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**Additional resources for Intro to Tensor Calculus**

**Sample text**

A third order system a mn with , m, n = 1, 2, 3 is said to be symmetric in two of its subscripts if the components are unaltered when these subscripts are interchanged. When a a mn = am n =a nm = amn = anm = an m. mn is completely symmetric then Whenever this third order system is completely symmetric, then: (i) How many components are there? (ii) How many of these components are distinct? Hint: Consider the three cases (i) = m = n 22. A third order system b mn (ii) =m=n (iii) = m = n. with , m, n = 1, 2, 3 is said to be skew-symmetric in two of its subscripts if the components change sign when the subscripts are interchanged.

Pk where δsr is the Kronecker delta. 123 eijk = δijk Show (b) Show ijk eijk = δ123 (c) Show ij δmn = eij emn (d) Define rs rsp δmn = δmnp and show (summation on p) rs r s s = δm δn − δnr δm δmn Note that by combining the above result with the result from part (c) we obtain the two dimensional form of the e − δ identity r rn = 12 δmn (e) Define δm 27. Let denote the cofactor of ari a11 in the determinant a21 a31 (a) Show erst Air = eijk asj atk 28. rst δpst = 2δpr (summation on n) and show rst δrst = 3!

59) and obtain the equation r Aqp Bkij ∂xq ∂xs ∂xk ∂xi ∂xj ∂xr ∂xs ∂xm ∂xl ∂xp ∂xs ∂xm . = Arqm Brql l ∂x ∂xp = l Cm Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j, q to i and r to k and write the above equation as ∂xs ∂xj Use inner multiplication by ∂xn ∂xs r Aqp m ∂xq ∂xk k ∂x − A im ∂xi ∂xr ∂xp Bkij = 0. and simplify this equation to the form r δjn Aqp r Aqp m ∂xq ∂xk k ∂x − A Bkij = 0 im ∂xi ∂xr ∂xp or m ∂xq ∂xk k ∂x − A Bkin = 0. im ∂xi ∂xr ∂xp Because Bkin is an arbitrary tensor, the quantity inside the brackets is zero and therefore r Aqp m ∂xq ∂xk k ∂x = 0.