Topology, nonlocality and duality in classical electrodynamics

Abstract

We have recently (Heras et al. in Eur. Phys. J. Plus 136:847, 2021) argued that classical electrodynamics can predict nonlocal effects by showing an example of a topological and nonlocal electromagnetic angular momentum. In this paper, we discuss the dual of this angular momentum which is also topological and nonlocal. We then unify both angular momenta by means of the electromagnetic angular momentum arising in the configuration formed by a dyon encircling an infinitely long dual solenoid enclosing uniform electric and magnetic fluxes and show that this electromagnetic angular momentum is topological because it depends on a winding number, is nonlocal because the electric and magnetic fields of this dual solenoid act on the dyon in regions for which these fields are excluded and is invariant under electromagnetic duality transformations. We explicitly verify that this duality-invariant electromagnetic angular momentum is insensitive to the radiative effects of the Liénard–Wiechert fields of the encircling dyon. We also show how duality symmetry of this angular momentum suggests different physical interpretations for the corresponding angular momenta that it unifies.

Appendices

Appendix A. Poynting theorem for two sets of Maxwell’s equations and the proof of (45)

Consider two independent sets of electromagnetic equations. The first set describes the time-dependent electric and magnetic fields ${\varvec{\mathcal {E}}}(\mathbf{x},t)$ and ${\varvec{\mathcal {B}}}(\mathbf{x},t)$ produced by the electric and magnetic charge densities $\varrho _e(\mathbf{x},t)$ and $\varrho _m(\mathbf{x},t)$ and the electric and magnetic current densities $\varvec{\mathcal {J}}_e(\mathbf{x},t)$ and $\varvec{\mathcal {J}}_m(\mathbf{x},t)$. The corresponding Maxwell equations read

$$\begin{aligned} \nabla \cdot {\varvec{\mathcal {E}}}=4\pi \varrho _e,\;\; \nabla \cdot {\varvec{\mathcal {B}}}=4\pi \varrho _m,\;\;\nabla \times {\varvec{\mathcal {E}}}+\frac{1}{c} \frac{\partial {\varvec{\mathcal {B}}}}{\partial t} = -\frac{4\pi }{c}\varvec{\mathcal {J}}_m,\;\;\nabla \times {\varvec{\mathcal {B}}}-\frac{1}{c} \frac{\partial {\varvec{\mathcal {E}}}}{\partial t} =\frac{4\pi }{c}\varvec{\mathcal {J}}_e. \end{aligned}$$

(A1)

The second set describes the electric and magnetic fields $\mathbf{E}(\mathbf{x},t)$ and $\mathbf{B}(\mathbf{x},t)$ produced by the charge and current densities $\rho _e(\mathbf{x},t), \rho _m(\mathbf{x},t), \mathbf{J}_e(\mathbf{x},t)$ and $\mathbf{J}_m(\mathbf{x},t)$ and satisfying the Maxwell equations

$$\begin{aligned} \nabla \cdot \mathbf{E}=4\pi \rho _e,\;\; \nabla \cdot \mathbf{B}=4\pi \rho _m,\;\; \nabla \times \mathbf{E}+\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t} =-\frac{4\pi }{c}\mathbf{J}_m,\;\; \nabla \times \mathbf{B}-\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} =\frac{4\pi }{c}\mathbf{J}_e, \end{aligned}$$

(A2)

We shall now obtain the Poynting theorem of the system formed by (A1) and (A2). Using these equations we can directly show that the divergence to the vector $\mathbf{E}\times \varvec{\mathcal {B}}+ \varvec{\mathcal {E}}\times \mathbf{B}$ yields the identity

$$\begin{aligned} \nabla \cdot (\mathbf{E}\times \varvec{\mathcal {B}}+ \varvec{\mathcal {E}}\times \mathbf{B})= -\frac{4\pi }{c}(\mathbf{E}\cdot \varvec{\mathcal {J}}_e +\varvec{\mathcal {E}}\cdot \mathbf{J}_e+\mathbf{B}\cdot \varvec{\mathcal {J}}_m +\varvec{\mathcal {B}}\cdot \mathbf{J}_m)-\frac{1}{c} \frac{\partial }{\partial t}\bigg ( \varvec{\mathcal {E}}\cdot \mathbf{E} + \varvec{\mathcal {B}}\cdot \mathbf{B} \bigg ), \end{aligned}$$

(A3)

which implies the Poynting theorem

$$\begin{aligned} \nabla \cdot \mathbf{S} + \frac{\partial U}{\partial t}=-\mathbf{E}\cdot \varvec{\mathcal {J}}_e -\varvec{\mathcal {E}}\cdot \mathbf{J}_e-\mathbf{B}\cdot \varvec{\mathcal {J}}_m -\varvec{\mathcal {B}}\cdot \mathbf{J}_m. \end{aligned}$$

(A4)

where the interaction energy density is given by

$$\begin{aligned} U= \frac{1}{4\pi }\bigg ( \varvec{\mathcal {E}}\cdot \mathbf{E} + \varvec{\mathcal {B}}\cdot \mathbf{B}\bigg ), \end{aligned}$$

(A5)

and the interaction Poynting vector by

$$\begin{aligned} \mathbf{S}= \frac{c}{4\pi }(\mathbf{E}\times \varvec{\mathcal {B}}+ \varvec{\mathcal {E}}\times \mathbf{B}). \end{aligned}$$

(A6)

Equations (A4)–(A6) are of general character. In particular, they hold when the electric and magnetic are time-independent ${\mathbf{E}}(\mathbf{x})$ and ${\mathbf{B}}(\mathbf{x})$ and produced by the current densities ${\mathbf{J}_e}(\mathbf{x})$ and ${\mathbf{J}_m}(\mathbf{x})$ satisfying the time-independent equations $\nabla \cdot \mathbf{E}=0,\nabla \times \mathbf{E}=-4\pi \mathbf{J}_m/c,\nabla \cdot \mathbf{B}=0,\nabla \times \mathbf{B}=4\pi \mathbf{J}_e/c.$ For this case, the corresponding electromagnetic momentum density is given by $\mathbf{g}=(\mathbf{E}\times \varvec{\mathcal {B}}+ \varvec{\mathcal {E}}\times \mathbf{B})/(4\pi c)$ and its associated electromagnetic angular momentum density by ${\varvec{l}}=\mathbf{x}\times (\mathbf{E}\times \varvec{\mathcal {B}}+ \varvec{\mathcal {E}}\times \mathbf{B})/(4\pi c),$ whose volume integration yields the electromagnetic angular momentum given in (45).

Appendix B. Proof of (46)

A direct vector calculation leads to the identities

$$\begin{aligned} \partial ^k\big [\varepsilon ^{sqi}x_q(\delta _{ik}{{{\mathcal {E}}}}_m A^m-{{{\mathcal {E}}}}_k A_i-{{{\mathcal {E}}}}_i A_k)\big ]= & {} \,-4\pi \varrho _e\varepsilon ^{sqi}x_q A_i+\varepsilon ^{sqi}x_q {{{\mathcal {E}}}}^k(\partial _k A_i-\partial _i A_k)\nonumber \\&+\varepsilon ^{sqi}x_q A^k(\partial _i{{{\mathcal {E}}}}_k-\partial _k{{{\mathcal {E}}}}_i)-\varepsilon ^{sqi}x_q{{{\mathcal {E}}}}_i\partial ^kA_k \nonumber \\= & {} -4\pi \big [\varrho _e\mathbf{x}\times \mathbf{A} \big ]^s +\big [\mathbf{x}\times (\varvec{\mathcal {E}}\times \mathbf{B}) \big ]^s\nonumber \\&-\big [\mathbf{x}\times \{(\nabla \times \varvec{\mathcal {E}})\times \mathbf{A}\} \big ]^s-\big [\mathbf{x}\times \varvec{\mathcal {E}}(\nabla \cdot \mathbf{A})\big ]^s. \end{aligned}$$

(B1)

$$\begin{aligned} \partial ^k\big [\varepsilon ^{sqi}x_q(\delta _{ik}{{{\mathcal {B}}}}_m C^m-{{{\mathcal {B}}}}_k C_i-{{{\mathcal {B}}}}_i C_k)\big ]= & {} \,-4\pi \varrho _m\,\varepsilon ^{sqi}x_q C_i+\varepsilon ^{sqi}x_q {{{\mathcal {B}}}}^k(\partial _k C_i-\partial _i C_k)\nonumber \\&+\varepsilon ^{sqi}x_q C^k(\partial _i{{{\mathcal {B}}}}_k-\partial _k{{{\mathcal {B}}}}_i)-\varepsilon ^{sqi}x_q{{{\mathcal {B}}}}_i\partial ^kC_k\nonumber \\= & {} -4\pi \big [\varrho _m\,\mathbf{x}\times \mathbf{C} \big ]^s +\big [\mathbf{x}\times (\mathbf{E}\times \varvec{\mathcal {B}}) \big ]^s\nonumber \\&-\big [\mathbf{x}\times \{(\nabla \times \varvec{\mathcal {B}})\times \mathbf{C}\} \big ]^s-\big [\mathbf{x}\times \varvec{\mathcal {B}}(\nabla \cdot \mathbf{C})\big ]^s. \end{aligned}$$

