Deriving Heaviside-Feynman formula for the electric field of an arbitrarily moving charge from Lienard-Wiechert potential

I've been trying to derive this (which Feynman warns takes a lot of work) for a couple of days now, without success. My current best derivation which however doesn't give the right answer is:

First, realizing that to go from derivatives with respect to time, $t$, to ones with respect to retarded time, $t'=t - \frac{r}{c}$, we need:

$$\frac{d t'}{d t} = 1 - \frac{\dot{r}}{c} \tag{1}$$

Where $r=|\vec{r}|=|\vec{r}_{1}-\vec{r}_{2}(t')|$ Where $\vec{r}_{1}$ and $\vec{r}_{2}(t')$ are the fixed (time-independent) position vector of the observation point and the retarded position vector of the charge (at time $t'$), respectively. And the dot represents derivation with respect to $t$

The Lienard-Wiechert potentials are:

$$\phi(\vec{r}_{1}, t) = \frac{q}{4\pi\epsilon_{0}(r-\frac{\vec{v}\cdot\vec{r}}{c})}$$ $$\vec{A}(\vec{r}_{1}, t) = \frac{q\vec{v}}{4\pi\epsilon_{0}c^{2}(r-\frac{\vec{v}\cdot\vec{r}}{c})}$$

Where $\vec{v} = \frac{d \vec{r}_2}{d t'}|_{t'=t - \frac{r}{c}}$; that is, the standard retarded velocity.

Now, it is useful to note:

$$\frac{1}{1-\frac{\vec{v}\cdot\vec{r}}{rc}} = \frac{1}{1+\frac{\frac{d r}{ dt'}}{c}} = \frac{1}{1+\frac{\dot{r}}{c-\dot{r}}} = 1-\frac{\dot{r}}{c} \tag{2}$$

Where we have used $(1)$ to transform the time derivative.

Then I rewrite the LW potentials as:

$$\phi(\vec{r}_{1}, t) = \frac{q}{4\pi\epsilon_{0}r}\Big(1 - \frac{\dot{r}}{c}\Big)$$ $$\vec{A}(\vec{r}_{1}, t) = \frac{q\dot{\vec{r}}}{4\pi\epsilon_{0}c^{2}r}$$

Finally, I can work out the electric field:

$$\vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} = \frac{-q}{4\pi\epsilon_{0}}\bigg(\frac{-\vec{r}}{r^{3}}\Big(1 - \frac{\dot{r}}{c}\Big)-\frac{1}{rc}\vec{\nabla}\dot{r}+\Big(\big(1 - \frac{\dot{r}}{c}\big)\frac{1}{r^{2}}\frac{dr}{dt'}-\frac{1}{r}\frac{d}{dt'}\big(1 - \frac{\dot{r}}{c}\big)\Big)\frac{\vec{\nabla}r}{c}+\frac{\ddot{\vec{r}}}{c^{2}r} - \frac{\dot{\vec{r}}\dot{r}}{c^{2}r^{2}}\bigg)$$

Where the spatial gradient is with respect to $\vec{r}_{1}$, and where I've had to derive with respect to $\vec{r}_{1}$ directly and then with respect to $t'$ because it too depends on $\vec{r}_{1}$ through $r$. Now, $\vec{\nabla}\dot{r} = \frac{\partial}{\partial t}(\vec{\nabla}r)=\frac{\partial}{\partial t}(\frac{\vec{r}}{r})$ because these partial derivatives commute. Finally, I can again convert the time derivatives using $(1)$ so:

