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

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!

• A particle traveling away from or towards you looks identical to a particle standing still, the field radiated looks unchanged from a stationary particle which is what the first term yields Jan 16, 2018 at 21:23
• We have also seen that if the velocity $\:\upsilon\:$ of a charge is always much less than c, and if we consider only points at large distances from the charge, so that only the last term of Eq. (21.1) is important, the fields can also be written as $$\boldsymbol{E}=\dfrac{q}{4\pi\epsilon_{0}c^{2}r'} \begin{bmatrix} \text{acceleration of the charge at } (t-r'/c) \\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\text{projected at right angles to } r' \end{bmatrix} \tag{21.1^\boldsymbol{\prime}}$$ (Feynman Lectures, Volume 2, $\S$21-1) Jan 16, 2018 at 22:51
• @Frobenius But the general solution to Maxwell's equation says that$$\mathbf{A}(1,t)=\int\frac{q\mathbf{v}(2,t-r_{12}/c)}{4\pi\epsilon_{0} c^2r_{12}}\,dV_2$$which means that$$\frac {\partial \mathbf{A}}{\partial t}=\int\frac{q}{4\pi\epsilon_{0} c^2}\frac{\partial \frac{\mathbf{v}}{r_{12}}}{\partial{t}}\,dV_2$$ Here, the acceleration resulting from $\frac{\partial \frac{\mathbf{v}}{r_{12}}}{\partial{t}}\,dV_2$will also have a component along $\mathbf{r'}$, not just perpendicular to it, as Feynman is telling. So, my question is- what happened to the component of $\mathbf{a}$ along $\mathbf{r'}$?
– Prem
Jan 17, 2018 at 5:52
• @Triatticus But the general solution of Maxwell's equation the equation appears to be saying that a moving charge simply radiates its scalar and vector potentials in all directions, and then an observer just takes the space and time derivatives of these vectors to obtain the electric fields. In this picture, even if the electron is accelerating directly towards the observer, $$\frac {\partial \mathbf{A}}{\partial t}=\int\frac{q}{4\pi\epsilon_{0} c^2}\frac{\partial \frac{\mathbf{v}}{r_{12}}}{\partial{t}}\,dV_2$$ Which means even the acceleration along the line of sight will act.
– Prem
Jan 17, 2018 at 6:00

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:

$$$$\mathbf{E}(\mathbf{x},t) = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01}$$$$ 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 $$$$\mathbf{E}(\mathbf{x},t) \approx -\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{(\mathbf{n}\boldsymbol{\times} \dot{\boldsymbol{\beta}})\times\mathbf{n}}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{04}$$$$ But $$\:(\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} }$$].

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

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 $$$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) = \underbrace{\frac{q}{4\pi\epsilon_0}\Biggl[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R^2} \Biggr]_{\mathrm{ret}}}_{\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{+}\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)} + \underbrace{\frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\times\left[(\mathbf{n}-\boldsymbol{\beta})\times \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\cdot\mathbf{n})^3 R}\right]_{\mathrm{ret}}}_{\boldsymbol{-}\frac {\partial \mathbf{A}}{\partial t}\boldsymbol{-}\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)} \tag{01}$$$$ where $$\mathbf{f}\left(\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}\right)$$ a vector function of $$\mathbf{R},\boldsymbol{\beta},\dot{\boldsymbol{\beta}}$$.

• Thank you for the answer! This means that the Heaviside-Feynman equation makes the small velocity approximation for the acceleration (second) term. Also, i was completely unaware about the Liénard-Wiechert potentials, so thanks for telling me about that!
– Prem
Jan 23, 2018 at 11:21
• @Prem kumar Welcome. I want to believe that I helped you somehow. It's very important to know about Liénard-Wiechert potentials. Jan 23, 2018 at 13:50
• Your partitioning of E(\vec{x},t) into the two parts as in footnote 1 is incorrect. Both -grad(\phi} and -\frac{\partial \vec{A}}{\partial t} should contribute to either terms. This is clearly seen from Griffith 3rd Eqns (10.62) and (10.63). Sep 6, 2019 at 19:35
• @Prem-kumar No, Heaviside-Feynman formula is exact in the sense it is consistent with the Lienard-Weichert potentials. There is no approximation made. Your observation in your question "that ∇ϕ part of the equation is a function only of the velocity of the source and not its acceleration" is not correct. You might have overlooked the fact that ϕ is a function of the retarded speed v. Thus the grad of it introduces "a"(acceleration) terms. Sep 6, 2019 at 19:40
• @verdelite : You are absolutely right. I edit my answer. Sep 6, 2019 at 21:07