# Gradient of function in terms of retarded time produces time derivative

In Griffith's introduction to electromagnetism on page 479, there is this following equation:

\begin{align*} \nabla V &\cong \nabla \left[\frac{1}{4\pi \epsilon_0} \frac{\boldsymbol{\hat r} \cdot \dot{\mathbf{p}}(t_0)}{rc} \right] \\ &\cong \left[\frac{1}{4\pi \epsilon_0} \frac{\boldsymbol{\hat r} \cdot \ddot{\mathbf{p}}(t_0)}{rc} \right] \nabla t_0 \\ &= -\frac{1}{4\pi \epsilon_0 c^2} \frac{[\boldsymbol{\hat r} \cdot \ddot{\mathbf{p}}(t_0)]}{r} \boldsymbol{\hat{r}} \end{align*}

given that $$t_0\equiv t - \dfrac{r}{c}$$. Why is it that one the second line the dipole moment -- the $$p$$ function -- gains an extra time derivative if the gradient is in terms of partial spatial derivatives?

$$t_0=t-r/c$$ is a function of $$r$$, so the $$\nabla$$ acting on it is the same as $$\partial_t$$ (complicated a bit by signs and $$c$$). It would be easier to see if he didn't use $$t_0$$, and put it all in terms of $$(t-r/c)$$ explicitly.
• Ohh, I think I get it. Since we are working with retarded time, time then also becomes a function of $r$ and so the gradient acts on it. Jul 23 '20 at 4:08