(B2)

The identities (B1) and (B2) imply

$$\begin{aligned}{}[\mathbf{x}\times (\varvec{\mathcal {E}}\times \mathbf{B} +\mathbf{E}\times \varvec{\mathcal {B}})]^s= & {} [\mathbf{x} \times ( \varrho _e\mathbf{A}+\varrho _m{\mathbf{C}})]^s+ \big [\mathbf{x}\times \{(\nabla \times \varvec{\mathcal {E}})\times \mathbf{A}+(\nabla \times \varvec{\mathcal {B}})\times \mathbf{C}\} \big ]^s\nonumber \\&+\;[\mathbf{x} \times \{\varvec{\mathcal {E}}(\nabla \cdot \mathbf{A})+\varvec{\mathcal {B}}(\nabla \cdot \mathbf{C})\}]^s\nonumber \\&+\;\partial ^k\big [\varepsilon ^{sqi}x_q(\delta _{ik}({{{\mathcal {E}}}}_m A^m+{{{\mathcal {B}}}}_m C^m) -{{{\mathcal {E}}}}_k A_i-{{{\mathcal {B}}}}_k C_i-{{{\mathcal {E}}}}_i A_k-{{{\mathcal {B}}}}_i C_k)\big ]^s. \end{aligned}$$

(B3)

Volume integration of (B3) gives

$$\begin{aligned} \frac{1}{4\pi c}\int _{V}\,[\mathbf{x}\times (\varvec{\mathcal {E}}\times \mathbf{B} +\mathbf{E}\times \varvec{\mathcal {B}})]^s\,d^3x= & {} \frac{1}{c}\int _{V}[\mathbf{x} \times ( \varrho _e\mathbf{A}+\varrho _m{\mathbf{C}})]^s d^3x\nonumber \\&+\frac{1}{4\pi c}\int _{V}\big [\mathbf{x}\times \{(\nabla \times \varvec{\mathcal {E}})\times \mathbf{A}+(\nabla \times \varvec{\mathcal {B}})\times \mathbf{C}\} \big ]^s d^3x\nonumber \\&+\frac{1}{4\pi c}\int _{V}[{\mathbf {x}} \times [\varvec{\mathcal {E}}(\nabla \cdot \mathbf{A})+\varvec{\mathcal {B}}(\nabla \cdot \mathbf{C})]^s\,d^3x\nonumber \\&+\frac{1}{4\pi c}\oint _{S} [{\mathbf {x}} \times \{\hat{\mathbf{n }}(\varvec{\mathcal {E}}\cdot \mathbf{A }+\varvec{\mathcal {B}}\cdot \mathbf{C })-\mathbf{A }(\hat{\mathbf{n }}\cdot \varvec{\mathcal {E}})-\mathbf{C }(\hat{\mathbf{n }}\cdot \varvec{\mathcal {B}})-\varvec{\mathcal {E}}(\hat{\mathbf{n }}\cdot \mathbf{A })-\varvec{\mathcal {B}}(\hat{\mathbf{n }}\cdot \mathbf{C })\}\big ]^s\,d{S},\nonumber \\ \end{aligned}$$

(B4)

where the volume integral of the last term on the right-hand side of (B3) has been transformed into a surface integral by making the replacements $\partial ^k\rightarrow (\hat{\mathbf{n}})^k$ and $\int _{V}d^3x\rightarrow \oint _{S}dS$. Equation (B4) is equivalent to (46).

Appendix C. The Liénard–Wiechert fields of a non-relativistic dyon in uniform circular motion

The explicit form of the Liénard–Wiechert fields of an arbitrarily moving dyon was first derived by one of us [19, 20]. These fields describe the retarded solutions of the Maxwell’s generalised Eq. (A1) for the case of a point dyon in arbitrary motion and can be expressed as

$$\begin{aligned} \varvec{\mathcal {E}}\!= & {} \! \Bigg [\frac{q({\widehat{\varvec{\mathcal {R}}}}\!-\!\varvec{\beta })- g\varvec{\beta }\times {\widehat{\varvec{\mathcal {R}}}}}{\gamma ^2{{{\mathcal {R}}}}^2(1\!-\!{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })^3}+\frac{q{\widehat{\varvec{\mathcal {R}}}}\times (({\widehat{\varvec{\mathcal {R}}}}\!-\!\varvec{\beta })\times {\dot{\varvec{\beta }}})}{c {{{\mathcal {R}}}}(1-{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })^3}+\frac{g(1\!-\!{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta }){\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}} + g({\widehat{\varvec{\mathcal {R}}}}\cdot {\dot{\varvec{\beta }}}){\widehat{\varvec{\mathcal {R}}}}\times \varvec{\beta }}{c {{{\mathcal {R}}}}(1-{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })^3}\Bigg ]_{\mathrm{r}},\quad \end{aligned}$$

(C1)

$$\begin{aligned} \varvec{\mathcal {B}}\!= & {} \! \Bigg [\frac{g({\widehat{\varvec{\mathcal {R}}}}\!-\!\varvec{\beta })+ q\varvec{\beta }\times {\widehat{\varvec{\mathcal {R}}}}}{\gamma ^2{{{\mathcal {R}}}}^2(1-{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })^3}+\frac{g{\widehat{\varvec{\mathcal {R}}}}\times (({\widehat{\varvec{\mathcal {R}}}}\!-\!\varvec{\beta })\times {\dot{\varvec{\beta }}})}{c {{{\mathcal {R}}}}(1-{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })^3}-\frac{q(1\!-\!{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta }){\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}} +q({\widehat{\varvec{\mathcal {R}}}}\cdot {\dot{\varvec{\beta }}}){\widehat{\varvec{\mathcal {R}}}}\times \varvec{\beta }}{c {{{\mathcal {R}}}}(1-{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })^3}\Bigg ]_{\mathrm{r}},\quad \end{aligned}$$

(C2)

where $\varvec{\mathcal {R}}=\mathbf{x} - \mathbf{x}_{qg}(t_r),$ with $\mathbf{x}$ being the field point and $\mathbf{x}_{qg}$ the position of the dyon, ${{{\mathcal {R}}}}= |\mathbf{x} - \mathbf{x}_{qg}(t_r)|,$ $\widehat{{\varvec{\mathcal {R}}}}= \varvec{\mathcal {R}}/{{{\mathcal {R}}}}$, $\varvec{\beta }= \dot{\mathbf{x}}_{qg}(t_r)/c=(1/c)d \mathbf{x}_{qg}/dt_r,$ ${\dot{\varvec{\beta }}}(t_r)= d \varvec{\beta }/dt_r,$ $\gamma =1/\sqrt{1-\varvec{\beta }^2},$ and the square brackets $[\,\,\,\,\,]_{\mathrm{r}}$ indicate that the enclosed quantities are to be evaluated at the retarded time $t_r = t- {{{\mathcal {R}}}}(t_r)/c.$ The terms varying as $1/{{{\mathcal {R}}}}^2$ are the velocity fields, and the terms varying as $1/{{{\mathcal {R}}}}$ are the acceleration fields. Let us now assume that the dyon is moving with non-relativistic velocity $\beta<<1$. In this case, $\gamma \approx 1,$ $(1-{\widehat{\varvec{\mathcal {R}}}}\cdot \varvec{\beta })\approx 1,$ and $({\widehat{\varvec{\mathcal {R}}}}-\varvec{\beta })\approx {\widehat{\varvec{\mathcal {R}}}},$ and the fields (C1) and (C2) reduce to the expressions

$$\begin{aligned} \varvec{\mathcal {E}}= & {} \Bigg [\frac{q{\widehat{\varvec{\mathcal {R}}}}- g\varvec{\beta }\times {\widehat{\varvec{\mathcal {R}}}} }{{{{\mathcal {R}}}}^2}+\frac{q{\widehat{\varvec{\mathcal {R}}}}\times ({\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}})}{c {{{\mathcal {R}}}}}+\frac{g{\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}} + g({\widehat{\varvec{\mathcal {R}}}}\cdot {\dot{\varvec{\beta }}}){\widehat{\varvec{\mathcal {R}}}}\times \varvec{\beta }}{c {{{\mathcal {R}}}}}\Bigg ]_{\mathrm{r}}, \end{aligned}$$

(C3)

$$\begin{aligned} \varvec{\mathcal {B}}= & {} \Bigg [\frac{g{\widehat{\varvec{\mathcal {R}}}}+ q\varvec{\beta }\times {\widehat{\varvec{\mathcal {R}}}}}{{{{\mathcal {R}}}}^2}+\frac{g{\widehat{\varvec{\mathcal {R}}}}\times ({\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}})}{c {{{\mathcal {R}}}}}-\frac{q{\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}} +q({\widehat{\varvec{\mathcal {R}}}}\cdot {\dot{\varvec{\beta }}}){\widehat{\varvec{\mathcal {R}}}}\times \varvec{\beta }}{c {{{\mathcal {R}}}}}\Bigg ]_{\mathrm{r}}. \end{aligned}$$

