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When studying the potential of an uniformly moving charge in vacuum, Feynman proposes to apply a Lorentz transformation on the Coulomb potential, which reads in the rest frame

$ \phi'(\mathbf r',t') = \frac{q}{4\pi\epsilon_0} \frac{1}{r'} $,

where $ |\mathbf r'| = r' $. In a frame with constant velocity $ \mathbf v $ along the x-axis, he obtains the following expression: $$ \phi(\mathbf r, t) = \frac{\gamma q}{4\pi\epsilon_0} \dfrac{1}{\sqrt{(\gamma(x-vt))^2+y^2+z^2}} \tag 1 $$

by transforming $ \phi = \gamma\left(\phi'+\dfrac{A'_xv}{c^2}\right) $, where $ \gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}} $ and the vector potential $ \mathbf A' $ vanishes within the rest frame. Another Lorentz transformation of the time and space coordinates $ (\mathbf r', t') \rightarrow (\mathbf r,t) $ yields (1). I suspect that (1) describes the potential at a given point for the instantaneous time t. What I am wondering is how this formula is connected to the expression of Liénard and Wiechert, namely $$ \phi(\mathbf r, t)=\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{|\mathbf r - \mathbf x(t_{ret})| - \frac{1}{c}\mathbf v(t_{ret})\cdot(\mathbf r - \mathbf x(t_{ret}))} \tag 2, $$

where $ \mathbf x(t_{ret}) $ describes the position of the charge and $ \mathbf v(t_{ret}) = \frac{d}{dt}\mathbf x(t)\bigg|_{t=t_{ret}} $ its velocity at the retarded time $ t_{ret}(\mathbf r,t) = t-\frac{|\mathbf r - \mathbf x(t_{ret})|}{c} $, respectively.

In the case of uniform motion, we have $ \mathbf x(t) = (vt,0,0)^\intercal $.

How do I get now from (2) to (1)?

My idea is to actually calculate an explicit expression for the retarded time and plug it into (2), which should yield (1) if I understand it correctly. By asserting that $ c^2(t-t_{ret})^2 = (x-vt_{ret})^2+y^2+z^2 $, $ t_{ret} $ can be found in terms of solving the quadratic equation, leading to the solutions

$ t_{ret}^\pm = \gamma\left(\gamma(t-\frac{vx}{c^2})\pm\sqrt{\gamma^2(t-\frac{vx}{c^2})^2-t^2+\frac{r^2}{c^2}}\right) = \gamma\left(\gamma t'\pm\sqrt{\gamma^2t'^2-\tau^2}\right)$ where $ t' $ is the Lorentz transformation of $ t $ and $ \tau = \frac{t}{\gamma} $ looks like some proper time. Plugging this into (2) looks nothing like (1), what am I missing?

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If you look at Feynman Volume II Section 21-6, he walks through this calculation. Your idea and initial assertion look good; the trick is to manage the algebra to get to the final form you want.

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Let the following symbols for a general curvilinear motion of the charge $\:q$ , see Figure-01.
\begin{align} \mathbf{r} & \equiv [\text{position 3-vector of field point}\: \mathrm A] =\left(x,y,z\right) \tag{01a}\\ \mathbf{x}\left(t\right) & \equiv [\text{equation of motion of charge}\: q] \tag{01b}\\ \boldsymbol{\upsilon}\left(t\right) & \equiv [\text{velocity vector of charge}\: q]=\dfrac{\mathrm d\mathbf{x}}{\mathrm d t} \tag{01c}\\ \mathbf{x}^{\boldsymbol{*}} & \equiv [\text{retarded position of charge}\: q]=\mathbf{x}\left(t^{\boldsymbol{*}}\right) \tag{01d}\\ t^{\boldsymbol{*}} & \equiv [\text{retarded time of charge}\:q] =t-\dfrac{\left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert}{c} \tag{01e}\\ \boldsymbol{\upsilon}^{\boldsymbol{*}} & \equiv [\text{velocity vector at retarded time }\: t^{\boldsymbol{*}}]=\boldsymbol{\upsilon}\left(t^{\boldsymbol{*}}\right) \tag{01f} \end{align} For given equation of motion $\:\mathbf{x}\left(t\right)\:$ the retarded quantities, position and time, are functions of the field point position vector $\:\mathbf{r}\:$ and present time $\:t$ : \begin{align} \mathbf{x}^{\boldsymbol{*}} & = \mathbf{x}^{\boldsymbol{*}}\!\left(\mathbf{r},t\right) \tag{02a}\\ t^{\boldsymbol{*}}& = \, t^{\boldsymbol{*}}\!\left(\mathbf{r},t\right) \tag{02b} \end{align} Always we have such a pair of retarded quantities if the charge $\:q\:$ exists far in the past from the present time $\:t$. Moreover this pair is unique (these conclusions fall under the derivation of the Lienard-Wiechert potentials).

