Amending the Maxwell’s Equations for Moving Media

1. Introduction

The classical electromagnetic theory is one of the most successful theories in history, in which the Maxwell’s equations play a central role [1,2,3]. However, there still exist a few inconsistencies or even controversies in it, especially concerning with the electrodynamics of moving objects and moving media [4].

There are several formulations of Maxwell’s equations for moving media. We take the Chu formulation [4,5,6,7] as an example to show the inconsistencies. When being exposed in an external electric field, the molecules or atoms in the medium will be polarized. In simple substances, the dominant molecule moment is the dipole and the polarization can be characterized with electric dipoles. We usually utilize the polarization vector $P\left(r,t\right)$ to represent the volume density of the electric dipoles. The electric polarization current $J_p^e\left(r,t\right)$ and the electric polarization charge ${\rho }_p^e\left(r,t\right)$ can be expressed in terms of the polarization vector $P\left(r,t\right)$ with $J_p^e\left(r,t\right)=\partial P\left(r,t\right)/\partial t$ and ${\rho }_p^e\left(r,t\right)=-\nabla \cdot P\left(r,t\right)$. When the medium moves with velocity $v$, the derivative terms with respect to time $t$ should be replaced by the corresponding substantial derivatives [8]. The continuity law for the polarization current is derived to be

$$\nabla \cdot \left[{\displaystyle \frac{\partial }{\partial t}}P\left(r,t\right)+\left(v\cdot \nabla \right)P\left(r,t\right)-v\nabla \cdot P\left(r,t\right)\right]+{\displaystyle \frac{\partial }{\partial t}}{\rho }_p^e\left(r,t\right)=0.$$

(1)

The terms in the square bracket amount to the electric polarization current in the moving medium, including an additional current $-v\nabla \cdot P\left(r,t\right)$ produced by the moving polarization charge. For the sake of simplicity, we still denote it as $J_p^e\left(r,t\right)$, which can be transformed as follows,

$$J_p^e\left(r,t\right)={\displaystyle \frac{\partial }{\partial t}}P\left(r,t\right)+\left(v\cdot \nabla \right)P\left(r,t\right)-v\nabla \cdot P\left(r,t\right)={\displaystyle \frac{\partial }{\partial t}}P\left(r,t\right)+\nabla \times \left[P\left(r,t\right)\times v\right].$$

(2)

On the other hand, when a medium is exposed in an external magnetic field, the magnetization will generate magnetic dipoles in the medium. The volume density of the magnetic dipoles is represented by the magnetization vector $M\left(r,t\right)$. The Lorentz model is used in Chu formulation [5,7], where a magnetic dipole is modelled with two closely located magnetic charges ${\rho }_m^m\left(r,t\right)=-{\mu }_0\nabla \cdot M\left(r,t\right)$, and are connected with a magnetic current $J_m^m\left(r,t\right)={\mu }_0\partial M\left(r,t\right)/\partial t$. In the symbols for current and charge, we use the superscript to indicate whether they are electric quantities or magnetic quantities, and the subscript to indicate whether they are caused by polarization or by magnetization.

The Chu formulation Maxwell’s equations for moving media [4] are summarized as follows

$$\left\{\begin{array}{l}{\mu }_0^{-1}\nabla \times E=-{\displaystyle \frac{\partial H}{\partial t}}-{\displaystyle \frac{\partial \left({\mu }_{0}M\right)}{\partial t}}-\nabla \times \left(M\times v\right)\\ \nabla \times H={\epsilon }_0{\displaystyle \frac{\partial E}{\partial t}}+{\displaystyle \frac{\partial P}{\partial t}}+\nabla \times \left(P\times v\right)+J_f\\ {\mu }_0\nabla \cdot H=-\nabla \cdot \left({\mu }_{0}M\right)\\ {\epsilon }_0\nabla \cdot E=-\nabla \cdot P+{\rho }_f\end{array}\right.,$$

(3)

where $J_f$ and ${\rho }_f$ are the current density and charge density associated with free charges.

