin \theta \sin \phi , \end{aligned} \end{aligned}$$
(46)
and the second term of \(E_{{abs,\perp }_y}({\mathbf{r}},t)\) is the following:
$$\begin{aligned} -\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}C_2 \sin \theta \sin \phi . \end{aligned}$$
(47)
Therefore, \(E_{{abs,\perp }_y}({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} E_{{abs,\perp }_y}=\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}(C_1-C_2)\sin \theta \sin \phi . \end{aligned}$$
(48)
The \(z\)-component of \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) is the following:
$$\begin{aligned} \begin{aligned}&E_{{abs,\perp }_z}({\mathbf{r}},t)\\ {}&\quad =-\frac{i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}\int _{-1}^{1}\Big [ \frac{1}{2}d(\cos \theta ')\\&\qquad \times\int _{0}^{2\pi } \frac{d\phi '}{2\pi }\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))e^{iu\cos \theta '}\{-\cos \theta ' +\cos ({\mathbf{r}}_k,{\mathbf{U}}(t))\cos \theta \}\Big ]. \end{aligned} \end{aligned}$$
(49)
The first term of \(E_{{abs,\perp }_z}({\mathbf{r}},t)\) is the following:
$$\begin{aligned} \begin{aligned}&\frac{i\omega q}{4\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}\int _{-1}^{1} \Big [ \frac{1}{2}d(\cos \theta ')\\&\qquad \times\int _{0}^{2\pi } \frac{d\phi '}{2\pi }\{\sin \theta \sin \theta '\cos (\phi -\phi ')+\cos \theta \cos \theta '\}e^{iu\cos \theta '}\cos \theta '\Big ] \\ {}&\quad = \frac{i\omega q}{8\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}C_3 \cos \theta , \end{aligned} \end{aligned}$$
(50)
where \(C_3\equiv \int _{-1}^{1} x^2 e^{iux}dx=2\frac{(u^2-2)\sin u+2u\cos u}{u^3}\). And the second term of \(E_{{abs,\perp }_z}({\mathbf{r}},t)\) is the following:
$$\begin{aligned} -\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}C_2 \cos \theta . \end{aligned}$$
(51)
Therefore, \(E_{{abs,\perp }_z}({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}(2C_3-C_2) \cos \theta . \end{aligned}$$
(52)
Hence, the resultant \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \begin{aligned} {\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)&= \frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}\{(C_1-C_2)\sin \theta (\cos \phi {\hat{x}}+\sin \phi {\hat{y}})\\&\quad +(2C_3-C_2)\cos \theta {\hat{z}}\}. \end{aligned} \end{aligned}$$
(53)
To express \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) in a coordinate system of which \({\hat{z}}\) is along \({\mathbf{r}}_s(t)\) (in the remainder of this Appendix 2, this coordinate system is used), (53) should be transformed by the following matrix:
$$\begin{aligned}\begin{pmatrix} \cos \theta \cos \phi &{} \cos \theta \sin \phi &{}-\sin \theta \\ -\sin \phi &{} \cos \phi &{} 0 \\ \sin \theta \cos \phi &{} \sin \theta \sin \phi &{}\cos \theta \end{pmatrix}=\begin{pmatrix} \cos \theta &{} 0 &{}-\sin \theta \\ 0 &{} 1 &{} 0\\ \sin \theta &{} 0 &{} \cos \theta \end{pmatrix} \begin{pmatrix} \cos \phi &{} \sin \phi &{} 0\\ -\sin \phi &{} \cos \phi &{}0\\ 0&{}0&{}1 \end{pmatrix}. \end{aligned}$$
