Why doesn't the acceleration of an electron along the line of sight from the observer contribute to the electric field?

Question

In Feynman Lectures on Physics, Volume 2, Feynman gives the general solution of the Maxwell's equations as following:

\begin{gather*} \begin{aligned} \end{aligned}\\ \mathbf{E}=-\nabla \phi - \frac {\partial \mathbf{A}}{\partial t}\\[1ex] \mathbf{B}=\nabla \times \mathbf{A}&\\[2.5ex] \phi(1,t)=\int\frac{\rho(2,t-r_{12}/c)}{4\pi\epsilon_{0} r_{12}}\,dV_2\\[1ex] \mathbf{A}(1,t)=\int\frac{\mathbf{j}(2,t-r_{12}/c)}{4\pi\epsilon_{0} c^2r_{12}}\,dV_2 \end{gather*}

What I understand from this equation is that an arbitrarily moving electron simply radiates scalar and vector potentials $\phi$ and $\mathbf{A}$ at the speed of light in all the directions, where the magnitude of the potential drops with increasing distance, and in the end one can evaluate $\mathbf{E}$ and $\mathbf{B}$ using the formula given above.

But in the same book, Feynman gives the following equation (Heaviside-Feynman Formula) for the electric field:

$$\mathbf{E} = \frac{-q}{4 \pi \epsilon_0} \left[ \frac{ \mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'} }{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right]$$ where $\mathbf{E}$ is the electric field at a point P due to a charge $q$, that is a distance $r$ away. $\mathbf{e}_{r'}$ is the unit vector from P in the direction of $q$.

In this equation, you can notice that the third term is the contribution of acceleration of the source charge to the electric field. But, interestingly, as Feynman himself notes in the book, only the acceleration of the source perpendicular to the line of sight from P has a contribution to the resulting electric field at P. In other words, the component of the acceleration of the source charge along the the observer will contribute nothing to the electric field.

Given my understanding of the general solution to Maxwell's equations, the $\frac {\partial \mathbf{A}}{\partial t}$ term in $\mathbf{E}=-\nabla \phi - \frac {\partial \mathbf{A}}{\partial t}$ should cause the entire acceleration of the source to add to the electric field at P, and not just the part perpendicular to the line of sight.

Also, while working out the details, I realized that $\nabla \phi$ part of the equation is a function only of the velocity of the source and not its acceleration. More precisely, I calculated that $$\nabla \phi= \frac {\mathbf{e_{r}}}{r^2(c-\mathbf{v_{r}})}$$, where $\mathbf{v_{r}}$ is the velocity of the source along the line of sight.

This equation, and another that I am getting, resembles nothing like the Heaviside-Feynman formula. So, what am I missing here? Do I understand the general solution to Maxwell's equation properly?

Edit: I seem to have a misconception somewhere in my interpretation of the general solution of the Maxwell's equation, as given in Feynman's book. So, asking my question in a slightly different way, what exactly are the first four equations given in my answer saying? Thanks in advance!

Answer 1

The Heaviside-Feynman Formula ( the 5fth equation in your question) is derived from the Liénard-Wiechert potentials, which in turn are derived from the first four equations of yours. But from the Liénard-Wiechert potentials you could derive the following equation which is more convenient for our case¹:

\begin{equation} \mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01} \end{equation} where

\begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{02c} \end{align} The electric field of equation (01) is given by equation 14.14 from Jackson's 'Classical Electrodynamics',3rd Edition.

For points at large distances from the charge [$R^{-2}\rightarrow 0$] and velocities $\:\upsilon\:$ of the charge always much less than c [$\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3\rightarrow 1$] the first term in the rhs of (01) tends to zero and it's ignored. For the second term with velocities $\:\upsilon\:$ of the charge always much less than c we have the following \begin{align} \beta & = \dfrac{\upsilon}{c} \ll 1 \tag{03a}\\ \mathbf{n}-\boldsymbol{\beta} & \approx \mathbf{n} \tag{03b} \end{align} so \begin{equation} \mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04} \end{equation} But $\:(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\boldsymbol{\times} \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\perp \mathbf{n} }\:$ is the vectorial projection of the acceleration vector $\:\dot{\boldsymbol{\beta}}\:$ on the direction normal to $\:\mathbf{n}\:$, that is normal to $\:\mathbf{r'}$ [while the vectorial projection on the direction $\:\mathbf{n}\:$ is derived by replacing in the previous expression the outer products by the inner ones $\:(\mathbf{n}\boldsymbol{\cdot} \dot{\boldsymbol{\beta}})\cdot \mathbf{n}\equiv \dot{\boldsymbol{\beta}}_{\parallel \mathbf{n} }$].

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$1 \quad$ Note that \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}} \tag{01-false} \end{equation}

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EDIT : Thanks to a comment by @verdelite : "Your partitioning of $\mathbf{E}(\mathbf{x},t)$ into the two parts as in footnote 1 is incorrect..." the correct one is \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{+}\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}\boldsymbol{-}\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)} \tag{01} \end{equation} where $\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)$ a vector function of $\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}$.