(C4)

The fields in (C3) and (C4) are suitable to obtain the fields due to a non-relativistic dyon in uniform circular motion. We assume that the dyon lies in the x-y plane and moving in a circle of fixed radius a around the origin. The position, velocity and acceleration of the dyon at time t are given by

$$\begin{aligned} \mathbf{x}_{qg}= & {} a[\cos (\omega t)\hat{\mathbf{x}} + \sin (\omega t)\hat{\mathbf{y}}],\,\,\,\,\quad \end{aligned}$$

(C5)

$$\begin{aligned} \dot{\mathbf{x}}_{qg}= & {} \omega a[- \sin (\omega t)\hat{\mathbf{x}} + \cos (\omega t)\hat{\mathbf{y}}],\,\, \end{aligned}$$

(C6)

$$\begin{aligned} \ddot{\mathbf{x}}_{qg}= & {} -\omega ^2 a[\cos (\omega t)\hat{\mathbf{x}} + \sin (\omega t)\hat{\mathbf{y}}], \end{aligned}$$

(C7)

where $\omega$ is the constant angular velocity. On the other hand, at time t we also have

$$\begin{aligned} \varvec{\mathcal {R}}=[x-a\cos (\omega t)]\hat{\mathbf{x}} + [y-a\sin (\omega t)]\hat{\mathbf{y}} + z\hat{\mathbf{z}},\qquad \,\,\, \end{aligned}$$

(C8)

$$\begin{aligned} {{{\mathcal {R}}}}= \sqrt{x^2 + y^2 +z^2 +a^2- 2a[x\cos (\omega t) + y \sin (\omega t)]}. \end{aligned}$$

(C9)

We also note that the fields (C3) and (C4) are invariant under the dual changes $q\rightarrow g$ and $g\rightarrow - q,$ a property that will later be used. The next step is to transform (C5)–(C9) in cylindrical coordinates and with respect to the field coordinates $\rho ,\phi$ and z. On the other hand, the source coordinates specifying the position of the dyon at time t read $a,\omega t,$ and 0. Inserting $x=\rho \cos \phi ,y=\rho \sin \phi ,\hat{\mathbf{x}}=\cos \phi \,\hat{\!\varvec{\rho }} - \sin \phi \,\hat{\!\varvec{\phi }},$ and $\hat{\mathbf{y}}=\sin \phi \,\hat{\!\varvec{\rho }}+\cos \phi \,\hat{\!\varvec{\phi }}$ in (C5)–(C9) and after some simplifications, we obtain

$$\begin{aligned} \mathbf{x}_{qg}=a[\cos (\phi - \omega t)\,\hat{\!\varvec{\rho }} -\sin (\phi -\omega t) \,\hat{\!\varvec{\phi }}],\qquad \qquad \qquad \end{aligned}$$

(C10)

$$\begin{aligned} \dot{\mathbf{x}}_{qg}=\omega a[\sin (\phi - \omega t)\,\hat{\!\varvec{\rho }} +\cos (\phi -\omega t) \,\hat{\!\varvec{\phi }}],\qquad \qquad \,\,\,\,\,\,\, \end{aligned}$$

(C11)

$$\begin{aligned} \ddot{\mathbf{x}}_{qg}=-\omega ^2 a[\cos (\phi - \omega t)\,\hat{\!\varvec{\rho }} -\sin (\phi -\omega t) \,\hat{\!\varvec{\phi }}],\qquad \quad \,\,\,\,\,\, \end{aligned}$$

(C12)

$$\begin{aligned} \varvec{\mathcal {R}}=[\rho - a \cos (\phi -\omega t)]\,\hat{\!\varvec{\rho }} + a\sin (\phi -\omega t)\,\hat{\!\varvec{\phi }} + z \hat{\mathbf{z}},\quad \end{aligned}$$

(C13)

$$\begin{aligned} {{{\mathcal {R}}}}= \sqrt{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)}.\qquad \qquad \quad \,\,\, \end{aligned}$$

(C14)

Using (C10)–(C14) and performing the specified operations, we obtain

$$\begin{aligned} \frac{{\widehat{\varvec{\mathcal {R}}}}}{{{{\mathcal {R}}}}^2}= & {} \frac{[\rho - a \cos (\phi -\omega t)]\,\hat{\!\varvec{\rho }} + a\sin (\phi -\omega t)\,\hat{\!\varvec{\phi }} + z \hat{\mathbf{z}}}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)]^{3/2}},\quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \,\,\,\,\, \end{aligned}$$

(C15)

$$\begin{aligned} \frac{\varvec{\beta }\times {\widehat{\varvec{\mathcal {R}}}}}{{{{\mathcal {R}}}}^2}= & {} \frac{\omega a}{c}\frac{z\cos (\phi -\omega t)\,\hat{\!\varvec{\rho }}-z\sin (\phi -\omega t)\,\hat{\!\varvec{\phi }} - [\rho \cos (\phi -\omega t)-a] \hat{\mathbf{z}}}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)]^{3/2}},\quad \qquad \qquad \qquad \qquad \qquad \,\, \end{aligned}$$

(C16)

$$\begin{aligned} \frac{{\widehat{\varvec{\mathcal {R}}}}\times ({\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}})}{{{{\mathcal {R}}}}}= & {} \frac{\omega ^2 a}{c}\frac{[a\rho \sin ^2(\phi -\omega t) +z^2\cos (\phi -\omega t)]\,\hat{\!\varvec{\rho }} - \sin (\phi -\omega t)[\rho ^2 - \rho a \cos (\phi - \omega t) + z^2]\,\hat{\!\varvec{\phi }}}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)]^{3/2}} \nonumber \\&-\frac{\omega ^2 a}{c}\frac{z[\rho \cos (\phi - \omega t) -a]\hat{\mathbf{z}}}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)]^{3/2}},\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(C17)

$$\begin{aligned} \frac{{\widehat{\varvec{\mathcal {R}}}}\times {\dot{\varvec{\beta }}}}{{{{\mathcal {R}}}}}= & {} -\frac{\omega ^2 a}{c}\frac{z\sin (\phi -\omega t)\,\hat{\!\varvec{\rho }} + z\cos (\phi -\omega t)\,\hat{\!\varvec{\phi }} -\rho \sin (\phi -\omega t)\hat{\mathbf{z}}}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)},\qquad \qquad \qquad \qquad \qquad \qquad \,\,\, \end{aligned}$$

(C18)

$$\begin{aligned} \frac{({\widehat{\varvec{\mathcal {R}}}}\cdot {\dot{\varvec{\beta }}}){\widehat{\varvec{\mathcal {R}}}}\times \varvec{\beta }}{{{{\mathcal {R}}}}}= & {} \frac{\omega ^3 a^2}{c^2}\frac{[\rho \cos (\phi \!-\!\omega t) -a][z\cos (\phi \!-\!\omega t)\,\hat{\!\varvec{\rho }}-z\sin (\phi \!-\!\omega t)\,\hat{\!\varvec{\phi }} - [\rho \cos (\phi \!-\!\omega t)-a] \hat{\mathbf{z}}]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t)]^{3/2}}.\qquad \end{aligned}$$

(C19)

Now, in cylindrical coordinates we can decompose the fields (C3) and (C4) in the following form:

$$\begin{aligned} \varvec{\mathcal {E}}= & {} ({{{\mathcal {E}}}}^{v}_{\rho } +{{{\mathcal {E}}}}^{a}_{\rho })\,\hat{\!\varvec{\rho }} + ({{{\mathcal {E}}}}^{v}_{\phi } +{{{\mathcal {E}}}}^{a}_{\phi })\,\hat{\!\varvec{\phi }} + ({{{\mathcal {E}}}}^{v}_{z} +{{{\mathcal {E}}}}^{a}_{z})\hat{\mathbf{z}},\,\, \end{aligned}$$

(C20)

$$\begin{aligned} \varvec{\mathcal {B}}= & {} ({{{\mathcal {B}}}}^{v}_{\rho } +{{{\mathcal {B}}}}^{a}_{\rho })\,\hat{\!\varvec{\rho }} + ({{{\mathcal {B}}}}^{v}_{\phi } +{{{\mathcal {B}}}}^{a}_{\phi })\,\hat{\!\varvec{\phi }} + ({{{\mathcal {B}}}}^{v}_{z} +{{{\mathcal {B}}}}^{a}_{z})\hat{\mathbf{z}}, \end{aligned}$$

