Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.