The \(-x\)-component of the transformed \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) is the \({\hat{\rho }}\) (axial unit vector in the cylindrical coordinate system)-component in the new coordinate system, while \(y\) and \(z\)-components vanish. The resultant \({\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \begin{aligned} {\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)&=\frac{i\omega q}{16\pi \varepsilon _0 c^3}U_0 e^{-i \omega t}(2C_3-C_1)\sin \theta \cos \theta {\hat{\rho }} \\ {}&=\frac{q\omega U_0}{4\pi \varepsilon _0 c^3}e^{-i\omega t}\frac{1}{u}\Big [i\sin u+i\frac{3}{u}\cos u-i\frac{3}{u^2}\sin u\Big ]\sin \theta \cos \theta {\hat{\rho }}, \end{aligned} \end{aligned}$$
(54)
and \({\mathbf{E}}_{abs,\parallel }\) of WFAT is given as the following:
$$\begin{aligned} \begin{aligned} {\mathbf{E}}_{abs,\parallel }({\mathbf{r}},t)&= \frac{q\omega U_0}{4\pi \varepsilon _0 c^3} e^{-i\omega t}\frac{1}{u}\Big \{-i\sin u \sin ^2 \theta +i\big (\frac{\cos u}{u}-\frac{\sin u}{u^2}\big )\\&\quad \times \,(3\cos ^2 \theta -1)\Big \}{\hat{z}}. \end{aligned} \end{aligned}$$
(55)
Therefore, \({\mathbf{E}}_{abs}({\mathbf{r}},t)={\mathbf{E}}_{abs,\parallel }({\mathbf{r}},t)+{\mathbf{E}}_{abs,\perp }({\mathbf{r}},t)\) becomes the following:
$$\begin{aligned} \begin{aligned} {\mathbf{E}}_{abs}({\mathbf{r}},t)&= \frac{q\omega U_0}{4\pi \varepsilon _0 c^3}\frac{e^{-i\omega t}}{ u}\Bigg [\bigg \{\big (\frac{i}{u}-\frac{1}{u^2}\big )e^{iu}+\big (\frac{i}{u}+\frac{1}{u^2}\big )e^{-iu}\bigg \}\cos \theta {\hat{r}} \\ {}&\quad +\frac{1}{2}\bigg \{\big (1+\frac{i}{u}-\frac{1}{u^2}\big )e^{iu}+\big (-1+\frac{i}{u}+\frac{1}{u^2}\big )e^{-iu}\bigg \}\sin \theta {\hat{\theta }}\Bigg ]. \end{aligned} \end{aligned}$$
(56)
Appendix 3: Two Radiating Charged Particles’ Power and Radiated Energy Flux
Minus of sum of two charged particles’ power, \(W_{two}\), can be decomposed into the two charged particles’ minus of powers as the following:
$$\begin{aligned} \begin{aligned}&W_{two}(t,\alpha ,\delta ,d)=W_1(t,\alpha ,\delta ,d)+W_2(t,\alpha ,\delta ,d). \end{aligned} \end{aligned}$$
(57)
For \(d\) longer than a few radiation wavelengths, minus of power of charged particle 1 is:
$$\begin{aligned} \begin{aligned} W_1(t,\alpha ,\delta ,d)&=-q\lim _{{\mathbf{r}}\rightarrow {\mathbf{r}}_1}\{{\mathbf{E}}_{abs,1}({\mathbf{r}},t)+{\mathbf{E}}_{ret,2}({\mathbf{r}},t)\}\cdot Re[\dot{{\mathbf{r}}}_1]\\&=-\frac{q^2U_0}{4\pi \varepsilon _0d^2\omega }\cos \alpha \sin (\omega t)\\ {}&\quad +\frac{q^2U_0^2}{6\pi \varepsilon _0 c^3} \Big [\sin ^2(\omega t) +3\sin (\omega t)\big \{X(d,\alpha )\sin (\omega t-\delta )\\&\quad +Y(d,\alpha )\cos (\omega t-\delta )\big \}\Big ] , \end{aligned} \end{aligned}$$
