0
$\begingroup$

Let $ \overrightarrow v=v_i \overrightarrow e_i \thinspace,\nabla \cdot v=0$ , show that $\nabla \cdot(\nabla v)=0 $

Gradient is defined as : $$\nabla v=\frac{\partial v}{\partial x_j} \otimes e_j=\frac{\partial (v_ie_i)}{\partial x_j}\otimes e_j=\frac{\partial v_i}{\partial x_j}e_i\otimes e_j$$

Divergence is defined as:

$$\nabla \cdot v=\frac{\partial v}{\partial x_j}\cdot e_j=\frac{\partial (v_ie_i)}{\partial x_j}\cdot e_j=\frac{\partial v_i}{\partial x_j}(e_i\cdot e_j)= \frac{\partial v_i}{\partial x_j}(\delta_{ij})=\frac{\partial v_i}{\partial x_i}$$ I got confused abut proving this in Einstein notation so It would be great if this made clear.

$\endgroup$
  • $\begingroup$ Are you sure you're stating the problem correctly? $\nabla \cdot (\nabla v)$ is the vector laplacian, which need not be $0$. Now the divergence of the curl of a vector field, or the curl of the gradient are both $0$. $\endgroup$ – Joe Feb 28 at 20:34
  • $\begingroup$ The idea is to get gradient of a divergence from that and then we use the assumption that the the divergence is zero to get zero. Could you get to that identity in Einstein notation which doesn’t seems to be easy $\endgroup$ – F.O Feb 28 at 20:57
  • $\begingroup$ Maybe I'm misunderstanding something here. In the picture you posted, what is $A$? $\endgroup$ – Joe Feb 28 at 21:47
  • $\begingroup$ Also I misspoke above. I meant the scalar Laplacian $\endgroup$ – Joe Feb 28 at 21:49
  • $\begingroup$ A is a vector as on wikipedia.en.wikipedia.org/wiki/Vector_Laplacian $\endgroup$ – F.O Feb 28 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.