# 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.

• vanhees71

Gold Member
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.

• vanhees71
Wannabe Physicist
Oh right! Thanks a lot!