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?
Answer
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
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