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 some other 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!

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!

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 some other 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?

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 some other 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!