We know that coulomb charge can't be applied on charges in motion. Why? What if they are moving in same velocity? Aren't they actually at rest for each other? ( yeah, it depends upon medium, so I'm assuming vaccum for all the cases). So what's the new coming in motion? Also, if that's false, doesn't that disprove that motion is relative because we can just find out in only 2 objects who's moving by calculating if coulomb's formula gave us right answer about the force.

  • $\begingroup$ Is this a question about two individual charges going in parallel, or two parallel conductors with identical current? $\endgroup$ Jul 18, 2021 at 2:40
  • 1
    $\begingroup$ About 2 charges moving exerting electrostatic force on each other $\endgroup$ Jul 18, 2021 at 4:39

3 Answers 3


$\texttt{C O N T E N T S}$


$\boldsymbol\S\texttt{ A. Two charges }q_1,q_2\texttt{ at rest (system S$'$)}$

$\boldsymbol\S\texttt{ B. Two charges }q_1,q_2\texttt{ in uniform translational motion (system S)}$

$\boldsymbol\S\texttt{ C. The Lorentz boost transformation ($\texttt S$$'\longrightarrow$ S) }$


A first case concerns two electric charges $\:q_1\:$ and $\:q_2\:$ at rest with respect to an inertial system $\:\texttt S'$. The electrostatic fields and forces using Coulomb's law are given in $\boldsymbol\S\,\texttt{A}$. A second case concerns two electric charges $\:q_1\:$ and $\:q_2\:$ in motion both with common uniform velocity $\,\boldsymbol{\upsilon}\,$ with respect to an inertial system $\:\texttt S$. The electromagnetic fields and forces are given in $\boldsymbol\S\,\texttt{B}\,$ using well-known expressions derived from the relativistic Liénard–Wiechert potentials. The two cases are studied independently of one another with no relation between them until $\boldsymbol\S\,\texttt{C}$ where it is assumed that the inertial system $\,\texttt S\,$ of the latter case arises from a Lorentz boost transformation with velocity $\,\boldsymbol{-\upsilon}\,$ with respect to the inertial system $\,\texttt S'\,$ of the former case. Now the electromagnetic fields and forces in system $\,\texttt S\,$ are derived from the Lorentz boost transformation of the electrostatic fields and forces in the rest system $\,\texttt S'$. The results from this application of the Lorentz transformation are in full agreement with those in $\boldsymbol\S\,\texttt{B}\,$ derived from the Liénard–Wiechert potentials. Moreover they provide relations connecting the two systems.

$\boldsymbol\S$ A. Two charges $\,q_1,q_2\,$ at rest

Consider two electric charges $\:q_1\:$ and $\:q_2\:$ at rest with respect to an inertial system $\:\texttt S'$, see Figure-01. One charge feels only the electrostatic Coulomb force of the other. Fields and forces are as follows \begin{align} \texttt{field of charge }q_1\texttt{ in }\texttt S' :\quad\mathbf E'_1 &\boldsymbol{=}\boldsymbol{+}\dfrac{q_1}{4\pi\epsilon_0}\dfrac{\mathbf r'}{\Vert \mathbf r' \Vert^3}\,, \quad \mathbf B'_1\boldsymbol{=0} \tag{A-01a}\label{A-01a}\\ \texttt{field of charge }q_2\texttt{ in }\texttt S' :\quad\mathbf E'_2 &\boldsymbol{=}\boldsymbol{-}\dfrac{q_2}{4\pi\epsilon_0}\dfrac{\mathbf r'}{\Vert \mathbf r' \Vert^3}\,, \quad \mathbf B'_2\boldsymbol{=0} \tag{A-01b}\label{A-01b}\\ \texttt{force on charge }q_1\texttt{ in }\texttt S' :\quad\mathbf F'_1 & \boldsymbol{=}q_1\mathbf E'_2 \boldsymbol{=}\boldsymbol{-}\dfrac{q_1q_2}{4\pi\epsilon_0}\dfrac{\mathbf r'}{\Vert \mathbf r' \Vert^3} \tag{A-01c}\label{A-01c}\\ \texttt{force on charge }q_2\texttt{ in }\texttt S' :\quad\mathbf F'_2 & \boldsymbol{=}q_2\mathbf E'_1 \boldsymbol{=}\boldsymbol{+}\dfrac{q_2q_1}{4\pi\epsilon_0}\dfrac{\mathbf r'}{\Vert \mathbf r' \Vert^3} \tag{A-01d}\label{A-01d} \end{align}

