The Lorenz-Transformation of the EM-Tensor F is given by the equation $$ F'^{\mu \nu} = \Lambda^{\mu}_{\ \ \rho} \Lambda^\nu_{\ \ \sigma} F^{\rho \sigma}$$ Then it says that this is equivalent to the matrix-equation $$ F' = \Lambda \ F \ \Lambda^T $$

How can i see that one of the Lambda-matrices is the transposed of the other, if the indices are at the same spots? And is it convention that the first index (regardless if upper or lower index) is the row of a matrix and the second index is the row? $$ \Lambda^1_{\ \ 2}$$ would be first row second column of Lambda, and $$ \Lambda_1^{\ \ 2}$$ as well?


1 Answer 1


One trick that you can use is that summed indices should be near each other (diagonally). In your case this amounts to $$F^{\prime\mu\nu} = \Lambda^\mu_{\ \ \rho}\Lambda^\nu_{\ \ \sigma} F^{\rho\sigma} = \Lambda^\mu_{\ \ \color{red}{\rho}}F^{\color{red}{\rho}\color{blue}{\sigma}}\Lambda_{\color{blue}{\sigma}}^{\ \ \nu}$$ Now, by definition $$\left(\Lambda^T\right)^\nu_{\ \ \sigma} \equiv \Lambda_\sigma^{\ \ \nu}$$ so you get the desired result.

  • $\begingroup$ Ok, thanks for the help! Do the indices I sum over always have to be on the oposite site? i.e. one covariant and one contravariant? $\endgroup$
    – Samuel
    Jul 22, 2020 at 11:08
  • $\begingroup$ And by the definition of the transpose of Lambda, my assumption about row and column should be truth as well, right? (first index row, second index column) $\endgroup$
    – Samuel
    Jul 22, 2020 at 11:10
  • $\begingroup$ @Samuel The answer is yes to both questions $\endgroup$ Jul 22, 2020 at 11:53
  • $\begingroup$ Could you explain further why the two indices have to be on opisite sites? I mean e.g. the dot-product of two normal (covariant) vectors can be written as $$ a^{\mu} b^\mu$$ Is it because the dotproduct needs to be independent of coordinate-transformations? So with a metric g this would read $$a^\mu g_{\mu \nu} b^{\nu}$$ which would be invariant under all transformations (not only orthogonal transformations)? $\endgroup$
    – Samuel
    Jul 22, 2020 at 14:10
  • $\begingroup$ @Samuel $a^\mu b^\mu$ is not the dot product of $a$ an $b$, it's a tracing operation on a rank-$2$ tensor. The dot product is defined as $$a\cdot b = g_{\mu\nu} a^\mu b^\nu$$ The fact that the indices have to be near each other is just a little trick. $\endgroup$ Jul 22, 2020 at 14:25

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