# Transformation of Electromagnetic Four-Tensor

I apologize if I am missing something obvious, but I am in my first class with tensors and I am still learning the notation. I am running into a problem with the transformation of the transformation of the four-tensor for electromagnetism that is given by

$$F^{\mu \nu} = \left[ \matrix { 0 & - \cal{E}_x/c & - \cal{E}_y /c & - \cal{E}_z /c \\ \cal{E}_x/c & 0 & -B_z & B_y \\ \cal{E}_y/c & B_z & 0 & -B_x \\ \cal{E}_z/c & -B_y & B_x & 0 } \right]$$

And the Lorentz transformation of this tensor is given by

$$F^{' \mu \nu} = \sum\limits^3_{\alpha, \beta = 0}{\Lambda^\mu_\alpha \Lambda^\nu_\beta F^{\alpha \beta}}$$

From what I understand, this is just a single sum and $\alpha = \beta$ over every iteration; however, by this $F^{00} = F^{11} = F^{22} = F^{33} = 0$, so the transformation of any element would result in a zero.

Where have I gone wrong here?

The notation means \begin{align} \sum^3_{\alpha, \beta = 0}{\Lambda^\mu_\alpha \Lambda^\nu_\beta F^{\alpha \beta}} &= \sum_{\alpha=0}^3\sum_{\beta = 0}^3\Lambda^\mu_\alpha \Lambda^\nu_\beta F^{\alpha \beta} \\ &= \Lambda^\mu_0 \Lambda^\nu_0 F^{00}+\Lambda^\mu_0 \Lambda^\nu_1 F^{0 1}+\Lambda^\mu_0 \Lambda^\nu_2 F^{0 2}+\Lambda^\mu_0 \Lambda^\nu_3 F^{0 3}+\text{12 more terms} \end{align} it's not just the "diagonal" terms with $\alpha = \beta$.
• @danielu13 Well I don't necessarily have any general tricks, but one technique that's often useful is to separate spatial components $\mu=1,2,3$ from the time components in which case terms might mix in nice ways. Other than that, whether or not you need to brute force depends on the specific calculations. Computer algebra systems are very useful for brute force calculations. Commented Feb 4, 2014 at 5:25