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$\boldsymbol\S$ B. Two charges $\,q_1,q_2\,$ in uniform translational motion

Consider two electric charges $\:q_1\:$ and $\:q_2\:$ in motion both with common uniform velocity $\,\boldsymbol{\upsilon}\,$ with respect to an inertial system $\:\texttt S$. The electromagnetic fields and forces are shown in Figure-02 and Figure-03 respectively. Derived from the relativistic Liénard–Wiechert potentials are as follows \begin{align} &\texttt{field of charge }q_1\texttt{ in }\texttt S : \nonumber\\ \mathbf E_1 & \boldsymbol{=}\boldsymbol{+}\dfrac{q_1}{4\pi \epsilon_0}\dfrac{\left(1\!\boldsymbol{-}\!\beta^2\right)}{\left(1\!\boldsymbol{-}\!\beta^2\sin^2\!\phi\right)^{3/2}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^3}\,, \quad\mathbf B_1\boldsymbol{=}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf E_1\right) \tag{B-01a}\label{B-01a}\\ &\texttt{field of charge }q_2\texttt{ in }\texttt S : \nonumber\\ \mathbf E_2 & \boldsymbol{=}\boldsymbol{-}\dfrac{q_2}{4\pi \epsilon_0}\dfrac{\left(1\!\boldsymbol{-}\!\beta^2\right)}{\left(1\!\boldsymbol{-}\!\beta^2\sin^2\!\phi\right)^{3/2}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^3}\,, \quad \mathbf B_2\boldsymbol{=}\dfrac{1}{c^2}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf E_2\right) \tag{B-01b}\label{B-01b}\\ &\texttt{Lorentz force on charge }q_1\texttt{ in }\texttt S : \quad\mathbf F_1 \boldsymbol{=}q_1\left(\mathbf E_2\boldsymbol{+}\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf B_2\right) \tag{B-01c}\label{B-01c}\\ &\texttt{Lorentz force on charge }q_2\texttt{ in }\texttt S : \quad\mathbf F_2 \boldsymbol{=}q_2\left(\mathbf E_1\boldsymbol{+}\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf B_1\right) \tag{B-01d}\label{B-01d} \end{align}
where \begin{equation} \beta\boldsymbol{=}\dfrac{\upsilon}{c} \tag{B-02}\label{B-02} \end{equation}

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$\boldsymbol\S$ C. The Lorentz boost transformation

Suppose now that the system $\,\texttt S\,$ of $\boldsymbol\S\,\texttt{B}\,$ arises from the system $\,\texttt S'\,$ of $\boldsymbol\S\,\texttt{A}\,$ by a Lorentz boost with velocity $\,\boldsymbol{-\upsilon}$. Then the electromagnetic fields of the two charges and the Lorentz forces between them $\mathbf E_\jmath,\mathbf B_\jmath, \mathbf F_\jmath$ could result from the Lorentz transformation of those $\mathbf E'_\jmath,\mathbf B'_\jmath, \mathbf F'_\jmath$ in the rest system $\,\texttt S'$. Our task in this paragraph is to prove that the results provided by this Lorentz transformation are in full agreement with those in $\boldsymbol\S\,\texttt{B}\,$ provided by the Liénard–Wiechert potentials.

