Proving a general formula for the boost transformation of the electromagnetic field

In inertial frame $\mathcal{O}$, a region of space-time is filled with constant electric field $\vec{E}$ and magnetic field $\vec{B}$. Another inertial frame $\mathcal{O}'$ has 3-veclocity $\vec{V}$ relative to $\mathcal{O}$. What is the electromagnetic field $\left(\vec{E}', \vec{B}'\right)$ measured in $\mathcal{O}'$? Express the result in terms of $\vec{E}$, $\vec{B}$, $\vec{V}$, dot product ($\cdot$) and cross product ($\times$).

This is to get the general formula for the boost transformation of the electromagnetic fields. I know the general form of the Lorentz boost transformation. So, obviously the solution for this problem seems to be applying this boost transformation to the electromagnetic field tensor $F^{uv}$. That is, for the boost transformation $\Lambda^u_v$, calculate $F'^{ab}=\Lambda^a_u \Lambda^b_vF^{uv}$. But this seems like a tremendous amount of work... Is there any more efficient solution than this? Could anyone suggest me?

• Look at how $A^a$ transforms. – J.G. Apr 1 '18 at 17:27
• Then I also must see how $\partial_a$ transforms. Right? – Keith Apr 1 '18 at 17:28
• You're making life too hard for yourself. Forget calculus altogether at the start: just write $\phi',\,\mathbf{A}'$ in terms of their unprimed counterparts. – J.G. Apr 1 '18 at 17:34
• Use w.l.o.g.w.c.a.t. (without loss of generality, we can assume that). For example, assume that the velocity is in the $z$-direction, derive some formulae, then generalize back to dots and crosses based on the patterns. The answer is also in Jackson's electromagnetism textbook. – Sean E. Lake Apr 1 '18 at 18:06

• using transformation of the electromagnetic field tensor $$F^{\mu\nu}$$ (which is TOO LONG!!!)
• using the transformation of 4-vector potential $$\mathbf A^\mu$$
• using the Lorentz force $$\mathbf{F}=q \mathbf{E}+q \mathbf{v}\times\mathbf{B}$$