In Ref. [1], the equations have been written in the Gauss system of units, for simplicity. However, due to the fact that in the Gauss units the Coulomb law is written as

$$\begin{array}{}{\displaystyle \mathbf{F}=q\phantom{\rule{thinmathspace}{0ex}}{q}^{\prime}\frac{\mathbf{r}}{{r}^{3}},}\end{array}$$(1)

in these units the charge dimension is $\begin{array}{}Q=(M\phantom{\rule{thinmathspace}{0ex}}{L}^{3}\phantom{\rule{thinmathspace}{0ex}}{T}^{-2}{)}^{\frac{1}{2}},\end{array}$ where *M*, *L* and *T* represent mass, length, and time, respectively. This would introduce an undesirable coupling when writing weak-field asymptotic expansions, for in fact charge varies independently of mass, length, and time. Therefore, while presenting now the main equations [1], we rewrite them in the so-called SI units, *i.e*., in the “meter, kilogram, second, Ampère” (MKSA) system. The electromagnetic field is defined by the antisymmetric space-time tensor *F*: *F*_{μν} = −*F*_{νμ}, that obeys the standard first group of the Maxwell equations:

$$\begin{array}{}{\displaystyle {F}_{\lambda \mu \phantom{\rule{thinmathspace}{0ex}},\nu}+{F}_{\mu \nu ,\lambda}+{F}_{\nu \lambda ,\mu}={F}_{\lambda \mu \phantom{\rule{thinmathspace}{0ex}};\nu}+{F}_{\mu \nu ;\lambda}+{F}_{\nu \lambda ;\mu}=0.}\end{array}$$(2)

The expression of the Lorentz force is ^{1}

$$\begin{array}{}{\displaystyle {F}^{i}=q\phantom{\rule{thinmathspace}{0ex}}c\left(\frac{{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0}^{i}}{\beta}+{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j}^{i}\phantom{\rule{thinmathspace}{0ex}}\frac{{v}^{j}}{c}\right)=q\phantom{\rule{thinmathspace}{0ex}}{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mu}^{i}\phantom{\rule{thinmathspace}{0ex}}\frac{\text{d}{x}^{\mu}}{\text{d}{t}_{\mathbf{x}}}.}\end{array}$$(3)

Here *v*^{j} ≡ d*x*^{j}/d*t*_{x} (*j* = 1,2,3) is the velocity of the particle, measured with the local time *t*_{x}, with

$$\begin{array}{}{\displaystyle \frac{\text{d}{t}_{\mathbf{x}}}{\text{d}t}=\beta (t,\mathbf{x}),}\end{array}$$(4)

where *t* = *x*^{0}/*c* is the coordinate time in a coordinate system (*x*^{μ}) adapted to (or bound with) the preferred reference fluid 𝓔 assumed by the theory, and such that the synchronization condition *γ*_{0j} = 0 is true [1]; **x** ≡ (*x*^{i}); and (in any such coordinates)

$$\begin{array}{}{\displaystyle \beta \equiv \sqrt{{\gamma}_{00}}.}\end{array}$$(5)

Equation (3) may be rewritten in space-vector form as

$$\begin{array}{}{\displaystyle \mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\wedge \phantom{\rule{thinmathspace}{0ex}}\mathbf{B}\right),\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}(\mathbf{a}\wedge \mathbf{b}{)}^{i}\equiv {e}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}jk}^{i}\phantom{\rule{thinmathspace}{0ex}}{a}^{j}\phantom{\rule{thinmathspace}{0ex}}{b}^{k},}\end{array}$$(6)

where the electric and magnetic vector fields are the spatial vector fields with components

$$\begin{array}{}{\displaystyle {E}^{i}\equiv \frac{c\phantom{\rule{thinmathspace}{0ex}}{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0}^{i}}{\beta},\phantom{\rule{1em}{0ex}}{B}^{k}\equiv -\frac{1}{2}{e}^{ijk}{F}_{ij}.}\end{array}$$(7)

In Eqs. (6) and (7), *e*_{ijk} is the usual antisymmetric spatial tensor, its indices being raised or lowered using the spatial metric *g* in the preferred frame 𝓔; in spatial coordinate systems whose natural basis is direct, we have

$$\begin{array}{}{\displaystyle {e}_{ijk}=\sqrt{g}\phantom{\rule{thinmathspace}{0ex}}{\epsilon}_{ijk},\phantom{\rule{2em}{0ex}}{e}^{ijk}=\frac{1}{\sqrt{g}}\phantom{\rule{thinmathspace}{0ex}}{\epsilon}_{ijk},}\end{array}$$(8)

with *ε*_{ijk} the signature of the permutation (*i j k*) and

$$\begin{array}{}{\displaystyle g\equiv \mathrm{d}\mathrm{e}\mathrm{t}({g}_{ij}).}\end{array}$$(9)

For a continuous charged medium, the electric charge density is defined as

$$\begin{array}{}{\displaystyle {\rho}_{\mathrm{e}\mathrm{l}}\equiv \delta q/\delta V,}\end{array}$$(10)

where *δV* is the volume element measured with the physical volume measure:

$$\begin{array}{}{\displaystyle \delta V=\sqrt{g}\phantom{\rule{thinmathspace}{0ex}}\text{d}{x}^{1}\text{d}{x}^{2}\text{d}{x}^{3}.}\end{array}$$(11)

The 4-current is

$$\begin{array}{}{\displaystyle {J}^{\phantom{\rule{thinmathspace}{0ex}}\mu}\equiv {\rho}_{\mathrm{e}\mathrm{l}}\phantom{\rule{thinmathspace}{0ex}}\text{d}{x}^{\mu}/\text{d}{t}_{\mathbf{x}}.}\end{array}$$(12)

