Kronecker delta and Levi-civita function PDF

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Introduction to kronecker delta & Levi-civita function along with solved conceptual questions...

Description

Kronecker delta and Levi-Civita symbol We define the Kronecker delta δij to be +1 if i = j and 0 otherwise, and the Levi-Civita symbol ϵijk to be +1 if i, j, and k are a cyclic permutation of (1, 2, 3) (that is, one of (1, 2, 3), (2, 3, 1) or (3, 1, 2)); -1 if an anticyclic permutation of (1, 2, 3) (that is, one of (3, 2, 1), (2, 1, 3) or (1, 3, 2)); and 0 if any two indices are equal. More formally, 1, 𝑖𝑓 𝑖 = 𝑗 𝑖𝑓 𝑖 ≠ 𝑗

δij = 0,

and

+1 , if (𝑖, 𝑗, 𝑘) is (1, 2, 3), (2, 3, 1) or (3, 1, 2);

ϵijk = 󰇱 −1, if (𝑖, 𝑗, 𝑘) is (3, 2, 1), (2, 1, 3) or (1, 3, 2); 0,

if i = j, or j = k, or k = i.

For convenience, we will use the Einstein summation convention when working with these symbols, where a repeated index implies summation over that index. For example, δii = δ 11 + δ 22 + δ 33 = 3; and ϵijk ϵijk = 6, where we have summed over i, j, and k. This latter expression contains a total of 3 3 = 27 terms in the sum, where six of the terms are equal to one and the remaining terms are equal to zero. There is a remarkable relationship between the product of Levi-Civita symbols and the Kronecker delta, given by the determinant 𝛿𝑖𝑙

ϵijk ϵlmn =  𝛿𝑗𝑙

𝛿𝑘𝑙

𝛿𝑖𝑚 𝛿𝑖𝑛 𝛿𝑗𝑚 𝛿𝑗𝑛  = 𝛿𝑘𝑚 𝛿𝑘𝑛

{δil ( δjm δkn - δjn δkm ) - δim ( δjl δkn – δjn δkl )

+ δin ( δjl δkm - δjm δkl)}

Important identities that follow from this relationship are

ϵijk ϵimn = δjm δkn - δjn δkm, ϵijk ϵijn = 2 δkn. derive some common vector identities. The dot-product is written as 𝐴 . 𝐵 = Ai the Levi-Civita symbol is used to write the ith component of the cross-product as

The Kronecker delta, Levi-Civita symbol, and the Einstein summation convention are used to

Bi, and

[𝐴 𝐴 x 𝐵]i = ϵijk Aj Bk, which can be made self-evident by explicitly writing out the three components. The Kronecker delta finds use as δij Aj = Ai. Problems 1. Prove the following cyclic and anticyclic permutation identities: a) ϵijk

= ϵjki = ϵkij ;

b) ϵijk

= -ϵjik, ϵijk = -ϵkji, ϵijk =- ϵikj

SOLUTION

a) If ijk is a cyclic permutation of (1, 2, 3), then ϵijk anticyclic permutation of (1, 2, 3), then ϵijk

= ϵjki

= ϵjki = ϵkij = 1. If ijk is an = ϵkij = -1. And if any two indices are

equal, then ϵijk = ϵjki = ϵkij = 0. The use is that we can cyclically permute the indices of the Levi-Civita tensor without changing its value. b) If ijk is a cyclic permutation of (1, 2, 3), then ϵijk

= 1 and ϵjik = ϵkji = ϵikj = -1. If ijk is an anticyclic permutation of (1, 2, 3), then ϵijk = -1 and ϵjik = ϵkji = ϵikj = 1. And if any two indices are equal, then ϵijk = ϵjik = ϵikj = 0. The use is that we can swap any two

indices of the Levi-Civita symbol if we change its sign .

2. Verify the cross-product relation [𝐴 𝐴 SOLUTION

x 𝐵]i = ϵijk Aj Bk by considering i = 1, 2, 3.

ϵ1jk AjBk = ϵ123 A2 B3 + ϵ132 A3 B2 = A2 B3 – A3 B2 = [𝐴 𝐴 x 𝐵 ] 1, ϵ2jk AjBk = ϵ231 A3 B1 + ϵ213 A1 B3 = A3 B1 – A1 B3 = [𝐴 𝐴 x 𝐵 ] 2, ϵ3jk AjBk = ϵ312 A1 B2 + ϵ321 A2 B1 = A1 B2 – A2 B1 = [𝐴 𝐴 x 𝐵 ] 3. We have

3. Prove the following Kronecker-delta identities: a) δij Aj = Ai;

b) δik kij

= δij .

SOLUTION

Aj = δi1 A1 + δi2 A2 + δi3 A3 . The only nonzero term has the index of A equal to i, therefore δij Aj = Ai.

a)

Now, δij

b) Now, δik δkj = δi1 δ1j + δi2 δ2j + δi3 δ3j . If i = j, then every term in the sum is zero. If i = j, then only one term is nonzero and equal to one. Therefore, δik δkj = δij . 4. Given the most general identity relating the Levi-Civita symbol to the Kronecker delta,

ϵijk ϵlmn = δil ( δjm δkn - δjn δkm ) - δim ( δjl δkn – δjn δkl ) + δin ( δjl δkm - δjm δkl) prove the following simpler and more useful relations: a) ϵijk ϵimn = δjm δkn – δjn δkm; b) ϵijk ϵijn

= 2 δkn.

SOLUTION

We make use of the identities δii = 3 and δik δjk of the indices doesn’t matter. We also use

= δij. For the Kronecker delta, the order

ϵijk ϵlmn = δil ( δjm δkn - δjn δkm ) - δim ( δjl δkn – δjn δkl ) + δin ( δjl δkm - δjm δkl) a)

ϵijk ϵimn = δii ( δjm δkn - δjn δkm ) - δim ( δji δkn – δjn δki ) + δin ( δji δkm - δjm δki) = 3 ( δjm δkn - δjn δkm ) - ( δjm δkn - δjn δkm ) + ( δjn δkm - δjm δkn ) = δjm δkn – δjn δkm. b) We use the result of a) and find

ϵijk ϵijn = δjj δkn – δjn δkj = 3 δkn - δkn = 2 δkn....