# Math Help - Question involving Levi-Civita symbol

1. ## Question involving Levi-Civita symbol

Can someone please explain to me why

$\epsilon_{ijk}\frac{\partial}{\partial x_i}\frac{\partial A_k}{\partial x_j} = 0$

where A is a constant vector field.

2. ## Re: Question involving Levi-Civita symbol

$\epsilon_{ijk}$ is a totally antisymmetric symbol and $\frac{\partial^2 A_{k}}{\partial x_{i}\partial x_{j}}$ is symmetric under the exchange of $i$ and $j$, because the partial derivatives commute.

In your particular case, as $\vec{A}$ is constant, the derivatives are always zero.

3. ## Re: Question involving Levi-Civita symbol

OK thanks. So which part of the expression is zero? The $\epsilon_{ijk}$ part?

4. ## Re: Question involving Levi-Civita symbol

$\displaystyle\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3 \epsilon_{ijk} \frac{\partial ^2 A_{k}}{\partial x_{i}\partial x_{j}}$

This is

$\epsilon_{111}\frac{\partial^2 A_{1}}{\partial x_{1}^2}+\epsilon_{112}\frac{\partial ^2 A_{1}}{\partial x_{1}\partial x_{2}}+\epsilon_{113}\frac{\partial ^2 A_{1}}{\partial x_{1}\partial x_{3}}+\epsilon_{121}\frac{\partial ^2 A_{1}}{\partial x_{2}\partial x_{1}}+\epsilon_{122}\frac{\partial ^2 A_{2}}{\partial x_{1}\partial x_{2}}+...$

The definition of $\epsilon_{ijk}$ gives you a lot of zeros, in particular the only non zero are $1=\epsilon_{123}=\epsilon_{231}=\epsilon_{312}$ and $-1=\epsilon_{321}=\epsilon_{132}=\epsilon_{213}$. From this fact and that

$\frac{\partial ^2}{\partial x_{i} \partial x_{j}}=\frac{\partial ^2}{\partial x_{j} \partial x_{i}}$

you will have pairs of terms in your sum like

$\epsilon_{123}\frac{\partial A_{3}}{\partial x_{1}\partial x_{2}}+\epsilon_{213}\frac{\partial A_{3}}{\partial x_{2}\partial x_{1}}$

which cancel and gives you zero.

5. ## Re: Question involving Levi-Civita symbol

Fantastic, problem solved.