Say now I have an arbitrary field strength tensor $F$, and I want to boost it according to a Lorentz transformation matrix $(\Lambda)$
The transformation is given by $$ F^{'\mu \nu} = \Lambda^{\mu}_{\;\alpha} \Lambda^{\nu}_{\;\beta} F^{\alpha\beta} $$
The question is, how do I actually calculate this with actual values?
Answer
Important Note.
As written, the transformation formula in the question is suppressing an important subtlety. The field strength tensor is a tensor field; it is a function on spacetime. As such, it's not just its indices that must transform under a Lorentz transformation, but also its argument. The full transformation is as follows: \begin{align} F'^{\mu\nu}(x') = \Lambda^\mu_{\phantom\mu\alpha}\Lambda^\nu_{\phantom\nu\beta}F^{\alpha\beta}(x), \qquad x' = \Lambda x \end{align} Or, put another way \begin{align} F'^{\mu\nu}(x) = \Lambda^\mu_{\phantom\mu\alpha}\Lambda^\nu_{\phantom\nu\beta}F^{\alpha\beta}(\Lambda^{-1}x) \end{align} Authors often suppress this fact in the notation, but I found it confusing when I first learned the subject, so I think it's dangerous to not be explicit.
As a terminological aside, the transformation of the tensor indices is often referred to as a target space transformation while the transformation of its argument is sometimes referred to as a spacetime transformation.
Target space transformation - efficient computation
An efficient way to compute an arbitrary target space transformation in this context is to note that when dealing with strings of contracted two-tensors, one can convert the expression into a sequence of matrix multiplications.
If we define matrices $\Lambda = (\Lambda^\mu_{\phantom\mu\nu})$ and $F = (F^{\mu\nu})$ where the left-most index is the "row" label and the right-most index is the "column" label, then we find that \begin{align} \Lambda^\mu_{\phantom\mu\alpha}\Lambda^\nu_{\phantom\nu\beta}F^{\alpha\beta} = \Lambda^\mu_{\phantom\mu\alpha}F^{\alpha\beta}\Lambda^\nu_{\phantom\nu\beta} = (\Lambda F\Lambda^T)^{\mu\nu} \end{align} where the expression in parenthesis denotes successive matrix multiplication.
Special Cases
The matrix multiplication above is made significantly easier provided the Lorentz transformation one is performing is special. In particular, suppose for instance that the Lorentz transformation is a boost along the $x$-direction. Then the matrix $\Lambda$ will have the following block-matrix form \begin{align} \Lambda = \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix} \end{align} where \begin{align} \lambda = \begin{pmatrix} \gamma & -\gamma\beta \\ -\gamma \beta & \gamma \end{pmatrix}, \qquad I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align} Therefore, if we write $F$ is 2-by-2 block matrix form; \begin{align} F = \begin{pmatrix} f_{11} & f_{12} \\ f_{12} & f_{22} \\ \end{pmatrix} \end{align} then we have \begin{align} F' &= \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix} \begin{pmatrix} f_{11} & f_{12} \\ f_{21} & f_{22} \\ \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix} \\ &= \begin{pmatrix} \lambda f_{11} & \lambda f_{12} \\ f_{21} & f_{22} \\ \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & I_2 \end{pmatrix}\\ &= \begin{pmatrix} \lambda f_{11}\lambda & \lambda f_{12} \\ f_{21}\lambda & f_{22} \\ \end{pmatrix} \end{align} so now one just needs do perform four 2-by-2 matrix multiplications, which is relatively easy. Boosts in the $y$- and $z$- directions are similarly straightforward.
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