With these symbols the Lienard-Wiechert scalar potential at field point $\:\mathrm A\:$ is \begin{equation} \phi(\mathbf r, t)=\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{\left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert - \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)} \tag{03} \end{equation} What we have to do is to prove that this equation in case of a charge moving with constant velocity $\:\boldsymbol{\upsilon}\left(t\right)=\boldsymbol{\upsilon}=\textbf{constant}\:$ is the Lorentz equation \begin{equation} \phi(\mathbf{r},t) = \dfrac{\gamma q}{4\pi\epsilon_0} \dfrac{1}{\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^2+z^2}} \tag{04} \end{equation} This will be accomplished if we eliminate the retarded quantities from (03) expressing them as functions of the present quantities. More precisely, we must find the vector $\:\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)\:$ and its norm $\:\left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert\:$ and replace them in the denominator of the rhs of (03).

So, from the triangle $\:\mathrm{Q^{\boldsymbol{*}}QA}\:$, Figure-01, we have for the general case \begin{equation} \left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)=\left(\mathbf{r}-\mathbf{x}\right)+\left(\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right) \tag{05} \end{equation} so \begin{equation} \left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert^{2}=\left\Vert\mathbf{r}-\mathbf{x}\right\Vert^{2}+\left\Vert\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right\Vert^{2}+2\left(\mathbf{r}-\mathbf{x}\right)\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right) \tag{06} \end{equation}

Now, according to the meaning of the retarded position and time, if the charge emitted a light signal, speed $\:c$, from the retarded position $\: \mathbf{x}^{\boldsymbol{*}}$ (point $\:\mathrm{Q^{\boldsymbol{*}}}$) at the retarded time $\: t^{\boldsymbol{*}}\:$ towards field point $\:\mathrm A\:$ then this signal and the charge $\:q\:$ arrive simultaneously at field point $\:\mathrm A\:$ and position $\: \mathbf{x}$ (point $\:\mathrm Q$) respectively at the present time moment $\:\mathrm t$. The common time duration of these travels is \begin{equation} \Delta t = t-t^{\boldsymbol{*}} \tag{07} \end{equation} That is, during the time interval $\:\Delta t\:$ the signal travels rectilinearly the distance $\:\left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert\:$ with constant speed $\:c$, so : \begin{equation} \left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert= c\,\Delta t \tag{08} \end{equation} while, on the other hand, the charge $\:q\:$ travels along its generally curvilinear trajectory from the position $\:\mathbf{x}^{\boldsymbol{*}}\:$ in the past to its position $\:\mathbf{x}\:$ in present time.