(58)
where \(X(d,\alpha )\) and \(Y(d,\alpha )\) are defined as the following:
$$\begin{aligned} X(d,\alpha )\equiv & {} \Big [\cos ^2 \alpha \Big \{-\frac{\cos (kd)}{(kd)^2}+\frac{\sin (kd)}{(kd)^3}\Big \} \nonumber \\&+\frac{\sin ^2 \alpha }{2}\Big \{\frac{\sin (kd)}{kd}+\frac{\cos (kd)}{(kd)^2}-\frac{\sin (kd)}{(kd)^3}\Big \}\Big ], \end{aligned}$$
(59)
$$\begin{aligned} Y(d,\alpha )\equiv & {} \Big [\cos ^2 \alpha \Big \{\frac{\sin (kd)}{(kd)^2}+\frac{\cos (kd)}{(kd)^3}\Big \} \nonumber \\&+\frac{\sin ^2 \alpha }{2}\Big \{\frac{\cos (kd)}{kd}-\frac{\sin (kd)}{(kd)^2}-\frac{\cos (kd)}{(kd)^3}\Big \}\Big ] . \end{aligned}$$
(60)
Similarly, minus of power of charged particle 2 is:
$$\begin{aligned} \begin{aligned} W_2(t,\alpha ,\delta ,d)&=-q\lim _{{\mathbf{r}}\rightarrow {\mathbf{r}}_2}\{{\mathbf{E}}_{abs,2}({\mathbf{r}},t)+{\mathbf{E}}_{ret,1}({\mathbf{r}},t)\}\cdot Re[\dot{{\mathbf{r}}}_2]\\&=\frac{q^2U_0}{4\pi \varepsilon _0d^2\omega }\cos \alpha \sin (\omega t-\delta ) \\ {}&\quad +\frac{q^2U_0^2}{6\pi \varepsilon _0 c^3} \Big [\sin ^2(\omega t-\delta ) +3\sin (\omega t-\delta )\big \{X(d,\alpha )\sin (\omega t)\\&\quad +Y(d,\alpha )\cos (\omega t)\big \}\Big ]. \end{aligned} \end{aligned}$$
(61)
In the lab frame, (58) and (61) are changed to [using the Lorentz transformed (from the average rest frame) electromagnetic field and charged particles’ velocity]:
$$\begin{aligned} W_1'(t,\alpha ,\delta ,d)= & {} W_1(t,\alpha ,\delta ,d)-\frac{q^2U_0^2}{4\pi \varepsilon _0 c^3}\sin (\omega t)\Big [\frac{\cos (\omega t-\delta -kd)}{kd}\nonumber \\&+\frac{\sin (\omega t-\delta -kd)}{(kd)^2}\Big ]+\frac{q^2c}{4\pi \varepsilon _0 d^2}, \end{aligned}$$
(62)
$$\begin{aligned} W_2'(t,\alpha ,\delta ,d)\,= & {} W_2(t,\alpha ,\delta ,d)+\frac{q^2U_0^2}{4\pi \varepsilon _0 c^3}\sin (\omega t-\delta )\Big [\frac{\cos (\omega t-kd)}{kd}\nonumber \\&+\frac{\sin (\omega t-kd)}{(kd)^2}\Big ]-\frac{q^2c}{4\pi \varepsilon _0 d^2}. \end{aligned}$$
(63)
For \(d\) shorter than a few radiation wavelengths, \(W_{1}\) and \(W_{2}\) become:
$$\begin{aligned} W_1(t,\alpha ,\delta ,d)= & {} \lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_1}\{{\mathbf{E}}_{abs,1}({\mathbf{r}},t)+{\mathbf{E}}_{abs,2}({\mathbf{r}},t)\}\cdot Re[\dot{{\mathbf{r}}}_1] \nonumber \\ {}= & {} \frac{q^2U_0^2}{6\pi \varepsilon _0 c^3} \Big \{\sin ^2(\omega t)+3X(d,\alpha )\sin (\omega t)\sin (\omega t-\delta )\Big \}, \end{aligned}$$
(64)
$$\begin{aligned} W_2(t,\alpha ,\delta ,d)= & {} \lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_2}\{{\mathbf{E}}_{abs,1}({\mathbf{r}},t)+{\mathbf{E}}_{abs,2}({\mathbf{r}},t)\}\cdot Re[\dot{{\mathbf{r}}}_2] \nonumber \\ {}= & {} \frac{q^2U_0^2}{6\pi \varepsilon _0 c^3} \Big \{\sin ^2(\omega t-\delta )+3X(d,\alpha )\sin (\omega t)\sin (\omega t-\delta )\Big \}. \end{aligned}$$
