# Help me understand eqn. 28.6 from Feynman

Feynman says that the electric field, $\bf{E}$, can be written as,

$$\mathbf{E} = \frac{-q}{4 \pi \epsilon_0} \left[ \frac{ \mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'} }{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right]$$

where $\mathbf{E}$ is the electric field at a point, P, from a charge, $q$, that is a distance $r$ apart. My impression is that $\mathbf{e}_{r'}$ is the unit vector from P in the direction of $q$, presumably $\frac{r'}{||r'||}$.

The first of the three terms is Coulomb's law, the second term is an apparent time-delay Coulomb field, and the third is the predominant factor contributing to the electric field for a large distance $r$.

Since radiation behaves as if it were inversely related to the distance and the distance is large, the first two terms are ignored leaving us with, $$\mathbf{E} = \frac{-q}{4 \pi \epsilon_0 c^2 } \frac{d^2 \mathbf{e}_{r'}}{dt^2}.$$

A further simplification is sought for when "charges are moving only a small distance at a relatively slow rate." Now he argues that the charge's motion is so tiny that its distance does not change and so the relay remains $r/c$. Now he says,

Then our rule becomes the following: If the charged object is moving in a very small motion and it is laterally displaced by the distance $x(t)$, then the angle that the unit vector $\mathbf{e}_{r'}$ is displaced is $x/r$, and since $r$ is practically constant, the $x$-component of $d^2 \mathbf{e}_{r'}/dt^2$ is simply the acceleration of $x$ itself at an earlier time, and so finally we get the law we want, which is $$\mathbf{E}_x(t) = \frac{-q}{4 \pi \epsilon_0 c^2 r}a_x \left(t - \frac{r}{c}\right).$$

See here for the relevant portion. My problem is that I do not understand how he arrived at this formula. How I picture this scenario is what I've drawn: I picture a unit vector for $e_{r'}$ sitting at the dark spot, point P or where we measure E, along the line of $r'$ and a corresponding one for $r$. The angle between the two unit vectors is small, so that $\sin \theta = \frac{x}{r}$, but the angle is small enough that the approximation can be used to give $\theta = \frac{x}{r}$.

What I have trouble now is making a formula of the form $\mathbf{e}_{r'}(t)$. I do think that $\mathbf{e}_r' = \frac{r'}{||r'||}$ makes sense as a starting point, but what I need to describe is the time-dependent behavior. It seems like $r = \sqrt{r'^2 + x^2}$ is also equally valid to write. We might say that $r(t) = \sqrt{ r'^2 + x(t)^2}$, where $x(t) = \dot{x} t$. Now writing, $r(t) = \sqrt{ r'^2 + \dot{x}t}$. So say I want to compute $\frac{d e'_x}{dt} =\lim_{h \rightarrow 0} \frac{\frac{r(t+h) - r(t)}{h}}{||r'||} = \lim_{h \rightarrow 0} \frac{\frac{\sqrt{r'^2 + \dot{x}(t+h)}) - \sqrt{r'^2 + \dot{x}(t)}}{h}}{||r'||}$, but this appears to be getting pretty messy already and I haven't even tackled the second derivative.

What am I missing here?

EDIT: In RE Art Brown Answer

The following image reflects my understanding of your description: Effectively we ignore any $y$ component of $\mathbf{e}_r$, but we still need to express the $x$-component of $\mathbf{e}_r$ before we can calculate its second derivative. Looking at the top view it seems that $\mathbf{e}_r (t) = \frac{r(t)}{||r(t)||}$. $x^2 + r(t)^2 = r'^2$ therefore $\sqrt{r'^2 -x^2} = r(t)$. We know that $x(t) = \dot{x}t$, thus $$r(t) = \sqrt{r'^2 - (\dot{x}t)^2}$$

$$\frac{d^2 \mathbf{e}_r}{dt^2} = \frac{d^2}{dt^2} \frac{\sqrt{r'^2 - (\dot{x}t)^2}}{||\sqrt{r'^2 - (\dot{x}t)^2}||}$$

Which is not pretty.