However, Chu formulation (3) is not in compatible with the Fresnel’s coefficient. Fresnel proposed in 1818 the famous Fresnel coefficient [9,10,11] when he attempted to explain the Arago’s experiments. His theory is based on a stationary ether. The light speed in a medium is $c=c_0/n$, where $c_0=1/\sqrt{{\mu }_0{\epsilon }_0}$ is the light speed in the vacuum and $n$ is the refractive index of the medium at rest. When the medium moves with velocity $v=\left|v\right|$, the light may be dragged by the medium and the light speed $u$ in the moving medium must change. If we determine the light speed according to the Newtonian mechanics, the two velocities should be added like vectors, namely, $u=c\pm v$, in which it takes positive sign when the light propagates in the same direction with the medium and negative when in the opposite direction. However, this prediction was found to be not in agreement with the experimental results. Therefore, Fresnel assumed that when a medium moves through the stationary ether, it only drags the light propagating through it with a fraction of the medium’s speed, that is, the light speed in the moving medium is of the form

$$u=c\pm fv=c\pm v\left(1-n^{-2}\right),$$

(4)

where $f=1-n^{-2}$ is the Fresnel’s dragging coefficient.

We are now to find the plane wave solution of (3) in a uniformly moving homogeneous nonmagnetic medium like water. In this situation, we have $M=0$, $P=\left(\epsilon -{\epsilon }_0\right)E$, and $J_f={\rho }_f=0$ in the medium. Assume that the plane wave propagates in parallel with the moving direction of the medium and its fields take the form of $E\left(r,t\right)=\widehat{x}{E}_0\exp (j\omega t-kz)$, $H\left(r,t\right)=\widehat{y}{H}_0\exp (j\omega t-kz)$. From the two curl equations in (3) we obtain the matrix equation

$$\left[\begin{array}{cc}k& -\omega {\mu }_0\\ \omega \epsilon -kv\left(\epsilon -{\epsilon }_0\right)& -k\end{array}\right]\left[\begin{array}{c}{E}_0\\ H_0\end{array}\right]=0.$$

(5)

The condition for (5) to have non-zero solutions is that its determinant vanishes, that is,

$$k^2+\omega {\mu }_0\left(\epsilon -{\epsilon }_0\right)vk-{\omega }^2{\mu }_0\epsilon =0.$$

(6)

Solving $k$ from (6), we get

$$k={\displaystyle \frac{-\omega {\mu }_0\left(\epsilon -{\epsilon }_0\right)v\pm \omega \sqrt{{\left({\mu }_{0}v\right)}^2{\left(\epsilon -{\epsilon }_0\right)}^2+4{\mu }_0\epsilon }}{2}}.$$

(7)

The corresponding solutions represent two plane waves propagating in the $\pm z$directions. Taking the plane wave in $+z$ direction ($k>0$) as example, we obtain its phase velocity as below

$$u={\displaystyle \frac{\omega }{k}}={\displaystyle \frac{c}{-0.5{\mu }_0\left(\epsilon -{\epsilon }_0\right)cv\pm 1}}\approx c\pm 0.5v\left(1-n^{-2}\right),$$

(8)

where $n=\epsilon /{\epsilon }_0$ is the refractive index of the nonmagnetic medium. Obviously, the dragging coefficient in (8) is different from the Fresnel’s dragging coefficient in (4) by a factor of 0.5.

Fresnel’s coefficient has been verified by the experiment of Fizeau with high accuracy [12,13]. This inconsistency implies that the Chu formulation of the Maxwell’s equations for moving media may have missed something in the issue.

2. The Inconsistency with the Fields of a Moving Oscillating Dipole

A moving oscillating dipole can be modelled in two ways. The first one is to take the Lorentz model, in which the dipole is modelled with two closely-located charges with densities of ${\rho }_{\pm }=\pm {\rho }_0\left(t\right)\delta \left(r-x\left(t\right)\mp 0.5l\widehat{p}\right)$, where $l$ is the small spacing between the two charges, $\widehat{p}$ is the unit vector of the electric polarization that points from ${\rho }_{-}$ to ${\rho }_{+}$, and $x\left(t\right)$ denotes the position of the center of the dipole on the trajectory. When $l\to 0$, the two charge densities can be expressed concisely with

$${\rho }_{dip}^e\left(r,t\right)=-\left(p\left(t\right)\cdot \nabla \right)\delta \left(r-x\left(t\right)\right),$$

