I Question about Cartesian Tensors

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Equation I-10 equals 1 when k equals i and 0 when k does not equal i due to the contributions of terms in the summation based on the indices. In the case where x'_0=1 and x'_1=x'_2=0, only the terms with k=0 contribute, leading to the conclusion that for i=0, the equation simplifies to 1, while for i=1, it results in 0. The variable "a" represents the cosine function, but it is clarified that the indices i, j, and k are not related to the cosine function directly. This distinction is crucial for understanding the behavior of the tensor equations in mechanics. The discussion emphasizes the importance of index notation in tensor calculus.
Worn_Out_Tools
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I am not a mathematician but an Engineer-in-Training studying mechanics. That being said why does equation I-10 equal 1 when k equals i and 0 when k does not equal i?

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Worn_Out_Tools said:
I am not a mathematician but an Engineer-in-Training studying mechanics. That being said why does equation I-10 equal 1 when k equals i and 0 when k does not equal i?
Consider the case of ##x'_0=1, x'_1=0, x'_2=0##. In this case, only the terms with ##k=0## contribute to the sum in the eqn. I-9, which becomes ##x'_i=a_{ij}a_{0j}##. For ##i=0## it becomes ##1=a_{0j}a_{0j}##, for ##i=1## it becomes ##0=a_{1j}a_{0j}##, etc. For all combinations of ##k## and ##i##, you get the eqn. I-10.
 
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“a” is defined as shorthand for the cosine function earlier in the book. So 1 = cos(0,j) * cos (0,j)?
 
Worn_Out_Tools said:
“a” is defined as shorthand for the cosine function earlier in the book. So 1 = cos(0,j) * cos (0,j)?
No. ##i, j, k## are indices.
 

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