What I really want to say is that $\frac{d^2 \mathbf{e}_r}{dt^2}$ is affected by $x$ and $\ddot{x}= a_x$, but this is delayed by $(t - r/c)$. This, however, seems hand-wavy to me.

## 1 Answer

The quantity $x$ is not the difference between $\boldsymbol{r}$ and $\boldsymbol{r'}$. Instead, Prof. Feynman uses $x$ to denote displacement of the charge in a direction perpendicular to $\boldsymbol{e}_{r'}$ (aka a transverse direction), where $\boldsymbol{e}_{r'}$ is defined as the unit vector pointing from the observation point towards the charge (at the retarded time $t-r/c$). Likewise, $a_x$ is the acceleration of the charge in the transverse direction. The actual value of the distance $r$ doesn't affect $x$ or $a_x$ (except through determining the retarded time).

So, if the charge happens to be accelerating parallel to $\boldsymbol{e}_{r'}$ (at the retarded time), then $\boldsymbol{e}_{r'}$ doesn't change, so $d^2 \boldsymbol{e}_{r'} / dt^2 = 0$. Clearly there's also no transverse acceleration in this case, so $a_x=0$, and the formula works.

On the other hand, suppose the charge is accelerating with amplitude $a_x$ transverse to $\boldsymbol{e}_{r'}$ (again at the retarded time). Now there will be a corresponding acceleration in $\boldsymbol{e}_{r'}$, also in the transverse direction. How much acceleration? Well, the farther away the charge (the distance $r$), the smaller the acceleration of $\boldsymbol{e}_{r'}$ needed to follow along. In fact, it's just $a_x/r$, and there's your formula.

• I thought he defined it as laterally displaced instead of transversely displaced. If it were moving perpendicular to the unit vector then wouldn't it appear on our projected screen as just a point to us and the unit vector would not move or have any acceleration at all? – sciencenewbie Jul 5 '13 at 16:33
• 1) In the first full paragraph after 28.5 he decomposes the acceleration of $\boldsymbol{e}_{r'}$ into "transverse" and "radial" components, and then argues away the radial piece as having an inverse square fall-off with distance. It is the remaining transverse component he is considering when he discusses "lateral displacement". 2) The "screen" plane is perpendicular to $\boldsymbol{e}_{r'}$, so transverse/lateral charge acceleration is visible (pun intended) on the screen, and $\boldsymbol{e}_{r'}$ accelerates to follow. – Art Brown Jul 5 '13 at 16:50
• Sorry, no. It looks like you're still defining x as the difference between r and r'. Forget about r; the observer "sees" the charge at r'. One way to proceed: Pick a z-axis along r'; then e_r'=(0,0,1). Suppose the charge has a velocity v in the x-direction at that (retarded) instant. Then de_r'/dt=(v/r',0,0) (neglecting the small radial (z) component that keeps it a unit vector). Now iterate for the acceleration. Finally, the screen device is at the observer, like a painter's canvas, not the charge. – Art Brown Jul 6 '13 at 5:23
• @TolgaYilmaz, (i) If you're referring to the units, the convention is the vector $\boldsymbol{r'} =|\boldsymbol{r'}| \boldsymbol{e}_{r'}= r' \boldsymbol{e}_{r'}$, so $\boldsymbol{e}_{r'}$ is unit-less. – Art Brown Dec 27 '15 at 4:51
• @TolgaYilmaz, (ii) When I calculate that second derivative of the unit vector (which I did a while ago, taking a very long time to work through it), the result is complicated, and in fact is non-zero even if the acceleration $a'=0$. What I wrote in my answer was only approximate (good at low velocities and large distances). However, Feynman's expression (28.6) for the electric field is exact: I was able to show its equivalence to the results in Jackson's and Schwartz's textbooks. – Art Brown Dec 27 '15 at 5:38