On the Validity of Swapping Dummy Indices in Tensor Manipulation

[Wannabe Physicist]

Homework Statement

Using property (a), show that a symmetric tensor $T_{i_1 i_2 }$ remains symmetric under all rotations.

Relevant Equations

(1) Transformation law under rotation: $T_{i_1 i_2 }' = r_{i_1 j_1} r_{i_2 j_2} T_{j_1 j_2}$
(2) Definition of symmetric tensor: $T_{i_1 i_2} - T_{i_2 i_1} = 0$

Property (a) simply states that a second rank tensor that vanishes in one frame vanishes in all frames related by rotations.

I am supposed to prove: $T_{i_1 i_2} - T_{i_2 i_1} = 0 \implies T_{i_1 i_2}' - T_{i_2 i_1}' = 0$

Here's my solution. Consider,

$$T_{i_1 i_2}' - T_{i_2 i_1}' = r_{i_1 j_1} r_{i_2 j_2} T_{j_1 j_2} - r_{i_2 j_1} r_{i_1 j_2} T_{j_1 j_2}$$

**Now consider this statement:** Because $j_1$ and $j_2$ are dummy indices and both are summed from 1 to 3, we can swap these indices exclusively for the second term in the above expression.

If I assume the above statement it is easy to obtain

$$T_{i_1 i_2}' - T_{i_2 i_1}' = r_{i_1 j_1} r_{i_2 j_2} T_{j_1 j_2} - r_{i_2 j_2} r_{i_1 j_1} T_{j_2 j_1}$$
$$T_{i_1 i_2}' - T_{i_2 i_1}' = r_{i_1 j_1} r_{i_2 j_2} [T_{j_1 j_2} - T_{j_2 j_1}]$$And then using property (a), I can prove the required statement.

But I am not sure if the statement of swapping indices is valid.

 

[anuttarasammyak]

Homework Statement:: Using property (a), show that a symmetric tensor $T_{i_1 i_2 }$ remains symmetric under all rotations.
Relevant Equations:: (1) Transformation law under rotation: $T_{i_1 i_2 }' = r_{i_1 j_1} r_{i_2 j_2} T_{j_1 j_2}$
(2) Definition of symmetric tensor: $T_{i_1 i_2} - T_{i_2 i_1} = 0$

But I am not sure if the statement of swapping indices is valid.

It simply says obviously
$$\sum_{i=1}^n A_i B_i=\sum_{j=1}^n A_j B_j=\sum_{\gamma=1}^n A_\gamma B_\gamma$$
where $\gamma=\{a,b,c,...,\alpha,\beta,...,\xi,\eta,\zeta,...\}$ any symbol you like.

 

[Wannabe Physicist]

Oh right! Thanks a lot!