The Lorentz force density is written, in accordance with (3), as

$$\begin{array}{}{\displaystyle {f}^{i}\equiv \frac{\delta {F}^{i}}{\delta V}={F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mu}^{i}\phantom{\rule{thinmathspace}{0ex}}{J}^{\mu}.}\end{array}$$(13)

Dynamics of a test particle is defined by an extension to curved spacetime of the special-relativistic form of Newton’s second law [2]. This applies to the non-interacting particles that constitute a *dust*. The following dynamical equation for a continuous medium with velocity field **v**, subjected to a non-gravitational external force density field **f**, has thus been *derived* for a dust from Newton’s second law, and has been assumed to stay valid for a general continuum [1]:

$$\begin{array}{}{\displaystyle {T}_{\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}\phantom{\rule{thinmathspace}{0ex}};\nu}^{0\nu}={b}^{0}({\mathit{T}}_{\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}})+\frac{\mathbf{f}\mathbf{.}\mathbf{v}}{c\beta},}\\ {T}_{\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}\phantom{\rule{thinmathspace}{0ex}};\nu}^{i\nu}={b}^{i}({\mathit{T}}_{\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}})+{f}^{i},\end{array}$$(14)

where *T*_{medium} is the energy-momentum tensor of the continuous medium and

$$\begin{array}{}{\displaystyle {b}^{0}(\mathit{T})\equiv \frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\gamma}^{00}\phantom{\rule{thinmathspace}{0ex}}{g}_{ij,0}\phantom{\rule{thinmathspace}{0ex}}{T}^{ij},\phantom{\rule{1em}{0ex}}{b}^{i}(\mathit{T})\equiv \frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{g}^{ij}{g}_{jk,0}\phantom{\rule{thinmathspace}{0ex}}{T}^{0k}.}\end{array}$$(15)

For a *charged continuum*, *f*^{i} is given by Eq. (13), and the energy-momentum tensor *T*_{charged medium} has to be substituted for *T*_{medium}. Then Eq. (14) for this medium can be rewritten as:

$$\begin{array}{}{\displaystyle {T}_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};\nu}^{\mu \nu}={b}^{\mu}({\mathit{T}}_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}})+{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu}^{\mu}\phantom{\rule{thinmathspace}{0ex}}{J}^{\nu}.}\end{array}$$(16)

It seems natural, almost obvious, to assume (i) that the *total* energy-momentum is the sum

$$\begin{array}{}{\displaystyle \mathit{T}={\mathit{T}}_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\mathrm{m}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{m}}+{\mathit{T}}_{\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}},}\end{array}$$(17)

where *T*_{field} is the energy-momentum tensor of the electromagnetic field [3, 4]:

$$\begin{array}{}{\displaystyle {T}_{\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}^{\phantom{\rule{thinmathspace}{0ex}}\mu \nu}\equiv \left(-{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\lambda}^{\mu}{F}^{\phantom{\rule{thinmathspace}{0ex}}\nu \lambda}+\frac{1}{4}{\gamma}^{\phantom{\rule{thinmathspace}{0ex}}\mu \nu}{F}_{\lambda \rho}{F}^{\lambda \rho \phantom{\rule{thinmathspace}{0ex}}}\right)/{\mu}_{0},}\end{array}$$(18)

or

$$\begin{array}{}{\displaystyle {\mathit{T}}_{\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}\equiv ({T}_{{}_{\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}\nu}^{\mu})=\left[{\mathit{F}}^{2}-\frac{1}{4}(\mathrm{t}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}{\mathit{F}}^{2})\mathit{I}\right]/{\mu}_{0},}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathit{F}\equiv ({F}_{\nu}^{\mu});\end{array}$$(19)

and (ii) that the total tensor *T* obeys the general equation (14) for continuum dynamics, without any non-gravitational external force:

$$\begin{array}{}{\displaystyle {T}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};\nu}^{\mu \nu}={b}^{\mu}(\mathit{T}).}\end{array}$$(20)

These two assumptions have indeed been made in Ref. [1]. Combining Eqs. (16) and (20) using Eqs. (17) and (18), one derives

$$\begin{array}{}{\displaystyle {F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\lambda}^{\mu}\phantom{\rule{thinmathspace}{0ex}}{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};\nu}^{\lambda \nu}={\mu}_{0}\left[{b}^{\mu}\left({\mathit{T}}_{\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}}\right)-{F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\lambda}^{\mu}\phantom{\rule{thinmathspace}{0ex}}{J}^{\lambda}\right],}\end{array}$$(21)

where *b*^{μ} (*T*_{field} is given by Eqs. (15) and (18). Under the two assumptions right above, this is the second group of the gravitationally-modified Maxwell equations in the investigated theory — at least for the generic case where the field tensor *F* is invertible (det $\begin{array}{}{\displaystyle \mathit{F}\equiv det({F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu}^{\mu})\ne 0).}\end{array}$

Indeed, if the matrix $\begin{array}{}{\displaystyle ({F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu}^{\mu})}\end{array}$ is invertible, (21) is equivalent to

$$\begin{array}{}{\displaystyle {F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};\nu}^{\mu \nu}={\mu}_{0}\left({G}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu}^{\mu}\phantom{\rule{thinmathspace}{0ex}}{b}^{\nu}({\mathit{T}}_{\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}})-{J}^{\mu}\right),\phantom{\rule{1em}{0ex}}({G}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu}^{\mu})\equiv ({F}_{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\nu}^{\mu}{)}^{-1}.}\end{array}$$(22)

(Note that *G*, like *F*, is an antisymmetric tensor, *G*_{νμ} = −*G*_{μν}.)

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