(9)

where $p\left(t\right)={\rho }_0\left(t\right)l\widehat{p}$ is the electric dipole moment. According to the current continuity law, a short current filament with density of $J_{dip}^e\left(r,t\right)=\dot{p}\left(t\right)\delta \left(r-x\left(t\right)\right)$ exists between the two charges. Besides this current, the moving charges also produce a current with density of

$$J_{\rho }^e\left(r,t\right)=-v\left(p\left(t\right)\cdot \nabla \right)\delta \left(r-x\left(t\right)\right),$$

(10)

where $v=\dot{x}\left(t\right)$ is the velocity of the dipole. We use dot on top of a quantity to represent the derivative with respect to time $t$. The total current density is the sum of the two currents,

$$J_{tot}^e\left(r,t\right)=J_{dip}^e+J_{\rho }^e=\dot{p}\left(t\right)\delta \left(r-x\left(t\right)\right)-v\left(p\left(t\right)\cdot \nabla \right)\delta \left(r-x\left(t\right)\right).$$

(11)

The second method for modelling the dipole is based on the polarization vector $P\left(r,t\right)$, like that in handling moving media. For the oscillating moving dipole, we may consider that it is the polarization vector that is moving. Therefore, we can express the polarization vector by $P\left(r,t\right)=p\left(t\right)\delta \left(r-x\left(t\right)\right)$. The charge density expressed in terms of $P\left(r,t\right)$ is the same as ${\rho }_p^e\left(r,t\right)$ that has been defined previously.

The current density of the moving dipole expressed in term of $P\left(r,t\right)$ consists of two terms,

$$J_p^e\left(r,t\right)=\dot{p}\left(t\right)\delta \left(r-x\left(t\right)\right)-\left(v\cdot \nabla \right)\left[p\left(t\right)\delta \left(r-x\left(t\right)\right)\right].$$

(12)

Obviously, $J_p^e\left(r,t\right)\ne J_{tot}^e\left(r,t\right)$. Since $J_{tot}^e\left(r,t\right)$ represents the total current generated by the moving current filament and the current induced by the two moving charges, it is basically correct because its physical meaning is clear and intuitive. However, $J_p^e\left(r,t\right)$ is derived from the electric polarization vector and may have not taken all effect into account. Making use of the vector identity of $\left(v\cdot \nabla \right)A=v\nabla \cdot A+\nabla \times \left(A\times v\right)$, (12) can be cast into

$$J_p^e\left(r,t\right)=\dot{p}\left(t\right)\delta \left(r-x\left(t\right)\right)-v\nabla \cdot \left[p\left(t\right)\delta \left(r-x\left(t\right)\right)\right]-\nabla \times \left[p\left(t\right)\times v\delta \left(r-x\left(t\right)\right)\right].$$

(13)

We can recognize that the first two terms in the righthand side of (13) is $J_{tot}^e\left(r,t\right)$. Therefore, (13) clearly indicates that $J_p^e\left(r,t\right)$ is only a part of $J_{tot}^e\left(r,t\right)$. Obviously, if we denote an equivalent magnetization vector defined by

$$M_P\left(r,t\right)=\left[p\left(t\right)\times v\delta \left(r-x\left(t\right)\right)\right]=P\left(r,t\right)\times v.$$

(14)

Then $J_m^e\left(r,t\right)=\nabla \times M_P\left(r,t\right)$ represents the magnetization electric current associated with the $M_P\left(r,t\right)$ [2]. We use subscript “P” to indicate that the magnetization is caused by the polarization of the moving electric dipole. The derivation demonstrates that a moving oscillating electric dipole also produces a magnetization [14], which is similar to that of a moving static electric dipole [15]. Reorganizing (13) we can get that the total current is the sum of the polarization current and the magnetization current,

$$J_{tot}^e\left(r,t\right)=J_p^e\left(r,t\right)+J_m^e\left(r,t\right).$$

(15)

The analysis reveals one fact: the motion of the polarization $P\left(r,t\right)$ will induce a magnetization $M_P\left(r,t\right)$. However, we can check that only the polarization vector has been taken into account of in the Chu formulation. The effect of the magnetization has been ignored. In the next section, we will show that we cannot get correct solutions of the electromagnetic fields for the moving Hertzian dipole without considering the contribution from the magnetization.

