Sunday 20 January 2019

special relativity - Gauss' theorem in relativistic conditions


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 $$f(\theta,\varphi)=k_e q\,\frac{1-\beta^{2}}{\left[1-\beta^{2}\sin^2\theta\right]^{\tfrac{3}{2}}}$$


enter image description here


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



Reference : My answer here Electric field associated with moving charge





enter image description here


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.


enter image description here


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.


enter image description here


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}


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...