$$\vec{E} = \frac{q}{4\pi\epsilon_{0}}\bigg(\frac{\vec{r}}{r^{3}}-\frac{\vec{r}\dot{r}}{r^{3}c}+\frac{1}{rc}\frac{\partial}{\partial t}\Big(\frac{\vec{r}}{r}\Big)+\frac{\ddot{\vec{r}}}{c^{2}r} - \frac{\dot{\vec{r}}\dot{r}}{c^{2}r^{2}} - \frac{\vec{r}}{rc}\Big(\frac{\dot{r}}{r^{2}} + \frac{\ddot{r}}{r(c-\dot{r})}\Big)\bigg) = \frac{q}{4\pi\epsilon_{0}} \bigg(\frac{\vec{r}}{r^{3}} + \frac{r}{c} \frac{\partial}{\partial t}\Big(\frac{\vec{r}}{r^{3}}\Big)+\frac{1}{c^{2}}\Big(\frac{\ddot{\vec{r}}}{r} - \frac{\dot{\vec{r}}\dot{r}}{r^{2}}-\frac{\vec{r}\ddot{rc}}{r^{2}(c-\dot{r})}\Big)\bigg)$$

The first two terms are right but the third, although close, isn't right (specially annoying is that $c-\dot{r}$ in the denominator). The actual equation is found in Feynman's Lectures on Physics. I've found a paper (pages 22-23) that says that the Heaviside-Feynman formula can't actually be derived from the LW potentials, but I don't know, I think I trust Feynman more. Has anyone here done this derivation?

I finally found my mistake!

As I commented in Art Brown's answer, after thinking more about it after a lecture we had today, I realized that I was computing my gradient with respect to $\vec{r}_{1}$ wrongly. That is, I thought, in my derivation above, that

$$\vec{\nabla} (r) = \frac{\vec{r}}{r}$$

However this is wrong, because I just differentiated with respect to the explicit $\vec{r}_{1}$ in the $\vec{r}=\vec{r_{1}}-\vec{r_{2}(t')}$. However, there's $\vec{r}_{1}$-dependence in the $\vec{r}_{2}$ too because $t'=t - \frac{r}{c}$ depends on $\vec{r}_{1}$!

To take this into account we must derive implicitly in order to get an expression for this gradient:

$$\vec{\nabla} (r) = \frac{\vec{r}}{r}-\frac{\vec{r}}{r}\cdot\frac{d\vec{r}_{2}}{d t'}\bigg(\frac{-\vec{\nabla} (r)}{c}\bigg)$$

Rearranging, and noting $\frac{d\vec{r}_{2}}{dt'}=\vec{v}$,

$$\vec{\nabla} (r) = \frac{\vec{r}}{r} \frac{1}{1-\frac{\vec{r}\cdot\vec{v}}{rc}} = \frac{\vec{r}}{r} \bigg(1 - \frac{\dot{r}}{c}\bigg)$$

Where I used equation $(2)$ in my question. Now, I can evaluate $-\vec{\nabla} \phi$ again:

$$-\frac{4\pi\epsilon_{0}}{q}\vec{\nabla} \phi = \frac{1}{r^{2}}\bigg(1-\frac{\dot{r}}{c}\bigg)\vec{\nabla}(r)+\frac{1}{rc}\frac{\partial}{\partial t}\vec{\nabla}(r) = \frac{1}{r^{2}}\frac{\vec{r}}{r}\bigg(1-\frac{\dot{r}}{c}\bigg)^{2}+\frac{1}{rc}\frac{\partial}{\partial t} \Bigg(\frac{\vec{r}}{r}\bigg(1-\frac{\dot{r}}{c}\bigg)\Bigg) = \frac{1}{r^{2}}\frac{\vec{r}}{r}\bigg(1-\frac{2\dot{r}}{c}+\frac{\dot{r}^{2}}{c^{2}}\bigg) + \frac{1}{rc}\bigg(1-\frac{\dot{r}}{c}\bigg)\frac{\partial}{\partial t}\bigg(\frac{\vec{r}}{r}\bigg)-\frac{\vec{r}}{r^{2}}\frac{\ddot{r}}{c^{2}} = \frac{\vec{r}}{r^{3}} + \frac{r}{c} \frac{\partial}{\partial t}\bigg(\frac{\vec{r}}{r^{3}} \bigg) +\frac{2\vec{r}\dot{r}^{2}}{r^{3}c^{2}}-\frac{\dot{r}\dot{\vec{r}}}{r^{2}c^{2}}-\frac{\vec{r}\ddot{r}}{r^{2}c^{2}}$$

