Simplifying Index Notation in Vector Calculus

[andrien]

(r×∇).(r×∇)=r.∇×(r×∇)
now in index notation it is written as,
=x_i∂_jx_i∂_j-x_i∂_jx_j∂_i
but when I tried to prove it ,it just came out twice.can anyone tell how it is correct(given is the correct form).i really mean that i was getting four terms which gave twice of above after reshuffling so prove it.

 

[vanhees71]

All your formulae are correct.

The left-hand side of your first equation reads in index notation
$$(\vec{r} \times \vec{\nabla}) \cdot (\vec{r} \times \vec{\nabla}) \phi= \epsilon_{jkl} r_k \partial_l (\epsilon_{jmn} r_m \partial_n \phi).$$
Now using
$$\epsilon_{jkl} \epsilon_{jmn}=\delta_{km} \delta_{ln}-\delta_{kn} \delta_{lm},$$
you indeed get
$$(\vec{r} \times \vec{\nabla}) \cdot (\vec{r} \times \vec{\nabla}) \phi = r_{k} \partial_l(r_k \partial_l \phi)-r_k \partial_l(r_l \partial_k \phi).$$

The right-hand side of your first equation is
$$\vec{r} \cdot [\vec{\nabla} \times (\vec{r} \times \vec{\nabla}) \phi] = r_j \epsilon_{jkl} \partial_k (\epsilon_{lmn} r_m \partial_n \phi).$$
Again we have
$$\epsilon_{jkl} \epsilon_{lmn}=\epsilon_{ljk} \epsilon_{lmn}=\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}.$$
Thus we have
$$\vec{r} \cdot [\vec{\nabla} \times (\vec{r} \times \vec{\nabla}) \phi] = r_j \partial_k (r_j \partial_k \phi)-r_j \partial_k (r_k \partial_j \phi).$$
This shows that indeed both expressions of your first equations are equal, because the only difference is the naming of the dummy-summation indices :-).

 

[andrien]

$$\epsilon_{jkl} \epsilon_{jmn}=\delta_{km} \delta_{ln}-\delta_{kn} \delta_{lm},$$
I was aware of it which I have seen in butkov an year ago.but this is the first use of it.so thanks,van I think i am just becoming lazy.