(65)
In the lab frame, (64) and (65) are changed to:
$$\begin{aligned} W_1'(t,\alpha ,\delta ,d)\,= & {} W_1(t,\alpha ,\delta ,d)-\frac{q^2U_0^2}{4\pi \varepsilon _0 c^3}\sin (\omega t)\cos (\omega t-\delta )\nonumber \\&\quad \times \,\Big \{\frac{\cos (kd)}{kd}-\frac{\sin (kd)}{(kd)^2}\Big \}, \end{aligned}$$
(66)
$$\begin{aligned} W_2'(t,\alpha ,\delta ,d)\,= & {} W_2(t,\alpha ,\delta ,d)+\frac{q^2U_0^2}{4\pi \varepsilon _0 c^3}\sin (\omega t-\delta )\cos (\omega t)\nonumber \\&\quad \times \,\Big \{\frac{\cos (kd)}{kd}-\frac{\sin (kd)}{(kd)^2}\Big \}. \end{aligned}$$
(67)
The radiated energy flux over the boundary of a sphere of radius \(R\) centered at the origin is the following:
$$\begin{aligned} \begin{aligned}&P_{two}(t,R,\alpha ,\delta ,d) =\frac{1}{\mu _0}\oiint _{\partial V} \Big [Re[{\mathbf{E}}_{ret,1}+{\mathbf{E}}_{ret,2}]\times Re[ {\mathbf{B}}_{ret,1}+{\mathbf{B}}_{ret,2}] \Big ]\cdot d{\mathbf{A}}. \end{aligned} \end{aligned}$$
(68)
Then, \(\lim \limits _{R\rightarrow \infty }P_{two}(t,R,\alpha ,\delta ,d)\) is the following [\(\varDelta u\equiv \frac{kd}{2}(\sin \theta \cos \phi \sin \alpha +\cos \theta \cos \alpha )\), and \(\theta\) and \(\phi\) are the polar and azimuthal angles of the spherical coordinate system, respectively]:
$$\begin{aligned} \begin{aligned}&\lim \limits _{R\rightarrow \infty }P_{two}(t,R,\alpha ,\delta ,d) \\ {}&\quad =\lim \limits _{R\rightarrow \infty }\frac{q^2 U_0^2}{32\pi ^2 \varepsilon _0 c^3} \int _{\phi =0}^{\phi =2\pi } \int _{\theta =0}^{\theta =\pi }d\phi d\theta \Bigg [\sin ^3 \theta \Big \{Re[e^{-2i\omega (t-\frac{R}{c})}\{e^{i\varDelta u}\\&\qquad +e^{i(-\varDelta u +\delta )}\}^2]+4\cos ^2 \big (\varDelta u-\frac{\delta }{2}\big )\Big \} \Bigg ] \\ {}&\quad =\lim \limits _{R\rightarrow \infty }\frac{q^2 U_0^2}{4\pi ^2 \varepsilon _0 c^3} \cos ^2 \left\{ \omega (t-\frac{R}{c})-\frac{\delta }{2}\right\} \\&\qquad \times \,\int _{\phi =0}^{\phi =2\pi } \int _{\theta =0}^{\theta =\pi }\sin ^3 \theta \cos ^2(\varDelta u-\frac{\delta }{2})d\phi d\theta . \end{aligned} \end{aligned}$$
(69)
The above integral expression can be calculated as the following [\(\text {J}_n(x)\) is Bessel function of the first kind, and \(C_n^{\nu }(x)\) is Gegenbauer polynomial]:
$$\begin{aligned} \begin{aligned}&\int _{\phi =0}^{\phi =2\pi } \int _{\theta =0}^{\theta =\pi }\sin ^3 \theta \cos ^2(\varDelta u-\frac{\delta }{2})d\phi d\theta \\&\quad =\pi \int _{0}^{\pi }d\theta \Big [\sin ^3 \theta \{1+ \text {J}_0 (kd\sin \alpha \sin \theta )\cos (kd \cos \alpha \cos \theta -\delta )\} \Big ]\\&\quad =\frac{4}{3}\pi \Big [1+\cos \delta \\&\qquad \times\int _{0}^{\frac{\pi }{2}}[\sin \theta \{C_0^\frac{1}{2}(\cos \theta )-C_2^\frac{1}{2}(\cos \theta )\} \cos (kd \cos \alpha \cos \theta )\text {J}_0 (kd\sin \alpha \sin \theta )d\theta] \Big ]\\&\quad =\frac{4}{3}\pi \Big [1+\cos \delta \sqrt{\frac{\pi }{2kd}}\Big \{\frac{1}{2}(3\cos ^2 \alpha -1)\text {J}_{\frac{5}{2}}(kd)+\text {J}_{\frac{1}{2}}(kd)\Big \} \Big ]\\&\quad =\frac{4}{3}\pi \big [1+3X(d,\alpha )\cos \delta \big ], \end{aligned} \end{aligned}$$
