- #1
SiennaTheGr8
- 497
- 195
I'm trying to get the hang of index gymnastics, but I think I'm confused about the relationship between rank-2 tensor components and their matrix representations.
So in Hartle's book Gravity, there's Example 20.7 on p. 428. We're given the following metric:
##g_{AB} = \begin{bmatrix} F & 1 \\ 1 & 0 \end{bmatrix} \qquad \qquad g^{AB} = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix}##
And we're given the following components of a tensor:
##t_{AB} = \begin{bmatrix} G & 1 \\ -1 & 0 \end{bmatrix}##
Then we're asked to calculate ##{t^A}_B## and ##t^{AB}## (etc.).
Okay:
##{t^A}_B = g^{AC} t_{CB} = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \begin{bmatrix} G & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix}##
That's correct.
Now, for ##t^{AB}## I was thinking:
##t^{AB} = g^{AC} {t_C}^B##,
which is correct, but I didn't have ##{t_A}^B## yet, so I had to approach it another way. I thought maybe:
##t^{AB} = g^{CB} {t^A}_C = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix}##.
But that's actually the wrong order for the matrix multiplication (though if I'm not mistaken, ##g^{CB} {t^A}_C = {t^A}_C g^{CB}## -- is that right?). Instead, the correct answer is:
##t^{AB} = \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \Big( \stackrel{?}{=} {t^A}_C g^{CB} \Big)##.
So I suppose what I'm asking is, first, whether it's indeed true that ##g^{CB} {t^A}_C = {t^A}_C g^{CB}##, and then second, what should have tipped me off that I needed to "reverse" the order of the matrices representing ##g^{CB}## and ##{t^A}_C## when calculating ##t^{AB}##?
Hope that's clear enough.
So in Hartle's book Gravity, there's Example 20.7 on p. 428. We're given the following metric:
##g_{AB} = \begin{bmatrix} F & 1 \\ 1 & 0 \end{bmatrix} \qquad \qquad g^{AB} = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix}##
And we're given the following components of a tensor:
##t_{AB} = \begin{bmatrix} G & 1 \\ -1 & 0 \end{bmatrix}##
Then we're asked to calculate ##{t^A}_B## and ##t^{AB}## (etc.).
Okay:
##{t^A}_B = g^{AC} t_{CB} = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \begin{bmatrix} G & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix}##
That's correct.
Now, for ##t^{AB}## I was thinking:
##t^{AB} = g^{AC} {t_C}^B##,
which is correct, but I didn't have ##{t_A}^B## yet, so I had to approach it another way. I thought maybe:
##t^{AB} = g^{CB} {t^A}_C = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix}##.
But that's actually the wrong order for the matrix multiplication (though if I'm not mistaken, ##g^{CB} {t^A}_C = {t^A}_C g^{CB}## -- is that right?). Instead, the correct answer is:
##t^{AB} = \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \Big( \stackrel{?}{=} {t^A}_C g^{CB} \Big)##.
So I suppose what I'm asking is, first, whether it's indeed true that ##g^{CB} {t^A}_C = {t^A}_C g^{CB}##, and then second, what should have tipped me off that I needed to "reverse" the order of the matrices representing ##g^{CB}## and ##{t^A}_C## when calculating ##t^{AB}##?
Hope that's clear enough.