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I have a tensor $T^{\mu\nu}$ that looks like this:

T^mu,nu = {{2,0,1,-1},{1,0,-3,2},{-1,1,0,0},{2,1,-1,2}}

I want to find $T^{\nu\mu}$.

If I swap the indices, what does that do to the matrix and for what instances am I able to say $T^{\mu\nu} = -T^{\nu\mu}$? I don't think that would work for this case.

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2 Answers 2

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Little correction: the matrix itself is $T$. So $$ T = \begin{pmatrix} 2&0&1&-1 \\1&0&-3&2 \\ -1&1&0&0 \\ 2&1&-1&2 \end{pmatrix} $$ On the other hand, $T^{\mu \nu}$ is the element of the matrix at the row $\mu$ and column $\nu$ (e.g. $T^{23} = -3$).

Now for the question, the matrix whose elements are the same as $T$ but with indices swapped is the transpose of $T$, denoted by $T^T$. We swap the rows with the columns, so e.g. $(T^T)^{32} = T^{23} = -3$. For this matrix,

$$ T^T = \begin{pmatrix} 2&1&-1&2 \\ 0&0&1&1 \\ 1&-3&0&-1 \\ -1&2&0&2 \end{pmatrix} $$ From this representation, we can see that $T \neq - T^T$.

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    $\begingroup$ One small correction. The (2 2) part of the transposed matrix should be (2 1). Otherwise, +1. $\endgroup$ Mar 11, 2021 at 19:21
  • $\begingroup$ Thank you for catching this typo! Just edited. $\endgroup$
    – megaleo
    Mar 11, 2021 at 19:23
  • $\begingroup$ Sei benvenuto sempre!! $\endgroup$ Mar 11, 2021 at 19:30
  • $\begingroup$ Awesome, thank you! $\endgroup$
    – Tom
    Mar 14, 2021 at 20:28
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You have all the components so you also have $T^{\nu\mu}$. You probably mean the transpose of T, but you also have that. Only for $\mu=2$ and $\nu=0$, or vice versa, you have $T^{\mu\nu}=-T^{\nu\mu}$.

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