Integral/Vector calculus $\oint_{\partial S} u \vec \nabla v \cdot d \vec \lambda=\int_S (\vec \nabla u)\times (\vec \nabla v)\cdot d\vec S.$

Question

I am trying to show that $$ \oint_{\partial S} u \vec \nabla v \cdot d \vec \lambda=\int_S (\vec \nabla u)\times (\vec \nabla v)\cdot d\vec S $$ using Levi Cevita notation methods only. The Levi Cevita tensor is given by $\epsilon_{ijk}$ and is a totally anti symmetric tensor. The functions u and v are dependent on the radius vector in 3 dimensions ($\vec r$). I am stuck on this I know the general idea of using this notation is to write things like $$ (\vec \nabla \times \vec A)_i= \epsilon_{ijk} \partial_j A_k $$ but I still am stuck. Thank you

Answer 1

We can prove the identity by using Stoke's theorem and tensor notation \begin{equation} \oint_{\partial S}u \vec{\nabla} v \cdot d\vec{\lambda}=\int_S \big(\vec{\nabla} u\big)\times \big( \vec{\nabla}v \big)\cdot d\vec{S}. \end{equation} We now use Stoke's theorem to relate line integral to surface integral by $$ \oint_{\partial S}u \vec{\nabla} v \cdot d\vec{\lambda}=\int_{S}\vec{\nabla}\times (u \vec{\nabla} v) \cdot \hat{n} dS. $$ Now I will use your desired $\epsilon_{ijk}$ notation to show that $$ \int_{S}\vec{\nabla}\times (u \vec{\nabla} v) \cdot \hat{n} dS=\int_S \big(\vec{\nabla} u\big)\times \big( \vec{\nabla}v \big)\cdot d\vec{S}. $$
We obtain $$ \big( \vec{\nabla}\times (u \vec{\nabla}v) \big)_i=\epsilon_{ijk}\partial_j (u\partial_k v)=\epsilon_{ijk}( \partial_j u \partial_k v+u\partial_j \partial_k v) $$ where I used the product rule. However the term $\partial_j \partial_k v$ is symmetric, so we know this will vanish! We can see this by using $$ \partial_j \partial_k v=\frac{1}{2} (\partial_j \partial_k +\partial_k \partial_j)v $$ since it is symmetric w.r.t j and k. Thus we can see by swapping the indices j and k we obtain $$ \epsilon_{ijk}\partial_j\partial_k v=-\epsilon_{ijk}\partial_k \partial_jv $$
which is only true if this term is zero. Also you know that a total antisymmetric tensor times a symmetric quantity is zero. We are left with $$ \big( \vec{\nabla}\times (u \vec{\nabla}v) \big)_i=\epsilon_{ijk}\partial_j u \partial_k v=(\vec{\nabla}u \times \vec{\nabla}v)_i $$ which concludes the desired proof.