Now, the first two terms are right again! Let's see if we can get the third when computing $\vec{E}$ :

$$\vec{E} = - \vec{\nabla} \phi - \frac{\partial \vec{A}}{\partial t} = \frac{q}{4\pi\epsilon_{0}} \bigg(\frac{\vec{r}}{r^{3}} + \frac{r}{c} \frac{\partial}{\partial t}\bigg(\frac{\vec{r}}{r^{3}} \bigg) +\frac{2\vec{r}\dot{r}^{2}}{r^{3}c^{2}}-\frac{\dot{r}\dot{\vec{r}}}{r^{2}c^{2}}-\frac{\vec{r}\ddot{r}}{r^{2}c^{2}} + \frac{\ddot{\vec{r}}}{c^{2}r} - \frac{\dot{\vec{r}}\dot{r}}{c^{2}r^{2}} \bigg) = \frac{q}{4\pi\epsilon_{0}} \bigg(\frac{\vec{r}}{r^{3}} + \frac{r}{c} \frac{\partial}{\partial t}\bigg(\frac{\vec{r}}{r^{3}} \bigg) + \frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}} \bigg(\frac{\vec{r}}{r}\bigg) \bigg)$$

Which is the right Heaviside-Feynman formula! :D

• I have double-checked every step of this answer. It seems alright. The only thing is I feel like \frac{\partial}{\partial t} may better be \frac{d}{d t}. Sep 6, 2019 at 15:29
• As $r_1$ isn't changing anyways, you are right. But, I just wanted to make explicit the fact that $r_1$ is fixed when taking the time derivative, which I think is standard. Sep 6, 2019 at 16:18

A few years ago I gave for and by my own a proof of this equation of Feynman Lectures, also known as Heaviside-Feynman equation, starting from the retarded scalar and vector potentials instead of the Lienard-Wiechert ones. The latter are appeared in the proof inevitably as an intermediate step(1). I make use of the Dirac $\:\delta-$function and Jacobian determinants. The proof is written in $\LaTeX$ and Figures are produced by GeoGebra software. But the proof is too lengthy to post in the permissible length of an PSE answer (30.000 characters or so I think)(2). So, I have uploaded the relevant Adobe Acrobat .pdf file about 1,5 year ago in the following link :

$\color{blue}{\textbf{A Feynman Lectures EM Equation}}$

Note that in his own (Feynman's) words :

$\rule[0.6 mm]{2 mm}{2 mm}\:$ When we studied light, we began by writing down equations for the electric and magnetic fields produced by a charge which moves in any arbitrary way. Those equations were $$\mathbf{E}=\dfrac{q}{4\pi\epsilon_{0}}\left[\dfrac{\mathbf{e}_{r^{\prime}}}{r^{\prime 2}}+\dfrac{r^{\prime}}{c}\dfrac{d}{dt}\biggl(\dfrac{\mathbf{e}_{r^{\prime}}}{r^{\prime 2}}\biggr)+\dfrac{1}{c^{2}}\dfrac{d^{2}}{dt^{2}}\mathbf{e}_{r^{\prime}}\right] \tag{21.1}$$ $$c\mathbf{B}=\mathbf{e}_{r^{\prime}}\boldsymbol{\times}\mathbf{E} \nonumber$$ If a charge moves in an arbitrary way, the electric field we would find now at some point depends only on the position and motion of the charge not now, but at an earlier time-at an instant which is earlier by the time it would take light, going at the speed $\:c$, to travel the distance $\:r^\prime\:$ from the charge to the field point. In other words, if we want the electric field at point ($1$) at the time $\:t$, we must calculate the location ($2^\prime$) of the charge and its motion at the time $\:(t-r^\prime/c)\:$, where $\:r^\prime\:$ is the distance to the point ($1$) from the position of the charge ($2^\prime$) at the time $\:(t-r^\prime/c)\:$. The prime is to remind you that $\:r^\prime\:$ is the so-called “retarded distance” from the point ($2^\prime$) to the point ($1$), and not the actual distance between point ($2$), the position of the charge at the time $\:t$, and the field point ($1$)(see Fig. 21-1)$\:\rule[0.6 mm]{2 mm}{2 mm}$