The Lorentz boost transformation with finite differences is \begin{align} \Delta\mathbf x & \boldsymbol{=} \Delta\mathbf x'\boldsymbol{+}\left(\gamma\boldsymbol{-}1\right)\left(\Delta\mathbf x'\boldsymbol{\cdot}\mathbf n\right)\mathbf n\boldsymbol{+}\gamma\boldsymbol{\upsilon} \Delta t' \tag{C-01a}\label{C-01a}\\ \Delta t & \boldsymbol{=} \gamma\left(\Delta t'\boldsymbol{+}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \Delta\mathbf x'}{c^{2}}\right) \tag{C-01b}\label{C-01b}\\ \gamma & \boldsymbol{=} 1\Bigg/\sqrt{1 \boldsymbol{-}\dfrac{\upsilon^2}{c^2}}\,,\quad \mathbf n\boldsymbol{=}\dfrac{\boldsymbol{\upsilon}}{\Vert\boldsymbol{\upsilon}\Vert}\boldsymbol{=}\dfrac{\boldsymbol{\upsilon}}{\upsilon} \tag{C-01c}\label{C-01c} \end{align} where $\Delta\mathbf x', \Delta t'$ the space and time separation of two events in the accented rest frame $\,\texttt S'\,$ and $\Delta\mathbf x, \Delta t$ the space and time separation of two events in the frame $\,\texttt S$.

