- #1
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I'm reading a text on tensor analysis (on R³), and I don't understand the following exemple...
[tex]P=\frac{1}{2}(a_{ij}+a_{ji})x_ix_j=\frac{1}{2}(a_{ij}x_ix_j+a_{ij}x_jx_i)=a_{ij}x_ix_j[/tex]
To pass from the second to the last equality, he commuted the second pair of [itex]x_jx_i[/itex] into [itex]x_ix_j[/itex]. But he can't do that for it then changes radically the nature of P, for if we redistribute the [itex]a_{ij}[/itex], we are no longer summing according to the Einstein notation.
If what I just said is not clear, consider this. I am asserting that the author did the following in order to pass from the second to the third equality:
[tex]a_{ij}x_jx_i=a_{ij}x_ix_j[/tex]
On the LHS, we are summing over j but not in the RHS. So this commutation changes the nature of the expression.
[tex]P=\frac{1}{2}(a_{ij}+a_{ji})x_ix_j=\frac{1}{2}(a_{ij}x_ix_j+a_{ij}x_jx_i)=a_{ij}x_ix_j[/tex]
To pass from the second to the last equality, he commuted the second pair of [itex]x_jx_i[/itex] into [itex]x_ix_j[/itex]. But he can't do that for it then changes radically the nature of P, for if we redistribute the [itex]a_{ij}[/itex], we are no longer summing according to the Einstein notation.
If what I just said is not clear, consider this. I am asserting that the author did the following in order to pass from the second to the third equality:
[tex]a_{ij}x_jx_i=a_{ij}x_ix_j[/tex]
On the LHS, we are summing over j but not in the RHS. So this commutation changes the nature of the expression.