(C21)

where ${{{\mathcal {E}}}}^v_\rho ,{{{\mathcal {E}}}}^v_\phi ,{{{\mathcal {E}}}}^v_z$ and ${{{\mathcal {B}}}}^v_\rho ,{{{\mathcal {B}}}}^v_\phi ,{{{\mathcal {B}}}}^v_z$ are the components of the velocity fields (those varying as $1/{{{\mathcal {R}}}}^2$) and ${{{\mathcal {E}}}}^a_\rho ,{{{\mathcal {E}}}}^a_\phi ,{{{\mathcal {E}}}}^a_z$ and ${{{\mathcal {B}}}}^a_\rho ,{{{\mathcal {B}}}}^a_\phi ,{{{\mathcal {B}}}}^a_z$ are the components of the acceleration fields (those varying as $1/{{{\mathcal {R}}}}$). Using (C15)–(C19) evaluated at the retarded time $t_r=t-{{{\mathcal {R}}}}(t_r)/c$ and (C3), we find the components of the electric field of the encircling dyon

$$\begin{aligned} {{{\mathcal {E}}}}^{v}_{\rho }= & {} q\frac{[\rho - a \cos (\phi -\omega t_r)]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}-g\frac{\omega a}{c}\frac{z\cos (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}},\qquad \qquad \,\, \end{aligned}$$

(C22)

$$\begin{aligned} {{{\mathcal {E}}}}^{v}_{\phi }= & {} q\frac{a\sin (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}+g\frac{\omega a}{c}\frac{z\sin (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}},\qquad \qquad \,\, \end{aligned}$$

(C23)

$$\begin{aligned} {{{\mathcal {E}}}}^{v}_{z}= & {} q\frac{z}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}+g\frac{\omega a}{c}\frac{[\rho \cos (\phi -\omega t_r)-a]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}},\qquad \qquad \,\, \end{aligned}$$

(C24)

$$\begin{aligned} {{{\mathcal {E}}}}^{a}_{\rho }= & {} q\frac{\omega ^2a}{c^2}\frac{a\rho \sin ^2(\phi -\omega t_r) +z^2\cos (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}} + g\frac{\omega ^3 a^2}{c^3}\frac{z\cos (\phi -\omega t_r)(\rho \cos (\phi - \omega t_r)-a)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}\qquad \, \nonumber \\&- g\frac{\omega ^2 a}{c^2}\frac{z\sin (\phi -\omega t_r)}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)},\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\,\,\, \end{aligned}$$

(C25)

$$\begin{aligned} {{{\mathcal {E}}}}^{a}_{\phi }= & {} -q\frac{\omega ^2 a}{c^2}\frac{\sin (\phi -\omega t_r)[\rho ^2 - \rho a \cos (\phi - \omega t_r) + z^2]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}-g\frac{\omega ^3 a^2}{c^3}\frac{z\sin (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}} \nonumber \\&-g\frac{\omega ^2 a}{c^2}\frac{z\cos (\phi -\omega t_r)}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)},\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\, \end{aligned}$$

(C26)

$$\begin{aligned} {{{\mathcal {E}}}}^{a}_{z}= & {} -q\frac{\omega ^2 a}{c^2}\frac{z[\rho \cos (\phi - \omega t_r) -a]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}} -g\frac{\omega ^3 a^2}{c^3}\frac{[\rho \cos (\phi -\omega t_r)-a]^2}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}\quad \,\,\, \nonumber \\&+g\frac{\omega ^2 a}{c^2}\frac{\rho \sin (\phi -\omega t_r)}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)}.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\,\, \end{aligned}$$

(C27)

Using the dual changes $q\rightarrow g$ and $g\rightarrow -q$ in (C22)–(C27), we obtain the corresponding components of the magnetic field of the encircling dyon

$$\begin{aligned} {{{\mathcal {B}}}}^{v}_{\rho }= & {} g\frac{[\rho - a \cos (\phi -\omega t_r)]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}+q\frac{\omega a}{c}\frac{z\cos (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}},\qquad \qquad \,\, \end{aligned}$$

(C28)

$$\begin{aligned} {{{\mathcal {B}}}}^{v}_{\phi }= & {} g\frac{a\sin (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}-q\frac{\omega a}{c}\frac{z\sin (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}},\qquad \qquad \,\, \end{aligned}$$

(C29)

$$\begin{aligned} {{{\mathcal {B}}}}^{v}_{z}= & {} g\frac{z}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}-q\frac{\omega a}{c}\frac{[\rho \cos (\phi -\omega t_r)-a]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}},\qquad \qquad \,\, \end{aligned}$$

(C30)

$$\begin{aligned} {{{\mathcal {B}}}}^{a}_{\rho }= & {} g\frac{\omega ^2a}{c^2}\frac{a\rho \sin ^2(\phi -\omega t_r) +z^2\cos (\phi -\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}} - q\frac{\omega ^3 a^2}{c^3}\frac{z\cos (\phi -\omega t_r)(\rho \cos (\phi - \omega t_r)-a)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}\qquad \, \nonumber \\&+ q\frac{\omega ^2 a}{c^2}\frac{z\sin (\phi -\omega t_r)}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)},\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\,\,\, \end{aligned}$$

(C31)

$$\begin{aligned} {{{\mathcal {B}}}}^{a}_{\phi }= & {} -g\frac{\omega ^2 a}{c^2}\frac{\sin (\phi -\omega t_r)[\rho ^2 - \rho a \cos (\phi - \omega t_r) + z^2]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}+q\frac{\omega ^3 a^2}{c^3}\frac{z\sin (\phi \!-\!\omega t_r)}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}} \nonumber \\&+q\frac{\omega ^2 a}{c^2}\frac{z\cos (\phi -\omega t_r)}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)},\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\, \end{aligned}$$

(C32)

$$\begin{aligned} {{{\mathcal {B}}}}^{a}_{z}= & {} -g\frac{\omega ^2 a}{c^2}\frac{z[\rho \cos (\phi - \omega t_r) -a]}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}} +q\frac{\omega ^3 a^2}{c^3}\frac{[\rho \cos (\phi -\omega t_r)-a]^2}{[\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)]^{3/2}}\quad \,\,\, \nonumber \\&-q\frac{\omega ^2 a}{c^2}\frac{\rho \sin (\phi -\omega t_r)}{\rho ^2 + z^2 +a^2-2 \rho a\cos (\phi -\omega t_r)}.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\,\, \end{aligned}$$

(C33)

Equations (C20)–(C33) describe the components of the fields of a non-relativistic dyon in uniform circular motion expressed in terms of the retarded time $t_r=t-{{{\mathcal {R}}}}(t_r)/c= t-|\mathbf{x} - \mathbf{x}_{qg}(t_r)|/c,$ which implicitly depends on the retarded time itself.

Appendix D. Eliminating the implicit dependence of the retarded time in the fields

In this appendix, we will specify the conditions that allow us to eliminate the implicit dependence of the retarded time in the fields of the dyon–solenoid configuration when the dyon is in uniform circular motion, i.e., $t_r= t-|\mathbf{x} - \mathbf{x}_{qg}(t_r)|/c \rightarrow t_r=t-|\mathbf{x}|/c$. Since the retarded time in the field components (C22)–(C33) is only present in the functions $\sin (\phi - \omega t_r)$ and $\cos (\phi -\omega t_r)$, these functions are our quantities of interest. We will see that under certain conditions we can make the replacements $\sin [\phi - \omega (t_r=t-|\mathbf{x} - \mathbf{x}_{qg}(t_r)|/c)]\rightarrow \sin [\phi - \omega (t_r=t-|\mathbf{x}|/c)]$ and $\cos [\phi - \omega (t_r=t-|\mathbf{x} - \mathbf{x}_{qg}(t_r)|/c)]\rightarrow \cos [\phi - \omega (t_r=t-|\mathbf{x}|/c)]$ and thus eliminate the implicit dependence of the retarded time in the fields.

We now follow an argument similar to that given by Eyges [21]. Using the identities $\sin (A - B)=\sin A\cos B - \cos A \sin B$ and $\cos (A-B)=\sin A\sin B+ \cos A\cos B$, it follows

$$\begin{aligned} \sin (\phi - \omega t_r)=\sin \phi \cos (\omega t_r) - \cos \phi \sin (\omega t_r), \end{aligned}$$

(D1)

$$\begin{aligned} \cos (\phi -\omega t_r)=\sin \phi \sin (\omega t_r)+ \cos \phi \cos (\omega t_r), \end{aligned}$$

(D2)

and inserting the retarded time $t_r=t-|\mathbf{x} - \mathbf{x}_{qg}(t_r)|/c$ we obtain

$$\begin{aligned} \sin (\omega t_r)= \sin \bigg [\omega \bigg (t -\frac{|\mathbf{x}-\mathbf{x}_{qg}(t_r)|}{c}\bigg ) \bigg ],\,\,\,\cos (\omega t_r)= \cos \bigg [\omega \bigg (t -\frac{|\mathbf{x}-\mathbf{x}_{qg}(t_r)|}{c}\bigg ) \bigg ]. \end{aligned}$$

(D3)

The size of the considered distribution is of order a which is the radius of the orbit of the dyon. If a is sufficiently small, then we can make a Taylor expansion in $\mathbf{x}_{qg}(t_r)$ to the first order in the trigonometric functions in (D3) obtaining

$$\begin{aligned} \sin (\omega t_r)\approx \sin \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ] + \mathbf{x}_{qg}(t_r)\cdot \nabla \sin \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ], \end{aligned}$$

(D4)

$$\begin{aligned} \sin (\omega t_r)\approx \cos \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ] + \mathbf{x}_{qg}(t_r)\cdot \nabla \cos \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]. \end{aligned}$$

(D5)