Consider now that Figure-01 represents a snapshot of the charges in their rest system $\,\texttt S'\,$ so that \begin{equation} \Delta\mathbf x'\boldsymbol{=}\mathbf x'_2\boldsymbol{-}\mathbf x'_1\boldsymbol{=}\mathbf r'\,,\qquad \Delta t'\boldsymbol{=}t'_2\boldsymbol{-}t'_1\boldsymbol{=}0 \tag{C-02}\label{C-02} \end{equation} Inserting above $\Delta\mathbf x', \Delta t'$ in the Lorentz boost equations \eqref{C-01a}-\eqref{C-01b} the space and time separations $\Delta\mathbf x, \Delta t$ of these events in frame $\,\texttt S\,$ are \begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\mathbf r^{\boldsymbol{*}}& \boldsymbol{=} \Delta\mathbf x \boldsymbol{=}\mathbf x_2\boldsymbol{-}\mathbf x_1 \boldsymbol{=} \mathbf r'\boldsymbol{+}\left(\gamma\boldsymbol{-}1\right)\left(\mathbf r'\boldsymbol{\cdot}\mathbf n\right)\mathbf n \boldsymbol{=} \mathbf r'\boldsymbol{+}\left(\gamma\boldsymbol{-}1\right)\mathbf r'_{\boldsymbol{\Vert}} \boldsymbol{=}\mathbf r'_{\boldsymbol{\perp}}\boldsymbol{+}\gamma\mathbf r'_{\boldsymbol{\Vert}} \tag{C-03a}\label{C-03a}\\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \Delta t & \boldsymbol{=}t_2\boldsymbol{-}t_1\boldsymbol{=} \gamma\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf r'}{c^{2}} \tag{C-03b}\label{C-03b} \end{align} The two events are not simultaneous in general in system $\,\texttt S$. More exactly the charge $\,q_2\,$ positioned at $\,\mathbf r^{\boldsymbol{*}}\,$ with respect to $\,q_1\,$ is appeared there at a time $\,t_2\,$ where \begin{equation} \left. \begin{cases} t_2 \boldsymbol{>}t_1 & \texttt{if }\left(\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf r'\right)\boldsymbol{>}0 \\ t_2 \boldsymbol{=}t_1 & \texttt{if }\left(\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf r'\right)\boldsymbol{=}0 \\ t_2 \boldsymbol{<}t_1 & \texttt{if }\left(\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf r'\right)\boldsymbol{<}0 \end{cases}\right\} \tag{C-04}\label{C-04} \end{equation} To see where the charge $\,q_2\,$ would be at the time moment $\,t_1\,$ in order to have a snapshot at this time moment in $\,\texttt S$, we must translate it by \begin{equation} \boldsymbol{\upsilon}\Delta t \boldsymbol{=} \boldsymbol{\upsilon}\gamma\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf r'}{c^{2}}\boldsymbol{=}\gamma\dfrac{\upsilon^2}{c^2}\left(\mathbf r'\boldsymbol{\cdot}\mathbf n\right)\mathbf n \boldsymbol{=}\dfrac{\gamma^2\boldsymbol{-}1}{\gamma}\left(\mathbf r'\boldsymbol{\cdot}\mathbf n\right)\mathbf n\boldsymbol{=}\dfrac{\gamma^2\boldsymbol{-}1}{\gamma}\mathbf r'_{\boldsymbol{\Vert}} \tag{C-05}\label{C-05} \end{equation} meaning that at the time moment $\,t_1\,$ the charge $\,q_2\,$ is positioned with respect to $\,q_1\,$ at \begin{equation} \mathbf r \boldsymbol{=} \mathbf r^{\boldsymbol{*}}\boldsymbol{-}\boldsymbol{\upsilon}\Delta t \boldsymbol{=} \mathbf r'\boldsymbol{+}\left(\gamma\boldsymbol{-}1\right)\left(\mathbf r'\boldsymbol{\cdot}\mathbf n\right)\mathbf n \boldsymbol{-}\dfrac{\gamma^2\boldsymbol{-}1}{\gamma}\left(\mathbf r'\boldsymbol{\cdot}\mathbf n\right)\mathbf n \tag{C-06}\label{C-06} \end{equation} so \begin{equation} \mathbf r \boldsymbol{=}\mathbf r'\boldsymbol{-}\dfrac{\gamma\boldsymbol{-}1}{\gamma}(\mathbf r'\boldsymbol{\cdot} \mathbf n)\mathbf n \boldsymbol{=}\mathbf r'_{\boldsymbol{\perp}}\boldsymbol{+}\dfrac{1}{\gamma}\mathbf r'_{\boldsymbol{\Vert}} \tag{C-07}\label{C-07} \end{equation} The vectors $\mathbf r',\mathbf r^{\boldsymbol{*}},\mathbf r$ and their relations are shown in Figure-04. Each one is composed by two components, one along the velocity direction $\,\mathbf n\,$ and the other normal to it \begin{equation} \mathbf a \boldsymbol{=}\mathbf a_{\boldsymbol{\Vert}}\boldsymbol{+}\mathbf a_{\boldsymbol{\perp}}\boldsymbol{=}\underbrace{\left(\mathbf n\boldsymbol{\cdot}\mathbf a\right)\mathbf n}_{\mathbf a_{\boldsymbol{\Vert}}}\boldsymbol{+}\overbrace{\underbrace{\mathbf a\boldsymbol{-}\left(\mathbf n\boldsymbol{\cdot}\mathbf a\right)\mathbf n}_{\mathbf a_{\boldsymbol{\perp}}}}^{\left(\mathbf n\boldsymbol{\times}\mathbf a\right)\boldsymbol{\times}\mathbf n}\;,\qquad \mathbf a \boldsymbol{=}\mathbf r',\mathbf r^{\boldsymbol{*}},\mathbf r \tag{C-08}\label{C-08} \end{equation}

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Let examine now how the electromagnetic fields and forces produced by the charges are Lorentz transformed from the rest frame $\,\texttt S'\,$ to the frame $\,\texttt S$. It's sufficient to determine the field and the force from one particle, say $\,q_1$, to the other $\,q_2$.