3. Verification of the Oscillating Dipole Model

With the first model, we have derived the fields of a Hertzian dipole when it moves uniformly in the direction perpendicular to $\widehat{p}$, as shown in Figure 1. Generally, we use $\left(r,t\right)$ to represent the space-time coordinate of the fields or potentials and $\left(r_1,t_1\right)$ for the sources.

The Hertzian dipole moves with its center at $x\left(t_1\right)=v{t}_1$ with electric moment $p\left(t_1\right)=p_0\cos \left({\omega }_0{t}_1\right)\widehat{p}$, where ${\omega }_0$ is the angular frequency. The densities of the two moving charges can be expressed by

$${\rho }_{\pm }\left(r_1,t_1\right)=\pm {\rho }_0\delta \left(r_1-x\left(t_1\right)\mp 0.5l\widehat{p}\right)\cos \left({\omega }_0{t}_1\right).$$

(16)

Because of the motion of the charge, two currents are induced with current densities of

$$J_{\pm }\left(r_1,t_1\right)=\pm {\rho }_{0}v\delta \left(r_1-x\left(t_1\right)\mp 0.5l\widehat{p}\right)\cos \left({\omega }_0{t}_1\right).$$

(17)

The short current between the two charges also moves along the trajectory, which becomes

$$J_{dip}\left(r_1,t_1\right)=-{\omega }_0{\rho }_{0}l\delta \left(r_1-x\left(t_1\right)\right)\sin \left({\omega }_0{t}_1\right)\widehat{p}.$$

(18)

When $v\cdot \widehat{p}=0$, the Lienard-Wiechert potentials [16,17] of the moving Hertzian dipole have been solved in [14] with the first model. Alternatively, we can derive them using the second model with charge density (9) and current density (15).

The electromagnetic fields of the uniformly moving Hertzian dipole are derived to be [14].

$$E_{dip}\left(r,t\right)=E_0\left\{\begin{array}{l}{\displaystyle \frac{1-{\beta }^2}{R^3{\left(1-\widehat{n}\cdot \beta \right)}^3}}\left[{\displaystyle \frac{3\left(1-{\beta }^2\right)\left(\widehat{n}\cdot \widehat{p}\right)\left(\widehat{n}-\beta \right)}{{\left(1-\widehat{n}\cdot \beta \right)}^2}}-\widehat{p}\right]\cos \left({\omega }_0{t}_1\right)\\ -{\displaystyle \frac{1-{\beta }^2}{R^2{\left(1-\widehat{n}\cdot \beta \right)}^3}}\left[{\displaystyle \frac{3\left(\widehat{n}\cdot \widehat{p}\right)\left(\widehat{n}-\beta \right)}{1-\widehat{n}\cdot \beta }}-\widehat{p}\right]k_0\sin \left({\omega }_0{t}_1\right)\\ -{\displaystyle \frac{\widehat{n}\times \left(\widehat{n}\times \widehat{p}\right)+\widehat{n}\times \left(\widehat{p}\times \beta \right)}{R{\left(1-\widehat{n}\cdot \beta \right)}^3}}{k}_0^2\cos \left({\omega }_0{t}_1\right)\end{array}\right\}$$

(19)

