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Hey I'm currently just starting out doing special relativity and I stumbled upon following problem:

If I have the faraday antisymmetric tensor $F^{\alpha \beta}$ and I perform a spatial rotation of my axes, let's say in the $(x_1,x_2)$ plane I would immediatly just apply the Rotation Matrix $R=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & cos\theta & -sin\theta & 0 \\ 0 & sin\theta & cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$. When I do this though I do not get the expected Result meaning that the new eletric (resp. magnetic) field is just the old field $E^i$ $(resp. B^i)$ spatially rotated with the matrix $ R'=\begin{pmatrix} cos\theta & -sin\theta & 0 \\ sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} $. This should lead to $\vec{B'}=\begin{pmatrix} B_xcos\theta-B_ysin\theta \\ B_xsin\theta+B_ycos\theta \\ B_z\end{pmatrix}$ and $\vec{E'}=\begin{pmatrix} E_xcos\theta-E_ysin\theta \\ E_xsin\theta+E_ycos\theta \\ B_z\end{pmatrix}$

Instead I get $F'^{\alpha\beta} =\begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_xcos\theta-E_ysin\theta & -B_zsin\theta & -B_zcos\theta & B_ycos\theta+B_xsin\theta \\ E_xsin\theta+E_ycos\theta & B_zcos\theta & -B_zsin\theta & B_ysin\theta-B_xcos\theta\\ E_z & -B_y & B_x & 0 \end{pmatrix}$

This isn't even close to what I expect. There must be something I am doing horribly wrong. Does anyone have any hints?

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Since the EM tensor is a tensor of rank two, the transformation requires two matrices:

$$F^{\alpha\beta} \to F'^{\alpha\beta} = R^\alpha_\mu R^\beta_\nu F^{\mu\nu}$$

Or, in matrix form,

$$F \to F' = RTR^T$$

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  • $\begingroup$ Thanks, I will try this, but as I can already see from my previous calculations, this should lead to the correct result. Is there any theorem that tells me how many matrices the transformation of a tensor needs? $\endgroup$ – qacwnfq q Nov 2 '15 at 22:04
  • $\begingroup$ @qacwnfqq: The indices tell you that. You can see that I put a rotation matrix for each index. You could even take this as the definition of a tensor. $\endgroup$ – Javier Nov 2 '15 at 22:30

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