- #1
Autodidact
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Hello all,
I have a quick question on Einstein notation. I'll write the tensors as a capital letter and the covariant indices as lower case letters (and not use anything that has contravariant indices). I'll also use != for not equal or not congruent to.
In Schaum's outline of tensor calculus, the author stipulates (page 3):
Aij (Xi + Yj) != AijXi + AijYj
but I don't understand how to interpret the right hand side, or to figure out exactly what it does equal in an Einstein-summation form without the parenthetical expression, without choosing some "n" and simply expanding it out entirely.
This becomes relevant because problem 1.28 part c on page 7 (which sets Ei = 1 for all i) asks for a proof of:
Aij (Xi + Xj) = 2 Aij Ei Xj (with Aij symmetric)
I was doing this by expanding the right hand side, but apparently I have not been able to figure out how to do that properly as I have been unable to demonstrate this equality.
I appreciate any insight!
Thanks.
I have a quick question on Einstein notation. I'll write the tensors as a capital letter and the covariant indices as lower case letters (and not use anything that has contravariant indices). I'll also use != for not equal or not congruent to.
In Schaum's outline of tensor calculus, the author stipulates (page 3):
Aij (Xi + Yj) != AijXi + AijYj
but I don't understand how to interpret the right hand side, or to figure out exactly what it does equal in an Einstein-summation form without the parenthetical expression, without choosing some "n" and simply expanding it out entirely.
This becomes relevant because problem 1.28 part c on page 7 (which sets Ei = 1 for all i) asks for a proof of:
Aij (Xi + Xj) = 2 Aij Ei Xj (with Aij symmetric)
I was doing this by expanding the right hand side, but apparently I have not been able to figure out how to do that properly as I have been unable to demonstrate this equality.
I appreciate any insight!
Thanks.