Under the Lorentz boost transformation expressed by equations \eqref{C-01a}-\eqref{C-01b} the electromagnetic field is transformed as follows \begin{align} \mathbf E_1 & \boldsymbol{=}\gamma \mathbf E'_1\boldsymbol{-}\left(\gamma\boldsymbol{-}1\right)\left( \mathbf E'_1\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\boldsymbol{-}\;\gamma\,\left(\boldsymbol{\upsilon}\boldsymbol{\times} \mathbf B'_1\right) \tag{C-09a}\label{C-09a}\\ \mathbf B_1 & \boldsymbol{=} \gamma \mathbf B'_1\boldsymbol{-}\left(\gamma\boldsymbol{-}1\right)\left(\mathbf B'_1\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf E'_1\right) \tag{C-09b}\label{C-09b} \end{align} Since $\,\mathbf B'_1\boldsymbol{=0}$, see equation \eqref{A-01a}, we have \begin{align} \mathbf E_1 & \boldsymbol{=}\gamma \mathbf E'_1\boldsymbol{-}\left(\gamma\boldsymbol{-}1\right)\left( \mathbf E'_1\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n} \tag{C-10a}\label{C-10a}\\ \mathbf B_1 & \boldsymbol{=} \dfrac{\gamma}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf E'_1\right) \tag{C-10b}\label{C-10b} \end{align} Note that since $\,\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf n\boldsymbol{=}\boldsymbol{\upsilon}\boldsymbol{\times}\boldsymbol{\upsilon}/\upsilon\boldsymbol{=0}\,$ \begin{equation} \mathbf B_1 \boldsymbol{=} \dfrac{\gamma}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf E'_1\right)\boldsymbol{=} \dfrac{1}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\gamma\mathbf E'_1\right) \boldsymbol{=}\dfrac{1}{c^{2}}\Bigl(\boldsymbol{\upsilon}\boldsymbol{\times}\overbrace{\bigl[\gamma \mathbf E'_1\boldsymbol{-}\left(\gamma\boldsymbol{-}1\right)\left( \mathbf E'_1\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\bigr]}^{\mathbf E_1}\Bigr)\boldsymbol{\implies} \nonumber \end{equation} so \begin{equation} \boxed{\:\:\mathbf B_1\boldsymbol{=} \dfrac{1}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf E_1\right)\:\:} \tag{C-11}\label{C-11} \end{equation} This is identical to the expression of $\,\mathbf B_1\,$ in equation \eqref{B-01a} derived by the Liénard–Wiechert potentials.

Inserting in \eqref{C-10a} the expression $\,\mathbf E'_1 \boldsymbol{=}\left(q_1/4\pi\epsilon_0\right)\bigl(\mathbf r'/\Vert\mathbf r' \Vert^3\bigr)$, see equation \eqref{A-01a}, we have \begin{align} \mathbf E_1 & \boldsymbol{=} \dfrac{q_1}{4\pi \epsilon_0}\biggl[\gamma \dfrac{\mathbf r'}{\Vert \mathbf r' \Vert^3}\boldsymbol{-}\left(\gamma\boldsymbol{-}1\right)\left( \dfrac{\mathbf r'}{\Vert \mathbf r' \Vert^3}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\biggr] \nonumber\\ & \boldsymbol{=} \dfrac{\gamma}{\Vert \mathbf r' \Vert^3}\underbrace{\left[\mathbf r'\boldsymbol{-}\dfrac{\gamma\boldsymbol{-}1}{\gamma}(\mathbf r'\boldsymbol{\cdot} \mathbf n)\mathbf n \right]}_{\mathbf r}\boldsymbol{=} \dfrac{\gamma\,\mathbf r}{\Vert \mathbf r' \Vert^3} \implies \nonumber\\ \mathbf E_1 & \boldsymbol{=} \dfrac{q_1}{4\pi \epsilon_0}\dfrac{\gamma\,\mathbf r}{\Vert \mathbf r' \Vert^3} \tag{C-12}\label{C-12} \end{align} From the relation \eqref{C-07} between $\,\mathbf r,\mathbf r'$ and the help of Figure-04 we could prove(1) that \begin{equation} \dfrac{\gamma}{\Vert \mathbf r' \Vert^3}\boldsymbol{=} \dfrac{\left(1\boldsymbol{-}\beta^2\right)}{\left(1\boldsymbol{-}\beta^2\sin^2\!\phi\right)^{3/2}}\dfrac{1}{\Vert\mathbf r\Vert^3} \tag{C-13}\label{C-13} \end{equation} so \eqref{C-12} yields \begin{equation} \boxed{\:\:\mathbf E_1 \boldsymbol{=} \dfrac{q_1}{4\pi \epsilon_0}\dfrac{\left(1\boldsymbol{-}\beta^2\right)}{\left(1\boldsymbol{-}\beta^2\sin^2\!\phi\right)^{3/2}}\dfrac{\mathbf r}{\Vert\mathbf r\Vert^3}\:\:} \tag{C-14}\label{C-14} \end{equation} This is identical to the expression of $\,\mathbf E_1\,$ in equation \eqref{B-01a} derived by the Liénard–Wiechert potentials.


