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Gradient of function in terms of retarded time produces time derivative

Physics Asked by Evan Dodson on August 27, 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.

Correct answer by Jerrold Franklin on August 27, 2021

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