(1) The scalar and vector Lienard-Wiechert potentials are shown in the .pdf file detailed as equations (4-2.24),(4-2.25) respectively and in a compact form as (4-2.26),(4-2.27) respectively.

(2) If there is interest from PSE users to upload the .pdf file in MathJax as an answer , I could do it but at the expense of the hosting provided to me by PSE, since may be needed the length of 3-4 answers and the frequent bothersome appearance of the question as active because of the necessarily heavy editing.

• This answer seems going through the same route as @guillefix 's answer, but in greater details. Specifically, guillefix's equation (2) is reached. In some places (section 4.2.2) I feel like some partial derivatives might better be written as whole derivatives. Sep 6, 2019 at 15:36

For the sake of documentation, I introduce here a proof I found today (11/13/2019), A note on “Measuring propagation speed of Coulomb fields” by R. de Sangro, G. Finocchiaro, P. Patteri, M. Piccolo, G. Pizzella, published in 2016.

Basically they followed Feynman's suggestion by deriving the electrical field from the Heaviside-Feynman formula and comparing the result to standard result obtained from the Lienerd-Wiechert potentials found in many textbooks.

For the sake of documentation, I introduce another proof I found today (11/13/2019), Retarded electric and magnetic fields of a moving charge: Feynman's derivation of the lienard -wiechert potentials revisited by J.H.Field (last updated 2015). It is in Appendix B of the paper.

Basically in Appendix B the author followed Feynman's suggestion by deriving the fields from the formula by completing the differentiations. I think I should state that I may not agree with the author's view in other parts of the paper but this Appendix B is independent.

Probably finally a derivation is made, in spite that Feynman said it could not be derived. I found it today (01/23/2021) in P.A. Davidson's textbook, "An Introduction to Electrodynamics", first edition, 2019. In section 17.3, "The Heaviside-Feynman equations for the field of a point charge", the author derived it from Jefimenko Equations.

Besically he followed the following steps,

1, re-write partial derivitives in retared time with partial derivitives in current time,
2, plug in single particle representation (delta-functions)
3, make the space integration and reach expressions with whole derivitives in current time,
4, re-write whole derivitives in current time with whole derivitives in retarded time,
5, simplify,
6, re-write whole derivitives in retarded time with whole derivitives in current time.

I am too lazy to write down the equations.

The following follows Fulvio Melia's approach in his Electrodynamcs text (using cgs units). Note that the non-relativistic potential gives, $$\Phi(\mathbf r,t)=\left[\frac{q}{\left(1-\hat{n}\cdot\boldsymbol\beta\right)r}\right]\equiv q\int\frac{\delta\left[t'-t+r(t')/c\right]}{r(t')}dt'\tag{1}$$ where $\boldsymbol\beta=\mathbf v/c$ and $\hat{n}=\mathbf r/r$ and we use a particular property of the Dirac delta function to get the equivalence on the right; a similar equivalence can be derived for the vector potential. The gradient of (1) and the partial time derivative of the vector potential returns (with some simple differential calculus), $$\mathbf E=q\left\{\frac{\left(\hat{n}-\boldsymbol\beta\right)\left(1-\beta^2\right)}{\left(1-\hat{n}\cdot\boldsymbol\beta\right)^3r^2}\right\}_{ret}+\frac{q}{c}\left\{\frac{\hat n\times\left[\left(\hat n-\boldsymbol\beta\right)\times\dot{\boldsymbol\beta}\right]}{\left(1-\hat n\cdot\boldsymbol\beta\right)^3r}\right\}_{ret}\tag{2}$$ where $ret$ indicates the retarded potential (you could also use $\gamma^{-2}=1-\beta^2$ in the first term). You should be able to massage (2) into the Feynman-Heaviside formula by noting

