Gauss' theorem in relativistic conditions

Question

Remembering the surface integrals, we suppose that a vectorial field $\mathbf{F}(\bar{r})$ let be of the form $$\mathbf{F}(\bar{r})=\frac{f(\theta,\varphi)}{r^3}\,\bar{r}$$ For the flux through $S$, when it is parametrized by a domain $D_{(\theta,\varphi)}=[0,\pi]\times [0,2\pi]$, we know that $$\int_{S}\mathbf{F}(\bar{r})\,\overline{da}=\iint_D f(\theta,\varphi) \sin \theta\, d\theta d\varphi=4\pi k_e q \tag 1$$

If $S$ is closed (but not enclosure the charge), where $$f(\theta,\varphi)=k_e q\,\frac{1-\beta^{2}}{\left[1-\beta^{2}\sin^2\theta\right]^{\tfrac{3}{2}}}$$

and the direction of the motion of the charge $q$ (supposing an horizontal direction) generate an angle $\theta_0$, the domain $D_{(\theta,\varphi)}$ is as $[0,\theta_0]\times[0,2\pi]$ (hence $0<\theta_0 <\pi$), $$\int_{S}\mathbf{F}(\bar{r})\,\overline{da}=\iint_{D} f(\theta,\varphi) \sin \theta\, d\theta d\varphi=\int_{0}^{2\pi}d\varphi\int_{0}^{\theta_0}f(\theta,\varphi) \sin \theta\, d\theta \color{red}{\neq 4\pi k_e q} \tag 2.$$

What are the possible physical considerations if the flux is different by $4\pi k_e q$ as into the formula $(2)$?

What happen in this situation?

Answer 1

I don't know how you convinced yourself that the integral doesn't equal $4\pi$. It does. Just to make sure, I checked using the open-source computer algebra system Maxima:

maxima -q --batch-string="2*%pi*integrate(sin(x)*(1-beta^2)*(1-beta^2*sin(x)^2)^(-3/2),x,0,%pi);"

The result is $4\pi$:

                                              2
                                4 %pi (1 - beta )
(%o1)                         - -----------------
                                        2
                                    beta  - 1

Answer 2

Reference : My answer here Electric field associated with moving charge

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As proved in my above referenced answer, the electric field of a point charge $\,q\,$ in uniform rectilinear motion is given by the following equation \begin{equation} \mathbf{E}\left(\mathbf{r}\right) \boldsymbol{=}\dfrac{q}{4\pi \epsilon_{0}}\dfrac{(1\boldsymbol{-}\beta^2)}{\left(1\!\boldsymbol{-}\!\beta^{2}\sin^{2}\!\theta\right)^{\frac32}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{3}}\quad \text{(uniform rectilinear motion)} \tag{01}\label{01} \end{equation} as shown in Figure-01 (in all subsequent Figures we suppose, without loss of generality, that the charge is positive $\,q>0$). In this Figure for the constant velocity vector of the charge \begin{equation} \boldsymbol{\beta} \boldsymbol{=} \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta\boldsymbol{=}\dfrac{\upsilon}{c}, \quad \boldsymbol{\dot{\beta}}\boldsymbol{=}\dfrac{\mathrm d\boldsymbol{\beta} }{\mathrm dt}\boldsymbol{=0} \tag{02}\label{02} \end{equation} The closed curve shown in Figure-01 is locus of constant electric field magnitude \begin{equation} \Vert\mathbf{E}\Vert \boldsymbol{=}\dfrac{\vert q \vert}{4\pi \epsilon_{0}}\dfrac{(1\boldsymbol{-}\beta^2)}{\left(1\!\boldsymbol{-}\!\beta^{2}\sin^{2}\!\theta\right)^{\frac32}r^{2}} \boldsymbol{=}\text{constant} \tag{03}\label{03} \end{equation} More exactly the set of points with this constant magnitude of the electric field is the surface generated by a half revolution of this closed curve around the $\;x\boldsymbol{-}$axis or a half revolution around the $\;y\boldsymbol{-}$axis.

In Figure-02 above it's shown the electric field on a circle, that is on the surface of a sphere generated by a half revolution of this circle around the $\;x\boldsymbol{-}$axis or the $\;y\boldsymbol{-}$axis. We note that the field is always normal to the spherical surface and get stronger as we approach directions normal to that of the motion of the charge.

In Figure-03 above it's shown the electric flux through the circular arc $\rm ABC$, that is through the spherical cap generated by a half revolution of this circular arc around the $\;x\boldsymbol{-}$axis. As proved in my referenced answer in the beginning, the electric flux is given by the following equation

\begin{equation} \Phi\left(\theta\right)\boldsymbol{=}\ \dfrac{q}{2\epsilon_{0}}\Biggl[1\boldsymbol{-}\dfrac{ \cos\theta}{\sqrt{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} \cos^2\theta\right)\vphantom{\frac12}}}\Biggr] \tag{04}\label{04} \end{equation} For $\theta\boldsymbol{=}\pi$ we could verify Gauss Law \begin{equation} \Phi\left(\pi\right)\boldsymbol{=}\ \dfrac{q}{\epsilon_{0}} \tag{05}\label{05} \end{equation}

Now, note that the solid angle $\,\Omega\left(\theta\right)\,$ generated by a complete revolution of the plane angle $\,\theta\,$ around the $\;x\boldsymbol{-}$axis in Figure-03 is given by
\begin{equation} \Omega\left(\theta\right)\boldsymbol{=}2\pi\left(1\boldsymbol{-}\cos\theta\right) \tag{06}\label{06} \end{equation} We verify that $\,\Omega\left(0\right)\boldsymbol{=}0\:,\:\Omega\left(\pi/2\right)\boldsymbol{=}2\pi\:,\:\Omega\left(\pi\right)\boldsymbol{=}4\pi\:$ as expected.

From equations \eqref{04} and \eqref{06} we note that there doesn't exist analogy between the flux $\,\Phi\left(\theta\right)\,$ and the solid angle $\,\Omega\left(\theta\right)\,$ since
\begin{equation} \dfrac{\Phi\left(\theta\right)}{\Omega\left(\theta\right)}\boldsymbol{=} \dfrac{q}{4\pi\epsilon_{0}\left(1\boldsymbol{-}\cos\theta\right)}\Biggl[1\boldsymbol{-}\dfrac{ \cos\theta}{\sqrt{\left(1\!\boldsymbol{-}\!\beta^{2}+\beta^{2} \cos^2\theta\right)\vphantom{\frac12}}}\Biggr]\boldsymbol{\ne}\text{constant}\,, \qquad \left(\boldsymbol{\beta} \boldsymbol{\ne 0}\right) \tag{07}\label{07} \end{equation} This is due to the fact that there is no spherical symmetry as shown in Figure-02.

To the contrary, the flux $\,\Phi\left(\theta\right)\,$ is proportional to the solid angle $\,\Omega\left(\theta\right)\,$ in case of a charge at rest ($\boldsymbol{\beta} \boldsymbol{=0}$), since equation \eqref{07} for $\,\beta \boldsymbol{=}0\,$ yields \begin{equation} \dfrac{\Phi\left(\theta\right)}{\Omega\left(\theta\right)}\boldsymbol{=} \dfrac{q}{4\pi\epsilon_{0}}\boldsymbol{=}\text{constant}\,, \qquad \left(\boldsymbol{\beta} \boldsymbol{= 0}\right) \tag{08}\label{08} \end{equation}