next up previous contents
Next: Tensors Up: &delta#delta;_ij and &epsi#epsilon;_ijk Previous: The Levi-Civita Tensor   Contents

The Epsilon-Delta Identity

A commonly occurring relation in many of the identities of interest - in particular the $ \vA \times (\vB \times \vC)$ triple product - is the so-called epsilon-delta identity:

$\displaystyle \epsilon_{ijk} \epsilon_{imn} = \delta_{jm}\delta_{kn} -
\delta_{jn}\delta_{km} $

Note well that this is the contraction3.2 of two third rank tensors.! The result has the remaining four indices. Also note well that one can use this identity when summing over two indices that do not ``line up'' according to this basic identity by permuting the indices in a cyclic or anticyclic (with an extra minus sign) way until they do. So one can evaluate:

$\displaystyle \epsilon_{jik} \epsilon_{mni} = -(\delta_{jm}\delta_{kn} -
\delta_{jn}\delta_{km}) $

by using $ \epsilon_{mni} = \epsilon_{imn}$ and $ \epsilon_{jik} = -
\epsilon_{ijk}$ .

An example of how to use this follows. Suppose we wish to prove that:

$\displaystyle \vA \cdot (\vB \times \vC) = \vB \cdot (\vC \times \vA) = \vC \cdot
(\vA \times \vB) $

Let's write the first term using our new notation

$\displaystyle \vA \cdot (\vB \times \vC) = A_i \delta_{ij} (\epsilon_{mnj} B_m
C_n) $

where I left in parentheses to make it comparatively easy to track the conversion. We can now use the delta function to eliminate the $ j$ in favor of the $ i$ :

$\displaystyle \vA \cdot (\vB \times \vC) = \epsilon_{mni} A_i B_m C_n = B_m
\epsilon_{mni} C_n A_i = B_m \epsilon_{nim} C_n A_i $

where we can now reorder terms and indices in the product freely as long as we follow the cyclic permutation rule above in the $ \epsilon $ tensor when we alter the tensor connecting them. Finally, we re-insert a (redundant) $ \delta $ function and parentheses:

$\displaystyle \vA \cdot (\vB \times \vC) = B_m \delta_{mj} (\epsilon_{nij} C_n A_i)
= \vB \cdot (\vC \times \vA) $

Obviously the third form follows just from applying this rule and renaming the vectors.

This same approach can be used to prove the BAC-CAB rule. There are a number of equivalent paths through the algebra. We will leave the proof to the student, after giving them a small push start. First:

$\displaystyle \vA \times(\vB \times \vC) $

has components, so we expect to have precisely one ``leftover'' index after contraction of suitable expressions using the rules and tensors developed above. Hence:

$\displaystyle \left(\vA \times(\vB \times \vC) \right)_k = A_i ( B_m C_n
\epsilon_{mnj})\epsilon_{ijk} $

where the term in parentheses is the $ j$ th component of $ \vB \times
\vC$ . We ignore the parentheses and permute the repeated index to the first slot:

$\displaystyle \left( \vA \times(\vB \times \vC) \right)_k = A_i B_m C_n
\epsilon_{jmn}\epsilon_{jki} $

Apply the identity above:

$\displaystyle \left( \vA \times(\vB \times \vC) \right)_k = A_i B_m C_n
(\delta_{mk}\delta_{ni} - \delta_{mi}\delta_{nk} )$

We apply the delta function rules to eliminate all of the $ m$ and $ n$ combinations in favor of $ i$ and $ k$ :

$\displaystyle \left( \vA \times(\vB \times \vC) \right)_k = A_i B_k C_i - A_i B_i
C_k = B_k(\vA\cdot\vB) - C_k(\vA\cdot\vB) $

which is true for all three components of the vectors represented on both sides, Q.E.D.

In case this last step is obscure, note that one way to ring a unit vector into Einstein notation is to use a general symbol for unit vectors. A common one is $ \he_i$ , where $ \he_1 = \hx = \hi$ , $ \he_2 =
\hy = \hk$ , $ \he_3 = \hz = \hk$ where one can see immediately the problem with using $ \hi, \hj, \hk$ in any cartesian tensor theory where one plans to use Einstein summation - one of several reasons I do not care for them (they also can conflict with e.g. $ i = \sqrt{-1}$ or $ k$ the wave number, where $ \hx$ is unambiguously associated with $ x$ or $ A_x$ ). The last step can now be summed as:

$\displaystyle \vA \times(\vB \times \vC) = \left( \vA \times(\vB \times \vC)
...\vC) - C_k(\vA\cdot\vB) \right\}
\he_k = \vB(\vA \cdot \vB) - \vC(\vA\cdot\vB) $

This general approach will prove very useful when one needs to prove the related vector differential identities later on. Without it, tracking and reordering indices is very tedious indeed.

We have at this point covered several kinds of ``vector'' products, but have omitted what in some ways is the most obvious one. The outer product where the product of $ \vA$ and $ \vB$ is just $ \vA\vB$ the same way the scalar product of $ a$ and $ b$ is $ ab$ . However, this form is difficult to interpret. What kind of object, exactly, is the quantity $ \vA\vB$ , two vectors just written next to each other?

It is a tensor, and it is time to learn just what a tensor is (while learning a bunch of new and very interesting things along the way).

next up previous contents
Next: Tensors Up: &delta#delta;_ij and &epsi#epsilon;_ijk Previous: The Levi-Civita Tensor   Contents
Robert G. Brown 2017-07-11