Valiantly Validating Vexing Vector Verities

1. —(A·B) = Aâ(—âB) + Bâ(—âA) + (A·—) B + (B·—) A

For many students, one of the most challenging vector problems is proving the identity :

— A ◊ B = Aâ —âB + Bâ —âA + A ◊— B + B ◊— A (1)

Many are perplexed how something so innocuous looking on the left side can generate something so complex on the right; moreover, people frequently question how can dot products and gradients, relatively friendly vector operatons, involve two expressions straight out the "BAC-CAB"H horrorL studio?H L H L H L H L

I think part of the problem arises from years of "training" in physics and math where we look at the left side of an equation and proceed by manipulating the left side until it looks like the right. This is a case where we will start on the right and evolve an equation that proves our identity in (1)

To work with this identity, I would first start on the right and expand the terms Aâ(—âB) and Bâ(—âA). The game plan will be to expand those terms, construct an equation based on those expansions, and then prove the identity in (1).

Expanding Aâ(—âB): ∑ ∑ Aä “äB m =emni An eijk Bk =emni ejki An Bk = (2) ∑xj ∑xj ∑ ∑ @ H LDdjm dkn An Bk -dkm djn An Bk (3) ∑xj ∑xj

In the first product of deltas, we realize that k = n and j = m; in the second product of deltas, k must equal m and j must equal n, so (3) becomes : ∑ ∑ Aä “äB m = An Bn - An Bm (4) ∑xm ∑xn Now, let' s expand Bä(“äA). We don't need to do another notation summation again, we merely note that Bä(“äA) is of the same form as Aä(“äB), so we just use the result@ Hfrom (4)LD and switch A and B: ∑ ∑ Bä “äA m = Bn An - Bn Am (5) ∑xm ∑xn Now, I will add the results of (4) and (5): @ H LD 2 vectorproof.nb

∑ ∑ ∑ ∑ Aä “äB m + Bä “äA m = An Bn - An Bm + Bn An - Bn Am (6) ∑xm ∑xn ∑xm ∑xn and regroup the terms on the right as : @ H LD @ H LD ∑ ∑ ∑ ∑ An Bn + Bn An - An Bm + Bn Am (7) ∑xm ∑xm ∑xn ∑xn Look at the terms in the first parenthesis of (7); these two terms result from applying the product rule to: ∑ ∑ ∑ An Bn = An Bn + Bn An (8) ∑xm ∑xm ∑xm

We recognize immediately that A B is the scalar A·B, and ∑ A B is just the mth component of “ A ◊ B (so we finally see n n ∑x n n H L m how “ A ◊ B has anything to do with curls and cross products.) H L H L ∑ The terms in the second parenthesis are a little easier to decipher, the combination of An is nothing more than the dot product H L ∑xn ∑ between the two quantities bearing the repeated index n; An is the summation notation way of writing A ÿ“, so the term ∑xn ∑ th th An Bmis just the m component of A ◊— B Similarly, the last term in (7) is just the m component of (B·—)A. ∑xn

Now, if we sum over all m components, we can show that eq. (6) can be written: H L Aâ —âB + Bâ —âA =“A ◊ B - A ◊— B - B ◊— A. (9)

At this point, the most trivial algebraic manipulation yields :

— AH◊ B =L Aâ —âH B +L Bâ —âH A L+ I A ◊—M B +H B ◊—L A. (10)

And we have proven our identity. Give yourself a QED on the back. H L H L H L I M H L 2. —·(AâB) = B·(—âA) - A·(—âB)

This is an interesting identity to study because it will give us a chance to investigate a subtlety of the permutation tensor.

Our first step, as ever, is to write the identity in summation notation. This yields : ∑ “ÿ AäB = eijk Aj Bk (11) ∑xi

th Let' s make sure we understand why (11) is written this way. The grouping eijk A j Bk represents the i term of the cross product AäB. When we multiply the ith component of the crossH productL by the ith component of the operator ∑ , we have a product ∑xi involving a repeated subscript, and we recognize that as the dot operation. In this case, the dot operation is between the differen- tial operator and a cross product, so eq. (11) accurately expresses the identity in summation notation.