(70)
where the \(\theta\)-integration is given in Ref. [27]. Therefore, \(\lim \limits _{R\rightarrow \infty }P_{two}(t,R,\alpha ,\delta ,d)\) becomes:
$$\begin{aligned} \begin{aligned} \lim \limits _{R\rightarrow \infty }P_{two}(t,R,\alpha ,\delta ,d)&=\lim \limits _{R\rightarrow \infty }\frac{q^2 U_0^2}{3\pi \varepsilon _0 c^3} \cos ^2 \{\omega (t-\frac{R}{c})-\frac{\delta }{2}\}\\&\quad \times \,\big [1+3X(d,\alpha )\cos \delta \big ].\end{aligned} \end{aligned}$$
(71)
Appendix 4: Time Integration of Minus of Charged Particles’ Power over a Radiation Period
In CED, for radiation of the single charged particle, the field at the charged particle can be decomposed as the following:
$$\begin{aligned} \begin{aligned} \lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_s} {\mathbf{E}}_{ret}({\mathbf{r}})=\lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_s}[ {\mathbf{E}}_{abs}({\mathbf{r}})+\frac{1}{2}\{{\mathbf{E}}_{ret}({\mathbf{r}})+{\mathbf{E}}_{adv}({\mathbf{r}})\}]. \end{aligned} \end{aligned}$$
(72)
As \(\lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_s} \frac{1}{2}\{{\mathbf{E}}_{ret}({\mathbf{r}})+{\mathbf{E}}_{adv}({\mathbf{r}})\}\) is divergent, and its direction is indeterminate, the charged particle’s power is indeterminate in CED. However, as
$$\begin{aligned} -\lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_s}\int _{t_0}^{t_0+T}q\frac{1}{2}Re[{\mathbf{E}}_{ret}({\mathbf{r}})+{\mathbf{E}}_{adv}({\mathbf{r}})] \cdot Re[\dot{{\mathbf{r}}}_s(t)] dt=0, \end{aligned}$$
(73)
the following relation can be obtained according to \(W(t)\) of (3) and (72):
$$\begin{aligned} \int _{t_0}^{t_0+T} W(t)dt=-\lim \limits _{{\mathbf{r}}\rightarrow {\mathbf{r}}_s}\int _{t_0}^{t_0+T}qRe[{\mathbf{E}}_{ret}({\mathbf{r}})] \cdot Re[\dot{{\mathbf{r}}}_s(t)] dt, \end{aligned}$$
(74)
which means that time integrations of minus of charged particle’s power over a radiation period are the same in CED and WFAT. Then, according to \({\mathbf{E}}^{(i)}(t)\) of (11)–(12) and the following relation,
$$\begin{aligned} \begin{aligned}&\int _{t_0}^{t_0+T}dt\frac{1}{2}\{Re[{\mathbf{E}}_{ret,2}({\mathbf{r}}_1,t)+{\mathbf{E}}_{adv,2}({\mathbf{r}}_1,t)]\cdot Re[\dot{{\mathbf{r}}}_1(t)]\\&\quad +Re[{\mathbf{E}}_{ret,1}({\mathbf{r}}_2,t)+{\mathbf{E}}_{adv,1}({\mathbf{r}}_2,t)]\cdot Re[\dot{{\mathbf{r}}}_2(t)]\}=0, \end{aligned} \end{aligned}$$
(75)
time integrations of minus of the power of periodically oscillating charged particles radiating into the free-space over a radiation period are shown to be the same in CED and WFAT, regardless of number of charged particles.
