I am trying different methods for finding the retarded time in an electromagnetism problem. I'm getting different results, but I'm not sure what I'm doing wrong.

Consider a charge at the origin that begins moving at time t=0 at uniform velocity in the positive x direction and that it has been moving for quite some time so we can neglect edge effects.

In all the below, $\vec{\beta}=\vec{u}/c$

Method 1:

In the frame of the charge, The 4-potential is $(V/c, 0)$ where $V=\frac{q}{4\pi\epsilon_0\sqrt{x^2+y^2+z^2}}$.

Apply the Lorentz Transform then $V'=\frac{q}{4\pi\epsilon_0\sqrt{(x'-ut')^2+(y'^2)/\gamma^2+(z'^2)/\gamma^2}}$.

V' is the potential due to the charge's position in the past, with the distance term in the denominator representing what the distance would be in a corresponding electro static problem. So that distance over c is the retarded time:


Method 2:

Let $\vec{x}$ be the observation point and $\vec{x'}=\vec{u}t$ be the source point.

Then the retarted time,$t_r$, solves $|\vec{x}-\vec{u}t_r|=c(t-t_r)$.

Squaring both sides and factoring out $c$ to pair it with the $t_r$ one gets:


The quadratic equation gives:

$$ct_r=\gamma^2(ct-\vec{\beta}\cdot\vec{x}) (-+) \gamma\sqrt{\gamma^2(x-\beta c t)^2+(y^2+z^2)}$$

So : $$c(t-t_r)=\gamma^2(\vec{\beta}\cdot \vec{x}-\beta^2 ct)(+-)\gamma\sqrt{\gamma^2(x-\beta c t)^2+(y^2+z^2)}$$

Method 3:

Griffith's coordinate change method:

Begin with $V=\int\int\int{\frac{q\delta(\vec{x'}-\vec{u}t')d\tau}{4\pi\epsilon_0|\vec{x}-\vec{x'}|}}$

Find the Jacbian associated with the coordinate change, then evaluate the integral:

$$V'=\frac{q}{4\pi\epsilon_0(1-\hat{n}\cdot \vec{\beta})R_{eq}}$$

Where $\tau$ is infinitesimal volume, $\vec{x}$ is observation point, $\vec{x'}$ is current location of the charge, i.e. $\vec{u}t'$, and $t'$ is the retarded time , $t_r$. $\vec{u}$ is the uniform velocity. $\vec{\beta}$ is the velocity vector divided by c.

$$\vec{R_{eq}}=\vec{x}-\vec{x'}=\vec{x}-\vec{u}t_r$$ $R_{eq}=|\vec{R_{eq}}|$, and $\hat{n}=\vec{R_{eq}}/R_{eq}$



but by definition, doesn't $c(t-t_r)=|\vec{x}-\vec{u}t_r|$?

If so then the dot product is extraneous.


1 Answer 1


Method 1 :

Your assertion that

$^{\prime\prime}$V' is the potential due to the charge's position in the past, with the distance term in the denominator representing what the distance would be in a corresponding electrostatic problem$^{\prime\prime}$

is correct (although I think you may be confuse it with the fact that the electric field intensity $ \mathbf{E} $ points from the present position and not from the retarded one to the field point).

But your assertion that this distance is equal to the distance $\;R\boldsymbol{=}\left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert\;$ from the retarded position of the charge $\;q$, say $\;\mathbf{x}^{\boldsymbol{*}}$, to the field point (or observation point as you call it in Method 2), say $\;\mathbf{r}$, is incorrect. With symbols as in an answer of mine(1) the scalar potential at field point $\;\mathbf{r}\,$ is \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{q}{4\pi\epsilon_0}\dfrac{1}{\kappa\cdot R} \tag{02}\label{02} \end{equation} where \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} \tag{03}\label{03} \end{equation} is a factor less than or equal to or greater than 1 if instantaneously the charge $\;q\;$ from its retarded position is coming closer, is running away or keep the same distance from the field point respectively(2),(3). As you could realize you are keeping this factor, so your result is incorrect. What I want to say is that you incorrectly find the time difference as \begin{equation} \Delta t\boldsymbol{=}t\!\boldsymbol{-}\!t^{\boldsymbol{*}}\boldsymbol{=}\dfrac{\kappa\cdot \left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert}{c}\boldsymbol{=}\dfrac{\kappa\cdot R}{c} \qquad \textbf{(incorrect)} \tag{04}\label{04} \end{equation} while \begin{equation} \Delta t\boldsymbol{=}t\!\boldsymbol{-}\!t^{\boldsymbol{*}}\boldsymbol{=}\dfrac{ \left\Vert\mathbf{r}\!\boldsymbol{-}\!\mathbf{x}^{\boldsymbol{*}}\right\Vert}{c}\boldsymbol{=}\dfrac{ R}{c} \qquad \textbf{(correct)} \tag{05}\label{05} \end{equation}

Method 2 :

For the time difference my result is equation (15) in my answer therein(1) : From Liénard-Wiechert to Feynman potential expression which I repeat here for convenience \begin{equation} \Delta t =\dfrac{\upsilon\left(x\boldsymbol{-}\upsilon\,t\,\right)+\sqrt{c^{2}\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}+\left(c^{2}-\upsilon^{2}\right)\left(y^{2}\boldsymbol{+}z^{2}\right)}}{c^{2}-\upsilon^{2}} \tag{06} \end{equation} while your result is \begin{equation} c\Delta t=c(t-t_r)=\gamma^2(\vec{\beta}\cdot \vec{x}-\beta^2 ct)(+-)\gamma\sqrt{\gamma^2(x-\beta c t)^2+(y^2+z^2)} \tag{07} \end{equation} From a first check I think that these results are identical and I want to believe that they are correct. That you must keep the plus sign in your equation I prove in equations (13) in this same answer of mine.


(1) From Liénard-Wiechert to Feynman potential expression

(2) See $\color{blue}{\textbf{Example B }}$ in my answer therein :Physical meaning of the Jacobian in relation to Dirac delta function

(3) Feynman had dedicated a whole discussion to explain successfully this factor in an elementary way (no use of Dirac delta function) in his Lectures in $\S\;$§21-5. The potentials of a moving charge; the general solution of Lienard and Wiechert (Volume II, Electromagnetism and Matter).

(4) Related : Electric field of uniformly moving charge

  • $\begingroup$ I think you have pointed out my mistake. I was assuming $c(t-t_r)=||\vec{x}-\vec{u}t_r||$ Is the distance in the denominator I have to use but it's actually your expression $\vec{\kappa} \cdot \vec{R}$, Now isn't $c(t-t_r)$ the distance between the observation point to the charge's position at the retarded time? It seems the distance you need in the potential is a fraction of this. Is this because of a time dilation or length contraction effect, or is it because $c(t-t_r)$ was no the actual distance between observation point and source point at retarded time? $\endgroup$
    – R. Romero
    Jan 28, 2019 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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