0
$\begingroup$

EDIT: Added Clarification.

By various methods, it's possible to determine that in the formula for Electric potential for the position of a charge in uniform motion, $V=\frac{q}{4\pi\epsilon_0\kappa \cdot c(t-t_r)}$

Where $c(t-t_r)$ is the distance between field point and charge at the retarded time, and $\kappa=1-\frac{\vec{\beta}\cdot (\vec{x}-\vec{u}t_r)}{c(t-t_r)}$

$\vec{\beta}$ is velocity divided by c. The position of the charge at a given time $t$ is $\vec{u}t$

and the following equalities hold.

$$D=\sqrt{(x-ut)^2+(y^2+z^2)/\gamma^2}$$

where $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ and $\beta=u/c$, where $v$ is the uniform velocity of the charge.

Also, $D=c(t-t_r)-\vec{\beta}\cdot(\vec{x}-\vec{u}t_r)=\kappa c(t-t_r).$

Now $D$ is not the distance between the field point $\vec{x}$ and the charge at retarded time $\vec{u}t_r$. Yet $V=\frac{q}{4\pi\epsilon_0D}$.

How is one to interpret $D$?

As a commenter suggested, it might be better to consider the change has happening to the charge. Instead of $q$ we have $q_{eff}=q/\kappa$ and use the retarded time distance in the denominator $V=\frac{q_{eff}}{4\pi \epsilon_0\cdot c(t-t_r)}$.

Now the distance used makes sense, its the distance to the charge between field point and charge location at the retarded time. But what is the physical meaning of this adjusted charge? While it makes the math work and allows for the use of a physically meaningful distance in the formula, it introduces additional questions. What is the physical meaning of adjusting the charge like this? Charge is a Lorentz invariant and and $\kappa \neq \gamma$.

That the formula takes this form can be explained by the need to rescale the charge density and component of displacement in the direction of motion, but this fails to explain the new formula in terms of the retarded time.

$\endgroup$
0
$\begingroup$

enter image description here


Under a general smooth curvilinear motion of the charge $\;q$, see above Figure-01, the scalar electromagnetic potential is given by \begin{equation} \phi(\mathbf r, t)\boldsymbol{=}\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{\left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert\!\boldsymbol{-}\! \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\left(\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right)}\boldsymbol{=}\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{\left[1\!\boldsymbol{-}\! \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\dfrac{\left(\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right)}{\left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert}\right]\cdot\left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert} \tag{01}\label{01} \end{equation}
where the superscript $\;^{\prime}\boldsymbol{*}^{\prime}\;$ refers to the retarded quantities time, position and velocity $\,t^{\boldsymbol{*}},\mathbf{x}^{\boldsymbol{*}},\boldsymbol{\upsilon}^{\boldsymbol{*}}\,$ respectively. Above equation could be expressed as \begin{equation} \phi(\mathbf r, t)\boldsymbol{=}\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{\kappa\cdot R} \tag{02}\label{02} \end{equation} where \begin{equation} R\boldsymbol{=}\left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert \tag{03}\label{03} \end{equation} is the distance from the retarded position of the charge $q$ (point $\rm Q^{\boldsymbol{*}}$ in the Figure) to the field or observation point (point $\rm A$ in the Figure) and \begin{equation} \kappa\boldsymbol{=}1\!\boldsymbol{-}\! \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\dfrac{\left(\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right)}{\left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert}\boldsymbol{=}1\!\boldsymbol{-}\! \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\mathbf{n}_{_R} \tag{04}\label{04} \end{equation} is a positive factor less than or equal to or greater than 1 if instantaneously the charge $\;q\;$ from its retarded position $\rm Q^{\boldsymbol{*}}\;$ is coming closer, is running away or keep the same distance from the field point $\rm A$ respectively.

Now, to interpret the result \eqref{02} in the frame of electrostatics you define an effective distance \begin{equation} R_{\rm eff}\boldsymbol{\equiv}\kappa\cdot R \tag{05}\label{05} \end{equation} and conclude that the scalar potential is that of the electrostatic one due to the presence of the charge not at the real distance $R$ but at a distance $R_{\rm eff}$ \begin{equation} \phi(\mathbf r, t)\boldsymbol{=}\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{R_{\rm eff}} \tag{06}\label{06} \end{equation}

But I think that a more suitable interpretation would be to define an effective charge \begin{equation} q_{\rm eff}\boldsymbol{\equiv}\dfrac{q}{\kappa} \tag{07}\label{07} \end{equation} and conclude that the scalar potential is that of the electrostatic one due to the presence of a particle at the real distance $R$ not with the real charge $q$ but with the effective charge $q_{\rm eff}$ \begin{equation} \phi(\mathbf r, t)\boldsymbol{=}\dfrac{1}{4\pi\epsilon_0}\dfrac{q_{\rm eff}}{R} \tag{08}\label{08} \end{equation}

These results have nothing to do with the time dilation and length contraction effects.


Related : Retarded time miscalculation?

$\endgroup$
  • $\begingroup$ That makes some sense, keeps the distance matching $c(t-t_r)$. And the charge density is supposed to change between frames rendering exactly that effective charge. Charge is a Lorentz Invariant. Is this consistent with using an effective charge? Perhaps the "effective charge" we see here is just a failure to account for the magnetic influences of the charge? $\endgroup$ – R. Romero Jan 31 at 16:31
  • $\begingroup$ @R. Romero : Again, I don't think there is a relation with charge Lorentz invariance between frames. The charges $q$ and $q_{\rm ef}$ has no relation similar to that between the volume charge densities $\varrho$ and $\varrho^\prime$ under a Lorentz transformation. $\endgroup$ – Frobenius Jan 31 at 17:23
  • $\begingroup$ The notion of an effective charge makes the math come out right, but it would seem to need further justification. What is its physical meaning? Keeping R as the distance associated with the relation and the factor effecting other components makes sense. That makes the relation a function of distance to the charge at the retarded time. $\endgroup$ – R. Romero Jan 31 at 18:40

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.