• $\dot{r}=-c\left(\hat n\cdot\boldsymbol\beta\right)$
• $\dot{\hat n}=\frac{c}{r}\left[\hat n\left(\hat n\cdot\boldsymbol\beta\right)-\boldsymbol\beta\right]$

This might be the easier method than going through the gradients you've done.

• Well, you need to carry out the derivation. That is what the OP is asking for.
– Hans
Jan 7, 2022 at 6:56

Doesn't the Feynman-Heaviside formula contain only total time derivatives instead of partial time derivatives as guillefix's final result? Wouldn't that be problematic? E.g. the total time derivative of r is the magnitude of the radial velocity component while the partial time derivative or r is just c (from t' = t - r/c). Here by r I mean the magnitude of the postion vector. Secondly, in obtaining the electric field he seems to have taken the total time derivative of the vector potential instead of the partial time derivative. This would also be problematic for the same reason mention above as the vector potential is a function of r. I'm probably missing something but I would really appreciate a claritication.

• If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review Aug 21 at 18:40

Use differential forms! That's what they're there for. So ... the potential 1-form is (with $$\left(x^0, x^1, x^2, x^3\right) = (t, x, y, z)$$ and using the summation convention): $$A = A_μ dx^μ = 𝐀·d𝐫 - φ dt,\quad d𝐫 = (dx, dy, dz).$$ For the retarded potential $$A = \frac{Z_0q}{4π}\frac{𝐯·d𝐫 - c^2 dt}{Rc - 𝐑·𝐯},$$ where $$Z_0 = μ_0 c = \sqrt\frac{μ_0}{ε_0} = \frac{1}{ε_0c},\quad 𝐑 = 𝐫 - 𝐫(T), \quad T = t - \frac{R}{c}, \quad R = |𝐑|, \quad 𝐯 = 𝐫'(T).$$ What? Didn't anybody ever see it that way before? That's what it actually looks like! Now, isn't that a lot cleaner and easier? Compared to the alternative, that is?

Now ... all those factors hanging out in the front: call that $$g = Z_0q/(4π)$$. That's a magnetic charge (in Webers) or - more accurately - it's "gauge". Did you ever wonder what units "gauge" are in? Webers. And: what are the units for The Fifth Dimension in Kaluza-Klein? That's Webers, too. That's right: The Fifth Dimension is in Webers. So, we may write: $$A = g \frac{𝐯·d𝐫 - c^2 dt}{Rc - 𝐑·𝐯}.$$

What's the field strength? It's: $$F = dA = 𝐁·d𝐒 + 𝐄·d𝐫∧dt, \quad d𝐒 = (dy∧dz, dz∧dx, dx∧dy) = \frac{d𝐫\hat{×}d𝐫}2.$$ A little differential form algebra: $$(adt + 𝐛·d𝐫)∧(cdt + 𝐝·d𝐫) = (𝐛c - a𝐝)·d𝐫∧dt + 𝐛×𝐝·d𝐒,$$ not to confuse this "$$c$$" with the "light-speed" $$c$$ above.