$$B_{dip}\left(r,t\right)={\displaystyle \frac{E_0}{c_0}}\left\{\begin{array}{l}{\displaystyle \frac{\left(1-{\beta }^2\right)\beta \times }{R^3{\left(1-\widehat{n}\cdot \beta \right)}^3}}\left[{\displaystyle \frac{3\left(1-\beta \cdot \beta \right)}{{\left(1-\widehat{n}\cdot \beta \right)}^2}}\left(\widehat{n}\cdot \widehat{p}\right)\widehat{n}-\widehat{p}\right]\cos \left({\omega }_0{t}_1\right)\\ +{\displaystyle \frac{\left(1-\beta \cdot \beta \right)\widehat{n}\times }{R^2{\left(1-\widehat{n}\cdot \beta \right)}^4}}\left[3\left(\widehat{n}\cdot \widehat{p}\right)\beta +\left(1-\widehat{n}\cdot \beta \right)\widehat{p}\right]k_0\sin \left({\omega }_0{t}_1\right)\\ +{\displaystyle \frac{\widehat{n}\times }{R{\left(1-\widehat{n}\cdot \beta \right)}^2}}\left({\displaystyle \frac{\widehat{n}\cdot \widehat{p}}{1-\widehat{n}\cdot \beta }}\beta +\widehat{p}\right)k_0^2\cos \left({\omega }_0{t}_1\right)\end{array}\right\}$$

(20)

They are generated by the Hertzian dipole at the position $r_1=x\left(t_1\right)=v{t}_1$ and the time instant $t_1$. In the expressions, $R=r-x\left(t_1\right)$ is the radius vector pointing from the source point to the observation point, $R=\left|R\right|$, $\widehat{n}=R/R$, and $\beta =v/c_0=\beta \widehat{v}=v/c_0\widehat{v}$.${\epsilon }_0$ and $k_0$ are respectively the permittivity and wavenumber in the vacuum. $E_0=p_0/\left(4\pi {\epsilon }_0\right)$ is a source-related constant. In the vacuum, the fields generated at $\left(r_1,t_1\right)$ propagates to the observer at $\left(r,t\right)$ with the light velocity $c_0$. Therefore, we obtain the governing equation for the wave propagation,

$$c_0\left(t-t_1\right)=R=\left|r-x\left(t_1\right)\right|,$$

(21)

from which we can solve $t_1$ based on the causality principle, and obtain that 14:

$${\omega }_0{t}_1={\displaystyle \frac{{\omega }_0\left(c_0^{2}t-v\cdot r\right)}{c_0^2\left(1-{\beta }^2\right)}}-{\displaystyle \frac{{\omega }_0{c}_0\sqrt{\left(1-{\beta }^2{\sin }^2\theta \right)r^2-2r\cdot vt+v^2{t}^2}}{c_0^2\left(1-{\beta }^2\right)}},$$

(22)

where $\theta$ is the angle between the vector $\widehat{r}$ and $\beta$.

To verify the solutions, we have derived the fields with frame-hopping technique. Denote $K$ as the laboratory frame in which the observer is at rest, and $K^{\prime }$ as the inertial frame in which the Hertzian dipole is at rest. Firstly, we transform the Hertzian dipole to the rest frame $K^{\prime }$ with the Lorentz transformations [3,18]

$$\left\{\begin{array}{l}{\mathbf{r}}^{\prime }=r_{\perp }+\gamma \left(r_{\parallel }-vt\right)\\ t^{\prime }=\gamma \left(t-v\cdot r/c_0^2\right)\end{array}\right.,$$

(23)

where the Lorentz factor is $\gamma =1/\sqrt{1-{\beta }^2}$. The subscript “⊥” and “||” represent the transversal and parallel component of the vector, respectively. The space-time coordinates of the center of the Hertzian dipole are transformed to be

$$\left\{\begin{array}{l}{\mathbf{r}}_1^{\prime }=r_{1\perp }+\gamma \left(r_{1\parallel }-vt\right)=0\\ t_1^{\prime }=\gamma \left(t_1-v\cdot r_1/c_0^2\right)={\gamma }^{-1}{t}_1\end{array}\right..$$

(24)

Obviously, the center of the dipole is transformed to the origin of the frame $K^{\prime }$. Since $t_1=\gamma t_1^{\prime }$, we have $\left({\omega }_0{t}_1\right)=\left(\gamma {\omega }_0{t}_1^{\prime }\right)$. The moving velocity $v$ of the sources is transformed to $v^{\prime }=0$. Consequently, the three source densities (16)-(18) can be transformed into $K^{\prime }$ as follows,

$${\rho }_{\pm }^{\prime }\left({\mathbf{r}}_1^{\prime },t_1^{\prime }\right)=\pm {\rho }_0\delta \left({\mathbf{r}}_1^{\prime }\mp 0.5l{\widehat{p}}^{\prime }\right)\cos \left(\gamma {\omega }_0{t}_1^{\prime }\right),$$

