Lorentz Transformation of the 4-Vector Potential

Given is $A_{\mu} = (V /c , A_i)$. Firstly I wanted to know how this vector transforms by Lorentz transformations and then check if I get the right transformation equations for $E$ and $B$.

$$\bar{A_{\mu}} = \gamma \cdot ((V-v_i \cdot A_i),(A_i - v_i V/c^2))$$ I found this as the transformed 4-vector where $v_i = (v_x,v_y,v_z)$. $i$ stands for the space components only.

Substituting $\bar{A_{i}}$ in the equation $(\nabla \times \bar{A})_i = \bar{B_i}$ got me:$$\bar{B_i} = \gamma \cdot (B_i - \frac{1}{c^2}(E \times v)_i).$$ This got me a little excited because I think it's right? Then I tried to do the same for $\bar{E_i} = -\partial_t \bar{A_{i}} - c \cdot\nabla \bar{A_0}$ but this didn't work out as hoped. Filling in, $$\bar{E_i} = -\gamma \cdot (\partial_t (A_i -v_i V/c^2) + \nabla (V - v_i \cdot A_i))$$ didn't get me anywhere. I should get something like $\bar{E_i} = \gamma \cdot (E_i + (v \times B)_i).$

Does someone have an idea on how to get to that solution? Thanks in advance!

• You have at least missed a factor $1/c$ in your formula for the transformed vector potential. It should read $(V - \vec v \cdot \vec A)/c$ in the zeroth component. (because $A_0 = V/c$ and the transformation for the spatial components to the zero component is $-\gamma \vec v \cdot A / c$). Commented Jun 6, 2018 at 15:56
• To get the cross product $\vec v \times \vec B = \vec v \times (\nabla \times \vec A)$ from the given form, you will have use formula for the double cross product. Commented Jun 6, 2018 at 15:57
• And don't forget to impose a conventient gauge condition. Commented Jun 6, 2018 at 16:05