Now, since εijk, A j, and Bk are scalars, we can rearrange them as: ∑ ∑ eijk Aj Bk =eijk Aj Bk (12) ∑xi ∑xi vectorproof.nb 3

In (12) we have a differential operator acting on a product, so we apply the product rule and differentiate as : ∑ ∑ ∑ eijk Aj Bk =eijk Bk Aj + Aj Bk (13) ∑xi ∑xi ∑xi Again, exploiting the properties of working with scalars, we rearrange (13) as : B F ∑ ∑ ∑ ∑ eijk Bk Aj + Aj Bk = Bk eijk Aj + Aj eijk Bk (14) ∑xi ∑xi ∑xi ∑xi

∑ th Consider the terms in (14). The grouping eijk Aj is the k component of the curl of A. You can see this because the e B ∑xi F symbol tells you there will be a cross product of the next term terms; since these two terms are the differential operator and A, ∑ th th eijk Aj represents curl A. Since we are crossing the i component of the differential operator and the j component of A, we ∑xi are producing the iâ j = k component of the curl.

We can now recognize that the first term on the right in (14) is the product of Bk “äA k We realize immediately that this is B ◊ —âA . since it is the product of a repeated index.

H L ∑ Now, consider the second term on the right in (14). Notice that the grouping eijk Bk consists of a permutation followed by two H L ∑xi terms; we recognize that this means the cross product of the differential operator and B, in other words, a component of curl B. ∑ Which component is this? Looking at the order of terms, we realize that eijk Bk will produce the iâk =-j term, so that this ∑xi grouping produces the - j component of curl B. We further recognize that we are multiplying the - jth component of curl B with the jth component of A, so that the final term on the right of (14) is -A ◊ —âB). And we have proven this identity.

3. —â —âA =— —◊A -—2 A H

We have encountered double cross products previously in deriving the "BAC-CAB" rule, i.e.,

Aâ BâC = B A ◊ C - C A ◊ B (15)

We begin byH setting FL =—âA andH G=—â(—âL A)=—âF

In summation notation, we write : H L H L H L ∑ ∑ Fi =eijk Ak and also Gm =emni Fi (16) ∑xj ∑xn

∑ Since Fi =eijk Ak, we have that: ∑x j ∑ ∑ ∑ ∑ ∑ Gm =emni Fi =emni eijk Ak =emni eijk Ak (17) ∑xn ∑xn ∑xj ∑xn ∑xj

Permuting the second epsilon so that we can use the e-d relationship: H L ∑ ∑ ∑ ∑ emni eijk Ak =emni ejki Ak = ∑xn ∑xj ∑xn ∑xj 4 vectorproof.nb

∑ ∑ ∑ ∑ ∑ ∑ dmj dkn -dmk djn Ak =dmj dkn Ak -dmk djn Ak ∑xn ∑xj ∑xn ∑xj ∑xn ∑xj

∑ ∑ Now, we focus our attention on the two terms at the end of eq. (18). In the first of these terms, dmj dkn Ak, we know I M ∑xn ∑xj that j=m and also k=n else the Kronecker deltas will force the term to zero, therefore, we make the substitutions j=m and k=n and get: ∑ ∑ ∑ ∑ dmj dkn Ak = An (19) ∑xn ∑xj ∑xn ∑xm

Since we can always interchange order of differentiation, (19) becomes : ∑ ∑ ∑ th An = “ÿA = the m component of “ “ÿA (20) ∑xm ∑xn ∑xm

∑ ∑ th We can recognize the validity of (20) by recalling that An = “·A, and also that “ÿA represents the m component of ∑xnH L ∑xm H L the gradient of “·A.

On making the substitutions k=m and j=n, the last term in (18) becomes: H L ∑ ∑ ∑ ∑ dmk djn Ak = Am (21) ∑xn ∑xj ∑xn ∑xn

Here, the repeated index, n, means to take the dot product of the two differential operators, so this term becomes : ∑ ∑ Am = “ÿ“ Am (22) ∑xn ∑xn or the mth component of (“ÿ“ A =“2 A H L If we combine our results from 20 and 22 , we find that the mth component of G is : L ∂ G = —◊A - —◊— A (23) H L H L m m ∂xm ` We can construct the complete vector G from this by multiplying Gm by the unit vector em and obtain : H L H L ` ∂ ` ` G = Gm em = —◊A em - —◊— Am em (24) ∂xm which becomes : H L H L G =——◊A -—2 A (25) QED H L vectorproof.nb 5

Remember, remember, the Kronecker Delta And tensors permuting in turn, With subscript location, Summation Notation Shall be all yours to learn.