Appendix 5: Non-relativistic Approximation in the Average Rest Frame for the Charged Particles in Available Undulators
If a charged particle is in available undulators with \(K\) (undulator deflection parameter) normally ranged between one and three, its motion in the lab frame is [\(\beta _z'=v_z'/c\), \(\beta _x'=v_x'/c\), \({\bar{\gamma }}=\frac{1}{\sqrt{1-{\bar{\beta }}^2}}\gg 1\), \(\gamma =\frac{1}{\sqrt{1-\beta '^2}}\gg 1\), and \({\bar{\beta }}=\frac{{\bar{v}}_x}{c}\), where \({\bar{v}}_x\) is the charged particle’s longitudinal average velocity in the lab frame. And the charged particle satisfies \(x'(t'=0)=0\)]:
$$\begin{aligned} \beta _x'= & {} {\bar{\beta }}+\frac{K^2}{4\gamma ^2}\cos (2k_u'x')+{\mathcal{O}}\Big (\frac{1}{\gamma ^4}\Big ), \end{aligned}$$
(76)
$$\begin{aligned} {\dot{\beta }}_x'= & {} -\frac{K^2\omega _u'}{2\gamma ^2}\sin (2k_u'x')+{\mathcal{O}}\Big (\frac{1}{\gamma ^4}\Big ), \end{aligned}$$
(77)
$$\begin{aligned} \beta _z'= & {} -\frac{K}{\gamma }\sin (k_u'x'), \end{aligned}$$
(78)
$$\begin{aligned} {\dot{\beta }}_z'= & {} -\frac{ K\omega _u'}{\gamma }\cos (k_u'x')\Big \{{\bar{\beta }}+\frac{K^2}{4\gamma ^2}\cos (2k_u'x')\Big \}+{\mathcal{O}}\Big (\frac{1}{\gamma ^5}\Big ). \end{aligned}$$
(79)
\(\omega _u'=ck_u'\) [\({\mathbf{B}}_u'=B_0'\cos (k_u' x'){\hat{y}}\)] is undulator’s angular frequency, and \(\varepsilon = \frac{K}{\sqrt{2+K^2}}< 1\) is treated as a small perturbation parameter. Then the charged particle’s motion in the charged particle’s average rest frame is:
$$\begin{aligned} \beta _z= & {} -\sqrt{2}\varepsilon \frac{\sin (\omega _u' t')}{1-\varepsilon ^2 \cos (2\omega _u' t')/2}, \end{aligned}$$
(80)
$$\begin{aligned} \beta _x= & {} \varepsilon ^2\frac{\cos (2\omega _u' t')}{2\{1-\varepsilon ^2 \cos (2\omega _u' t')/2\}}. \end{aligned}$$
(81)
Using \(t'= t\gamma +{\mathcal{O}}(\varepsilon ^{2})\) which is obtained from
$$\begin{aligned} t={\bar{\gamma }}(t'-\frac{{\bar{\beta }}}{c}x')=\frac{1}{\gamma \sqrt{1-\varepsilon ^2}}\{t'-\frac{\varepsilon ^2}{4\omega _u'}\sin (2\omega _u't')\}+{\mathcal{O}}\Big (\frac{1}{\gamma ^3}\Big ), \end{aligned}$$
(82)
if terms only up to \({\mathcal{O}}(\varepsilon ^1)\) are preserved in (80) and (81) longitudinal motion vanishes and transverse motion becomes sinusoidal (\(\omega \equiv \gamma \omega _u'\));
$$\begin{aligned} \dot{\varvec{\beta }}=-\sqrt{2}\varepsilon \omega \cos (\omega t){\hat{z}}, \end{aligned}$$
(83)
which is consistent with the model of charged particles’ motion used in Sects. 2.1 and 2.2. Fixing the resonant frequency \(\omega\) does not dictate the spectral width of time integration of radiated energy flux at large distance from the charged particle, but length of interval of the time integration dictates the spectral width. The full radiated energy flux from the relativistic Larmor formula [1] in the lab frame is:
