Question 6.22: In Prob. 6.4, you calculated the force on a dipole by “brute...

(because $\oint d l = 0$ ). Now 

$=I\left[\left( \nabla _{0}\right)_{i}\left( a \cdot B _{0}\right)-a_{i}\left( \nabla _{0} \cdot B _{0}\right)\right]$ .

But  $\nabla _{0} \cdot B _{0}=0(\text { Eq. } 5.50), \text { and } m =I a (\text { Eq. } 5.86), \text { so } F = \nabla _{0}\left( m \cdot B _{0}\right)$ (the subscript just reminds us to take the derivatives at the point where m is located). qed

∇ · B = 0                                     (5.50) 

$m \equiv I \int d a =I a$                             (5.86)

Proof of Lemma 1:

Eq. 1.108 says $\oint( c \cdot r ) d l = a \times c =- c \times a$ . The jth component is $\sum_{p} \oint c_{p} r_{p} d l_{j}=-\sum_{p, m} \epsilon_{j p m} c_{p} a_{m}$ .  Pick

$\oint( c \cdot r ) d l = a \times c$                        (1.108)

$c_{p}=\delta_{p l}$ (i.e. 1 for the lth component, zero for the others). Then $\oint r_{l} d l_{j}=-\sum_{m} \epsilon_{j l m} a_{m}=\sum_{m} \epsilon_{l j m} a_{m}$ .  qed

Proof of Lemma 2:

$\epsilon_{i j k} \epsilon_{l j m}=0 \text { unless } i j k \text { and } l j m$ are both permutations of 123. In particular, i must either be l or m, and k must be the other, so

$\sum_{j} \epsilon_{i j k} \epsilon_{l j m}=A \delta_{i l} \delta_{k m}+B \delta_{i m} \delta_{k l}$ .

To determine the constant A, pick i = l = 1, k = m = 3; the only contribution comes from j = 2:

$\epsilon_{123} \epsilon_{123}=1=A \delta_{11} \delta_{33}+B \delta_{13} \delta_{31}=A \Rightarrow A=1$ .

To determine B, pick i = m = 1, k = l = 3:

$\epsilon_{123} \epsilon_{321}=-1=A \delta_{13} \delta_{31}+B \delta_{11} \delta_{33}=B \Rightarrow B=-1$ .

So

$\sum_{j} \epsilon_{i j k} \epsilon_{l j m}=\delta_{i l} \delta_{k m}-\delta_{i m} \delta_{k l}$ .  qed