Reference (1) : Magnetic field due to a single moving charge.

Reference (2): Electric field associated with moving charge.


(1) Proof of equation \eqref{C-13}

From equation \eqref{C-07} and Figure-04 we have \begin{align} \Vert\mathbf r\Vert^2 & \boldsymbol{=}\Vert\mathbf r'\Vert^2\boldsymbol{+}\left(\dfrac{\gamma\boldsymbol{-}1}{\gamma}\right)^2\underbrace{\Vert\left(\mathbf r'\boldsymbol{\cdot} \mathbf n\right)\mathbf n\Vert^2}_{\Vert\mathbf r'\Vert^2\cos^2\phi'} \boldsymbol{-}2\left(\dfrac{\gamma\boldsymbol{-}1}{\gamma}\right)\underbrace{\left(\mathbf r'\boldsymbol{\cdot} \mathbf n\right)^2}_{\Vert\mathbf r'\Vert^2\cos^2\phi'} \boldsymbol{\implies} \nonumber\\ \Vert\mathbf r\Vert^2 & \boldsymbol{=}\Vert\mathbf r'\Vert^2\left[1\boldsymbol{-}\left(1\boldsymbol{-}\dfrac{1}{\gamma^2}\right)\cos^2\phi'\right] \boldsymbol{\implies} \nonumber \end{align} \begin{equation} \Vert\mathbf r\Vert^2\boldsymbol{=} \Vert\mathbf r'\Vert^2\left(1\boldsymbol{-}\beta^2\cos^2\phi'\right) \tag{Pr-01}\label{Pr-01} \end{equation} From the normal components equality $\:\mathbf r_{\boldsymbol{\perp}}\boldsymbol{=}\mathbf r'_{\boldsymbol{\perp}}\:$ we have \begin{equation} \Vert\mathbf r\Vert^2\sin^2\phi\boldsymbol{=} \Vert\mathbf r'\Vert^2\sin^2\phi' \tag{Pr-02}\label{Pr-02} \end{equation} that is \begin{equation} \cos^2\phi'\boldsymbol{=}1\boldsymbol{-}\dfrac{\Vert\mathbf r\Vert^2}{\Vert\mathbf r'\Vert^2}\sin^2\phi \tag{Pr-03}\label{Pr-03} \end{equation} Inserting this expression of $\:\cos^2\phi'\:$ in equation \eqref{Pr-01} and solving with respect to $\:\Vert\mathbf r\Vert^2/\Vert\mathbf r'\Vert^2\:$ we find \begin{equation} \dfrac{\Vert\mathbf r\Vert^2}{\Vert\mathbf r'\Vert^2}\boldsymbol{=}\dfrac{1\boldsymbol{-}\beta^2}{1\boldsymbol{-}\beta^2\sin^2\phi} \tag{Pr-04}\label{Pr-04} \end{equation} so \begin{equation} \dfrac{\Vert\mathbf r\Vert^3}{\Vert\mathbf r'\Vert^3}\boldsymbol{=}\dfrac{\left(1\boldsymbol{-}\beta^2\right)^{3/2}}{\left(1\boldsymbol{-}\beta^2\sin^2\phi\right)^{3/2}}\boldsymbol{=}\dfrac{\sqrt{1\boldsymbol{-}\beta^2}\left(1\boldsymbol{-}\beta^2\right)}{\left(1\boldsymbol{-}\beta^2\sin^2\phi\right)^{3/2}}\boldsymbol{=}\dfrac{\gamma^{\boldsymbol{-}1}\left(1\boldsymbol{-}\beta^2\right)}{\left(1\boldsymbol{-}\beta^2\sin^2\phi\right)^{3/2}} \tag{Pr-05}\label{Pr-05} \end{equation} or \begin{equation} \dfrac{\gamma}{\Vert \mathbf r' \Vert^3}\boldsymbol{=} \dfrac{\left(1\boldsymbol{-}\beta^2\right)}{\left(1\boldsymbol{-}\beta^2\sin^2\!\phi\right)^{3/2}}\dfrac{1}{\Vert\mathbf r\Vert^3} \tag{Pr-06}\label{Pr-06} \end{equation} proving equation \eqref{C-13}.
  • $\begingroup$ Kindly you are my biggest help. Could you please look at and/or edit my question physics.stackexchange.com/questions/665634/… to make it clearer for someone to answer? I thank you in advance. Thank you so much. I hope for an answer like this. $\endgroup$
    – Sebastiano
    Sep 13, 2021 at 22:04