So, now: write down the differentials for the various quantities: $$d𝐑 = d𝐫 - 𝐫'(T)dT = d𝐫 - 𝐯 dT,\quad dT = dt - \frac{dR}{c},\quad dR = \frac{𝐑·d𝐑}{R}.$$ The acceleration creeps into the picture because of: $$d𝐯 = d(𝐫'(T)) = 𝐫''(T) dT = 𝐚 dT \quad (𝐚 = 𝐫''(T)).$$

These are algebraic relations for $$dR$$, $$dT$$ and $$d𝐑$$. You can apply them directly, substituting and rewriting the potential as: $$A = g \frac{𝐯·(d𝐑 + 𝐯dT) - c (c dT + dR)}{Rc - 𝐑·𝐯} = g \frac{𝐯·d𝐑 - c dR + (|𝐯|^2 - c^2) dT}{Rc - 𝐑·𝐯}.$$ By the way: $$\frac{𝐯·d𝐑 - c dR}{Rc - 𝐑·𝐯} = -\frac{𝐑·d𝐯 + d(Rc - 𝐑·𝐯)}{Rc - 𝐑·𝐯} = -\frac{𝐑·d𝐯}{Rc - 𝐑·𝐯} - d\ln|𝐑·𝐯 - Rc|.$$ Total differentials don't count in potential one-forms, so we just "re-gauge" it away: $$A = g \frac{\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯}{Rc - 𝐑·𝐯} - d(g \ln|𝐑·𝐯 - Rc|)\quad⇒\quad A = g \frac{\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯}{Rc - 𝐑·𝐯}.$$

Thus, $$F = dA = g \frac{d\left(\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯\right)}{Rc - 𝐑·𝐯} - g \frac{d(Rc - 𝐑·𝐯)}{Rc - 𝐑·𝐯} ∧ \frac{\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯}{Rc - 𝐑·𝐯}.$$

For the first term, noting that $$d𝐯∧dT = 𝐚dT∧dT = 𝟬$$, we have: $$d\left(\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯\right) = 2𝐯·d𝐯∧dT - d𝐑\hat{·}d𝐯 = -𝐚·d𝐑∧dT.$$

For the second term, the contribution of $$d𝐯$$ from $$d(Rc - 𝐑·𝐯)$$ cancels, since: $$d𝐯∧\left(\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯\right) = 𝐚\left(|𝐯|^2 - c^2 - 𝐑·𝐚\right)dT∧dT = 0.$$ So, we can write the term as: $$-g \frac{d(Rc - 𝐑·𝐯)}{Rc - 𝐑·𝐯} ∧ \frac{\left(|𝐯|^2 - c^2\right) dT - 𝐑·d𝐯}{Rc - 𝐑·𝐯} = g \frac{(c dR - 𝐯·d𝐑) ∧ (𝐑·𝐚 + c^2 - |𝐯|^2)dT}{(Rc - 𝐑·𝐯)^2}.$$ Thus, $$F = -g \frac{𝐚·d𝐑∧dT}{Rc - 𝐑·𝐯} + g \frac{(𝐑·𝐚 + c^2 - |𝐯|^2)(c dR ∧ dT - 𝐯·d𝐑 ∧ dT)}{(Rc - 𝐑·𝐯)^2}.$$

To convert $$d𝐑∧dT$$ and $$dR∧dT$$ to coordinate differentials, we can first solve the algebraic relations between the differentials for $$dR$$ and $$dT$$: $$dR = \frac{𝐑c·(d𝐫 - 𝐯dt)}{Rc - 𝐑·𝐯},\quad dT = dt - \frac{dR}{c} = \frac{Rcdt - 𝐑·d𝐫}{Rc - 𝐑·𝐯}.$$ Then, we have, for an arbitrary vector $$𝐛$$: $$𝐛·d𝐑∧dT = 𝐛·(d𝐫 - 𝐫dT)∧dT = 𝐛·d𝐫∧dT = 𝐛·d𝐫∧\frac{Rcdt - 𝐑·d𝐫}{Rc - 𝐑·𝐯},$$ or, after a little differential form algebra: $$𝐛·d𝐑∧dT = \frac{Rc𝐛·d𝐫∧dt - 𝐛×𝐑·d𝐒}{Rc - 𝐑·𝐯}.$$

In addition, we have: $$dR∧dT = dR∧\left(dt - \frac{dR}{c}\right) = dR∧dt = \frac{𝐑c·(d𝐫 - 𝐯dt)}{Rc - 𝐑·𝐯}∧dt = \frac{𝐑c·d𝐫∧dt}{Rc - 𝐑·𝐯}.$$ The $$d𝐒$$ term in $$d𝐑∧dT$$ is the only place you're getting any magnetic field out of. There's no magnetic field coming out of $$dR∧dT$$.

