MOTIVATING A PROOF OF THE e - d RELATIONSHIP
In class we went over briefly a quick motivation for the e - d relationship that we employ in proving vector identities via summation notation. This relationship shows :
eijk eist =djs dkt -djt dks (1)
Our first step in motivating this proof will be to show how we can write determinants in terms of Kronecker deltas and permutation tensors. Let' s consider the very simple determinant consisting of Kronecker deltas :
d11 d12 d13
d21 d22 d23 =d11 d22 d33 -d23 d32 -d12 ... +d13 ... = 1 d31 d32 d33 I leave parentheses in the last two terms since we already know those terms are zero because of the value of the leading Kronecker delta.
Now, if we allow the first index in the Kronecker deltas to vary (i.e., we permute the rows), we have :
di1 di2 di3
dj1 dj2 dj3 =di1 dj2 dk3 -dj3 dk2 -di2 dj1 dk3 -dj3 dk1 +di3 dj1 dk2 -dj2 dk1 dk1 dk2 dk3 Look at the subscripts in the expression above. Notice that if i = 1, the second two terms must be zero because d12 and d13 will cause the second two terms to vanish. Further, if i=1, the first term will equal 1 if {i,j,k}={1,2,3}; -1 if {i,j,k}={1,3,2} or 0 for any other combination of i, j, k.
Similar analyses for i=2 and i=3 will confirm that the value of the determinant can only be 1, 0, or -1 depending on the choices of {i,j,k}. In aggregate, these results confirm that:
di1 di2 di3
dj1 dj2 dj3 =eijk dk1 dk2 dk3 Now, we allow the second index in the Kroneckers to vary (i.e., we permute the columns) : 2 proofepsilondelta.nb
dir dis dit
djr djs djt =dir djs dkt -djt dks -dis djr dkt -djt dkr +dit djr dks -djs dkr dkr dks dkt
Using the same reasoning we employed above will enable us to show that the value of this deter- minant is zero whenever any two indices of {i, j, k} or {r, s, t} are identical. See if you can work through the logic of this. A short Mathematica program might help elucidate this result. After clearing variables, I define a function, kd to be the Kronecker delta of any two indices. I do this to avoid having to write out "KroneckerDelta" 15 times. Inputting the values of the indices (i,j,k,r,s,t) as a list, we can evaluate the value of this determinant by easily varying any of the indices. Then, I evaluate the determinant. I show here the evaluation when {i,j,k}={1,2,3} ={r,s,t}. I invite you to copy this program into your own notebooks and try any combination of values.
Clear kd,a,b,i,j,k,r,s,t kd a_, b_ : KroneckerDelta a, b i, j, k, r, s, t 1, 2, 3, 1, 2, 3 ; det kd i, r kd j, s kd k, t kd j, t kd k, s kd i, s kd j, r kd k, t kd j, t kd k, r kd i, t kd j, r kd k, s kd j, s kd k, r 1
After studying the various determinants explained above as well as the results of the Mathemat- ica program, we can see that the following relationship holds :
dir dis dit
djr djs djt =eijk erst dkr dks dkt Now, on to determing the epsilon delta relationship. We know this relationship requires that there be a repeated index the e terms, and that the repeated index must occupy the same slot in the permuation tensor. Therefore, we can meet this criterion by setting r = i and obtaining the determinant :
dii dis dit
dji djs djt =dii djs dkt -djt dks -dis dji dkt -djt dki +dit dji dks -djs dki (2) dki dks dkt We can evaluate the terms on the right by applying summation notation to Kronecker deltas : dii =d11 +d22 +d33 = 3
dmn dnp =dmp contraction
proofepsilondelta.nb 3
Finally, the order of indices in a Kronecker delta does not matter, since dab = dba. Using these rules, the right hand side of eq. (2) becomes :
3 djs dkt -djt dks -djs dkt +djt dks +djt dks -djs dkt =djs dkt -djt dks and we have reproduced the relationship shown in equation (1) at the outset.