  • $\begingroup$ In figure 2 when $\theta = 90°$ your equation gives $ E_1= q_1 /{ 4 \pi \epsilon r^2 \gamma}$ which means that $r \sqrt{\gamma}$ is the distance that fields travel before acting on $q_2$. Can you help in deriving your value cause I derived it to be $r \gamma$. I found this by thinking that electric fields released by $q_1$ at any specific point A can only affect $q_2$ at B point only. So i used only one field line not multiple as working on $q_2$ at any location of $q_2$. $\endgroup$
    – James Webb
    Feb 24 at 2:42
  • $\begingroup$ @James Webb : The results could not interpreted using distances. It's not so simple. For example, may be it could interpreted as the Coulomb force but the particle $\:q_2\:$ $''$feeling$''$ an effective charge $\:q_1/\gamma$. $\endgroup$
    – Frobenius
    Feb 24 at 7:34
  • $\begingroup$ @frobenius But charges are invariant, so how can you have a $q_1/\gamma$ charges in effect. This was not in my question. $\endgroup$
    – James Webb
    Feb 24 at 8:38
  • $\begingroup$ @frobenius i calculated the total force on $q_2$ where both charges are equidistant from observer and moving with $v$ velocity relative to him. $F = {q_1 q_2 sin\theta}/{ 4 \pi \epsilon r^2 \gamma^3} $ $= {q_1 q_2}/{4 \pi \epsilon r^2 \gamma^4}$. This component of force is perpendicular to their motion. But it is different from the force in the $q_2$'s frame by $1/\gamma^4$ factor even time dilation cannot remove it. A part of this factor still remains. $\endgroup$
    – James Webb
    Feb 24 at 9:09

If two objects move in parallel in vacuum with the same velocity, and you found an intertial frame of reference where both of them are at rest, then yes, you are perfectly fine to consider the situation in that frame of reference.

  • $\begingroup$ So coulomb's charge and hence force changes due to change of frame? But we have read ghat force do not depends upon frame of reference $\endgroup$ Jul 28, 2021 at 14:08
  • $\begingroup$ @Rajataggarwal Does this answer your question: physics.stackexchange.com/questions/126518 Coulomb charge doesn't depend on frame of reference, however forces do in special relativity. Electromagnetism is a manifestation of relativistic effects. $\endgroup$ Jul 28, 2021 at 15:04

In the inertial frame of reference in which both charges are at rest, you can apply Coulomb's law to find the force exerted by one charge on the other.

But then, to return to the "fixed" frame of reference, you must not forget that, in relativity, the forces are not invariant when you change the inertial frame of reference.

Sorry for my poor english. My native language is french.

  • $\begingroup$ Shall you please explain me 'force are not invariant in relativity'. I'm sorry. I could not understand that. Just a high school guy $\endgroup$ Jul 28, 2021 at 14:13

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