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Let see what is happening in the special case of rectilinear motion of the charge. Without loss of generality we suppose that the charge is moving along the positive $\:x-$axis, unit vector $\:\mathbf{i}$, with constant velocity \begin{equation} \boldsymbol{\upsilon}\left(t\right)=\boldsymbol{\upsilon}=\upsilon \mathbf{i}\,,\, \quad \upsilon \in \left(0,\boldsymbol{+}c\right) \tag{09} \end{equation} and at time $\:t=0\:$ is on the origin of the coordinate system $\:\mathrm O$, Figure-02, so : \begin{align} \mathbf{x}\left(t\right) & = \left(\upsilon\,t\,\right)\,\mathbf{i}\,,\quad \mathbf{x}^{\boldsymbol{*}}= \mathbf{x}\left(t^{\boldsymbol{*}}\right) =\left(\upsilon\,t^{\boldsymbol{*}}\right)\,\mathbf{i} \tag{10a}\\ \mathbf{x}-\mathbf{x}^{\boldsymbol{*}} & =\upsilon\,\left(t-t^{\boldsymbol{*}}\right)\,\mathbf{i}=\left(\upsilon\,\Delta t\right) \,\mathbf{i} \tag{10b} \end{align} and \begin{equation} \left\Vert\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right\Vert= \upsilon \,\Delta t \tag{11} \end{equation} From (06) \begin{equation} \underbrace{\left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert^{2}}_{c^{2}\left(\Delta t\right)^{2}}=\underbrace{\left\Vert\mathbf{r}-\mathbf{x}\right\Vert^{2}}_{\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}}+\underbrace{\left\Vert\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right\Vert^{2}}_{\upsilon^{2}\left(\Delta t\right)^{2}}+\underbrace{2\left(\mathbf{r}-\mathbf{x}\right)\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right)}_{2\,\left(x\boldsymbol{-}\upsilon\,t\,\right)\,\upsilon\,\Delta t} \tag{12} \end{equation} that is \begin{equation} \left[c^{2}-\upsilon^{2}\right]\left(\Delta t\right)^{2}-\left[2\,\upsilon\left(x\boldsymbol{-}\upsilon\,t\,\right)\,\right]\left(\Delta t\right)-\left[\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}\right]=0 \tag{13} \end{equation} with acceptable the non-negative root(1) with respect to $\:\Delta t\:$ \begin{equation} \Delta t =\dfrac{\upsilon\left(x\boldsymbol{-}\upsilon\,t\,\right)+\sqrt{\upsilon^{2}\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}+\left[c^{2}-\upsilon^{2}\right]\left[\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}\right]}}{c^{2}-\upsilon^{2}} \tag{14} \end{equation} or \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{15} \end{equation} From (08) \begin{equation} \left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert=c\Delta t =\sqrt{\gamma^{2}\!-\!1}\,\gamma\left(x\boldsymbol{\!-\!}\upsilon\,t\,\right)+\gamma\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}} \vphantom{\dfrac{\dfrac{1}{1}}{\dfrac{1}{1}}} \tag{16} \end{equation} For the retarded time we have \begin{equation} t^{\boldsymbol{*}}=t-\Delta t =\dfrac{\left(c^{2}t-\upsilon\,x\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{17} \end{equation} so(2) \begin{equation} t^{\boldsymbol{*}}=t-\Delta t =\gamma^{2}\left(t-\dfrac{\upsilon}{c^{2}}\,x\right)-\dfrac{\gamma\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c} \tag{18} \end{equation} For the retarded position \begin{equation} \mathbf{x}^{\boldsymbol{*}}=\left(\upsilon\,t^{\boldsymbol{*}}\right) \,\mathbf{i}=\Biggl[\gamma^{2}\left(\upsilon\, t-\dfrac{\upsilon^{2}}{c^{2}}\,x\right)-\dfrac{\gamma\,\upsilon\,\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c}\Biggr]\,\mathbf{i} \tag{19} \end{equation} and from this for the $x-$component of $ \left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)$ \begin{align} \left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)_{x} & =x-\Biggl[\gamma^{2}\left(\upsilon\, t-\dfrac{\upsilon^{2}}{c^{2}}\,x\right)-\dfrac{\gamma\,\upsilon\,\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c}\Biggr] \nonumber\\ & = \underbrace{\left(1+\dfrac{\gamma^{2}\upsilon^{2}}{c^{2}}\right)}_{\gamma^{2}}x-\gamma^{2}\upsilon\, t+\dfrac{\gamma\,\upsilon\,\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c} \nonumber\\ & =\gamma^{2}\left(x-\upsilon t\right)+\dfrac{\gamma\,\upsilon\,\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c} \tag{20} \end{align} that is \begin{equation} \left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)_{x}=\gamma^{2}\left(x-\upsilon t\right)+\dfrac{\gamma\,\upsilon\,\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c} \tag{21} \end{equation} Next \begin{align} \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right) & = \dfrac{\:\upsilon\:}{c}\cdot\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)_{x} \nonumber\\ & = \gamma^{2}\dfrac{\:\upsilon\:}{c}\left(x-\upsilon t\right)+\gamma\dfrac{\upsilon^{2}}{c^{2}}\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}} \nonumber\\ & = \sqrt{\gamma^{2}\!-\!1}\,\gamma\left(x\boldsymbol{\!-\!}\upsilon\,t\,\right)+\left(\gamma-\dfrac{\:1\:}{\gamma}\right)\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}} \vphantom{\dfrac{\dfrac{1}{1}}{\dfrac{1}{1}}} \tag{22} \end{align} so \begin{equation} \dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)=\sqrt{\gamma^{2}\!-\!1}\,\gamma\left(x\boldsymbol{\!-\!}\upsilon\,t\,\right)+\left(\gamma-\dfrac{\:1\:}{\gamma}\right)\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}} \vphantom{\dfrac{\dfrac{1}{1}}{\dfrac{1}{1}}} \tag{23} \end{equation} Subtracting (23) from (16) side by side \begin{equation} \left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert-\dfrac{\:\boldsymbol{\upsilon}^{\boldsymbol{*}}}{c}\boldsymbol{\cdot}\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)=\dfrac{\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{\gamma} \vphantom{\dfrac{\dfrac{1}{1}}{\dfrac{1}{1}}} \tag{24} \end{equation} Inserting this expression in the denominator of (03) we prove the Lorentz equation (04).


