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## Main Question or Discussion Point

Just to remind you, the Lienard-Wiechert potential, as far as the scalar potential is concerned, is:

[tex] \phi(\textbf{r},t)=\frac{q}{|\textbf r-\textbf r'(t')|\left(1-\frac{v'(t')\cdot \textbf{u}}{c}\right)}[/tex]

, where [itex]\textbf u[/itex] is the unit vector in the direction of [itex]\textbf r-\textbf r'[/itex]. This equation gives the scalar potential at position [itex]\textbf{r}[/itex] and at time t (the prime indicates that those quantities are to be evaluated at the retarded time). From here, and the similar expression for the vector potential, we can derive the electric and magnetic fields. To do so, we need to evaluate the gradient of [itex]t'[/itex] where:

[tex]t'=t-\frac{|\textbf r-\textbf r'|}{c}[/tex]

This i can't do; in fact, I don't understand with respect to what I have to take the gradient. Answer should be:

[tex]grad (t')=\frac{-\textbf u}{c-\textbf v'\cdot \textbf u}[/tex]

Any help is much appreciated!

[tex] \phi(\textbf{r},t)=\frac{q}{|\textbf r-\textbf r'(t')|\left(1-\frac{v'(t')\cdot \textbf{u}}{c}\right)}[/tex]

, where [itex]\textbf u[/itex] is the unit vector in the direction of [itex]\textbf r-\textbf r'[/itex]. This equation gives the scalar potential at position [itex]\textbf{r}[/itex] and at time t (the prime indicates that those quantities are to be evaluated at the retarded time). From here, and the similar expression for the vector potential, we can derive the electric and magnetic fields. To do so, we need to evaluate the gradient of [itex]t'[/itex] where:

[tex]t'=t-\frac{|\textbf r-\textbf r'|}{c}[/tex]

This i can't do; in fact, I don't understand with respect to what I have to take the gradient. Answer should be:

[tex]grad (t')=\frac{-\textbf u}{c-\textbf v'\cdot \textbf u}[/tex]

Any help is much appreciated!