Retarded time miscalculation?

Question

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:

$$c(t-t_r)=\sqrt{(x'-ut')^2+(y'^2)/\gamma^2+(z'^2)/\gamma^2}$$

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:

$$(\beta^2-1)c^2t_r^2+2(ct-\vec{\beta}\cdot\vec{x})ct_r+(x^2-c^2t^2)=0$$

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 $\hat{n}=\vec{R_{eq}}/R_{eq}$

Yielding:

$$c(t-t_r)=|\vec{x}-\vec{u}t_r|-\vec{\beta}\cdot(\vec{x}-\vec{u}t_r)$$

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

If so then the dot product is extraneous.

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}

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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}