(1) The roots of (13) with respect to $\:\left(\Delta t\right)\:$ are \begin{equation} \left(\Delta t\right)_{\boldsymbol{\pm}}=\dfrac{\upsilon\left(x\boldsymbol{-}\upsilon\,t\,\right)\boldsymbol{\pm}\sqrt{\upsilon^{2}\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}+\left[c^{2}-\upsilon^{2}\right]\left[\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}\right]}}{c^{2}-\upsilon^{2}} \tag{13a} \end{equation} Defining for convenience the real variable $\:f=\upsilon\left(x\boldsymbol{-}\upsilon\,t\,\right)\:$ we have \begin{align} \left(\Delta t\right)_{\boldsymbol{+}} & \ge \dfrac{f\boldsymbol{+}\vert f\vert}{c^{2}-\upsilon^{2}} \ge 0 \tag{13b}\\ \left(\Delta t\right)_{\boldsymbol{-}} & \le \dfrac{f\boldsymbol{-}\vert f\vert}{c^{2}-\upsilon^{2}} \le 0 \tag{13c} \end{align} that is a non-negative and a non-positive root. Note that from (13) \begin{equation} \left(\Delta t\right)_{\boldsymbol{+}}\cdot \left(\Delta t\right)_{\boldsymbol{-}} =-\dfrac{\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}}{c^{2}-\upsilon^{2}}\le 0 \tag{13a} \end{equation} Acceptable is the non-negative one, equation (14).


(2) There exists an interpretation of equation (18) via the Lorentz transformation. This equation is expressed as follows \begin{equation} t^{\boldsymbol{*}}=\gamma\Biggl[\gamma\left(t-\dfrac{\upsilon}{c^{2}}\,x\right)-\dfrac{\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c}\Biggr] \tag{25} \end{equation} We agreed previously, see after (09), that $\:t=t_{0}=0\:$ when the charge $\:q\:$ is on the origin $\:\mathrm O\:$ of the frame $\:\mathrm Oxyz$. Now, we agree also to set $\:\tau=\tau_{0}=0\:$ when $\:q\:$ on the origin $\:\mathrm O$, where $\:\tau\:$ the time in the rest frame of the charge $\:q\:$, that is its proper time.

