# Lorentz transformation of four potential

In the following, $$c=1$$ (and so $$\beta=v$$) and the signature is $$(-+++)$$.

The four-potential is $$A_\mu=(-\phi,\mathbf{A})$$. It transforms as $$A'_{\mu'}=\Lambda_{\mu'}^{\ \ \ \ \mu}A_{\mu}$$ so that, under a boost in the $$x^1=x$$ direction, it becomes $$A'_{\mu'}=\begin{pmatrix}A'_0\\A'_1\\A'_2\\A'_3\end{pmatrix}=\begin{pmatrix}\gamma&v\gamma&0&0\\ v\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\cdot\begin{pmatrix}-\phi\\ A_1\\ A_2\\ A_3\end{pmatrix}=\begin{pmatrix}\gamma(-\phi+vA_1)\\\gamma(-v\phi+A_1)\\ A_2\\ A_3\end{pmatrix}$$ Throughout, the unprimed quantities are quantities from before the boost. From $$\mathbf{B}=\nabla\times\mathbf{A}$$, it is clear that $$B'_1=(\nabla\times\mathbf{A}')_1=\partial_yA_3'-\partial_zA'_2=\partial_yA_3-\partial_zA_2=B_1$$ i.e. it is unchanged. However, I then find that $$\tag{1}B'_2=(\nabla\times\mathbf{A}')_2=\partial_zA_1'-\partial_xA'_3=-\gamma\partial_z\phi+\gamma v\partial_zA_1-\partial_xA_3.$$

I know the answer is supposed to be $$B_2'=\gamma(B_2+vE_3)$$, and I also know that $$E_3=(-\nabla\phi-\partial_0\mathbf{A})_3=-\partial_z\phi-\partial_0A_3$$ so that I can write my answer (1) as $$\tag{1}B'_2=(\nabla\times\mathbf{A}')_2=\gamma E_3+\gamma\partial_0A_3+\gamma v\partial_zA_1-\partial_xA_3.$$ I don't really know how to proceed towards the correct answer from here. I suspect I might have made a mistake. Any help would be greatly appreciated.

• you have to "transform" your differentials as well, i.e. $\partial\mapsto\partial'=\Lambda\partial$. Apr 23, 2020 at 10:41
• @Phoenix87 that's true... thanks! Apr 23, 2020 at 10:45