(25)

$${\mathbf{J}}_{\pm }^{\prime }\left({\mathbf{r}}_1^{\prime },t_1^{\prime }\right)=0,$$

(26)

$${\mathbf{J}}_{dip}^{\prime }\left({\mathbf{r}}_1^{\prime },t_1^{\prime }\right)=-\gamma {\omega }_0{\rho }_{0}l\delta \left({\mathbf{r}}_1^{\prime }\right)\sin \left(\gamma {\omega }_0{t}_1^{\prime }\right){\widehat{p}}^{\prime }.$$

(27)

Note that in the case of $v\cdot \widehat{p}=0$, $l$ has no Lorentz contraction. The angular frequency of the source is transformed from ${\omega }_0$ to $\gamma {\omega }_0$.

The space-time coordinate $\left(r,t\right)$ of the observation position can be transformed into the space-time coordinate $\left(r^{\prime },t^{\prime }\right)$ in the frame $K^{\prime }$ with the Lorentz transformations. The fields generated by the motionless Hertzian dipole in the frame $K^{\prime }$ are solved to be

$${\mathbf{E}}_{dip}^{\prime }\left(r^{\prime },t^{\prime }\right)=E_0\left\{\begin{array}{l}{\displaystyle \frac{1}{R^{\prime 3}}}\left[3{\widehat{n}}^{\prime }\left({\widehat{n}}^{\prime }\cdot {\widehat{p}}^{\prime }\right)-{\widehat{p}}^{\prime }\right]\cos \left(\gamma {\omega }_0{t}_1^{\prime }\right)-{\displaystyle \frac{1}{R^{\prime 2}}}\left[3{\widehat{n}}^{\prime }\left({\widehat{n}}^{\prime }\cdot {\widehat{p}}^{\prime }\right)-{\widehat{p}}^{\prime }\right]\gamma k_0\sin \left(\gamma {\omega }_0{t}_1^{\prime }\right)\\ -{\displaystyle \frac{{\widehat{n}}^{\prime }\times \left({\widehat{n}}^{\prime }\times {\widehat{p}}^{\prime }\right)}{R^{\prime }}}{\gamma }^2{k}_0^2\cos \left(\gamma {\omega }_0{t}_1^{\prime }\right)\end{array}\right\},$$

(28)

$${\mathbf{B}}_{dip}^{\prime }\left(r^{\prime },t^{\prime }\right)={\mu }_0{\mathbf{H}}_{dip}^{\prime }\left(r^{\prime },t^{\prime }\right)={\displaystyle \frac{E_0}{c_0}}\left\{{\displaystyle \frac{{\widehat{n}}^{\prime }\times {\widehat{p}}^{\prime }}{R^{\prime 2}}}\gamma k_0\sin \left(\gamma {\omega }_0{t}_1^{\prime }\right)+{\displaystyle \frac{{\widehat{n}}^{\prime }\times {\widehat{p}}^{\prime }}{R^{\prime }}}{\gamma }^2{k}_0^2\cos \left(\gamma {\omega }_0{t}_1^{\prime }\right)\right\}.$$

(29)

In the expressions, $R^{\prime }=r^{\prime }-{\mathbf{r}}_1^{\prime }=\left|r^{\prime }\right|$, $R^{\prime }=\left|r^{\prime }-{\mathbf{r}}_1^{\prime }\right|=r^{\prime }$, ${\widehat{p}}^{\prime }=\widehat{p}$. The electromagnetic pulse generated at $\left({\mathbf{r}}_1^{\prime },t_1^{\prime }\right)$ propagates to $\left(r^{\prime },t^{\prime }\right)$ with velocity $c_0$, hence, $R^{\prime }=\left|r^{\prime }-{\mathbf{r}}_1^{\prime }\right|=r^{\prime }=c_0\left(t^{\prime }-t_1^{\prime }\right)$, from which we can obtain that

$$t_1^{\prime }=t^{\prime }-{\displaystyle \frac{\left|r^{\prime }\right|}{c_0}}=\gamma \left(t-v\cdot r/c_0^2\right)-{\displaystyle \frac{\left|r_{\perp }+\gamma \left(r_{\parallel }-vt\right)\right|}{c_0}}.$$

