Write the following in Cartesian tensor form

$$(1) \nabla (\operatorname{div} G) \times \nabla\Omega$$ $$(2) (\operatorname{curl}(F)\times G)\cdot \nabla(Φ)$$

I have answers for these two questions, however I do not understand how to show it step by step.

  • $\begingroup$ Please LaTeX it up.Its too hard too read. $\endgroup$ – Devarsh Ruparelia Dec 22 '14 at 16:18
  • $\begingroup$ what do you need to prove? $\endgroup$ – alexjo Dec 22 '14 at 17:17

On the first expression.

Let $G=G(x_1,x_2,x_3)$ be a vector field in $\mathbb R^3$, with $$G(x_1,x_2,x_3)=\left(G_1(x_1,x_2,x_3), G_2(x_1,x_2,x_3), G_3(x_1,x_2,x_3))\right), $$

and $\Omega=\Omega(x_1,x_2,x_3)$ be a differentiable function on $\mathbb R^3$. Then, by definition of divergence

$$\operatorname{div}G:=\sum_{i=1}^3\frac{\partial G_i}{\partial x_i}(x_1,x_2,x_3) $$

and so the $j$-th component $\nabla_j(\operatorname{div}G)$ of $\nabla(\operatorname{div}G)$ is given by

$$\nabla_j(\operatorname{div}G)= \sum_{i=1}^3\frac{\partial^2 G_i}{\partial x_i\partial x_j}(x_1,x_2,x_3),$$

for all $j=1,2,3$. The gradient $\nabla\Omega$ of $\Omega$ has components $\nabla_j\Omega=\frac{\partial\Omega}{\partial x_j}(x_1,x_2,x_3).$

It remains to compute the cross product $\nabla(\operatorname{div}G)\times \nabla\Omega$; this can be done either using the notation

$$\left(\nabla(\operatorname{div}G)\times \nabla\Omega\right)_i = \sum_{j,k=1}^3\epsilon_{ijk} \nabla_j(\operatorname{div}G) \nabla_k\Omega = \text{using the above results} = \sum_{j,k=1}^3\epsilon_{ijk} \left( \sum_{l=1}^3\frac{\partial^2 G_i}{\partial x_l\partial x_j}\right)\frac{\partial\Omega}{\partial x_k},~~~i=1,2,3 $$

where $\epsilon_{ijk}$ is the Levi-Civita antisymmetric tensor, or applying the expansion -component by component - that can be found in this link at the end of the "Computing cross product" section. The two methods are equivalent.

On the second expression.

The methods for the second expression are analogous to the ones shown above. The definition of curl of a vector field is needed: it can be found here.

  • $\begingroup$ @Emily Peers did it help? Do you need more explanations? $\endgroup$ – Avitus Dec 24 '14 at 20:14

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