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Gradient of function in terms of retarded time produces time derivative

Physics Asked on December 4, 2021

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

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

given that $t_0equiv t – dfrac{r}{c}$. 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?

One Answer

$t_0=t-r/c$ 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.

Answered by Jerrold Franklin on December 4, 2021

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