From (C10), it follows $\mathbf{x}_{qg}(t_r)=a[\cos (\phi - \omega t_r)\,\hat{\!\varvec{\rho }} -\sin (\phi -\omega t_r) \,\hat{\!\varvec{\phi }}]$, which is used together with $\nabla = [\partial /\partial \rho ]\,\hat{\!\varvec{\rho }} + (1/\rho )[\partial /\partial \phi ]\,\hat{\!\varvec{\phi }} + [\partial /\partial z]\hat{\mathbf{z}}$ and $|\mathbf{x}| = \sqrt{\rho ^2+z^2}$ to obtain

$$\begin{aligned} \sin (\omega t_r)\approx \sin \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ] -\frac{\rho a \omega }{c |\mathbf{x}|}\cos \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c} \bigg ) \bigg ]\cos (\omega t_r), \end{aligned}$$

(D6)

$$\begin{aligned} \cos (\omega t_r)\approx \cos \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ] + \frac{\rho a \omega }{c|\mathbf{x}|}\sin \bigg [\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]\sin (\omega t_r). \end{aligned}$$

(D7)

Using (D6) and (D7) in (D1) and (D2) together with the identities $\sin (A - B)=\sin A\cos B - \cos A \sin B$ and $\cos (A-B)=\sin A\sin B+ \cos A\cos B$, we obtain

$$\begin{aligned} \sin (\phi - \omega t_r)\approx \sin \bigg [\phi -\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]+\frac{\rho \beta }{|\mathbf{x}|}\cos \bigg [\phi -\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]\cos (\phi -\omega t_r), \end{aligned}$$

(D8)

$$\begin{aligned} \cos (\phi - \omega t_r)\approx \cos \bigg [\phi -\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]-\frac{\rho \beta }{|\mathbf{x}|}\sin \bigg [\phi -\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]\cos (\phi -\omega t_r), \end{aligned}$$

(D9)

where $\beta = |\dot{\mathbf{x}}_{qg}|/c=\omega a/c$ with $|\dot{\mathbf{x}}_{qg}|= \omega a$ being the magnitude of the velocity of the dyon. The ratio of the second terms on the right-hand side of (D8) and (D9) compared with their corresponding first terms is of order $\beta .$ For a non-relativistic dyon, $\beta<<1$ and therefore the second terms on the right-hand side are much smaller compared to the first terms. Accordingly, we can write write the approximation

$$\begin{aligned} \sin (\phi -\omega t_r)\approx \sin \bigg [\phi -\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ],\,\,\,\cos (\phi -\omega t_r)\approx \sin \bigg [\phi -\omega \bigg (t -\frac{|\mathbf{x}|}{c}\bigg ) \bigg ]. \end{aligned}$$

(D10)

Using Eqs. (D10) and (C20)–(C33), it follows that in the non-relativistic approximation of the fields due to a dyon in uniform circular motion, the retarded time can be expressed as $t_r= t- |\mathbf{x}|/c$ where now $\mathbf{x}$ does not depend on the retarded time itself. This property will allow us to integrate the associated fields given in Appendix E.

Appendix E. Proof of ${\varvec{L}}_{\mathrm{R}}=0$ for the dyon–solenoid configuration

Using the generalised Maxwell equations $\nabla \times \varvec{\mathcal {E}}=-(1/c)\partial \varvec{\mathcal {B}}/\partial t - (4\pi /c) \varvec{\mathcal {J}}_{m}$ and $\nabla \times \varvec{\mathcal {B}}= (1/c)\partial \varvec{\mathcal {E}}/\partial t + (4\pi /c)\varvec{\mathcal {J}}_{e}$, we can write Eq. (48) as

$$\begin{aligned} {{\varvec{L}}}_{\mathrm{R}}= & {} \frac{1}{4\pi c^2}\frac{\partial }{\partial t}\int _{V}\mathbf{x} \times \!\big (\varvec{\mathcal {E}}\times \mathbf{C}\big )\,d^3x+\frac{1}{c^2}\int _{V}\mathbf{x} \times \big (\varvec{\mathcal {J}}_{e}\times \mathbf{C} \big )d^3x\qquad \nonumber \\&-\frac{1}{4\pi c^2}\frac{\partial }{\partial t}\int _{V}\mathbf{x} \times \!\big (\varvec{\mathcal {B}}\times \mathbf{A}\big )\,d^3x-\frac{1}{c^2} \int _{V}\mathbf{x} \times \big (\varvec{\mathcal {J}}_{m}\times \mathbf{A} \big )d^3x.\,\,\, \end{aligned}$$

(E1)

We can now apply (D1) to the dyon–solenoid configuration covering all space expect the dual solenoid. Inserting the position vector $\mathbf{x}= \rho {\hat{\varvec{\rho }}} + z \hat{\mathbf{z}},$ the potentials outside the dual solenoid $\mathbf{A}=\mathbf{A}_{\mathrm{out}}(\mathbf{x})=\varPhi _{m}\, \hat{\!\varvec{\phi }}/(2\pi \rho )$ and $\mathbf{C}=\mathbf{C}_{\mathrm{out}}(\mathbf{x})=-\varPhi _{e}\, \hat{\!\varvec{\phi }}/(2\pi \rho ),$ and $d^3x=\rho d\rho d\phi dz$ in (D1), we obtain

$$\begin{aligned} {{\varvec{L}}}_{\mathrm{R}}= & {} -\frac{\varPhi _{e}}{8\pi ^2 c^2}\frac{\partial }{\partial t}\bigg [ \int _{V}\rho \,\hat{\!\varvec{\rho }}\times \!\big (\varvec{\mathcal {E}}\times \,\hat{\!\varvec{\phi }}\big )\,d\rho d\phi dz + \int _{V}z\hat{\mathbf{z}}\times \!\big (\varvec{\mathcal {E}}\times \,\hat{\!\varvec{\phi }}\big )\, d\rho d\phi dz \bigg ] \,\,\, \nonumber \\&-\frac{\varPhi _{e}}{2\pi c^2}\bigg [\int _{V} \rho \,\hat{\!\varvec{\rho }}\times \!\big (\varvec{\mathcal {J}}_{e}\times \,\hat{\!\varvec{\phi }}\big )\, d\rho d\phi dz+\int _{V}z\hat{\mathbf{z}}\times \!\big (\varvec{\mathcal {J}}_{e}\times \,\hat{\!\varvec{\phi }}\big )\, d\rho d\phi dz \bigg ]\quad \nonumber \\&- \frac{\varPhi _{m}}{8\pi ^2 c^2}\frac{\partial }{\partial t} \bigg [\int _{V}\rho \,\hat{\!\varvec{\rho }}\times \!\big (\varvec{\mathcal {B}}\times \,\hat{\!\varvec{\phi }}\big )\,d\rho d\phi dz+\int _{V}z\hat{\mathbf{z}}\times \!\big (\varvec{\mathcal {B}}\times \,\hat{\!\varvec{\phi }}\big )\,d\rho d\phi dz\bigg ]\, \nonumber \\&- \frac{\varPhi _{m}}{2\pi c^2}\bigg [\int _{V} \rho \,\hat{\!\varvec{\rho }}\times \!\big (\varvec{\mathcal {J}}_{m}\times \,\hat{\!\varvec{\phi }}\big )\, d\rho d\phi dz + \int _{V}z\hat{\mathbf{z}}\times \!\big (\varvec{\mathcal {J}}_{m}\times \,\hat{\!\varvec{\phi }}\big )\, d\rho d\phi dz \bigg ]. \end{aligned}$$

(E2)

The fields are Liénard–Wiechert fields produced by the encircling dyon which from (C20) and (C21) can be decomposed as $\varvec{\mathcal {E}}= ({{{\mathcal {E}}}}^{v}_{\rho } +{{{\mathcal {E}}}}^{a}_{\rho })\,\hat{\!\varvec{\rho }} + ({{{\mathcal {E}}}}^{v}_{\phi } +{{{\mathcal {E}}}}^{a}_{\phi })\,\hat{\!\varvec{\phi }} + ({{{\mathcal {E}}}}^{v}_{z} +{{{\mathcal {E}}}}^{a}_{z})\hat{\mathbf{z}}$ and $\varvec{\mathcal {B}}= ({{{\mathcal {B}}}}^{v}_{\rho } +{{{\mathcal {B}}}}^{a}_{\rho })\,\hat{\!\varvec{\rho }} + ({{{\mathcal {B}}}}^{v}_{\phi } +{{{\mathcal {B}}}}^{a}_{\phi })\,\hat{\!\varvec{\phi }} + ({{{\mathcal {B}}}}^{v}_{z} +{{{\mathcal {B}}}}^{a}_{z})\hat{\mathbf{z}}$ where ${{{\mathcal {E}}}}^v_\rho ,{{{\mathcal {E}}}}^v_\phi ,{{{\mathcal {E}}}}^v_z$ and ${{{\mathcal {B}}}}^v_\rho ,{{{\mathcal {B}}}}^v_\phi ,{{{\mathcal {B}}}}^v_z$ are the components of the velocity fields and ${{{\mathcal {E}}}}^a_\rho ,{{{\mathcal {E}}}}^a_\phi ,{{{\mathcal {E}}}}^a_z$ and ${{{\mathcal {B}}}}^a_\rho ,{{{\mathcal {B}}}}^a_\phi ,{{{\mathcal {B}}}}^a_z$ are the components of the acceleration fields . The currents of the encircling dyon can be written as $\varvec{\mathcal {J}}_{e}=({{{\mathcal {J}}}}_e)_\rho \,\hat{\!\varvec{\rho }}+({{{\mathcal {J}}}}_e)_\phi \,\hat{\!\varvec{\phi }}+({{{\mathcal {J}}}}_e)_z\hat{\mathbf{z}}$ and $\varvec{\mathcal {J}}_{m}=({{{\mathcal {J}}}}_m)_\rho \,\hat{\!\varvec{\rho }}+({{{\mathcal {J}}}}_m)_\phi \,\hat{\!\varvec{\phi }}+({{{\mathcal {J}}}}_m)_z\hat{\mathbf{z}}$. Using these results in (E2) and performing the specified operations, we obtain

