1
$\begingroup$

If I have a tensor $X^{\mu}{}_{\nu} = \begin{bmatrix} a & b & c \\ d & e& f\\ g&h&i\\ \end{bmatrix}$ then what is $X^{\mu}_{\;\;\mu}$?

From what I understand it would be $(a,b,c)\cdot(a,b,c) +(d,e,f)\cdot(d,e,f)+(g,h,i)\cdot(g,h,i). $

Is this correct?

$\endgroup$
4
  • $\begingroup$ The left most index indicates the row number, the right most index indicates the column number. So would it be $(a,b,c)\cdot(a,d,g) +(d,e,f)\cdot(b,e,h)+(c,f,i)\cdot(g,h,i). $ $\endgroup$
    – Tim
    Mar 21, 2017 at 22:03
  • 2
    $\begingroup$ Related: math.stackexchange.com/q/1254041/72459 By the way, why do you think that you can count the indices up independently? When $\mu=1$, you have $X^1_1$, there's no way to get e.g. $X^1_2$ in $X^\mu_\mu$. $\endgroup$ Mar 21, 2017 at 22:12
  • $\begingroup$ The above mentioned is not a tensor: it is the components of a tensor in some particular basis. $\endgroup$
    – gented
    Mar 21, 2017 at 22:12
  • 4
    $\begingroup$ So then it would just be $a+e+i$? $\endgroup$
    – Tim
    Mar 21, 2017 at 22:40

1 Answer 1

1
$\begingroup$

$$ {\mathrm tr}\left(X\right) = X^{\mu}{}_{\mu} = X^{0}{}_{0} + X^{1}{}_{1} + X^{2}{}_{2} = a + e + i $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.