Combining terms, and setting (with a little vector algebra $$𝐑·𝐚𝐯 - 𝐚𝐑·𝐯 = 𝐑×(𝐯×𝐚)$$): $$F_0 = g \frac{𝐑·𝐚 + c^2 - |𝐯|^2}{(Rc - 𝐑·𝐯)^2},\\ 𝐅_1 = g \frac{𝐚}{Rc - 𝐑·𝐯} + F_0 𝐯 = g\frac{𝐑×(𝐯×𝐚) + Rc𝐚 + \left(c^2 - |𝐯|^2\right)𝐯}{(Rc - 𝐑·𝐯)^2},$$ we get: \begin{align} F &= c F_0 dR ∧ dT - 𝐅_1·d𝐑 ∧ dT\\ &= \frac{c\left(𝐑cF_0 - R𝐅_1\right)·d𝐫∧dt + 𝐅_1×𝐑·d𝐒}{Rc - 𝐑·𝐯}. \end{align}

Extracting the components (and using more vector algebra $$𝐑𝐑·𝐚 - R^2𝐚 = 𝐑×(𝐑×𝐚)$$): \begin{align} 𝐄 &= c \frac{𝐑cF_0 - R𝐅_1}{Rc - 𝐑·𝐯}\\ &= gc \frac{𝐑×((𝐑c - R𝐯)×𝐚) + \left(c^2 - |𝐯|^2\right)(𝐑c - R𝐯)}{(Rc - 𝐑·𝐯)^3},\\ 𝐁 &= \frac{𝐅_1×𝐑}{Rc - 𝐑·𝐯}\\ &= g\frac{𝐑×(𝐯×𝐚)×𝐑 + Rc𝐚×𝐑 + \left(c^2 - |𝐯|^2\right)𝐯×𝐑}{(Rc - 𝐑·𝐯)^3}. \end{align}

With the normalizations $$𝐑 = R\hat{𝐧},\quad 𝐯 = 𝝱c,\quad 𝐚 = \dot{𝝱}c,\quad γ = \frac{1}{\sqrt{1 - |𝐯|^2/c^2}}$$ noting that $$gc = q/(4πε_0)$$, this reduces to: $$𝐄 = \frac{q}{4πε_0cR} \frac{\hat{𝐧}×((\hat{𝐧} - 𝝱)×\dot{𝝱})}{(1 - \hat{𝐧}·𝝱)^3} + \frac{q}{4πε_0R^2} \frac{\hat{𝐧} - 𝝱}{γ^2(1 - \hat{𝐧}·𝝱)^3},\\ 𝐁 = \frac{μ_0q}{4πR}\frac{\hat{𝐧}×(𝝱×\dot{𝝱})×\hat{𝐧} + \dot{𝝱}×\hat{𝐧}}{(1 - \hat{𝐧}·𝝱)^3} + \frac{μ_0cq}{4πR^2}\frac{𝝱×\hat{𝐧}}{γ^2(1 - \hat{𝐧}·𝝱)^3} = \frac{\hat{𝐧}×𝐄}{c}.$$

There: see how easy that was? Now, at this point you're probably going, "That's not easy at all - it's a mess! I'm just going to go over onto The Wikipedia and check their derivation" (Search, search, Ding! Wikipedia.) "So, like I was saying: wow!, yes, that really is easier!" And, you might also go: "But theirs looks different. They got extra brackets on the cross-products. You can't remove the brackets, that's illegal: it's not associative!" and I'll do a Palpatine and go: "I'll make it legal", $$(𝐚×𝐛)×𝐚 = 𝐚×𝐛×𝐚 = 𝐚×(𝐛×𝐚).$$