$$\begin{aligned} {{\varvec{L}}}_{\mathrm{R}}= & {} \frac{\varPhi _{e}\,\hat{\!\varvec{\phi }}}{8\pi ^2 c^2}\bigg [\frac{\partial I_1}{\partial t} + \frac{\partial I_2}{\partial t}+\frac{\partial I_3}{\partial t} + \frac{\partial I_4}{\partial t}\bigg ]+\frac{\varPhi _{e}\,\hat{\!\varvec{\phi }}}{2\pi c^2}[I_5 + I_{6}]+ \frac{\varPhi _{m}\,\hat{\!\varvec{\phi }}}{8\pi ^2 c^2}\bigg [\frac{\partial I_7}{\partial t} + \frac{\partial I_8}{\partial t}+\frac{\partial I_9}{\partial t} + \frac{\partial I_{10}}{\partial t}\bigg ] \nonumber \\&+ \frac{\varPhi _{m}\,\hat{\!\varvec{\phi }}}{2\pi c^2}[I_{11} + I_{12}], \end{aligned}$$

(E3)

where the integrals $I_{1}-I_{12}$ are defined as

$$\begin{aligned} I_1= & {} \int _{V}\rho {{{\mathcal {E}}}}^{v}_{\rho }\,d\rho d\phi dz,\,\, I_2=\int _{V}\rho {{{\mathcal {E}}}}^{a}_{\rho }\,d\rho d\phi dz,\,\, I_3= \int _{V}z {{{\mathcal {E}}}}^{v}_{z}\, d\rho d\phi dz,\,\,\,I_4= \int _{V}z {{{\mathcal {E}}}}^{a}_{z}\, d\rho d\phi dz,\,\,\,\,\, \end{aligned}$$

(E4)

$$\begin{aligned} I_5= & {} \int _{V}\rho ({{{\mathcal {J}}}}_{e})_{\rho }\,d\rho d\phi dz,\,\, I_{6}= \int _{V}z ({{{\mathcal {J}}}}_{e})_{z}\, d\rho d\phi dz,\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \,\,\,\,\, \end{aligned}$$

(E5)

$$\begin{aligned} I_7= & {} \int _{V}\rho {{{\mathcal {B}}}}^{v}_{\rho }\,d\rho d\phi dz,\,\, I_8=\int _{V}\rho {{{\mathcal {B}}}}^{a}_{\rho }d\rho d\phi dz,\,\, I_9= \int _{V}z {{{\mathcal {B}}}}^{v}_{z}\, d\rho d\phi dz,\,\,I_{10}= \int _{V}z {{{\mathcal {B}}}}^{a}_{z}\, d\rho d\phi dz,\,\,\,\, \end{aligned}$$

(E6)

$$\begin{aligned} I_{11}= & {} \int _{V}\rho ({{{\mathcal {J}}}}_{m})_{\rho }\,d\rho d\phi dz,\quad I_{12}= \int _{V}z ({{{\mathcal {J}}}}_{m})_{z}\, d\rho d\phi dz,\qquad \qquad \qquad \qquad \qquad \qquad \qquad \,\,\,\,\,\,\, \end{aligned}$$

(E7)

The field components ${{{\mathcal {E}}}}^v_\rho ,{{{\mathcal {E}}}}^v_\phi ,{{{\mathcal {E}}}}^v_z,{{{\mathcal {E}}}}^a_\rho ,{{{\mathcal {E}}}}^a_\phi ,{{{\mathcal {E}}}}^a_z$ and ${{{\mathcal {B}}}}^v_\rho ,{{{\mathcal {B}}}}^v_\phi ,{{{\mathcal {B}}}}^v_z,{{{\mathcal {B}}}}^a_\rho ,{{{\mathcal {B}}}}^a_\phi ,{{{\mathcal {B}}}}^a_z$ as well as the current components $({{{\mathcal {J}}}}_e)_\rho ,({{{\mathcal {J}}}}_e)_\phi ,({{{\mathcal {J}}}}_e)_z$ and $({{{\mathcal {J}}}}_m)_\rho ,({{{\mathcal {J}}}}_m)_\phi ,({{{\mathcal {J}}}}_m)_z$ should now be specified. For simplicity, we consider the fields and currents due to a dyon encircling the dual solenoid in non-relativistic uniform circular motion along the x-y plane. More specifically, we assume that the dyon is moving with constant angular velocity $\omega$ in a circle of fixed radius a around the dual solenoid. Field coordinates are denoted as $\rho ,\phi ,z$ and the source coordinates specifying the position of the dyon are denoted as $a,\omega t,0.$ In this case, we can use the field components in (C23)–(C33) together with the approximation in (D10). To find the components of the current densities, we use $\varvec{\mathcal {J}}_{e}= q \dot{\mathbf{x}}_{qg}\delta [\mathbf{x}- \mathbf{x}_{qg}(t)]$ and $\varvec{\mathcal {J}}_{m}= g \dot{\mathbf{x}}_{qg}\delta [\mathbf{x}- \mathbf{x}_{qg}(t)]$ together with (C11) and $\delta [\mathbf{x}- \mathbf{x}_{qg}(t)]=\delta (\rho - a)\delta (\phi - \omega t)\delta (z)/\rho$ to obtain $\varvec{\mathcal {J}}_{e}=[q\omega a/\rho ][\sin (\phi - \omega t)\,\hat{\!\varvec{\rho }} +\cos (\phi -\omega t) \,\hat{\!\varvec{\phi }}]\delta (\rho - a)\delta (\phi - \omega t)\delta (z)$ and $\varvec{\mathcal {J}}_{m}=[g\omega a/\rho ][\sin (\phi - \omega t)\,\hat{\!\varvec{\rho }} +\cos (\phi -\omega t) \,\hat{\!\varvec{\phi }}]\delta (\rho - a)\delta (\phi - \omega t)\delta (z)$. Therefore,

$$\begin{aligned} ({{{\mathcal {J}}}}_{e})_{\rho }=\frac{q \omega a}{\rho } \sin (\phi - \omega t)\delta (\rho - a)\delta (\phi - \omega t)\delta (z),\,\,\,\,\, \end{aligned}$$

(E8)

$$\begin{aligned} ({{{\mathcal {J}}}}_{e})_{\phi }=\frac{q \omega a}{\rho } \cos (\phi - \omega t)\delta (\rho - a)\delta (\phi - \omega t)\delta (z),\,\,\,\, \end{aligned}$$

(E9)

$$\begin{aligned} ({{{\mathcal {J}}}}_{e})_{z}=0,\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \,\,\,\, \end{aligned}$$

(E10)

$$\begin{aligned} ({{{\mathcal {J}}}}_{m})_{\rho }=\frac{g \omega a}{\rho } \sin (\phi - \omega t)\delta (\rho - a)\delta (\phi - \omega t)\delta (z),\,\,\,\,\, \end{aligned}$$

(E11)

$$\begin{aligned} ({{{\mathcal {J}}}}_{m})_{\phi }=\frac{g \omega a}{\rho } \cos (\phi - \omega t)\delta (\rho - a)\delta (\phi - \omega t)\delta (z),\,\,\,\, \end{aligned}$$

(E12)

$$\begin{aligned} ({{{\mathcal {J}}}}_{m})_{z}=0.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \end{aligned}$$

(E13)

Let us now proceed to evaluate the components within the brackets in (E3).