(30)

With some vector operations, we can further check that

$$\gamma {\omega }_0{t}_1^{\prime }={\displaystyle \frac{{\omega }_0\left(c_0^{2}t-v\cdot r\right)-{\omega }_0{c}_0\sqrt{r^2\left(1-{\beta }^2{\sin }^2\theta \right)-2r\cdot vt+v^2{t}^2}}{c_0^2{\gamma }^{-2}}}.$$

(31)

Substituting $v\cdot r=c_0\beta r\cos \theta$ and ${\gamma }^{-2}=1-{\beta }^2$ into (31), we can verify that it is exactly the same as (22).

Secondly, we derive the fields $E_{dip}\left(r,t\right)$ and $H_{dip}\left(r,t\right)$ in the frame $K$ with the inverse Lorentz transformations for the electromagnetic fields [19,20],

$$\left[\begin{array}{c}{\mathbf{E}}_{dip}\left(r,t\right)\\ {\mathbf{H}}_{dip}\left(r,t\right)\end{array}\right]=\mathcal{L}\mathcal{T}\left(-v\right)\left[\begin{array}{c}{\mathbf{E}}_{dip}^{\prime }\left(r^{\prime },t^{\prime }\right)\\ {\mathbf{H}}_{dip}^{\prime }\left(r^{\prime },t^{\prime }\right)\end{array}\right],$$

(32)

where the inverse Lorentz transformations $\mathcal{L}\mathcal{T}\left(-v\right)$ is defined by,

$$\mathcal{L}\mathcal{T}\left(-v\right)=\left[\begin{array}{cc}\left[\left(1-\gamma \right)\widehat{\mathbf{v}}\widehat{\mathbf{v}}+\gamma \overline{I}\right]\cdot & -{\mu }_0\gamma v\times \\ {\epsilon }_0\gamma v\times & \left[\left(1-\gamma \right)\widehat{\mathbf{v}}\widehat{\mathbf{v}}+\gamma \overline{I}\right]\cdot \end{array}\right].$$

(33)

$\overline{I}$ is the identity operator. With the definitions of the vectors, we can verify that 1117

$$\left\{\begin{array}{l}{R}^{\prime }=\gamma R\left(1-\widehat{n}\cdot \beta \right)\\ {\widehat{n}}^{\prime }={\displaystyle \frac{\widehat{n}+\left(\gamma -1\right)\widehat{n}\cdot \widehat{v}\widehat{v}-\gamma \beta }{\gamma \left(1-\widehat{n}\cdot \beta \right)}}\end{array}\right..$$

With these relationships, it is straightforward to verify that the fields obtained with (32) are exactly the same as that expressed by (19) and (20). Therefore, we have demonstrated that the radiation fields of the moving Hertzian dipole can either be solved in the laboratory frame through the Lienard-Wiechert potentials, or be solved using the frame-hopping technique with Lorentz transformations. The agreement of the results shows that the magnetization $M_P\left(r,t\right)$ is necessary for correctly modelling the oscillating moving electric dipoles, hence, it is reasonable to speculate that the Chu formulation needs amending.

4. Amending the Chu Formulation

The analogy between modelling a moving dipole and modelling the polarization in a moving medium is sketched in Figure 2. For moving nonmagnetic media, we can amend the Chu formulation by directly taking the contribution of the motion-induced magnetization into account.

