How to transform Electromagnetic fields for Lorentz Boost in arbitrary direction?

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The discussion focuses on transforming electromagnetic fields under a Lorentz boost in an arbitrary direction rather than along a specific Cartesian axis. It explains that the Lorentz transformation for a 4-vector A is given by A'^{\mu}=\Lambda^\mu_\nu A^{\nu}, while a 4-tensor B transforms as B'^{\mu\nu}=\Lambda^\mu_\alpha B^{\alpha \beta }(\Lambda^{-1})^\nu_\beta. To achieve a boost in any direction, the transformation matrix can be expressed as a product of a rotation matrix R and a boost matrix, allowing for the adjustment of the boost direction. The discussion references Wikipedia for further clarification on vector transformations. Understanding these transformations is crucial for accurately modeling electromagnetic fields in relativistic contexts.
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Homework Statement
In an inertial frame $\boldsymbol{S}$, components of electric and magnetic fields are $\vec{E}=2 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{B}=\overrightarrow{0}$ respectively, in some chosen system of unit. In another inertial frame $\boldsymbol{S}^{\prime}$, which is moving with uniform speed with respect to the former, components of magnetic field are $\overrightarrow{B^{\prime}}=2 \hat{i}+1 \hat{j}+3 \hat{k}$. What may be the components of the electric field $\overrightarrow{E^{\prime}}$ in the second frame of reference?
Relevant Equations
The Field Strength tensor is goven by
$$
F^{\alpha \beta}=\partial^\alpha A^\beta-\partial^\beta A^\alpha=\left[\begin{array}{cccc}
0 & -E_x & -E_y & -E_z \\
E_x & 0 & -B_z & B_y \\
E_y & B_z & 0 & -B_x \\
E_z & -B_y & B_x & 0
\end{array}\right]
$$
I know the Lorentz matrix for a boost along any of the Cartesian axes. But, since all the components of the magnetic field has changed in the frames, the boost is along an arbitrary direction, rather tahn a particular axis. How do I solve the problem in this case?
 
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Say 4-vector A is Lorentz transformed as
A'^{\mu}=\Lambda^\mu_\nu A^{\nu}
which is product of matrix and vector, 4-tensor B with form of matrix is Lorentz transformaed as
B'^{\mu\nu}=\Lambda^\mu_\alpha B^{\alpha \beta }(\Lambda^{-1})^\nu_\beta
which is product of matrices. In order that we can chose arbitrary boost direction
\Lambda^\mu_\nu=R^\mu_\alpha \mathbf{\Lambda}^\alpha_\beta (R^{-1})^{\beta}_\nu
where ##\mathbf{\Lambda}^\alpha_\beta## is e.g, z boost and ##R^\mu_\alpha## is matrix for a rotation so that the boost direction be z axis.
 
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At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...