• (i) Proof of ${\partial I_1/\partial t=0}$ Using (C22) and $I_1$ in (E4), it follows

$$\begin{aligned} I_1= & {} q\int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty } dz \oint _{C}\frac{\left[ \rho - a \cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] \,d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \,\,\, \nonumber \\&-g\frac{\omega a}{c}\int ^{\infty }_{R}\rho d\rho \oint _{C}d\phi \int ^{+\infty }_{-\infty }\frac{z\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}.\qquad \end{aligned}$$

(E14)

We observe that the integrand in the second term on the right-hand side is an odd function of z. Therefore, it directly follows

$$\begin{aligned} \int ^{+\infty }_{-\infty }\frac{z\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}=0, \end{aligned}$$

(E15)

which is used in (E14) to obtain

$$\begin{aligned} I_1= q\int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty } dz \oint _{C}\frac{\left[ \rho - a \cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] \,d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}. \end{aligned}$$

(E16)

Inserting the operator $\partial /\partial t$ in (E16), we obtain

$$\begin{aligned} \frac{\partial I_1}{\partial t}= q\int ^{\infty }_{R}\rho ^2 d\rho \int ^{+\infty }_{-\infty } dz \oint _{C}\frac{\partial }{\partial t}\left[ \frac{\left[ \rho - a \cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] \,d\phi . \end{aligned}$$

(E17)

The partial derivative gives

$$\begin{aligned}&\frac{\partial }{\partial t}\left[ \frac{\left[ \rho - a \cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] \qquad \qquad \qquad \qquad \qquad \qquad \qquad \nonumber \\&\quad = - \frac{a\omega \sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \qquad \qquad \quad \,\,\,\,\,\,\, \nonumber \\&\qquad +\frac{3a\rho ^2\omega \sin \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}\qquad \qquad \qquad \qquad \qquad \nonumber \\&\qquad -\frac{3a^2\rho \omega \sin \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}},\qquad \quad \qquad \end{aligned}$$

(E18)

which is used in (E17) to obtain

$$\begin{aligned} \frac{\partial I_1}{\partial t}= & {} -qa \omega \int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty } dz \oint _{C}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \quad \,\,\,\, \nonumber \\&+3 q a\omega \int ^{\infty }_{R}\rho ^3 d\rho \int ^{+\infty }_{-\infty } dz \oint _{C}\frac{\sin \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}\qquad \qquad \,\,\,\,\nonumber \\&-3 q a^2\omega \int ^{\infty }_{R}\rho ^2 d\rho \int ^{+\infty }_{-\infty }\!\! dz \oint _{C}\frac{\sin \left( \phi \!-\!\omega \left( t\!-\! \sqrt{\rho ^2 \!+\! z^2}/c\right) \right) \cos \left( \phi \!-\!\omega \left( t\!-\! \sqrt{\rho ^2 \!+\! z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}.\quad \,\,\,\,\,\,\,\,\,\,\,\, \end{aligned}$$

(E19)

The azimuthal integrals can be evaluated via a Contour integration. Here, we use Mathematica to calculate them

$$\begin{aligned} \int ^{2\pi n}_{0}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}=0,\qquad \qquad \end{aligned}$$

(E20)

$$\begin{aligned} \int ^{2\pi n}_{0}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}=0,\qquad \qquad \end{aligned}$$

(E21)

$$\begin{aligned} \int ^{2\pi n}_{0}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \cos \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t- \sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}=0,\quad \end{aligned}$$

(E22)

where n is the winding number. Using (E20)–(E22) in (E19), we obtain

$$\begin{aligned} \frac{\partial I_1}{\partial t}=0. \end{aligned}$$

(E23)

• (ii) Proof of ${\partial I_2/\partial t=0}$ Using (C25) and $I_2$ in (E4), we obtain

$$\begin{aligned} I_2= & {} q\frac{\omega ^2a^2}{c^2} \int ^{\infty }_{R}\rho ^2 d\rho \int ^{+\infty }_{-\infty } dz \oint _{C} \frac{\sin ^2\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \nonumber \\&+q\frac{\omega ^2a}{c^2} \int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C} \frac{\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\nonumber \\&+ g\frac{\omega ^3 a^2}{c^3}\int ^{\infty }_{R}\rho ^2 d\rho \oint _{C}d\phi \int ^{+\infty }_{-\infty } \frac{z\cos ^2\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}} \nonumber \\&-g\frac{\omega ^3 a^3}{c^3}\int ^{\infty }_{R}\rho d\rho \oint _{C}d\phi \int ^{+\infty }_{-\infty }\frac{z \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}} \nonumber \\&- g\frac{\omega ^2 a}{c^2}\int ^{\infty }_{R}\rho d\rho \oint _{C}d\phi \int ^{+\infty }_{-\infty } \frac{z\sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }.\qquad \end{aligned}$$

(E24)

Using (E15), the fourth term on the right-hand side of (E24) vanishes. We observe that the integrand in the fifth term of (24) is an odd function of z and therefore it directly follows

$$\begin{aligned} \int ^{+\infty }_{-\infty }\frac{z\sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }=0. \end{aligned}$$

(E25)

On the other hand, the integrand in the second term of the right-hand side of (E24) is also an odd function of z and therefore

$$\begin{aligned} \int ^{+\infty }_{-\infty }\frac{z\cos ^2\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}=0. \end{aligned}$$

(E26)

Using (E15), (E25) and (E26) in (E24), we obtain

$$\begin{aligned} I_2= & {} q\frac{\omega ^2a^2}{c^2} \int ^{\infty }_{R}\rho ^2 d\rho \int ^{+\infty }_{-\infty } dz \oint _{C} \frac{\sin ^2\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \nonumber \\&+q\frac{\omega ^2a}{c^2} \int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C} \frac{\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}. \end{aligned}$$

(E27)

Inserting the operator $\partial /\partial t$ in (E27), we obtain

$$\begin{aligned} \frac{\partial I_2}{\partial t}= & {} q\frac{\omega ^2a^2}{c^2} \int ^{\infty }_{R}\rho ^2 d\rho \int ^{+\infty }_{-\infty } dz \oint _{C} \frac{\partial }{\partial t}\left[ \frac{\sin ^2\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] d\phi \qquad \nonumber \\&+q\frac{\omega ^2a}{c^2} \!\int ^{\infty }_{R}\rho d\rho \!\int ^{+\infty }_{-\infty }\!\!z^2 dz \!\oint _{C} \frac{\partial }{\partial t}\left[ \frac{\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 \!+\! z^2 \!+\!a^2\!-\!2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] d\phi .\quad \,\,\,\,\,\, \end{aligned}$$

(E28)

The partial derivatives in (E28) give

$$\begin{aligned}&\frac{\partial }{\partial t}\left[ \frac{\sin ^2\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] \qquad \qquad \qquad \qquad \qquad \qquad \qquad \,\,\,\,\nonumber \\&\quad = -\frac{2\omega \sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \,\,\,\,\nonumber \\&\qquad +\frac{3 a \rho \sin ^3\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}},\qquad \qquad \qquad \quad \,\,\, \end{aligned}$$

(E29)

$$\begin{aligned}&\frac{\partial }{\partial t}\left[ \frac{\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 \!+\! z^2 \!+\!a^2\!-\!2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \nonumber \\&\quad =\frac{\omega \sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 \!+\! z^2 \!+\!a^2\!-\!2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \qquad \qquad \qquad \nonumber \\&\qquad + \frac{3a\rho \omega \sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 \!+\! z^2 \!+\!a^2\!-\!2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}},\qquad \quad \,\,\, \end{aligned}$$

(E30)

which are used in (E28) to obtain

$$\begin{aligned} \frac{\partial I_2}{\partial t}= & {} -2q\omega \frac{\omega ^2a^2}{c^2} \!\int ^{\infty }_{R}\rho ^2\! d\rho \!\int ^{+\infty }_{-\infty }\!\! dz \oint _{C} \frac{\sin \left( \phi \!-\!\omega \left( t\!-\!\sqrt{\rho ^2 \!+\! z^2}/c\right) \right) \cos \left( \phi \!-\!\omega \left( t\!-\!\sqrt{\rho ^2 \!+\! z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \nonumber \\&+q\frac{3\omega ^2a^3}{c^2} \int ^{\infty }_{R}\rho ^3 d\rho \int ^{+\infty }_{-\infty } dz \oint _{C}\frac{\sin ^3\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}\qquad \,\,\,\nonumber \\&+q\frac{\omega ^3a}{c^2} \!\int ^{\infty }_{R}\rho d\rho \!\int ^{+\infty }_{-\infty }\!\!z^2 dz \!\oint _{C} \frac{\sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 \!+\! z^2 \!+\!a^2\!-\!2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \,\,\,\nonumber \\&+3q\frac{\omega ^3a^2}{c^2} \!\!\int ^{\infty }_{R}\rho ^2 d\rho \!\int ^{+\infty }_{-\infty }\!\!z^2 dz \!\oint _{C}\frac{\sin \left( \phi \!-\!\omega \left( t\!-\!\sqrt{\rho ^2 \!+\! z^2}/c\right) \right) \cos \left( \phi \!-\!\omega \left( t\!-\!\sqrt{\rho ^2 \!+\! z^2}/c\right) \right) d\phi }{\left[ \rho ^2 \!+\! z^2 \!+\!a^2\!-\!2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}.\quad \,\,\, \end{aligned}$$

(E31)

The third term on the right-hand side of (E31) vanishes on account of (E20), and the fourth term on the right-hand side vanishes on account of (E22). The remaining azimuthal integrals may be evaluated via a Contour integration. Here, we use Mathematica to calculate them

$$\begin{aligned}&\int ^{2\pi n}_{0} \frac{\sin \left( \phi \!-\!\omega \left( t\!-\!\sqrt{\rho ^2 \!+\! z^2}/c\right) \right) \cos \left( \phi \!-\!\omega \left( t\!-\!\sqrt{\rho ^2 \!+\! z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}=0, \end{aligned}$$

