Vector Identities

As an example, we will derive the simple vector identities using $\epsilon_{ijk}$.

First, lets do the scalar triple product of vectors. Note that this product is completely symmetric among the three vectors once its written in our notation. Its simply cyclic combinations have a plus sign and anticyclic have a minus sign.

$$\vec{A}\cdot\left(\vec{B}\times\vec{C}\right) = A_kB_iC_j\epsilon_{ijk} = A_iB_jC_k\epsilon_{ijk}$$

$$=\vec{B}\cdot\left(\vec{C}\times\vec{A}\right) = \vec{C}\cdot\left(\vec{A}\times\vec{B}\right)$$

$$= -\vec{A}\cdot\left(\vec{C}\times\vec{B}\right) = -\vec{B}\cdot\left(\vec{A}\times\vec{C}\right) = -\vec{C}\cdot\left(\vec{B}\times\vec{A}\right)$$

The triple vector product will require us to ``derive'' an identity involving the product of two $\epsilon_{ijk}$s.

$$\left[\vec{A}\times\left(\vec{B}\times\vec{C}... ...ht)\epsilon_{ijk} = A_iB_mC_n \left(\epsilon_{mnj}\epsilon_{ijk}\right)$$

Since the $\epsilon$s are well defined and have simple properties, we must be able to compute a simple expression for $\epsilon_{mnj}\epsilon_{ijk}$. Note that the index $j$ is repeated and therefore summed over. We just need to do the sum but some thought will be required. Nevertheless, you could do this thinking if stranded on a dessert island and in need of vector identities.

We now do the sum $\epsilon_{mnj}\epsilon_{ijk}=-\epsilon_{mnj}\epsilon_{ikj}$. First of all $\epsilon_{ijk}$ will be zero if any of the indices are repeated. So me must have $m\neq n$ and $i\neq k$. We will sum over $j$, but starting from the first $\epsilon$, $j$ must be the other index besides the two used up for $m$ and $n$ so there is only one non-zero term in the sum. This is also true for the second $\epsilon$. Since the index $j$ in the two tensors is the same we must have $m$ and $n$ using up the same two indices that $i$ and $k$ do. So the two possibilities are

$$m=i n=k$$

$${\mathrm or}$$

$$m=k n=i$$

There is only one term in the sum. We can write this result in terms of Kronecker $\delta$s.

$$\epsilon_{mnj}\epsilon_{ikj} = \delta_{mi}\delta_{nk} - \delta_{mk}\delta_{ni}$$

$$-\epsilon_{mnj}\epsilon_{ikj} = \delta_{mk}\delta_{ni} - \delta_{mi}\delta_{nk}$$

We can now easily compute the triple vector product identities.

$$\left[\vec{A}\times\left(\vec{B}\times\vec{C}\right)\right]_k = A_iB_mC_n \left(\epsilon_{mnj}\epsilon_{ijk}\right) = A_iB_mC_n... ...ht) = B_k\left(\vec{A}\cdot\vec{C}\right) - C_k\left(\vec{A}\cdot\vec{B}\right)$$

$$\vec{A}\times\left(\vec{B}\times\vec{C}\right) =\vec{B}\left(\vec{A}\cdot\vec{C}\right) - \vec{C}\left(\vec{A}\cdot\vec{B}\right)$$

For expressions involving derivatives in 3D, these tools become even more useful.