Let the two events \begin{align} E_{1} & = q\:\:\text{on the origin}\:\: \mathrm O \tag{26.1}\\ E_{2} & = \text{the light signal emitted from retarded position arrives at field point }\:\mathrm A \tag{26.2} \end{align} The space-time intervals that separate these two events are : first in $\:\mathrm Oxyz$ \begin{align} \overset{\boldsymbol{-}}{\Delta} x & = x_{2}-x_{1}=x-0=x\,,\quad x=\text{coordinate of field point}\:\: \mathrm A \tag{27.1}\\ \overset{\boldsymbol{-}}{\Delta} t & = t_{2}-t_{1}=t-0=t\,,\quad t=\text{present time} \tag{27.2} \end{align} and second in the rest frame of $\:q\:$ \begin{align} \overset{\boldsymbol{-}}{\Delta} x^{(q)} & = x^{(q)}_{2}-x^{(q)}_{1}=x^{(q)}-0=x^{(q)}\,,\quad x^{(q)}=\text{coordinate of field point}\:\: \mathrm A \tag{28.1}\\ \overset{\boldsymbol{-}}{\Delta} \tau & = \tau_{2}-\tau_{1}=\tau-0=\tau\,,\quad \tau=\text{present proper time} \tag{28.2} \end{align} where $\:x^{(q)},y^{(q)},z^{(q)}\:$ the coordinates in the rest frame of the charge $\:q$.

The Lorentz transformation equations are \begin{align} \overset{\boldsymbol{-}}{\Delta} x^{(q)} & = \gamma\left(\overset{\boldsymbol{-}}{\Delta} x-\upsilon\,\overset{\boldsymbol{-}}{\Delta} t\right) \tag{29.1}\\ \overset{\boldsymbol{-}}{\Delta} y^{(q)} & = \hphantom{\left(\!a\right)} \overset{\boldsymbol{-}}{\Delta} y \tag{29.2}\\ \overset{\boldsymbol{-}}{\Delta} z^{(q)} & = \hphantom{\left(\!a\right)}\overset{\boldsymbol{-}}{\Delta} z \tag{29.3}\\ \overset{\boldsymbol{-}}{\Delta} \tau & = \gamma\left(\overset{\boldsymbol{-}}{\Delta} t-\dfrac{\upsilon}{c^{2}}\,\overset{\boldsymbol{-}}{\Delta} x\right) \tag{29.4} \end{align} and inserting the space-time intervals (27),(28) \begin{align} x^{(q)} & = \gamma\left( x-\upsilon\,t\right) \tag{30.1}\\ y^{(q)} & = \hphantom{\left(\!a\right)} y \tag{30.2}\\ z^{(q)} & = \hphantom{\left(\!a\right)} z \tag{30.3}\\ \tau & = \gamma\left(t-\dfrac{\upsilon}{c^{2}}\, x\right) \tag{30.4} \end{align} We recognize at once that the first term in the bracket of the rhs of (25) is the present proper time $\:\tau\:$, so \begin{equation} t^{\boldsymbol{*}}=\gamma\Biggl[\tau-\dfrac{\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c}\Biggr] \tag{31} \end{equation} We proceed now to the interpretation of the second term in the brackets, that with the square root.