$$\begin{aligned} \begin{aligned}&P_{L,R}'=\frac{q^2\gamma ^6}{6\pi \varepsilon _0c}\{{\dot{\beta }}'^2-|\varvec{\beta }' \times \dot{\varvec{\beta }}'|^2\}. \end{aligned} \end{aligned}$$
(84)
The \({\dot{\beta }}'^2\) term is:
$$\begin{aligned} \begin{aligned} {\dot{\beta }}'^2&={{\dot{\beta }}_x}'^2+{{\dot{\beta }}_z}'^2\\&=\Big (\frac{K\omega _u'}{\gamma }\Big )^2\Big [\frac{1}{4}\Big (\frac{K}{\gamma }\Big )^2\sin ^2(2k_u'x') +\cos ^2(k_u'x')\Big \{1-\frac{1}{{\bar{\gamma }}^2}+\frac{K^2}{2\gamma ^2}\cos (2k_u'x')\Big \}\Big ]\\&\quad +{\mathcal{O}}\Big (\frac{1}{\gamma ^5}\Big ), \end{aligned}\end{aligned}$$
(85)
and the \(|\varvec{\beta }' \times \dot{\varvec{\beta }}'|^2\) term is:
$$\begin{aligned} \begin{aligned} |\varvec{\beta }' \times \dot{\varvec{\beta }}'|^2&= \Big [ -\frac{K\omega _u'}{\gamma }\cos (k_u'x')\Big \{1-\frac{1}{{\bar{\gamma }}^2}+\frac{K^2}{2\gamma ^2}\cos (2k_u'x')\Big \}\\&\quad -\frac{K^3\omega _u'}{2\gamma ^3}\sin (k_u'x')\sin (2k_u'x')+{\mathcal{O}}\Big (\frac{1}{\gamma ^5}\Big )\Big ]^2 \\ {}&=\Big (\frac{K\omega _u'}{\gamma }\Big )^2\Big [\frac{1}{2}\Big (\frac{K}{\gamma }\Big )^2\sin ^2(2k_u'x')\\&\quad +\cos ^2(k_u'x')\Big \{1-\frac{2}{{\bar{\gamma }}^2}+\frac{K^2}{\gamma ^2}\cos (2k_u'x')\Big \}\Big ]+{\mathcal{O}}\Big (\frac{1}{\gamma ^5}\Big ). \end{aligned} \end{aligned}$$
(86)
Hence, \({\dot{\beta }}'^2-|\varvec{\beta }' \times \dot{\varvec{\beta }}'|^2\) becomes:
$$\begin{aligned} {\dot{\beta }}'^2-|\varvec{\beta }' \times \dot{\varvec{\beta }}'|^2=\frac{\omega _u'^2}{{\bar{\gamma }}^4}\cos ^2(k_u'x')2\varepsilon ^2(1-\varepsilon ^2), \end{aligned}$$
(87)
and \(P_{L,R}'\) of (84) becomes:
$$\begin{aligned} \begin{aligned}&P_{L,R}'=\frac{q^2 \omega ^2}{3\pi \varepsilon _0c}\varepsilon ^2\cos ^2(\omega _u't')\frac{1}{1-\varepsilon ^2}. \end{aligned} \end{aligned}$$
(88)
On the other hand, from the non-relativistic Larmor formula [1] the radiated energy flux for charged particle approximated to be non-relativistic in the charged particle’s average rest frame as shown in (83) is:
$$\begin{aligned} P_{L,NR}=\frac{q^2}{6\pi \varepsilon _0c}{\dot{\beta }}^2=\frac{q^2\omega ^2}{3\pi \varepsilon _0c}\varepsilon ^2 \cos ^2(\omega _u' t'). \end{aligned}$$
(89)
Then as radiated energy fluxes for the average rest frame and the lab frame are the same, ratio of the approximated non-relativistic radiated energy flux to the full relativistic one becomes:
$$\begin{aligned} R_{NR/R}\equiv \frac{P_{L,NR}}{P_{L,R}} =1-\varepsilon ^2=\frac{2}{K^2+2}. \end{aligned}$$
(90)
Therefore, the motion of charged particles in available undulators can be approximated by the model of non-relativistic oscillation used in Sects. 2.1 and 2.2, which can account for noticeable fraction of the full radiated energy.
