0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

$t_0=t-r/c$ 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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Jul 23 '20 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.