We extend the modification to magnetic media. We can check that a moving magnetic dipole with magnetization $M\left(r,t\right)$ produces a polarization $P_M\left(r,t\right)={\mu }_0{\epsilon }_{0}v\times M\left(r,t\right)$, which is observed in [21] for static magnetic dipole. Therefore, the amended Chu formulation Maxwell’s equations are

$$\left\{\begin{array}{l}{\mu }_0^{-1}\nabla \times E=-{\displaystyle \frac{\partial H}{\partial t}}-{\displaystyle \frac{\partial \left(M+M_P\right)}{\partial t}}-\left(v\cdot \nabla \right)M\\ \nabla \times H={\epsilon }_0{\displaystyle \frac{\partial E}{\partial t}}+{\displaystyle \frac{\partial \left(P+P_M\right)}{\partial t}}+\left(v\cdot \nabla \right)P+J_f\\ {\mu }_0\nabla \cdot H=-{\mu }_0\nabla \cdot \left(M+M_P\right)\\ {\epsilon }_0\nabla \cdot E=-\nabla \cdot \left(P+P_M\right)+{\rho }_f\end{array}\right.,$$

(34)

with $M_P=P\times v$, and $P_M={\mu }_0{\epsilon }_{0}v\times M$. At a fixed point in a simple moving medium, we have the relationship of $P=\left(\epsilon -{\epsilon }_0\right)E$ and $M=\left(\mu /{\mu }_0-1\right)H$.

To find the plane wave in a homogeneous moving medium with $J_f={\rho }_f=0$, we again assume that the wave propagates in parallel with the moving direction of the medium and the fields take the form of $E\left(r,t\right)=\widehat{x}{E}_0\exp (j\omega t-jkz)$, $H\left(r,t\right)=\widehat{y}{H}_0\exp (j\omega t-jkz)$. From the two curl equations in (34) we obtain

$$\left[\begin{array}{cc}k+\omega {\mu }_0\left(\epsilon -{\epsilon }_0\right)v& kv\left(\mu -{\mu }_0\right)-\omega \mu \\ \omega \epsilon -kv\left(\epsilon -{\epsilon }_0\right)& -\omega {\epsilon }_0\left(\mu -{\mu }_0\right)v-k\end{array}\right]\left[\begin{array}{c}{E}_0\\ H_0\end{array}\right]=0.$$

(35)

The determinant of the coefficient matrix must be zero in order that (35) has non-zero fields. Discarding the terms that contain $v$ with orders higher than one, we get the equation,

$$k^2+2\omega v\left(\mu \epsilon -{\mu }_0{\epsilon }_0\right)k-{\omega }^2\mu \epsilon \approx 0.$$

(36)

Solving equation (35) we get

$$\begin{array}{c}k=-\omega \left(\mu \epsilon -{\mu }_0{\epsilon }_0\right)v\pm \omega \sqrt{v^2{\left(\mu \epsilon -{\mu }_0{\epsilon }_0\right)}^2+\mu \epsilon }\\ \approx -\omega \left(\mu \epsilon -{\mu }_0{\epsilon }_0\right)v\pm \omega \sqrt{\mu \epsilon }\end{array}.$$

(37)

From which we can get the phase velocity of the $+z$ plane wave,

$$u\approx c\left[1\pm v\left(\mu \epsilon -{\mu }_0{\epsilon }_0\right)\right]=c\pm v\left(1-n^{-2}\right),$$

(38)

where $c=1/\sqrt{\mu \epsilon }$,$n^2=\left({\mu }_0{\epsilon }_0\right)/\left(\mu \epsilon \right)$. Obviously, (38) exactly agrees with the Fresnel’s coefficient (4).

In other formulations that the magnetization in the medium is described using the magnetization current with density of $J_m^e\left(r,t\right)=\nabla \times M_P\left(r,t\right)$ and zero electric magnetization charge, the inconsistency with the Fresnel’s coefficient also exists. Basically, they can be amended in the same way as we have proposed in amending the Chu formulation. It is straightforward to verify that the plane wave solutions of the corresponding amended formulations are compatible with the Fresnel’s coefficient.

5. Conclusions

We have clearly revealed that both the two inconsistencies in the conventional formulations of the Maxwell’s equations for moving media are caused by the fact that we have ignored the effect of the magnetization in moving electric dipoles and the polarization in moving magnetic dipoles in modelling the moving media. The two inconsistencies are closely related to each other and have existed in the classical electromagnetic theory for century long. The proposed modification of the Maxwell’s equations for moving media is simply a remedy for the two inconsistencies. With the amended formulations, we can correctly derive the Fresnel’s dragging coefficient from the Maxwell’s equations.