Vector to scalar potential, transformation of fields

[OhNoYaDidn't]

Hey guys. So, as i was going through Griffith's Electrodynamics, and i came across this problem:

In the solutions:

How to they actually get to that expression for V = (V(bar)+vAx(bar) )Ɣ? I understand everything after that, but this just made me very confused. How do they get this from the inverse Lorentz transformations?

Thank you.

 

[kontejnjer]

The Lorentz transformation that goes from $S$ to $\bar S$ is given by the matrix:
$$\Lambda(S\rightarrow \bar S)=\begin{pmatrix}\gamma& -\beta\gamma & 0 & 0\\ -\beta\gamma &\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\
\end{pmatrix} $$
The inverse is simply:

$$\Lambda(\bar S\rightarrow S)=\begin{pmatrix}\gamma& \beta\gamma & 0 & 0\\ \beta\gamma &\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\
\end{pmatrix} $$
You can easily verify this by multiplying these two in any order and using $\gamma=1/\sqrt{1-\beta^2}$. Basically it's the transformation with the velocity in the opposite direction, so $\beta\rightarrow -\beta$, while $\gamma$ doesn't change since it's quadratic. If you take into account that
$$
\bar A^\mu=\begin{pmatrix}\bar V/c\\ \bar A_x \\ \bar A_y \\ \bar A_z\\
\end{pmatrix} $$

and $A=\Lambda(\bar S \rightarrow S)\bar A$, you should be able to get the desired result.

 

[OhNoYaDidn't]

Thank you so much, it all makes sense now, even how to apply those!