(E32)

$$\begin{aligned}&\int ^{2\pi n}_{0}\frac{\sin ^3\left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}=0. \end{aligned}$$

(E33)

Using (E20), (E22) and (E32)–(E33) in (E31), we obtain

$$\begin{aligned} \frac{\partial I_2}{\partial t}=0. \end{aligned}$$

(E34)

• (iii) Proof of ${\partial I_3/\partial t=0}$ Using (C24) and $I_3$ in (E4), we obtain

$$\begin{aligned} I_3= & {} q\int ^{\infty }_{R} d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \nonumber \\&+g\frac{\omega a}{c} \int ^{\infty }_{R}d\rho \oint _{C}d\phi \int ^{+\infty }_{-\infty } \frac{z\left[ \rho \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}. \end{aligned}$$

(E35)

The integrand in the second term of the right-hand side is an odd function of z, and therefore

$$\begin{aligned} \int ^{+\infty }_{-\infty } \frac{z\left[ \rho \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}=0. \end{aligned}$$

(E36)

Using (E36) in (E35), we obtain

$$\begin{aligned} I_3 = q\int ^{\infty }_{R} d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}. \end{aligned}$$

(E37)

Inserting the operator $\partial /\partial t$ in (E37), we obtain

$$\begin{aligned} \frac{\partial I_3}{\partial t}= q\int ^{\infty }_{R} d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{\partial }{\partial t}\left[ \frac{1}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] d\phi . \end{aligned}$$

(E38)

The partial derivative gives

$$\begin{aligned}&\frac{\partial }{\partial t}\left[ \frac{1}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] \qquad \qquad \qquad \nonumber \\&\quad =\frac{3 a \rho \omega \sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}, \end{aligned}$$

(E39)

which is used in (E38) to obtain

$$\begin{aligned} \frac{\partial I_3}{\partial t}= 3qa\omega \int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C} \frac{\sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}. \end{aligned}$$

(E40)

The azimuthal integral vanishes on account of (E21), and therefore we obtain

$$\begin{aligned} \frac{\partial I_3}{\partial t}=0. \end{aligned}$$

(E41)

• (iv) Proof of ${\partial I_4/\partial t=0}$. Using (C27) and $I_4$ in (E4), we obtain

$$\begin{aligned} I_4= & {} -q\frac{\omega ^2 a}{c^2}\int ^{\infty }_{R} d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C} \frac{\left[ \rho \cos \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}} \nonumber \\&-g\frac{\omega ^3 a^2}{c^3}\int ^{\infty }_{R} d\rho \oint _{C}d \phi \int ^{+\infty }_{-\infty } \frac{z\left[ \rho \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] ^2 dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\nonumber \\&+g\frac{\omega ^2 a}{c^2}\int ^{\infty }_{R}\rho d\rho \oint _{C}d\phi \int ^{+\infty }_{-\infty } \frac{z\sin \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) dz}{\rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) },\,\,\,\, \end{aligned}$$

(E42)

The third term on the right-hand side of (E42) vanishes on account of (E25). The integrand in the second term is an odd function of z, and therefore

$$\begin{aligned} \int ^{+\infty }_{-\infty }\frac{z\left[ \rho \cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] ^2 dz}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}=0, \end{aligned}$$

(E43)

which is used in (E42) together with (E25) to obtain

$$\begin{aligned} I_4=-q\frac{\omega ^2 a}{c^2}\int ^{\infty }_{R} d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C} \frac{\left[ \rho \cos \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}. \end{aligned}$$

(E44)

Inserting the operator $\partial /\partial t$ in (E44), we obtain

$$\begin{aligned} \frac{\partial I_4}{\partial t}= -q\frac{\omega ^2 a}{c^2}\int ^{\infty }_{R} d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{\partial }{\partial t} \left[ \frac{\left[ \rho \cos \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a\right] }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] d\phi .\qquad \end{aligned}$$

(E45)

The partial derivative gives

$$\begin{aligned}&\frac{\partial }{\partial t} \left[ \frac{\left[ \rho \cos (\phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) -a]}{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\right] \qquad \qquad \qquad \qquad \qquad \qquad \qquad \nonumber \\&\quad =\frac{\omega \rho \sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \qquad \qquad \nonumber \\&\qquad +\frac{3a\omega \rho ^2\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \cos \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}\qquad \,\,\,\nonumber \\&\qquad -\frac{3a^2\omega \rho \sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}},\qquad \qquad \qquad \,\,\,\,\,\,\, \end{aligned}$$

(E46)

which is used in (E45) to obtain

$$\begin{aligned}&\frac{\partial I_4}{\partial t}= -q\frac{\omega ^3 a}{c^2}\int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{3/2}}\qquad \qquad \nonumber \\&\qquad -3q\frac{\omega ^3 a^2}{c^2}\int ^{\infty }_{R}\rho ^2 d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \cos \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}\nonumber \\&\qquad +3q\frac{\omega ^3 a^3}{c^2}\int ^{\infty }_{R}\rho d\rho \int ^{+\infty }_{-\infty }z^2 dz \oint _{C}\frac{\sin \left( \phi - \omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) d\phi }{\left[ \rho ^2 + z^2 +a^2-2 \rho a\cos \left( \phi -\omega \left( t-\sqrt{\rho ^2 + z^2}/c\right) \right) \right] ^{5/2}}.\qquad \,\,\,\, \end{aligned}$$

(E47)

The three azimuthal integrals in the right-hand side of (E47) vanish on account of (E20)–(E22), and therefore we obtain

$$\begin{aligned} \frac{\partial I_4}{\partial t}=0. \end{aligned}$$

(E48)

• (v) Proof of ${I_5=0}$ Using (E9) and $I_5$ in (E5), we obtain

$$\begin{aligned} I_5=q \omega a \int ^{\infty }_{R}\delta (\rho - a) d\rho \int ^{+\infty }_{-\infty }\delta (z)dz \oint _{C} \sin (\phi - \omega t)\delta (\phi - \omega t)d\phi . \end{aligned}$$

(E49)

The first two integrals give

$$\begin{aligned} \int ^{\infty }_{R}\delta (\rho - a) d\rho =\varTheta (a-R)=1\,\,\,(a>R),\quad \int ^{+\infty }_{-\infty }\delta (z)dz=1, \end{aligned}$$

(E50)

which are used in (E49) to obtain

$$\begin{aligned} I_5=q \omega a\oint _{C} \sin (\phi - \omega t)\delta (\phi - \omega t)d\phi . \end{aligned}$$

(E51)

Using Mathematica, the azimuthal integral gives

$$\begin{aligned} \int ^{2\pi n}_{0}\sin (\phi - \omega t)\delta (\phi - \omega t)d\phi =0. \end{aligned}$$

(E52)

Using (E52) in (E51), we obtain

$$\begin{aligned} I_5=0. \end{aligned}$$

(E53)

• (vi) Proof of ${I_6=0}$. From (E10), we have $({{{\mathcal {J}}}}_{e})_{z}=0$ which is used in $I_6$ in (E5) to obtain

$$\begin{aligned} I_6=0. \end{aligned}$$

(E54)

• (vii–xii) Proofs of ${\partial I_7/\partial t=0, \partial I_8/\partial t=0, \partial I_9/\partial t=0, \partial I_{10}/\partial t=0, I_{11}=0,}$ and ${I_{12}=0}$ The integrals $I_{7}-I_{12}$ are of the same form as those in $I_1-I_6.$ In fact by making the dual changes $q\rightarrow -g$ and $g\rightarrow -q$ in the integrals $I_1-I_6$, we obtain the integrals $I_{7}-I_{12}$. Therefore,

$$\begin{aligned} \frac{\partial I_7}{\partial t}=0,\,\,\,\frac{\partial I_8}{\partial t}=0,\,\,\,\frac{\partial I_9}{\partial t}=0,\,\,\,\frac{\partial I_{10}}{\partial t}=0,\,\,\, I_{11}=0, \,\,\,I_{12}=0. \end{aligned}$$

(E55)

Using (E23), (E34), (E41), (E48), and (E53)–(E54) in (E3), we finally obtain

$$\begin{aligned} {{\varvec{L}}}_{\mathrm{R}}=0. \end{aligned}$$

(E56)

Since this relation is valid for a dyon with charges $q\not =0$ and $g\not =0$ and for a dual solenoid with fluxes $\varPhi _e\not =0$ and $\varPhi _m\not =0$, it is valid for the particular case in which the dyon has the charges $q=0$ and $g\not =0$ (a magnetic monopole) and the dual solenoid has the fluxes $\varPhi _m=0$ and $\varPhi _e\not =0$ (an electric solenoid):

$$\begin{aligned} {{\varvec{L}}}_{\mathrm{R}}(q=0, \varPhi _m=0)=0, \end{aligned}$$

(E57)

which corresponds to the monopole–solenoid configuration (see (20)).