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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.

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    $\begingroup$ you have to "transform" your differentials as well, i.e. $\partial\mapsto\partial'=\Lambda\partial$. $\endgroup$ – Phoenix87 Apr 23 at 10:41
  • $\begingroup$ @Phoenix87 that's true... thanks! $\endgroup$ – martin Apr 23 at 10:45
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It might help in this case to consider that the electric and magnetic fields are the "components" of the electromagnetic tensor. Then one can use the transformation properties of a rank-2 tensor to derive the expression of the transformed fields.

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  • $\begingroup$ Or just look up the transformation in Wikipedia. $\endgroup$ – my2cts Apr 23 at 12:45

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