I was given to prove $$∇·(s∇×(s\vec V))=(∇×\vec V)·∇(s^2/2)$$ by using index notations i.e. Kronecker delta and Levi-Civita symbol. But I cannot figure out how I have to tackle the problem to get that $s^2/2$ is right hand side of the equation.


EXE : If $V$ is vector field on $\mathbb{R}^3$, then recall

$$\nabla \cdot (sV)=\nabla s\cdot V + s\nabla\cdot V$$

EXE : $ \nabla \times (sV)= \nabla s\times V +s\nabla\times V $

EXE : $\nabla \cdot (V\times W)=( \nabla \times V)\cdot W - V\cdot (\nabla\times W) $

EXE : $\nabla\cdot (\nabla\times V)=0$

EXE : \begin{align*}&\ \ \nabla\cdot (s\nabla\times (sV)) \\&= \nabla s\cdot (\nabla\times (sV)) +s\nabla\cdot ( \nabla\times (sV)) \\&= \nabla s\cdot (\nabla s\times V +s\nabla \times V) +s\nabla\cdot ( \nabla s\times V +s\nabla \times V ) \\&= \nabla s\cdot (s\nabla \times V) + s \{ ( \nabla\times \nabla s )\cdot V - (\nabla\times V)\cdot \nabla s \} +s\{ \nabla s\cdot (\nabla \times V) +s \nabla\cdot(\nabla\times V) \} \\&= s V\cdot (\nabla\times \nabla s) + s\nabla s\cdot (\nabla \times V)\end{align*}

  • $\begingroup$ What does "EXE" mean/stand for here? +1, Thanks. $\endgroup$ – Robert Lewis Oct 9 '18 at 1:20
  • 2
    $\begingroup$ I mean that EXE (exercise) is a preparatory fact. $\endgroup$ – HK Lee Oct 9 '18 at 1:26

$\nabla \cdot(s \nabla \times (s\vec V)) \overset{?}{=}(\nabla \times\vec V) \cdot \nabla (s^2/2) \tag 1$

I'm afraid I'm not sufficiently adept with Levi-Civita symbols etc., to easily offer much by way of a derivation using them; I can, however, like my colleague HK Lee, validate the result by means of classic vector analysis:

$s \nabla \times (s \vec V) = s(\nabla s \times \vec V + s\nabla \times \vec V) = s\nabla s \times \vec V + s^2 \nabla \times \vec V; \tag 2$

$\nabla \cdot ( s \nabla \times (s \vec V)) = \nabla \cdot (s\nabla s \times \vec V + s^2\nabla \times \vec V)$ $= \nabla \cdot (s \nabla s \times \vec V) + \nabla \cdot (s^2\nabla \times \vec V); \tag 3$

we make use of the following identity from https://en.m.wikipedia.org/wiki/Vector_calculus_identities:

$\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B); \tag 4$


$\nabla \cdot (s \nabla s \times \vec V) = \vec V \cdot (\nabla \times s\nabla s) - s\nabla s \cdot (\nabla \times \vec V); \tag 5$.


$\nabla \times s\nabla s = \nabla \times \nabla \dfrac{s^2}{2} = 0, \tag 6$

since the curl of a gradient always vanishes, identically; thus,

$\nabla \cdot (s \nabla s \times \vec V) = -s\nabla s \cdot (\nabla \times \vec V) = -\nabla(\dfrac{s^2}{2}) \cdot \nabla \times \vec V; \tag 7$

this leaves us with

$\nabla \cdot (s^2 \nabla \times \vec V) = \nabla s^2 \cdot \nabla \times \vec V + s^2 \nabla \cdot (\nabla \times \vec V) = \nabla s^2 \cdot \nabla \times \vec V, \tag 8$

since the divergence of a curl vanishes, always.

Combining (7) and (8) together into (3) we conclude

$\nabla \cdot ( s \nabla \times (s \vec V)) =-\nabla(\dfrac{s^2}{2}) \cdot \nabla \times \vec V + \nabla s^2 \cdot \nabla \times \vec V = \nabla(\dfrac{s^2}{2}) \cdot (\nabla \times \vec V), \tag 9$

as per request.

  • 1
    $\begingroup$ How we can conclude s∇s=∇(s^2/2) ? $\endgroup$ – Alex Parker Oct 9 '18 at 1:44
  • 1
    $\begingroup$ @AliRahmany: recall that $\partial s^2 / \partial x = ds^2 / ds \partial s / \partial x =2s \partial x / \partial x$ by the chain rule; This is an example of the more general rule $\nabla f(u) = df(u)/du \nabla u$ which itself follows from $df(u) / du \langle X, \nabla u \rangle = df(u) / du X[u] = X[f(u)] = \langle X, \nabla f(u) \rangle.$ $\endgroup$ – Robert Lewis Oct 9 '18 at 2:00

Summing over repeated indices, the left-hand side is $$\epsilon_{ijk}\partial_i (s\partial_j (sV_k))=\epsilon_{ijk}(\partial_i s\partial_j sV_k+s\partial_i\partial_j sV_k+\partial_i(s^2) \partial_j V_k+s^2 \partial_i\partial_j V_k).$$In the above calculation, which you can follow as an exercise, I've pulled the Levi-Civita symbol outside the derivative and repeatedly used the product rule. Symmetry arguments drop most terms, and we're left with $\partial_i(s^2)\epsilon_{ijk}\partial_j V_k$ as required.


Your Answer

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

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