# Lienard Wienchert Potentials

• Jul 17th 2010, 04:02 PM
Kiwi_Dave
Lienard Wienchert Potentials
I would post on the Physics help forum, but Latex does not work there! Perhaps a mathematician needs to help them out?

My text asks me to obtain the Lienard-Wiechert Potentials. The potentials are quoted as:

$\displaystyle \phi= \left[\frac q {r(1+u_r/c)}\right]_{ret}$ and $\displaystyle \underline w= \left[\frac {q \underline u} {r(1+u_r/c)}\right]_{ret}$

This seems wrong to me, I get the following

$\displaystyle \underline w= \left[\frac {q \underline u/c} {r(1+u_r/c)}\right]_{ret}$

Is my text incorrect?
• Jul 17th 2010, 04:58 PM
Ackbeet
Two thoughts:

1. Is your textbook using a c=1 convention? If not, then
2. You should be able to distinguish between those two potentials using dimensional analysis. The one you put forth has an extra sec/m multiplying it, but is otherwise the same.
• Jul 17th 2010, 09:07 PM
Kiwi_Dave
$\displaystyle \phi$ and $\displaystyle \underline w$ can be combined into the 4-vector $\displaystyle (-\underline w,\phi)$ so I expect that $\displaystyle \phi$ and $\displaystyle \underline w$ must have the same units (coulombs per meter). This is why I think the additional c is required and the text is wrong?

The text (Rindler; Relativity Special, General and Cosmological) does not use c=1.
• Jul 19th 2010, 04:58 AM
Ackbeet
I would agree that the components of a 4-vector must all have the same units. Otherwise, when you compute the dot product of a 4-vector with itself (thus producing a Lorentz invariant), you'd be adding up terms that didn't have the same units.

So the only question left is this: what are the units of $\displaystyle \underline{u}$? If its units are, as I suspect, [m/s], then you are correct and the book must be wrong. If, on the other hand, $\displaystyle \underline{u}$ is dimensionless, then the book is correct.

That's my verdict.