Proving the Vector Identity: curl(r x curlF) + (r . ∇)curlF + 2curlF = 0

[chipotleaway]

Homework Statement

Show that:
curl(r \times curlF)+(r.\nabla)curlF+2curlF=0, where r is a vector and F is a vector field.

(Or letting G=curlF=\nabla \times F
i.e. \nabla \times (r \times G) + (r.\nabla)G+2G=0)

The Attempt at a Solution

I used an identity to change it to reduce (?) it to
(\nabla.G)r+(G.\nabla)r-(\nabla.r)G-(r.\nabla)G+(r.\nabla)G+2G
(\nabla.G)r+(G.\nabla)r-(\nabla.r)G+2G

I'm not sure where to go from here to show that it's equal to zero. At the moment the only approach I know of is to compute all the components an hope they sum up to zero but surely there's another identity that can simplify this a bit further.

 

[Simon Bridge]

Sub back G=curl.F

div.curl.F=?

 

[chipotleaway]

0!
Which gives
((\nabla \times F).\nabla)r-(\nabla.r)(\nabla \times F)+2\nabla \times F
or
(G.\nabla)r-(\nabla.r)G+2G

One term less = a bunch of less components to deal with - I'll try expanding it out now and see where I get.

 

[chipotleaway]

Now I've got:

(G_1\frac{\partial r}{\partial x}+G_2\frac{\partial r}{\partial y}+G_3\frac{\partial r}{\partial z})-(G\frac{\partial r_1}{\partial x}+G\frac{\partial r_2}{\partial y}+G\frac{\partial r_3}{\partial z})+2G_1+2G_2+2G_3.

When I expand out the vectors(\frac{\partial r}{\partial x} into \frac{\partial r_1}{\partial x}, etc. and G into G_1, G_2, G_3), the diagonal terms cancel, i.e. G_1\frac{\partial r_1}{\partial x}i, G_2\frac{\partial r_2}{\partial y}j, G_3\frac{\partial r_3}{\partial z}k . What I'm left with doesn't look like it sums to zero, however.

 

[vela]

Homework Statement

Show that:
curl(r \times curlF)+(r.\nabla)F+2curlF=0, where r is a vector and F is a vector field.

(Or letting G=curlF=\nabla \times F
i.e. \nabla \times (r \times G) + (r.\nabla)G+2G=0)

Why did you change $(\vec{r}\cdot\nabla)\vec{F}$ into $(\vec{r}\cdot\nabla)\vec{G}$?

The Attempt at a Solution

I used an identity to change it to reduce (?) it to
(\nabla.G)r+(G.\nabla)r-(\nabla.r)G-(r.\nabla)G+(r.\nabla)G+2G
(\nabla.G)r+(G.\nabla)r-(\nabla.r)G+2G

I'm not sure where to go from here to show that it's equal to zero. At the moment the only approach I know of is to compute all the components an hope they sum up to zero but surely there's another identity that can simplify this a bit further.

 

[chipotleaway]

Why did you change $(\vec{r}\cdot\nabla)\vec{F}$ into $(\vec{r}\cdot\nabla)\vec{G}$?

Sorry, my mistake. IT should be $(\vec{r}\cdot\nabla)\nabla \times \vec{F}$

 

[vela]

I think you're supposed to use the fact that $\vec{r}$ is not just any vector but that it's equal to $\vec{r} = (x, y, z)$.