anonanon 152k1212 gold badges233233 silver badges398398 bronze badges $endgroup$ 4 $begingroup$ That's excellent thanks a bunch! I also posted the same concern in this article: math.stackexchange.com/questions/2135276/… that you might have some insight on? $endgroup$
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δ i l δ i m δ i n δ j l δ j m δ j n δ k l δ k m δ k n
I am new to Blender, I need to develop this sure condition of the shampoo bottle, any help is appreciated!
These Houses are important for knowledge the habits with the tensor in several coordinate techniques.
From this actions, we can certainly deduce what $epsilon^ ij4k $ might be without having undergoing the computations Again: it is simply the results of exchanging the last two indices of $epsilon^ ijk4 $,
then still left facet is zero, and suitable side is also zero because two of its rows are equal. In the same way for j c = j c + 1 displaystyle j_ c =j_ c+1
This tensor shouldn't be bewildered Using the tensor density area mentioned above. The presentation With this portion intently follows Carroll 2004.
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Utilizing the funds pi notation Π for common multiplication of numbers, an explicit expression for that symbol is:[citation essential]
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The covariant Levi-Civita tensor (often called the Riemannian quantity form) in any coordinate technique that matches the chosen orientation is
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Do you see how this can have an affect on the worth of $epsilon^ alphabetagammadelta epsilon_ alpha'beta'gamma'delta' $? $endgroup$