In frame $\:\mathrm Oxyz\:$ the signal emitted from the retarded position $\:\mathbf{x}^{\boldsymbol{*}}\:$ (point $\:\mathrm Q^{\boldsymbol{*}}$) towards the field point $\:\mathrm A\:$ travels from the start to the end of the vector $\:\left(\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right)\:$ component-wise \begin{equation} \mathbf{r}-\mathbf{x}^{\boldsymbol{*}}= \begin{bmatrix} x-\upsilon\, t^{\boldsymbol{*}} \vphantom{\dfrac12}\\ y\vphantom{\dfrac12}\\ z\vphantom{\dfrac12} \end{bmatrix} \tag{32} \end{equation} spending time, see (14) \begin{align} \Delta t &=\dfrac{\left\Vert\mathbf{r}-\mathbf{x}^{\boldsymbol{*}}\right\Vert}{c}= \dfrac{\sqrt{\left(x\boldsymbol{-}\upsilon\,t^{\boldsymbol{*}}\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}}}{c} \tag{33}\\ &=\dfrac{\upsilon\left(x\boldsymbol{-}\upsilon\,t\,\right)+\sqrt{\upsilon^{2}\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}+\left[c^{2}-\upsilon^{2}\right]\left[\left(x\boldsymbol{-}\upsilon\,t\,\right)^{2}\boldsymbol{+}y^{2}\boldsymbol{+}z^{2}\right]}}{c^{2}-\upsilon^{2}} \nonumber \end{align} But in the rest frame of the charge the signal seems to travel from its present position $\:\mathbf{x}\:$ (point $\:\mathrm Q$) towards the field point $\:\mathrm A\:$, that is from the start to the end of the vector $\:\left(\mathbf{r}-\mathbf{x}\right)\:$ component-wise in the frame $\:\mathrm Oxyz$ \begin{equation} \mathbf{r}-\mathbf{x}= \begin{bmatrix} x-\upsilon\, t \vphantom{\dfrac12}\\ y\vphantom{\dfrac12}\\ z\vphantom{\dfrac12} \end{bmatrix} \tag{34} \end{equation} and according to the above Lorentz transformation component-wise in the rest frame of the charge \begin{equation} \left(\mathbf{r}-\mathbf{x}\right)^{(q)}= \begin{bmatrix} \gamma\left(x-\upsilon\, t\right) \vphantom{\dfrac12}\\ y\vphantom{\dfrac12}\\ z\vphantom{\dfrac12} \end{bmatrix} \tag{35} \end{equation} spending proper time \begin{equation} \Delta \tau =\dfrac{\left\Vert \left(\mathbf{r}-\mathbf{x}\right)^{(q)} \right\Vert}{c}= \dfrac{\sqrt{\bigl[\gamma\left(x-\upsilon t\right)\bigr]^2+y^{2}\boldsymbol{+}z^{2}}}{c}=\tau-\tau^{\boldsymbol{*}} \tag{36} \end{equation} and (31) yields \begin{equation} t^{\boldsymbol{*}}=\gamma\tau^{\boldsymbol{*}} \tag{37} \end{equation} Note : Because of the very much misunderstanding meaning of time dilation, I don't dare to refer to (37) as a time dilation one, since it could be expressed as \begin{equation} t^{\boldsymbol{*}}\boldsymbol{-}\underbrace{t_{0}}_{=0}=\gamma\left(\tau^{\boldsymbol{*}}\boldsymbol{-}\underbrace{\tau_{0}}_{=0}\right) \tag{38} \end{equation}

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  • $\begingroup$ +1, for the answer. How do you create such marvellous images? $\endgroup$ – exp ikx Apr 23 at 7:34
  • $\begingroup$ @exp ikx : It's Geogebra. One of its advantages is that you can insert symbols and equations in LaTeX. $\endgroup$ – Frobenius Apr 23 at 8:03
  • $\begingroup$ Thanks, I thought it was GeoGebra, but didn't know we can use LaTeX. Btw, I have used one of your images in a question of mine. Due credit is given, hope you don't mind. $\endgroup$ – exp ikx Apr 23 at 8:06
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    $\begingroup$ @exp ikx : Of course I don't mind. This stuff is property of PSE and its users. $\endgroup$ – Frobenius Apr 23 at 8:18

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